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17300 | https://math.libretexts.org/Courses/Cosumnes_River_College/Math_420%3A_Differential_Equations_(Breitenbach)/02%3A_First_Order_Equations/2.04%3A_Solving_Differential_Equations_by_Substitutions | 2.4.1
2.4.2
2.4.3
2.4.4
2.4.2
dydx=f(Ax+By+C)
2.4.3
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2.4: Solving Differential Equations by Substitutions
Last updated
: Jun 10, 2023
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2.3E: Linear First Order Equations (Exercises)
2.4E: Solving Differential Equations by Substitution (Exercises)
Page ID
: 103474
William F. Trench
Trinity University
( \newcommand{\kernel}{\mathrm{null}\,})
Bernoulli Equations
Definition 2.4.1
A Bernoulli equation is an equation of the form
dydx+p(x)y=f(x)yr,
dydx+p(x)y=f(x)yr,(2.4.1)
where rr can be any real number other than 00 or 11. (Note that Equation 2.4.12.4.1 is linear if and only if r=0r=0 or r=1r=1.)
Theorem 2.4.12.4.1
The substitution u=y1−r,u=y1−r, will turn the Bernoulli Equation 2.4.12.4.1 into a linear equation.
Proof
Dividing 2.4.12.4.1 by yryr yields y−rdydx+p(x)y1−r=f(x).
y−rdydx+p(x)y1−r=f(x).
If we make the substitution u=y1−r
u=y1−r
and differentiate with respect to x we get dudx=(1−r)y−rdydx.
dudx=(1−r)y−rdydx.
From this we can see that 2.4.12.4.1 becomes 11−rdudx+p(x)u=f(x)
11−rdudx+p(x)u=f(x)
which is linear and can be solved by the methods of section 2.3.
Example 2.4.1
Solve the Bernoulli equation
y′−y=xy2.
y′−y=xy2.(2.4.2)
Solution
Dividing by y2y2 we get
y−2y′−y−1=x.
y−2y′−y−1=x.
If we now let u=y−1,u=y−1, we get dudx=−y−2dydxdudx=−y−2dydx and substituting into 2.4.22.4.2 we get
−dudx−u=x
−dudx−u=x
which is a linear equation. Putting it in standard linear form we get
dudx+u=−x
dudx+u=−x
and using the method of section 2.3 we get the solution
u=−(x−1)+ce−x
u=−(x−1)+ce−x
and substituting back in for u we get
y−1=−(x−1)+ce−x
y−1=−(x−1)+ce−x
and solving for y we get our final solution
y=−1x−1+ce−x.
y=−1x−1+ce−x.
Figure 2.4.1
shows the direction field and some integral curves of Equation 2.4.22.4.2.
Homogeneous Equations
Definition 2.4.2
We say that a function f(x,y)f(x,y) is homogeneous of degree n if
f(tx,ty)=tnf(x,y)
f(tx,ty)=tnf(x,y)
Example 2.4.22.4.2
Determine if f(x,y)=x5+x2y3
f(x,y)=x5+x2y3
is a homogeneous function and, if so, of what degree.
Solution
Substituting txtx for xx and tyty for yy we get f(tx,ty)=(tx)5+(tx)2(ty)3=t5(x5+x2y3)=t5f(x,y).
f(tx,ty)=(tx)5+(tx)2(ty)3=t5(x5+x2y3)=t5f(x,y).
Therefore, f(x,y)f(x,y) is a homogeneous function of degree 5.
Example 2.4.32.4.3
Determine if f(x,y)=x2+xy2
f(x,y)=x2+xy2
is a homogeneous function and, if so, of what degree.
Solution
Substituting txtx for xx and tyty for yy we get f(tx,ty)=(tx)2+(tx)(ty)2=t2(x2+txy2)≠tnf(x,y).
f(tx,ty)=(tx)2+(tx)(ty)2=t2(x2+txy2)≠tnf(x,y).
for any number nn. Therefore, f(x,y) is not a homogeneous function.
Note: For this type of differential equation, as in section 2.2, it is convenient to write first order differential equations in the form
M(x,y)dx+N(x,y)dy=0.
Definition 2.4.3
We say that a differential equation of the form M(x,y)dx+N(x,y)dy=0 is homogeneous if M(x,y) and N(x,y) are homogeneous functions of the same degree.
Example 2.4.4
Determine if (x2+xy)dx+y2dy=0
is a homogeneous differential equation.
Solution
Substituting tx for x and ty for y in M(x,y) we get M(tx,ty)=(tx)2+(tx)(ty)=t2(x2+xy)=t2M(x,y)
and M(x,y) is a homogeneous function of degree 2.
Substituting tx for x and ty for y in N(x,y) we get N(tx,ty)=(ty)2=t2y2=t2N(x,y)
and N(x,y) is a homogeneous function of degree 2.
Since both M(x,y) and N(x,y) are homogeneous functions of degree 2, the differential equation is homogeneous.
Theorem 2.4.2
The substitution y=ux, where u is a function of x, or x=vy, where v is a function of y will make a homogeneous equation separable.
Proof
Note: we will only prove this for the substitution y=ux and leave the substitution x=vy to the reader.
Letting y=ux
and differentiating with respect to x we get
dydx=u+xdudx
which can be rewritten as dy=udx+xdu.
Substituting y and dy into 2.4.3 we get M(x,ux)dx+N(x,ux)[udx+xdu]=0.
Using the fact that both M(x,y) and N(x,y) are homogeneous of the same degree we get
xnM(1,u)dx+xnN(1,u)[udx+xdu]=0.
Expanding and rewriting this we eventually come to
xn[M(1,u)+uN(1,u)]dx=−xn+1N(1,u)du
which is clearly a separable differential equation:
1xdx=−N(1,u)M(1,u)+uN(1,u)du.
Example 2.4.5
Solve
xdy=(y+xe−y/x)dx.
First verify that M(x,y) and N(x,y) are homogeneous functions of the same degree.
Substituting y=ux into Equation 2.4.4 yields
x(udx+xdu)=(ux+xe−ux/x)dx.
Simplifying and separating variables yields
eudu=1xdx.
Integrating yields eu=ln|x|+c. Therefore u=ln(ln|x|+c) and y=ux=xln(ln|x|+c).
Figure 2.4.2
shows a direction field and integral curves for Equation 2.4.4.
Equations of the form dydx=f(Ax+By+C)
Theorem 2.4.3
The substitution u=Ax+By+C
will make equations of the form dydx=f(Ax+By+C)separable.
Proof
Consider a differential equation of the form 2.4.5.
Let u=Ax+By+C
Taking the derivative with respect to x we get dudx=A+Bdydx.
Substituting into 2.4.5 we get 1B(dudx−A)=f(u)
which is clearly a separable differential equation: duBf(u)+A=dx
Example 2.4.6
Solve
y′=sec2(y−x−3).
Solution
Letting u=y−x−3
and differentiating we get dudx=dydx−1
Substituting into Equation 2.4.6 we get
dudx+1=sec2u
This is now a separable equation which can be solved using the method of section 2.1, resulting in a solution of
−cotu−u=x+c
and substituting in for u we get
−cot(y−x−3)−(y−x−3)=x+c
and simplifying we get
cot(y−x−3)=−y+c
2.3E: Linear First Order Equations (Exercises)
2.4E: Solving Differential Equations by Substitution (Exercises) |
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17302 | https://undergroundmathematics.org/pervasive-ideas/constraints/download/constraints.pdf | Constraints underground mathematics Most exercises in mathematics can be seen as construction tasks, in that we are asked to construct a mathematical object that meets certain constraints. Finding a locus is an example, as is solving an equation. To solve the equation 3𝑥+ 7 = 5 is to construct a number meeting the constraint that multiplying by 3 and adding 7 results in 5. Similarly, the graph of 𝑦= 3𝑥+ 7 is the set of all coordinates of points (𝑥,𝑦) meeting the constraint that 𝑦= 3𝑥+ 7.
Any statement about (or property of) particular mathematical objects can be regarded as a constraint when we focus on the objects for which the statement is true — the objects that satisfy the constraint. For example, a circle could be described as a closed curve formed by the set of points in a plane that are equidistant from a given point. We can interpret this as a constraint on curves or as a constraint on points in a plane, restricting us to those points that are the same distance from a given point. There are other constraints that define a unique circle and one is featured in Circle of Apollonius…coordinate edition where coordinate geometry is used to explore and solve an ancient geometry problem involving loci. Finding circles also explores which pieces of information are sufficient to determine a single circle.
From Finding circles When we want to understand how something works, we might play with it a bit to see how changing different variables affects behaviour and how behaviour is constrained. In Logarithm lattice, for example, we explore how changing 𝑎or 𝑏affects the value of log𝑎𝑏. In How high am I? we explore how an angular change on a Ferris wheel affects the height of a rider and in Sector spirals we consider how changing the radius or angle affects the arc length and area of a circular sector. The basic idea is to learn how a small change in one feature is connected to a change in another feature. When we can quantify these features as variables, we can talk about covariation.
When thinking about graphs of functions, we may ask ‘How does a small increase in the 𝑥variable relate to the change in the 𝑦variable?’ Covariation is at the root of our understanding of gradient, rate of change and optimisation. Relationships between Page 1 of 3 Copyright © University of Cambridge. All rights reserved.
Last updated 03-Nov-16 a function and its gradient function can be explored in Gradient match and Can you find…curvy cubics edition.
From Gradient match As noted previously, an equation is an example of a constraint. We can use this to think about what it could mean to solve equations and inequalities. For example, solving 3𝑥+ 4 = 10 gives 𝑥= 2, which is a simpler way to express the same constraint. Rewriting an equation offers different ways to see the constraint and is central to Summing to one where Problem 2 asks how many ways we can choose integers 𝑥and 𝑦that satisfy log6 𝑥+ log6 𝑦= 1.
As well as equations, we can consider inequalities as constraints. Which is bigger? asks ‘Which is bigger 2𝑥 5 or 𝑥 3?’ and other similar questions. In Two-way algebra we ask students to identify equations or inequalities that satisfy some further property. This is one of several resources based on incomplete two-way tables where we ask students to recognise shared properties among examples and then find further examples. In doing this, students are organising and characterising objects within a larger family, thereby exploring freedom and constraints together.
These examples bring us to the idea of dimension. The 𝑥,𝑦-plane is a 2-dimensional space.
Adding an equation such as 𝑦= 𝑥or 𝑥2 + 𝑦2 = 1 restricts us to a 1-dimensional space, and imposing another equation restricts us to individual points (or possibly no points). So we see that adding the equational constraints reduces the dimension. If instead we use an inequality such as 𝑥2 + 𝑦2 < 1 or 𝑦> 𝑥, we still have a 2-dimensional space, unless the inequality cannot be satisfied.
It can also be interesting to think about the dimension of the space of all objects of a particular kind. For example, if we think of all quadratics of the form 𝑎𝑥2 + 𝑏𝑥+ 𝑐then there are three free parameters or degrees of freedom, and therefore the space of all such quadratics is 3-dimensional. Identifying the number of degrees of freedom is an important idea that lies behind the family of Can you find…? resources, including Can you find… cubic edition and Can you find… trigonometry edition in which we work backwards from constraints to examples or general solutions.
From Can you find… cubic edition Page 2 of 3 Copyright © University of Cambridge. All rights reserved.
Last updated 03-Nov-16 When seeking to construct an object meeting several constraints, it can help to remove one or more constraints. Prime triangles asks which triangular numbers are also prime and how many ways we can justify our answer. Making sense of each constraint separately offers different ways to think about the constraints before combining them.
From Prime triangles Another example involving multiple constraints is Straight line pairs in which we are asked to construct a pair of straight lines whose 𝑥–intercepts differ by 2, whose 𝑦-intercepts differ by 3 and whose slopes differ by 1. In order to do this, it helps to work with one constraint, then add each constraint on to the class of objects already constructed.
From Straight line pairs We have described several situations where we are required to construct an object that satisfies one or more constraints. But there are constraints that are not met by any objects — for example, 𝑥2 = −1 is not satisfied by any real number. Often mathematicians explore the possibility of including more objects so that an object meeting a constraint can be constructed. Thus the whole numbers are extended to the integers (by including 0 and negatives), to the rationals, to the reals, to the complex numbers and beyond.
Constraints is one of the ideas that we have chosen to highlight in our pervasive ideas.
Page 3 of 3 Copyright © University of Cambridge. All rights reserved.
Last updated 03-Nov-16 |
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Drag on a Sphere
Drag on a Sphere
Discover the fascinating characteristics and principles of Drag on a Sphere in the realm of engineering. This comprehensive guide will delve into the definition, practical examples, and diverse applications of this vital concept, especially pertaining to fluid mechanics. Understand how to accurately calculate drag force on a sphere and interpret the complexities of the formula used. The article also lays emphasis on recognising the effects of turbulence on Drag on a Sphere, a subject of high relevance in environmental and fluid engineering. Acquire an in-depth understanding of this essential dynamics principle that holds significance in various engineering fields.
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Last Updated: 06.10.2023
Published at: 06.10.2023
25 min reading time
Aerospace Engineering
Artificial Intelligence & Engineering
Audio Engineering
Automotive Engineering
Chemical Engineering
Civil Engineering
Design Engineering
Design and Technology
Electrical Engineering
Engineering Fluid Mechanics
Aerofoil
Atmospheric Drag
Atmospheric Pressure
Atmospheric Waves
Axial Flow Pump
Bernoulli Equation
Boat Hull
Boundary Layer
Boussinesq Approximation
Buckingham Pi Theorem
Capillarity
Cauchy Equation
Cavitation
Centrifugal Pump
Circulation in Fluid Dynamics
Colebrook Equation
Compressible Fluid
Continuity Equation
Continuous Matter
Control Volume
Convective Derivative
Coriolis Force
Couette Flow
Density Column
Dimensional Analysis
Dimensional Equation
Dimensionless Numbers in Fluid Mechanics
Dispersion Relation
Drag on a Sphere
Dynamic Pump
Dynamic Similarity
Dynamic Viscosity
Eddy Viscosity
Energy Equation Fluids
Equation of Continuity
Euler's Equation Fluid
Eulerian Description
Eulerian Fluid
Flow Over Body
Flow Regime
Flow Separation
Fluid Bearing
Fluid Density
Fluid Dynamic Drag
Fluid Dynamics
Fluid Fundamentals
Fluid Internal Energy
Fluid Kinematics
Fluid Mechanics Applications
Fluid Pressure in a Column
Fluid Pumps
Fluid Statics
Froude Number
Gas Molecular Structure
Gas Turbine
Hagen Poiseuille Equation
Heat Transfer Fluid
Hydraulic Press
Hydraulic Section
Hydrodynamic Stability
Hydrostatic Equation
Hydrostatic Force
Hydrostatic Force on Curved Surface
Hydrostatic Force on Plane Surface
Hydrostatics
Impulse Turbine
Incompressible Fluid
Internal Flow
Internal Waves
Inviscid Flow
Inviscid Fluid
Ion Thruster
Irrotational Flow
Jet Propulsion
Kinematic Viscosity
Kutta Joukowski Theorem
Lagrangian Description
Lagrangian Fluid
Laminar Flow in Pipe
Laminar vs Turbulent Flow
Laplace Pressure
Lift Force
Linear Momentum Equation
Liquid Molecular Structure
Mach Number
Magnetohydrodynamics
Manometer
Mass Flow Rate
Material Derivative
Momentum Analysis of Flow Systems
Moody Chart
Navier Stokes Cartesian
Navier Stokes Cylindrical
Navier Stokes Equation in Spherical Coordinates
No Slip Condition
Non Newtonian Fluid
Nondimensionalization
Nozzles
Open Channel Flow
Orifice Flow
Pascal Principle
Pathline
Piezometer
Pipe Flow
Piping
Pitot Tube
Plasma
Plasma Parameters
Plasma Uses
Pneumatic Pistons
Poiseuille Flow
Positive Displacement Pump
Positive Displacement Turbine
Potential Flow
Prandtl Meyer Expansion
Pressure Change in a Pipe
Pressure Drag
Pressure Field
Pressure Head
Pressure Measurement
Propeller
Pump Characteristics
Pump Performance Curve
Pumps in Series vs Parallel
Reaction Turbine
Relativistic Fluid Dynamics
Reynolds Experiment
Reynolds Number
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Wind Tunnel
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Wing Aerodynamics
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Engineering Mathematics
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Materials Engineering
Mechanical Engineering
Professional Engineering
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What is Engineering
Contents
Aerospace Engineering
Artificial Intelligence & Engineering
Audio Engineering
Automotive Engineering
Chemical Engineering
Civil Engineering
Design Engineering
Design and Technology
Electrical Engineering
Engineering Fluid Mechanics
Aerofoil
Atmospheric Drag
Atmospheric Pressure
Atmospheric Waves
Axial Flow Pump
Bernoulli Equation
Boat Hull
Boundary Layer
Boussinesq Approximation
Buckingham Pi Theorem
Capillarity
Cauchy Equation
Cavitation
Centrifugal Pump
Circulation in Fluid Dynamics
Colebrook Equation
Compressible Fluid
Continuity Equation
Continuous Matter
Control Volume
Convective Derivative
Coriolis Force
Couette Flow
Density Column
Dimensional Analysis
Dimensional Equation
Dimensionless Numbers in Fluid Mechanics
Dispersion Relation
Drag on a Sphere
Dynamic Pump
Dynamic Similarity
Dynamic Viscosity
Eddy Viscosity
Energy Equation Fluids
Equation of Continuity
Euler's Equation Fluid
Eulerian Description
Eulerian Fluid
Flow Over Body
Flow Regime
Flow Separation
Fluid Bearing
Fluid Density
Fluid Dynamic Drag
Fluid Dynamics
Fluid Fundamentals
Fluid Internal Energy
Fluid Kinematics
Fluid Mechanics Applications
Fluid Pressure in a Column
Fluid Pumps
Fluid Statics
Froude Number
Gas Molecular Structure
Gas Turbine
Hagen Poiseuille Equation
Heat Transfer Fluid
Hydraulic Press
Hydraulic Section
Hydrodynamic Stability
Hydrostatic Equation
Hydrostatic Force
Hydrostatic Force on Curved Surface
Hydrostatic Force on Plane Surface
Hydrostatics
Impulse Turbine
Incompressible Fluid
Internal Flow
Internal Waves
Inviscid Flow
Inviscid Fluid
Ion Thruster
Irrotational Flow
Jet Propulsion
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Last Updated: 06.10.2023
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Understanding Drag on a Sphere
Practical Drag on a Sphere Examples
Diverse Applications of Drag on a Sphere
How to Calculate Drag Force on a Sphere
An In-depth Look at the Drag on a Sphere Formula
Recognising Drag on a Sphere in Turbulent Flow
Drag on a Sphere - Key takeaways
Similar topics in Engineering
Related topics to Engineering Fluid Mechanics
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What effects are brought about due to a sphere moving through a fluid under turbulent flow conditions?
A. In turbulent flow, the drag force experienced by a sphere varies compared to laminar flow. The drag coefficient is not constant - but fluctuates, due to turbulent flow's dramatic nature. The drag coefficient initially decreases with an increase in Reynolds number, reaches a minimum, and then subtly increases. B. In turbulent flow, the movement of the sphere becomes linear and predictable as the drag force and the drag coefficient become constant and immobile. C. While moving under turbulent flow, the sphere experiences a steady increase in the drag coefficient which is directly proportional to the Reynolds number. D. Turbulent flow conditions do not account for any changes in the sphere's drag force and the drag coefficient remains constant, just like in the laminar flow conditions.
1/3
How does the concept of drag on a sphere apply to aerospace engineering?
A. In aerospace engineering, drag on a sphere isn't considered as it has no impact on the design of aircraft or spacecraft. Fuel efficiency and smoother rides are only related to the engine's functionality. B. In aerospace engineering, the sphere's shape, often representing the aircraft's nose or spacecraft, is designed for effective aerodynamics. Reducing drag can lead to fuel economy, a smoother ride, and less atmospheric heating during re-entry for spacecraft. C. The concept of drag on a sphere doesn't apply to aerospace engineering since aircraft and spacecraft aren't spherical objects. Therefore, their designs aren't influenced by aerodynamics or drag. D. In aerospace engineering, increasing drag is the key to achieving fuel economy and a smoother ride. High levels of atmospheric heating during re-entry for spacecraft are also preferred.
1/3
Why is understanding drag on a sphere important in engineering?
A. Understanding drag on a sphere is vital in engineering for the creation of stationary structures such as buildings and bridges. B. Understanding drag on a sphere is primarily important in engineering because it helps in the design of aesthetic structures and decorations. C. The concept of drag on a sphere has limited application in engineering; it is mostly used in artistic sculpting and pottery. D. Understanding drag on a sphere is important in engineering as it allows for more efficient design of vehicles and sporting equipment, prediction of weather patterns and simulation of pollution dispersion in environmental engineering.
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Understanding Drag on a Sphere
In the realm of Physics and Engineering, you might often encounter the concept of drag on a sphere. This term is central to the science of fluid mechanics, specially when it comes to predicting how objects move through fluid mediums.
Definition of Drag on a Sphere
Drag on a sphere, as the name suggests, refers to the resistive force that a spherical object encounters when it moves through a fluid. This can include liquids, such as water, or gases like air.
It can be defined by the drag equation: F D=1 2×ρ×v 2×C D×A where F D is the drag force, ρ is the density of the fluid, v is the speed of the object relative to the fluid, C D is the drag coefficient, and A is the cross-sectional area of the object (in this case, a sphere).
The Link between Engineering Fluid Mechanics and Drag on a Sphere
Engineering Fluid Mechanics is a branch of engineering that deals with the behaviour, control and mechanism of fluid, either in motion or at rest. And the concept of drag on a sphere is significantly tied to it. The aforementioned equation for drag on a sphere is a cornerstone of fluid mechanics as it helps engineers predict how fluid will interact with physical bodies.
For instance, in aircraft design, engineers often need to account for the drag that will be exerted on the aircraft in the air. Similarly, nautical engineers use this concept to design efficient ship hulls that minimize drag in water.
Different Types of Drag Affecting a Sphere
You'll find that there are various types of drag forces that can exert an influence on a sphere moving through a fluid. The primary types include:
Pressure drag
Friction drag
Compressibility drag
Pressure drag, also known as form drag, arises due to the air pressure variation around the object. Friction drag, as implied by its name, results from the friction between the fluid and the surface of the object. Compressibility drag occurs in cases where the speed of the object is comparable to the speed of sound in the fluid, leading to compression effects.
The Role of Fluid Dynamics in Determining Drag on a Sphere
Fluid dynamics, a sub-discipline of fluid mechanics, deals with fluid flow—the science of liquids and gases in motion. When considering drag on a sphere, fluid dynamics plays a crucial role.
Fluid dynamics helps in understanding how the movement and behavior of the fluid will affect the overall drag experienced by the sphere. This is governed by various factors such as viscosity, density, and flow velocity of the fluid, as well as the size and speed of the sphere.
For example, a small, slow-moving sphere in a highly viscous liquid will experience significant drag, while a large, fast-moving sphere in a low-viscosity gas may experience relatively less drag.
As you progress in your understanding of fluid mechanics and engineering, the comprehension of drag on a sphere and its related factors become crucial. They unfold the complexities of real-world fluid systems and offer valuable insights into designing and optimizing various mechanical and engineering systems.
Practical Drag on a Sphere Examples
In the observable world, countless practical examples of drag on a sphere manifest daily. These instances provide fundamental insight into the application of concepts like fluid dynamics and fluid mechanics.
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Everyday Instances of Drag on a Sphere
Drag on a sphere isn't just confined to engineering or aviation. It permeates everyday phenomena too. For instance, when you kick a football, it doesn’t travel indefinitely but eventually stops moving - a clear illustration of drag at work. The air resists the motion of the ball, gradually reducing its speed until it stops.
The same effect can be observed with bubbles rising in a liquid. Have you ever watched bubbles in a fizzy drink or in a fish tank? They don’t shoot straight to the surface at a constant speed. They instead move more slowly as they get closer to the surface. Again, this is drag at work: the liquid acts as a resistive force, slowing down the bubble's ascent. These examples represent drag on a sphere at work in daily life. They highlight how the resistive force exerted by a fluid medium, be it a gas or a liquid, impacts the motion of spherical objects.
Analysing Drag on a Sphere in Real-Life Scenarios
To accurately analyse drag on a sphere in real-life scenarios, you have to consider various factors. These can be seen explicitly by applying the drag equation: F D=1 2×ρ×v 2×C D×A In each of the previous examples – the football or the rising bubble – every component of the equation plays a role.
ρ – the density of the fluid: In both cases, the density of the fluid regulates the extent of drag. Air, being less dense than a liquid like water, creates less drag.
v – the velocity of the object relative to the fluid: A faster football or a swiftly rising bubble will encounter more drag due to the square of the velocity in the drag equation.
C D – the drag coefficient: This quantity is more complex, as it depends on factors like the shape of the object and the properties of the fluid. For a sphere, however, this is a standard value.
A – the cross-sectional area of the sphere: A larger football or a bigger bubble faces more drag due to the larger cross-sectional area in the fluid’s path.
By assessing these parameters, you can directly infer how the drag on a sphere emerges and behaves in real-life contexts.
Case Studies: Understanding Drag on a Sphere's Impact in Engineering
A glance at engineering presents cases where comprehending and utilising the principles of drag on a sphere is indispensable. For instance, the aviation and nautical sectors heavily rely on this concept in their vehicle designs. Aircraft and ship designers work extensively with models to predict and minimise drag. For example, the design of an airplane's nose is spherical to better negotiate the drag while moving through air. By correctly understanding and predicting drag forces, engineers can design more efficient aircraft, reducing fuel consumption and thereby lowering operational costs and environmental impact. In the case of nautical engineering, understanding drag is critical for improving ship fuel efficiency. Investigating how water resistance - or hydrodynamic drag - impacts the ship's movement is essential. Naval architects design the hull shape to minimise drag. The optimal shape reduces the energy required to move the ship through water, enhancing fuel efficiency and reducing operating costs. In both case studies, the application of the drag equation and a deep understanding of drag on a sphere enable engineers to predict, analyse and optimise their designs, contributing to the continual evolution of technology and engineering efficiencies. The phenomenal power of physics thus interweaves with significant practical applications, highlighting the relevance of drag on a sphere in engineering design.
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Diverse Applications of Drag on a Sphere
Drag on a sphere unveils a broad spectrum of applications across varied fields, from aerospace engineering to environmental management. It’s not just confined to academic equations and theoretical considerations. By understanding the dynamics that shape the drag on a sphere, you'll be able to appreciate its profound influence on various facets of real-world applications and technological advancements.
Roles of Drag on a Sphere in Various Engineering Fields
Engineering has several subdivisions, each with unique responsibilities and goals, but many rely on the understanding of drag on a sphere as part of their fundamental principles. In fields like civil engineering and architectural design, the principles concerning a sphere’s drag help in the optimisation of structural designs. Engineers account for air resistance when planning skyscrapers or suspension bridges, for example. Analyzing how air might flow around these structures helps predict and mitigate potential stress factors, ensuring the structure’s resilience and longevity. In biomedical engineering, drag on a sphere can be a crucial factor. For instance, medical devices such as stents or catheters need to navigate the fluid-filled environment of the human body with ease. The principles that govern drag on a sphere can thus guide design processes to make these devices as efficient and least disruptive as possible.
For instance, drag considerations come into play when designing artificial heart valves, an area where fluid dynamics intersect with biomedical engineering. Given that the blood's fluid characteristics and the valve's design can significantly affect the heart's pumping efficiency, engineers must carefully consider drag forces.
Crucially, drag elements are integral to automotive engineering. Aerodynamics plays a vital role in creating vehicles that are fuel-efficient and safe. Reducing drag can contribute to a vehicle's improved fuel economy, which isn't just economical but also environmentally friendly. The list goes on. From sports engineering to materials science, virtually every engineer will encounter situations where understanding drag on a sphere aids in predicting, understanding, and optimising system performance.
The Impact of Drag on a Sphere in Aerodynamics and Fluid Engineering
Aerodynamics and fluid engineering are fields that pivot centrally on the concept of drag on a sphere. Both of these fields deal extensively with fluid flow and resistance - air, in the case of aerodynamics, and any fluid (including air and water) regarding fluid engineering. In aerodynamics, one of the primary objectives is designing vehicles, namely aircraft, that can efficiently travel through air. Here, the drag on a sphere principle heavily dictates the structural design. Observe an airplane; you'll notice that many components, such as the nose cone or even the body structure, mimic the form of a sphere. The aircraft’s design is intentional to negotiate the air resistance--the drag--effectively during flight.
Aerodynamics is the study of how gases interact with moving bodies. Given that gases are fluid, aerodynamics is a branch of fluid dynamics. It involves studying fluid flow around a body, the forces acting on a body moving through a fluid and the effects of the body on the fluid.
Similarly, fluid engineering (often incorporated under Mechanical engineering) utilises the principles of drag on a sphere to design efficient systems. Engineers often employ the drag coefficient when designing pipelines ensured to facilitate the smooth flow of fluids with minimized resistance. In turbine design, for example, fluid engineers work to minimize fluid resistance around the turbine blades for optimal energy generation. These instances highlight the interdependence between these engineering branches and the understanding of drag on a sphere. By mastering this concept, engineers within these fields can develop efficient, safe, and high-performing designs.
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The Relevance of Drag on a Sphere in Environmental Engineering
In environmental engineering, a proper understanding of the drag on a sphere finds relevance in several areas. These range from designing wastewater treatment systems to studying the movement of pollutants in the atmosphere or bodies of water. In the design of treatment plants, for instance, the movement of solid particles in tanks or through pipelines needs to be effectively managed. By understanding the drag principles governing these particles, engineers can design systems that maximise treatment efficiencies while minimising energy use. Moreover, when studying the dispersion and settling of pollutants, whether in air or water bodies, drag principles come into play. Predicting the movement of pollution requires an understanding of how these particles or droplets will interact with the fluid environment, which ultimately comes down to factors such as size (which affects the drag force), buoyancy, and fluid currents. The understanding of drag also aids in the analysis and prediction of erosion rates caused by wind or water flows. Predicting and mitigating soil erosion are critical components of environmental conservation efforts and can guide land management decisions. In each case, the principles of drag on a sphere serve as foundational tools, allowing environmental engineers to analyse, predict, and design systems that protect and sustain our environment.
How to Calculate Drag Force on a Sphere
The process of calculating the drag force on a sphere highlights an interesting intersection of physics and mathematics, calling for a firm understanding of fluid mechanics principles. The key determinant of drag force is the drag coefficient, particular to the sphere in this case. It's important to establish that the calculation of drag force hinges primarily on the following factors: fluid density, velocity of the sphere relative to the fluid, the sphere's cross-sectional area, and the drag coefficient.
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Steps to Determine Drag Force on a Sphere
To calculate the drag force on a sphere, you need to follow specific steps. These steps relate directly to the drag equation, which forms the cornerstone of the calculations. It's crucial to get each step right to ensure an accurate calculation: 1. Identify the parameters required: The four essential variables you need to determine the drag force on a sphere include fluid density, relative velocity, the sphere's cross-sectional area, and the drag coefficient. 2. Measure or determine the values of these parameters: Each variable's value can be obtained through measuring (for instance, fluid density and relative velocity) or referring to standard physics tables (for the drag coefficient). 3. Insert the values into the drag equation: Once you have determined all the required parameters, you can substitute them into the drag equation: F D=1 2×ρ×v 2×C D×A Keep in mind that these steps provide an ideal approach towards understanding and calculating drag force on a sphere. However, in real-life applications, there could be additional variables to consider such as viscous drag or buoyant forces, especially when dealing with highly viscous fluids or for objects moving in a fluid at very high velocities.
Advanced Calculations: The Coefficient of Drag on a Sphere
When it comes to calculating the coefficient of drag for a sphere, more advanced calculations come into play. The coefficient of drag (C D) for a sphere depends on the Reynolds number, a dimensionless quantity derived from the physical characteristics of the fluid and the sphere. The Reynolds number (R e) is given by the equation: R e=ρ×v×D μ where ρ is the fluid density, v is the velocity of the sphere relative to the fluid, D is the diameter of the sphere, and μ is the dynamic viscosity of the fluid. If you have to calculate the coefficient of drag on a sphere explicitly, you would utilise the Reynolds number and refer to a standard drag curve chart for spheres that provides a relationship between the coefficient of drag and the Reynolds number. However, often a standard value is used for the drag coefficient of a smooth sphere in a particular range of Reynolds number, which simplifies calculations greatly.
The Importance of Calculating Accurate Drag Forces
Accurate calculation of the drag force on a sphere can have a critical impact in numerous real-world contexts. Errors or inaccuracies in determining this force could lead to faulty designs in engineering scenarios that could have potentially significant consequences. The principle of drag is used extensively in vehicle engineering, with designs aimed to minimise drag and improve efficiency. An accurate knowledge of the drag force helps design streamlined shapes that can move easily through a fluid medium, buses, delivery trucks, high-speed trains or race cars, for instance. In aircraft engineering, accurate drag force calculations are absolutely essential. Inaccuracies can directly affect flight dynamics, including stability, control and fuel efficiency. Even small errors can potentially prove to be disastrous. In sports engineering, the importance of accurate drag force calculations is undeniable. Whether it’s designing a golf ball or predicting the trajectory of a football, understanding drag forces is crucial.
In one instance, when golf ball manufacturers started adding dimples to the ball's surface, they noticed the balls reached further distances. The explanation lies in the fact that dimples on a sphere drastically alter the flow behaviour around the sphere, reducing the drag force and allowing the ball to travel further.
In environmental engineering, we use the understanding of drag forces to analyse pollutants' dispersion in the air or water bodies, aiding in environmental protection initiatives. The necessity for concern about accurately calculating drag forces extends well beyond, including urban planning, meteorology, and even certain biomedical applications. Therefore, achieving precise drag predictions underscores the need for accuracy, reinforcing its indisputable importance.
An In-depth Look at the Drag on a Sphere Formula
Taking an in-depth look at the drag on a sphere formula is like peeling apart the layers of a complex blockbuster movie — the more closely you examine it, the more fascinating it becomes. This particular equation unveils the underlying physics employed to determine the fuerza de arrastre, or drag force, acting on a spherical object in a fluid medium.
Understanding and Interpreting the Drag on a Sphere Formula
When it comes to the drag on a sphere formula, it's essential to appreciate the interpretive aspects, as they provide a comprehensive picture of the physics phenomenon involved. Here's the magical formula: F D=1 2×ρ×v 2×C D×A This equation describes the drag force (F D) acting on a spherical object immersed in a fluid. Each variable has a distinct meaning and impact. 1. ρ: This is the fluid density in which the sphere is present. Fluid density directly influences the resulting drag force. A greater density yields a stronger drag force. This principle explains why moving in water is significantly harder than moving in air, given that water is denser. 2. v: This is the relative velocity of the sphere against the fluid. This relationship is squared, which means the resulting drag force increases quadratically with the velocity. 3. C D: This is the drag coefficient, which depends on the Reynolds number, the flow regime, and the sphere's physical characteristics. It's through C D that the complexity and nuance of fluid mechanics come into play. 4. A: This is the projected area of the sphere, the cross-sectional area seen by the oncoming fluid flow. Naturally, a bigger area induces greater drag. Given these variables, one can appreciate how the formula elegantly captures the nuanced interaction between an object and the fluid it's moving through, enabling engineers to make calculated predictions and take appropriate action.
Exploring Variables Within the Drag on a Sphere Formula
Now that we have a broader understanding of the drag on a sphere formula, let's delve deeper into two noteworthy variables: the drag coefficient C D and fluid velocity v. 1. The Drag Coefficient C D: The drag coefficient is an expression of the sphere's shape and surface roughness. For a smooth sphere, the drag coefficient typically ranges from 0.2 to 0.6 under normal conditions. However, intriguingly, this value isn't constant and can change with flow conditions represented by the Reynolds number (R e) — a dimensionless number used to predict flow patterns. This intriguing relationship allows engineers to deliberately alter circumstances to achieve the desired effect. 2. The Fluid Velocity v: The velocity here is relative — it's just as valid for a stationary sphere in a moving fluid as a moving sphere in a stationary fluid. The quadratic relationship between fluid velocity and drag force is an important characteristic of this formula. It means that if you double the velocity, the drag force increases by a factor of four, a sobering fact for any engineer trying to design high-speed vehicles or machines.
Application of the Drag on a Sphere formula in Solving Practical Problems
The drag on a sphere formula finds extensive application in practical problems. Solving initial-value problems involving drag often calls for a grasp of differential equations. However, a lot can be gleaned from simpler experiments too. For instance, determining an object's terminal velocity can be a useful experiment. Terminal Velocity Experiment: Consider an experiment where a small ball bearing is dropped into a tall, viscous fluid-filled cylinder. Once dropped, the ball bearing will initially accelerate due to gravity. However, as its speed increases, it will experience an increasingly large drag force opposing its motion. Eventually, the drag force will equal the gravitational force on the object, resulting in zero net force. At this point, the ball bearing will stop accelerating and continue to move at a constant velocity - the terminal velocity. Here’s the interesting part. The terminal velocity of the ball bearing can be related to the drag force formula. When the ball bearing reaches terminal velocity, the gravitational force (F g=m×g) and the drag force are equal, hence: m×g=1 2×ρ×v t 2×C D×A Rearranging this equation to solve for the terminal velocity (v t) yields: v t=2 m×g ρ×C D×A This equation illuminates how the terminal velocity is affected by fluid density, drag coefficient, and the sphere's size A, as well as the effects of gravity. This experiment really shows how useful the drag on a sphere formula can be in a practical scenario, allowing us to make accurate predictions and design smarter systems.
Recognising Drag on a Sphere in Turbulent Flow
Understanding drag force in turbulent flow can make a world of difference when you're trying to grasp the mechanics behind fast-flowing rivers, air flow around vehicles, or even blood flow in arteries. Turbulence significantly complicates fluid flow, leading to changes in drag experienced by an object - in this case, a sphere.
Explanation of Drag on a Sphere within Turbulent Flow
When a sphere moves through a fluid medium such as air or water, it faces resistance, which we term as drag. In the realm of turbulent flow - where the fluid particles move in random and chaotic motions, creating eddies and swirls - the drag experienced by a sphere undergoes significant changes compared to steady or laminar flow. At the core of understanding this phenomenon is the transition from laminar to turbulent flow. This transition is often defined by a dimensionless number known as the Reynolds Number (R e). The Reynolds number, defined as R e=ρ v d μ, where ρ is fluid density, v is fluid velocity, d is diameter of the sphere, and μ is dynamic viscosity of the fluid, predicts the onset of turbulence. For a sphere, the critical Reynolds number is approximately 2100. Below this value, the flow is laminar; above it, the flow turns turbulent. In turbulent flow, because of the chaotic and random motion of fluid particles, pressure and shear stress fluctuations cause the drag force to increase abruptly - a key characteristic feature that distinguishes it from laminar flow.
Observing the Effects of Turbulence on Drag on a Sphere
To observe the effects of turbulence on the drag experienced by a sphere, we need to delve deeper into the role played by the drag coefficient during turbulent flow. Given the fluctuant nature of turbulence, there is an alteration to the drag coefficient, C D, used in the drag force formula. During turbulence, irregular vortices and eddies formed around the sphere cause chaotic pressure fluctuations and increased shear stresses acting on the sphere. This erratic behaviour is reflected in the value of C D, which notably increases with growing turbulence. To put this into perspective, recall that the drag force is given by F D=1 2×ρ×v 2×C D×A. In turbulent flow scenarios, C D increases, leading to a greater drag force. The exact value of C D in a turbulent regime can be experimentally determined or estimated using previously established correlations. It is noteworthy to mention that these correlations are usually functions of the Reynolds number, reinforcing the intimate link between turbulence and R e.
Application of Drag on a Sphere Concept in Turbulent Flow Conditions
Understanding the drag on a sphere in turbulent flow conditions is crucial in many engineering and scientific applications. It helps in designing vehicles for efficient fuel consumption, improving sports equipment designs, or even better understanding natural phenomena like sedimentation in rivers. Aerodynamic Vehicle Design: Engineers employ the concepts of drag in turbulent flow when designing vehicles, especially aircraft and high-speed cars. By minimising drag forces, these vehicles can achieve higher speeds and consume less fuel. Sports Equipment: Sports equipment, such as golf balls, are an interesting application of the drag on a sphere concept. A golf ball's unique dimpled design helps it navigate the turbulence, reducing drag and allowing it to travel farther. Natural Phenomena: Understanding drag in turbulent flow aids in interpreting natural scenarios such as sediment transport in rivers and blood flow through arteries. This understanding informs better predictions and modelling, thereby offering insights for environmental conservation and medical treatment strategies. In summary, knowing how to apply the concept of drag on a sphere in turbulent flow conditions truly bridges the theoretical world of fluid mechanics with practical, real-world applications.
Drag on a Sphere - Key takeaways
Drag on a Sphere is a significant concept in various fields including, aerospace and nautical engineering, civil engineering, biomedical engineering, and environmental engineering. It helps to predict, analyse and optimise designs, and contributes to the continual evolution of technology and engineering efficiencies.
Drag on a Sphere is integral in designing structures like skyscrapers and bridges, medical devices like stents and catheters, and fuel-efficient vehicles. Understanding the concept can help in predicting, understanding and optimising system performance across these diverse domains.
In fields such as aerodynamics and fluid engineering, the drag on a sphere principle greatly influences the structural designs. In these fields, the aim is to design systems that can efficiently deal with fluid flow and resistance.
Environmental engineering also finds relevance of drag on a sphere in multiple areas, from designing wastewater treatment systems to studying the movement of pollutants in the atmosphere or bodies of water. Understanding of drag principles aids in the analysis and prediction of erosion rates as well.
Calculating the drag force on a sphere involves understanding of the drag equation and use of variables such as fluid density, velocity of the sphere relative to the fluid, sphere's cross-sectional area, and the drag coefficient. Getting the calculations right is critical in various real-world contexts including vehicle and aircraft engineering, sports engineering, and environmental protection initiatives.
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Flashcards in Drag on a Sphere ------------------------------ 24
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What effects are brought about due to a sphere moving through a fluid under turbulent flow conditions? In turbulent flow, the drag force experienced by a sphere varies compared to laminar flow. The drag coefficient is not constant - but fluctuates, due to turbulent flow's dramatic nature. The drag coefficient initially decreases with an increase in Reynolds number, reaches a minimum, and then subtly increases.
How does the concept of drag on a sphere apply to aerospace engineering? In aerospace engineering, the sphere's shape, often representing the aircraft's nose or spacecraft, is designed for effective aerodynamics. Reducing drag can lead to fuel economy, a smoother ride, and less atmospheric heating during re-entry for spacecraft.
Why is understanding drag on a sphere important in engineering? Understanding drag on a sphere is important in engineering as it allows for more efficient design of vehicles and sporting equipment, prediction of weather patterns and simulation of pollution dispersion in environmental engineering.
What is the meaning of "drag on a sphere"? "Drag on a sphere" refers to the force exerted by fluid particles on the surface of a sphere moving through it, which resists the sphere's forward motion. This usually occurs when an object moves in a fluid medium such as air or water.
What are some practical examples of drag on a sphere in the real world? Examples include a cricket ball experiencing air resistance changing its speed and trajectory, the design of an airplane's nose-cone for better aerodynamics, and the spherical shape of raindrops to reduce atmospheric drag.
How does the fluid's density and viscosity affect the drag on a sphere? Fluid density and viscosity play major roles in determining the extent of the drag force. A sphere falling through a highly viscous fluid like honey experiences more drag compared to the same sphere falling through water.
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Frequently Asked Questions about Drag on a Sphere
What is drag on a sphere? Please write in UK English.
Drag on a sphere refers to the resistance or force opposing the motion of a spherical object moving through a fluid (like air or water). It's caused by friction and differences in pressure and is impacted by factors like the sphere's size, velocity, and the fluid's properties.
How can one calculate the air drag on a sphere?
To calculate air drag on a sphere, use the drag equation: Fd = 0.5 Cd A r V^2. Here, Fd is the drag force, Cd is drag coefficient, A is the cross-sectional area of the sphere, r is air density, and v is velocity of the sphere.
What is an example of drag on a sphere? Please write in UK English.
An example of drag on a sphere is a football being kicked and experiencing air resistance as it travels. The air resistance pushing against the ball as it moves is the drag force.
How do you determine the coefficient of drag on a sphere?
The coefficient of drag on a sphere can be calculated using empirical formulas or computational fluid dynamics. It is dependent on factors such as the Reynolds number, the fluid properties and the sphere's size and speed. Generally, it involves complex mathematical modelling and experiments.
What is the drag on a sphere in turbulent flow?
The drag on a sphere in turbulent flow refers to the resistance, or force, that the sphere experiences when moving through a fluid (e.g., air or water) under turbulent conditions. The magnitude of this force is primarily determined by the sphere's speed, size and the fluid's properties.
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17305 | https://saspublishers.com/article/302/download/ | 94
SAS Journal of Surgery ISSN 2454 -51 04
SAS J . Su rg., Volume -2; Issue -3 (May -Jun , 201 6); p -94-97
Available online at
Modes of presentation and outcome of gastric ulcer perforation
Dr. Mayakonda Krishnamurthy Ramesh 1, Dr. Niyaz Ahmed 2, Praveen Kumar Mallangoudapatil 3
1
Professor, 2Senior Resident, 3Post graduate, Department of General Surgery, Bangalore Medical College, Bangalore, Karnataka, India
Corresponding author
Dr. Mayakonda Krishnamurthy Ramesh
Email: ramspsm@gmail.com
Abstract: The treatment of perforation still continues to be controversial. Just closure of perforation may save life, but chance of recurrence of ulcer is too high and patient may not turn up for a second curative surgery. The study was conducted in Victoria Hospital, BMCRI, Bangalore, from NOV 2011 to MAY 2013. During this period the number of cases admitted and selected for analyzing the data was 30. These 30 cases were studied thoroughly according to the proforma. The details of 30 patients were arranged in the master chart for convenience of presentation. Distension was most commonly observed in cases with >24hrs history of symptoms in all cases and in all cases after 48hrs.Rigidiry was present in all cases. Abdomen was board like and tense in all the cases. Size of perforation, has a significant role in prognosis.
Keywords: Gastric ulcer, Perforation, Clinical Presentation
INTRODUCTION:
Gastric ulcer is also more likely to occur in elderly patients, and admissions for bleeding gastric ulcers have increased during the past several years .
Ulcer disease continues to exact a heavy personal and financial toll. Currently, the personal toll of ulcer disease is seen mainly in the complications of perforation and bleeding . Gastric ulcer has a higher mortality than duodenal ulcer because of its increased prevalence in the elderly. Recent studies have shown an increase in the rates of hospitalization and mortality in elderly patients for the peptic ulcer complications like bleeding and perforation. Presumably this is due to the increasingly common use of NSAIDs and aspirin in this elderly cohort, many of whom have H. pylori infection
.
The treatment of perforation still continues to be controversial. Just closure of perforation may save life, but chance of recurrence of ulcer is too high and patient may not turn up for a second curative surgery. So, there is a school of thought, which recommends definitive surgery in a perforated gastric perforation. This may to a certain extent reduce the mortality and morbidity of the patient, because patients have to risk a major operation when the general condition is not good. On the other hand it saves the patient of further surgery.
When acute or chronic gastric ulcer perforates into peritoneal cavity, three components require treatment viz., the ulcer, the perforation and the resultant peritonitis. The perforation and resultant peritonitis are immediate threats to the life; the ulcer in itself is not. The therapeutic priorities thus are treatment of peritonitis and securing the closure of perforation, which may be achieved with surgical procedure.
In spite of better understanding of disease, effective resuscitation and prompt surgery under modern anesthesia techniques, there is high morbidity (36%) and mortality (6%). Hence, attempt has been made to analyze the various factors, which are affecting the morbidity / mortality of patients with gastric perforations.
Gastric perforation is now a common complication. It was rare until the end of the 19 th
century, but since then its frequency has increased progressively. Moreover, there was a curious change in incidence in the 19 th century, most perforations were gastric perforations and the majority affected women, especially girls aged from 10-28 years. By 1959, duodenal perforations greatly exceeded gastric, men were affected more than women and most cases occurring between 25-45 years
METHODOLOGY:
The study was conducted in Victoria Hospital, BMCRI, Bangalore, from NOV 2011 to MAY 2013. During this period the number of cases admitted and selected for analyzing the data was 30. These 30 cases were studied thoroughly according to the proforma. The
Original Research Article Ramesh et al ., SAS J. Surg., 2016; 2(3):94-97
95
details of 30 patients were arranged in the master chart for convenience of presentation.
The diagnosis was made on clinical findings supported by investigations like plain x-ray abdomen erect posture. In cases managed surgically, confirmation was made on the operation table only and intra-operative edge biopsy taken to look for malignancy and H.Pylori.
A detailed history was taken when the condition of the patient is stable. In critically ill patients, the patients were resuscitated and history was taken after the patient was stabilized. The hospital records were also reviewed to obtain appropriate epidemiological information regarding age, sex, occupation, and clinical presentation, duration of symptoms, past history of chronic gastric ulcer, investigations and mode of treatment.
For selecting a case for definitive surgery most times general condition of the patient taken up for surgery and also operating findings were taken into consideration. In those cases, where both these conditions were satisfactory, definitive surgery was performed, giving weightage to the choice of the surgeon. In all other cases of perforation, surgery was done to close the perforation expect where condition of the patient was very poor (shock at the time of
Examination:
All the patients with suspected peptic ulcer perforation were examined thoroughly and base line findings were recorded, repeated examination of the patient was done resuscitation and till the diagnosis is confirmed.
Tachycardia associated with fever, tenderness in the epigastrium and abdominal rigidity pointed towards the diagnosis of peritonitis. I examined all the patents as per the proforma. In the all patients, with peptic ulcer perforation complete physical examination to rule out associated disease was done.
Investigations:
Relevant investigations were done like plain x-ray erect abdomen, blood grouping and typing, CBC, BT, CT, blood urea, serum Creatinine, Serum. Electrolytes, ultrasound of abdomen.
Paracentesis:
Diagnostic peritoneal tap was done. Fluid drawn was found to be turbid and bile stained indicating peptic ulcer perforation and in few cases with pus and flakes as tap content. Intra operative biopsy of ulcer for H.Pylori infection and malignancy
Prognostic scoring system to mention about the general condition of the patient
General Condition:
1) Good-Patient is conscious and cooperative.
-Pulse rate < 90/min
-BP 120/80mm Hg.
Urine output good
No associated medical problems like-hypertension, diabetes mellitus, tuberculosis or Myocardial infraction.
2) Average- Patient conscious
Pulse rate 90-110/min.
BP 120/80mm Hg.
-Urine-Oliguria
No anyone associated medical illness.
3) Poor - Patient conscious and poor orientation
Hippocratic facies
Pulse rate-tachycardia > 120/min and low volume
BP-systolic < 80mm Hg or not recordable
Urine – anuria
Medical illness may or may not be present
Outcome of the patient (recovery):
Good – Discharge at 7 th – 9th postoperative day, without intra or postoperative complications.
Average – Intraoperative anaesthesia complication – Postoperative complications like brouncho pneumonia, wound gaping, wound infection, but recovery before discharge.
Poor – Patient survived with burst abdomen / enterocutaneous fistula / severe malnutrition.
Death – In the postoperative period.
RESULTS:
Pain was the presenting symptom in all cases and onset was acute in all of them. In most cases pain was situated at the epigastrium, and right hypochondrium. Vomiting was present in 18 cases started along with the pain abdomen and contained food particles and bile. Distension was most commonly observed in cases with >24hrs history of symptoms in all cases and in all cases after 48hrs.
Rigidity was present in all cases. Abdomen was board like and tense in all the cases. Liver dullness was obliterated in 28 cases. In two cases there was no obliteration of liver dullness. None obliterated of liver dullness may be due to adhesion to some inflammatory pathology earlier. Bowel sounds were either sluggish or absent in most of the cases
Table1: Mode of presentation: Symptoms:
Symptoms No.of cases
Pain abdomen 30
Distension of abdomen 27
Vomiting 20
Fever 10 Ramesh et al ., SAS J. Surg., 2016; 2(3):94-97
96
Table 2: Mode of presentation: signs
Signs No.of cases
Dehydration 16
Shock 11
Pallor 15
Distension 30
Tenderness 30
Rigidity 30
Obliteration of liver dullness 28
Absent bowel sounds 30
INVESTIGATIONS:
The higher hemoglobin could be due to hemo concentration. Total count was raised above 11,000 cell/mm3 in 18 patients. Six patients were in pre renal type of acute renal failure. Altered liver function test was demonstrated in two patients. In erect abdomen X-ray Gas under the diaphragm was observed in 24 patients (80%).
Table 3: X ray
Plain x-ray abdomen (erect) No.of cases
Positive 24
Negative 06
All patients were put on drip and suction, antibiotics consisting of a Cephalosporin, aminoglycoside and an antimicrobial against anaerobes (Metronidazole). A watch was kept on vital signs and abdominal girth. All patients were taken up for emergency laparotomy.
Table 4: Type of Anesthesia
Type of Anesthesia No of cases
General Anesthesia 27
Epidural Anesthesia 03
Midline incision made in all 30 cases.
Table 5: Peritoneal Fluid
Type of Peritoneal fluid No of cases
Greenish 24
Feculent None
Purulent 02
Flakes 04
Peritoneal fluid varied from 500ml to 2 liters.
Table 6: Site Of Perforation
Site Of Gastric
Perforation
No Of
Cases Percentage
Pyulorus 10 33.33
Prepyloric 12 40
Antral 07 23.33
Lesser Curvature 01 3.33
Table 7: Size of Perforation
SIZE NUMBER OF
CASES
SHOCK
<0.5cm 12 01
0.6 -1cm 15 06
1cm 03 02
POST OPERATIVE MANAGEMENT:
Ryle’s tube aspirate is average for 2 days for
all the patients. V-fluids given are dextrose, dextrose with saline, ringer lactate, normal saline, isolate G.
Antibiotics used – Cephalosporins, Anti anerobics were used. Electrolytes imbalance-ten patients developed electrolytes imbalance, two developed acute renal failure, managed conservatively and recovered.
Table 8: Complications
Complications No Of Cases.
Average days of hospitalization
Bronchopneumonia 04 15
Wound infection 05 14
Residual abscess 01 21
Burst abdomen 01 21
No complications 19 08
Among 30 patients studied, 11 patients developed complication and remaining 19 patients had smooth recovery. Most common postoperative complication was wound infection in about 5 cases. 4 patients had broncho pneumonia, one had residual abscess managed by ultrasound guided and one patient had burst abdomen. 2 patients died within 48-72hrs of postoperative period. These patients presented with severe shock and septicemia and died because of multiorgan failure.
Out of 30 patients studied in our series, 2 patients died. All 28 patients were advised anti-H-pylori treatment with omeprazole, amoxicillin and metranidazole for one week followed by omeprazole 20mg OD. For 3 months and follow up every month. In the follow up of 3 months period only 4 patients out 28 patients who came for follow-up complained of pain abdomen, suggestive of peptic ulcer disease. They were advised endoscopy and definitive surgery and put on medical line of treatment. Out of 4, only 1 turned up for definitive surgery.
DISCUSSION:
Walgenbach S and Bernhard G analyzed that time interval between onset of acute symptoms and surgery was less than or equal to 2 hours mortality rate is 12% and if more than 24 hours the mortality rate is 21%. The mortality risk for a patient who is operated on more than 24 hours after the onset of acute symptoms is 4.9 times to that of a patient operated within 24 hours. Ramesh et al ., SAS J. Surg., 2016; 2(3):94-97
97
So the interval between the time of perforation and surgery has a very strong significance in deciding the mode of treatment i.e. type of surgery to be planned and outcome of the disease . 60% of patients reached the hospital >24hrs after the onset of symptoms. Most of our patients are from rural area, probably be the reason for the delay.
Table 9: Duration of symptoms before presentation of hospital
Duration (in hours)
De Bakey Series
Bharati CRamesh et
al.;
Present series
0-6 50.83% 12.00% 0%
7-12 13.02% 12% 10%
13 -24 4.73% 24% 30%
24 13.60% 64.00% 60%
Tsugawa K et al.; reviewed that three risk factors: pre-operative shock, delay to surgery over 24 hours and medical illness, was shown by the progressive rise in the mortality rate with the increasing number of risk factors . Boey John et al.; revealed concurrent medical illness, pre-operative shock and delayed operation (>48hours) as significant risk factors that increase mortality in patients with perforated duodenal ulcers . In the present study (2005) we reported that age, site of perforation, size of perforation, duration of perforation, H-pylori infection, pre-operative shock are the risk factors for the outcome of perorated peptic ulcer. The mortality and morbidity are increased whenever, perforation exceed 12 hours because of the peritoneal infection [9, 10]. In the presence of gross contamination, late exploration (after 48hours) carried a high mortality i.e.50% et al.; . The importance of the peritoneal spoilage and duration of perforation is mentioned as a risk in the outcome of the perforation of duodenal ulcer . Bharati CRamesh et al.; reported than 12% of patients reached the hospital within 12 hours, 40% reached hospital within 25-48 hours and 24% after 48 hours .
In the present series (2005) 35% patients presented to hospital after 24 hours and the mortality in patients who presented to hospital after 24 hours is found to be 8.5%.
CONCLUSION:
Perforation of more than 0.5cm size has high a morbidity of, which indicates that size of perforation, has a significant role in prognosis
REFERENCES:
WM David, KR Emily; Stomach. Townsend, Beauchamp, Mark Evers, Mattox. Sabiston textbook of surgery. 18 th edition. Vo12.Philadelphia, Elsevier; 2008; 1236-56.
Daniel TD; Stomach.Brunicardi, Andersen, Billian,
Dunn, Hunter, EP Raphel. Achwartz’s principles of
surgery.8 th edition. USA, McGraw Hill; 2005:933-70.
Walgen bach S, Bernhard G, Dürr HR, Weis C;
Perforation of gastroduodenal ulcer: a risk analysis.
Med Klin (Munich ) 1983 ; 87(8):403 -407 .
Walgen bach S, Bernhard G, Dürr HR, Weis C;
Perforation of gastroduodenal ulcer: a risk analysis.
Med Klin (Munich) 1983 ; 87(8):403 -407 .
DeBakey ME; Acute perforated gastroduodenal ulceration. A statistical analysis and review of the literature. Surgery 1940; 8(5):852-884.
Bharti C Ramesh, Marwaha Dc; Immediate definitive surgery in perforated duodenal ulcer: a comparative study between surgery and simple closure. Indian J Surg 1996; 275-279.
Williams L Peter; The development of the alimentary and respiratory apparatus, Embryology,
Gray’s Anatomy, 37 th Edition, 1989; The
Abdomen, splanchnology, Gray’s Anatomy, 37 th
Edition; 1989; 1333-1416: 227-248.
David M.Mahvi, Seth B. Krantz; stomach, sabiston Textbook of surgery, 19 th Edition, 2013; 1182-1226.
Chalapathy Rao PV ; Experiences tic ulcer. Postgraduate medicine, 1962; 32: 119-126. With emergency vagotomy and drainage procedure for perforated duodenal ulcer, Indian Journal of Surgery, 1981; 419-423.
Lawal O.O, Oluwole S.F, Fadiran O.A, Campbell
B; Clinical pattern of perforated prepyloric and
duodenal ulcer at Ile -Ife, Ne \igeria. Trop.Doc,
1998; 28(3):152 -155.
Fombellids J Deus ; Risk factors in the surgical management of perforated peptic ulcer. Rev. Esp. Enferm. Dig., 1998 ;( 7):502-513. |
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Tangent (geometry)
A tangent line is a line that closely approximates a curve at a point. That is, if you zoom in very closely, the tangent line and the curve will become indistinguishable from each other at a certain point where they intersect.
Contents
1 Intersection
2 Tangents to Circles
3 Generalizations
4 See also
Intersection
Locally, a tangent line intersects a curve in a single point. However, if a curve is neither convex nor concave, it is possible for a tangent line to intersect a curve in additional points. For instance, the tangent line of the curve at intersects it in 1 point, while the tangent line at intersects it in 2 points and the tangent line at intersects it in infinitely many points (and is in fact the tangent line at each point of intersection).
At a given point, a curve may have either 0 or 1 tangent lines. For the graph of a function, the condition "having a tangent line at a point" is equivalent to "being a differentiable function at that point." It is a fairly strong condition on a function -- only continuous functions may have tangent lines, and there are many continuous functions which fail to have tangent lines either at some points (for instance, the absolute value function at ) or even at all points!
Tangents to Circles
When the curve being considered is a circle, the tangent has many nice properties. For example, it is perpendicular to the radius that passes through the point of tangency. Any two disjoint circles have four tangents in common, two internal and two external.
Generalizations
Within the realm of differential geometry, the generalization of a tangent is a tangent space. Generally, the tangent space to a k-form is also a k-form. This may not always be the case though as you can have tangent “planes” to a volume or surface form.
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Calculus
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17307 | https://convert-formula.com/mah-hours | Convert mAh (milliamp hours) to Time in Hours
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The formula is (mAh)/(Amps1000) = (Hours). For example, if you have a 3000 mAh battery that runs at 0.2 Amps (0.2Amps = 200mA), then the time that the battery will last for is (3000)/(0.21000) = (3000)/(200)= 15 Hours.
Option B) Enter the mAh, the Volts and the Watts below, and click on the button to calculate the hours:
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If you have voltage and watts, then the formula is (mAhVolts)/(Watts1000) = (Hours). For example, if you have a 3000 mAh battery that runs at 5 Volts and 15 Watts, then the time that the battery will last for is (30005)/(151000) = 1 Hours.
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Liberty University
General Chemistry II
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Calculate Ka value for NH4NH4
Liberty University
General Chemistry II
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Anonymous Student
1 year ago
Calculate Ka�� value for NH4+NH4+.
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Answer Created with AI
1 year ago
Calculation of Ka Value for NH4+
The Ka value, or acid dissociation constant, is a measure of the strength of an acid in solution. It is calculated using the concentrations of the reactants and products at equilibrium. For the ammonium ion (NH4+), the dissociation reaction is as follows:
js
NH4+ <=> NH3 + H+
The Ka expression for this reaction is:
js
Ka = [NH3][H+] / [NH4+]
To calculate the Ka value, you need the equilibrium concentrations of NH3, H+, and NH4+. These values are typically given in the problem or can be determined from initial concentrations and the stoichiometry of the reaction.
However, for NH4+, it's more common to refer to the Kb of NH3 (ammonia) and use the relationship between Ka, Kb, and Kw (the ion product of water) to find the Ka. This is because NH3 is a weak base and its Kb is more commonly tabulated. The relationship is:
js
Ka Kb = Kw
Where Kw is the ion product of water, which is 1.0 x 10^-14 at 25 degrees Celsius. If you know the Kb of NH3, you can rearrange this equation to solve for Ka:
js
Ka = Kw / Kb
For example, if the Kb of NH3 is 1.8 x 10^-5, the Ka of NH4+ would be:
js
Ka = (1.0 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10
This is a simplified example and actual calculations may require taking into account reaction stoichiometry and initial concentrations. Always make sure to check the conditions of your specific problem.
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Showing that if f′′(x)=0 f″(x)=0, f(x)=a x+b f(x)=a x+b without integrating
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Assume that f:A→R f:A→R is two times differentiable with f′′(x)=0 f″(x)=0 for all x∈A x∈A, with A A an interval.
Show, (not by integration), that f f is of the form f(x)=a x+b f(x)=a x+b for some constants a,b a,b.
My answer: I will use the Mean Value Theorem:
Take x,y∈A,x<y x,y∈A,x<y. Applying the Mean Value Theorem on [x,y][x,y] gives:
f′′(c)=f′(y)−f′(x)y−x f″(c)=f′(y)−f′(x)y−x for some c∈A c∈A. If f′′(x)=0 f″(x)=0 for all x∈A x∈A, then f′′(c)=0 f″(c)=0, wich means that f′′(c)=f′(y)−f′(x)y−x=0 f″(c)=f′(y)−f′(x)y−x=0, which implies that f′(y)−f′(x)=0 f′(y)−f′(x)=0. So f′(y)=f′(x)f′(y)=f′(x). Set k k equal to this common value. Because x and y are arbitrary, it follows that f′(x)=a f′(x)=a for all x∈A x∈A
2. We are now given that f′(x)=a f′(x)=a for all x∈A x∈A. This means that, if we take x,y∈A,x<y,x,y∈A,x<y,, by applying the Mean Value Theorem on [x,y],
f′(c)=f(y)−f(x)y−x f′(c)=f(y)−f(x)y−x for some c∈A c∈A. We know that f′(x)=a f′(x)=a for all x∈A x∈A, which means that f′(c)=f(y)−f(x)y−x=a⟹f(y)−f(x)=a(y−x)f′(c)=f(y)−f(x)y−x=a⟹f(y)−f(x)=a(y−x)
What's next?
I could also use the definition of derivative, considering:
f′(x)=a=lim x→c f(x)−f(c)x−c f′(x)=a=lim x→c f(x)−f(c)x−c
How can I show that f(x)=a x+b f(x)=a x+b ?
real-analysis
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edited Oct 22, 2012 at 19:08
Applied mathematicianApplied mathematician
asked Oct 22, 2012 at 17:26
Applied mathematicianApplied mathematician
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1 Change to y−x y−x your denominator in (1)-(2)DonAntonio –DonAntonio 2012-10-22 17:35:26 +00:00 Commented Oct 22, 2012 at 17:35
1 At the end of point 1., you write that f′(x)=f′(y)=0 f′(x)=f′(y)=0 and then you write that f′(x)=k f′(x)=k for all x∈A x∈A. Am I not seeing why those two statements don't conflict? Did you mean to write f′(x)−f′(y)=0 f′(x)−f′(y)=0?Todd Wilcox –Todd Wilcox 2012-10-22 17:36:56 +00:00 Commented Oct 22, 2012 at 17:36
@ToddWilcox Ty. I updated it. Do you guys know how to go further with this proof? :-)Applied mathematician –Applied mathematician 2012-10-22 17:39:40 +00:00 Commented Oct 22, 2012 at 17:39
1 you can't say ∀c∈A∀c∈A. It has to be ∀c∈]x,y[∀c∈]x,y[.user31280 –user31280 2012-10-22 18:15:11 +00:00 Commented Oct 22, 2012 at 18:15
1 @EmmadKareem No, because this is an exercise from Real Analysis. I have to understand simple things deeply :)Applied mathematician –Applied mathematician 2012-10-22 18:20:33 +00:00 Commented Oct 22, 2012 at 18:20
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Using the taylor expansion of f(x+0)=∑k≥0 x k f(k)(0)k!=f(0)+x f′(0)+x 2 f′′(0)2!+⋯f(x+0)=∑k≥0 x k f(k)(0)k!=f(0)+x f′(0)+x 2 f″(0)2!+⋯ Since f′′(x)=0 f″(x)=0 then every subsequent derivative is 0 0. Thus we have f(x)=f(0)+x f′(0)f(x)=f(0)+x f′(0) with a=f′(0)a=f′(0) and b=f(0)b=f(0).
EDIT: But the Taylor series of a function does not always converge to the function, in this case the Taylor Series is a polynomial. Check this proof to see how such functions equal to their Taylor series.
You see, no integration at all. This is one of the most important uses of Taylor Series, solving differential equations. Check here for futher details.
Lemme restart your work:A=[c,d]A=[c,d]
Given a function f(x)∈C 2[c,d]|f′′(x)=0,∀x∈[c,d]f(x)∈C 2[c,d]|f″(x)=0,∀x∈[c,d]. Using the Mean Value Theorem, since the second derivative is 0 0 then f′(x)=a∈[c,d]f′(x)=a∈[c,d].
Using the theorem again on f′(x)f′(x), we have that ∀x∈[c,d],∃b 1∈[c,x]|f′(b 1)=f(x)−f(c)x−c=a∀x∈[c,d],∃b 1∈[c,x]|f′(b 1)=f(x)−f(c)x−c=a and ∃b 2∈[x,d]|f′(b 2)=f(d)−f(x)d−x=a∃b 2∈[x,d]|f′(b 2)=f(d)−f(x)d−x=a. Since the function is two times derivable then both the right and left derivative should be equal. Thus we have f(d)−f(x)=a(d−x)f(d)−f(x)=a(d−x) and f(x)−f(c)=a(x−c)f(x)−f(c)=a(x−c) which implies
f(x)=a x+f(d)−a d f(x)=a x+f(c)−a c f(x)=a x+f(d)−a d f(x)=a x+f(c)−a c
Now we have a problem where we have to prove that f(d)−a d=f(c)−a c f(d)−a d=f(c)−a c. Assuming it is true then f(d)−a d=f(c)−a c⇒a=f(d)−f(c)d−c f(d)−a d=f(c)−a c⇒a=f(d)−f(c)d−c. Since a=f′(x)=f(d)−f(c)d−c,∀x∈[c,d]a=f′(x)=f(d)−f(c)d−c,∀x∈[c,d], then our proof is complete and we can conclude that f(d)−a d=f(c)−a c=b∈[c,d]f(d)−a d=f(c)−a c=b∈[c,d] and that
f(x)=a x+b f(x)=a x+b
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edited Apr 13, 2017 at 12:20
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answered Oct 22, 2012 at 17:50
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Thank you for your help. But I have to use the theory of derivatives, especially the mean value theorem :)Applied mathematician –Applied mathematician 2012-10-22 18:05:57 +00:00 Commented Oct 22, 2012 at 18:05
1 check now! I've used the Mean Value Theorem user31280 –user31280 2012-10-22 19:00:47 +00:00 Commented Oct 22, 2012 at 19:00
very good, thanks!!!Applied mathematician –Applied mathematician 2012-10-22 19:09:25 +00:00 Commented Oct 22, 2012 at 19:09
I'm just asking: Is it clear that the Taylor Series really is equal to the function, i.e. that f is analytical?AndreasS –AndreasS 2012-10-22 19:13:15 +00:00 Commented Oct 22, 2012 at 19:13
2 @Hempo: It's probably worth noting that these two proofs are in some sense the same proof expanded differently -- the way you prove Taylor's theorem is by repeatedly applying the MVT to the derivatives of your function...Micah –Micah 2012-10-22 20:47:45 +00:00 Commented Oct 22, 2012 at 20:47
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You're effectively done once you have f(y)−f(x)y−x=a f(y)−f(x)y−x=a. For this can be rewritten as
f(y)−f(x)=a(y−x)f(y)−f(x)=a(y−x)
or just
f(x)=a x+a y−f(y)f(x)=a x+a y−f(y)
This holds for all x x and y y in your interval. So you can fix one such y y, and define b=a y−f(y)b=a y−f(y). Then the above equation is exactly
f(x)=a x+b f(x)=a x+b
Since this holds for all x x in the interval you are done.
By the way, if you want to use Taylor series, you can use the form with remainder which says that for fixed x 0 x 0 in the interior of your interval, for any x x in the interval you have
f(x)=f(x 0)+f′(x 0)(x−x 0)+1 2 f′′(c x)(x−x 0)2 f(x)=f(x 0)+f′(x 0)(x−x 0)+1 2 f″(c x)(x−x 0)2
Here c x c x is between b b and x x. Since f′′(c x)=0 f″(c x)=0, this means that for all x x you have
f(x)=f(x 0)+f′(x 0)(x−x 0)f(x)=f(x 0)+f′(x 0)(x−x 0)
This is of the form a x+b a x+b as needed.
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edited Oct 22, 2012 at 21:39
answered Oct 22, 2012 at 21:24
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Just finish it from your last line. Fix any y 0 y 0. Then, f(x)=a(x−y 0)+f(y 0)f(x)=a(x−y 0)+f(y 0) for all x x, i.e. f(x)f(x) is a line.
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answered Oct 22, 2012 at 19:34
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鼻咽的位置：
ˉ鼻咽（Nasopharynx）又稱「後鼻部」，位於鼻腔正後方，咽喉的上方，腦部下方，本身像個洞穴。兩旁各有一個耳咽管開口，而中央接後鼻孔，是呼吸道必經之地。 鼻咽身處深處，既看不著也摸不著，因此鼻咽有問題時很難自我察覺。
中國人特別好發
ˉ根據統計，男性每十萬人每年得鼻咽癌(NPC)的人數如下表：
| 種族 | 中國 | 中國 | 中國 | 中國 | 中國 | 美國黑人 | 美國白人 | 日本 | 英國 |
--- --- --- --- --- |
| 地方 | 香港 | 新加坡 | 洛杉磯 | 臺灣 | 上海 | 洛杉磯 | 洛杉磯 | 大阪 | 伯明罕 |
| NPC | 25.8 | 19.4 | 14.6 | 8.8 | 5.6 | 0.8 | 0.7 | 0.4 | 0.4 |
ˉ可見鼻咽癌特別好發於中國人，尤其是東南沿海各省，故又稱「廣東瘤」。日、韓雖亦為蒙古種，但比率甚低。中國人第二代移居美國鼻咽癌比例雖略降，但仍遠高於外國人。在西方世界，鼻咽癌非常罕見。 在臺灣，鼻咽癌是男性癌症死亡原因第六位，女性第九位，每年約有1000位鼻咽癌新病例，每年約有800位死於鼻咽癌。（1993年資料）
ˉ鼻咽癌發病年齡平均為40歲，較其他癌症平均提早20年發病。男女比例為3比1。
然而現今，十到二十幾歲的病例越來越多了，事實上，從三個月的嬰兒到八十多歲老人都有過鼻咽癌的病例報告。
＊病因：仍未明瞭，可能原因是：
(一)種族、遺傳這是目前病因中唯一最肯定的因素。可從三方面證明：
(1)中國南方人好發：ˉ即使遷移海外罹病率亦比白人多20倍。一般認為「南蒙古人種」(長江以南)比「北蒙古人種」(長江以北、日、韓)易得鼻咽癌。
(2)HLA（組織配對抗原）：ˉ位於人體第六對染色體上的一組基因群，像有A2-BW46組合者罹病率是3.4倍，AW19-B17是2.2倍。
(3)鼻咽癌家庭史：ˉ家族中有人得鼻咽癌，則自己得鼻咽癌是常人的6-10倍，若一等親有人得鼻咽癌，則自己得鼻咽癌是常人的60-100倍。
(二)EB病毒:ˉ鼻咽癌病人血清中EB病毒抗體指數都很高，因此認為鼻咽癌和EB病毒有某種關係，但確實關聯性仍未明瞭。有95%的中國人帶有EB病毒抗體（感染過），而外國人則很少發現有EB病毒感染。
(三)飲食
ˉ從小時候就常食用醃製品、發酵品，則長大後得鼻咽癌比率大增，如常攝食發酵類食物（每週二餐）得鼻咽癌危險度為常人五倍，醬菜類食物（每週五餐）危險度為七倍。
(四)工作環境
ˉ在不流通的密閉空間或有空氣污染的環境下工作，「可能」易得鼻咽癌，但未證實。如85年4月電信局3員工疑似因長期接觸「硫酸氣」而致鼻咽癌。
(五)吸煙、喝酒、嚼檳榔
ˉ長期刺激鼻咽表皮可能易致癌，但未證實。
(六)鼻竇炎病史
有鼻竇炎病史者得鼻咽癌是常人的5.3倍，可能是細菌感染長期刺激鼻咽表皮所致。故有鼻竇炎者應儘速治療。
早期症狀
鼻咽位於深處，早期症狀常不明顯而被忽略。但早期發現的治癒率極高(90%以上)，應特別重視。
頸部腫瘤(40%) 是最常見的症狀。不明原因的頸部腫瘤，尤其是長在耳垂下方，上頸部兩側，不痛不癢，三星期以上未消腫者，要懷疑是鼻咽癌。
流鼻血（鼻水中帶血、痰中帶血）(30%) ˉ一般鼻中膈、鼻前庭流血都為鮮血，量可多可少，而鼻咽癌組織破裂初期只會微量流血。因此若只是鼻水中略帶血絲，或咳痰中有血絲，通常是暗紅色或帶血塊時，應就醫檢查！
單耳的耳鳴、重聽(25%)ˉ因癌細胞侵入耳咽管，造成中耳積水引起的。若有耳塞、耳鳴或重聽，甚至覺得耳內有水流聲，尤其是單耳，且沒有感冒或鼻竇炎時，應就醫檢查鼻咽！
中晚期症狀
頭痛(15%)ˉ癌細胞已侵入顱骨底。
眼皮下垂、臉麻、嘴角歪一邊、吞嚥困難、聲音沙啞：(腦神經麻痺，5%)癌細胞侵犯顏面神經會有臉麻、嘴角歪一邊的現象。
鼻塞、流鼻涕,ˉ癌細胞已侵入鼻腔，併發鼻竇炎，為晚期症狀。
診斷：
[主要診斷]
咽鏡切片檢查：ˉ傳統是以一個圓形小鏡子從口腔裡檢查鼻咽；以纖維性鼻咽鏡由鼻孔內檢查會更清楚，懷疑時作病理切片，以做正確診斷。
[輔助診斷]
電腦斷層檢查(CT SCAN)
ˉ若鼻咽鏡未看見腫瘤，但臨床上懷疑時可作電腦斷層，以評估腫瘤位置及大小。
EB病毒血清
ˉEB病毒血清不可作為診斷依據，只可作診斷參考，因EB指數高者不一定有鼻咽癌，只是機率較高而已，正確診斷仍需鼻咽切片。
在鼻咽癌病人治癒後，EB指數會下降；復發時，EB指數又會上升。因此EB病毒血清可當做鼻咽癌病人治療前後的反應指標。
EB病毒血清種類很多，針對鼻咽癌常見的有：
(1)VCA IgG, IgA（Viral Capsid Antigen, 病毒外鞘抗原抗體）：最常用。
(2)EA IgG, IgA（Early Antigen, 早期抗原抗體）
(3)DNAase（抗病毒DNA抗體）：仍在發展中。判讀如下：
| EB指數 | EB VCA | EB EA | EB DNAase |
--- --- |
| 易得鼻咽癌 | >1：40 | >1：40 | ？ |
（註：如結果為1：80, 1：1600等，表示>1：40，容易得鼻咽癌）
[遠方轉移]ˉ鼻咽癌的癌細胞遠方轉移率為30%，一半以上是骨轉移，其次依序為肝、肺、腦。以下檢查是在證實罹患鼻咽癌後，檢查是否轉移：骨核醫掃瞄(Bone Scan),肝超音波,胸部Ｘ光
鼻咽癌的治療原則
放射治療為主
鼻咽癌細胞對放射治療的效果極佳，早期（第一期）治癒率可達90%以上，第二期70%，第三期50%，第四期20%。
化學治療為輔
ˉ對於中晚期病人放射治療時同時做化學治療可增加治癒率、減少轉移率。
4. 若有遠方轉移（骨、肝、肺、腦）
ˉ無法治癒，只能追加放射治療、化學治療減慢病程。
放射治療(Radiotherapy, 俗稱「電療」，以下簡稱「放療」）
材料：「鈷60」或「直線加速器」
療程「分段式放療」：較常採用。若以總照射量7000雷得(rad)為例，則以每天照射200雷得，每週一至週五各一次，共七週完成。
「高分段式放療」：為上下午各照射一次，可縮短總療程，但早期併發症較大，較難忍受，且不適於同時做化療。
3. 照射範圍：由鼻咽到下頸部。
輔助性化學治療(Chemotherapy,以下簡稱「化療」)
化學治療本身不能治癒鼻咽癌，但可縮小癌細胞範圍，並減低遠方轉移率！（因一但轉移就很難治癒了！）。
多使用在中、晚期病人，可在放射治療前、中、後做化療一次或多次。現在以「同時放療＋化療(Concomitant Chemo-radiotherapy)較常採用，但併發症也較大。
治療後的追蹤
一般若五年內無復發或轉移現象，大都表示已治癒了！放療後請配合醫生做定期追蹤檢查。
追蹤項目：
問診、視診
EB病毒血清: 在鼻咽癌病人治癒後，EB指數會下降；復發時，EB指數又會上升。因此EB病毒血清可當做鼻咽癌病人治療前後的反應指標。 鼻咽鏡電腦斷層檢查
放射治療及化學治療之早期併發症
指放療中及放療後數月內產生的，大都是暫時的，除「口乾舌燥」外，大都會恢復。
口腔破皮(100%):會發紅、破皮、流血，同時做化療會更明顯，影響飲食。應多漱口，注意口腔衛生，少吃辛辣及粗糙食物，勿塗抹軟膏。
口乾舌燥(100%): 因放療破壞唾液腺分泌，口水漸漸減少，形成乾性咽喉炎，是永久的後遺症。應多補充水分。
味覺喪失(30%)：約2-4個月後會恢復。
皮膚炎:皮膚發紅、發黑、脫皮，約在放療一個月後恢復。應忌日曬，外出要穿高領衣服。
嘔吐感、掉頭髮：化療較常見。頭髮會在兩個月內長出。
放射治療的晚期併發症：
是放療後數月到數年後才產生的後遺症，都是不易恢復的。化療則很少有晚期併發症。
中耳積水：最常見因放療破壞耳咽管功能造成。可重覆用細針抽水治療。儘量勿放置中耳通氣管，因為會增加感染率。
牙齒病變：因放療後牙齒缺乏口水保護，蛀牙處易加速損壞。故放療前應將蛀牙做根管治療或拔除。
萎縮性鼻炎、鼻竇炎鼻粘膜被破壞所致，易有鼻痂、流鼻血、鼻涕，應定時請醫生清除。
口乾舌燥：如前。
牙關緊閉:是因咬肌放療後纖維化所致，預防法應在放療期多做咀嚼運動（如嚼口香糖）。
頸部僵硬：因頸部照射量大時引起皮下組織硬化。
感音性重聽：一年後才產生，高音頻較受影響。
骨壞死：很少發生，因蛀牙未處理而感染會導致下頜骨壞死，而鼻咽照射量較大時會使顱底骨壞死造成鼻出血。
內分泌功能低下:生長激素、甲狀腺素、性賀爾蒙等都可能降低，症狀不明顯，應抽血檢驗，若需要則補充賀爾蒙。
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17311 | https://www.mav.vic.edu.au/Tenant/C0000019/00000001/downloads/Resources/annual-conferences/2008/Leong/3%20A%20Leaky%20Tank%20-%20calculus%20problem.pdf | (Adapted from: Elementary Principles of Chemical Processes 3rd edition, Felder and Rousseau, pp547-549) A Leaky Tank Prior knowledge - Differentiation of polynomials, finding stationary points and intercepts. Task: A 12.5 m3 tank is being filled at a rate of 0.05 m3/s. The moment the tank reaches 120m3 of water a bottom leak forms and gets progressively worse with time. The rate of leakage can be approximated as 0.0025t (m3/s), where t is the time in seconds from the moment the leak begins. 1. Write a differential equation for the rate the volume changes (dV / dt). Give initial conditions for your equation. 2. Give an expression for V (solve for V ) 3. Graph V versus t, and explain what it means, and find any other critical information. The solutions: Conditions: Time is taken from the moment the leak begins : i.e.: when t = 0 , V = 0 1. dV dt = rate water in - rate water out = 0.05 – 0.0025t 2. IntegratingV = 0.05t – 0.00125 t 2 + C Substituting in to find C: t = 0 , V = 0, we get C = 1.2 Thus V = 0.05t – 0.00125 t 2 + 1.2 3. Will the City Reservoir dry up? A more advanced problem similar to the above. Involves integrating ex TASK: The water level in a city reservoir has been decreasing steadily during a dry spell, and there is concern it could dry up in the next 60 days if the drought continues. The local water company estimates the consumption rate is approximately 107 L/day and the State Conservation Service estimates the rainfall and stream drainage into the reservoir coupled with evaporation from the reservoir should yield a net water input rate of 106exp(-t/100) L/day, where t is time in days from the beginning of the drought, at which time the reservoir contained 109 L of water. Solutions: let V = volume in reservoir (Litres), t = time (days) dV dt = 10 6 e – t 100 – 10 7 , integrate to get V = -10 8 e – t 100 – 10 7 t + 10 9 + 10 8where t = 0, V = 109 , Evaluating V when t = 60 days, we get V = 4.45 x 108 L t (s) 10 20 30 40 50 60 V (m3/s) 1 2 The graph indicates that the water initially rises in the tank, as the leak gets larger the tank begins to drain. The maximum volume can be found to be 1.7m3, well below the capacity of 12.5m3. At 57 seconds the tanks is completely dry (x-intercept), and the function gives negative values, but the volume actually remains zero so the actual solution for V is a piece-wise function V = 1.2 + 0.05t – 0.00125t 2 0 0 t 57s t > 57s w ater in 0.05m3/s w ater out 0.0025m3/s tank capacity 12.5 m3 Volume V t=0, V=1.2m3 |
17312 | https://www.omnicalculator.com/conversion/seconds-to-years-conversion | Board
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Seconds to Years Conversion
If you're wondering what it's like to do a seconds to years conversion or vice-versa, you're not alone. Our seconds to years calculator is here to simplify the process for you.
Whether you're doing scientific research, tracking a long-term project, or converting your age from years into seconds, we've got you covered.
Together, let's explore the fascinating world of time conversion with our handy tool, which is designed to convert seconds to years along with many other time measuring units, and turn those years back into 31,536,000 seconds or 31,557,600 seconds for better accuracy and a new and intriguing perspective.
Using seconds to years conversion tool
You can quickly do seconds to years conversion or vice-versa using our calculator:
Input the desired number of seconds or years in our seconds to years conversion tool.
Next, wait; there is no next step... You have your result. Our seconds to years calculator handles the rest and immediately returns the result.
Let's suppose you enter 500,000 seconds out of sheer curiosity. The result is 0.015844 years.
Since it is quite useful for scientific, educational, and personal purposes, let's also learn how to do it manually next, without using any converters.
Learn how to convert years to seconds
Converting seconds to years is quite straightforward, so let's dive in and see how we can do it in situations where we can't take assistance from a tool.
🙋 Generally, people calculate a year with 365 days, but because we have a leap year every four years, the figure we use is closer to 365.25 days. Leap year rules get even more complex than this, but they won't affect your calculation unless you are dealing with many trillions of seconds.
Now, let's calculate this with better precision:
1 year = 31,557,600 seconds.
Inversely:
1 second = 0.00000003169 years.
Thus, to convert seconds into years, we can use the following formula:
Years = seconds × 0.00000003169
For instance, to convert 10,000,000 seconds into years, multiply it by 0.00000003169, which is ≈ 0.3169 years.
And, to convert years to seconds, we can use:
Seconds = years × 31557600
For instance, to convert four years into seconds, multiply it by 31557600, which is = 126,230,400 seconds.
Other useful date and time calculators
Here's a list of our other useful time calculators to help you stay on track:
Time unit converter;
Hours to week conversion;
Days to weeks calculator;
Weeks to months calculator;
Minutes to years conversion;
Hours to years calculator;
Days to years calculator; and
Weeks to years converter.
FAQs
How can I find the number of seconds in a year?
There are 31,557,600 seconds in a year.
You can find this number by multiplying the total number of days in a year with 24 hours, 60 minutes, and 60 seconds as follows:
1 Year = 365.25 × 24 × 60 × 60
1 year = 31,557,600 seconds
How many years are there in a trillion seconds?
There are approximately 31,688 years in one trillion seconds.
To calculate this, multiply the total number of given seconds, which is one trillion, with the number of years in one second, that is 0.00000003169, i.e.:
Years = seconds × 0.00000003169
1,000,000,000,000 × 0.00000003169 = 31,688
How can I convert 100 years into seconds?
To convert 100 years into seconds:
Multiply the number of years by 31,557,600 seconds, as 1 year is equal to 365.25 days, which is equal to 31,557,600 seconds, i.e.,
Seconds = years × 31557600
2. Placing the values, we get:
100 × 31557600 = 3155760000 seconds
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17313 | https://www.quora.com/What-is-a-critical-point-in-calculus | What is a critical point in calculus? - Quora
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What is a critical point in calculus?
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1y
In calculus, a critical point of a function is a point in its domain where the derivative is either zero or undefined. Critical points are important because they can indicate locations of local maxima, local minima, or saddle points on the graph of the function.
Definition:
Let f(x)f(x) be a function defined on an interval. A point c c in that interval is a critical point if:
f′(c)=0 f′(c)=0 (the derivative is zero), or
f′(c)f′(c) is undefined.
Importance of Critical Points:
Finding Extrema: Critical points are used to find local and absolute extrema of a function. By analyzing the behavior of the function around th
Continue Reading
In calculus, a critical point of a function is a point in its domain where the derivative is either zero or undefined. Critical points are important because they can indicate locations of local maxima, local minima, or saddle points on the graph of the function.
Definition:
Let f(x)f(x) be a function defined on an interval. A point c c in that interval is a critical point if:
f′(c)=0 f′(c)=0 (the derivative is zero), or
f′(c)f′(c) is undefined.
Importance of Critical Points:
Finding Extrema: Critical points are used to find local and absolute extrema of a function. By analyzing the behavior of the function around these points, one can determine if they correspond to local maxima or minima.
First Derivative Test: This test involves checking the sign of the derivative before and after the critical point to determine the nature of the critical point (increasing or decreasing behavior).
Second Derivative Test: This involves evaluating the second derivative at the critical point to determine concavity and confirm whether it's a local maximum or minimum.
Example:
For the function f(x)=x 2−4 x+3 f(x)=x 2−4 x+3:
Find the derivative: f′(x)=2 x−4 f′(x)=2 x−4.
Set the derivative to zero: 2 x−4=0 2 x−4=0 ⇒ x=2 x=2.
Check if the derivative is undefined: In this case, it is not undefined.
Thus, x=2 x=2 is a critical point. To determine its nature, you could use the first or second derivative test.
Overall, critical points are essential for understanding the behavior of functions in calculus.
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Related questions
More answers below
What does " critical point " mean?
What is the difference between stationary points and critical points?
What is the difference between critical point, critical number, and critical value in calculus?
Why are critical points important in calculus?
Does calculus have a point?
Fabio García
MSc in Mathematics, CIMAT (Graduated 2018) · Upvoted by
John George
, PhD Mathematics, The University of Alabama (1963) · Author has 247 answers and 623.3K answer views
·9y
You're correct to notice that "the interior of the domain" doesn't mean the same thing as "the domain". The interior of a set is its largest open subset; that is, the interior of a closed interval is the open interval that excludes the endpoints, while the interior of an open interval is the open interval itself.
The term "critical point" is pretty much another way to say "the derivative is zero here and that means I could have an extreme value". It's not very well-defined because the whole purpose of its existence is to make it easier to speak about finding maxima and minima of functions over
Continue Reading
You're correct to notice that "the interior of the domain" doesn't mean the same thing as "the domain". The interior of a set is its largest open subset; that is, the interior of a closed interval is the open interval that excludes the endpoints, while the interior of an open interval is the open interval itself.
The term "critical point" is pretty much another way to say "the derivative is zero here and that means I could have an extreme value". It's not very well-defined because the whole purpose of its existence is to make it easier to speak about finding maxima and minima of functions over an interval, and the technical details are of little practical relevance toward finding such extreme points.
The theorem that guarantees that a function achieves its maximum and minimum over an interval has several assumptions. The interval must be closed, the function must be differentiable over the interior and continuous over the whole interval, including the endpoints. Then you're guaranteed that the function attains its extreme values somewhere in the interval, and because local extremes can only occur where the derivative is zero, you can be sure that the maximum and minimum can be found only at the endpoints or at the points where the derivative is zero.
Some sources go on to abbreviate this as "critical points and endpoints [of the interval being studied"
which is not the same as "critical points and endpoints [of the domain of the function]"
which might be where your confusion is coming from.
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Bruce Balden
Experience in teaching at college level in the US and Canada · Upvoted by
John George
, PhD Mathematics, The University of Alabama (1963) · Author has 1.4K answers and 3.1M answer views
·9y
Let me just expand a little on the excellent response of Fabio García.
First, derivatives in the classic sense only exist for a point in the interior of the domain of a function. So if we are searching for extrema of f(x)f(x), then calculus helps to the extent that we include places where f′(x)=0 f′(x)=0 and that in turn is not defined the edges of the domain. But we want to define critical points to be all possible extrema. So what are the possible locations of an extremum on a closed interval?
endpoints (simple example f(x)=x f(x)=x which is steadily increasing on the whole real line. So for this function,
Continue Reading
Let me just expand a little on the excellent response of Fabio García.
First, derivatives in the classic sense only exist for a point in the interior of the domain of a function. So if we are searching for extrema of f(x)f(x), then calculus helps to the extent that we include places where f′(x)=0 f′(x)=0 and that in turn is not defined the edges of the domain. But we want to define critical points to be all possible extrema. So what are the possible locations of an extremum on a closed interval?
endpoints (simple example f(x)=x f(x)=x which is steadily increasing on the whole real line. So for this function, extrema for a closed interval occur at the endpoints).
classic critical points (interior points where the derivative is zero). Depending on other considerations, such a point may be a local extremum only, or nothing special.
Places where the derivative does not exist. The absolute value of otherwise smooth functions can (usually does) create "corners" where the function suddenly changes directions. It's customary to add such points to the list of critical points. Having done so, we have an exhaustive list of where to look for extrema of a piece-wise smooth function. The usual functions are very smooth, but absolute value is a typical way that corners are introduced. Example f(x)=|(x−1)(x−2)(x−3)|f(x)=|(x−1)(x−2)(x−3)| on the interval [0,4]
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Jony Valenz
BA in Pure Mathematics, University of California, Los Angeles (Graduated 2014) · Author has 120 answers and 108.8K answer views
·7y
a critical point is a point “x” in the Domain space of a function F:
F: Domain → Range
such that that dF/dx(“x”) = 0.
in other words, it is a point where the slope of a function F is equal to zero, and we recall that the slope can be found by differentiating a function to get a new function
dF/dx = F’
This is something that Wikipedia probably explains very well. Just google it.
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Related questions
More answers below
What is a critical point?
What is a critical point in aviation?
Where can we use basic calculus?
Why was calculus ever needed?
In calculus, if a critical value is not a min nor max, would it still be a critical point if I group it with a Y value?
Alex Sadovsky
Studied Mathematics&Biomechanics at University of California, Irvine · Author has 13.4K answers and 7.8M answer views
·7y
Asked and answered here: What does " critical point " mean?
Also, a quick Google search leads to this: Critical point (mathematics) - Wikipedia.
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Vishakh Rajendran
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What is the difference between a stationary point and a critical point in calculus?
A stationary point of a function is a point on the graph where the function’s derivative is zero.
A stationary point in one where there is a change in the slope behavior from being positive to negative (or) negative to positive. Informally, it is a point where the function "stops" increasing or decreasing.
Critical point is one at which
a. Either the function is not differentiable (or)
b. The derivative equals 0.
We already saw that a point where the derivative = 0 is the stationary point. Thus, stationary point is a subset under critical points.
If you have an equation of the form a/b, then the der
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A stationary point of a function is a point on the graph where the function’s derivative is zero.
A stationary point in one where there is a change in the slope behavior from being positive to negative (or) negative to positive. Informally, it is a point where the function "stops" increasing or decreasing.
Critical point is one at which
a. Either the function is not differentiable (or)
b. The derivative equals 0.
We already saw that a point where the derivative = 0 is the stationary point. Thus, stationary point is a subset under critical points.
If you have an equation of the form a/b, then the derivative does not exist if b = 0.
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9 7
Alec Brady
1 year of Pure Mathematics at university, and a lifelong fascination with it. · Author has 210 answers and 309.9K answer views
·8y
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How important is calculus?
I think the clearest answer to this question is this fact: that the Industrial Revolution only began after calculus became widespread. Newton's De Quadrature Curvarum was published in 1676, and his Methodus Fluxionum in 1736; Savery's steam engine was produced in 1698 and Watt's in 1778. These machines were the fruit of a massive improvement in the understanding of physics that would have been impossible without calculus.
The same is true today. Engineering would be simply impossible without calculus. It is the foundation of any approach to optimisation, of any study of equilibrium states, and
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I think the clearest answer to this question is this fact: that the Industrial Revolution only began after calculus became widespread. Newton's De Quadrature Curvarum was published in 1676, and his Methodus Fluxionum in 1736; Savery's steam engine was produced in 1698 and Watt's in 1778. These machines were the fruit of a massive improvement in the understanding of physics that would have been impossible without calculus.
The same is true today. Engineering would be simply impossible without calculus. It is the foundation of any approach to optimisation, of any study of equilibrium states, and of all dynamical systems. It appears in all branches of engineering, whether civil, mechanical, electrical, electronic, nautical or aeronautical. It is used in game theory and economics and statistics. It is essential for quantum theory, without which we wouldn't have lasers or computers.
So its importance for society is immeasurable. Without it we would gradually sink back to the technological level of the seventeenth century.
What about its personal importance? This is less easy to demonstrate. After all, billions of people who couldn't tell a Taylor expansion from a tailor's dummy still travel by air and have cellphones and televisions. Still, for them the worlds of game theory and economics are a closed book. As a result they are potentially gullible when it comes to understanding how the world works (though they may imagine they understand these things; the usual clue is that they refer to “Economics 101” as though it meant something).
An understanding of calculus is never wasted, and will form your intuitions in valuable ways - for example, you will learn the properties of maxima and minima, which will guide your thinking even in areas where numerical measurement is not possible.
And it will be an inexhaustible source of pleasure and entertainment. Check out Essence of calculus - YouTube to see how.
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Brandon R.
J.D. from Western Michigan University (Graduated 2007) · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 22.2K answers and 108.8M answer views
·Updated 3y
Related
Why is calculus called calculus?
The word “calculus” just means some method of computation. You can apply it to many different things.
From latin, the “calculus” refers to a small pebble by which you count or measure something else. Latin borrowed the root word, calx, from Greek khaliks.
Here is a line from A Civil Action, played by Jon Travolta.
“It's like this. A dead plaintiff is rarely worth as much as a living, severely maimed plaintiff. However, if it's a long agonizing death as opposed to a quick drowning or car wreck, the value can rise considerably. A dead adult in his 20's is generally worth less than one who is middle
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The word “calculus” just means some method of computation. You can apply it to many different things.
From latin, the “calculus” refers to a small pebble by which you count or measure something else. Latin borrowed the root word, calx, from Greek khaliks.
Here is a line from A Civil Action, played by Jon Travolta.
“It's like this. A dead plaintiff is rarely worth as much as a living, severely maimed plaintiff. However, if it's a long agonizing death as opposed to a quick drowning or car wreck, the value can rise considerably. A dead adult in his 20's is generally worth less than one who is middle-aged, a dead woman less than a dead man, a single adult less than one who is married, black less than white, poor less than rich. The perfect victim is a while male professional, 40 years old, at the height of his earning power, struck down in his prime. And the most imperfect? Well, in the calculus of personal injury law, a dead child is worth the least of all...”
Here, the character is performing a calculation about the monetary value of a personal injury case.
When people usually say calculus—as in a mathematics class or subject—they mean what was once called “infinitesimal calculus”.
Or more commonly said today, differential calculus or integral calculus. Originally, two separate areas. But now known to be fundamentally related.
They methods of computing “infinitely” small quantities—often to study things which undergo continuous change.
The real world is a very dynamic place. Change happens constantly. Or—to avoid the pun—change happens. Use calculus to understand it.
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Harry Seitz
Upvoted by
Erik Bergland
, PhD Mathematics, Brown University (2024) and
Michael Jørgensen
, PhD in mathematics · Author has 706 answers and 4.4M answer views
·Updated 5y
Related
What started calculus?
Frustration.
Imagine you’re Leibniz or Newton in 17th century Europe. There are gravity defying Baroque cathedrals fronted by city squares tinkling with fountains. Children snack on candy canes as their servants pressure cook quail and pheasant for supper back at the manor. They might not have ventured out of doors if not for the reassurance of fair weather from the trusty barometer.
Gentlemen sip champagne from fluted glasses and synchronize their pocket watches with the pendulum clock on the mantle as they discuss Drebbel’s submarine and how Guericke’s air pumps might allow a man to enter and
Continue Reading
Frustration.
Imagine you’re Leibniz or Newton in 17th century Europe. There are gravity defying Baroque cathedrals fronted by city squares tinkling with fountains. Children snack on candy canes as their servants pressure cook quail and pheasant for supper back at the manor. They might not have ventured out of doors if not for the reassurance of fair weather from the trusty barometer.
Gentlemen sip champagne from fluted glasses and synchronize their pocket watches with the pendulum clock on the mantle as they discuss Drebbel’s submarine and how Guericke’s air pumps might allow a man to enter and egress the vessel whilst still submerged! It’s a long shot, but Giovanni Branca’s steam turbine might someday be reconfigured to animate the conveyance and a host of others.
Apothecaries are finally approaching a consensus as to how the four fundamental humors govern health, and have even figured out how to transfuse blood from the robust to the pallid. A gentleman might very well retain his youthful vigor well into his late twenties.
Perhaps most remarkable, William Oughtred has constructed an analog computing device that allows even the dimmest of gentlemen to perform complex mathematical operations quickly and accurately with no need of quill or paper. It’s so simple, even a woman could use it.
But if you want to figure out the area under a curve, you have to count row after column of ever shrinking squares and triangles. With a good quill, a steady hand, and several hours of tedious coloring, one might produce an answer neither consistent or accurate enough to be of much use in a world more complex with each swing of the pendulum.
Newton and Leibniz (and probably others) finally got fed up. The world demanded a better way, and it was obvious that there had to be one.
The answer had been staring mathematicians in the face for 2,000 years. If you look at the equations for the circumference, area, and volume of circles and spheres, you can begin to see a pattern of integration. In retrospect, it seems almost impossible that no one had thought to apply this to a function.
The confluence of necessity, innovation, and advancements across the board are responsible for the development of calculus. The time was ripe.
People were more sexist back then and it’s widely believed Newton died a virgin.
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JH Simon
Author of 'How To Kill A Narcissist'
·Updated Fri
How do we become aware of and heal the broken parts of us that attract(ed) and allow(ed) narcissistic abuse?
Isolation is where you should begin. Like being a fish in water, entering into a narcissistic relationship happens seamlessly. By spending extended periods alone without a narcissist to muddy your mental and emotional space, you can grow your awareness of the deep longing which drives you into the arms of a narcissist in the first place. Also, you can see how this longing is deeply rooted in pain from your past.
Go to the movies alone, go for a long walk alone, travel alone for extended periods, sit alone in your room and meditate, go shopping alone, go to an event alone. During all of these ac
Continue Reading
Isolation is where you should begin. Like being a fish in water, entering into a narcissistic relationship happens seamlessly. By spending extended periods alone without a narcissist to muddy your mental and emotional space, you can grow your awareness of the deep longing which drives you into the arms of a narcissist in the first place. Also, you can see how this longing is deeply rooted in pain from your past.
Go to the movies alone, go for a long walk alone, travel alone for extended periods, sit alone in your room and meditate, go shopping alone, go to an event alone. During all of these activities, be mindful of how you feel. Shame, fear and uncertainty will all show up. Accepting them and being with them in a mindful way is at first unsettling and deeply painful. But if you pay attention, you’ll see how these echoes from the past are actually the wounds which lead you into abusive relationships as a form of escape. When they come up, don’t panic. Breathe deeply and focus on them. Don’t just pay attention to your thoughts about them. Instead, really go into them. Where are they in your body? In your chest? In your belly? In your legs as tension? What qualities do they have? Is the pain dull or sharp? Does the pain originate in one place or is it all over your body? Breathe deep into your belly and stay with the feeling. Does the feeling lead to total confusion in the mind and make you think you are going crazy? Right, stay with that confusion. Go deeper into it for as long as you can. You won’t go mad.
Does the impulse to run away from your inner state arise? Go deep into that impulse. Do 1001 judgements and reasons to distract yourself come up? Pay careful attention to each one of them. Don’t react. Don’t do anything. Just be mindful. For as long as you can tolerate.
When you have had enough, back away and do something else. Then come back to it. The longer you stay with the feelings and confusion, the more you strengthen your capacity to tolerate them, and the closer you come to transcendence. Once you transcend these feelings and mind states, you can then see how they map out into unhealthy behaviour. You’ll also come to learn about your belief system and on occasion you will realise that your beliefs about yourself are not gospel.
Secondly, pay attention to who you are as a person when you are alone. Beyond the shame, fear and uncertainty, what else do you notice? At first you may not notice much else. The longing for someone to save you might be all you have. But as you practice solitude, something else will enter that abundant spiritual space inside you. If you focus on it long enough, your relationship with it will transform, and you will notice new states of being arise. For example, if you travel alone and have a day full of fun and adventure, you’ll realise that you were the one who did that. By challenging yourself in this way, you will call on new resources to help you cope. The ‘you’ which emerges from this challenge is the you which doesn’t need a special someone to make things happen. You learn that it is ok to be ‘you,’ pain and all. Best of all, you learn that you can handle the pain, embrace it, and use it to transform your being.
In short, go at it alone in as many challenging ways as you can think of, be mindful, be with your painful emotions, try to look past the longing, and trust that you will evolve to meet every challenge. With enough of these experiences, you will have developed a ‘you’ which has nothing to do with a narcissist. In this way, you will finally have a choice in the matter. You will have a deep knowledge that life is far richer without a narcissist in your life, and that richness is deep within you.
If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist.
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Logan Wayne
Mathematics and Chemistry Teacher · Author has 893 answers and 2.5M answer views
·9y
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What does " critical point " mean?
As the other writers have stated, a critical point is a point where f'(x)=0 or does not exist. I thought I would lend a different perspective. I should mention now that I am not sure how or why they became known as critical points. Hopefully, after this answer you'll see how they're important to the entire graph.
Assume for a second that all you knew about f(x) was its derivative f'(x). If f(x) describes the position of an object then the critical point is known as a fixed point. The rate of change is zero, hence fixed. Therefore, an object starting at a fixed point remains there for eter
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As the other writers have stated, a critical point is a point where f'(x)=0 or does not exist. I thought I would lend a different perspective. I should mention now that I am not sure how or why they became known as critical points. Hopefully, after this answer you'll see how they're important to the entire graph.
Assume for a second that all you knew about f(x) was its derivative f'(x). If f(x) describes the position of an object then the critical point is known as a fixed point. The rate of change is zero, hence fixed. Therefore, an object starting at a fixed point remains there for eternity. Now this fixed point, or critical point, will either repel or attract things that start near it. These points are called stable or unstable fixed points respectively. This depends upon the change of the sign of the derivative relative to the fixed point. You'll recall from the intermediate value theorem that a critical point is required when the derivative changes sign. In higher dimensions, more complex behavior is possible. So these points are very important in defining the behavior of a system in their local area.
Consider for a second the concept of a max or a min on a graph. If you know anything about physics then this should be intuitive. If you start at the very top of a hill (a maximum) and you're at rest you will not move. If you're at the bottom of a valley (a min) you also will not move. These are sometimes called equilibrium points.
Perhaps this strayed a bit too far from your initial question. If you're interested, then you should google fixed points of dynamical systems.
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Colette Hinds
Studied Mathematics at University of Kansas (Graduated 1992) · Upvoted by
José Ilhano Silva
, M. S. Mathematics & Differential Geometry, Federal University of Ceará (2017)
·7y
Related
Why was calculus invented?
Calculus wasn’t invented. It was discovered.
In a nutshell, it was discovered because of the need to calculate changing quantities. Calculus is a tool that enables us to do that.
Most notably, the theory that the planets orbited the sun was finally signed, sealed, and delivered by calculus.
Chiefly on Isaac Newton’s mind while he formulated his theories was astronomy. Nobody had been able to describe with precision the motion of the planets. The planets are continuously in motion, therefore continuously changing.
Newton developed the basic ideas of calculus, and they aren’t called Newton’s Laws of
Continue Reading
Calculus wasn’t invented. It was discovered.
In a nutshell, it was discovered because of the need to calculate changing quantities. Calculus is a tool that enables us to do that.
Most notably, the theory that the planets orbited the sun was finally signed, sealed, and delivered by calculus.
Chiefly on Isaac Newton’s mind while he formulated his theories was astronomy. Nobody had been able to describe with precision the motion of the planets. The planets are continuously in motion, therefore continuously changing.
Newton developed the basic ideas of calculus, and they aren’t called Newton’s Laws of Motion for no reason.
I like to use this example to explain calculus: we can use easy-to-define things we already know, such as the area of a rectangle, to calculate the area of oddly-shaped regions. The way to do this is to draw a lot of rectangles in that area and add up the areas of the rectangles to get the area of the whole region.
Well, the more rectangles we use, the less “wasted space” we have that can’t be measured. So, we “theoretically” draw an “infinite” number of rectangles inside it to calculate the area of it almost exactly (“infinitely” close).
This diagram shows a bunch of rectangles inside an oddly-shaped area under a curve. It is easy to see that we can add up all the rectangles to measure the oddly-shaped area, and that the smaller and smaller rectangles we draw, the more we can fit in there to get a more exact calculation.
One thing calculus does is give us a tool to add up very large numbers of rectangles with a few formulas instead of sitting there drawing smaller and smaller rectangles and adding them all up.
So, we can use infinite numbers of easy-to-define things to measure hard-to-define things, such as changing quantities.
Other changing quantities calculus enables us to measure are the tides, acceleration of flying objects such as cannonballs, changes in volume, such as in hydrodynamics and thermodynamics, etc. etc. etc.
And that was what the mathematicians were looking for when calculus was discovered.
Quod erat faciendum.
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David Joyce
Professor Emeritus of Mathematics at Clark University · Upvoted by
Aditya Garg
, M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) and
Tom McFarlane
, M.S. Mathematics, University of Washington (1994) · Author has 9.9K answers and 68.4M answer views
·5y
Related
What started calculus?
The problem of varying velocities started calculus, and it is quite old. The Greek geometers, even Archimedes, couldn’t deal with it. They had no difficulty with constant speeds either when an object moved in a straight line or in a circle, but they couldn’t deal with changing velocities.
In Europe, it was Bradwardine, Heytesbury, Swineshead, and Dumbleton who tackled the problem of changing velocities in the first half of the 1300s at Oxford. They didn’t solve it for all changing velocities, but they did manage one case, that case being when the velocities themselves had a constant rate of cha
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The problem of varying velocities started calculus, and it is quite old. The Greek geometers, even Archimedes, couldn’t deal with it. They had no difficulty with constant speeds either when an object moved in a straight line or in a circle, but they couldn’t deal with changing velocities.
In Europe, it was Bradwardine, Heytesbury, Swineshead, and Dumbleton who tackled the problem of changing velocities in the first half of the 1300s at Oxford. They didn’t solve it for all changing velocities, but they did manage one case, that case being when the velocities themselves had a constant rate of change, that is, when the acceleration was constant. (They called that “uniformly difform motion”.) They discovered the “mean speed theorem”: the distance travelled by an object undergoing constant acceleration or constant deceleration was the same distance as that traveling at a constant velocity equal to the average of the initial and final velocities.
A few years later, about 1350, Oresme, in Paris, tackled the general problem. He developed a theory which answered the question of how far an object goes when its velocity changes. Draw a graph of the velocity where the horizontal axis is time and at each point in time, draw a vertical line above the axis representing the velocity at that instant. That produces a figure that Oresme called a “form”. The area of that figure, then, is the distance travelled. That is the Fundamental Theorem of Calculus, which, when written in modern notation says
∫b a f′(x)d x=f(b)−f(a).∫a b f′(x)d x=f(b)−f(a).
The Black Plague hit Europe just at that time, and that severely held back progress. Nonetheless, others including Galileo, Cavalieri, and Torricelli in Italy, Wallace and Gregory in Britain, Pascal, Descartes, and Fermat in France, as well as others in Germany and Spain developed the concepts and proved theorems during the succeeding 300 years. By 1650 mathematical analysis was fairly well developed. It was helped along by the development of symbolic algebra in the 1500s and by the analytic geometry of Fermat and Descartes in the couple of decades before 1650.
It is often said that Leibniz and Newton invented calculus, but much of it was developed before them.
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Steven Haddock
Studied at York University (Canada) · Author has 35K answers and 568M answer views
·2y
Related
How hard is Calculus II compared to Calculus I? I got a C+ in Calc. I?
Calculus II is very hard.
Calc 1 was pretty much a repeat of high school calculus but they made sure you could actually do things like work out trigonometric derivatives from first principles rather than memorize which one went with which.
Calc 2 introduces you to concepts like using calculus for three dimensional functions, so instead of calculating the area under a curve you’re calculating the volume under a plane. Algebra and trig in Calc 1 are pretty straightforward. In Calc 2 it gets pretty trippy. I started to lose it when they introduced imaginary numbers to the calculations (which appare
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Calculus II is very hard.
Calc 1 was pretty much a repeat of high school calculus but they made sure you could actually do things like work out trigonometric derivatives from first principles rather than memorize which one went with which.
Calc 2 introduces you to concepts like using calculus for three dimensional functions, so instead of calculating the area under a curve you’re calculating the volume under a plane. Algebra and trig in Calc 1 are pretty straightforward. In Calc 2 it gets pretty trippy. I started to lose it when they introduced imaginary numbers to the calculations (which apparently does have some real world applications in engineering). Then there were the functions which immediately went from one state to another instantaneously.
I would recommend Calc 1 for anyone who has decent math skills. I would only recommend Calc 2 for math and engineering majors who do a lot of other math. Back in high school I did Calculus, Algebra, and Relations & Functions and you can draw on each one for either of the other two. Calculus gets very confusing unless you’re doing other math at the same time that you can lean back on.
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Related questions
What does " critical point " mean?
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Why are critical points important in calculus?
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Laplace Transform - Wikipedia, The Free Encyclopedia
The Laplace transform is a widely used integral transform in mathematics. It is a linear operator of a function f(t) that transforms it to a function F(s) with a complex argument s. The tran…
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Laplace transform
From Wikipedia, the free encyclopedia
In mathematics, the
Laplace transform
is a widely used integral transfor m. Denoted , it is a linear operat or of a function
f
(
t
)with a real argument
t
(
t
≥ 0) t hat transforms it to a function
F
(
s
) with a complex argument
s
. This transformation is essentially bijective for the majority of practical uses; the respective pairs of
f
(
t
) and
F
(
s
) are matched in tables. The Laplace transform has the useful property that many relations hips and operations over the orig inals
f
(
t
) correspond to simpler relationships and operations over the images
F
(
s
).
The Laplace transform has many important applications throughout the sciences. It is named for Pierre-Simon Laplace who introduced the transform in his work on probability theory.The Laplace transform is related to the Fourier transform, but whereas the Fourier transform resolves a function or signal into its modes of vibration, the Laplace transform resolves a function into its moments. Like the Fourier transform, the Laplace transform is used for solving differential and integral equations. In physics and engineering, it is used for analysis of linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems. In this analysis, the Laplace transform is often interpreted as a transform ation from the
time-domain
, in which inputs and outputs are functions of time, to the
freque ncy-domain
, where the same inputs and outputs are functions of complex angular frequency, in radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.
Contents
1 History 2 Formal definition 2.1 Probability theory 2.2 Bilateral Laplace transform 2.3 Inverse Laplace t ransform 3 Region of convergence 4 Properties and theorems 4.1 Proof of the Laplace transform of a function's deri vative 4.2 Evaluating improper integrals 4.3 Relationship to other transforms 4.3.1 Laplace–Stieltjes transform 4.3.2 Fourier transform 4.3.3 Mellin transform 4.3.4 Z-transform 4.3.5 Borel transform 4.3.6 Fundamental relationships 5 Table of selected Laplace transforms 6 s-Domain equivalent circuits and impedances 7 Examples: How to apply the properties and theorems 7.1 Example 1: Solving a differential equation 7.2 Example 2: Deriving the complex impedance for a capacitor 7.3 Example 3: Method of partial fraction expansion 7.4 Example 4: Mixing sines, cosines, and exponentials 7.5 Example 5: Phase delay 8 See also 9 Notes 10 References 10.1 Modern 10.2 Historical 11 External links
History
The Laplace transform is named in honor of mathematician and astronomer Pierre-Simon Laplace, who used the transform in his work on probabi lity theory. From 1744, Leonh ard Euler in vestig ated integr als of the form L a p l a c e t r a n s f o r m - W i k i p e d i a, t h e f r e e e n c y c l o p e d i a h t t p://e n.w i k i p e d i a.o r g/w i k i/L a p l a c e _ t r a n s f o r m 1 o f 1 7 9/3 0/2 0 1 1 3:2 2 P M
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as solutions of differential equations but did not pursue the matter very far.
Joseph Louis Lagrange was an admirer of Euler and, in his work on integrating probability density functions, investigated expressions of the form which some modern historians have interpreted within modern Laplace transform theory.
These types of integrals seem first to have attracted Laplace's attention in 1782 where he was following in the spirit of Euler in using the integrals themselves as solutions of equations.
However, in 1785, Laplace took the critical step forward when, rather than just looking for a solution in the form of an integral, he started to apply the transforms in the sense that was later to become popular. He used an integral of the form:akin to a Mellin transform, to transform the whole of a difference equation, in order to look for solutions of the transformed equation. He then went on to apply the Laplace transform in the same way and started to derive some of its properties, beginning to appreciate its potential po wer.
Laplace also recognised that Joseph Fourier's method of Fourier series for solving the diffusion equation could only apply to a limited region of space as the solutions were periodic. In 1809, Laplace applied his transform to find solutions that diffused indefinitely in space.
Formal definition
The Laplace transform of a function
f
(
t
), defined for all real numbers
t
≥ 0, is the function
F
(
s
), defined by:The parameter
s
is a complex number:with real numbers σ and ω.The meaning of the integral depends on types of functions of interest. A necessary condition for existence of the integral is that
ƒ
must be locally integrable on [0,∞). For locally integrable functions that decay at infinity or are of exponential type, the integral can be understood as a (proper) Lebesgue integral. However, for many applications it is necessary to regard it as a conditionally convergent improper integral at ∞. Still more generally, the integral can be understood in a weak sense, and this is dealt with below.One can define the Laplace transform of a finite Borel measure µ by the Lebesgue integral
An important special case is where µ is a probability measure or, even more specifically, the Dirac delta function. In operational calculus,the Laplace transform of a measure is often treated as though the measure came from a distribution function
ƒ
. In that case, to avoid potential con fusion, one often wri tes where the lower limit of 0
−
is short notation to mean L a p l a c e t r a n s f o r m - W i k i p e d i a, t h e f r e e e n c y c l o p e d i a h t t p://e n.w i k i p e d i a.o r g/w i k i/L a p l a c e _ t r a n s f o r m 2 o f 1 7 9/3 0/2 0 1 1 3:2 2 P M
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This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace transform. Although with the Lebesgue integral,it is not necessary to take such a limit, it does appear more naturally in connection with the Laplace–Stieltjes transform.
Probability theory
In pure and applied probability, the Laplace transform is defined by means of an expectation value. If
X
is a random variable with probabi lity dens ity function
ƒ
, then the La place transform of
ƒ
is given by the expectation By abuse of language, this is referred to as the Laplace transform of the random variable
X
itself. Replacing
s
by −
t
gives the moment generating function of
X
. The Laplace transform has applications throughout probability theory, including first passage times of stochastic processes such as M arkov chains, and renewal theory.
Bilateral Laplace transform
Main arti cle: Tw o-sided Laplace tr ansfor m
When one says "the Laplace transform" without qualification, the unilateral or one-sided transform is normally intended. The Laplace transform can be alternatively defined as the
bilateral Laplace transform
or two-sided Laplace transform by extending the limits of integration to be the entire real axis. If that is done the common unilateral transform simply becomes a special case of the bilateral transform where the definition of the function being transformed is multiplied by the Heaviside step function.The bilateral Laplace transform is defined as follows:
Inverse Laplace transform
For more det ails on this topic, see Inver se Laplace tr ansform.
The inverse Laplace transform is given by the following complex integral, which is known by various names (the
Bromwich integral
, the
Fourier-Mellin integral
, and
Mellin's inverse formu la
):where
γ
is a real number so that the contour path of integration is in the
region of convergence
of
F
(
s
). An alternative formula for the inverse Laplace transform is given by Post's inversion formula.
Region of convergence
If
ƒ
is a locally integrable function (or more generally a Borel measure locally of bounded variation), then the Laplace transform
F
(
s
) of
ƒ
converges provided that the limit exists. The Laplace transform converges absolutely if the integral exists (as proper Lebesgue integral). The Laplace transform is usually understood as conditionally convergent, meaning that it converges in the former instead of the latter sense.The set of values for which
F
(
s
) converges absolutely is either of the form Re{
s
} >
a
or else Re{
s
} ≥
a
, where
a
is an extended real constant, −∞ ≤
a
≤ ∞. (This follows from the dominated convergence theorem.) The constant
a
is known as the abscissa of absolute convergence, and depends on the growth behavior of
ƒ
(
t
).
Analogously, the two-sided transform converges absolutely in a strip of the L a p l a c e t r a n s f o r m - W i k i p e d i a, t h e f r e e e n c y c l o p e d i a h t t p://e n.w i k i p e d i a.o r g/w i k i/L a p l a c e _ t r a n s f o r m 3 o f 1 7 9/3 0/2 0 1 1 3:2 2 P M
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form
a
< Re{
s
} <
b
, and possibly including the lines Re{
s
} =
a
or Re{
s
} =
b
.
[10
The subset of values of
s
for which the Laplace transform converges absolutely is called the region of absolute convergence or the domain of absolute convergence. In the two-sided case, it is sometimes called the strip of absolute convergence. The Laplace transform is analytic in the region of absolute convergence.Similarly, the set of values for which
F
(
s
) converges (conditionally or absolutely) is known as the region of conditional convergence, or simply the
region of convergence
(ROC). If the Laplace transform converges (conditionally) at
s
=
s
0
, then it automatically converges for all
s
with Re{
s
} > Re{
s
0
}. Therefore the region of convergence is a half-plane of the form Re{
s
} >
a
, possibly including some points of the boundary line Re{
s
} =
a
. In the region of convergence Re{
s
} > Re{
s
0
}, the Laplace t ransform of
ƒ
can be expressed by integrating by parts as the integral That is, in the region of convergence
F
(
s
) can effectively be expressed as the absolutely convergent Laplace transform of some other function. In particular, it is analytic.A variety of theorems, in the form of Paley–Wiener theorems, exist concerning the relationship between the decay properties of
ƒ
and the properties of the Laplace transform within the region of convergence.In engineering applications, a function corresponding to a linear time-invariant (LTI) system is
stable
if every bounded input produces a bounded o utput. Thi s is eq uivale nt to the absolute con verge nce of the Laplace transfo rm of the im puls e respons e function i n the regi on Re{
s
} ≥ 0. As a result, LTI systems are stable provided the poles of the Laplace transform of the impulse response function have negati ve real part.
Properties and theorems
The Laplace transform has a number of properties that make it useful for analyzing linear dynamical systems. The most significant advantage is that differentiation and integration become multiplication and division, respectively, by
s
(similarly to logarithms cha nging multiplication of numbers to addition of their logarithms). Because of this property, the Laplace variable
s
is also known as
operator variable
in the L domain: either
derivative operator
or (for
s
−1
)
integration operator
. The transform turns integral equations and differential equations to polynomial equations, which are much easier to solve. Once solved, use of the inverse Laplace transform reverts back to the time d omain.Given the functions
f
(
t
) and
g
(
t
), and their respective Laplace transforms
F
(
s
) and
G
(
s
):the following table is a list of properties of unilateral Laplace transform:
L a p l a c e t r a n s f o r m - W i k i p e d i a, t h e f r e e e n c y c l o p e d i a h t t p://e n.w i k i p e d i a.o r g/w i k i/L a p l a c e _ t r a n s f o r m 4 o f 1 7 9/3 0/2 0 1 1 3:2 2 P M
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17315 | https://openstax.org/books/chemistry-2e/pages/16-exercises | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Chemistry 2e
Exercises
Chemistry 2eExercises
Search for key terms or text.
## 16.1 Spontaneity
1.
What is a spontaneous reaction?
What is a nonspontaneous reaction?
3.
Indicate whether the following processes are spontaneous or nonspontaneous.
(a) Liquid water freezing at a temperature below its freezing point
(b) Liquid water freezing at a temperature above its freezing point
(c) The combustion of gasoline
(d) A ball thrown into the air
(e) A raindrop falling to the ground
(f) Iron rusting in a moist atmosphere
A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
5.
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
## 16.2 Entropy
In Figure 16.8 all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
7.
In Figure 16.8 all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, for the system when it is converted from distribution (b) to distribution (d).
How does the process described in the previous item relate to the system shown in Figure 16.4?
9.
Consider a system similar to the one in Figure 16.8, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to What does this comparison tell us about even larger systems?
Consider the system shown in Figure 16.9. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?
11.
Consider the system shown in Figure 16.9. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.
(a) H2(g), HBrO4(g), HBr(g)
(b) H2O(l), H2O(g), H2O(s)
(c) He(g), Cl2(g), P4(g)
13.
At room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.
Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature but somewhat higher pressure).
Is ΔS positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?
15.
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
(a) C2H5OH(l) or C3H7OH(l)
(b) C2H5OH(l) or C2H5OH(g)
(c) 2H(g) or H(g)
Predict the sign of the entropy change for the following processes.
(a) An ice cube is warmed to near its melting point.
(b) Exhaled breath forms fog on a cold morning.
(c) Snow melts.
17.
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
(a)
(b)
(c)
Write the balanced chemical equation for the combustion of methane, CH4(g), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether ΔS is positive or negative for this chemical reaction.
19.
Write the balanced chemical equation for the combustion of benzene, C6H6(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process?
## 16.3 The Second and Third Laws of Thermodynamics
What is the difference between ΔS and ΔS° for a chemical change?
21.
Calculate for the following changes.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Determine the entropy change for the combustion of liquid ethanol, C2H5OH, under the standard conditions to give gaseous carbon dioxide and liquid water.
23.
Determine the entropy change for the combustion of gaseous propane, C3H8, under the standard conditions to give gaseous carbon dioxide and water.
“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
25.
Using the relevant values listed in Appendix G, calculate for the following changes:
(a)
(b)
From the following information, determine for the following:
27.
By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
(a)
(b)
(c)
(d)
(e)
(f)
29.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
(a)
(b)
(c)
(d)
(e)
(f)
## 16.4 Free Energy
What is the difference between ΔG and ΔG° for a chemical change?
31.
A reaction has = 100 kJ/mol and Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
33.
Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
(a)
(b)
(c)
(d)
(e)
(f)
Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
(a)
(b)
(c)
(d)
(e)
(f)
35.
Given:
(a) Determine the standard free energy of formation, for phosphoric acid.
(b) How does your calculated result compare to the value in Appendix G? Explain.
Is the formation of ozone (O3(g)) from oxygen (O2(g)) spontaneous at room temperature under standard state conditions?
37.
Consider the decomposition of red mercury(II) oxide under standard state conditions.
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
(a) Ammonia:
(b) Diborane:
(c) Hydrazine:
(d) Hydrogen peroxide:
39.
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
(a)
(b)
(c)
(d)
(e)
(f)
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
(a)
(b)
(c)
(d)
(e)
(f)
41.
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
(a)
(b)
(c)
(d)
(e)
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
(a)
(b)
(c)
(d)
(e)
43.
Calculate the equilibrium constant at the temperature given.
(a)
(b)
(c)
(d)
(e)
Calculate the equilibrium constant at the temperature given.
(a)
(b)
(c)
(d)
(e)
45.
Consider the following reaction at 298 K:
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
Determine the normal boiling point (in kelvin) of dichloromethane, CH2Cl2. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
47.
Under what conditions is spontaneous?
At room temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.00 10−14. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
49.
Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
Consider the decomposition of CaCO3(s) into CaO(s) and CO2(g). What is the equilibrium partial pressure of CO2 at room temperature?
51.
In the laboratory, hydrogen chloride (HCl(g)) and ammonia (NH3(g)) often escape from bottles of their solutions and react to form the ammonium chloride (NH4Cl(s)), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH3 in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
Benzene can be prepared from acetylene. Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
53.
Carbon dioxide decomposes into CO and O2 at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO2 was 1.15 atm?
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
55.
Acetic acid, CH3CO2H, can form a dimer, (CH3CO2H)2, in the gas phase.
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
At 25 °C, the equilibrium constant for the dimerization is 1.3 103 (pressure in atm). What is ΔS° for the reaction?
56.
Determine ΔGº for the following reactions.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
(b) Chlorine molecules dissociate according to this reaction:
1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.
Given that the for Pb2+(aq) and Cl−(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).
58.
Determine the standard free energy change, for the formation of S2−(aq) given that the for Ag+(aq) and Ag2S(s) are 77.1 kJ/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 10−51.
Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
60.
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
(b) Determine the equilibrium constant, KP, for this physical process.
(c) By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, is 0.011 atm.
(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of in the air?
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
In this process, ATP becomes ADP summarized by the following equation:
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
62.
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
(a)
(b)
(c)
(d)
64.
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the following equation:
(a) Determine for the decomposition of Cu2S(s).
(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine for the process.
(c) The production of copper from chalcocite is performed by roasting the Cu2S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
66.
What happens to (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
(a)
(b)
(c)
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17316 | https://flexbooks.ck12.org/cbook/ck-12-cbse-maths-class-11/section/6.2/primary/lesson/fundamental-principles-of-counting/ | Skip to content
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6.2
Fundamental Principles of Counting
Written by:Neha Khandelwal
Fact-checked by:The CK-12 Editorial Team
Last Modified: Jul 04, 2025
Counting is an essential mathematical skill that we use daily, whether it's arranging books on a shelf, forming teams for a sports event, or creating passwords for online accounts. However, as the number of possibilities increases, manually listing each case becomes inefficient. This is where the Fundamental Principles of Counting help simplify complex problems.
These principles form the foundation for permutations and combinations, which are widely used in probability, data analysis, and decision-making. By applying these principles, we can determine the number of ways an event can occur without listing every possibility.
There are two fundamental principles of counting:
The Fundamental Principle of Multiplication - Used when multiple actions occur in sequence.
The Fundamental Principle of Addition - Used when choosing between different independent actions.
Understanding these principles allows us to solve counting problems systematically and efficiently.
Fundamental Principle of Multiplication
The Fundamental Principle of Multiplication states that if an event consists of two or more independent choices, where:
The first choice can be made in @$\begin{align}m\end{align}@$ ways, and
The second choice can be made in @$\begin{align}n\end{align}@$ ways,
Then the total number of ways to perform both actions in sequence is:
@$$\begin{align}\text{Total Ways} = m \times n\end{align}@$$ This principle can be extended to multiple independent choices. If there are @$\begin{align}k\end{align}@$ independent choices where the first can be done in @$\begin{align}m_1\end{align}@$ ways, the second in @$\begin{align}m_2\end{align}@$ ways, and so on, then the total number of ways to perform all tasks is:
@$$\begin{align}m_1 \times m_2 \times m_3 \times ... \times m_k\end{align}@$$ Let us take a look at an example.
At the local ice cream shop, there are 5 flavors of homemade ice cream - vanilla, chocolate, strawberry, cookie dough, and coffee. A customer can choose to have their ice cream in a dish or in a cone. How many different combinations of ice cream orders can be made?
If you list them all out, you will see that there are 10 ice cream orders. For each of the 5 flavours, there are 2 choices for how the ice cream is served (dish or cone).
Applying the multiplication principle:
@$$\begin{align}5 \times 2=10\end{align}@$$
This means that there are 10 different ways to order the ice cream.
Fundamental Counting Principle: For independent events @$\begin{align}A\end{align}@$ and @$\begin{align}B,\end{align}@$ if there are @$\begin{align}n\end{align}@$ outcomes in event @$\begin{align}A\end{align}@$ and @$\begin{align}m\end{align}@$ outcomes in event @$\begin{align}B\end{align}@$, then there are @$\begin{align}n \cdot m\end{align}@$ outcomes for events @$\begin{align}A\end{align}@$ and @$\begin{align}B\end{align}@$ together.
The Fundamental Counting Principle works similarly for more than two events: multiply the number of outcomes in each event to find the total number of outcomes.
Consider a scenario where you are shopping for a camping trip. You need to select one tent and one backpack.
There are 2 choices for tents.
There are 3 choices for backpacks.
Using the Fundamental Counting Principle, the total number of combinations is:
@$$\begin{align}2 \times 3 = 6\end{align}@$$ These examples demonstrate how the Fundamental Principle of Multiplication helps count arrangements efficiently, eliminating the need to list all possibilities manually.
Fundamental Principle of Addition
The Fundamental Principle of Addition states that if an event can be performed in @$\begin{align}m\end{align}@$ ways and another event, which is mutually exclusive from the first, can be performed in @$\begin{align}n\end{align}@$ ways, then the number of ways to perform either of the two events is:
@$$\begin{align}\text{Total Ways} = m + n\end{align}@$$ This principle is used when a choice must be made between one option or another, rather than a combination of both.
Let us take a look at an example.
Suppose you are selecting a mode of transport to go to school. If you can either take a bus (3 routes available) or a metro (2 routes available), then the total number of ways to travel to school is:
@$$\begin{align}3 + 2 = 5\end{align}@$$
This means you can take any one of the 3 bus routes or any one of the 2 train routes, from a total of 5 choices.
Let us look at another example.
In a class, there are 10 boys and 8 girls. The teacher wants to select either a boy or a girl to represent the class in a function. Here, the teacher has two independent choices:
Selecting a boy (10 options).
Selecting a girl (8 options).
Since the selection is either a boy or a girl, the number of ways the teacher can make a selection is:
@$$\begin{align}10 + 8 = 18\end{align}@$$ Thus, there are 18 ways to make the selection.
Difference Between the Two Fundamental Principles
The key difference between the Principle of Multiplication and the Principle of Addition lies in the type of event being considered.
The Multiplication Principle is used when multiple independent actions are performed together, meaning the choices are linked.
The Addition Principle is used when choosing between separate options, where only one of the choices can be made.
Let's look at an example to understand the difference between the two.
Consider a student choosing subjects in an examination:
If the student must select one domain from Math (4 domains) and one from Science (3 domains), the total choices available are:
@$$\begin{align}4 \times 3 = 12\end{align}@$$
(Multiplication Principle: since both subjects must be chosen together).
If the student can choose either Math (4 domains) or Science (3 topics), but not both, the total choices available are:
@$$\begin{align}4+3=7\end{align}@$$(Addition Principle: since only one subject can be chosen).
Understanding the difference between these two principles is crucial in solving problems related to permutations, combinations, and probability.
Case Study: Lab Paths & Plant Rules - Conditional Combinations in Biotechnology
In a biotechnology lab, researchers are analysing DNA sequences using automated machines. To prepare a DNA sample, they must choose one of 3 different enzymes for cutting the DNA and then one of 4 different dyes to tag the sample for identification. Additionally, the lab offers two types of analysis: electrophoresis or chromatography. The researchers must choose exactly one enzyme, one dye, and one analysis method. However, in a special case, if the DNA is from a plant sample, only two enzymes are suitable, and only electrophoresis is used for analysis. Understanding how many different procedures are possible under different conditions helps the lab plan time and resources effectively.
Based on this passage, answer the following questions:
Examples of Fundamental Principles of Counting
Example 1
In a school, there are 28 boys and 18 girls. The teacher wants to select one boy and one girl to represent the class in an inter-school event. In how many ways can the teacher make this selection?
Number of ways to select one boy out of 28 = 28
Number of ways to select one girl out of 18 = 18
Since both selections occur independently, we use the Fundamental Principle of Multiplication (FPM):
@$$\begin{align}28 \times 18 = 504.\end{align}@$$ Thus, there are 504 ways to select one boy and one girl.
Example 2
If repetition is allowed, how many three-digit numbers can be formed using only the digits 1 and 3?
A three-digit number consists of hundreds, tens, and unit places.
The hundreds place can be filled in 2 ways (1 or 3).
The tens place can be filled in 2 ways (1 or 3).
The units can be filled in 2 ways (1 or 3).
Since each place is selected independently, we apply the Fundamental Principle of Multiplication (FPM):
@$$\begin{align}2 \times 2 \times 2 = 8.\end{align}@$$ The possible numbers are: 111, 113, 131, 133, 311, 313, 331, 333.
Example 3
There are 7 vacant seats on a bus, and 4 passengers are entering the bus. How many ways can the 4 passengers take their seats?
The first passenger can sit in 7 ways.
The second passenger can sit in 6 ways (since one seat is already occupied).
The third passenger can sit in 5 ways.
The fourth passenger can sit in 4 ways.
By the Fundamental Principle of Multiplication (FPM), the total number of ways the passengers can be seated is:
@$$\begin{align}7 \times 6 \times 5 \times 4 = 840.\end{align}@$$ Thus, there are 840 ways to seat the passengers.
Example 4
A sports club has 7 boys and 5 girls on one team and 6 boys and 4 girls on another team. If a match is arranged so that one boy from each team and one girl from each team compete against each other, how many different matches can be formed?
Ways to select one boy from the first team = 7
Ways to select one boy from the second team = 6
Thus, the number of boy's matches(FPM):
@$$\begin{align}7 \times 6 = 42.\end{align}@$$ Ways to select one girl from the first team = 5
Ways to select one girl from the second team = 4
Thus, the number of girl's matches(FPM):
@$$\begin{align}5 \times 4 = 20.\end{align}@$$ Now, since both types of matches (boys and girls) are arranged separately and do not affect each other, the total number of matches is the sum of both results:
@$$\begin{align}42 + 20 = 62.\end{align}@$$ Thus, there are 62 different matches.
Example 5
A customer forgets a four-digit ATM PIN but remembers that it consists of the digits 2, 4, 5, and 8, with each digit used only once. How many PINs could the customer try?
Ways to choose the first digit = 4.
Ways to choose the second digit = 3.
Ways to choose the third digit = 2.
Ways to choose the fourth digit = 1.
By the Fundamental Principle of Multiplication (FPM), total possible PINs:
@$$\begin{align}4 \times 3 \times 2 \times 1 = 24.\end{align}@$$ Thus, the customer may have to try 24 different PINs.
Example 6
A teacher gives a 6-question quiz, where each question has two possible answers: True or False. No student has answered all questions correctly, and no two students have given the same sequence of answers. What is the maximum number of students in the class?
Each question has 2 possible answers (True/False).
Ways to answer 1st question = 2.
Ways to answer 2nd question = 2.
Ways to answer 3rd question = 2.
Ways to answer 4th question = 2.
Ways to answer 5th question = 2.
Ways to answer 6th question = 2.
By the Fundamental Principle of Multiplication (FPM), the total number of unique answer sequences:
@$$\begin{align}2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = 64.\end{align}@$$ Since there is one correct answer sequence, but no student has answered all questions correctly, the maximum number of students who can have unique answer sequences is:
@$$\begin{align}64 - 1 = 63.\end{align}@$$ Thus, the maximum number of students in the class is 63.
Example 7
By using the digits 1, 2, 3, 4, 5, and 6 (without repetition), how many non-zero numbers can be formed using any number of digits?
We need to count all possible numbers of different lengths (1-digit, 2-digit, ..., 6-digit). The total count of such numbers is calculated by summing all possible cases.
1-digit numbers: Any of the 6 digits can be used @$\begin{align}→\end{align}@$ 6 ways.
2-digit numbers: First digit = 6 choices, second digit = 5 choices @$\begin{align}→\end{align}@$ 6 × 5 = 30 ways.
3-digit numbers: 6 × 5 × 4 = 120 ways.
4-digit numbers: 6 × 5 × 4 × 3 = 360 ways.
5-digit numbers: 6 × 5 × 4 × 3 × 2 = 720 ways.
6-digit numbers: 6 × 5 × 4 × 3 × 2 × 1 = 720 ways.
By the Fundamental Principle of Multiplication (FPM) within each case and the Fundamental Principle of Addition (FPA) to sum all cases:
@$$\begin{align}6 + 30 + 120 + 360 + 720 + 720 = 1956.\end{align}@$$ Thus, the total number of non-zero numbers that can be formed is 1956.
| |
| Summary of Fundamental Principles of Counting |
| Fundamental Principles of Counting + Multiplication Principle – Used for sequential independent actions. If an event consists of multiple independent choices where the first choice can be made in @$\begin{align}m\end{align}@$ ways and the second in @$\begin{align}n\end{align}@$ ways, the total ways to perform both actions is @$\begin{align}m × n.\end{align}@$ This extends to multiple choices: @$\begin{align}m₁ × m₂ × … × mₖ.\end{align}@$ + Addition Principle – Used when choosing between mutually exclusive options. If one event can occur in @$\begin{align}m\end{align}@$ ways and another in @$\begin{align}n\end{align}@$ ways, the total ways to perform either event is @$\begin{align}m + n.\end{align}@$ Key Difference Between the Two Principles + Multiplication Principle applies when multiple actions occur together (linked choices). + Addition Principle applies when choosing between separate options (only one choice is made). |
Review Questions of Fundamental Principles of Counting
How many three-digit numbers are there?
How many three-digit numbers are there with no digit repeated?
How many three-digit numbers are there, with distinct digits, with each digit odd?
How many different five-digit number licence plates can be made if (i) the first digit cannot be zero and repetition of digits is not allowed, (ii) the first digit cannot be zero, but repetition of digits is allowed?
A restaurant offers 4 types of pasta and 3 types of sauces. How many different pasta dishes can be ordered?
A company is designing an ID system where a password consists of 3 letters followed by 2 digits. If there are 26 letters and 10 digits available, how many different passwords can be created?
A clothing store sells 6 different shirts and 3 types of pants. How many unique outfits can a customer create by choosing one shirt and pair of pants?
A new smartphone is available in 4 colours with 2 storage options. How many different models can a customer choose from?
A college professor must choose one assistant from either 6 undergraduate students or 4 graduate students. How many ways can the professor make a selection?
A bookstore has 7 different novels and 5 magazines available. If a customer wants to buy either a novel or a magazine but not both, how many choices does the customer have?
A sports coach needs to select one player for a relay race. There are 8 male and 7 female athletes. In how many ways can the coach select a participant?
A number lock on a suitcase has 3 wheels each labeled with ten digits 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible? Also, find the number of unsuccessful attempts to open the lock.
How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7, and 9 when no digit is repeated? How many of them are divisible by 10?
A customer forgets a four-digit code for an Automatic Teller Machine (ATM) in a bank. However, he remembers that this code consists of digits 3, 5, 6, and 9. Find the largest possible number of trials necessary to obtain the correct code.
In how many ways can three jobs I, II, and III be assigned to three persons A, B, and C if one person is assigned only one job and all are capable of doing each job?
If three six-faced dice each marked with numbers 1 to 6 on six faces, are thrown, find the total number of possible outcomes.
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A Complete Guide to the Retail Inventory Method (RIM)
The retail inventory method can help save you time and money on physical inventory counts by using a simple calculation. Learn more in this article.
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The retail inventory method is an accounting approach used to estimate the value of your store’s ending inventory for a specific time period.
Doing physical inventory counts is time-consuming, and can be disruptive to business operations if it requires you to close your shop. The retail inventory method (RIM) is an alternative solution that lets you estimate how much stock is remaining during a given period, using a simple calculation.
You probably have a lot of cash invested in stock, which makes sense, as buying inventory is essential to any retailer. That’s why it’s important to monitor your inventory so you can choose wisely when it comes to ordering stock, carrying more products, and deciding what types of merchandise you should invest in to increase sales.
There are many ways to keep a pulse on the value of your inventory, but the retail inventory method is one of the fastest and most cost-efficient ways to do it on a monthly basis.
In this article, you’ll learn more about RIM, how to run the calculation, and the types of businesses that benefit most from using the retail inventory method.
What is the retail inventory method?
The retail inventory method calculates the value of your inventory over time. It measures the cost of your inventory in relation to the retail price of the products and uses the cost-to-retail ratio.
While it’s a quick way to count inventory, it’s not 100% accurate. Physical inventory counts or cycle counts should still be part of your inventory management strategy to ensure your year-end financial statements are correct.
How to calculate ending inventory with the retail inventory method
Follow these steps to use the retail inventory method to calculate your monthly ending inventory.
Cost-to-retail ratio
Cost of goods available for sale
Calculate the cost of goods available for sale by adding your cost of purchases to your beginning inventory cost.
Cost of sales
Calculate your cost of sales by adding up all your sales and then multiplying the total by your cost-to-retail ratio.
Ending inventory
Calculate ending inventory by subtracting the cost of sales from the cost of goods available for sale.
💡 PRO TIP: With Shopify POS, you can skip the manual calculations and see your inventory cost, ending quantity, and total value for each month. To get started, view the Month-end inventory snapshot report in Shopify admin.
Example of the retail inventory method
Let’s take a look at a hypothetical example of the retail inventory method calculation:
If a skincare retailer sells face cream for $25 and purchases each bottle for $5, the cost-to-retail ratio would be 20%.
Let’s say the retailer’s beginning at-cost inventory was $5,000 (1,000 units = $5000 / $5), and it purchased $2,500 (500 units = $2500 / $5) worth of additional inventory during the month for a total of $7500 (1,500 units). In the same period, it sold $10,000 worth of products (400 units = $10,000 / $25).
Here’s how we can calculate its ending inventory cost using the retail inventory method:
Using this calculation, you can measure your ending inventory cost and also estimate your physical inventory counts. If you divide the ending inventory cost ($5,500) by the cost per unit of face cream ($5), we can assume there are 1,100 bottles left of the total 1,500 units that were purchased throughout the month.
Assuming there were no at-cost or mark-up changes during the month, this calculation would provide a good level of accuracy.
Advantages and disadvantages of the retail inventory method
The main advantage of the retail inventory method is saving time on doing physical inventory counts. The disadvantage is that, depending on certain variables, it’s not always the most accurate technique. Let’s take a closer look:
Advantages of RIM
When Malcolm McNair, a Harvard Business School professor, invented the retail inventory method in the early 1900s, he aimed to make tracking inventory and bookkeeping less time-consuming and more cost-effective.
RIM was created with three main goals in mind, which today translate into the benefits of using the retail inventory method:
Disadvantages of RIM
As with any solution, you should also consider the drawbacks of the retail inventory method:
Who should use the retail inventory method?
There are many types of businesses that can benefit from using RIM to calculate ending inventory, including the following:
Multi-location retailers
If you have many store locations, RIM is a great way to get a quick overview of your inventory across all locations without having to manually count each stock unit and potentially disrupt business operations.
Retailers with warehouses
Using the retail inventory method to calculate stock you have in a warehouse saves you the hassle of going to the warehouse to count it (or spending money to have the warehouse employees count it). Also, the cost of merchandise in your warehouse is usually consistent—there are no sales or price cuts like you have in-store—which leads to more accurate results with the RIM calculation.
Retailers that use markups consistently
If your products consistently have the same cost-to-retail ratio, meaning your product markups are mostly the same across all your merchandise, the retail inventory method is a reliable solution. But if you sell a range of products that have varying markups, RIM may not be the best inventory management approach for your business.
Retailers that don’t run a lot of sales
If you operate your retail business on a model where there are few to no sales promotions, the retail inventory method is a great option, since you don’t run the risk of discrepancies due to inconsistent markups.
Wholesalers with large volumes of similar products
If you sell your products to other retailers in large quantities and all the items have the same or a similar markup, the retail inventory method is a great way to reduce the time spent on physical inventory counts.
Is RIM right for your store?
The retail inventory method is time-saving and cost-effective, but it’s not flawless. This approach works best when it’s part of your overall inventory management strategy. Use RIM in tandem with other techniques like a powerful retail management system or POS, physical inventory or cycle counts, and consistently reviewing your sales performance and stock.
Read more
Retail inventory method FAQ
Is the retail inventory method LIFO or FIFO?
What is the retail inventory method quizlet?
Who uses retail inventory method?
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17318 | https://www.geogebra.org/m/hgw9wgff | Exterior Angles of Regular Polygons – GeoGebra
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17319 | https://www.youtube.com/watch?v=qz6aYAvUrZI | Converting units of measure with grids
owigger
738 subscribers
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Posted: 21 Aug 2012
In this video you will learn how to convert units of measure using grids. All equalities can be made into fractions. Those fractions can be used to convert any unit of measure to another.
Transcript:
we'll be doing A New Concept for most people and that is how to use unit cancellation unit cancellation relies upon the use of ratios and just to get us started here I'm going to start with the problem here in letter b and I'm going to show you that for all problems that you do in this class you're going to have a given and a fine if you're doing homework and you have to do a homework problem where you're supposed to copy the question if it's a calculation problem all you need to do is copy down the given and find and you're you're done you go ahead and work the problem after that so the given in this particular case is what anybody yes two point I'm sorry you're looking at letter A good okay four thousand three hundred twenty mils and then what were we supposed to find leaders okay you're supposed to find how many liters and you see that is adequate information to be able to solve the problem are we good the thing that you don't have is what I call a sidebar a sidebar is information that's related to the problem that you will need to be able to solve it now can anybody see something up here that we might be able to use to solve this particular problem some ratio of Mills to liters anybody yeah the fact that one thousand Mills is equal to one liter is good information and I'm going to show you how to use it now in this particular problem you can probably do it in your head at least some of you can but I'm going to show you the way I want you to work the problems and how I want you to show your work Okay so in all of these problems we start with the given over one you're going to find out as you work with this over and over and over again that really makes it nice because you always know how to start a problem okay so we're going to start with 4320 mils which is of course are given that's going over one and solving these problems is called the way I'm solving these problems is called Unit cancellation and unit cancellation is just involved in doing uh two different two three or four fractions to do the calculations that you need to get the answer that you're asked to get and of course the answer we're asked to get is liters so I know I'm going to be working with fractions so let's go ahead and watch what I'm doing do this right along with me I'm going to circle my Mills and I'm going to put that same thing down here sidebar that's the information that I had to have to be able to solve the problem in the first place and now in my sidebar I have how many mils is equal to a liter now who can tell me what a coefficient is anyone with all Heaven algebra haven't you let us go fishing give it a try actually you have it backwards but you have the right idea it's the number yes it is it's the number next to the letter like 2A is equal to 3B 2 is the coefficient of a and three is the coefficient of B okay so what we're going to do here is we look at the Mills and um valine if I get it did I say that right salary oh okay I didn't see your your they are in there properly I apologize all right okay what is the coefficient of ml in my little sidebar area excellent you got it right cool okay it's the number to the left of the letters so here was the letters and the Thousand was right next to it so that thousand is what we're going to put down here to in this area to match the Mills up now what do you think we're going to put on the top what uh-huh no look at the other side of the equality and so what do I put on the top the coefficient and the letter okay what's the other side of the equality the left side was 1000 mils what's the right side because that's the ratio you see it's 1 000 milliliters is equal to one liter now do you think I could cancel anything here yes Mills here Mills here yep just they're gone they're toast no longer there cancel and the only unit of measure that's left is my liters so now all I have to do is divide the four thousand three twenty by the one thousand and that gives me 4.320 and what do you think the unit of measures leaders yes the leaders and there's our answer does that match what the find has yeah it matches perfectly all right so this is kind of the basic start and all the problems are going to be just like this so let's look up here at letter C that's what we'll do next convert using grids who has an idea of what to find a given is I'm sorry what's the given and anyway go what um you got it right right on okay so the given for number c is 529 was it yeah 29 and what what's the unit 526. okay 526 and what's the end of the measure milligrams okay and what are we supposed to find you can find as well okay first two grams then to kilograms right okay so let's go ahead and do our fine uh first of all grams then how many kilograms as well let's go and so we look up here at the things that I gave you to put your little sidebars and which two of those are we going to need the grams at top there yeah and the last one milligrams and grams okay so let's go ahead and put that in as a sidebar we got one thousand milligrams is equal to one gram that's good information and then we have one thousand grams is equal to a kilogram okay because kilo stands for a thousand so that means kilogram means one thousand grams where milligrams means one one thousandth of a gram one one thousandth is equal to a thousand milligrams okay so now we're ready to go ahead and work our problem now there's going to be two things that we're going to do here so let's let's start at the beginning and how about we do this in a plot so we have you have an over one so what am I going to start with 52 526 milligrams or and I'm going to circle this and what do I put on the bottom of that fraction see now I look over here and I'm looking for Milli milligrams okay uh this video You Wanna Give us a shot on this what am I going to put with milligrams here yes because you see the coefficient of milligrams is a thousand so that's what's going to go down here and what do you think we'd have on top one gram you got it you're right on excellent and now I could stop here and figure the kit the grams now let's go ahead and do that but we didn't have to we could have just done the next fraction and you can do that five six seven eight nine ten fractions doesn't matter I certainly do the question was do I want to cancel out milligrams I definitely do and now if I divide that by a thousand that moves the decimal over 3 to the left is everybody okay with that yeah so we're at zero point five two six grams now it's not milligrams anymore and that's my first answer now my first answer is going to become my given yeah it's going to become my given for the next dude so we're going to start with 0.526 grams over one and what's at the bottom of that next fraction bottom of the next fraction go for it no well don't don't give me the Thousand yet just the grams okay so we got grams here we got grams here now the number the coefficient for the grams yeah is a thousand okay so one thousand goes down here and what do we have on top kilogram and that that coefficient is one okay so we got one kilogram and can we cross out anything yeah we can we can cross out the grams and we move the decimal three more to the left perfect and so now we have zero point three zeros and then the five two six and that would be the kilos and that would be our final answer if you do this on your calculator you're going to see this 5.26 oops let the six out okay 5.26 Five Points two six there we go okay 5.26 times 10 to the minus 4. that's what it'll say on your calculator now is that fully the same number okay let's watch that's why I have this little pointer here for this number right here are original oops boy uh there there we are so 5.26 times 10 to the minus three or ten to the minus four so that means the decimal started right here yes and are we four over from there okay so if we start right here now let's check it out this is where we're starting we're gonna go one two three four and the decimal is here and there's our zeros one two three zeros so yeah so 5.26 once I put those zeros there this goes away it's no longer there because it's done shot and that's called scientific notation by the way a number a decimal and however many sig figs are after it and we will learn more about scientific notation tomorrow for right now just want you to see the method so far so good okay now now let's go to tonight's homework and let's see how that little guy is going to work would you all get out that sheet and let's see what it looks like here it is you know I think I'm going to stop right here and we'll make it a second video uh |
17320 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5455754/ | Structure Determination of Au on Pt(111) Surface: LEED, STM and DFT Study - PMC
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Materials (Basel)
. 2015 May 27;8(6):2935–2952. doi: 10.3390/ma8062935
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Structure Determination of Au on Pt(111) Surface: LEED, STM and DFT Study
Katarzyna Krupski
Katarzyna Krupski
1 Department of Physics, University of Warwick, Coventry CV4 7AL, UK; E-Mail: k.j.krupski@warwick.ac.uk
Find articles by Katarzyna Krupski
1, Marco Moors
Marco Moors
2 Peter Grünberg Institut, Forschungszentrum Jülich, Wilhelm-Johnen-Str., 52425 Jülich, Germany; E-Mail: m.moors@fz-juelich.de
Find articles by Marco Moors
2, Paweł Jóźwik
Paweł Jóźwik
3 Department of Advanced Materials and Technologies, Faculty of Advanced Technologies and Chemistry, Military University of Technology, Kaliskiego 2 Str., 00-908 Warszawa, Poland; E-Mail: pjozwik@wat.edu.pl
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3, Tomasz Kobiela
Tomasz Kobiela
4 Faculty of Chemistry, Warsaw University of Technology, ul. Noakowskiego 3, 00-664 Warsaw, Poland; E-Mail: kobiela@ch.pw.edu.pl
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4, Aleksander Krupski
Aleksander Krupski
1 Department of Physics, University of Warwick, Coventry CV4 7AL, UK; E-Mail: k.j.krupski@warwick.ac.uk
3 Department of Advanced Materials and Technologies, Faculty of Advanced Technologies and Chemistry, Military University of Technology, Kaliskiego 2 Str., 00-908 Warszawa, Poland; E-Mail: pjozwik@wat.edu.pl
5 Faculty of Science, SEES, University of Portsmouth, Portsmouth PO1 3QL, UK
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1,3,5,
Editor: Marco Salerno
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1 Department of Physics, University of Warwick, Coventry CV4 7AL, UK; E-Mail: k.j.krupski@warwick.ac.uk
2 Peter Grünberg Institut, Forschungszentrum Jülich, Wilhelm-Johnen-Str., 52425 Jülich, Germany; E-Mail: m.moors@fz-juelich.de
3 Department of Advanced Materials and Technologies, Faculty of Advanced Technologies and Chemistry, Military University of Technology, Kaliskiego 2 Str., 00-908 Warszawa, Poland; E-Mail: pjozwik@wat.edu.pl
4 Faculty of Chemistry, Warsaw University of Technology, ul. Noakowskiego 3, 00-664 Warsaw, Poland; E-Mail: kobiela@ch.pw.edu.pl
5 Faculty of Science, SEES, University of Portsmouth, Portsmouth PO1 3QL, UK
Author to whom correspondence should be addressed; E-Mail: aleksander.krupski@port.ac.uk.
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Marco Salerno: Academic Editor
Received 2015 Jan 20; Accepted 2015 May 8; Collection date 2015 Jun.
© 2015 by the authors;
licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution license (
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PMCID: PMC5455754
Abstract
Low-energy electron diffraction (LEED), scanning tunneling microscopy (STM) and density functional theory (DFT) calculations have been used to investigate the atomic and electronic structure of gold deposited (between 0.8 and 1.0 monolayer) on the Pt(111) face in ultrahigh vacuum at room temperature. The analysis of LEED and STM measurements indicates two-dimensional growth of the first Au monolayer. Change of the measured surface lattice constant equal to 2.80 Å after Au adsorption was not observed. Based on DFT, the distance between the nearest atoms in the case of bare Pt(111) and Au/Pt(111) surface is equal to 2.83 Å, which gives 1% difference in comparison with STM values. The first and second interlayer spacing of the clean Pt(111) surface are expanded by +0.87% and contracted by −0.43%, respectively. The adsorption energy of the Au atom on the Pt(111) surface is dependent on the adsorption position, and there is a preference for a hollow fcc site. For the Au/Pt(111) surface, the top interlayer spacing is expanded by +2.16% with respect to the ideal bulk value. Changes in the electronic properties of the Au/Pt(111) system below the Fermi level connected to the interaction of Au atoms with Pt(111) surface are observed.
Keywords: density functional theory calculations, scanning tunneling microscopy, low-energy electron diffraction, surface structure, metallic surfaces, gold, platinum, metal-metal interfaces, low index single crystal surface
1. Introduction
A large number of studies on epitaxy have been carried out for many years. Ultra-thin epitaxial film systems exhibit a variety of interesting properties due to the strong correlation between the electronic structure of the film and its morphology, strain, and defect structure [1,2,3,4,5,6,7,8,9,10]. Structural studies of fcc/fcc systems provide a great deal of information on the connection between the geometrical properties of the adsorbed atomic layers and the atomic arrangements of the substrates. Various fields are concerned with epitaxial growth; these range from basic research on the growth mechanism of thin films to advanced research on the development of devices. Platinum is widely used as a catalyst in the chemical and petrochemical industries [11,12]. For example, in oil refineries, platinum catalysts are employed in processes that involve the reforming of paraffin and the hydrogenation of unsaturated hydrocarbons [11,12,13,14].
The clean Pt(111) surface itself has been the subject of several structural determinations with Low Energy Electron Diffraction (LEED) [15,16,17,18,19,20,21,22,23], medium energy ion scattering (MEIS) , high energy ion scattering (HEIS) and surface X-ray diffraction (SXRD) [26,27]. Differently from the other noble metal (111) surfaces, clean Pt(111) normally has an unreconstructed bulk-periodic surface [23,28,29,30]. Adams et al. found the first layer spacing to be possible to expand by 0.04 ± 0.10 Å, while Hayek et al. found an unrelaxed surface with 0.05 Å. Materer et al. found the first and second interlayer spacing expanded by 0.04 ± 0.10 Å and 0.005 ± 0.03 Å, respectively. MEIS and HEIS experiments [24,25] support the LEED results. Namely, the ion scattering data indicate that the Pt(111) structure deviates from the bulk geometry by a possible small outward expansion of the top interlayer spacing of 0.03 ± 0.02 Å or 0.03 ± 0.02 Å . No deviation from the bulk position was found in the direction parallel to the surface, with a small accuracy of about 0.01 ± 0.02 Å. The surface geometry of clean Pt(111) has been the subject of surface X-ray diffraction investigations . These investigations gave an outward relaxation of the topmost layer of 0.045 ± 0.005 Å (+2.0%) with respect to the ideal bulk termination.
Properties of ultrathin gold layers deposited on the Pt(111) face were investigated in a number of works [31,32,33,34,35,36,37,38,39,40,41,42]. Studies on single crystalline Au-Pt(111) model surfaces, for instance, have provided detailed information on the catalytic properties of these surfaces [31,32,33]. Davies et al. studied the growth and chemisorptive properties of gold and silver monolayers on platinum (111) and (553) single crystal surfaces using Auger electron spectroscopy (AES), LEED, and temperature-programmed desorption (TPD). The AES results suggested that the growth of Au proceeds via a Stranski-Krastanov mechanism at room temperature, and that at temperatures above 800 K gold dissolves into the Pt crystal bulk. No extra LEED order spots or spot streaking was observed. In contrast, Shatler et al. with the use of AES, LEED, and TPD found that deposition of gold on Pt(111) near T = 300 K indicates a layer-by-layer (Frank-van der Merwe) growth mechanism up to three gold monolayers. The analysis of AES measurements showed that two-dimensional islands growth below one monolayer took place. Furthermore, with increasing coverage, the gold islands grew until the monolayer is completed, before the second layer begins to form. In additional studies by Sachtler et al. , the activity for conversion of n-hexane as a function of Au surface concentration on Pt(111) was monitored. The Au-covered crystal was then annealed at elevated temperatures to allow Au intermixing with the Pt substrate. The formed Au-Pt(111) surface alloy showed a much higher activity for n-hexane isomerization than pure Pt. Moreover, it has been reported that Au in a dispersed state exhibits a high activity for some reactions at low temperatures (e.g., CO oxidation) and that this feature depends on the preparation conditions, size and shape of the Au nanostructures . Adsorption experiments with CO as a titration agent showed a significantly lower affinity of the Au-Pt surface alloy in comparison to the clean Pt surface . Salmeron et al. used photoelectron spectroscopy techniques (UPS (ultraviolet) and XPS (X-ray)), LEED and AES to study the electronic structure of Au and Ag overlayers deposited on Pt(111), Pt(100), and Pt(997). Between 0 and 1 monolayer, the valence bands of Au and Ag show changes in the form of shifts of the most tightly bound peaks and the appearance of the new structures around a coverage (θ Au) of one monolayer. The Au 5 d 3/2 peak shifts 0.6 eV towards higher binding energies when coverage varies from 0.1 to 1 monolayer and 0.5 eV more when coverage varies from one to six monolayers. These shifts are explained as due to the changing contributions of the Au atoms in island edges for surface (θ Au< 1) monolayer and bulk (θ Au> 1) coordination positions. Using AES, they found that gold on Pt(111) grows layer-by-layer. Below θ Au< 1, no extra LEED spots were observed. In addition, the work function decreased upon gold deposition from its initial value of 6.08 ± 0.15 eV for clean Pt(111) down to 5.8 ± 0.15 eV. That value was reached at the monolayer and remained constant thereafter up to five monolayers and is clearly larger than the 5.31 eV value reported by Potter et al. for bulk Au(111). The work function for the Pt(111) surface compares only fairly with that reported by Ertl et al. of 6.40 eV. Its smaller value might reflect a less perfect surface with larger number of residual steps. It should be pointed out here that Pt(111) surface presents the highest work function value among other metals surfaces. Vogt et al. studied Au/Pt(111) system by spin-, angle- and energy-resolved photoemission with normal incident circularly polarized synchrotron radiation of BESSY and normal photoelectron emission for different Au coverages. The prepared layers were characterized by AES and LEED and turned out to grow up two-dimensionally and epitaxially. LEED spots did not show any changes in geometry during the evaporation time up to the coverage of a thick Au layer . Later, the electrodeposition of Au on Pt(111) from electrolytes containing µM concentrations of was investigated by in situ electrochemical scanning tunneling microscopy (EC-STM) by Sibert et al. [41,42]. Under conditions of high Au surface mobility, multilayer growth proceeds via a typical Stranski-Krastanov growth mode, with layer-by-layer growth of a pseudomorphic Au film up to two monolayers and three-dimensional growth of structurally relaxed islands at higher coverage, indicating thermodynamic control under these conditions.
In the present work, in order to study the structural and electronic properties during the initial adsorption process of gold on Pt(111) surface at room temperature, we have performed low-energy electron diffraction, scanning tunneling microscopy measurements in ultrahigh vacuum and density functional theory calculations with the use of CASTEP code.
2. Experimental Details
The measurements were carried out in a stainless steel ultra-high vacuum chamber with a base pressure of 2.0 × 10−8 Pa. The chamber was equipped with a reverse-view LEED optics, which was used for low-energy electron diffraction measurements, and also with a variable-temperature scanning tunneling microscopy stage. The Pt(111) single crystal was supplied by MaTeck . The surface of the Pt(111) single crystal was cleaned by repeated cycles of sputtering with 3 keV Argon ions at T = 300 K and annealing at T = 1100 K. After annealing at 1100 K, the residual carbon was removed in 7.0 × 10−4 Pa of oxygen, followed by desorption of any remaining oxygen at 1200 K. This procedure was repeated until the LEED pattern of a clean Pt(111) surface with sharp spots and low background was obtained. The deposition of Au (99.999%) on the Pt(111) sample was achieved by vaporization from a Knudsen cell and the coverage of gold was determined via STM. Film coverages are described in monolayers (ML), where a 1 ML Pt(111) film corresponds to an atomic packing density of 1.503 × 10 15 atoms/cm 2 obtained from a bulk lattice constant a Pt = 3.9239 Å (for comparison the atomic packing density of Au(111) equals 1.387 × 10 15 atoms/cm 2 for a Au = 4.0785 Å ). This cell had been constructed from an Al 2 O 3 crucible from Friatec with a diameter of 5 mm. It was filled with a 0.5 mm thick Au wire from Goodfellow and closed by a two-hole ceramic. The Knudsen cell was heated by a tungsten wire from Goodfellow (diameter 0.3 mm) wound around the crucible and thermally shielded by a water-cooled jacket. In order to control the deposition time, a rotatable shutter was placed in front of the cell opening. The working pressure during Au deposition was below 1.0 × 10−7 Pa. All STM measurements were performed with the use of electrochemically etched W (99.99%) tips (diameter 0.5 mm, length 3.5 mm). For the potassium hydroxide electrolyte, a 4 V p−p square wave voltage (f = 100 Hz) was applied to the tip. In the electrochemical cell, a tungsten wire is used as the working electrode (anode) and a Pt (99.999%) loop (diameter 10 mm) is used as the counter electrode (cathode). A 3 M KOH solution from Sigma Aldrich is used as the electrolyte. The following reactions take place:
Cathode Pt (Reduction Reaction):
(1)
Anode W (Oxidation Reaction):
(2)
(3)
Total Reaction:
(4)
All presented STM images were recorded in constant current mode and processed by the WSXM image-processing software . Before starting experimental investigations of the Pt(111) and Au-Pt(111) surfaces, the experimental system was calibrated with the use of well know Si(111)-(7 × 7) reconstructed surface [51,52,53,54] (Figure 1). Si(111)-(7 × 7) surface was prepared by twice direct current flashing (I = 4.0 A) an p-type Si(111) substrate (size: 1 × 10 × 0.5 mm, resistivity ρ ≈ 1–10 Ω cm) at 1220 K, after degassing at 970 K for two hours by joule heating with a current of 1 A. Atomically resolved STM images of the empty and filed states of Si(111)-(7 × 7) are presented in Figure 1b,c, respectively. The measured surface unit cell is characterized by two diagonals of the diamond (a 1 = 46.6 Å and a 2 = 26.9 Å). Silicon adatoms (12 per unit cell) are marked in red in Figure 1b, and they occur as bright “dots” in empty-state STM image. Visible in STM images deep holes (depth ~ 2 Å) are called corner hols.
Figure 1.
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Si(111)-(7 × 7) surface at T = 300 K: (a) LEED patterns recorded at normal electron incidence for E = 40 eV; (b) STM image of empty states (150 Å × 150 Å, I T = 0.5 nA, U bias = +1.6 V); and (c) STM image of filled states (150 Å × 150 Å, I T = 0.5 nA, U bias = −1.6 V). Au deposited on Pt(111) at T = 300 K at a coverage θ Au ≤ 1.0 ML: (a) θ Au ≈ 0.8 ML (5000 Å × 5000 Å, I T = 4.0 nA, U bias = 1.0 V); (d–f) line scans along the lines A, B, and C from the image in (b). The unit cell is indicated by the blue diamond (diagonals: a 1 = 46.6 Å, a 2 = 26.9 Å). Si adatoms (12 per surface unit cell) cell are marked as red dots.
3. Calculation Details
All calculations were performed based on the pseudo-potential plane-wave within the density functional theory [55,56], using the Cambridge serial total energy package (CASTEP) . The effects of exchange correlation interaction are treated with the generalized gradient approximation (GGA) of Perdew-Burke-Ernzerhof (PBE) [58,59]. The ultra-soft pseudo-potentials describe this electron-ion interaction system to high accuracy with a plane wave energy cutoff of 600 eV. The energy calculations in the first irreducible Brillouin-zone were conducted by using the (4 × 4 × 1) k-point grid of the Monkhorst-Pack scheme . Spin polarization of platinum was included in the calculations to correctly account for its magnetic properties. All atomic positions have been relaxed according to the total energy and force using the BFGS scheme based on the cell optimization criterion RMS force of 0.03 eV/Å, stress of 0.05 GPa, and displacement of 0.001 Å. The calculation of total energy and electronic structure is followed by cell optimization with SCF tolerance of 1 × 10−6 eV/atom. The Pt(111) surface was modeled using a slab containing 7 (=15.84 Å) and 10 (=22.63 Å) layers of Pt atoms with a vacuum gap in the direction equal to 20.57 Å and 30.37 Å, respectively. Full slab relaxation was performed in both cases.
4. Results and Discussion
4.1. LEED and STM
Gold atoms on the Pt(111) face form an ordered structure after evaporation onto the crystal face. Typical LEED pattern observed before and after deposition of gold on the Pt(111) face in normal electron incidence are shown in Figure 2. In this figure, the unit cell of the platinum lattice is indicated. The lattice constant of the platinum surface unit cell is 2.775 Å (primitive fcc (111) unit cell) . The patterns are shown to demonstrate the quality of the structural order at the surface. It should be pointed out that the positions of the LEED spots associated with the Pt(111) substrate remains unchanged during the gold deposition at 300 K (Figure 2c), as was previously reported by Sachtler and Samorjai and Vogt et al. . Thus, the lattice constant of the first substrate layer remains constant, too, and suggests a two-dimensional growth of the first gold layer.
Figure 2.
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LEED patterns observed during the growth of Au on the Pt(111) surface recorded at normal electron incidence for E = 120 eV, and T = 300 K: (a) clean Pt(111) for E = 120 eV. k X and k Y denote axes in the reciprocal lattice; (b) 1 ML of Au on Pt(111); and (c) line profile along the lines from the image in (a,b) demonstrating that the position of LEED spots remain unchanged after gold deposition. The unit cell is outlined.
The results of our STM measurements on the clean Pt(111) surface are presented in Figure 3. Figure 3a displays an STM image, taken on a low-index Pt(111) substrate with terraces between 100 and 300 nm width separated by monoatomic steps. The height of the steps on the Pt(111) surface was measured by STM to be 2.26 ± 0.3 Å (Figure 3b). However, one need to remember that the observed by STM step height includes geometric and electronic factors. Figure 3c presents atomic resolution of the Pt(111) face. The obtained topography shows a hexagonal lattice arrangement of Pt atoms with the nearest neighbor distance of 2.80 Å. This value describes the dimension of the surface unit cell and is 0.90% higher compared to the literature value (=2.775 Å) . Figure 3c demonstrates that the surface structure seen in the obtained STM image has a clear long-range character. Figure 4 shows STM images of the Pt(111) surface with varying Au coverage in order to illustrate the morphology of the Au layers deposited on Pt at room temperature. Figure 4a shows a typical STM image corresponding to a submonolayer coverage of θ Au ≈ 0.8 ML. An analysis of the STM measurements indicates that for coverage less than 1 ML, two-dimensional growth of gold layer is observed. This is in agreement with photoelectron spectroscopy study , our present and previous AES/LEED measurements [32,39]. The darker features in Figure 4a represent still visible platinum substrate as predicted in the previous studies . Similar to the results observed by us, two-dimensional gold monolayer was obtained by electrodeposition of Au on Pt(111) from electrolytes containing µM concentrations of of . The line scan in Figure 4b shows that the height of the first gold layer corresponds to the height of a single Pt step height equal to 2.26 Å. As the Au coverage is close to 1 ML, Au wets the Pt(111) surface completely, as can be seen in Figure 4c. This is not easy to confirm with STM, whether the surface is wetted or not. However, the reason for the perfect wetting is because of the high-specific surface free energy of the Pt(111) surface (2.299 J/m 2< γ Pt(111)< 2.489 J/m 2) [64,65,66,67,68] as compared with that of the Au(111) surface (1.283 J/m 2< γ Au(111)< 1.506 J/m 2) [64,65,66,67,68]. Since the total specific surface free energy should be minimized, a covered Pt(111) surface is favored . Closer view of the STM image topography in Figure 4d reveals the presence of well-ordered gold structures. STM images indicate a long-range order in the surface system. The obtained topography shows a hexagonal lattice arrangement of Au atoms with a nearest neighbor distance of 2.80 Å, which is exactly the same value as mentioned above in the case of Pt atoms. The same value of the surface unit cell after adsorption of gold could suggest that the gold atoms are adsorbed in sites (hollow fcc or hcp), which are a direct continuation of the Pt lattice ABCABCA. This is in good agreement with the supposition from the spin-resolved photoemission studies of Au-Pt(111) system , where the best fit of experimental results and theoretical model was achieved on that basis.
Figure 3.
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STM images of the clean Pt(111) surface: (a) T = 25 K (3734 Å × 3734 Å, I T = 141 pA, U bias = +50 mV); (b) line-scan corresponding to line drawn in (a); and (c) T = 300 K (40 Å × 40 Å, I T = 49 pA, U bias = +48 mV). The unit cell is outlined. STM image evidences a hexagonal lattice arrangement of Pt atoms with measured nearest neighbor distance of 2.80 Å.
Figure 4.
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STM images of Au deposited on Pt(111) at T = 300 K at a coverage θ Au ≤ 1.0 ML: (a) θ Au ≈ 0.8 ML (5000 Å × 5000 Å, I T = 4.0 nA, U bias = +1.0 V); (b) line scan along the line from the image in (a) demonstrating that the height of the gold layer corresponds to the height of a single Pt step height; (c) θ Au ≈ 1.0 ML (5000 Å × 5000 Å, I T = 2.0 nA, U bias = +1.0 V); and (d) (30 Å × 30 Å, I T = 4.65 nA, U bias = +159 mV). The unit cell is outlined. STM image evidences a hexagonal lattice arrangement of Au atoms with measured nearest neighbor distance of 2.80 Å.
4.2. DFT
4.2.1. Structure of Clean Pt(111)
In the theoretical part of our work, we have calculated multilayer relaxations of the Pt(111) system using the slab with 7 and 10 atomic layers. Figure 5a shows the schematic view of relaxed slab structure for the seven platinum layers. The platinum low-index surface was modeled by repeated slabs with a (1 × 1) surface unit cell with four atoms in each layer. The calculated atomic layer distances for seven and ten planes are shown in Table 1. defines the distance along the surface normal direction between the X atom at the i atomic layer and the Y atom at the j atomic layer. Surface relaxation is characterized as the percent of change of the spacing between layers i and j versus the bulk layer spacing (d 0). Bulk value (d 0) is taken from our GGA calculations and describes average distance between atomic planes of seven (=2.30 Å) and ten (=2.29 Å) platinum layers, respectively. Further calculations of gold adsorption on Pt(111) surface has been performed on seven platinum layers.
Figure 5.
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(a) Side view of the relaxed Pt(111) surface for seven layers. Values of denoted characteristic inter plane distances are given in Table 1. (b) Considered positions of Au adsorption on the Pt(111) surface: A—on top; B—hollow fcc; C—hollow hcp; and D—bridge. The unit cell is outlined. The nearest neighbor Pt-Pt distance of 2.83 Å is obtained from our theoretical calculations.
Table 1.
Distances () between the atomic planes of the relaxed Pt(111) system, and their percentage changes () with respect to the bulk value (d 0), calculated for the slab with 7 and 10 atomic layers and compared with experimental [15,16,18,20,21,22,23,26,72] and theoretical [30,70,71] literature data. Notation of inter-plane distances are the same as in Figure 1. denotes the interlayer spacing between layers i and j for the X and Y atoms type. d 0—average distance between atomic planes of seven and ten layers, respectively. a 0—lattice constant of Pt. GGA—generalized gradient approximation, LDA—local density approximation, LEED—low energy electron diffraction, SXRD—surface X-ray diffraction.
| Present Work GGA (CASTEP) | Previous DFT study | LEED | SXRD g |
:---: :---: |
| (Å) | 7 layers | (%) | 10 layers | (%) | (Å) | (%) | (Å) | (Å) | (%) |
| | 2.32 | +0.87 | 2.33 | +1.75 | 2.766 h(LDA) | | 2.26 a | 2.31 ± 0.05 | +2.0 |
| +0.44 h(LDA) | 2.29 ± 0.1 b |
| +0.85 i(LDA) | 2.2713 c |
| +1.14 j(LDA) | 2.26 ± 0.05 d |
| +0.85 j(LDA) | 2.2655 ± 0.025 e |
| | 2.29 ± 0.001 f |
| | 2.29 | −0.43 | 2.29 2.746 h(LDA) | −0.31 h(LDA) | 2.26 a | | |
| −0.56 i(GGA) | 2.2405 ± 0.025 e | | |
| −0.29 j(LDA) | 2.27 ± 0.003 f | | |
| −0.22 j(LDA) | | | |
| | 2.30 2.29 | −0.15 i(GGA) | 2.26 a 2.2655 ± 0.05 e | | |
| −0.21 j(LDA) | | |
| −0.17 j(LDA) | | |
| | 2.30 2.30 | +0.43 | | | 2.26 a | | |
| | 2.29 | −0.43 | 2.29 | | | | |
| | 2.30 2.30 | +0.43 | | | | | |
| | | | 2.29 | | | | |
| | | | 2.29 | | | | |
| | | | 2.30 | +0.43 | | | | | |
| | 2.30 | | 2.29 | | 2.75 h(LDA) | | 2.26 c | | |
| | 2.2655 e | 2.26 | |
| | 2.265 f | | |
| | 3.99 | | 3.99 | | 3.99 i(GGA) | 3.92 k(EXP) | 3.92 a 3.9231 d | | |
| 3.97 j(LDA) | | |
| 3.89 j(LDA) | | |
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a Ref. [15,16]; b Ref. ; c Ref. ; d Ref. ; e Ref. ; f Ref. ; g Ref. ; h Ref. ; i Ref. ; j Ref. ; k Ref. .
Our calculations for the clean Pt(111) show very good agreement with the above-presented STM results and with the other experimental and theoretical literature studies [15,16,18,20,21,22,23,26,30,70,71] presented in Table 1. Obtained lateral geometrical properties of Pt(111) surface and distances between the nearest Pt atoms in the structure (=2.83 Å) are very close to STM measurements (=2.80 Å), with the difference about 1%. The first and second interlayer spacings of the clean Pt(111) surface were determined to be 2.32 Å and 2.29 Å, respectively, in case of calculated slab with seven atomic layers. This corresponds to a +0.87% expansion and −0.43% contraction of the first and second metal layer spacings of the ideally terminated Pt(111) clean surface (=2.30 Å), respectively. These values and the value obtained in our calculations of lattice constant of the bulk Pt (=3.99 Å) are in excellent agreement with the previous GGA calculations and surface X-ray diffraction results (Table 1). However, comparison of our calculations to the quantitative low-energy electron diffraction value of the first interlayer spacing shows that our theoretical value (=2.32 Å) is slightly larger (+2%) then the average value of 2.27 Å observed experimentally [15,16,18,20,21,22,23].
4.2.2. Structure of the Au/Pt(111) System
Figure 5b shows the four possible gold adsorption sites on the Pt(111) surface with one on-top site (labeled as A), two hollow sites: hollow fcc (labeled as B), hollow hcp (labeled as C), and one bridge site (labeled as D), In our calculations, we define one monolayer of adsorbed Au atoms corresponding to the same atoms as the atomic sites in the surface layer. One Au atom adsorbing on the Pt(111) surface corresponds to an adsorption coverage of 0.25 ML. The minimum adsorption energy (E ads) was calculated by means of the following total energy difference:
(5)
where E T is the total energy of the system and , , and refer to the atom-on-metal system, the free Au atom, and the bare Pt surface, respectively.
Table 2 displays the predicted adsorption energies of Au on the Pt(111) surface and the distance between the Au atom and its nearest (r NN) and next nearest neighbors (r NNN). A, B, C and D describe positions of Au atom on the Pt(111) surface before starting calculations. As one can see, only in the case of bridge position D displacement of gold atom towards hollow fcc position B is observed, while the other gold adsorption positions described as A, B and C remain unchanged. The comparison of the calculated adsorption energies reveals that the preferred position of the Au on the Pt(111) surface is the hollow fcc with the E ads = −0.578 eV. At this favorable position, the nearest to the nearest (NN) and next-nearest (NNN) neighbor distance is equal to 2.58 Å and 3.76 Å, respectively. The adsorption energy of one gold atom in hollow fcc site is negative, which indicates in addition that this adsorption position is the most stable. Similar conclusion was obtained in case of a quantitative LEED analysis of the structure of Pt(111) (√3 × √3) R30°-S, where the best agreement between experiment and theory has been found for a model with a sulfur atom in the three-fold hollow fcc site . Moreover, our theoretical studies are in agreement with the spin-resolved photoemission predictions where the Au is adsorbed in sites, which are a direct continuation of the Pt lattice . In contrast to very stable hollow fcc site, the on-top adsorption position is the most unstable place with E ads = +0.580 eV. Next, taking into account our experimental STM results, we have considered structural model of the Au/Pt(111) surface reproducing in the best way the topography of the obtained STM images. Structural relaxation has shown that such a model is stable. The lateral positions of all gold atoms in the relaxed structure remained the same as in the starting configuration. This model assumes that the gold structure is built up by Au hollow fcc and hollow hcp atoms (Figure 6). The obtained lateral geometrical properties of this Au/Pt(111) model and distances between the nearest gold atoms in the structure (=2.83 Å) are almost the same as those following STM measurements (=2.80 Å) with the difference close to 1%. Table 3 presents obtained changes in the Pt(111) geometry induced by presence of a two-dimensional gold layer. Namely, we find the top interlayer spacing noticeably expanded by +2.16% with respect to the ideal platinum bulk value (=2.31 Å). The calculated value of the surface free energy of gold layer equals to γ Au = 1.481 J/m 2, and it is in very good agreement with the value of the surface free energy of Au(111) mentioned in literature (1.283 J/m 2< γ Au(111)< 1.506 J/m 2) [64,65,66,67,68].
Table 2.
Calculation results of one Au atom adsorption on the Pt(111) surface. E ads—adsorption energy; r NN and r NNN describe the distance to the nearest (NN) and next-nearest (NNN) neighbors. D → B means that after calculations gold atom has moved from the bridge position D towards the most favorable hollow fcc position B.
| (111) | Site | E ads (eV) | r NN (Å) | r NNN (Å) |
:---: :---:
| Au on top | A | +0.580 | 2.58 | 3.76 |
| Hollow fcc | B | −0.578 | 2.74 | 3.89 |
| Hollow hcp | C | −0.518 | 2.75 | 3.92 |
| Bridge | D → B | −0.578 | 2.74 | 3.89 |
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Figure 6.
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(a) Side view of the calculated most stable hollow fcc position of a relaxed Au atom on the Pt(111) surface; (b) top view of the calculated Au/Pt(111) surface. The unit cell is outlined. The nearest neighbor Au-Au distance of 2.83 Å is obtained from our theoretical calculations.
Table 3.
Calculated distances () between the atomic planes of the relaxed Au-Pt(111) system, and their percentage changes () with respect to the ideal Pt bulk value (d 0), for the slab with eight atomic layers (see slab and top view of the considered structure in Figure 6).
| (Å) | Adsorption Site B |
:---: |
| 8 Layers (%) |
| | 2.36 | +2.16 |
| | 2.34 | +1.30 |
| | 2.31
| | 2.32 | +0.43 |
| | 2.31
| | 2.30 | −0.43 |
| | 2.33 | +0.86 |
| d 0 | 2.31
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4.2.3. Density of States
The calculated electronic structure (density of states—DOS) for the studied adsorption system is presented in Figure 7. The DOS curve for bare Pt(111) and Pt(111) covered by Au is displayed in Figure 7a as red dotted and black line, respectively. In case of Au/Pt(111) surface, the DOS curve was obtained by considering gold atoms sitting in the most stable hollow fcc positions. Density of states distributions of Pt(111) and Au/Pt(111) systems were calculated for seven (clean platinum) and eight (one gold monolayer on platinum) atomic layers, respectively. In the case of density of states for clean Pt(111) surface, our results are in very good agreement with previous theoretical studies [73,74,75,76,77]. Changes in the electronic properties of our Au/Pt(111) system, compared to Pt(111), are visible. In particular, noticeable increase in the intensity of occupied states in the energy range between −5 and −1 eV, and slight change of the DOS shape after including of one gold layer into calculations. Both alterations, mainly attributed to the interaction of Au atoms with Pt(111) surface [39,40], are represented by the projection of the adsorbed gold density of states in Figure 7a. Density of states distribution calculated for the bulk platinum presented in Figure 7b, confirms well that the electronic structure of platinum is dominated by d state within the whole considered energy range.
Figure 7.
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Density of states curves for gold on Pt(111): (a) Clean Pt(111) (red dotted line); 1 ML of Au on Pt(111) (black line). The projection of the adsorbed gold density of states is shaded in blue. (b) Clean bulk platinum (black line) and its components associated with s (orange line), p (green line) and d (blue line) orbitals. E F denotes Fermi level.
5. Conclusions
In this work, experimental and theoretical studies of the geometrical and electronic properties of (111) surface of the ordered Au-Pt adsorption system have been presented. The analysis of LEED and STM measurements indicates that for a coverage below 1 ML, two-dimensional growth of the first Au monolayer takes place. Based on LEED results, no change of the lattice constant after gold adsorption was observed. The topography of the obtained STM images of Pt(111) and Au/Pt(111) surfaces on the level of the atomic resolution demonstrate that the surface structures have hexagonal arrangement of atoms and that the surface lattice constant is equal to the distance between the nearest platinum surface atoms (=2.80 Å). This is in very good agreement (close to 1%) with our presented DFT calculations, where the distances between the nearest atoms in the case of bare Pt(111) and Au/Pt(111) surface equal to 2.83 Å. It was shown that the first and second interlayer spacings of the clean Pt(111) surface were determined to be expanded by +0.87% and contracted by −0.43%, respectively. The calculated adsorption energy of the Au atom on the Pt(111) surface is dependent on the adsorption site, and there is a preference for a hollow fcc site (E ads = −0.578 eV). In the presence of gold layer on the Pt(111) surface, the top interlayer spacing was found expanded by +2.16% with respect to the ideal bulk value. Density of states for the Pt(111) surface present very good agreement with previous literature studies, while observed changes in the electronic properties of the Au/Pt(111) system below the Fermi level are mainly connected to the interaction of Au atoms with Pt(111) surface.
Acknowledgments
Katarzyna Krupski acknowledges funding from the University of Warwick Chancellor’s scholarship. Work of Paweł Jóźwik and his research stay at the University of Warwick (IX 2014) was supported by the Military University of Technology in Warsaw under Grant No. 853/2013/MUT. Work of Tomasz Kobiela and his research stay at the University of Warwick (VII 2014) was supported by the Warsaw University of Technology. Katarzyna and Aleksander Krupski would like to thank Beata and Marek Chirek for useful discussions. Theoretical part of research was partly supported by PL-Grid Infrastructure .
Author Contributions
Aleksander and Katarzyna Krupski developed the concept and designed the manuscript. Marco Moors, Tomasz Kobiela, Paweł Jóźwik, Katarzyna and Aleksander Krupski have done STM measurements. Katarzyna Krupski has done DFT calculations. Katarzyna and Aleksander Krupski prepared the manuscript. Aleksander Krupski edited the English language. Katarzyna Krupski, Marco Moors, Tomasz Kobiela, Paweł Jóźwik, and Aleksander Krupski discussed the manuscript at all stages.
Conflicts of Interest
The authors declare no conflict of interest.
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17322 | https://arxiv.org/pdf/2308.02742 | arXiv:2308.02742v3 [math.NT] 18 Jul 2024
PELL EQUATION: A GENERALIZATION OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS USING THE LLL-ALGORITHM
JOS´ E I. LIBERATI ∗
Dedicated to my parents, H´ ector and Sheila, with heartfelt gratitude
Abstract. We introduce a generalization of the continued fraction and Chakravala algorithms for solving the Pell equation, utilizing the LLL-algorithm for rank 2 lattices.
Introduction
In 628, Brahmagupta was the first to discover the identity (1.1), which states that if the triples ( a, b, k ) and ( m, l, s ) satisfy the equations
a2 − db 2 = k and m2 − dl 2 = s,
then (a2 − db 2)( m2 − d l 2) = ( am + db l )2 − d(al + b m )2 = ks. (1.1) This allows the composition of two solution triples ( a, b, k ) and ( m, l, s ) into a new triple (am + db l, al + b m, ks ). (1.2) In the Chakravala method, discovered by Bhaskara II in the 12th century, the core idea is that given a triple ( a, b, k ) (which satisfies a2 − db ,2 = k), we can compose it with the trivial triple (m, 1, m 2 − d) (by setting l = 1 in (1.2)) to obtain a new triple ( am + db, a + b, m, k (m2 − d)) which can be scaled down by k to yield
( am + db
k
)2
− d
( a + b, m
k
)2
= m2 − d
k (1.3) and then we choose m to be a positive integer for which the three quotients in (1.3) are integers and minimize the absolute value of m2 − d. This results in a new triple of integers and the process is continued until a stage is reached at which the equation has the desired form
a2 − db 2 = 1 ,
solving the Pell equation. The fundamental concept of our algorithm is straightforward: we can replicate the entire argument with the inclusion of ” l” (instead of setting l = 1), and in each step, we need to choose the two variables m and l that satisfy specific conditions. One of the principal outcomes of this
Date : 16 jul, 2024. 2000 Mathematics Subject Classification. Primary 11D09; Secondary 11A55.
∗
Ciem - CONICET, Medina Allende y Haya de la Torre, Ciudad Universitaria, (5000) C´ ordoba - Argentina. e-mail: joseliberati@gmail.com Keywords: Pell equation, Chakravala, continued fraction, LLL-algorithm. ORCID Number: 0000-0002-5422-4056 .
12JOS ´E I. LIBERATI
study consists of two variations of this generalization, each imposing different conditions. These are defined as the First and Second Algorithms with L in Section 4. The other primary results of this research involve the implementations and enhancements of the Second Algorithm with L presented in Section 5. Notably, if we assume knowledge of a close lower bound of the regulator (not necessarily its integer part), we have developed an algorithm employing the LLL-algorithm with rank 2 lattices to identify the fundamental solution of the Pell equation. This is referred to as the Second Algorithm with LLL , as detailed in Subsection 5.4. Here, the LLL-algorithm refers to the algorithm as defined in . Utilizing analogous concepts, we introduce a generalization of the continued fraction algorithm in Section 6. We identify the study of convergence and its computational complexity as open problems. We believe that the concepts introduced in this work warrant further exploration and detailed examination. Sections 2 and 3 provide concise overviews of the simple continued fraction algorithm (Section 2) and the Chakravala method (Section 3) for solving the Pell equation. These sections serve to illustrate the natural extension of the algorithm, as we derived corresponding analogs to various equations presented therein (refer to Section 7). In Section 4, we introduce two novel algo-rithms that generalize the Chakravala method, accompanied by illustrative examples. Section 5 introduces several variants of these algorithms, incorporating the LLL-algorithm with rank 2 lattices. The discovery of these algorithms dates back to September 2015. In 2018, we further developed the details and compiled this work, excluding Subsection 5.5. The initial version of this work was published in August 2023, as referenced in . 2. Simple continued fractions and Pell equation
To create a self-contained work, in this section, we introduce the concept of a simple continued fraction and its application in solving the Pell equation. The subsequent definitions and results are standard, as found in , , , and . Let ⌊x⌋ denote the greatest integer less than or equal to the real number x.The simple continued fraction expansion of a real number φ (which we denote as φ0) is an expression of the form
φ = φ0 = q0 + 1
q1 + 1
q2 + 1
q3 + . . .(2.1) expressed as φ0 = [ q0; q1, q 2, q 3, . . . ], where q0 = ⌊φ0⌋, and the numbers qn for n > 0 are positive integers defined recursively by
φn+1 = 1
φn − qn
, qn+1 = ⌊φn+1 ⌋ for all n ≥ 0, (2.2) where φn is the n-th complete quotient . Note that φ0 = [ q0; q1, . . . , q n, φ n+1 ]. We define two sequences of integers recursively as A0 = q0, B0 = 1, A1 = q1q0 + 1, and
B1 = q1. Additionally, we have
An+1 = qn+1 An + An−1 (2.3)
Bn+1 = qn+1 Bn + Bn−1 for n ≥ 1.PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 3
It is straightforward to establish by induction (refer to Theorem 7.5 and Theorem 7.4 of ) that:
AnBn−1 − BnAn−1 = ( −1) n−1, (2.4) and
An
Bn
= [ q0; q1, q 2, q 3, . . . q n] = q0 + 1
q1 + 1
. . . + 1
qn−1 + 1
qn
The quotient An/B n is denoted as the n-th convergent of the continued fraction (2.1). We now focus on the simple continued fraction expansion of √d for a positive integer d that is not a perfect square. The simple continued fraction expansion for √d provides all the necessary tools to solve Pell’s equation x2 − dy 2 = 1, as demonstrated by Euler and Lagrange. The following result is derived from Theorem 8.1 in or page 346 in .
Proposition 2.1. Let d be a positive integer and not a perfect square. In the continued fraction of φ0 = √d, we have
φn = Pn + √d
Qn
, for n ≥ 0 (2.5)
where Pn and Qn are the integers defined recursively by P0 = 0 , Q0 = 1 and
Pn+1 = qnQn − Pn, and Qn+1 = d − P 2
n+1
Qn
for n ≥ 0, (2.6)
with q0 = ⌊√d⌋ and
qn =
⌊
Pn + ⌊√d⌋
Qn
⌋
. (2.7)
We also have that
0 < Q n < 2√d, 0 < P n < √d. (2.8) In the subsequent proposition, we present some valuable identities as they offer a clear con-nection between the simple continued fraction, the Chakravala method, and the generalizations provided in Section 4 (refer to equations (3.5) and (4.6), as well as Section 6).
Proposition 2.2. (p.46 in ) Let d be a positive integer and not a perfect square. In the continued fraction of φ0 = √d, the following relationships hold:
A2
n
− dB 2
n
= ( −1) n+1 Qn+1 (2.9)
and
An = Pn+1 An−1 + dB n−1
Qn
, Bn = An−1 + Pn+1 Bn−1
Qn
, (2.10)
along with
dB n = Pn+1 An + Qn+1 An−1 (2.11)
An = Pn+1 Bn + Qn+1 Bn−1.4 JOS ´E I. LIBERATI
The continued fraction algorithm stops when A2
n
− dB 2
n
= 1, i.e., when ( −1) n+1 Qn+1 = 1 (see (2.9)). The fundamental solution of x2 − y2d = 1 is the solution ( X, Y ) in the smallest positive integers, denoted as ǫ = X + Y √d. The number R10
d
= log 10 (X + Y √d) is referred to as the
regulator (with base 10) . We have used logarithm with base 10, even if it is not usual in number theory. We only use this notation in Section 5.4.
Theorem 2.3. (Theorem 7.26 of ) Let d be a positive integer, not a perfect square. If
ǫ = X + Y √d is the fundamental solution of x2 − dy 2 = 1 , then all positive solutions are given by xn, y n, for n = 1 , 2, . . . , where xn and yn are the integers defined by
xn + yn
√d = ( X + Y √d)n.
Observe that the equations in (2.10) can be rewritten as
An + √d B n = ( A0 + √dB 0)
n
∏
j=1
( Pj+1 + √d
Qj
)
. (2.12) where Pi and Qi are defined in Proposition 2.1. This is similar to what is referred to as the
”power product” in . Therefore, we have a product representation of the convergent and the fundamental solution of the Pell equation. 3. Chakravala or cyclic method
The Indian mathematician Bhaskara II described the first method to solve the Pell equation, known as the Chakravala (or cyclic) method , specifically addressing the case x2 − 61 y2 = 1 (among other examples). Here, we present a variant of the algorithm, considering the version provided in , , and . (Note: There are several variants depending on the signs; refer to Remark 3.1 for additional details.) As mentioned in the introduction, in the Chakravala method, the primary concept involves taking a triple ( a, b, k ) satisfying a2 −db 2 = k and composing it with the trivial triple ( m, 1, m 2 −
d) to obtain a new triple ( am + db, a + bm, k (m2 − d)). This can be scaled down by k to yield
( am + db
k
)2
− d
( a + bm
k
)2
= m2 − d
k . (3.1) We then choose m as a positive integer such that a + b m is divisible by k and minimizes the absolute value of m2 − d, and hence that of ( m2 − d)/k . Assuming ( a, b ) = 1 (and therefore (k, b ) = 1), we observe that m2 − d and am + db are also multiples of k using the following equations: (m2 − d)b2 = k − (a2 − m2b2) = k − (a + mb )( a − mb ) (3.2) and
am + db = ( a + mb )m − (m2 − d)b. (3.3) This leads to a new triple of integers
˜a = am + db
|k| , ˜b = a + bm
|k| , ˜k = m2 − d
k ,
satisfying ˜ a2 − d˜b2 = ˜k. The process continues until a stage is reached at which the equation has the desired form a2 − db 2 = 1, i.e., ˜k = 1. PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 5
More precisely, given a non-square positive integer d, the algorithm produces sequences of integers Ai, B i, Q i, and Pi according to the following recipe: we start with A0 = 1 , B 0 = 0 , Q 1 =1, P 1 = 0. Given integers An−1, B n−1, Q n and Pn where ( An−1, B n−1) = 1 such that
A2
n−1
− dB 2
n−1
= Qn,
we choose Pn+1 to be a positive integer for which An−1 + Bn−1Pn+1 is divisible by Qn and minimizes the absolute value of P 2
n+1
− d. Then we take (cf. (2.6))
Qn+1 = P 2
n+1
− d
Qn
(3.4) and (cf. (2.10))
An = An−1Pn+1 + dB n−1
|Qn| , Bn = An−1 + Bn−1Pn+1
|Qn| . (3.5) Using (3.2), (3.3), and (3.4), we obtain that An and Qn+1 are integers. By using (3.1) we get (cf.(2.9))
A2
n
− dB 2
n
= Qn+1 , (3.6) and ( An, B n) = 1 is obtained by observing that |AnBn−1 −BnAn−1| = 1. The method terminates when Qn+1 = 1 for some n, and it is possible to show that in this case, An + Bn
√d is the fundamental solution of the Pell equation. We shall not prove it here, see p.35 in .
Remark 3.1 . On page 33 of , and in , another version of the Chakravala method is employed, where all the elements Qn are always positive integers. Specifically, given integers An−1, Bn−1,
Qn, and Pn such that ( An−1, B n−1) = 1 and satisfying
|A2
n−1
− dB 2
n−1
| = Qn,
they select Pn+1 to be a positive integer such that An−1 + Bn−1Pn+1 is divisible by Qn and minimizes the absolute value of P 2
n+1
− d. Then, they define
Qn+1 = |P 2
n+1
− d|
Qn
and
An = An−1Pn+1 + dB n−1
Qn
, Bn = An−1 + Bn−1Pn+1
Qn
.
As mentioned earlier, this procedure ensures that An and Qn+1 are integers, and
|A2
n
− dB 2
n
| = Qn+1 .
The method terminates when, for some n, Qn+1 = 1 , 2, 4 (i.e. A2
n
− dB 2
n
= ±1, ±2, ±4). At this point, Brahmagupta’s composition method is used to construct the fundamental solution. However, we will not delve into the details of this method here (see for further information).
Remark 3.2 . The first step in the Chakravala algorithm: We choose P2 as a positive integer to minimize |P 22 − d|. Consequently, P2 can take the value of ⌊√d⌋ or ⌊√d⌋ + 1. Therefore
∗ if ( ⌊√d⌋ + 1) 2 − d < d − ⌊ √d⌋2 , then P2 = ⌊√d⌋ + 1 , A 1 = ⌊√d⌋ + 1 , B 1 = 1 , Q 2 = A21 − d B 21 ,
∗ else: P2 = ⌊√d⌋, A 1 = ⌊√d⌋, B 1 = 1 , Q 2 = A21 − d B 21 .From this point onward in the examples, we will start from this initial condition .By induction and using the ideas of the proof of Proposition 1 in , it is possible to demon-strate the following result, cf. (2.8). 6 JOS ´E I. LIBERATI
Proposition 3.3. The sequence of integers Qn satisfies
|Qn| < √d for all n ≥ 1. (3.7) By utilizing (3.7) and (3.6), in conjunction with the following proposition, we establish that the quotients An/B n generated by the Chakravala method are indeed convergent in the simple continued fraction expansion of √d.
Proposition 3.4. (Theorem 7.24 in ) If A and B are positive integers with (A, B ) = 1 , d
is a non-square positive integer and Q ∈ Z such that |Q| < √d and
A2 − dB 2 = Q,
then there is an i > 0 for which A = Ai and B = Bi, where Ai/B i is the i-th convergent in the simple continued fraction expansion of √d.
The exercises on page 35 of show that the cyclic method efficiently finds the Pell equa-tion’s fundamental solution, often skipping non-solution steps inherent in the continued fraction method. Computational evidence suggests the Chakravala method needs roughly 69% of the steps required by the continued fraction approach. Now, we present an implementation of the algorithm following p.34 in . The goal is to transform the conditions on Pn+1 , namely An−1 +Bn−1Pn+1 being divisible by Qn and |P 2
n+1
−d|
being minimal, into simpler ones, avoiding the use of the large numbers An−1 and Bn−1.Using (3.5) and (3.4), the following expression is derived:
PnBn−1 − An−1 = Pn(An−2 + Bn−2Pn) − (An−2Pn + dB n−2)
|Qn−1| = Bn−2
(P 2
n
− d)
|Qn−1| (3.8) = Bn−2Qn sign( Qn−1).
Hence, we have that PnBn−1 − An−1 ≡ 0 (mod |Qn|). Also, by construction, Qn|(Pn+1 Bn−1 +
An−1). Therefore, ( Pn+1 + Pn)Bn−1 ≡ 0 (mod |Qn|). Since ( Qn, B n−1)|(An−1, B n−1), it is obtained that ( Qn, B n−1) = 1, and
Pn+1 ≡ − Pn (mod |Qn|). (3.9) Hence, using the fact that Pj ’s are positive integers, the following relation holds:
Pn+1 = −Pn + qn|Qn|
for some positive integer qn (cf. (2.6)). It should be noted that Pn+1 must satisfy |P 2
n+1
− d| ≤ |P 2 − d| for any positive P congruent to −Pn modulo |Qn|. By Theorem 2.2 on page 34 in , at each step, the following expression is chosen (cf. (2.7)):
q =
⌊ Pn + √d
|Qn|
⌋
,
and the possible values of qn are q or q + 1, obtaining Pn+1 . Now, An, B n, and Qn+1 are defined as in (3.4) and (3.5). If Qn+1 = 1, then the algorithm is stopped, obtaining the fundamental solution. Observe that in the continued fraction algorithm, qn in (2.7) corresponds to selecting Pn+1
as the integer congruent to −Pn modulo Qn (where all Qn are positive) such that √d − Pn+1
is positive and minimum (refer to (2.6) and Section 6), or equivalently, such that d − P 2
n+1
is
positive and minimum. Similar to the continued fraction algorithm (see Proposition 2.1), we define the numbers Pn+1 and Qn+1 without involving An−1 and Bn−1. Finally, by using this PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 7
implementation of the Chakravala method, we can express the fundamental solution as a power product, akin to the situation observed in (2.12). 4. Two algorithms that generalize the Chakravala method, that incorporate an additional variable
In this section, we present one of the main results of this work: two algorithms that gener-alize the Chakravala method. In Subsection 4.1 we define the First Algorithm with L, and in Subsection 4.2 we define the Second Algorithm with L.The basic idea for both algorithms is very simple: in the Chakravala method, we composed a triple ( a, b, k ) with the trivial triple ( m, 1, m 2 − d) (that is, we put l = 1 in (1.2)) to get a new triple ( am + db, a + bm, k (m2 − d)) which can be scaled down by k as explained in the Introduction and Section 3. However, we can repeat the entire argument with the ” l” included. More precisely, as it was pointed out in (1.1) and (1.2), given a triple ( a, b, k ) (that satisfies a2 −db 2 = k), we can compose it with the triple ( m, l, m 2 − d l 2) (see (1.2)) to get a new triple ( am + db l, al + bm, k (m2 − dl 2)) which can be scaled down by k to get
( am + db l
k
)2
− d
( al + b m
k
)2
= m2 − dl 2
k . (4.1) We now have to choose two variables, m and l. First of all, we choose m and l to be positive integers for which al + b m is divisible by k. Then, assuming that ( a, b ) = 1 (and therefore (k, b ) = 1), we see that m2 − dl 2 is also a multiple of k. This is established by using the equation: (m2 − dl 2)b2 = kl 2 − (a2l2 − b2m2) = kl 2 − (al + bm )( al − bm ). (4.2) Furthermore, using (4.1), we have that am + db l is also a multiple of k. This results in a new triple of integers
˜a = am + db l
|k| , ˜b = al + b m
|k| , ˜k = m2 − dl 2
k ,
where ˜ a2 − d˜b2 = ˜k.Now, we are free to impose an additional condition of minimization, as in the Chakravala method. 4.1. First Algorithm with L. In this first case, we impose that m and l are chosen to minimize the absolute value of m2 − dl 2 and, consequently, that of ( m2 − dl 2)/k , with 1 ≤ l ≤ L for some fixed L. The process continues until a stage is reached at which the equation has the desired form a2 − db 2 = 1. More precisely, given a non-square positive integer d and a positive integer L, the algorithm produces sequences of integers ai, b i, k i, m i, and li using the following recipe. We start with the first step in the Chakravala algorithm, from Remark 3.2: if (⌊√d⌋ + 1) 2 − d < d − ⌊ √d⌋2, then a1 = ⌊√d⌋ + 1 , b 1 = 1 , k 2 = a21 − d b 21, (4.3) else : a1 = ⌊√d⌋, b 1 = 1 , k 2 = a21 − d b 21.8 JOS ´E I. LIBERATI
Now, given integers ai−1, b i−1 and ki where ( ai−1, b i−1) = 1 such that
a2
i−1
− d b 2
i−1
= ki, (4.4) we choose mi+1 and li+1 as positive integers for which ai−1li+1 + bi−1mi+1 is divisible by ki and minimizes the absolute value of m2
i+1
− d l 2
i+1
for 1 ≤ li+1 ≤ L. Then we take (cf. (3.4))
ki+1 = m2
i+1
− d l 2
i+1
ki
(4.5) and (cf. equations (2.10) and (3.5))
ai = ai−1mi+1 + d b i−1li+1
|ki| , bi = ai−1li+1 + bi−1mi+1
|ki| . (4.6) As before, using (4.2), (4.1) and ( ai−1, b i−1) = 1, we obtain that ki+1 and ai are integers. By (4.1) we get (cf. (2.9) and (3.6))
a2
i
− d b 2
i
= ki+1 ,
and to complete the recursive definition, we need to prove that ( ai, b i) = 1, see Proposition 4.1. The method terminates when kn = 1 for some n. In the examples, we always obtained the fundamental solution , except for extremely large values of L relative to d (cf. the analysis in Table 3). We refer to this algorithm as the First Algorithm with L .
Proposition 4.1. The integers ai and bi defined by (4.6) are coprime. Proof. Using (4.6) and (4.4), we have:
ai bi−1 − bi ai−1 = (ai−1mi+1 + d b i−1li+1 ) bi−1 − (ai−1li+1 + bi−1mi+1 ) ai−1
|ki|
= (db 2
i−1
− a2
i−1
)
|ki| li+1 = − ki
|ki| li+1 .
Hence, we obtain that |aibi−1 − biai−1| = li+1 . If we define h = ( ai, b i), then h| li+1 . Using this together with (4.6), we obtain that h|(bi−1mi+1 ) and h|(ai−1mi+1 ). But ( ai−1, b i−1) = 1, hence
h|mi+1 . Suppose h > 1, and we will demonstrate that this assumption leads to a contradiction. Using all the previous results, we can write li+1 = h ˜li+1 , m i+1 = h ˜mi+1 , and bi = h ˜bi. Since
bi|ki| = ai−1li+1 + bi−1mi+1 , then ˜bi|ki| = ai−1 ˜li+1 + bi−1 ˜mi+1 , and therefore, the positive integers ˜ mi+1 and ˜li+1 satisfy that ai−1 ˜li+1 + bi−1 ˜mi+1 is divisible by ki and:
|m2
i+1
− d l 2
i+1
| = h2 | ˜m2
i+1
− d ˜l 2
i+1
| > | ˜m2
i+1
− d ˜l 2
i+1
|
which is a contradiction due to the minimization property satisfied by the pair mi+1 and li+1 ,finishing the proof.
Note that this algorithm, with L = 1 (or li = 1 for all i > 0) corresponds to the Chakravala algorithm. Based on examples, we observe that (3.7) holds, that is
|ki| < √d for all i > 0, (4.7) and, using Proposition 3.4, we conclude that in this algorithm, ai/b i are convergent in the simple continued fraction expansion of √d.PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 9
4.2. Second Algorithm with L. Now, we present a second version of this algorithm by chang-ing the additional condition of minimization. In our second case, we choose to impose that m
and l are positive integers chosen such that they minimize the absolute value of m − √d l , with 1 ≤ l ≤ L for some fixed L. The process is continued until a stage is reached at which the equation has the desired form a2 − db 2 = 1, under certain conditions on L that depend on d
(cf. Table 3). More precisely, given a non-square positive integer d and a positive integer L, the algorithm produces sequences of integers ai, b i, k i, m i, and li by the following recipe: as before, we prefer to start with the first step in the Chakravala algorithm as in Remark 3.2 and (4.3): if (⌊√d⌋ + 1) 2 − d < d − ⌊ √d⌋2, then a1 = ⌊√d⌋ + 1 , b 1 = 1 , k 2 = a21 − d b 21, (4.8) else : a1 = ⌊√d⌋, b 1 = 1 , k 2 = a21 − d b 21.
Now, given integers ai−1, b i−1 and ki where ( ai−1, b i−1) = 1 such that
a2
i−1
− db 2
i−1
= ki,
we choose mi+1 and li+1 to be a positive integers for which ai−1li+1 + bi−1mi+1 is divisible by
ki and minimizes the absolute value of mi+1 − √d l i+1 for 1 ≤ li+1 ≤ L. Then we take ki+1 , a i
and bi as in (4.5) and (4.6). As before, using (4.2), (4.1) and ( ai−1, b i−1) = 1, we obtain that ki+1 and ai are integers. By using (4.1) we get
a2
i
− db 2
i
= ki+1 ,
and in order to complete the recursive definition, we need to prove that ( ai, b i) = 1. However, this follows immediately from the same arguments given in the proof of Proposition 4.1. The method terminates when kn = 1 for some n. In all the examples, we obtain the fundamental solution , except for extremely large values of L, see the analysis in Table 3.
Claim 4.2 . (Based on examples) In this algorithm, ai/b i are also convergent in the simple continued fraction expansion of √d, and |ki| < 2√d for all i > 0. Observe that this algorithm is not equivalent to Chakravala when restricted to the case L = 1. 4.3. Examples. Now, we present some illustrative examples. In Table 1, the initial example corresponds to d = 61, where we display the sequences (generated by the four algorithms) of three terms ( a, b, k ) such that a2 − db 2 = k, until reaching the fundamental solution of the Pell equation: 10 JOS ´E I. LIBERATI
Table 1. Computing the fundamental solution when d = 61
Cont. Frac. Chakravala 1st algorithm with L= 9 2nd algorithm with L= 9
(7, 1, -12)
(8, 1, 3) (8, 1, 3) (8, 1, 3) (8, 1, 3)
(39, 5, -4) (39, 5, -4) (39, 5, -4)
(125, 16, 9)
(164, 21, -5) (164, 21, -5) (164, 21, -5) (164, 21, -5)
(453, 58, 5) (453, 58, 5)
(1070, 137, -9)
(1523, 195, 4) (1523, 195, 4) (1523, 195, 4) (1523, 195, 4)
(5639, 722, -3) (5639, 722, -3)
(24079, 3083, 12)
(29718, 3805, -1) (29718, 3805, -1) (29718, 3805, -1) (29718, 3805, -1)
(440131, 56353, 12)
(469849, 60158, -3) (469849, 60158, -3) (469849, 60158, -3)
(2319527, 296985, 4) (2319527, 296985, 4) (2319527, 296985, 4) (2319527, 296985, 4)
(7428430, 951113, -9)
(9747957, 1248098, 5) (9747957, 1248098, 5) (9747957, 1248098, 5)
(26924344, 3447309, -5) (26924344, 3447309, -5)
(63596645, 8142716, 9) (63596645, 8142716, 9)
(90520989, 11590025, -4) (90520989, 11590025, -4) (90520989, 11590025, -4)
(335159612, 42912791, 3) (335159612, 42912791, 3) (335159612, 42912791, 3)
(1431159437, 183241189, -12)
(1766319049, 226153980, 1) (1766319049, 226153980, 1) (1766319049, 226153980, 1) (1766319049, 226153980, 1)
Observe, that we need 22 steps with the continued fraction, 14 steps with Chakravala, and only 10 and 8 steps in both algorithms with L = 9, respectively. We always obtained convergent of the continued fraction of √61. Note that in Chakravala and the First Algorithm with L = 9, we have |ki| < √61, but in the other two algorithms, we have |ki| < 2√61. In Table 2, we present the number of steps required for each value of d in each algorithm.
Table 2. Number of steps required for each value of d
d Cont. Frac. Chakravala 1st algorithm with L = 9 2nd algorithm with L = 9
46 12 8 4 4
61 22 14 10 8
97 22 12 8 6
109 30 22 15 11
313 34 26 14 14
541 78 56 32 27
Based on examples, when using L = 9 in the Second Algorithm, the number of steps needed is 0 .34 times the number of steps using continued fractions. Also, with L = 100, it is 0 .2 times the number of steps. We also observe an asymptotic trend in the number of steps in the following example. Taking
d = 132901, in this case, we need 422 steps with the continued fractions algorithm. The table below illustrates the steps for different values of L with the Second Algorithm: PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 11
Table 3. Number of steps with L when d = 132901
L 9 100 1000 1500 1625 1687 1698 1699 ≥ 1700
steps 141 83 63 60 60 59 59 59 diverges
In this case, ”diverges” and ” ≥ 1700” mean that for several values of L greater or equal to 1700, the algorithm does not stop after more than 300 steps. 5. Implementation and improvements of both algorithms
The other main results of this research pertain to the implementations and enhancements of the Second Algorithm with L. Notably, the Second Algorithm with LLL , as detailed in Subsection 5.4, stands out as the most significant. As in the implementation of the Chakravala algorithm presented in Section 3, the aim of this section is to transform the conditions on mi+1 and
li+1 into simpler ones, avoiding the use of the very large numbers ai−1 and bi−1. Observe that in both algorithms, we have to find at each step mi+1 and li+1 satisfying that ki divides
ai−1li+1 + bi−1mi+1 . In Subsection 5.1 and Subsection 5.5, we present an equivalent condition without involving ai−1 and bi−1. In Subsection 5.2, we study the minimization condition in the First Algorithm with L. In Subsection 5.3, we describe the minimization condition in the Second Algorithm with L, and we give the implementation and some improvements of the Second Algorithm with L, presenting two variants of it. In Subsection 5.4, we define the Second Algorithm with LLL. 5.1. An equivalent condition for mi+1 and li+1 such that ki divides ai−1li+1 + bi−1mi+1 .
The following results hold for both algorithms with L. Since ( ai−1, b i−1) = 1, we have that (bi−1, k i) = 1. Therefore, the condition that ki divides ai−1li+1 + bi−1mi+1 is equivalent to
mi+1 ≡ − ai−1
bi−1
li+1 (mod |ki|). (5.1) Now, we define Mi such that 0 ≤ Mi < |ki| and
Mi ≡ − ai−1
bi−1
(mod |ki|). (5.2) Hence, we have that ki divides ai−1li+1 + bi−1mi+1 if and only if
mi+1 = Mi li+1 + ri |ki|, (5.3) for some integer ri. In Subsection 5.5, we present how to get Mi in terms of mj , l j , k j , and Mj−1
for j ≤ i, that is, without involving the large numbers ai−1 and bi−1, and this can be used in the implementation of any version of the algorithm. Equation (5.3) will be used to impose the minimization conditions of both algorithms. 5.2. Minimization Condition of the First Algorithm with L. Recall that in the First Algorithm with L we have to minimize |m2
i+1
− d l 2
i+1
| for 1 ≤ li+1 ≤ L. Using (5.3), we have
∣∣m2
i+1
− d l 2
i+1
∣∣ = |mi+1 − √d l i+1 || mi+1 + √d l i+1 | (5.4) = |ki|2
∣∣∣∣ri −
( √d − Mi
|ki|
)
li+1
∣∣∣∣∣∣∣∣ri +
(√d + Mi
|ki|
)
li+1
∣∣∣∣.12 JOS ´E I. LIBERATI
For each li+1 with 0 < l i+1 ≤ L, we have two options for ri: r+
i
as one of the integers near to
(√d−Mi)
|ki|
li+1 or r−
i
as one of the integers near to (−√d−Mi)
|ki|
li+1 . If we take r−
i
, then we have mi+1
as an integer near to Mili+1 + |ki| (−√d−Mi)
|ki|
li+1 = −√d l i+1 which is negative, contradicting our assumptions. Therefore, we must take r+
i
, which is always positive because, by (4.7), we have Mi < |ki| <√d. Thus, we clearly need to take
r+
i
= floor
( (√d − Mi)
|ki| li+1
)
or r+
i
= ceil
( (√d − Mi)
|ki| li+1
)
.
Then, li+1 is the integer that produces the minimum of (5.4), for 1 ≤ li+1 ≤ L. Unfortunately, we do not know how to determine this minimum in a simple and fast way. With this, we conclude the analysis of the First Algorithm in this work.
5.3. Implementation of the Second Algorithm with L. This subsection is one of the main parts of this work. In each step of the Second Algorithm with L, we need to choose mi+1 and
li+1 as positive integers for which ai−1li+1 + bi−1mi+1 is divisible by ki and | mi+1 − √d l i+1 | is minimal for 1 ≤ li+1 ≤ L. We have seen in (5.3) that the condition ki | (ai−1li+1 + bi−1mi+1 ) is equivalent to mi+1 = Mi li+1 + ri |ki| for some integer ri, where Mi can be computed without involving ai−1 and bi−1 as in Subsection 5.5. Recall that in this algorithm |ki| < 2 √d, but we obtain the following claim
Claim 5.1 . (Based on examples) In this case, we also have Mi < √d for all i > 0. Since mi+1 − li+1
√d = |ki|ri − li+1 (√d − Mi), we are looking for positive integers r and l
such that ∣∣|ki|r − l(√d − Mi)∣∣ is minimum for l from 1 to L or equivalently, that minimize
∣∣r − l( √d−Mi
|ki|
)∣ ∣. Observe that for each l, the integer r is uniquely determined, that is r =round
(
l (√d−Mi)
|ki|
)
, where ”round” is the closest integer.
Summary of the implementation of the Second Algorithm with L without involving ai and b i. We take a1, b 1, k 2 as in (4.8). Then, we define l3, m 3 and k3 using Step 2 and Step 3, with M2 ≡ − a1
b1
(mod |k2|). Then, given ( li, m i, k i), we define ( li+1 , m i+1 , k i+1 ) as follows:
Step 1. Compute Mi as in Subsection 5.5.
Step 2. Take li+1 as the integer that satisfies the following minimization min
l=1 ,...,L
∣∣∣∣∣ l (√d − Mi)
|ki| − round
(
l (√d − Mi)
|ki|
) ∣∣∣∣∣, (5.5) and
ri = round
(
li+1
(√d − Mi)
|ki|
)
. (5.6)
Step 3. Define
mi+1 = Mi li+1 + |ki| ri, (5.7)
ki+1 = m2
i+1
− d l 2
i+1
ki
.
It is clear, by using (5.6) and (5.7), that mi+1 ≃ li+1
√d, proving that mi+1 is positive. The method terminates when kn = 1 for some n. In all the examples, we obtain the fundamental solution, except for extremely large values of L.PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 13
In this way, the integers li+1 , m i+1 , and ki+1 are defined without using the large numbers ai
and bi. Still, if we want to include them, we must take
ai = ai−1mi+1 + db i−1li+1
|ki| , b i = ai−1li+1 + bi−1mi+1
|ki| . (5.8) Observe that in both algorithms with L, (5.8) produces a product representation of the conver-gent and the fundamental solution of the Pell equation as follows:
ai + √d b i = (a1 + √d b 1
) i∏
j=2
( mj+1 + √d l j+1
|kj |
)
(5.9) and (5.9) is similar to what is called ”power product” in .
Computation of the minimum in Step 2. We present in detail some options in the im-plementation of the algorithm depending strictly on different ways to compute the minimum in Step 2. First, observe that in (5.5), we are basically looking for a couple of positive integers li+1
and ri that produce the following minimization min
l=1 ,...,L
all r
∣∣ r − l α i
∣∣, (5.10) where
αi =
√d − Mi
|ki| .
Now, we recall a basic result in the theory of simple continued fractions (see p.340 in ): A pair of positive integers a and b is called a good approximation to the positive irrational number
ξ if
| b ξ − a | = min
y=1 ,...,b
all x
∣∣ y ξ − x ∣∣.
A classical result is that the pair a and b is a good approximation of ξ if and only if a/b is a convergent of ξ. Using this result, we present two options for computing Step 2. First, for a fixed L, we can apply the method of continued fractions to αi in order to find li+1
and ri. If ˜ an/˜b n is the n-th convergent of αi with ˜b n ≤ L < ˜b n+1 , then, we take li+1 = ˜b n and
ri = ˜ an. These numbers minimize (5.10) by Theorem 7.13 in . This variant of our algorithm is denoted as the Second Algorithm with CF and L . We shall not present any example of this implementation. Secondly, we have the option to set (or adjust, if necessary) the number of steps s used in applying the continued fractions algorithm to αi. Then, we set Li = li+1 = bs and ri = as.Notably, L is not fixed in this scenario. This particular variation of our algorithm is referred to as the Second Algorithm with CF and s . Once more, we will refrain from providing an example of this implementation.
Remark 5.2 . Observe that the Second Algorithm with CF and L (resp. with CF and s), can also be used with different values of L (resp. s) as we did with the following LLL version. 14 JOS ´E I. LIBERATI
5.4. Second Algorithm with LLL. We present one of the most important parts of this work. This improvement in the implementation of the second algorithm with L consists of replacing Step 2 with an approximation to αi using the LLL-algorithm. The LLL-algorithm is the common notation for the Lenstra–Lenstra–Lov´ asz (LLL) lattice basis reduction algorithm defined in . It is a polynomial time lattice reduction algorithm. We only need to use it for lattices of rank 2, see Section 9 in for details. By Proposition 1.39 in , we have the following result: given rational numbers α and ε, satisfying 0 < ε < 1, the LLL-algorithm (for rank 2 lattices) finds integers p and q for which
| p − qα | ≤ ε and 1 ≤ q ≤
√2
ε .
Now, for a fixed positive integer LLL , we set ε =
√2
LLL
and choose α as a rational approximation of αi. Therefore, the LLL-algorithm finds integers p and q such that
| p − qα | ≤
√2
LLL and 1 ≤ q ≤ LLL.
In our examples, we applied the ”lattice” command in Maple 8 to the lattice generated by (1 , 0) and ( − α, ε2
√2
), to produce the positive integers li+1 = q and ri = p that we use to replace the Step 2. However, it’s important to note that these values may not necessarily minimize (5.10). For detailed information, refer to pages 139-140 in . Consequently, in some cases, we may not obtain the fundamental solution, as we will observe in certain examples. This variant of our algorithm is denoted as the Second Algorithm with LLL .The following table provides an example of the different algorithms applied to d = 1234567890, where the cases with L correspond to the Second Algorithm with L. In this case, the fundamental solution ǫ has 1935 decimal digits (or R10
d
= 1935, using our notation for the regulator (with base 10) introduced after Proposition 2.2).
Table 4. Number of steps when d = 1234567890
CF Chak. L= 9 L= 100 L= 200 LLL = 10 6LLL = 10 18 LLL = 10 20 LLL = 10 25
steps 3772 2611 1302 768 690 304 105 95 76
Observe that in all the cases with LLL , the number of digits of the fundamental solution is related to the value of LLL and the number of steps. Specifically, the fundamental solution has 1935 decimal digits, and with LLL = 10 20 the number of steps is 95, hence we have 1935
95
≃
The same holds for the other exponents with LLL : 1935
105
≃ 18. Again, the study of the convergence of the different algorithms remains as an open problem, as well as the running time of them. In the following example, we take d = 130940879. In this case, the continued fraction algo-rithm needs 5259 steps, and the fundamental solution has 2727 decimal digits. Observe that
⌊√R10
d
⌋ = 52. In this case, with the Second Algorithm with LLL = 10 52 , we needed 52 steps to obtain the fundamental solution. Now, in the same example, we use the Second Algorithm with LLL, but we apply it with 2 different speeds , that is, we use a value of LLL for a certain fixed number of steps, and then we continue with another value of LLL for the remaining steps. More precisely, PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 15
• 27 steps with LLL = 10 75 + more than 500 steps with LLL = 10 10 : diverges.
• 27 steps with LLL = 10 75 + 128 steps with LLL = 10 5: 155 steps, we get solution ǫ.
• 35 steps with LLL = 10 75 + more than 500 steps with LLL = 10 10 : diverges.
• 35 steps with LLL = 10 75 + 523 steps with LLL = 10 5: 558 steps, we get solution ǫ2.
• 27 steps with LLL = 10 100 + 3 steps with LLL = 10 5: 30 steps, we get solution ǫ.
• 18 steps with LLL = 10 150 + 510 steps with LLL = 10 5: 528 steps, we get solution ǫ2.
• 18 steps with LLL = 10 150 + 3 steps with LLL = 10 6: 21 steps, we get solution ǫ.
• 9 steps with LLL = 10 250 + more than 500 steps with LLL = 10 10 : diverges.
• 9 steps with LLL = 10 250 + 87 steps with LLL = 10 5: 96 steps, we get solution ǫ.
• 11 steps with LLL = 10 250 + 501 steps with LLL = 10 5: 512 steps, we get solution ǫ2.
• 9 steps with LLL = 10 300 + 4 steps with LLL = 10 5: 13 steps, we get solution ǫ.
• 9 steps with LLL = 10 300 + 2 steps with LLL = 10 10 : 11 steps, we get solution ǫ.
• 9 steps with LLL = 10 300 + 1 step with LLL = 10 20 : 10 steps, we get solution ǫ.
• 6 steps with LLL = 10 452 + 1 step with LLL = 10 10 : 7 steps, we get solution ǫ.
• 5 steps with LLL = 10 542 + more than 500 steps with LLL = 10 10 : diverges.
• 5 steps with LLL = 10 542 + 3 steps with LLL = 10 5: 8 steps, we get solution ǫ.Observe that in some cases the algorithm diverges, and in some other cases, we get the square of the fundamental solution. In most cases, we took the number of steps for the first value of
LLL to approximate to the floor of the regulator R10
d
, which is equal to 2727. For example, in the last item in the previous list, we took 5 steps with LLL = 10 542 since 5 · 542 = 2710, and if we add 3 steps with LLL = 10 5, we have 5 · 542 + 3 · 5 = 2725 ≃ 2727. If we know a close lower bound of the regulator R10
d
, then we can take a few steps with a big
LLL to approximate it and then continue with a lower value of LLL to get the fundamental solution. The approximation of the regulator that we need is not necessarily the floor of the regulator, as in Section 6 in . It is enough to know a lower bound that should follow from the analysis of convergence and the running time. What we essentially have done was the following: if R10
d
≃ kd hd +cd, where kd, h d, c d are positive integers, then we apply kd steps with LLL = 10 hd
and cd steps LLL = 10, to obtain the fundamental solution. Continuing with the same example, and considering that 272 · 10 = 2720 ≃ 2727, we have the following case: if we take 10 steps with LLL = 10 272 , then the algorithm converges to ǫ.If we accept a weak version of the folklore conjecture given in p.7 in , then we may assume the first inequality in √d
(log d)q < R 10
d
· log(10) < √d (log(4 d) + 2) ,
for some positive integer q. In the examples presented on pages 345-348 in , the regulators satisfy this inequality with q = 1. Therefore, we may apply the previous idea to
√d
log(10) ·(log d)q
, a value that is close to R10
d
, and take positive integers kd and hd such that
√d
log(10) ·(log d)q
≃ kd hd.The most important part of the running time is given by the implementation of the LLL-algorithm at each step. If it were necessary, for each αi, we can take the appropriate LLL i to reduce the running time, by using the remarks on pages 140 and 147 in about the implemen-tation of the LLL-algorithm in the simple and special case of lattices of rank 2. 5.5. Computation of Mi without involving ai−1 and bi−1. The results of this section hold for the First and Second Algorithms with L. Recall that we defined Mi such that 0 ≤ Mi < |ki|
and
Mi ≡ − ai−1
bi−1
(mod |ki|).16 JOS ´E I. LIBERATI
To obtain Mi in a different way, we need some results. Using the idea in (3.8), together with (4.5) and (4.6), we have
bi−1 mi − ai−1li = sign( ki−1) bi−2 ki. (5.11) Similarly, one can see that
ai−1mi − d b i−1li = sign( ki−1) ai−2 ki. (5.12) Hence
mi bi−1 ≡ li ai−1 (mod |ki|), (5.13) or
mi ≡ − Mi li (mod |ki|). (5.14) Observe that, from (5.1) and (5.13), we obtain
limi+1 ≡ − li+1 mi (mod |ki|) , (5.15) and we recover (3.9) by taking 1 = L = li for all i in (5.15), that is the main equation in the implementation of the Chakravala algorithm, cf. (3.9). In certain steps within the examples, there might occur situations where li and ki are not coprime. Hence sometimes we cannot take Mi ≡ − mi
li
(mod |ki|) in (5.14). Recall from (5.3), that we have
mi+1 = Mi li+1 + ri |ki|, (5.16) for some integer ri. Hence, by definition we have that ri−1 = mi−Mi−1li
|ki−1|
, and now we define
si−1 := d l i − Mi−1mi
|ki−1| .
Proposition 5.3. The following properties hold: (a) si−1 is an integer. (b) ai−1ri−1 ≡ bi−1si−1 (mod |ki|) . (c) (li, r i−1) = 1 .(d) Mi ≡ Mi−1 (mod (li, k i)) .Proof. (a) By (5.16), we have that Mi−1mi ≡ M 2
i−1
li (mod |ki−1|) . Using that a2
i−2
− d b 2
i−2
=
ki−1, we obtain that M 2
i−1
≡
( ai−2
bi−1
)2
≡ d (mod |ki−1|). Therefore, Mi−1mi ≡ d l i (mod |ki−1|), finishing the proof of (a). (b) Using (5.11) and (5.12), observe that
ai−1ri−1 − bi−1si−1 = ai−1
(mi − Mi−1li
|ki−1|
)
− bi−1
( d l i − Mi−1mi
|ki−1|
)
= 1
|ki−1|
(
Mi−1
(bi−1mi − ai−1li
) + (ai−1mi − d b i−1li
))
= ki
ki−1
(Mi−1bi−2 + ai−2
),
but ( Mi−1bi−2 + ai−2
) is a multiple of ki−1 by the definition of Mi−1, finishing the proof of (b). (c) Suppose that ni = ( li, r i−1) > 1. Since mi = Mi−1li +ri−1|ki−1|, then ni divides mi. Define
˜mi = mi/n i, ˜li = li/n i and ˜ ri−1 = ri−1/n i. Then ˜mi and ˜li satisfy ˜mi = Mi−1˜li + ˜ ri−1|ki−1|, i.e.,
ki−1 divides ai−2 ˜li + bi−2 ˜mi, and
| mi − √d l i| = ni | ˜mi − √d ˜li|,PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 17
which contradicts the minimality of | mi − √d l i| in the Second Algorithm with L. Similarly, a contradiction is obtained for the First Algorithm with L, concluding the proof of (c). (d) Finally, observe that ( li, k i)| mi by (5.14). Hence, using that mi = Mi−1li + ri−1|ki−1|,we obtain that ( li, k i)|(ri−1|ki−1|). Therefore, using part (c), we have that ( li, k i)|ki−1 and (li, k i) = ( li, k i, k i−1). Now, by (7.1), we have ai−1bi−2 ≡ bi−1ai−2 (mod li). In particular,
ai−1bi−2 ≡ bi−1ai−2 (mod ( li, k i, k i−1)). Hence we can take the quotients to get
ai−1
bi−1
≡ ai−2
bi−2
(mod ( li, k i)) .
From this, (d) is obtained.
Summary of the computation of Mi:
∗ If ( li, k i) = 1, then by (5.14), we take
Mi ≡ − mi
li
(mod |ki|).
Suppose that ( li, k i) > 1, then
Mi ≡ − ˜mi
˜li
(mod |˜ki|), (5.17) where ˜mi = mi
(li,k i)
, ˜li = li
(li,k i)
and ˜ki = ki
(li,k i)
. Observe that ˜mi is an integer since ( li, k i)| mi by (5.14).
∗ If ( li, k i) > 1 and
(
(li, k i), ki
(li,k i)
)
= 1, then using (5.17) together with Mi ≡ Mi−1 (mod ( li, k i)), and the Chinese remainder theorem, we can get Mi.From now on, we suppose that ( li, k i) > 1 and
(
(li, k i), ki
(li,k i)
)
1
∗ If ( ri−1, k i) = 1, then using Proposition 5.3 (b), we take
Mi ≡ − si−1
ri−1
(mod |ki|).
∗ If ( ri−1, k i) > 1, then we have
Mi ≡ − ̂ si−1̂
ri−1
(mod |̂ki|), (5.18) where ̂ ri−1 = ri−1
(ri−1,k i)
,̂ si−1 = si−1
(ri−1,k i)
and ̂ ki = ki
(ri−1,k i)
. Using Proposition 5.3, that is ( li, r i−1) = 1, we have that ki divides ̂ ki · ˜ki. Then we can take ̂ Ki (resp. ˜Ki) a divisor of ̂ ki (resp. ˜ki) such that ( ̂Ki, ˜Ki) = 1 and ki =̂ Ki · ˜Ki. Finally, using these divisors in (5.17) and (5.18), we can apply the Chinese remainder theorem to obtain the value of Mi.This concludes the computation of Mi without using ai−1 and bi−1. There might be a simpler approach to achieve this. Additionally, it remains uncertain whether an implementation exists where one can ensure, at each step, that ki divides ai−1li+1 + bi−1mi+1 without relying on the values of Mi.6. Generalization of the Continued Fraction Algorithm with an Additional Variable
In this section, we introduce a generalization of the continued fraction algorithm, drawing inspiration from the ideas of the First Algorithm with L. Note that in the continued fraction 18 JOS ´E I. LIBERATI
algorithm, qn in (2.7) corresponds to selecting Pn+1 as the positive integer congruent to −Pn
module Qn (where all Qn are positive) satisfying
√d − Pn+1 > 0 and minimizing this value ,
or equivalently
d − P 2
n+1
0 and minimizing this value .
With this motivation, we can attempt to define a kind of continued fraction algorithm with L by replacing the minimization conditions in the First and Second Algorithms with L with:
li+1
√d − mi+1 > 0 and minimizing this value , (6.1) or
l 2
i+1
d − m2
i+1
0 and minimizing this value . (6.2) More precisely, we will use (6.2) to define the Continued Fraction Algorithm with L as follows: given a non-square positive integer d and a positive integer L, the algorithm produces the sequences of integers ai, b i, k i, m i, l i by the following recipe: we start with the first step in Chakravala algorithm as in Remark 3.2: if (⌊√d⌋ + 1) 2 − d < d − ⌊ √d⌋2, then a1 = ⌊√d⌋ + 1 , b 1 = 1 , k 2 = a21 − d b 21,
else : a1 = ⌊√d⌋, b 1 = 1 , k 2 = a21 − d b 21.
Now, given integers ai−1, b i−1 and ki where ( ai−1, b i−1) = 1 such that
a2
i−1
− d b 2
i−1
= ki,
we choose mi+1 and li+1 to be a positive integers for which ai−1li+1 + bi−1mi+1 is divisible by
ki and
l 2
i+1
d − m2
i+1
0 and minimizing this value for 1 ≤ li+1 ≤ L. (6.3) Then we take
ki+1 = m2
i+1
− d l 2
i+1
ki
and
ai = ai−1mi+1 + d b i−1li+1
|ki| , bi = ai−1li+1 + bi−1mi+1
|ki| .
As before, using (4.2), (4.1) and ( ai−1, b i−1) = 1, we obtain that ki+1 and ai are integers. By a direct computation, we have
a2
i
− d b 2
i
= ki+1 ,
and to complete the recursive definition, we need to prove that ( ai, b i) = 1. This is immediate by the ideas in the proof of Proposition 4.1. The method terminates when kn = 1 for some n.
In the examples, we always obtained the fundamental solution, except for extremely large values of L relative to d. It needs more steps than the First Algorithm with L. As before, we identify the study of convergence and its computational complexity as open problems. Finally, if we replace (6.3) by (6.1) in the previous algorithm, the algorithm does not work. PELL EQUATION: GENERALIZATIONS OF CONTINUED FRACTION AND CHAKRAVALA ALGORITHMS 19
Some Formulas and Final Remarks
In this section, we present some formulas analogous to well-known formulas for continued fractions written in Section 2. These formulas apply to the Second Algorithm with L. By definition and a simple computation, we have
a2
i
− d b 2
i
= ki+1
ai ai−1 − d b i bi−1 = sign( ki) mi+1
which are the analogs to equations (3.13) in . By the proof of Proposition 4.1, we have
ai bi−1 − bi ai−1 = −sign( ki) li+1 (7.1) that corresponds to (2.4). The following formulas were proved in examples. These are the version with L of (2.3):
li+1 ai+1 = qi+1 ai − sign( ki) sign( ki+1 ) li+2 ai−1
li+1 bi+1 = qi+1 bi − sign( ki) sign( ki+1 ) li+2 bi−1,
where qi is defined by li mi+1 + mi li+1 = qi |ki|, which is a positive integer by (5.15). These are the version with L of (2.11):
li+1 d b i = mi+1 ai − sign( ki) ki+1 ai−1
li+1 ai = mi+1 bi − sign( ki) ki+1 bi−1.
Now, we present a formula analogous to (2.2), that is, a kind of interpretation of this algorithm in terms of a generalized continued fraction expansion of √d. We define (cf. (2.5))
φi = (mi + li
√d)
|ki| li+1 .
By the standard computation, one can prove that
φi+1 = s l i li+2
φi − qi
where qi was defined above, and s = −sign( ki) sign( ki+1 ).
Acknowledgements. Dedicated to my parents, H´ ector and Sheila, with heartfelt gratitude.
References
Barbeau, Edward J. Pell’s equation. Problem Books in Mathematics. Springer-Verlag, New York, 2003. xii+212 pp. Bauval, Anne. An elementary proof of the halting property for Chakravala algorithm , preprint (2014). Edwards, Harold M. Fermat’s last theorem. A genetic introduction to algebraic number theory. Graduate Texts in Mathematics, 50. Springer-Verlag, New York-Berlin, 1977. xv+410 pp. Hua, Loo Keng. Introduction to number theory. Translated from the Chinese by Peter Shiu. Springer-Verlag, Berlin-New York, 1982. xviii+572 pp. Jacobson, Michael J., Jr.; Williams, Hugh C. Solving the Pell equation . CMS Books in Mathematics/Ouvrages de Math´ ematiques de la SMC. Springer, New York, 2009. xx+495 pp. Lenstra, Hendrik W., Jr. Solving the Pell equation. Algorithmic number theory: lattices, number fields, curves and cryptography, 1–23, Math. Sci. Res. Inst. Publ., 44, Cambridge Univ. Press, Cambridge, 2008. Lenstra, Hendrik W., Jr. Lattices . Algorithmic number theory: lattices, number fields, curves and cryptog-raphy, 127–181, Math. Sci. Res. Inst. Publ., 44, Cambridge Univ. Press, Cambridge, 2008. Liberati, Jose I. Pell equation: A generalization of continued fraction and Chakravala algorithms using the LLL-algorithm , August 2023. Preprint. 20 JOS ´E I. LIBERATI
Lenstra, A. K.; Lenstra, H. W., Jr.; Lov´ asz, L. Factoring polynomials with rational coefficients . Math. Ann.
261 (1982), no. 4, 515–534. Niven, Ivan; Zuckerman, Herbert S.; Montgomery, Hugh L. An introduction to the theory of numbers. Fifth edition. John Wiley & Sons, Inc., New York, 1991. xiv+529 pp. Wagstaff, Samuel S., Jr. The joy of factoring. Student Mathematical Library, 68. American Mathematical Society, Providence, RI, 2013. xiv+293 pp. https : //en.wikipedia.org/wiki/Chakravala − method.
Ciem - CONICET, Medina Allende y Haya de la Torre, Ciudad Universitaria, (5000) C´ ordoba - Argentina. e-mail: joseliberati@gmail.com |
17323 | https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al.)/07%3A_Differential_Equations/7.04%3A_Separable_Differential_Equations | 7.4: Separable Differential Equations - Mathematics LibreTexts
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7: Differential Equations
Book: Active Calculus (Boelkins et al.)
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Wed, 09 Nov 2022 17:00:15 GMT
7.4: Separable Differential Equations
107841
107841
Joshua Halpern
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[ "article:topic", "Separable Differential Equations", "license:ccbysa", "showtoc:no", "authorname:activecalc", "autonumheader:yes2", "licenseversion:40", "source@ ]
[ "article:topic", "Separable Differential Equations", "license:ccbysa", "showtoc:no", "authorname:activecalc", "autonumheader:yes2", "licenseversion:40", "source@ ]
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7.4: Separable Differential Equations
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7.3: Euler's Method
7.5: Modeling with Differential Equations
Page ID 107841
Matthew Boelkins, David Austin & Steven Schlicker
Grand Valley State University via ScholarWorks @Grand Valley State University
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Motivating Questions
2. Preview Activity 7.4.1
3. Solving separable differential equations
1. Example 7.4.1
2. Example 7.4.2
3. Activity 7.4.2
4. Activity 7.4.3
5. Activity 7.4.4
Summary
Motivating Questions
What is a separable differential equation?
How can we find solutions to a separable differential equation?
Are some of the differential equations that arise in applications separable?
In Sections 7.2 and 7.3, we have seen several ways to approximate the solution to an initial value problem. Given the frequency with which differential equations arise in the world around us, we would like to have some techniques for finding explicit algebraic solutions of certain initial value problems. In this section, we focus on a particular class of differential equations (called separable) and develop a method for finding algebraic formulas for their solutions.
A separable differential equation is a differential equation whose algebraic structure allows the variables to be separated in a particular way. For instance, consider the equation
dy dt=ty.
We would like to separate the variables t and y so that all occurrences of t appear on the right-hand side, and all occurrences of y appear on the left, multiplied by dy/dt. For this example, we divide both sides by y so that
1 ydy dt=t.
Note that when we attempt to separate the variables in a differential equation, we require that one side is a product in which the derivative dy/dt is one factor and the other factor is solely an expression involving y.
Not every differential equation is separable. For example, if we consider the equation
dy dt=t−y,
it may seem natural to separate it by writing
y+dy dt=t.
As we will see, this will not be helpful, since the left-hand side is not a product of a function of y with dy dt.
Preview Activity 7.4.1
In this preview activity, we explore whether certain differential equations are separable or not, and then revisit some key ideas from earlier work in integral calculus.
Which of the following differential equations are separable? If the equation is separable, write the equation in the revised form g(y)dy dt=h(t).
dy dt=−3y.
dy dt=ty−y.
dy dt=t+1.
dy dt=t 2−y 2.
Explain why any autonomous differential equation is guaranteed to be separable.
Why do we include the term “+C” in the expression ∫x dx=x 2 2+C?
Suppose we know that a certain function f satisfies the equation ∫f′(x)dx=∫x dx.
What can you conclude about f?
Solving separable differential equations
Before we discuss a general approach to solving a separable differential equation, it is instructive to consider an example.
Example 7.4.1
Find all functions y that are solutions to the differential equation
dy dt=t y 2.
Answer
We begin by separating the variables and writing
y 2dy dt=t.
Integrating both sides of the equation with respect to the independent variable t shows that
∫y 2dy dt dt=∫t dt.
Next, we notice that the left-hand side allows us to change the variable of antidifferentiation 1 from t to y. In particular, dy=dy dt dt, so we now have
∫y 2 dy=∫t dt.
This is why we required that the left-hand side be written as a product in which dy/dt is one of the terms.
This equation says that two families of antiderivatives are equal to each other. Therefore, when we find representative antiderivatives of both sides, we know they must differ by an arbitrary constant C. Antidifferentiating and including the integration constant C on the right, we find that
y 3 3=t 2 2+C.
It is not necessary to include an arbitrary constant on both sides of the equation; we know that y 3/3 and t 2/2 are in the same family of antiderivatives and must therefore differ by a single constant.
Finally, we solve the last equation above for y as a function of t, which gives
y(t)=3 2 t 2+3C 3.
Of course, the term 3C on the right-hand side represents 3 times an unknown constant. It is, therefore, still an unknown constant, which we will rewrite as C. We thus conclude that the funtion
y(t)=3 2 t 2+C 3
is a solution to the original differential equation for any value of C.
Notice that because this solution depends on the arbitrary constant C, we have found an infinite family of solutions. This makes sense because we expect to find a unique solution that corresponds to any given initial value.
For example, if we want to solve the initial value problem
dy dt=t y 2,y(0)=2,
we know that the solution has the form y(t)=3 2 t 2+C 3 for some constant C. We therefore must find the appropriate value for C that gives the initial value y(0)=2. Hence,
2=y(0)=3 2 0 2+C 3=C 3,
which shows that C=2 3=8. The solution to the initial value problem is then
y(t)=3 2 t 2+8 3.
The strategy of Example 7.4.1 may be applied to any differential equation of the form dy dt=g(y)⋅h(t), and any differential equation of this form is said to be separable. We work to solve a separable differential equation by writing
1 g(y)dy dt=h(t),
and then integrating both sides with respect to t. After integrating, we try to solve algebraically for y in order to write y as a function of t.
Example 7.4.2
Solve the differential equation
dy dt=3y.
Answer
Following the same strategy as in Example 7.4.1, we have
1 ydy dt=3.
Integrating both sides with respect to t,
∫1 ydy dt dt=∫3 dt,
and thus
∫1 y dy=∫3 dt.
Antidifferentiating and including the integration constant, we find that
ln|y|=3t+C.
Finally, we need to solve for y. Here, one point deserves careful attention. By the definition of the natural logarithm function, it follows that
|y|=e 3t+C=e 3te C.
Since C is an unknown constant, e C is as well, though we do know that it is positive (because e x is positive for any x). When we remove the absolute value in order to solve for y, however, this constant may be either positive or negative. To account for a possible + or −, we denote this updated constant by C to obtain
y(t)=Ce 3t.
There is one more technical point to make. Notice that y=0 is an equilibrium solution to this differential equation. In solving the equation above, we begin by dividing both sides by y, which is not allowed if y=0. To be perfectly careful, therefore, we should consider the equilibrium solutions separately. In this case, notice that the final form of our solution captures the equilibrium solution by allowing C=0.
Activity 7.4.2
Suppose that the population of a town is growing continuously at an annual rate of 3% per year.
Let P(t) be the population of the town in year t. Write a differential equation that describes the annual growth rate.
Find the solutions of this differential equation.
If you know that the town's population in year 0 is 10,000, find the population P(t).
How long does it take for the population to double? This time is called the doubling time.
Working more generally, find the doubling time if the annual growth rate is k times the population.
Activity 7.4.3
Suppose that a cup of coffee is initially at a temperature of 105∘ F and is placed in a 75∘ F room. Newton's law of cooling says that
dT dt=−k(T−75),
where k is a constant of proportionality.
Suppose you measure that the coffee is cooling at one degree per minute at the time the coffee is brought into the room. Use the differential equation to determine the value of the constant k.
Find all the solutions of this differential equation.
What happens to all the solutions as t→∞? Explain how this agrees with your intuition.
What is the temperature of the cup of coffee after 20 minutes?
How long does it take for the coffee to cool to 80∘?
Activity 7.4.4
Solve each of the following differential equations or initial value problems.
dy dt−(2−t)y=2−t
1 tdy dt=e t 2−2y
y′=2y+2,y(0)=2
y′=2y 2,y(−1)=2
dy dt=−2ty t 2+1,y(0)=4
Summary
A separable differential equation is one that may be rewritten with all occurrences of the dependent variable multiplying the derivative and all occurrences of the independent variable on the other side of the equation.
We may find the solutions to certain separable differential equations by separating variables, integrating with respect to t, and ultimately solving the resulting algebraic equation for y.
This technique allows us to solve many important differential equations that arise in the world around us. For instance, questions of growth and decay and Newton's Law of Cooling give rise to separable differential equations. Later, we will learn in Section 7.6 that the important logistic differential equation is also separable.
This page titled 7.4: Separable Differential Equations is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform.
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7.3: Euler's Method
7.5: Modeling with Differential Equations |
17324 | https://collab.its.virginia.edu/access/content/group/f85bed6c-45d2-4b18-b868-6a2353586804/2/Ch06_Turissini_J_Carboxypeptidase_A-_-/Ch06_Turissini_J_Carboxypeptidase_A_Carboxypeptidase_A.html | Carboxypeptidase A
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Carboxypeptidase_A
Carboxypeptidase A
created by Jon Turissini
Carboxypeptidase A(PDB ID 1hdu), found as a pancreatic digestive enzyme in Bos Taurus,is a zinc exopeptidase that is the cornerstone protein enzyme for the study of metalloproteases (PDB, 1). It contains 307 residues per subunit, has a molecular weight of 137,580.37 Da, and has an isoelctric point (pI) of 7.47 (Artimo, 1). It is fully crystallized at a pH of 7.5 (PDB, 1). Carboxypeptidase A , when bound a by zinc ion (Zn 2+), hydrolyzes the carboxy-terminus (C-terminus) peptide, cleaving it from its polypeptide substrate. Carboxypeptidase A breaks down proteins in the pancreas of a cow into amino acids to later be formed into compulsory proteins for the cow. Carboxypeptidase A also is a key component in the uncovering of the primary structure of all proteins. Due to limitations in substrate binding, it is used in conjunction with carboxypeptidase B and carboxypeptidase Y to cleave the C-terminus of proteins. The rate of cleavage is used to determine which amino acid on the protein substrate was cleaved (Grisham, 115-116). The rate of cleavage of peptides for carboxypeptidase A is near 10 2 s-1 (Christianson, 62). If a protein is used with only carboxypeptidase A and there is no reaction, the peptide in question is one that carboxypeptidase A cannot cleave, such as proline, arginine, lysine, aspartic acid, and glutamic acid (Grisham, 115-116). At this point, another carboxypeptidase can be used to properly postulate the unknown peptide.
Carboxypeptidase A exists in nature as a homotetramer,four subunitsthat are all identical. These subunits are separated into two important domains, S’1 and S1, that each have a distinct function in the catalytic cleavage of the C-terminus amino acid (Christianson, 63).Although these domains are not entirely hydrophobic or hydrophilic, each subunit haspertinent residuesof both characteristics that carry out a specific function.
TheS’1 domaindetermines the specificity for the substrate. Although carboxypeptidase A is a hydrolytic peptidase, the peptides in S’1 prohibit the cleavage of particular peptides that reside on the C-terminus of a given polypeptide substrate. This is because the S’1 domain uses three main residues,Tyr-248,Arg-145, andAsn-144, and a hydrophobic “pocket” to ensure specificity (Christianson, 63). Specificity of carboxypeptidase A is focused towards the carboxylate moiety of a peptide. When interacting with carboxypeptidase A, the carboxylate of the substrate forms a salt link with the guanidinium moiety of Arg-145. Both Arg-145 and Asn-144 perform in hydrogen bonding with this carboxylate, as well (Christianson, 63). These interactions increase the affinity of the polypeptide substrate for the binding site. Competition in binding at this subunit ensures specificity by tolerating only molecules with such intermolecular interactions to successfully bind.
In accordance with the induced fit model that was postulated by Koshland, when a ligand is bound, the phenolic moiety in Tyr-248 and its associated loop undergo a drastic conformational change to include this residue in S’1 (Cho, 2015). The phenol of this residue “flips down” exposing it to hydrogen bond interactions. This allows Tyr-248 to donate a hydrogen bond to the polypeptide substrate’s C-terminus carboxylate and accept a hydrogen bond from an amide of the preceding peptide on the substrate (Cho 2015). Removal of the ligand, Zn 2+, yields an inactive enzyme, showing the importance of the Tyr-248 residue in specificity (Anfinsen, 4). In the absence of a zinc ion ligand, Tyr-248 is unable to change conformation and thus, the reaction will not occur.
Due to the hydrophobic pocket in the S’1 domain, both peptide substrates with aliphatic and aromatic side chains have a higher binding affinity (Christianson, 63). Benzylic or aromatic moieties also exploit this hydrophobic pocket, specifically the tyrosine and phenylalanine peptides, to increase their affinity via “edge to face” aromatic interactions (Jennings, 885).
TheS1 domainis the catalytic domain where hydrolysis occurs. Once the polypeptide substrate is successfully bound to the active site, hydrolysis can occur. For hydrolysis to occur, Zn 2+ must be present. Because it bears an open coordination sphere, Zn 2+ is able to bind to a water molecule and also perform inmetallic interactionswith the carboxylate inGlu-72and the nitrogen atoms inHis-69andHis-196(McCall, 1437). Because the zinc ion has a full d orbital, it acts as a Lewis acid and becomes the electrophilic catalyst in this reaction. Due to these properties, it is able to aid in the deprotonation of the water molecule for nucleophilic attack and electrostatically stabilize the negative charges of the carbonyl on the polypeptide substrate.
To begin hydrolysis, the carboxylate anion onGlu-270deprotonates the water molecule bound to zinc, forming a hydroxide, a more suitable molecule for nucleophilic attack. The hydroxide attacks the carbonyl of the peptide substrate forming a tetrahedral transition state, which is stabilized by ionic interactions from the zinc ion and the hydrogen bonds of the guanidinium moiety ofArg-127(Cho 2015). Arg-127 also activates the scissile peptide bond via hydrogen bonding. Proton donation by Glu-270 and the desired formation of a more stable neutral carbonyl on the polypeptide substrate causes cleavage of the peptide bond at the C-terminus (Christianson, 68).
The secondligand componentfor carboxypeptidase A isD-[(amino)carbonyl]phenylalanine. This ligand looks and functions very much like a phenylalanine residue on a polypeptide substrate, however instead of binding to carboxypeptidase A and being cleaved, it binds and inhibits carboxypeptidase A. D-[(amino)carbonyl]phenylalanine utilizes competitive inhibition, having an affinity more than three fold of normal polypeptide substrates, to cease hydrolyzation (PDB, 1). In addition to the hydrogen bonding and ionic interactions that a polypeptide substrate has, the phenyl group on D-[(amino)carbonyl]phenylalanine is fitted into the substrate recognition pocket at the S’1, domain and it also has supplementary hydrogen bonds due to its additional amide moiety, further increasing it affinity for the binding site (Cho 2015).
Thesecondary structureofeach subunitof carboxypeptidase A is 38% alpha helices and 17% beta sheets (PDB, 1). There are, in total, 12 beta sheets strands (PDB) with about an equal amount of parallel and anti-parallel strands (Anfinsen, 9). These beta sheets form a large beta sheet that extends through the center of the molecule and twists by 120 degrees (Anfinsen, 9). This twist is due to the packing of hydrophobic side chains from adjacent strands. On one side of the beta sheet that extends through the molecule there are more alpha helices, while the other side contains the active site, including both S’1 and S1 domains, and the sole disulfide bond between Cys-138 and Cys-161 (Anfinsen, 9). This protein has a comparatively large surface area. 53.4% of all of the side chains, with a preference of aspartic acid and glutamic acid, make contact with non-isolated water molecules, and are considered “outside” (Anfinsen, 10). Many of the key peptide residues are located on random coil. These residues and many others that make up the S’1 and S1 domains mainly project in towards the center of the protein, forming the pocket that is the active site.
Procarboxypeptidase A2, a serine protease found in Homo Sapiens, is very similar to carboxypeptidase A in both structure and function. It contains 401 residues, has a molecular weight of 44,956.60 Da, and has a pI of 5.52 (Artimo, 1). Procarboxypeptidase A2 has an approximate 66% sequence similarity to carboxypeptidase A and has a nearly identical structure (Altschul, 1). PSI-BLAST is a database used to find other proteins with similar primary structure to a given protein. The Dali Server is a database used to find other proteins with similar tertiary structure to a given protein by measuring intramolecular distances given by the Z score. A Z-score above two means the compared proteins have similar tertiary structure. Shown by the results of DALI (Z= 52.0, rmsd= 1.7) (Holm, 1) and protein BLAST (E= 8e-148) (Altschul, 1), these two proteins have an analogous primary and tertiary structure. E-value is calculated by assigning gaps in the total sequence homology. Gaps increase the E-value and homology decreases it. Although these proteins have asimilar structure, procarboxypeptidase A2 has only one subunit, whereas carboxypeptidase A has four identical subunits. However, when procarboxypeptidase A2 issuperimposedon a subunit of carboxypeptidase A, the differences in their tertiary structures are very subtle (Holm, 1).
Both carboxypeptidase A and procarboxypeptidase A2 are pancreatic zinc C-terminus peptidases, so many of the key residues are conserved to perform the same function under similar conditions. All three residues that interact with the zinc ion, Glu-72, His-69, and His-196, areconservedand reside in the same position. Glu-270 and Arg-127, two residues that are imperative for catalysis, are also conserved. Tyr-248, Arg-145, and Asn-144, three residues essential for substrate specificity, are also all preserved (Catasus, 6651). The first 96 residues on procarboxypeptidase A2 are generally excised from the protein to abdicate its zymogen form, however, additional and different residues in procarboxypeptidase A2 can be explained by the increased specificity of a serine protease in relation to carboxypeptidase A (Garcia-Saez, 6906). Changes in the active site such as a substitution of proline for alanine at position 142 allows for less bulky residues on the substrate, such as the serine, to enter, however, steric hindrance prevents the larger residues, such as phenylalanine, to enter or bind. |
17325 | https://artofproblemsolving.com/wiki/index.php/Incircle?srsltid=AfmBOopOJ0ETgF3z_hCS7bVJyjKVag6ySsWvUA14euXhNQf0BpWKR4D5 | Art of Problem Solving
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Incircle
An incircle of a convexpolygon is a circle which is inside the figure and tangent to each side. Every triangle and regular polygon has a unique incircle, but in general polygons with 4 or more sides (such as non-squarerectangles) do not have an incircle. A quadrilateral that does have an incircle is called a Tangential Quadrilateral. For a triangle, the center of the incircle is the Incenter, where the incircle is the largest circle that can be inscribed in the polygon. The Incenter can be constructed by drawing the intersection of angle bisectors.
Formulas
The radius of an incircle of a triangle (the inradius) with sides and area is
The area of any triangle is where is the Semiperimeter of the triangle.
The formula above can be simplified with Heron's Formula, yielding
The radius of an incircle of a right triangle (the inradius) with legs and hypotenuse is .
For any polygon with an incircle, , where is the area, is the semi perimeter, and is the inradius.
The coordinates of the incenter (center of incircle) are , if the coordinates of each vertex are , , and , the side opposite of has length , the side opposite of has length , and the side opposite of has length .
The formula for the semiperimeter is .
The area of the triangle by Heron's Formula is .
See also
Circumradius
Inradius
Kimberling center
Circumcircle
Click here to learn about the orthocenter, and Line's Tangent
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17326 | https://physics.stackexchange.com/questions/684869/mechanical-power-as-p-f-v | Stack Exchange Network
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Mechanical power as $P = F v$
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I understand that, in the particular case of a constant force $F$ applied to an object, the speed increases linearly, both the instantaneous power $P$ and the kinetic energy also increase linearly.
Often, we deal with the situation in which power $P$ is set to be constant and we hear that if the force $F$ is large the speed $v$ is low and vice versa. Even if this makes sense mathematically, I have to admit that it is still does not make full sense conceptually.
Constant power $P$ means that we supply the object with energy at a constant rate (ex: say 2 Joule every second). Why would a large force imply a small velocity and vice versa?
I think that the underlying assumptions are:
a) the net power is zero: a constant positive input power that injects energy in the system per unit time is matched by a constant negative power (due to resistive forces) that removes an equal amount of energy from the system
b) The speed is constant since the net power is zero and there is not kinetic energy change
I just cannot wrap my head around the idea that when the object moves at small constant speed $v$ the force $F$ on it is large and vice versa. I am stuck with thinking that a large force has to correlate with a large speed.
newtonian-mechanics
forces
velocity
power
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edited Dec 23, 2021 at 3:51
Qmechanic♦
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asked Dec 22, 2021 at 23:46
Brett CooperBrett Cooper
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$\endgroup$
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$\begingroup$ "the speed increases linearly, both the instantaneous power P and the kinetic energy also increase linearly." This wouldn't be possible. the kinetic energy depends quadratically on the speed, so the speed can't be changing both linearly and quadratically at the same time. $\endgroup$
Steeven
– Steeven
2021-12-23 00:02:37 +00:00
Commented Dec 23, 2021 at 0:02
$\begingroup$ Sorry, the change in KE is linear. That is what I meant. $\endgroup$
Brett Cooper
– Brett Cooper
2021-12-23 01:18:27 +00:00
Commented Dec 23, 2021 at 1:18
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4 Answers 4
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$\rm power = force \times velocity$ so if the power is constant a large force implies a small velocity.
The power equation comes from $\rm work \,\,done = force \times displacement$ and dividing both sides of the equation by $\rm time$. So if a body is travelling fast it undergoes a large displacement per second and so to keep the power at some constant value the force must be small.
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answered Dec 23, 2021 at 0:06
FarcherFarcher
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$\begingroup$ I see. Your reply is clear. If the displacement is small over a second, the speed would be low and the force would be large. I get that must be the case to keep power constant. But why does a large force not manage to increase to speed to a larger value and acts on the body while it only moves slowly? Is it simply because the force I am discussing is perfectly matched by another opposing force? Both forces produce equal and constant amounts of power but opposite in sign. The net power must be zero.... $\endgroup$
Brett Cooper
– Brett Cooper
2021-12-23 01:29:37 +00:00
Commented Dec 23, 2021 at 1:29
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Constant power ð means that we supply the object with energy at a constant rate (ex: say 2 Joule every second). Why would a large force imply a small velocity and vice versa?
The engine can't cause velocity. All it can do is supply a force. This relation (assuming it holds) tells us that a constant power engine can only provide small forces at high velocity. We can't pick both the power and the force.
Remember we don't need a large force to travel quickly. A small force over time can do that if drag is small. Forces cause changes in velocity, not velocity itself.
So, say we start from rest with our car in low gear (1s gear) pressing the gas pedal all the way. The car speeds up (i.e. accelerates, changes velocity). What happens if we don't shift and stay in the 1st low gear? Does the car reach a constant speed for the 1st gear because the force on the ground is matched by the net resistive force? I don't think the car speeds up any further...Why?
You can keep accelerating for a while, but normally you don't want to because of damage to the engine. But if your engine can't blow up, you'll accelerate until your RPMs get so high that the power output of the engine drops. Above a certain point, the engine can no longer deliver much force. When this happens, it's balanced against the sources of drag and the acceleration stops.
For a simpler scenario, imagine a merry-go-round. While it's stopped, you can push pretty hard on it. Your force is able to accelerate it quite rapidly. As it goes faster, you can no longer push as hard. The same thing happens in any moving system. As the velocity increases, the force you can develop to push it decreases.
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edited Dec 23, 2021 at 6:07
answered Dec 23, 2021 at 1:26
BowlOfRedBowlOfRed
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$\endgroup$
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$\begingroup$ Thanks. So, say we start from rest with our car in low gear (1s gear) pressing the gas pedal all the way. The car speeds up (i.e. accelerates, changes velocity). What happens if we don't shift and stay in the 1st low gear? Does the car reach a constant speed for the 1st gear because the force on the ground is matched by the net resistive force? I don't think the car speeds up any further...Why? $\endgroup$
Brett Cooper
– Brett Cooper
2021-12-23 02:37:14 +00:00
Commented Dec 23, 2021 at 2:37
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3
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Just a short answer but maybe this helps:Your question is equivalent to asking why kinetic energy $E=\frac{1}{2}mv^2$ increases with $v^2$ instead of $v$ because $dE/dt=ma\cdot v=P$. So the amount of energy needed to increase the velocity of an object $dv$ at high velocities is greater than it is for small velocities because $dE(v)=mv\cdot dv$. Vice versa this means that the power for a given force (that tells you how much the object changes its velocity per time unit) is greater at high velocities because $dE/dt=P=Fv$. This is kind of a circular argument but usually one does not question the fact that kinetic energy increases with $v^2$ (which also simply follows from $F=ma$ and $dE:=F\cdot ds$) and maybe you are more used to this.
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answered Dec 23, 2021 at 6:18
Physics newbiePhysics newbie
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I think the confusion is treating everything constant between power, force, and velocity, but this isn't possible.
If power is constant, then $P=Fv\to F=P/v$, meaning $F$ is a monotonically decreasing function of $v$. So when $v$ is "small", $F$ is "large". This causes an acceleration, which increases $v$ and decreases $F$. Note that this has to happen for constant power. There are no other forces present. You have said the power is constant for this process (whatever it is), and so the force has to be a function of $v$ as $F=P/v$.
If it helps, you can work out the velocity as a function of time in the case of a constant power:
$$F=\frac Pv=m\frac{\text dv}{\text dt}$$ $$\int_{v_0}^{v}v'\,\text dv'=\int_0^t\frac Pm\,\text dt'$$ $$\frac12\left(v^2-v_0^2\right)=\frac Pmt$$ $$v(t)=\sqrt{\frac{2P}{m}t+v_0^2}$$
And we can determine the force as well
$$F(t)=m\frac{\text dv}{\text dt}=\frac{P}{\sqrt{(2P/m)t+v_0^2}}$$
which is just $P/v$ as we come full circle.
So as you can see, over time the velocity increases and the force decreases in such a way that the power is constant. No other forces or $0$ net power is needed to explain this. Note that all of the above assumes a positive power.
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edited Dec 23, 2021 at 11:57
answered Dec 23, 2021 at 6:25
BioPhysicistBioPhysicist
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$\begingroup$ thanks @BioPhysicist. I am be starting to get it. Constant power $P$ means providing the body energy at a constant rate. If power is constant ( ex: body gets 2 J every second), the force we apply to the body must change and be inversely proportional to the instantaneous speed $v$ of the body: $F= \frac {P} {v}$. Say the body starts with $v=0$, the force must be initially very large (technically infinite) and decreases as the speed $v$ increases (which it does). The work done by the force remains constant: the force decreases but it gets applied to a larger displacement during each second.... $\endgroup$
Brett Cooper
– Brett Cooper
2021-12-23 14:44:16 +00:00
Commented Dec 23, 2021 at 14:44
$\begingroup$ @BrettCooper Yep, that is correct :) $\endgroup$
BioPhysicist
– BioPhysicist
2021-12-23 14:45:38 +00:00
Commented Dec 23, 2021 at 14:45
$\begingroup$ My recent comment considers the body speed to change, KE increasing while power is constant. However, the common case of the body moving at constant speed (bike in low gear and cyclist pedaling) implies zero net work, zero net power and multiple (2 or more) forces acting on the body each producing equal amounts of power (oppositely signed). The cyclist manages to generate constant input power power. The net resistive force creates an equal amount of power. The propulsive force by the wheel on the ground, the net resistive force would be constant forces (the power and speed are constant) $\endgroup$
Brett Cooper
– Brett Cooper
2021-12-23 14:59:23 +00:00
Commented Dec 23, 2021 at 14:59
$\begingroup$ @BrettCooper Yes, that is a scenario you can consider. I fail to see the point of your above comment though. $\endgroup$
BioPhysicist
– BioPhysicist
2021-12-23 15:48:07 +00:00
Commented Dec 23, 2021 at 15:48
$\begingroup$ @BiPhysicist, I know, that has been my frustrating thought: a bike moving at a constant speed and with constant input power. The applied force $F$ is also constant. And if $v$ is small, $F$ must be large. The "coexistence" of a constant large force with a constant speed has been bothering me. $\endgroup$
Brett Cooper
– Brett Cooper
2021-12-23 16:52:26 +00:00
Commented Dec 23, 2021 at 16:52
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17327 | https://cs.uwaterloo.ca/journals/JIS/VOL21/Wong2/wong8.pdf | 23 11 Article 18.6.3 Journal of Integer Sequences, Vol. 21 (2018), 2 3 6 1 47 A Probabilistic Take-Away Game Tony W. H. Wong and Jiao Xu Department of Mathematics Kutztown University of Pennsylvania Kutztown, PA 19530 USA wong@kutztown.edu jxu399@live.kutztown.edu Abstract Alice and Bob are playing a very simple game. Each of them starts with a pile of n chips, and they take turns to remove 1 or 2 chips from their own pile randomly and independently with equal probability. The first player who removes all chips from their pile is the winner. In this paper, we find the winning probability for Bob and analyze a new integer sequence. We also show that this game is highly disadvantageous to the second player, which is counter-intuitive. Furthermore, we study several variations of this game and determine the winning probability for Bob in each case.
1 Introduction Take-away games are mathematical games that involve two players, who take turns to remove items from a pile or multiple piles of objects. The winner is the first player achieving a certain predefined goal.
Here is a classical single-pile take-away game: Alice and Bob take turns to remove 1 or 2 chips from a pile of finitely many chips, with Alice going first. Whoever removes the last chip is the winner. Such a game and many variations are well-studied, for example by Schwenk .
Another similar take-away game is the game of Nim. It involves multiple piles, and each player may remove one or more chips from one of the piles. Again, whoever removes the 1 last chip is the winner. Berlekamp, Conway, and Guy as well as Bouton thoroughly discussed the strategies of this game.
In recent years, mathematicians revisited many well-studied deterministic problems in a probabilistic setting. In this paper, we begin with a basic version of a probabilistic take-away game: Alice and Bob each have a pile of n chips, and they take turns to remove 1 or 2 chips from their pile, with Alice going first. Whether to remove 1 or 2 chips is decided by flipping a fair coin. In other words, every move is independent, and the probability of removing 1 or 2 chips is 1 2 each. If there is only 1 chip left in the pile, then it is removed with probability 1 in the next move. The first player who removes all chips from their pile is the winner.
Equivalently, the game can be played in the following manner: Alice and Bob start with no chips, and they take turns to collect 1 or 2 chips randomly and independently with equal probability. The first person who collects at least n chips is the winner. This model of the game avoids the complication when there is only 1 chip left, thus it is preferred and will be adopted in this paper.
Let pn denote the probability that Bob wins by collecting at least n chips before Alice does. At first glance, one may suggest that pn = 1 2 since the game seems to be fair to both Alice and Bob. With more thought, one may realize that pn ≤1 2 since Bob is in a disadvantageous position by having one fewer move half the time. It is also reasonable to predict that pn converges to 1 2 as n goes to infinity. In this paper, we study precisely what pn is for each n, and how this sequence behaves.
It is apparent that p1 = 0, since Alice wins on the first move. When n = 2, Bob can only win if Alice collects 1 chip in her first move and Bob collects 2 in his first move, implying that p2 = 1 4. To study pn in general, we need some formal definitions.
Let X1, X2, . . . , Xn, Y1, Y2, . . . , Yn be independent identically distributed random variables such that P(Xi = 1) = P(Xi = 2) = P(Yi = 1) = P(Yi = 2) = 1 2, where Xi and Yi represent the number of chips collected by Alice and Bob in the i-th move respectively.
Before we proceed to derive the formula, let us first study the initial terms of pn. As mentioned, p1 = 0 and p2 = 1 4. The following calculations find pn when n = 3, 4, 5, 6, and 7.
When n = 3, p3 = P (X1, X2) = (1, 1) · P (Y1, Y2) = (1, 2), (2, 1), or (2, 2) = 1 4 · 3 4 = 3 16.
2 When n = 4, p4 = P (X1, X2) = (1, 1), (1, 2), or (2, 1) · P (Y1, Y2) = (2, 2) + P (X1, X2, X3) = (1, 1, 1) · P (Y1, Y2, Y3) = (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 2, 2), or (2, 1, 2) = 3 4 · 1 4 + 1 8 · 5 8 = 17 64.
When n = 5, p5 = P (X1, X2, X3) = (1, 1, 1), (1, 1, 2), (1, 2, 1), or (2, 1, 1) · P (Y1, Y2, Y3) = (1, 2, 2), (2, 1, 2), (2, 2, 1), or (2, 2, 2) + P (X1, X2, X3, X4) = (1, 1, 1, 1) · P (Y1, Y2, Y3, Y4) = (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1), (1, 1, 2, 2), (1, 2, 1, 2), or (2, 1, 1, 2) = 1 2 · 1 2 + 1 16 · 7 16 = 71 256.
When n = 6, p6 = P (X1, X2, X3) ̸= (2, 2, 2) · P (Y1, Y2, Y3) = (2, 2, 2) + P (X1, X2, X3, X4) = (1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), or (2, 1, 1, 1) · P (Y1, Y2, Y3, Y4) = (1, 1, 2, 2), (1, 2, 1, 2), (1, 2, 2, 1), (2, 1, 1, 2), (2, 1, 2, 1), (2, 2, 1, 1), (1, 2, 2, 2), (2, 1, 2, 2) or (2, 2, 1, 2) + P (X1, X2, X3, X4, X5) = (1, 1, 1, 1, 1) · P (Y1, Y2, Y3, Y4, Y5) = (1, 1, 1, 1, 2), (1, 1, 1, 2, 1), (1, 1, 2, 1, 1), (1, 2, 1, 1, 1), (2, 1, 1, 1, 1), (1, 1, 1, 2, 2), (1, 1, 2, 1, 2), (1, 2, 1, 1, 2), or (2, 1, 1, 1, 2) = 7 8 · 1 8 + 5 16 · 9 16 + 1 32 · 9 32 = 301 1024.
3 When n = 7, p7 = P (X1, X2, X3, X4) ̸= (1, 2, 2, 2), (2, 1, 2, 2), (2, 2, 1, 2), (2, 2, 2, 1), and (2, 2, 2, 2) · P (Y1, Y2, Y3, Y4) = (1, 2, 2, 2), (2, 1, 2, 2), (2, 2, 1, 2), (2, 2, 2, 1), or (2, 2, 2, 2) + P (X1, X2, X3, X4, X5) = (1, 1, 1, 1, 1), (1, 1, 1, 1, 2), (1, 1, 1, 2, 1), (1, 1, 2, 1, 1), (1, 2, 1, 1, 1), or (2, 1, 1, 1, 1) · P (Y1, Y2, Y3, Y4, Y5) = (1, 1, 1, 2, 2), (1, 1, 2, 1, 2), (1, 1, 2, 2, 1), (1, 2, 1, 1, 2), (1, 2, 1, 2, 1), (1, 2, 2, 1, 1), (2, 1, 1, 1, 2), (2, 1, 1, 2, 1), (2, 1, 2, 1, 1), (2, 2, 1, 1, 1), (1, 1, 2, 2, 2), (1, 2, 1, 2, 2), (1, 2, 2, 1, 2), (2, 1, 1, 2, 2), (2, 1, 2, 1, 2), or (2, 2, 1, 1, 2) + P (X1, X2, X3, X4, X5, X6) = (1, 1, 1, 1, 1, 1) · P (Y1, Y2, Y3, Y4, Y5, Y6) = (1, 1, 1, 1, 1, 2), (1, 1, 1, 1, 2, 1), (1, 1, 1, 2, 1, 1), (1, 1, 2, 1, 1, 1), (1, 2, 1, 1, 1, 1), (2, 1, 1, 1, 1, 1), (1, 1, 1, 1, 2, 2), (1, 1, 1, 2, 1, 2), (1, 1, 2, 1, 1, 2), (1, 2, 1, 1, 1, 2), or (2, 1, 1, 1, 1, 2) = 11 16 · 5 16 + 6 32 · 16 32 + 1 64 · 11 64 = 1275 4096.
From these calculations, it seems that the denominator of pn is 4n−1. As for the nu-merator, we observe that the sequence 3, 17, 71, 301, 1275 satisfies the recurrence relation xn+2 = 4xn+1 + xn. However, when we use these two observations to project p100, we get p100 ≈64.4353, which is absurd since pn ≤1. In other words, the pattern developed from small cases is not helpful, so we need to generalize our calculations to obtain a formula for pn. Nevertheless, the fact that these five terms satisfy an elegant recurrence relation is interesting.
In Section 2, we present a complete solution for this basic probabilistic take-away game.
In Section 3, we discuss two integer sequences obtained from this result.
In Section 4, we extend our calculations to some general versions of this probabilistic take-away game.
Finally, in Section 5, we discuss some numerical computations and approximations of the probabilities, as well as some open questions.
2 Results for the basic version Let Sk = Pk i=1 Xi, and Tk = Pk i=1 Yi. It is clear that pn = n X k=1 P(Sk < n) · P(Tk ≥n and Tk−1 < n) .
(1) Let sk = |{i : Xi = 2, 1 ≤i ≤k}|, and let tk = |{i : Yi = 2, 1 ≤i ≤k}|. Notice that 4 sk + k = Sk and tk + k = Tk. Hence, equation (1) becomes pn = n X k=1 P(sk < n −k) · P(tk ≥n −k and tk−1 < n −(k −1)) .
Note that P(sk < n −k) = 1 2k n−k−1 X i=0 k i . Let hk = n−k−1 X i=0 k i , so P(sk < n −k) = hk 2k .
Then P(tk ≥n −k and tk−1 < n −(k −1)) = P(tk ≥n −k) −P(tk−1 ≥n −(k −1)) = 1 2k k X i=n−k k i − 1 2k−1 k−1 X i=n−(k−1) k −1 i = 1 −hk 2k − 1 −hk−1 2k−1 = 2hk−1 −hk 2k .
Hence, pn = n X k=1 hk(2hk−1 −hk) 4k . If we rearrange the terms, we have pn = 2h1h0 4 + n X k=2 −4h2 k−1 + 2hk−1hk 4k −h2 n 4n = 1 + n X k=2 2hk−1(hk −2hk−1) 4k , since h0 = 1, h1 = 2, and hn = 0. By adding the two expressions for pn obtained in the last two lines and dividing the sum by 2, we have pn = 1 2 −1 2 n X k=2 (2hk−1 −hk)2 4k = 1 2 −1 2 n X k=1 (2hk−1 −hk)2 4k .
(2) Finally, we have the following lemma.
Lemma 1. Let hk = n−k−1 X i=0 k i . Then 2hk−1 −hk = k n −k + k −1 n −k .
Proof.
2hk−1 −hk = 2 n−(k−1)−1 X i=0 k −1 i − n−k−1 X i=0 k i = 2 n−k X i=0 k −1 i − n−k−1 X i=1 k −1 i + k −1 i −1 − k 0 = k −1 0 + 2 k −1 n −k + k −1 n −k −1 − k 0 = k −1 n −k + k n −k .
5 Remark 2. The above proof of Lemma 1 is purely algebraic. Here is an alternate proof that is more combinatorial.
2hk−1 −hk 2k = P(tk ≥n −k and tk−1 < n −(k −1)) = P(Tk ≥n and Tk−1 < n) = P Tk = n or (Tk−1 = n −1 and Xk = 2) = P tk = n −k or (tk−1 = n −k and Xk = 2) = k n−k 2k + k−1 n−k 2k−1 · 1 2 = k n−k + k−1 n−k 2k , which yields our desired result.
Applying Lemma 1 to equation (2), we obtain the following theorem.
Theorem 3. Alice and Bob take turns to collect 1 or 2 chips randomly and independently with equal probability. The probability that Bob collects at least n chips before Alice is pn = 1 2 −1 2 n X k=1 1 4k k n −k + k −1 n −k 2 .
It is worth noting that the summation can start from k = ln 2 m , since all the summands are zero when k < n 2. However, we decide to start the summation from k = 1 since it is more elegant.
3 Observations on the sequences Let pn = cn dn , where cn and dn are nonnegative integers that are relatively prime. Notice that 4n−1pn = 4n−1 2 −1 2 n X k=1 4n−k−1 k n −k + k −1 n −k 2 is an odd integer, since 1 2 n X k=n−1 4n−k−1 k n −k + k −1 n −k 2 = 1 2 n −1 1 + n −2 1 2 + 4−1 n 0 + n −1 0 2!
= 1 2 (2n −3)2 + 1 = 2n2 −6n + 5 is an odd integer, and all other summands are even.
In other words, dn = 4n−1, and cn = 1 8 4n − n X k=1 4n−k k n −k + k −1 n −k 2!
. As mentioned in Section 1, we observed 6 that c3, c4, c5, c6, and c7 satisfy the recurrence relation xn+2 = 4xn+1+xn, but this recurrence relation does not hold from c8 onwards. The integer sequence cn is added to the On-Line Encyclopedia of Integer Sequences (OEIS), listed as A265919. The integer sequence formed by the numerators of the winning probabilities for Alice is listed as A265920.
Apart from the integer sequences that we discovered, we are also very interested in the sequence of fractions pn. The disadvantage that Bob has due to Alice going first should be less significant as n increases, so it is reasonable to believe that pn should increase with n.
This trend is mostly true: with the exception of n = 3, pn−1 < pn for all n up to 5000.
Another interesting observation about pn is that its convergence rate towards 1 2 is slow.
For example, p100 ≈44.83%, which is very surprising since most people would believe that the disadvantage of Bob is insignificant by the time n reaches 100. In fact, p1000 ≈48.36%, p10000 ≈49.48%, and p100000 ≈49.84%. In other words, the fact that Bob starts second puts him in a significantly disadvantageous position, which is counter-intuitive.
Theorem 3 solves this basic probabilistic take-away game, but there are several variations of the game that one can study. For instance, what is the winning probability for Bob in the following scenarios?
1. Alice and Bob take turns to collect a or b chips randomly and independently with equal probability, where a < b are positive integers.
2. Alice and Bob take turns to collect a or b chips randomly and independently with probabilities p and 1 −p respectively, where a < b are positive integers and p ∈[0, 1].
3. Alice and Bob take turns to collect a1, a2, . . . , or am chips randomly and independently with probabilities p1, p2, . . . , pm respectively, where a1 < a2 < · · · < am are positive integers, p1, p2, . . . , pm ∈[0, 1], and p1 + p2 + · · · + pm = 1.
4 Results for general cases Before we proceed, let us introduce a different proof of Theorem 3. This proof is due to Taoye Zhang (personal communication, October 2015). Let X1, X2, . . . , Xn, Y1, Y2, . . . , Yn be independent identically distributed integer-valued random variables, where Xi and Yi represent the number of chips collected by Alice and Bob in the i-th move respectively. Let Sk = Pk i=1 Xi, and Tk = Pk i=1 Yi. Let rk = P(Sk ≥n and Sk−1 < n), and let a be the minimum positive integer such that P(X1 = a) > 0.
Lemma 4. The probability that Bob wins the game is 1 2 −1 2 ⌈n a ⌉ X k=1 r2 k.
7 Proof. Since {Sk ≥n and Sk−1 < n}1≤k≤⌈n a ⌉are mutually exclusive events and their union covers the entire sample space, we have X 1≤k≤⌈n a ⌉ rk = 1.
Hence, the probability that Bob wins the game is ⌈n a ⌉ X k=1 P(Sk < n)P(Tk ≥n and Tk−1 < n) = ⌈n a ⌉ X k=1 1 − k X i=1 ri !
rk = ⌈n a ⌉ X k=1 ⌈n a ⌉ X i=k+1 ri rk = X 1≤k<i≤⌈n a ⌉ rirk = 1 2 ⌈n a ⌉ X k=1 rk 2 − ⌈n a ⌉ X k=1 r2 k = 1 2 −1 2 ⌈n a ⌉ X k=1 r2 k.
Second Proof of Theorem 3. Since Alice and Bob collect 1 or 2 chips randomly and indepen-dently with equal probability, P(Xi = 1) = P(Xi = 2) = P(Yi = 1) = P(Yi = 2) = 1 2. Let sk = |{i : Xi = 2, 1 ≤i ≤k}|, and let tk = |{i : Yi = 2, 1 ≤i ≤k}|. Notice that sk + k = Sk and tk + k = Tk. Hence, rk = P(Sk = n) + P(Sk−1 = n −1 and Xk = 2) = P(sk = n −k) + P(sk−1 = n −k and Xk = 2) = 1 2k k n −k + 1 2k−1 k −1 n −k · 1 2 = 1 2k k n −k + k −1 n −k .
Substituting this into the winning probability for Bob in Lemma 4 with a = 1 yields the desired result.
Next, we consider the three scenarios listed in Section 3. Let x y int = x y if y is a nonnegative integer, and x y int = 0 otherwise.
8 Theorem 5. Alice and Bob take turns to collect a or b chips randomly and independently with probabilities p and q = 1−p respectively, where a < b are positive integers and p ∈[0, 1].
Then the probability that Bob collects at least n chips before Alice is 1 2 −1 2 · ⌈n a⌉ X k=1 k n−ak b−a int p bk−n b−a q n−ak b−a + a−1 X i=1 k −1 n−a(k−1)−i b−a int p b(k−1)−n+i b−a q n−a(k−1)−i b−a + b−1 X i=a k −1 n−a(k−1)−i b−a int p b(k−1)−n+i b−a q n+b−ak−i b−a !2 .
Proof. In this case, P(Xi = a) = P(Yi = a) = p and P(Xi = b) = P(Yi = b) = 1 −p = q.
Let sk = |{i : Xi = b, 1 ≤i ≤k}|, and let tk = |{i : Yi = b, 1 ≤i ≤k}|. Notice that sk(b −a) + ka = Sk and tk(b −a) + ka = Tk. Hence, rk = P(Sk = n) + a−1 X i=1 P(Sk−1 = n −i) + b−1 X i=a P(Sk−1 = n −i and Xk = b) = P sk = n −ak b −a + a−1 X i=1 P sk−1 = n −a(k −1) −i b −a + b−1 X i=a P sk−1 = n −a(k −1) −i b −a and Xk = b = k n−ak b−a int p bk−n b−a q n−ak b−a + a−1 X i=1 k −1 n−a(k−1)−i b−a int p b(k−1)−n+i b−a q n−a(k−1)−i b−a + b−1 X i=a k −1 n−a(k−1)−i b−a int p b(k−1)−n+i b−a q n−a(k−1)−i b−a · q = k n−ak b−a int p bk−n b−a q n−ak b−a + a−1 X i=1 k −1 n−a(k−1)−i b−a int p b(k−1)−n+i b−a q n−a(k−1)−i b−a + b−1 X i=a k −1 n−a(k−1)−i b−a int p b(k−1)−n+i b−a q n+b−ak−i b−a .
Substituting this into the winning probability for Bob in Lemma 4 yields the desired result.
In particular, if p = 1 2, then we have the following corollary.
Corollary 6. If Alice and Bob take turns to collect a or b chips randomly and independently with equal probability, where a < b are positive integers, then the probability that Bob collects 9 at least n chips before Alice is 1 2 −1 2 · ⌈n a⌉ X k=1 1 4k k n−ak b−a int + 2 · a−1 X i=1 k −1 n−a(k−1)−i b−a int + b−1 X i=a k −1 n−a(k−1)−i b−a int !2 .
Proof. Substituting p = q = 1 2 into the probability found in Theorem 5, we have 1 2−1 2· ⌈n a⌉ X k=1 k n−ak b−a int 1 2 k + a−1 X i=1 k −1 n−a(k−1)−i b−a int 1 2 k−1 + b−1 X i=a k −1 n−a(k−1)−i b−a int 1 2 k!2 .
This yields our desired result.
Theorem 7. Alice and Bob take turns to collect a1, a2, . . . , or am chips randomly and independently with probabilities p1, p2, . . . , pm respectively, where a1 < a2 < · · · < am are positive integers, p1, p2, . . . , pm ∈[0, 1], and p1 + p2 + · · · + pm = 1. The probability that Bob collects at least n chips before Alice is 1 2 −1 2 · l n a1 m X k=1 X (sk1,sk2,...,skm)∈Zm ≥0: Pm g=1 skgag=n Pm g=1 skg=k k sk1, sk2, . . . , skm m Y ℓ=1 pskℓ ℓ + m X j=1 aj−1 X i=aj−1 X (sk1,sk2,...,skm)∈Zm ≥0: Pm g=1 skgag=n−i Pm g=1 skg=k−1 k −1 sk1, sk2, . . . , skm m Y ℓ=1 pskℓ ℓ m X α=j pα !
2 .
Proof. In this case, P(Xi = aj) = P(Yi = aj) = pj for all j = 1, 2, . . . , m. Let skj = |{i : Xi = aj, 1 ≤i ≤k}|, and let tkj = |{i : Yi = aj, 1 ≤i ≤k}|, where j = 1, 2, . . . , m. Define 10 a0 = 1. Notice that Pm j=1 skjaj = Sk and Pm j=1 tkjaj = Tk. Hence, rk = P(Sk = n) + m X j=1 aj−1 X i=aj−1 P(Sk−1 = n −i and Xk = aj, aj+1, . . . , or am) = X (sk1,sk2,...,skm)∈Zm ≥0: Pm g=1 skgag=n Pm g=1 skg=k k sk1, sk2, . . . , skm m Y ℓ=1 pskℓ ℓ + m X j=1 aj−1 X i=aj−1 X (sk1,sk2,...,skm)∈Zm ≥0: Pm g=1 skgag=n−i Pm g=1 skg=k−1 k −1 sk1, sk2, . . . , skm m Y ℓ=1 pskℓ ℓ m X α=j pα !
.
Substituting this into the winning probability for Bob in Lemma 4 with a = a1 yields the desired result.
5 Further observations and open questions The numerical results given by the formulae in Theorems 3, 5, and 7 can also be computed by using recursive formulae and dynamic programming. This is particularly useful for the case in Theorem 7, since it significantly reduces the computational complexity.
Let position (i, j) denote the instance when Alice and Bob have already collected n −i and n −j chips respectively. In other words, Alice and Bob need to collect i and j chips respectively to win the game. Let Pi,j = P(Bob wins from position (i, j) | Alice starts).
Then P(Alice wins from position (i, j) | Alice starts) = 1 −Pi,j, P(Alice wins from position (i, j) | Bob starts) = Pj,i, and P(Bob wins from position (i, j) | Bob starts) = 1 −Pj,i.
In each case, the winning probability for Bob is given by Pn,n.
Under the conditions given in Theorem 3, Pi,j satisfies the recurrence relation Pi,j = 1 2(1 −Pj,i−1) + 1 2(1 −Pj,i−2) = 1 −1 2(Pj,i−1 + Pj,i−2).
Under the conditions given in Theorem 5, Pi,j satisfies the recurrence relation Pi,j = p(1 −Pj,i−a) + q(1 −Pj,i−b) = 1 −(pPj,i−a + qPj,i−b).
11 Finally, under the conditions given in Theorem 7, Pi,j satisfies the recurrence relation Pi,j = m X g=1 pg(1 −Pj,i−ag) = 1 − m X g=1 pgPj,i−ag.
In all three cases, the initial conditions for Pi,j are 1. Pi,j = 0 for all i, j ∈Z such that i ≤a and j ≥1; 2. Pi,j = 1 for all i, j ∈Z such that i ≥1 and j ≤0, where a is the least number of chips that Alice or Bob can collect in one move. Note that we do not need to specify Pi,j if i, j ≤0.
Another observation is related to the numerical approximation of pn in the basic proba-bilistic take-away game. Recall from Theorem 3 that pn = 1 2 −1 2 n X k=1 1 4k k n −k + k −1 n −k 2 .
The following nonrigorous arguments suggest that pn ∼1 2 − r 27 32πn.
(3) First, 1 2 n X k=1 1 4k k n −k + k −1 n −k 2 = 1 2 n X k=⌈n 2 ⌉ 1 4k k!
(n −k)!(2k −n)!
1 + 2k −n k 2 .
By Stirling’s approximation, when n is large, this is approximately 1 2 n X k=⌈n 2 ⌉ 1 4k √ 2πk k e k p 2π(n −k) n−k e n−k p 2π(2k −n) 2k−n e 2k−n 3k −n k !2 , which can be simplified to 1 π n X k=⌈n 2 ⌉ k2k+2 22k+2 · 1 k(n −k)2(n−k)+1(2k −n)2(2k−n)+1 · (3k −n)2 k2 , and can be transformed into 1 π n X k=⌈n 2 ⌉ 3 −n k 2 k 2n k −2 2(n−k)+1 4 −2n k 2(2k−n)+1.
12 We are going to approximate the last sum by turning it into an integral, with the discrete variable k replaced by a continuous variable u. The integral is 1 π Z n n 2 3 −n u 2 u 2n u −2 2(n−u)+1 4 −2n u 2(2u−n)+1du.
We can perform a substitution x = 2n u , which yields 1 π Z 2 4 3 −x 2 2 2n x (x −2)2(n−2n x )+1(4 −x)2(2 2n x −n)+1 −2n x2 dx.
This integral simplifies to 1 4π Z 4 2 (6 −x)2 x(x −2) 2n x (x−2)+1(4 −x) 2n x (4−x)+1dx, and we further transform it into 1 4π Z 4 2 (6 −x)2 x(x −2)(4 −x) · en· −2 x ln((x−2)x−2(4−x)4−x)dx.
(4) Let I(n) = Z 4 2 f(x)enφ(x)dx, where f(x) = (6 −x)2 x(x −2)(4 −x) and φ(x) = −2 x ln ((x −2)x−2(4 −x)4−x). This is a Laplace integral. Note that φ′(x) = 2 x2 ln ((x −2)−2(4 −x)4), whose only zero on the interval (2, 4) is x = 3. Note also that f(3) = 3 ̸= 0 and φ′′(3) = −4 3 < 0. Although I(n) is an improper integral, since lim x→2 φ(x) = −ln 4 and lim x→4 φ(x) = −ln 4 2 , the techniques on a Laplace integral [1, pp. 261–267] still apply. Hence, we have I(n) ∼ √ 2πf(3)enφ(3) p −nφ′′(c) = √ 2π · 3 · 1 q −n · −4 3 = r 27π 2n .
(5) Substituting (5) into (4) yields (3).
Finally, after studying various scenarios in Section 4, there are still some related open questions. One such question is the following.
Question 8. Find the winning probability for Bob if Alice and Bob take turns to lose 1 chip (i.e., collect −1 chip) or collect 2 chips randomly and independently with equal probability.
13 6 Acknowledgement The authors would like to thank the referee for their extensive help and useful suggestions, especially on the dynamic programming and numerical approximation of the probabilities.
We would also like to thank Brooks Emerick for his help with the approximation of the Laplace integral. The second author, Jiao Xu, was supported by the Carole and Ray Neag Undergraduate Research Fund at Kutztown University of Pennsylvania.
References Carl M. Bender and Steven A. Orszag, Advanced Mathematical Methods for Scientists and Engineers: Asymptotic Methods and Perturbation Theory, Springer, 1999.
Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for Your Mathematical Plays, Academic Press, 1982.
Charles L. Bouton, Nim, a game with a complete mathematical theory, Ann. of Math.
(2) 3 (1901), 35–39.
Allen J. Schwenk, Take-away games, Fibonacci Quart. 8 (1970), 225–234.
2010 Mathematics Subject Classification: Primary 97A20; Secondary 11Y55, 05A10.
Keywords: take-away game, probability.
(Concerned with sequences A265919 and A265920.) Received May 11 2016; revised versions received May 30 2017; January 21 2018. Published in Journal of Integer Sequences, August 22 2018.
Return to Journal of Integer Sequences home page.
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in a previous video we defined this random variable X it's a discret random variable it can only take on a finite number of values and I defined it as the number of workouts I might do in a week and we calculated the expected value of our random variable X which you could also denote as the mean of X and we use the Greek letter mu which we use for population mean and all we did is it's the probability weighted sum of the various outcomes and we got for this random variable with this probability distribution we got an expected value or a mean of 2.1 what we're going to do now is extend this idea to measuring spread and so we're going to think about what is the variance of this random variable and then we could take the square root of that to find out what is the standard deviation the way we are going to do this has parallels with the way that we've calculated variance in the past so the variance of our random variable x what we're going to do is take the difference between each outcome and the mean square that difference and then we're going to multiply it by the probability of that outcome so for example for this first data point you're going to have 0 - 2.1 squared times the probability of getting 0 time 0.1 then you are going to get Plus + 1 - 2.1 2 time the probability that you get 1 0.15 then you're going to get + 2 - 2.1 squared times the probability that you get a two 0.4 then you have plus 3 - 2.1 2ar 0.25 and then last but not least you have + 4 - 2.1 sared times 0.1 so once again the difference between each outcome and the mean we Square it and we multiply times the probability of that outcome so this is going to be -2.1 sared which is just 2.1 squar so I'll just write this as 2.1 sared time .1 that's the first term and then we're going to have plus 1 - 2.1 is -1.1 and then we're going to square that so that's just going to be the same thing as 1.1 squared which is 1.21 but I'll just write it out 1.1 squared time15 and then this is going to be 2 minus 2.1 is .1 when you you square it is going to be equal to so plus .1 if you have .1 .1 that's 01 0.4 times4 and then plus we this is going to be 0.9 squared so that is 81 time2 25 and then we're almost there this is going to be plus 1.9 2ar 1.9 2ar time .1 and we get 1.19 so this is all going to be equal to 1.19 and if we want to get the standard deviation for this random variable and we would denote that with the Greek letter Sigma the standard deviation for the random variable X is going to be equal to the square root of the variance the square root of 1.19 which is equal to let's just get the calculator back here so we are just going to take the square root of what we just I'll just type it again 1.19 and that gives us so it's approximately 1.09 approximately 1.09 so let's see if this makes sense me put this all on a number line right over here so you have the outcome 0o 1 two three and four so you have a 10% chance of getting a zero so I will draw that like this let's just say this is a height of 10% you have a 15% chance of getting one so that'll be one and a half times higher so it look something like this this you have a 40% chance of getting a two so that's going to be like this so you get a 40% chance of getting a two you have a 25% chance of getting a three so it look like this and then you have a 10% chance of getting a four so it look like that so this is a visualization of this discrete probability distribution where I didn't draw the vertical axis here but this would be 0.1 this would be 0.15 this is 0.25 and that is0 4 and then we see that the mean is at 2.1 the mean is the mean is at 2.1 which makes sense even though this random variable only takes on Integer values you can have a mean that takes on a non- integer value and then the standard deviation is 1.09 so 1.09 above the mean is going to get get us close to 3.2 and 1.09 below the mean is going to get us close to one and so this all at least intuitively feels reasonable this mean does seem to be indicative of the central tendency of this distribution and the standard deviation does seem to be a decent measure of the spread |
17329 | https://math.stackexchange.com/questions/4763918/determining-convergence-for-sequence-a-1-3-a-n1-sqrt3a-n-2 | real analysis - Determining convergence for sequence $a_1 = 3, \ a_{n+1} = \sqrt{3a_n - 2}$ - Mathematics Stack Exchange
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Determining convergence for sequence a 1=3,a n+1=3 a n−2−−−−−−√a 1=3,a n+1=3 a n−2
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Problem
Given the sequence where a 1=3 a 1=3 and a n+1=3 a n−2−−−−−−√a n+1=3 a n−2, show that it converges, and find the limit lim n→∞a n lim n→∞a n.
Progress
The way we are taught to solve these, is to say that if we let x=lim n→∞a n x=lim n→∞a n
lim n→∞a n+1=3 x−2−−−−−−√=x lim n→∞a n+1=3 x−2=x
since a n+1=a n a n+1=a n as n→∞n→∞.
Solving this equation, we get that x=1∨x=2 x=1∨x=2, but from there I have to figure out which.
A quick Python-script seems to show that it progresses asymptotically towards 2, and no lower. This is no proof, but a basis for a hypothesis at least.
Induction?
My next idea is a proof by induction, which summarizes as:
Hypothesis: n≥1⇒a n>2 n≥1⇒a n>2. Holds for n=1,2 n=1,2 by trial.
Then I want to show that a n>2⇒a n+1>2 a n>2⇒a n+1>2, which is to say that if 3 a n−1−2−−−−−−−−√>2 3 a n−1−2>2 then 3 a n−2−−−−−−√>2 3 a n−2>2.
For the latter inequality, we can expand: 3 a n−2−−−−−−√=3 3 a n−1−2−−−−−−−−√−2−−−−−−−−−−−−−−√=9 a n−1−8−−−−−−−−√=3 a n−1−2−−−−−−−−√⋅9 a n−1−8 3 a n−1−2−−−−−−√>2 3 a n−2=3 3 a n−1−2−2=9 a n−1−8=3 a n−1−2⋅9 a n−1−8 3 a n−1−2>2
Here is where I need some validation. In this final product, the first factor is greater than 2 by the initial trials, and the second factor is greater than 1 for all a n−1>1 a n−1>1 (since it has a root at 8/9, and is strictly increasing).
Question
Does this proof hold? And are there other, simpler ways to conclude the same? Or was the hypothesis wrong in the first place?
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asked Sep 5, 2023 at 12:40
AlecAlec
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1 This seems incomplete. Proving that a n>2 a n>2 certainly eliminates x=1 x=1 as a possible solution, but it doesn't prove that x=2 x=2 is the solution. After all, the limit might not exist. The usual approach is to show that {a n}{a n} decreases and is bounded below (you've already proven it is bounded below). That proves existence.lulu –lulu 2023-09-05 12:45:14 +00:00 Commented Sep 5, 2023 at 12:45
Related: math.stackexchange.com/q/2708229/42969Martin R –Martin R 2023-09-05 12:47:09 +00:00 Commented Sep 5, 2023 at 12:47
1 Note also that f(x)=3 x−2−−−−−√f(x)=3 x−2 is a strictly function, so a n>2 a n>2 implies a n+1=f(a n)>f(2)=2 a n+1=f(a n)>f(2)=2.Martin R –Martin R 2023-09-05 12:49:16 +00:00 Commented Sep 5, 2023 at 12:49
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a 1=3 a 1=3 and a 2=7–√≤a 1 a 2=7≤a 1, assume if a k+1≤a k a k+1≤a k then
3 a k+1≤3 a k⟹3 a k+1−2≤3 a k−2⟹(3 a k+1−2)−−−−−−−−−√≤(3 a k−2)−−−−−−−−√⟹a k+2≤a k+1 3 a k+1≤3 a k⟹3 a k+1−2≤3 a k−2⟹(3 a k+1−2)≤(3 a k−2)⟹a k+2≤a k+1
so by induction, a n a n is decreasing.
if a 1=3≥2 a 1=3≥2 assume a k≥2⟹3 a k−2≥4⟹a k+1≥2.a k≥2⟹3 a k−2≥4⟹a k+1≥2. So a n a n is monotonically decreasing and bounded below so by the Monotone convergent theorem it is convergent.
Let lim n→∞(a n+1)=l=lim n→∞(a n)⟹l=2.lim n→∞(a n+1)=l=lim n→∞(a n)⟹l=2.
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edited Sep 5, 2023 at 16:09
answered Sep 5, 2023 at 12:59
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In the second part, you're using ≤≤ everywhere, but did you mean to use ≥≥?Alec –Alec 2023-09-05 14:28:13 +00:00 Commented Sep 5, 2023 at 14:28
@Alec oops it's just by mistake. Thanks Maths –Maths 2023-09-05 15:09:59 +00:00 Commented Sep 5, 2023 at 15:09
A note on MathJax: use \sqrt{3a_k-2} to get 3 a k−2−−−−−−√3 a k−2 jjagmath –jjagmath 2023-09-05 15:26:39 +00:00 Commented Sep 5, 2023 at 15:26
Thanks! I hadn't heard of that theorem by name before. Nice to have it validated.Alec –Alec 2023-09-05 21:19:57 +00:00 Commented Sep 5, 2023 at 21:19
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The question can be put in a general scheme, resembling the Banach fixed point theorem. This approach implies the exponential rate of the convergence of the sequence to its limit.
Let a n+1=f(a n),a n+1=f(a n), where 0<f′(x)≤r<1 0<f′(x)≤r<1 for c<x<d.c<x<d. Assume f(x 0)=x 0 f(x 0)=x 0 for some c<x 0<d.c<x 0<d. If c<a 1<d c<a 1<d then c<a n<d c<a n<d and a n a n is convergent to x 0.x 0.
Indeed, there are two cases a 1∈(x 0,d)a 1∈(x 0,d) or a 1∈(c,x 0].a 1∈(c,x 0]. We will consider the first case, the second can be dealt with similarly. We will prove by induction that a n∈(x 0,d).a n∈(x 0,d). Assume a n∈(x 0,d).a n∈(x 0,d). We have
a n+1−x 0=f(a n)−f(x 0)=f′(ξ n)(a n−x 0)≤r(a n−x 0)(∗)a n+1−x 0=f(a n)−f(x 0)=f′(ξ n)(a n−x 0)≤r(a n−x 0)(∗)
As f′>0 f′>0 we get a n+1>x 0.a n+1>x 0. Moreover
a n+1≤r a n+(1−r)x 0<d a n+1≤r a n+(1−r)x 0<d
This concludes the induction step.
The inequality (∗)(∗) implies
0<a n+1−x 0≤r n(a 1−x 0)0<a n+1−x 0≤r n(a 1−x 0)
Thus lim a n=x 0.lim a n=x 0.
In the OP case f(x)=3 x−2−−−−−−√f(x)=3 x−2, c=3 2 c=3 2 , d=∞d=∞ and x 0=2.x 0=2. Moreover
0<f′(x)=3 2 3 x−2−−−−−−√<3 2 2.5−−−√=:r<1 0<f′(x)=3 2 3 x−2<3 2 2.5=:r<1
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edited Sep 6, 2023 at 19:26
answered Sep 5, 2023 at 13:51
Ryszard SzwarcRyszard Szwarc
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0Show by induction that {a n}∞n=1{a n}n=1∞ is a decreasing sequence
Related
5Proving by induction that the sequence a n+1=3 a n−1−−−−−−√a n+1=3 a n−1 is increasing
1Convergence of root sequence.
1Proof of the sequence a n+1=4 a n 3 a n+3 a n+1=4 a n 3 a n+3 is limited.
0Show by induction that {a n}∞n=1{a n}n=1∞ is a decreasing sequence
2Understanding the proof: If a 1=1 a 1=1 and a n+1=a 1+a 2+⋯+a n−−−−−−−−−−−−−−−√.a n+1=a 1+a 2+⋯+a n. Then, lim n→∞a n n=1 2 lim n→∞a n n=1 2
12Given the sequence a 1=1 a 1=1,a n+1=1+n a n a n+1=1+n a n, does the sequence a n+n−a 2 n a n+n−a n 2 converges?
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Biology for AP® Courses
22.2 Structure of Prokaryotes
Biology for AP® Courses22.2 Structure of Prokaryotes
Contents Contents
Highlights
Table of contents
Preface
The Chemistry of Life
The Cell
Genetics
Evolutionary Processes
Biological Diversity
21 Viruses
22 Prokaryotes: Bacteria and Archaea
Introduction
22.1 Prokaryotic Diversity
22.2 Structure of Prokaryotes
22.3 Prokaryotic Metabolism
22.4 Bacterial Diseases in Humans
22.5 Beneficial Prokaryotes
Key Terms
Chapter Summary
Review Questions
Critical Thinking Questions
Test Prep for AP®Courses
Science Practice Challenge Questions
Plant Structure and Function
Animal Structure and Function
Ecology
A | The Periodic Table of Elements
B | Geological Time
C | Measurements and the Metric System
Index
Search for key terms or text.
Close
Learning Objectives
In this section, you will explore the following questions:
What are similarities in the structures of the prokaryotes, Archaea and Bacteria?
What are examples of structural differences between Archaea and Bacteria?
Connection for AP® Courses
Domains Archaea and Bacteria contain single-celled organisms lacking a nucleus and other membrane-bound organelles. The two groups have substantial biochemical and structural differences. Most have a cell wall external to the plasma cell membrane, the composition of which can vary among groups, and many have additional structures such as flagella and pili. Prokaryotes also have ribosomes, where protein synthesis occurs. For the purpose of AP®, you do not have to memorize the various groups of bacteria. You should, however, be able to distinguish between prokaryotes and eukaryotes and know the domains.
Teacher Support
Provide students with multiple opportunities to summarize the similarities and differences between prokaryotic and eukaryotic cells and between cells in the three domains (Eukarya, Archaea, Bacteria). You may wish to ask students to sketch typical cells of each class or domain, create tables comparing and contrasting the cellular and genomic organization in each, or complete other short activities. When discussing similarities and differences, be sure to offer or ask for qualifying details where it makes sense to do so. (For example, cell walls are found in prokaryotes and some eukaryotes; the material of which they are made is quite different.)
When reviewing prokaryotic reproduction, take time to connect new information to students’ previous knowledge. For example, remind students of the importance of genetic diversity as discussed in chapters on evolutionary theory. Emphasize that although new mutations are a major source of variation (as they learned in previous chapters), additional diversity arises in prokaryotic populations from genetic recombination. Stress that while eukaryotes carry out the sexual processes of meiosis and fertilization that combine DNA from two individuals, prokaryotes uses other processes (transformation, transduction, and conjugation) to bring together DNA from different individuals. You may wish to ask students to consider the advantages of several modes of genetic recombination for a population.
Information presented and the examples highlighted in the section support concepts outlined in Big Idea 2 and Big Idea 3 of the AP® Biology Curriculum Framework. The AP® Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A learning objective merges required content with one or more of the seven science practices.
Big Idea 2Biological systems utilize free energy and molecular building blocks to grow, to reproduce, and to maintain dynamic homeostasis.
Enduring Understanding 2.BGrowth, reproduction and dynamic homeostasis require that cell create and maintain internal environments that are different form their external environment.
Essential Knowledge2.B.3Archaea and Bacteria generally lack internal membranes and organelles.
Science Practice1.4The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively.
Learning Objective2.14The student is able to use representations and models to describe differences in prokaryotic and eukaryotic cells.
Big Idea 3Living systems store, retrieve, transmit and respond to information essential to life processes.
Enduring Understanding 3.CThe processing of genetic information is imperfect and is a source of genetic variation.
Essential Knowledge3.C.2Prokaryotes contain circular chromosomes and plasmid DNA.
Science Practice6.2The student can construct explanations of phenomena based on evidence produced through scientific practices.
Learning Objective3.27The student is able to compare and contrast processes by which genetic variation is produced and maintained in organisms from multiple domains.
Essential Knowledge3.C.2Prokaryotes contain circular chromosomes and plasmid DNA.
Science Practice7.2The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas.
Learning Objective3.28The student is able to construct an explanation of the multiple processes that increase variation within a population.
The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards:
[APLO 2.5][APLO 2.13][APLO 2.14][APLO 4.9]
There are many differences between prokaryotic and eukaryotic cells. However, all cells have four common structures: the plasma membrane, which functions as a barrier for the cell and separates the cell from its environment; the cytoplasm, a jelly-like substance inside the cell; nucleic acids, the genetic material of the cell; and ribosomes, where protein synthesis takes place. Prokaryotes come in various shapes, but many fall into three categories: cocci (spherical), bacilli (rod-shaped), and spirilli (spiral-shaped) (Figure 22.9).
Figure 22.9 Prokaryotes fall into three basic categories based on their shape, visualized here using scanning electron microscopy: (a) cocci, or spherical (a pair is shown); (b) bacilli, or rod-shaped; and (c) spirilli, or spiral-shaped. (credit a: modification of work by Janice Haney Carr, Dr. Richard Facklam, CDC; credit c: modification of work by Dr. David Cox; scale-bar data from Matt Russell)
The Prokaryotic Cell
Recall that prokaryotes (Figure 22.10) are unicellular organisms that lack organelles or other internal membrane-bound structures. Therefore, they do not have a nucleus but instead generally have a single chromosome—a piece of circular, double-stranded DNA located in an area of the cell called the nucleoid. Most prokaryotes have a cell wall outside the plasma membrane.
Figure 22.10 The features of a typical prokaryotic cell are shown.
Recall that prokaryotes are divided into two different domains, Bacteria and Archaea, which together with Eukarya, comprise the three domains of life (Figure 22.11).
Figure 22.11 Bacteria and Archaea are both prokaryotes but differ enough to be placed in separate domains. An ancestor of modern Archaea is believed to have given rise to Eukarya, the third domain of life. Archaeal and bacterial phyla are shown; the evolutionary relationship between these phyla is still open to debate.
The composition of the cell wall differs significantly between the domains Bacteria and Archaea. The composition of their cell walls also differs from the eukaryotic cell walls found in plants (cellulose) or fungi and insects (chitin). The cell wall functions as a protective layer, and it is responsible for the organism’s shape. Some bacteria have an outer capsule outside the cell wall. Other structures are present in some prokaryotic species, but not in others (Table 22.2). For example, the capsule found in some species enables the organism to attach to surfaces, protects it from dehydration and attack by phagocytic cells, and makes pathogens more resistant to our immune responses. Some species also have flagella (singular, flagellum) used for locomotion, and pili (singular, pilus) used for attachment to surfaces. Plasmids, which consist of extra-chromosomal DNA, are also present in many species of bacteria and archaea.
Characteristics of phyla of Bacteria are described in Figure 22.12 and Figure 22.13; Archaea are described in Figure 22.14.
Figure 22.12 Phylum Proteobacteria is one of up to 52 bacteria phyla. Proteobacteria is further subdivided into five classes, Alpha through Epsilon. (credit “Rickettsia rickettsii”: modification of work by CDC; credit “Spirillum minus”: modification of work by Wolframm Adlassnig; credit “Vibrio cholera”: modification of work by Janice Haney Carr, CDC; credit “Desulfovibrio vulgaris”: modification of work by Graham Bradley; credit “Campylobacter”: modification of work by De Wood, Pooley, USDA, ARS, EMU; scale-bar data from Matt Russell)
Figure 22.13 Chlamydia, Spirochetes, Cyanobacteria, and Gram-positive bacteria are described in this table. Note that bacterial shape is not phylum-dependent; bacteria within a phylum may be cocci, rod-shaped, or spiral. (credit “Chlamydia trachomatis”: modification of work by Dr. Lance Liotta Laboratory, NCI; credit “Treponema pallidum”: modification of work by Dr. David Cox, CDC; credit “Phormidium”: modification of work by USGS; credit “Clostridium difficile”: modification of work by Lois S. Wiggs, CDC; scale-bar data from Matt Russell)
Figure 22.14 Archaea are separated into four phyla: the Korarchaeota, Euryarchaeota, Crenarchaeota, and Nanoarchaeota. (credit “Halobacterium”: modification of work by NASA; credit “Nanoarchaeotum equitans”: modification of work by Karl O. Stetter; credit “korarchaeota”: modification of work by Office of Science of the U.S. Dept. of Energy; scale-bar data from Matt Russell)
The Plasma Membrane
The plasma membrane is a thin lipid bilayer (6 to 8 nanometers) that completely surrounds the cell and separates the inside from the outside. Its selectively permeable nature keeps ions, proteins, and other molecules within the cell and prevents them from diffusing into the extracellular environment, while other molecules may move through the membrane. Recall that the general structure of a cell membrane is a phospholipid bilayer composed of two layers of lipid molecules. In archaeal cell membranes, isoprene (phytanyl) chains linked to glycerol replace the fatty acids linked to glycerol in bacterial membranes. Some archaeal membranes are lipid monolayers instead of bilayers (Figure 22.14).
Figure 22.15 Archaeal phospholipids differ from those found in Bacteria and Eukarya in two ways. First, they have branched phytanyl sidechains instead of linear ones. Second, an ether bond instead of an ester bond connects the lipid to the glycerol.
The Cell Wall
The cytoplasm of prokaryotic cells has a high concentration of dissolved solutes. Therefore, the osmotic pressure within the cell is relatively high. The cell wall is a protective layer that surrounds some cells and gives them shape and rigidity. It is located outside the cell membrane and prevents osmotic lysis (bursting due to increasing volume). The chemical composition of the cell walls varies between archaea and bacteria, and also varies between bacterial species.
Bacterial cell walls contain peptidoglycan, composed of polysaccharide chains that are cross-linked by unusual peptides containing both L- and D-amino acids including D-glutamic acid and D-alanine. Proteins normally have only L-amino acids; as a consequence, many of our antibiotics work by mimicking D-amino acids and therefore have specific effects on bacterial cell wall development. There are more than 100 different forms of peptidoglycan. S-layer (surface layer) proteins are also present on the outside of cell walls of both archaea and bacteria.
Bacteria are divided into two major groups: Gram positive and Gram negative, based on their reaction to Gram staining. Note that all Gram-positive bacteria belong to two phyla (Firmicutes and Actinobacteria); bacteria in the other phyla (Proteobacteria, Chlamydias, Spirochetes, Cyanobacteria, and others) are Gram-negative. The Gram staining method is named after its inventor, Danish scientist Hans Christian Gram (1853–1938). The different bacterial responses to the staining procedure are ultimately due to cell wall structure. Gram-positive organisms typically lack the outer membrane found in Gram-negative organisms (Figure 22.15). Up to 90 percent of the cell wall in Gram-positive bacteria is composed of peptidoglycan, and most of the rest is composed of acidic substances called teichoic acids. Teichoic acids may be covalently linked to lipids in the plasma membrane to form lipoteichoic acids. Lipoteichoic acids anchor the cell wall to the cell membrane. Gram-negative bacteria have a relatively thin cell wall composed of a few layers of peptidoglycan (only 10 percent of the total cell wall), surrounded by an outer envelope containing lipopolysaccharides (LPS) and lipoproteins. This outer envelope is sometimes referred to as a second lipid bilayer. The chemistry of this outer envelope is very different, however, from that of the typical lipid bilayer that forms plasma membranes.
Visual Connection
Figure 22.16 Bacteria are divided into two major groups: Gram positive and Gram negative. Both groups have a cell wall composed of peptidoglycan: in Gram-positive bacteria, the wall is thick, whereas in Gram-negative bacteria, the wall is thin. In Gram-negative bacteria, the cell wall is surrounded by an outer membrane that contains lipopolysaccharides and lipoproteins. Porins are proteins in this cell membrane that allow substances to pass through the outer membrane of Gram-negative bacteria. In Gram-positive bacteria, lipoteichoic acid anchors the cell wall to the cell membrane. (credit: modification of work by "Franciscosp2"/Wikimedia Commons)
Which of the following statements is true?
Gram-positive bacteria have a cell wall anchored to the cell membrane by lipoteichoic acid.
Porins allow entry of substances into both Gram-positive and Gram-negative bacteria.
The cell wall of Gram-negative bacteria is thick, and the cell wall of Gram-positive bacteria is thin.
Gram-negative bacteria have a cell wall made of peptidoglycan, whereas Gram-positive bacteria have a cell wall made of lipoteichoic acid.
Archaean cell walls do not have peptidoglycan. There are four different types of Archaean cell walls. One type is composed of pseudopeptidoglycan, which is similar to peptidoglycan in morphology but contains different sugars in the polysaccharide chain. The other three types of cell walls are composed of polysaccharides, glycoproteins, or pure protein.
Structural Differences and Similarities between Bacteria and Archaea
| Structural Characteristic | Bacteria | Archaea |
---
| Cell type | Prokaryotic | Prokaryotic |
| Cell morphology | Variable | Variable |
| Cell wall | Contains peptidoglycan | Does not contain peptidoglycan |
| Cell membrane type | Lipid bilayer | Lipid bilayer or lipid monolayer |
| Plasma membrane lipids | Fatty acids | Phytanyl groups |
Table 22.2
Reproduction
Reproduction in prokaryotes is asexual and usually takes place by binary fission. Recall that the DNA of a prokaryote exists as a single, circular chromosome. Prokaryotes do not undergo mitosis. Rather the chromosome is replicated and the two resulting copies separate from one another, due to the growth of the cell. The prokaryote, now enlarged, is pinched inward at its equator and the two resulting cells, which are clones, separate. Binary fission does not provide an opportunity for genetic recombination or genetic diversity, but prokaryotes can share genes by three other mechanisms.
In transformation, the prokaryote takes in DNA found in its environment that is shed by other prokaryotes. If a nonpathogenic bacterium takes up DNA for a toxin gene from a pathogen and incorporates the new DNA into its own chromosome, it too may become pathogenic. In transduction, bacteriophages, the viruses that infect bacteria, sometimes also move short pieces of chromosomal DNA from one bacterium to another. Transduction results in a recombinant organism. Archaea are not affected by bacteriophages but instead have their own viruses that translocate genetic material from one individual to another. In conjugation, DNA is transferred from one prokaryote to another by means of a pilus, which brings the organisms into contact with one another. The DNA transferred can be in the form of a plasmid or as a hybrid, containing both plasmid and chromosomal DNA. These three processes of DNA exchange are shown in Figure 22.17.
Reproduction can be very rapid: a few minutes for some species. This short generation time coupled with mechanisms of genetic recombination and high rates of mutation result in the rapid evolution of prokaryotes, allowing them to respond to environmental changes (such as the introduction of an antibiotic) very quickly.
Figure 22.17 Besides binary fission, there are three other mechanisms by which prokaryotes can exchange DNA. In (a) transformation, the cell takes up prokaryotic DNA directly from the environment. The DNA may remain separate as plasmid DNA or be incorporated into the host genome. In (b) transduction, a bacteriophage injects DNA into the cell that contains a small fragment of DNA from a different prokaryote. In (c) conjugation, DNA is transferred from one cell to another via a mating bridge that connects the two cells after the sex pilus draws the two bacteria close enough to form the bridge.
Evolution Connection
The Evolution of Prokaryotes
How do scientists answer questions about the evolution of prokaryotes? Unlike with animals, artifacts in the fossil record of prokaryotes offer very little information. Fossils of ancient prokaryotes look like tiny bubbles in rock. Some scientists turn to genetics and to the principle of the molecular clock, which holds that the more recently two species have diverged, the more similar their genes (and thus proteins) will be. Conversely, species that diverged long ago will have more genes that are dissimilar.
Scientists at the NASA Astrobiology Institute and at the European Molecular Biology Laboratory collaborated to analyze the molecular evolution of 32 specific proteins common to 72 species of prokaryotes.2 The model they derived from their data indicates that three important groups of bacteria—Actinobacteria, Deinococcus, and Cyanobacteria (which the authors call Terrabacteria)—were the first to colonize land. (Recall that Deinococcus is a genus of prokaryote—a bacterium—that is highly resistant to ionizing radiation.) Cyanobacteria are photosynthesizers, while Actinobacteria are a group of very common bacteria that include species important in decomposition of organic wastes.
The timelines of divergence suggest that bacteria (members of the domain Bacteria) diverged from common ancestral species between 2.5 and 3.2 billion years ago, whereas archaea diverged earlier: between 3.1 and 4.1 billion years ago. Eukarya later diverged off the Archaean line. Stromatolites are some of the oldest fossilized organisms on Earth at around 3.5 million years ago. There is evidence that these prokaryotes were also some of the first photosynthesizes on Earth. In fact, bacterial prokaryotes were likely responsible for the first accumulation of oxygen in our atmosphere through photosynthesis. The group Terrabacteria possessed many adaptations for living on land, such as resistance to drying. Some of these adaptations were also related to photosynthesis, such as compounds that protect cells from excess light. These early prokaryotic pathways related to photosynthesis were the foundation for photosynthesis in eukaryotic cells. This is evidenced by the similarity in structure and function between some photosynthetic prokaryotes and eukaryotic chloroplasts.
Science Practice Connection for AP®Courses
Think About It
What features and metabolic processes do all cells, both prokaryotes and eukaryotes, have in common? How do prokaryotes and eukaryotes differ?
Teacher Support
Students should identify similarities between prokaryotes and eukaryotes such as the use of DNA, ribosomes, and ATP. Prokaryotes and eukaryotes are different in that prokaryotes do not have membrane-bound organelles such as mitochondria and nuclei. Both questions above are applications of AP® Learning Objective 2.14 and Science Practice 1.4 because students are asked to describe differences and similarities in prokaryotic and eukaryotic cells in addition to differences between bacteria and archaea.
Footnotes
2Battistuzzi, FU, Feijao, A, and Hedges, SB. A genomic timescale of prokaryote evolution: Insights into the origin of methanogenesis, phototrophy, and the colonization of land. BioMed Central: Evolutionary Biology 4 (2004): 44, doi:10.1186/1471-2148-4-44.
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17331 | https://math.usu.edu/schneit/StatsStuff/Inference/confidenceintervals.html | Statistical Inference
Everything we care about lies somewhere in the middle, where pattern and randomness interlace.
What is Statistics?
Is Stats Math?
Data Collection Introduction
Controlled Experiments
Observational Studies
Descriptive Statistics Introduction
Categorical Data Summary
Histograms
Measures of Center
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Bivariate Data
Bivariate Data 2
Correlation
Linear Regression
Probability Introduction
Categorical Data Summary
Histograms
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Bivariate Data
Correlation
Linear Regression
Random Variables Introduction
Discrete RV's
Continuous RV's
EV and Var
RV Combinations
Discrete RV Families
Continuous RV Families
The Normal Distribution
The Central Limit Theorem
Normal Approximations
Inference Introduction
Sampling Distributions
Estimation
Confidence Intervals
Hypothesis Tests Introduction
Hypothesis Tests Process
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Numerical Data Summary
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Confidence Intervals
On Memorial Day, 2020, Netflix had 834 movies in its library. An estimate of the mean run time of these movies, based on a sample of size 32 is 104.5
minutes with a standard error of 3.2 minutes. A 95% confidence interval for the mean run time of all 834 movies available on that date is (98, 111) minutes.
A (1−α)100%(1−α)100% confidence interval gives a set of plausible values for a parameter. It is formed from the point estimate
and a margin of error associated with a particular level of confidence.
Confidence intervals that are based on symmetric distributions such as the normal or t distributions usually
have the same basic form: point estimate ±± the margin of error.
The margin of error is determined by the standard error and a desired level of confidence.
Many confidence intervals have the form point estimate ±± the margin of error.
A 95% confidence interval for the mean runtime of Netflix movies is (98, 111). The point estimate obtained from the
sample is ˉX=104.5¯X=104.5 minutes and the margin of error is 6.5 minutes. 104.5±6.5:(98,111)104.5±6.5:(98,111)
The confidence level (1−α)100%(1−α)100% indicates the probability that the interval will cover the parameter of interest. The most common confidence level
is 95% in which case α=0.05α=0.05: (1−0.05)100%=95%(1−0.05)100%=95%. 90% and 99% are also common confidence levels but the level can be anything the researcher chooses. 100% confidence
levels are possible but usually not at all useful. We'll discuss this further when we talk about interpretation below.
The Confidence level, (1−α)100%(1−α)100%, specifies the probability that the confidence interval covers the true parameter value.
Constructing Confidence Intervals
A confidence interval for a mean has the form introduced previously: point estimate ±± the margin of error. The margin of error itself is
composed of a critical value that gives the desired confidence level, and the standard error thus the general form of the confidence interval can be re-written as:
point estimate ±(critical value)×(standard error)point estimate ±(critical value)×(standard error)
Point Estimate
The point estimate is the observed value of the statistic computed from the data. To construct a confidence interval for μμ, the point
estimate is the observed value of the sample mean, ˉx¯x. Note that the lowercase x indicates that this is an observed value rather than a random variable.
The point estimate for the mean runtime of movies available from Netflix, μμ, was computed from a random sample of 32 movies.
The data are shown here.
The point estimate for the mean runtime is ˉx=104.5¯x=104.5 minutes.
Critical Values
A critical value indicates how many standard errors either side of the mean a confidence interval must extend in order to obtain the desired level
of confidence. It's value depends on the distribution of the relevant statistic.
When computing a confidence interval for a population mean, the
relevant statistic is the standardized sample mean, ˉX−μs.e.(ˉX)¯X−μs.e.(¯X).
In the discussion of
sampling distributions
we learned that the distribution of this standarized statistic depends on the population distribution and whether the population variance is known.
If the population is normally distributed and σ2σ2 is known ˉX−μs.e.(ˉX)=ˉX−μσ/√n∼N(0,1)¯X−μs.e.(¯X)=¯X−μσ/√n∼N(0,1)
If the population is not normally distributed, but the sample size is large:
If σ2σ2 is known ˉX−μs.e.(ˉX)=ˉX−μσ/√n∼N(0,1)¯X−μs.e.(¯X)=¯X−μσ/√n∼N(0,1).
If σ2σ2 is unknown ˉX−μs.e.(ˉX)≈ˉX−μS/√n⋅∼N(0,1)¯X−μs.e.(¯X)≈¯X−μS/√n⋅∼N(0,1).
If the population is normally distributed, σ2σ2 is unknown, ˉX−μs.e.(ˉX)≈ˉX−μS/√(n)∼tn−1¯X−μs.e.(¯X)≈¯X−μS/√(n)∼tn−1
In most cases, the population variance is not known. If the distribution of the population is approximately normal, the t-distribution will be useful regardless of the
sample size (as the sample size increases, t→N(0,1)t→N(0,1)). However, if the population distribution is very different from normal, the
Central Limit Theorem
indicates that the distribution of ˉX−μS/√n¯X−μS/√n will be approximately standard normal if the sample size is large enough.
A critical value is a point under a distribution that marks a cut-off corresponding to a region with specified probability.
We denote a critical value as zαzα if ˉX−μs.e.(ˉX)∼N(0,1)¯X−μs.e.(¯X)∼N(0,1) and tα,νtα,ν if ˉX−μs.e.(ˉX)∼tν¯X−μs.e.(¯X)∼tν where αα
is the area to the right of the critical value under the relevant distribution. For our purposes, we'll usually think about Zα2Zα2 rather than ZαZα. If we
construct a two-sided (1−α)100%(1−α)100% confidence interval (one-sided intervals exist but are not as widely used), then αα is the probability that the interval
does not cover the parameter and half of this area should be on either end of the interval.
In the standard normal curve shown, the area to the right of zα2zα2 is α2α2. Since the curve is symmetric around 0,
the area to the left of −zα2−zα2 is also α2α2, that means that the area between those two points is 1−(2⋅α2)=1−α1−(2⋅α2)=1−α.
Drag the triangle at Zα2Zα2 to see how these values change.
In the tνtν curve shown, the area to the right of tα2,νtα2,ν is α2α2. Since the curve is symmetric around 0,
the area to the left of −tα2,ν−tα2,ν is also α2α2, that means that the area between those two points is 1−(2⋅α2)=1−α1−(2⋅α2)=1−α.
Drag the triangle at tα2,νtα2,ν and use the slider to adjust the degrees of freedom to see how these values change.
The population standard deviation, that is, the standard deviation of the runtime of all Netflix movies, is unknown but our sample size
is large enough that, from the Central Limit Theorem, we would expect the distribution of the sample mean to be approximately normal so we will use the normal distribution to
obtain a critical value. Since 95% is the most common confidence level, we will find the critical value for constructing a 95% confidence interval.
For a 95% confidence interval, α=1−0.95=0.05α=1−0.95=0.05, thus α2=0.025α2=0.025. Using the 'Normal Critical Values' applet above, we find that when
α2=0.025α2=0.025, zα2=1.96zα2=1.96.
Standard Error
We know from previous discussions that if a sample is drawn from a population with variance σ2σ2 then the standard error of ˉX¯X is
σ√nσ√n. Since σ2σ2 is generally unknown, we typically replace it with the estimated standard error S√nS√n.
Example: Compute the standard error needed to contruct a confidence interval for the mean runtime of Netflix movies.
Since σσ, the population standard deviation, is unknown we will use S, the sample standard deviation to find the estimated standard error of ˉX¯X.
From the data we find that s=18.22s=18.22. Since the sample size is 32, estimated s.e.(ˉX)=18.22√32=3.22(¯X)=18.22√32=3.22.
Putting It Together
A confidence interval for a population mean μμ has the form: point estimate ±(critical value)×(standard error)point estimate ±(critical value)×(standard error)
When the distribution of the population is approximately normal, the critical value can be obtained from a t-distribution so the (1−α)100%(1−α)100% confidence interval is
(ˉx−tα2,νs√n,ˉx+tα2,νs√n)(¯x−tα2,νs√n,¯x+tα2,νs√n)
If the population distribution is not normal, but the
sample size is large (around 30 is typically large enough) then the Central Limit Theorem justifies us in using the standard normal distribution to obtain the critical values thus
the form of a (1−α)100%(1−α)100% confidence interval for μμ is
(ˉx−zα2⋅s√(n),ˉx+zα2⋅s√(n))(¯x−zα2⋅s√(n),¯x+zα2⋅s√(n))
Note that, when the sample size is large, the critical values from a t-distribution will be approximately equal to those from a normal distribution. Thus, in practice,
the t-distribution is often used in both of the above cases to construct confidence intervals for a mean.
From the definition of a critical value,
P(−tα/2≤T≤tα/2)=1−α where T∼tn−1. Thus, since ˉX−μs/√n∼tn−1,
P(−tα/2≤ˉX−μs/√n≤tα/2)=1−α. The form of the
confidence interval, comes from solving this inequality for μ.
1−α=P(−tα/2≤ˉX−μs/√n≤tα/2)=P(−tα/2⋅s√(n)≤ˉX−μ≤tα/2⋅s√n)=P(−ˉX−tα/2⋅s√n≤−μ≤−ˉX+tα/2⋅s√n)=P(ˉX−tα/2⋅s√n)≤μ≤ˉX+tα/2⋅s√n)
The form of the other confidence intervals discussed in this section can be derived in a similar fashion.
For a sample of 10 cows that had recently given birth, the mean protein content in 40oz of milk was 3.4g with a sample standard
deviation of 0.4 g. Find a 95% confidence interval for the mean protein content of the milk.
Since the sample is small, use the t-distribution to obtain the critical value for the confidence interval.
For a 95% confidence interval, (1−α)⋅100%=95%, so α=0.05 and α/2=0.025
ˉx=3.4
s=0.4
t0.025,9=2.26
95% CI: (3.4−2.26⋅0.4√10,3.4+2.26⋅0.4√40)=(3.114,3.686).
On Memorial Day, 2020, Netflix had 834 movies in its library. A point estimate for the mean run time of these movies, based on a
sample of size 32 is 104.5 minutes with a standard error of 3.2 minutes. Show that a 95% CI for the mean runtime of the Netflix movies is (97,111).
For a 95% confidence interval, (1−α)⋅100%=95%, so α=0.05 and α/2=0.025
Since the sample size is larger than 30 and the population standard deviation is unknown, we'll use the normal distribution to obtain critical values.
ˉx=104.5
s.e.(ˉX)=s√32=3.2
z0.025=1.96
95% CI (rounded to the nearest minute): (104.5−1.96⋅3.2,104.5+1.96⋅3.2)=(98,111).
Note: Since the t-distribution converges to the normal distribution as the sample size increases, using critical values from the t-distribution
yields a similar result.
t0.025,31=2.04
95% CI (rounded to the nearest minute): (104.5−2.04⋅3.2,104.5+2.04⋅3.2)=(98,111).
Confidence Intervals for a Proportion
As with confidence intervals for a mean, a confidence interval for a proportion, p, has this familiar form: point estimate ±(critical value)×(standard error)
Point Estimate
A point estimate for a population proportion, p, is the observed value of the sample proportion, ˆp.
Critical Values
The critical value for a confidence interval for a proportion comes from the normal distribution, that is zα2.
The critical value depends
on the distribution of the the standardized sample proportion:
ˆp−p√p(1−p)n. The distribution of this standarized statistic is approximately standard normal when the
sample size is large.
Standard Error
The standard error of ˆp is √p(1−p)n. Notice that this depends on the value of the parameter. Since p is unknown
(or it wouldn't need to be estimated), use ˆp to find the estimated standard error: ˆp−p√ˆp(1−ˆp)n.
Putting it Together
When the sample size is large, a (1−α)⋅100% CI for a proportion is (ˆp−zα/2⋅√ˆp(1−ˆp)n,ˆp+zα/2⋅√ˆp(1−ˆp)n).
(ˆp−zα/2⋅√ˆp(1−ˆp)n,ˆp+zα/2⋅√ˆp(1−ˆp)n).
In 2009, the camera company, Nikon, released the results
of a survey called "Picture Yourself". For the survey, they obtained a sample of 1000 US adults. They found that the proportion of respondents said that they look better in person than they do in
photographs was 0.79. Find a 90% confidence interval for the true proportion of US adults who think they look better in person than in photographs.
For a 90% confidence interval, (1−α)⋅100%=90%, so α=0.1 and α/2=0.05
ˆp=0.79
z0.05=1.65
90% CI: (0.79−1.65⋅√0.79(0.21)1000,0.79+1.65⋅√0.79(0.21)1000)=(0.77,0.81).
Interpreting Confidence Intervals
A confidence interval is a set of plausible values for the parameter of interest.
From the derivation, we also saw that the confidence level
gives the probability that the interval covers the parameter. It is typical to say "we are 95% confident that the interval covers the parameter"
for a 95% confidence interval.
The applet below
illustrates another way to think about confidence intervals. If the process of sampling and computing a CI were repeated 100 times, we'd expect that about 95 (99, 90, etc.) of the resulting
95% (99%, 90%, etc.) confidence intervals would cover the true parameter.
Use the applet to simulate drawing 100 samples from the jar of balls and computing a CI interval for the proportion
of blue balls based on each sample.
Use the 'show p' checkbox to show the true proportion of blue balls. When this box is selected, the intervals that miss the parameter are colored red.
When the confidence level is set to 95, on average, 5 of the intervals will miss.
Consider:
How does the confidence level affect the width of the confident interval? Try changing the confidence level in the applet, how do the intervals change?
How does the sample size affect the width of the confident interval? Try changing the sample size in the applet and generating a new set of intervals,
how do these intervals compare?
Increasing the confidence level of a confidence interval widens the interval. When more values are admitted, there is a greater probability of that one of them is correct.
Increasing the confidence level results in wider confidence intervals.
Assuming the sample standard deviation remains the same, increasing the sample size results in a more narrow interval. The margin of error of a confidence interval
is the product of the critical value and the standard error. The standard error, in turn, is the standard deviation divided by the sample size thus the larger the sample size, the smaller the
standard error and the narrower the interval.
All else being equal, a larger sample size results in a narrower confidence interval.
Issues of Sample Size
Sometimes a researcher desires a certain level of precision in the confidence interval she constructs. For instance, she might like to have a
confidence interval for the protein in milk that is no more than 0.1 grams wide. How large would the sample need to be to attain this?
Each confidence interval on this page has the same basic form: point estimate ±margin of error. Since the margin of error
is subtracted from the point estimate to get the lower endpoint of the interval and added to the point estimate to get the upper endpoint, the width, W,
of the CI is twice the margin of error. That is W=2×margin of error=2×(critical value)×(standard error).
It is possible to use this fact to find the sample size needed to obtain a certain level of precision in the interval.
W=2×(critical value)×(standard error)
In a survey of 1000 adults, 0.79 said they think they look better in person than in photographs.
What sample size would we need to obtain a 95% confidence interval for the proportion of
adults who think they look better in person than in photographs that is no longer than 0.02?
W=2×(critical value)×(standard error)=2×zα2×(√ˆp(1−ˆp)n)
ˆp=0.79
z0.025=1.96
s.e.(ˆp)=√0.79(0.21)n
Thus 0.02=1.96⋅√0.79(0.21)n
Solving for n we get n=2048.5. Since we're talking about sampling people, we'll round up to the nearest whole number. That is, we'd need a sample of size 2049 to obatain a
95% confidence interval that is no longer than 0.02.
For a sample of 10 cows that had recently given birth, the mean protein content in 40oz of milk was 3.4g with a sample standard
deviation of 0.4 g. What sample size would we need to obtain a 95% confidence interval for the mean protein content that is 0.25g wide?
W=2×(critical value)×(standard error)=2×(tα2,n−1)×(s√n)
Solving the equation for n gives n=4(tα/2,n−1SW)2.
Notice that the critical value depends on n.
As n increases, the critical value decreases. In aprevious prompt, we found that a sample size of 10
yielded a CI with width 0.42 thus the sample size must be greater than 10 and the corresponding critical value will be smaller than
t0.025,9=2.26. So n=4(2.26(0.4)0.25)2=52.3. A sample of size 53 is sufficient.
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17332 | https://www.bakertilly.com/insights/revenue-recognition-uninstalled-materials | Have questions about the Moss Adams combination? We're here to help. Submit your inquiry.
Article
Revenue recognition: what to know about uninstalled materials
Feb. 3, 2022 · Authored by Michael C. Malinoski
Published in Construction Financial Management Association's January/February 2022 issue of Building Profits magazine.
When the Financial Accounting Standards Board (FASB) issued Revenue from Contracts with Customers (Topic 606), some of the accounting practices commonly used by contractors under ASC 605-35, which codified the previous guidance known as SOP 81-1, were no longer allowed. One such item of interest was the accounting for materials purchased or allocated to a project that was not yet installed — commonly referred to as uninstalled materials.
What qualifies as uninstalled materials?
Since contractors can, for the most part, recognize revenue under Topic 606 in a manner that is largely consistent with the percentage-of-completion method (PCM) of accounting under Topic 605 (reporting contract revenues based upon the ratio of the cost incurred on the project to the total estimated cost), some of the changes to revenue recognition addressed by Topic 606 have been occasionally overlooked. One of those changes relates to materials that have been purchased for a specific project but have not yet been installed. When examining the intricacies of Topic 606, it is easy to see how there can be a lack of understanding among construction contractors regarding what specifically qualifies as uninstalled materials.
One reason some contractors may not view uninstalled materials as a pertinent issue is because they typically buy materials as they use them. So when they apply the concepts of Topic 606, the amount of time it takes to install the material after it is received is so limited that the risk of a significant amount of uninstalled materials included in the cost-to-cost method is low. Also, most small- to mid-sized construction companies or independent contractors buy materials on a project-by-project basis and may view the uninstalled material guidelines as something that does not have a meaningful effect on their business since they already incurred the cost of the material.
However, in ASC 606-10-55-21, FASB specifically noted that a shortcoming of the PCM of accounting was that there may not be a direct relationship between the entity’s inputs (cost incurred) and the transfer of the good or service to the customer (satisfaction of the performance obligation). Certain types of contractors are more prone to this shortcoming than others.
A GC that subcontracts all the tasks in a project probably didn’t lose a lot of sleep thinking about Topic 606 and how to apply uninstalled materials in the PCM calculation; however, the contractor hired to build the elevator system probably did. While the purchasing agent who was able to get a deal on all the drywall isn’t thinking about the accounting ramifications, the financial manager should be thinking about whether the procurement of the drywall is:
Source: “Revenue from Contracts with Customers (Topic 606).” Financial Accounting Standards Board. May 2014. asc.fasb.org/imageRoot/00/51801400.pdf
## Michael C. Malinoski
Principal
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Cost incurred, which under Topic 606 is considered progress toward satisfying the performance obligation;
Cost incurred but uninstalled materials to exclude from the PCM calculation; or
Inventory to be used on future backlog and not yet a cost of the contract.
Therefore, it is important to thoroughly understand the details surrounding the material used to complete any construction project. When assessing how to account for this material, the financial manager should understand the specifications — not in an engineering sense — of the contract. No one is expecting the financial manager to run the project; however, the financial manager should know whether the contract is unique and how this uniqueness can affect the timing of the revenue recognition.
The financial manager and project manager should be able to answer where the material is located and who owns it. The material may be earmarked for a certain job, but if it’s sitting in a warehouse (or even on the jobsite), then who technically owns the material and is responsible for it at that point in time? Has the customer already paid for the material and taken control? Is the material being used on the job highly specialized, of a higher cost, or have any other unique specifications that can change how revenue may be recognized?
Sometimes the project’s contract clarifies these details, but many times it does not. In fact, it’s possible to have the same exact material and amounts on two different jobsites, yet the material needs to be handled differently for each scenario.
Uninstalled materials: who is affected?
Any contractor that needs to purchase substantial materials for projects could potentially be affected by the intricacies of Topic 606 relating to uninstalled materials. Subcontractors may especially be affected due to the amount of materials they have on hand to assist contactors on jobs.
Consider the case of an electrical contractor that buys a significant amount of wire or a mason who purchases a large volume of bricks. Contractors that purchase more material than they need at the moment could potentially have to make adjustments to the cost-to-cost calculations due to having uninstalled materials. Those contractors that purchase limited quantities of materials for a specific, short-term project are less likely to be impacted.
The intricacies of uninstalled materials: how revenue changes
Previously, when determining revenue recognition under Topic 605, a contractor would use all the costs incurred on the project, which ran through its job cost records and made progress to the extent the contract was complete. The uninstalled materials are not part of this cost and therefore excluded in revenue. However, under Topic 606, more information must be considered to determine how much progress has been made toward completion. In other words, how much of the contract value can be recognized as revenue? Sometimes the calculation is not as easy as it used to be.
Recognizing uninstalled materials in calculations
In order to calculate revenue under ASC 606-10-55-189 and 190, you must first understand that the amount of revenue to recognize is a combination of the progress completed and the uninstalled material. First, the contractor must identify the costs incurred but not yet installed and carve out the amount of uninstalled materials from the cost-to-date amount to determine costs incurred for the portion of revenue to recognize based upon the measurement of progress toward completion. Second, the cost of the uninstalled materials is added to determine the total amount of revenue recognized. Therefore, revenue recognized on uninstalled materials is recognized at a zero margin.
For a contractor with a significant amount of uninstalled materials, its work-in-progress (WIP) schedule can present job-to-date gross profit percentages that do not match the final estimated profit amount; this may take some getting used to. Some contractors may look at how to present their WIP schedules in a way that easily depicts the effects of uninstalled materials. Work with your accountant on the best way to present uninstalled materials on the WIP schedule.
A presentation that allows the users to recalculate the revenues earned on each project is recommended. Surety and banking relationships may have input on the presentation as well. And, overall, it is better to be upfront with the users than to explain at the last hour.
Example
To consider exactly how revenue and gross profit timing will change under Topic 606, consider a contractor that has a job with a final estimated 10% gross profit. Previously under ASC 605-35, the contractor would exclude the uninstalled materials from the cost-to-date to determine the percentage of total estimated cost complete and, consequently, the percentage of the total contract to recognize as revenue. Now, under ASC 606-10-55-189 and 190, uninstalled material is also excluded from the measurement of the portion of revenues recognized based upon the progress toward completion, but the cost is added to the revenue calculated without any margin. As such, a job with a 10% total gross profit may end up reporting a job-to-date gross profit of only 4%.
The example in Exhibit 1 demonstrates this scenario and how the changes can occur if uninstalled materials are present on a contract.
In that example, this job had $200,000 in costs incurred to date before uninstalled materials and a gross profit of $22,222 (see the Topic 605 row). However, not included in the cost is an additional $400,000 related to material acquired, which does not increase the amount of progress made toward job completion. As such, this cost is included on the balance sheet as inventory and is not run through the contract’s profitability.
Under Topic 606, these same costs were incurred, but the treatment of the $400,000 of uninstalled materials is different. This cost is added to revenue without a margin after the measurement of progress toward completion. In this case, the amount of revenue and cost are increased by the amount of uninstalled materials and the gross profit recognized changes. Acknowledging uninstalled materials leaves the total job profit unchanged, but it changes the revenue and profit recognition process.
Four criteria of Topic 606 for progress toward completion
The intent of Topic 606 is to eliminate the recognition of gross profit on costs incurred, which do not advance progress toward completion. Just because the material was obtained doesn’t mean the project is closer to completion. FASB developed four criteria (included in ASC 606-10-55-21) to determine when costs incurred are not proportionate to the entity’s progress toward completion.
Using the previous example, here is a review of these four criteria and how they would apply in practice.
The good is not distinctWhen looking at the example in Exhibit 1, the $400,000 of material is not distinct from the rest of the contract. Assuming the contractor in this example is an electrician, the material is necessary to complete the performance obligation. The material is part of an integrated set of activities to complete the electrical aspects of the contract.
The customer is expected to obtain control of the good significantly before receiving services related to the goodWhen exactly does the customer control the materials? The contractor needs to make this determination. For example, the contractor acquires the goods needed for the specific project and has billed the customer for this cost. In this situation, it would appear that the customer controls the material, but what if the contractor can receive a discount by buying goods in bulk and having the materials delivered to the project site? Judgment needs to be made on whether the customer controls the goods. If the contractor can move goods to another job due to a schedule change, then the customer probably doesn’t control the goods. In a case like this, it would appear that the costs qualify as inventoriable items under Topic 330, Inventory.
The cost of the transferred good is significantly relative to the total expected costs to completely satisfy the performance obligationIn Exhibit 1, $400,000 of the $600,000 in costs incurred to date under the contract are related to uninstalled materials. Having a ratio of 67% materials to total job costs appears to not be proportionate to the progress of the project toward completion. A good practice to use in analyzing jobs is to monitor the components of the cost on a completed job. If the contractor sees a trend where material on similar contracts is a consistent percentage of the job cost, then any job where the material cost percentage exceeds this trend is a good candidate to review for uninstalled material issues.
The entity procures the good from a third party and is not significantly involved in designing and manufacturing the goodA contractor that obtains parts for their customers for a project is probably not involved in the manufacturing. The contractor understands the specifications for the job and orders the materials; however, all the contractor did was order the material. They did not have a significant role in designing or manufacturing the materials.
There are certain contractors that have a manufacturing component to their operations, and they often incur significant material cost prior to the delivery and installation. As noted in Step 5 of Topic 606 regarding the contract process, revenue is recognized when (or as) the entity satisfies a performance obligation. In this situation, when does this type of contractor satisfy the performance obligation? How do they account for all of the material made for the project?
As previously noted in ASC 606-10-55-21(b), the fourth criteria indicates that the cost of materials is considered uninstalled materials based on whether the entity procures the goods from a third party and is not significantly designing and manufacturing the goods. A contractor that fabricates goods as part of a construction contract usually has both the engineering and manufacturing components. As such, these contractors are typically involved in the design and manufacturing of the materials to be installed as part of their performance obligation under the contract.
Therefore, all costs incurred to date that have gone through the manufacturing process are considered progress toward satisfying the performance obligation. Revenue recognition has become a hot topic in analyzing a company’s performance. To start, everyone usually looks to grow their business. For example, a contractor was able to double its revenue by simply changing its contracts. Previously, this contractor only contracted with the project owners to manage their projects. When looking at the job sites, this contractor was clearly managing the projects. Each of the trades working on the job contracted directly with the project owners as well. In this situation the contractor is only recognizing a fee to manage the job. The value of all of the subcontracts are not part of the contractor’s financial statements. Later, this contractor began to directly contract with the trades. This changes the entire value of the contract being recognized in the contractor’s financial statements. Now, all of the value of the subcontractors’ work is recognized on the contractor’s financial statements.
What was once an “agent-like” construction management contract became an “at-risk” construction contract, where the contractor was now acting as a principal in the arrangement. You can look at the jobsite under either of these scenarios and it will look the same. However, a simple change to the nature of the contract can substantially change the company’s revenue. This is one such example of how a contract can change the amount and timing of revenue recognition. A significant change in procurement practices can have an effect on the timing of revenue recognition for a contractor that is material intensive.
It is imperative to understand whether the cost of material that meets the requirements of ASC 606-10-55-21(b) should be excluded when determining progress toward completion. By not doing this, a contractor can misstate revenue if these costs happen to be significant.
Tax considerations
Although this article is focused on the issue of uninstalled materials and their application to recognition of revenue under Topic 606, there may also be some unintended tax consequences. Accelerating revenue recognition can increase taxable income, and a contractor can create an unplanned tax burden by trying to support the material cost as progress toward completion. A good practice that all contractors should follow is to review any changes to revenue recognition with their tax accountants. Not all U.S. GAAP changes are automatically applied to the Internal Revenue Code; it is better to know how these changes may affect your company rather than be surprised with an unexpected tax bill the day before your payment is due.
What should contractors do to monitor uninstalled materials?
Contractors should continually monitor their contracts and have a thorough understanding of what is considered truly uninstalled materials. Each type of contractor has different issues to consider, and to take it even further, each contract may have unique considerations. A GC that subcontracts most of its work probably does not have a significant risk of misstating revenue. However, the subcontractor that has material-intensive work should develop processes and procedures to quantify the effect of potential uninstalled material issues.
Contractors must consider if they are misstating profits on the job by choosing to leave uninstalled materials in their calculations. The reported profit needs to be in line with the contractor’s progress toward satisfying the performance obligation.
While performing financial statement audits since the adoption of Topic 606, Baker Tilly noted that uninstalled materials were an issue that required some attention. Although we did not see many material adjustments for uninstalled materials, the contractors that self-performed their work and were material intensive had to consider this issue. These contractors were able to support the nature of the work being done, and the materials purchased for jobs were installed so quickly following their arrival on the jobsite that any issues that may have existed were not material.
In some cases, it was noted that costs on specific projects seemed too high and not in line with expectations, but further analysis revealed the costs incurred were of such specialized material that the material portion of the cost structure was higher for those contracts. Essentially, these contractors did not blindly assume there were no issues to consider. Rather, they developed processes and procedures and were able to support their positions.
Some of the red flags that may help a contractor identify the existence of uninstalled materials are comparing the percentage of job cost for materials to the history of the type of contract and identifying large underbilled amounts on contracts.
No one wants to have a conversation regarding an inaccurate revenue amount. Whatever type of work is done by the contractor, the financial manager should work with their outside accountant to develop specific tools and processes to help mitigate the risk of improper revenue recognition.
Revenue recognition issues are different for each type of contractor; however, there are simple checks that financial professionals should put in place to ensure uninstalled materials are properly included in revenue recognition. These checks and balances are necessary to monitor for potential uninstalled materials and should not be an arduous task.. Remember, an ounce of prevention is worth a pound of cure.
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17333 | http://www.geom.uiuc.edu/~huberty/math5337/groupe/rearrange.html | Rearrangement Method
Rearrangement Method
Take a circle and cut it into four equal sectors. Then rearrange the sectors into a quasi-parallelogram as shown in the figure below. The area of the quasi-parallelogram can be (under)estimated by the area of the true parallelogram "inscribed" in the sectors. The "inscribed" parallelogram's area is easily found by the product of its width and height. This gives an estimate of exactly 2 for . It is apparent from the figure that this estimate will be poor, since much of the rearranged area remains outside of the "inscribed" parallelogram.
To get a better estimate, cut the circle into eight equal sectors and rearrange as shown in the figure below. It is apparent that the true parallelogram "inscribed" contains more of the circle's area than in the previous case. The estimate for is 2.829 in this case - better, but still not very accurate.
If the circle is cut into more sectors, the estimate for will improve since the rearranged sectors become more like a parallelogram. Try this using Geometer's Sketchpad rearrangment script. A Japanese document from 1689 uses this rearrangement method by slicing the circle into 32 sectors (shown below) [2, p.19]. What estimate for should the document have found?
Return to Historical Overview of Pi
Go to A Slice of Pi Home Page
Created: March 1996 ---- Last Modified: July 6, 1997
Copyright © 1996-1997 Michael D. Huberty, Ko Hayashi & Chia Vang |
17334 | https://www.theexpertta.com/book-files/OpenStaxBio2e/28.3%20Superphylum%20Lophotrochozoa%20-%20Flatworms,%20Rotifers,%20and%20Nemerteans.pdf | Figure 28.12 Hydrozoans. The polyp colony Obelia (a), siphonophore colonies Physalia (b) physalis, known as the Portuguese man o‘ war and Velella bae (c), and the solitary polyp Hydra (d) have different body shapes but all belong to the family Hydrozoa. (credit b: modification of work by NOAA; scale-bar data from Matt Russell) 28.3 | Superphylum Lophotrochozoa: Flatworms, Rotifers, and Nemerteans By the end of this section, you will be able to do the following: • Describe the unique anatomical and morphological features of flatworms, rotifers, and Nemertea • Identify an important extracoelomic cavity found in Nemertea • Explain the key features of Platyhelminthes and their importance as parasites Animals belonging to superphylum Lophotrochozoa are triploblastic (have three germ layers) and unlike the cnidarians, they possess an embryonic mesoderm sandwiched between the ectoderm and endoderm. These phyla are also bilaterally symmetrical, meaning that a longitudinal section will divide them into right and left sides that are superficially symmetrical. In these phyla, we also see the beginning of cephalization, the evolution of a concentration of nervous tissues and sensory organs in the head of the organism—exactly where a mobile 802 Chapter 28 | Invertebrates This OpenStax book is available for free at bilaterally symmetrical organism first encounters its environment.
Lophotrochozoa are also protostomes, in which the blastopore, or the point of invagination of the ectoderm (outer germ layer), becomes the mouth opening into the alimentary canal. This developmental pattern is called protostomy or “first mouth.” Protostomes include acoelomate, pseudocoelomate, and eucoelomate phyla. The coelom is a cavity that separates the ectoderm from the endoderm. In acoelomates, a solid mass of mesoderm is sandwiched between the ectoderm and endoderm and does not form a cavity or “coelom,” leaving little room for organ development; in pseudocoelomates, there is a cavity or pseudocoelom that replaces the blastocoel (the cavity within the blastula), but it is only lined by mesoderm on the outside of the cavity, leaving the gut tube and organs unlined; in eucoelomates, the cavity that obliterates the blastocoel as the coelom develops is lined both on the outside of the cavity (parietal layer) and also on the inside of the cavity, around the gut tube and the internal organs (visceral layer).
Eucoelmate protostomes are schizocoels, in which mesoderm-producing cells typically migrate into the blastocoel during gastrulation and multiply to form a solid mass of cells. Cavities then develop within the cell mass to form the coelom. Since the forming body cavity splits the mesoderm, this protostomic coelom is termed a schizocoelom. As we will see later in this chapter, chordates, the phylum to which we belong, generally develop a coelom by enterocoely: pouches of mesoderm pinch off the invaginating primitive gut, or archenteron, and then fuse to form a complete coelom. We should note here that a eucoelomate can form its “true coelom” by either schizocoely or enterocoely. The process that produces the coelom is different and of taxonomic importance, but the result is the same: a complete, mesodermally lined coelom.
Most organisms placed in the superphylum Lophotrochozoa possess either a lophophore feeding apparatus or a trochophore larvae (thus the contracted name, “lopho-trocho-zoa”). The lophophore is a feeding structure composed of a set of ciliated tentacles surrounding the mouth. A trochophore is a free-swimming larva characterized by two bands of cilia surrounding a top-like body. Some of the phyla classified as Lophotrochozoa may be missing one or both of these defining structures. Nevertheless their placement with the Lophotrochozoa is upheld when ribosomal RNA and other gene sequences are compared. The systematics of this complex group is still unclear and much more work remains to resolve the cladistic relationships among them.
Phylum Platyhelminthes The flatworms are acoelomate organisms that include many free-living and parasitic forms. The flatworms possess neither a lophophore nor trochophore larvae, although the larvae of one group of flatworms, the Polycladida (named after its many-branched digestive tract), are considered to be homologous to trochophore larvae. Spiral cleavage is also seen in the polycladids and other basal flatworm groups. The developmental pattern of some of the free-living forms is obscured by a phenomenon called "blastomere anarchy," in which a sort of temporary feeding larva forms, followed by a regrouping of cells within the embryo that gives rise to a second-stage embryo. However, both the monophyly of the flatworms and their placement in the Lophotrochozoa has been supported by molecular analyses.
The Platyhelminthes consist of two monophyletic lineages: the Catenulida and the Rhabditophora. The Catenulida, or "chain worms," is a small clade of just over 100 species. These worms typically reproduce asexually by budding. However, the offspring do not fully detach from the parents and the formation resembles a chain in appearance. All of the flatworms discussed here are part of the Rhabditophora ("rhabdite bearers").
Rhabdites are rodlike structures discharged in the mucus produced by some free-living flatworms; Eucoelmate protostomes are schizocoels, in which mesoderm-producing cells typically migrate into the blastocoel during gastrulation likely serve in both defense and to provide traction for ciliary gliding along the substrate. Unlike free-living flatworms, many species of trematodes and cestodes are parasitic, including important parasites of humans.
Chapter 28 | Invertebrates 803 Figure 28.13 Flatworms exhibit significant diversity. (a) A blue Pseudoceros flatworm (Pseudoceros bifurcus); (b) gold speckled flatworm (Thysanozoon nigropapillosum). (credit a: modification of work by Stephen Childs; b: modification of work by Pril Fish.) Flatworms have three embryonic tissue layers that give rise to epidermal tissues (from ectoderm), the lining of the digestive system (from endoderm), and other internal tissues (from mesoderm). The epidermal tissue is a single layer of cells or a layer of fused cells ( syncytium) that covers two layers of muscle, one circular and the other longitudinal. The mesodermal tissues include mesenchymal cells that contain collagen and support secretory cells that produce mucus and other materials at the surface. Because flatworms are acoelomates, the mesodermal layer forms a solid mass between the outer epidermal surface and the cavity of the digestive system.
Physiological Processes of Flatworms The free-living species of flatworms are predators or scavengers. Parasitic forms feed by absorbing nutrients provided by their hosts. Most flatworms, such as the planarian shown in Figure 28.14, have a branching gastrovascular cavity rather than a complete digestive system. In such animals, the “mouth” is also used to expel waste materials from the digestive system, and thus also serves as an anus. (A few species may have a second anal pore or opening.) The gut may be a simple sac or highly branched. Digestion is primarily extracellular, with digested materials taken into the cells of the gut lining by phagocytosis. One parasitic group, the tapeworms (cestodes), lacks a digestive system altogether, and absorb digested food from the host.
Flatworms have an excretory system with a network of tubules attached to flame cells, whose cilia beat to direct waste fluids concentrated in the tubules out of the body through excretory pores. The system is responsible for the regulation of dissolved salts and the excretion of nitrogenous wastes. The nervous system consists of a pair of lateral nerve cords running the length of the body with transverse connections between them. Two large cerebral ganglia—concentrations of nerve cell bodies at the anterior end of the worm—are associated with photosensory and chemosensory cells. There is neither a circulatory nor a respiratory system, with gas and nutrient exchange dependent on diffusion and cell-to-cell junctions. This necessarily limits the thickness of the body in these organisms, constraining them to be “flat” worms. Most flatworm species are monoecious (both male and female reproductive organs are found in the same individual), and fertilization is typically internal.
Asexual reproduction by fission is common in some groups.
804 Chapter 28 | Invertebrates This OpenStax book is available for free at Figure 28.14 Planaria, a free-living flatworm. The planarian is a flatworm that has a gastrovascular cavity with one opening that serves as both mouth and anus. The excretory system is made up of flame cells and tubules connected to excretory pores on both sides of the body. The nervous system is composed of two interconnected nerve cords running the length of the body, with cerebral ganglia and eyespots at the anterior end.
Diversity of Flatworms The flatworms have been traditionally divided into four classes: Turbellaria, Monogenea, Trematoda, and Cestoda (Figure 28.15). However, the relationships among members of these classes has recently been reassessed, with the turbellarians in particular now viewed as paraphyletic, since its descendants may also include members of the other three classes. Members of the clade or class Rhabditophora are now dispersed among multiple orders of Platyhelminthes, the most familiar of these being the Polycladida, which contains the large marine flatworms; the Tricladida (which includes Dugesia [“planaria”] and Planaria and its relatives); and the major parasitic orders: Monogenea (fish ectoparasites), Trematoda (flukes), and Cestoda (tapeworms), which together form a monophyletic clade.
Chapter 28 | Invertebrates 805 Figure 28.15 Traditional flatworm classes. Phylum Platyhelminthes was previously divided into four classes. (a) Class Turbellaria includes the free-living polycladid Bedford’s flatworm (Pseudobiceros bedfordi), which is about 8 to 10 cm in length. (b) The parasitic class Monogenea includes Dactylogyrus spp, commonly called gill flukes, which are about 0.2 mm in length and have two anchors, indicated by arrows used to attach the parasite on to the gills of host fish. (c) The class Trematoda includes Fascioloides magna (right) and Fasciola hepatica (two specimens on left, also known as the common liver fluke). (d) Class Cestoda includes tapeworms such as this Taenia saginata, infects both cattle and humans, and can reach 4 to 10 meters in length; the specimen shown here is about four meters long. (credit a: modification of work by Jan Derk; credit d: modification of work by CDC) Most free-living flatworms are marine polycladids, although tricladid species live in freshwater or moist terrestrial environments, and there are a number of members from other orders in both environments. The ventral epidermis of free-living flatworms is ciliated, which facilitates their locomotion. Some free-living flatworms are capable of remarkable feats of regeneration in which an individual may regrow its head or tail after being severed, or even several heads if the planaria is cut lengthwise.
The monogeneans are ectoparasites, mostly of fish, with simple life cycles that consist of a free-swimming larva that attaches to a fish, prior to its transformation to the ectoparasitic adult form. The parasite has only one host and that host is usually very specific. The worms may produce enzymes that digest the host tissues, or they may simply graze on surface mucus and skin particles. Most monogeneans are hermaphroditic, but the male gametes develop first and so cross-fertilization is quite common.
The trematodes, or flukes, are internal parasites of mollusks and many other groups, including humans.
Trematodes have complex life cycles that involve a primary host in which sexual reproduction occurs, and one or more secondary hosts in which asexual reproduction occurs. The primary host is usually a vertebrate and the secondary host is almost always a mollusk, in which multiple larvae are produced asexually. Trematodes, which attached internally to the host via an oral and medial sucker, are responsible for serious human diseases including schistosomiasis, caused by several species of the blood fluke, Schistosoma spp. Various forms of schistosomiasis infect an estimated 200 million people in the tropics, leading to organ damage, secondary infection by bacteria, and chronic symptoms like fatigue. Infection occurs when the human enters the water and 806 Chapter 28 | Invertebrates This OpenStax book is available for free at metacercaria larvae, released from the snail host, locate and penetrate the skin. The parasite infects various organs in the body and feeds on red blood cells before reproducing.
Many of the eggs are released in feces and find their way into a waterway, where they are able to reinfect the snail host. The eggs, which have a barb on them, can damage the vascular system of the human host, causing ulceration, abscesses, and bloody diarrhea, wherever they reside, thereby allowing other pathogens to cause secondary infections. In fact, it is the parasite’s eggs that produce most of the main ill effects of schistosomiasis. Many eggs do not make the transit through the veins of the host for elimination, and are swept by blood flow back to the liver and other locations, where they can cause severe inflammation. In the liver, the errant eggs may impede circulation and cause cirrhosis. Control is difficult in impoverished areas in unsanitary, crowded conditions, and prognosis is poor in people with heavy infections of Schistosoma japonicum, without early treatment.
The cestodes, or tapeworms, are also internal parasites, mainly of vertebrates (Figure 28.16). Tapeworms, such as those of Taenia spp, live in the intestinal tract of the primary host and remain fixed using a sucker or hooks on the anterior end, or scolex, of the tapeworm body, which is essentially a colony of similar subunits called proglottids. Each proglottid may contain an excretory system with flame cells, along with reproductive structures, both male and female. Because they are so long and flat, tapeworms do not need a digestive system; instead, they absorb nutrients from the food matter surrounding them in the host’s intestine by diffusion.
Proglottids are produced at the scolex and gradually migrate to the end of the tapeworm; at this point, they are “mature” and all structures except fertilized eggs have degenerated. Most reproduction occurs by cross-fertilization between different worms in the same host, but may also occur between proglottids. The mature proglottids detach from the body of the worm and are released into the feces of the organism. The eggs are eaten by an intermediate host, typically another vertebrate. The juvenile worm infects the intermediate host and takes up residence, usually in muscle tissue. When the muscle tissue is consumed by the primary host, the cycle is completed. There are several tapeworm parasites of humans that are transmitted by eating uncooked or poorly cooked pork, beef, or fish.
Figure 28.16 Tapeworm life cycle. Tapeworm (Taenia spp.) infections occur when humans consume raw or undercooked infected meat. (credit: modification of work by CDC) Chapter 28 | Invertebrates 807 Phylum Rotifera The rotifers ("wheel-bearer") belong to a group of microscopic (about 100 µm to 2 mm) mostly aquatic animals that get their name from the corona—a pair of ciliated feeding structures that appear to rotate when viewed under the light microscope (Figure 28.17). Although their taxonomic status is currently in flux, one treatment places the rotifers in three classes: Bdelloidea, Monogononta, and Seisonidea. In addition, the parasitic “spiny headed worms” currently in phylum Acanthocephala, appear to be modified rotifers and will probably be placed into the group in the near future. Undoubtedly the rotifers will continue to be revised as more phylogenetic evidence becomes available.
The pseudocoelomate body of a rotifer is remarkably complex for such a small animal (roughly the size of a Paramecium) and is divided into three sections: a head (which contains the corona), a trunk (which contains most of the internal organs), and the foot. A cuticle, rigid in some species and flexible in others, covers the body surface. They have both skeletal muscle associated with locomotion and visceral muscles associated with the gut, both composed of single cells. Rotifers are typically free-swimming or planktonic (drifting) organisms, but the toes or extensions of the foot can secrete a sticky material to help them adhere to surfaces. The head contains a number of eyespots and a bilobed “brain,” with nerves extending into the body.
Figure 28.17 Rotifers. Shown are examples from two of the three classes of rotifer. (a) Species from the class Bdelloidea are characterized by a large corona. The whole animals in the center of this scanning electron micrograph are shown surrounded by several sets of jaws from the mastax of rotifers. (b) Polyarthra, from the largest rotifer class Monogononta, has a smaller corona than bdelloid rotifers, and a single gonad, which give the class its name. (credit a: modification of work by Diego Fontaneto; credit b: modification of work by U.S. EPA; scale-bar data from Cory Zanker) Rotifers are commonly found in freshwater and some saltwater environments throughout the world. As filter feeders, they will eat dead material, algae, and other microscopic living organisms, and are therefore very important components of aquatic food webs. A rotifer's food is directed toward the mouth by the current created from the movement of the coronal cilia. The food particles enter the mouth and travel first to the mastax—a muscular pharynx with toothy jaw-like structures. Examples of the jaws of various rotifers are seen in Figure 28.17a. Masticated food passes near digestive and salivary glands, into the stomach, and then to the intestines.
Digestive and excretory wastes are collected in a cloacal bladder before being released out the anus.
Watch this video ( to see rotifers feeding.
About 2,200 species of rotifers have been identified. Figure 28.18 shows the anatomy of a rotifer belonging to class Bdelloidea. Some rotifers are dioecious organisms and exhibit sexual dimorphism (males and females have different forms). In many dioecious species, males are short-lived and smaller with no digestive system 808 Chapter 28 | Invertebrates This OpenStax book is available for free at and a single testis. Many rotifer species exhibit haplodiploidy, a method of sex determination in which a fertilized egg develops into a female and an unfertilized egg develops into a male. However, reproduction in the bdelloid rotifers is exclusively parthenogenetic and appears to have been so for millions of years: Thus, all bdelloid rotifers and their progeny are female! The bdelloids may compensate for this genetic insularity by borrowing genes from the DNA of other species. Up to 10% of a bdelloid genome comprises genes imported from related species. Some rotifer eggs are capable of extended dormancy for protection during harsh environmental conditions.
Figure 28.18 A bdelloid rotifer. This illustration shows the anatomy of a bdelloid rotifer.
Phylum Nemertea The Nemertea are colloquially known as ribbon worms or proboscis worms. Most species of phylum Nemertea are marine and predominantly benthic (bottom dwellers), with an estimated 900 known species. However, nemerteans have been recorded in freshwater and very damp terrestrial habitats as well. Most nemerteans are carnivores, feeding on worms, clams, and crustaceans. Some species are scavengers, and some, like Malacobdella grossa, have also evolved commensal relationships with mollusks. Economically important species have at times devastated commercial fishing of clams and crabs. Nemerteans have almost no predators and two species are sold as fish bait.
Morphology Nemerteans vary in size from 1 cm to several meters. They show bilateral symmetry and remarkable contractile properties. Because of their contractility, they can change their morphological presentation in response to environmental cues. Animals in phylum Nemertea are soft and unsegmented animals, with a morphology like a flattened tube. (Figure 28.19).
Chapter 28 | Invertebrates 809 Figure 28.19 A proboscis worm. The proboscis worm (Parborlasia corrugatus) is a scavenger that combs the sea floor for food. The species is a member of the phylum Nemertea. The specimen shown here was photographed in the Ross Sea, Antarctica. (credit: Henry Kaiser, National Science Foundation) A unique characteristic of this phylum is the presence of an eversible proboscis enclosed in a pocket called a rhynchocoel (not part of the animal's actual coelom). The proboscis is located dorsal to the gut and serves as a harpoon or tentacle for food capture. In some species it is ornamented with barbs. The rhynchocoel is a fluid-filled cavity that extends from the head to nearly two-thirds of the length of the gut in these animals (Figure 28.20). The proboscis may be extended by hydrostatic pressure generated by contraction of muscle of the rhynchocoel and retracted by a retractor muscle attached to the rear wall of the rhynchocoel.
Figure 28.20 The anatomy of a Nemertean is shown.
Watch this video ( to see a nemertean attack a polychaete with its proboscis.
Digestive System The nemerteans, which are primarily predators of annelids and crustaceans, have a well-developed digestive system. A mouth opening that is ventral to the rhynchocoel leads into the foregut, followed by the intestine. The intestine is present in the form of diverticular pouches and ends in a rectum that opens via an anus. Gonads are interspersed with the intestinal diverticular pouches and open outward via genital pores.
Nemerteans are sometimes classified as acoels, but because their closed circulatory system is derived from the coelomic cavity of the embryo, they may be regarded as coelomic. Their circulatory system consists of a closed loop formed by a connected pair of lateral blood vessels. Some species may also have a dorsal vessel or cross-connecting vessels in addition to lateral ones. Although the circulatory fluid contains cells, it is often colorless.
However, the blood cells of some species bear hemoglobin as well as other yellow or green pigments. The blood 810 Chapter 28 | Invertebrates This OpenStax book is available for free at vessels are contractile, although there is usually no regular circulatory pathway, and movement of blood is also facilitated by the contraction of muscles in the body wall. The circulation of fluids in the rhychocoel is more or less independent of the blood circulation, although blind branches from the blood vessels into the rhyncocoel wall can mediate exchange of materials between them. A pair of protonephridia, or excretory tubules, is present in these animals to facilitate osmoregulation. Gaseous exchange occurs through the skin.
Nervous System Nemerteans have a "brain" composed of four ganglia situated at the anterior end, around the rhynchocoel.
Paired longitudinal nerve cords emerge from the brain ganglia and extend to the posterior end. Additional nerve cords are found in some species. Interestingly, the brain can contain hemoglobin, which acts as an oxygen reserve. Ocelli or eyespots are present in pairs, in multiples of two in the anterior portion of the body. It is speculated that the eyespots originate from neural tissue and not from the epidermis.
Reproduction Nemerteans, like flatworms, have excellent powers of regeneration, and asexual reproduction by fragmentation is seen in some species. Most animals in phylum Nemertea are dioecious, although freshwater species may be hermaphroditic. Stem cells that become gametes aggregate within gonads placed along the digestive tract.
Eggs and sperm are released into the water, and fertilization occurs externally. Like most lophotrochozoan protostomes, cleavage is spiral, and development is usually direct, although some species have a trochophore-like larva, in which a young worm is constructed from a series of imaginal discs that begin as invaginations from the body surface of the larva.
28.4 | Superphylum Lophotrochozoa: Molluscs and Annelids By the end of this section, you will be able to do the following: • Describe the unique anatomical and morphological features of molluscs and annelids • Describe the formation of the coelom • Identify an important extracoelomic cavity in molluscs • Describe the major body regions of Mollusca and how they vary in different molluscan classes • Discuss the advantages of true body segmentation • Describe the features of animals classified in phylum Annelida The annelids and the mollusks are the most familiar of the lophotrochozoan protostomes. They are also more “typical” lophotrochozoans, since both groups include aquatic species with trochophore larvae, which unite both taxa in common ancestry. These phyla show how a flexible body plan can lead to biological success in terms of abundance and species diversity. The phylum Mollusca has the second greatest number of species of all animal phyla with nearly 100,000 described extant species, and about 80,000 described extinct species. In fact, it is estimated that about 25 percent of all known marine species are mollusks! The annelids and mollusca are both bilaterally symmetrical, cephalized, triploblastic, schizocoelous eucoeolomates They include animals you are likely to see in your backyard or on your dinner plate!
Phylum Mollusca The name “Mollusca” means “soft” body, since the earliest descriptions of molluscs came from observations of “squishy,” unshelled cuttlefish. Molluscs are predominantly a marine group of animals; however, they are also known to inhabit freshwater as well as terrestrial habitats. This enormous phylum includes chitons, tusk shells, snails, slugs, nudibranchs, sea butterflies, clams, mussels, oysters, squids, octopuses, and nautiluses. Molluscs display a wide range of morphologies in each class and subclass, but share a few key characteristics (Figure 28.21). The chief locomotor structure is usually a muscular foot. Most internal organs are contained in a region called the visceral mass. Overlying the visceral mass is a fold of tissue called the mantle; within the cavity formed by the mantle are respiratory structures called gills, that typically fold over the visceral mass. The mouths of most mollusks, except bivalves (e.g., clams) contain a specialized feeding organ called a radula, an abrasive tonguelike structure. Finally, the mantle secretes a calcium-carbonate-hardened shell in most molluscs, although Chapter 28 | Invertebrates 811 |
17335 | http://www.hormones.gr/pdf/1116414588.pdf | 100 C.D. SCOPA HORMONES 2004, 3(2):100-110 Review Histopathology of Thyroid Tumors. An Overview Chrisoula D. Scopa Dept. of Pathology, University of Patras, School of Health Sciences, Faculty of Medicine, Patras, Greece INTRODUCTION Thyroid cancer accounts for approximately 1% of total cancer cases in developed countries. It affects all age groups, although it is rare in children. Thyroid tumors are more frequent in women than in men.
Despite their relative rarity they exhibit a wide range of morphological patterns and biological behavior which may explain the great interest in these neo-plasms of both pathologists and clinicians.
Another issue of considerable interest is the mo-lecular abnormalities involved in thyroid tumor pa-thology. In this review, following the histologic classi-fication of thyroid neoplasms with their histologic and cytologic variants, a brief discussion of some of the most relevant molecular alterations recently described will be undertaken.
A. HISTOPATHOLOGY The thyroid gland contains two major types of ep-ithelial cells: the follicular cells, which convert iodine into thyroxine and triiodothyronine, and the parafol-licular or C-cells, which secrete calcitonin. Thyroid tumors can originate from these very different kinds of cells or from nonepithelial stromal elements, and architectural, cytologic and histogenetic features have been taken into consideration for neoplasms classifi-cation. According to the World Health Organization (WHO)1 primary thyroid tumors are classified as epi-thelial and nonepithelial, benign or malignant, with a separate category for lymphomas and miscellaneous neoplasms (Table 1). A slightly different classification scheme has been adopted by the Armed Forces Insti-tute of Pathology (AFIP)2, giving priority to the cell of origin and incorporating, in each cell type, special tumor types and subtypes designated as variants (Ta-ble 2). This review is based on the AFIP classifica-tion.
The traditional classification of thyroid cancer as well differentiated carcinomas (papillary and follicu-lar) characterized by relatively good prognosis, or poorly differentiated carcinomas (follicular, anaplas-tic) associated with aggressive behavior, metastases and death, is no longer applicable since certain mor-phologic variants of papillary carcinoma are associat-ed with poor prognosis. In addition, the existence of true mixed forms of papillary and follicular cancers has been disproved, while new entities such as mixed follicular-parafollicular carcinoma have emerged3.
The criteria for the recognition of follicular and papillary carcinomas have changed in recent years but both denominations have been retained. Papillae no longer seem to be necessary for the diagnosis of pap-illary carcinoma, and cytologic features such as onco-cytic, clear cell, squamous and mucinous changes have resulted in the designation of special tumor types and subtypes. The practical importance of these special types resides in differential diagnosis rather than their biological behavior2.
Key words: Thyroid, Thyroid histopatholo-gy, Tumors, Oncogenes Address correspondence and requests for reprints to: Chrisoula D. Scopa, M.D., P.O. Box 1174, 261 10 Patras, Greece, Tel: +30-2610- 990 038, Fax: +30-2610- 990 775, e-mail: cdscopa@med.upatras.gr Received 04-02-04, Revised 05-03-04, Accepted 10-03-04 101 Histopathology of Thyroid Tumors A.1. TUMORS OF FOLLICULAR CELLS AND THEIR VARIANTS A.1.1. Follicular Adenoma Follicular adenoma is defined as a benign encap-sulated tumor with follicular cell differentiation show-ing a uniform pattern throughout the confine nodule (Figure 1A). The fibrous capsule varies in thickness, but is usually thin. Follicular adenomas are solitary tumors with a solid, homogeneous cut surface, but hemorrhage and cystic degeneration are not uncom-mon. Their size is highly variable, ranging from 1 cm to over 10 cm. On the basis of microscopic features, several variants have been described, including onco-cytic adenoma (Hürthle cell adenoma), adenoma with clear cell change, atypical adenoma, hyalinizing tra-becular adenoma, adenoma with bizarre nuclei and rare types such as adenoma with adipose (adenolipo-ma) or cartilagenous (adenochondroma) metaplasia2.
Recent studies suggest that hyalinizing trabecular ad-enoma, also described as hyalinizing trabecular tumor Table 1. Histologic Classification of Thyroid Tumors (WHO)1 I. Epithelial Tumors A. Benign 1. Follicular adenoma 2. Others B. Malignant 1. Follicular carcinoma 2. Papillary carcinoma 3. Medullary carcinoma 4. Undifferentiated (anaplastic) carcinoma 5. Others II. Nonepithelial Tumors A. Benign B. Malignant III. Malignant lymphomas IV. Miscellaneous V. Secondary tumors VI. Unclassified tumors VII. Tumor-like lesions Table 2. Classification of Thyroid Tumors (AFIP)2 PRIMARY TUMORS 1. Epithelial tumors A. Tumors of follicular cells Benign: Follicular adenoma (conventional, variants) Malignant: carcinoma differentiated follicular carcinoma papillary carcinoma (conventional, variants) poorly differentiated (insular carcinoma, others) undifferentiated or anaplastic B. Tumors of C-cells (and related neuroendocrine cells) medullary carcinoma others C. Tumors of follicular and C-cells mixed medullary-follicular carcinoma 2. Sarcomas 3. Malignant lymphoma (and related hematopoietic neoplasms) 4. Miscellaneous neoplasms SECONDARY TUMORS TUMOR-LIKE LESIONS oncocytic, clear cell changes, atypical adenoma, hyalinizing trabecular, with bizarre nuclei microcarcinoma, encapsulated, follicular, tall/columnar cell, diffuse sclerosing, solid/trabecular 102 C.D. SCOPA Malignant thyroid tumors composed exclusively or predominately (over 75%) of oncocytes (Hürthle cell tumors) share some similarities with follicular carcinomas as regards the clinical presentation, the architectural features and the degree of invasiveness, and therefore should be considered as a variant of follicular carcinoma2,8,9. However, some authors have suggested that the morphologic features and natural history of these tumors are distinctive enough that they be considered as a separate entity13,14.
A.1.3. Papillary Carcinoma Papillary carcinoma is the most common type of thyroid cancer, comprising approximately 80% of all primary thyroid malignancies15. Classical or non-other-wise specified (NOS) papillary carcinoma is characte-rized by the formation of papillae and a set of distincti-ve nuclear features (optically clear appearance, over-lapping, pseudoinclusions and nuclear grooves) (Fig-ure 1C, 1D)2,16-18. The size of papillary carcinoma is extremely variable with a mean diameter of 2-3 cm2.
A clinically detected tumor is usually confined to the thyroid, is presented as a fairly well circumscribed or infiltrative neoplasm and has an indolent course. Its mode of spread is most commonly via lymphatics with-in the thyroid leading to multifocal disease and to cervical node metastases2,9. Indeed, 50% or more of papillary carcinomas have nodal metastases at initial diagnosis19.
There are several histologic variants of papillary carcinoma, some of which are associated with a more guarded prognosis (Table 3)10.
A.1.3.a. Variants of papillary carcinoma a.1. Papillary microcarcinoma: The term refers to papillary carcinomas measuring 1cm or less in di-ameter and replaces the older designation of occult sclerosing carcinoma, also known as nonencapsulated sclerosing tumor and occult papillary carcinoma2. Re-and paraganglioma-like adenoma of the thyroid, may be related to papillary carcinoma at the molecu-lar level4,5, while others propose a multidirectional dif-ferentiation from pluripotent primitive cells6. Some of these controversial issues are challenged by Lloyd7 who concludes that additional studies are needed to clearly define this entity. Until then, it would seem appropriate to regard and treat hyalinizing trabecu-lar adenoma as a benign neoplasm2. Toxic adenoma is a clinical rather than a pathologic entity, defining only those hyperfunctioning lesions in which clinical manifestations occur, and not any hot adenoma2.
A.1.2. Follicular Carcinoma Most authors agree that only follicular tumors that exhibit vascular and/or capsular invasion should be regarded as follicular carcinomas8. Depending on the degree of their invasiveness, follicular carcinomas have been divided into two major categories: minimally invasive or encapsulated (the most common), and widely invasive. The frequency of follicular carcino-ma among thyroid malignancies ranges from 5-10% in non-iodine-deficient areas to 30-40% in iodine-de-ficient areas2. Macroscopically, follicular carcinomas do not dif-fer appreciably from follicular adenomas. The fibrous capsule surrounding the tumor tends to be thicker and more irregular than in adenomas2. Minimally invasive follicular carcinoma is an encapsulated tumor show-ing capsular and/or vascular invasion only on micro-scopic evaluation, while the widely invasive neoplasm shows lack of complete encapsulation, extensive ar-eas of invasion to the adjacent thyroid tissue and/or widespread blood vessels infiltration2,8. Immunohistochemistry, morphometry, ploidy analysis, cytogenetic and oncogene markers have failed to provide reliable information concerning the distinc-tion between follicular carcinoma and follicular ade-noma. The current diagnostic criteria for malignancy are still the histologic assessment of true capsular in-filtration (the tumor must penetrate the entire thick-ness of the capsule) and/or invasion of blood vessels in or beyond the capsule (Figure 1B)2,8-10. It is appar-ent that minimally invasive follicular tumors cannot be accurately diagnosed by fine needle aspiration (FNA) cytology since the crucial diagnostic criteria are missing2,9,11. Similar problems exist in evaluating such lesions by frozen section2,11,12.
Table 3. Variants of Papillary Carcinoma and their Prognosis10 Good Variable Guarded Microcarcinoma Oxyphilic cell Diffuse sclerosing Encapsulated Follicular Tall/columnar cell Macrofollicular Solid sclerosing Diffuse follicular Solid/trabecular With nodular fasciitis-like stroma 103 Histopathology of Thyroid Tumors a.6. Other variants: Variants such as solid vari-ant, clear cell and oxyphilic variant, papillary carcino-ma with lipomatous stroma, Warthins-like tumor or with nodular fasciitis-like stroma and cribriform pap-illary carcinoma have been reported, but they are too few in number for an adequate assessment of their prognostic implication2,15,31. The term solid and/or tra-becular variant is used when a NOS papillary carcino-ma has a solid and/or trabecular pattern throughout the tumor2.
A.1.4. Poorly Differentiated Carcinoma Poorly differentiated thyroid carcinoma represents a heterogeneous group of malignant neoplasms, with various histologic patterns of growth and different biologic behavior, that lie somewhere between well-differentiated and undifferentiated carcinomas6,31-33.
The heterogeneity of these tumors reflects the differ-ent terms used to describe and diagnose this entity, including solid, trabecular, insular, poorly differenti-ated, intermediate type, less well-differentiated and follicular carcinoma with insular component2,32-35. Macroscopically, poorly differentiated carcinomas usually present as large (>5 cm in diameter), solid, unencapsulated, nodular or multinodular, grey-white tumors that tend to invade perithyroidal tissues2,15,33.
Microscopically, the majority of these tumors show a trabecular, solid or insular growth pattern10. A combi-nation of other histologic features such as follicles, columnar cell carcinoma, papillary, follicular and Hürthle cell carcinoma foci have been found2,31-33. Recent studies on molecular and genetic features of poorly differentiated carcinomas provide evidence that there is a link between these tumors and the pap-illary thyroid carcinoma, and they support the con-cept that poorly differentiated carcinoma represents an immediate step in the progression from well-dif-ferentiated to undifferentiated carcinomas36,37. How-ever, a divergent histogenesis must also be considered since poorly differentiated carcinomas exhibit, in ad-dition to the usual thyroglobulin immunoreactivity, focal immunoreactivity to neuroendocrine markers, such as calcitonin, neurotensin and somatostatin33,35,38. Apart from tumor stage, clinicopathologic features such as tumor necrosis, mitotic count (>3/10HPF) and the age (>45 years) of the patient, have been report-ed as being significantly associated with the clinically aggressive behavior of these tumors32,33.
cently, at the 12th Annual Cancer Meeting held at Por-to, Portugal, a consensus was reached by a group of experts to rename this entity as papillary microtumor20.
Papillary microcarcinomas are frequently detected as incidental findings in autopsy or in surgical specimens and are associated with an excellent prognosis despite occasional regional lymph node metastases. The re-ported incidence in autopsy material has ranged from 4% to 35.6%2,21-23.
a.2. Encapsulated variant: The tumor is totally surrounded by a fibrous capsule which may be intact or focally infiltrated by the tumor. These tumors have an exceptionally good prognosis and, although some lesions have shown lymph node involvement, distant metastases or death due to tumor are practically non-existent24.
a.3. Follicular variant: Papillary carcinomas hav-ing an exclusive or almost exclusive follicular pattern are designated as a follicular variant of papillary car-cinoma. The biologic behavior of this variant is anal-ogous to that of conventional papillary carcinoma. The metastases may have a mixed papillary and follicular formation. A diffuse or widely invasive form of the follicular variant and macrofollicular variant of papil-lary carcinoma have also been described25,26.
a.4. Tall and columnar cell variant: The main his-tologic feature of the tall cell variant of papillary car-cinoma is the presence of tall cells (the height be-ing twice the width), with an intense eosinophilic cy-toplasm, lining well-developed papillae (Figure 1E).
In the columnar cell variant, there is a marked nucle-ar stratification and the cytoplasm is clear, sometimes with subnuclear vacuolization2,9. Both the tall cell and columnar cell variant are said to be more aggressive than classical papillary carcinomas27,28. However, re-cent studies suggest that the clinical behavior of these rare types of papillary carcinoma depends on tumor size, extrathyroidal invasion and distant metastases29,30.
a.5. Diffuse sclerosing variant: This is an unusual form of papillary carcinoma first described by Vick-ery et al22, who noticed that it more frequently affects children and is associated with a poor prognosis. This tumor is characterized by diffuse involvement of one or two lobes and clinically may be misdiagnosed as Hashimotos thyroiditis2. Its hallmark, microscopical-ly, is the presence of widespread intrathyroid lymphatic permeation by numerous neoplastic micropapillae.
104 C.D. SCOPA A.1.5. Undifferentiated (Anaplastic) Carcinoma Undifferentiated thyroid carcinomas account for 5-10% of all primary malignant tumors of the thyroid2.
These tumors, usually present in elderly patients (mean age 60-65 years), are rapidly growing, with massive local invasion and early distant metastases, most frequently to lung, adrenals and bone2,9. Undifferentiated thyroid carcinoma exhibits a wide spectrum of morphologic types, singly or in com-bination. The three major patterns of growth are squa-moid (morphologic similarity to nonkeratinizing squa-mous cell carcinoma), spindle cell (sarcoma-like growth) and giant cell (numerous osteoclast-like multi-nucleated giant cells, resembling giant cell tumor of the bone or soft tissues) (Figure 1F). Features com-mon to all three types are high mitotic activity, exten-sive necrosis and a marked degree of invasiveness with-in the gland as well as to the extrathyroidal structures2,9. Areas of pre-existing well-differentiated thyroid tumor (more often follicular or papillary carcinoma) can be seen in many, if not most, undifferentiated carcinomas. On the other hand, the undifferentiated tumor may develop months or years after the remov-al of a well-differentiated thyroid neoplasm. These findings support the hypothesis that undifferentiated thyroid tumors arise from pre-existing well-differen-tiated thyroid carcinomas2,39.
A.2. TUMORS OF C-CELLS AND THEIR VARIANTS A.2.1. Medullary Carcinoma Medullary thyroid carcinoma is a malignant tumor of the thyroid which shows evidence of C-cell differ-entiation and usually contains calcitonin (Figure 1G)2.
It accounts for up to 10% of all malignant thyroid tu-mors2,15. The variants of medullary carcinoma are: glan-dular (composed in part of tubular or follicular struc-tures and may resemble follicular carcinoma), papil-lary (exhibiting true papillary pattern of growth), small cell (resembling the intermediate variant of small cell carcinoma of the lung), and giant cell (occasionally present or focal areas with giant cell formation). Less common are the clear cell, melanotic (pigmanted), oncocytic (oxyphilic), squamous, amphicrine (calcito-nin and mucin-producing cells) and paraganglioma-like variants2,40.
A.2.2. Mixed follicularparafollicular Carcinoma Mixed medullary and follicular carcinoma are rare neoplasms which show morphologic features of both follicular and C-cell differentiation41. The dual differ-entiation has also been noted in their metastatic sites.
These neoplasms must be distinguished from the fol-licular variant of medullary carcinoma and from med-ullary carcinoma with entrapped normal follicles. Thus, WHO is very strict in defining them as tumors show-ing both the morphologic features of medullary carci-noma together with immunoreactivity for calcitonin and the morphologic features of follicular carcinoma together with immunoreactivity for thyroglobulin42. Whether these tumors represent collision tumors or arise from a stem cell capable of dual differentia-tion into follicular and C-cell elements is the subject of several excellent reports in the recent literature3,43,44.
A.3. CYTOPLASMIC CHANGES IN THYROID TUMORS A.3.1. Tumors with oncocytic features: Oncocytes (oxiphilic cells, Hürthle cells) are derived from fol-licular epithelium, characterized morphologically by large size, distinct cell borders and abundant granular acidophilic cytoplasm, large nucleus and prominent nucleolus. The cytoplasmic granularity is produced by an increased number of huge mitochondria. Oncocyt-ic thyroid tumors (oncocytic adenoma/carcinoma) are composed exclusively or predominately of follicular cells exhibiting oncocytic features2. However, isolated cells or groups of follicular cells with oncocytic fea-tures can be seen in other conditions such as radiated thyroids, aging thyroids, nodular goiter, nonspecif-ic chronic thyroiditis or Hashimotos thyroiditis and Graves disease. In addition, several neoplasms of the thyroid (oncocytic papillary neoplasms, Warthin-like tumor of the thyroid, tall cell variant of papillary car-cinoma) exhibit oncocytic features9,45.
A.3.2. Tumors with clear cell features: Clear cell changes can occur in any of the major histologic types of benign and malignant thyroid neoplasms. They have also been observed in nodular hyperplasia and in Hash-imotos thyroiditis2. The clearing of the cytoplasm may be the consequence of intracytoplasmic accumulation of vesicles (of mitochondrial or other origin: Hürthle cell tumors, follicular tumors), glycogen (papillary 105 Histopathology of Thyroid Tumors Figure 1. Histology of Thyroid Tumors. A. Follicular adenoma: note the sharp separation of a follicular tumor from the surrounding tissue by a uniform fibrous capsule. B. Follicular carcinoma with capsular penetration. C. Papillary carcinoma metastatic to a lymph node: typical appearance of papillary carcinoma with complex and branching papillae. D. Higher magnification showing optical clear, overlapping and grooved (arrow) nuclei. E. Tall cell variant papillary carcinoma, lined by tall cells (arrow). F. Undifferentiated carcinoma with elongated tumor cells. G. Medullary carcinoma. H. Papillary carcinoma with clear cell changes: typical intranuclear inclusion (inset).
106 C.D. SCOPA carcinoma) (Figure 1H), lipid (undifferentiated car-cinoma), thyroglobulin and mucin2.
A.3.3. Tumors with squamous features: Squamous cells in the thyroid can originate from a remnant of the thyroglossal duct or ultimobranchial body or may be the result of squamous metaplasia in Hashimotos thyroiditis, papillary carcinoma or other conditions9.
Focal or extensive squamous differentiation has been described in papillary carcinoma, undifferentiated carcinoma, follicular neoplasms and medullary carci-noma2.
A.3.4. Tumors with mucinous features: Mucosub-stances have been detected in several thyroid tumors, either in the cytoplasm of the tumor cells or extracel-lulary46,47. Primary thyroid tumors that have been found to contain mucin include signet ring follicular ade-noma, mucoepidermoid carcinoma and sclerosing mucoepidermoid carcinoma with eosinophilia, papil-lary, undifferentiated and medullary carcinoma2.
A.4. THYROID SARCOMAS Sarcomas arising in the thyroid are extremely rare.
Various microscopic types have been reported in the form of isolated case reports, including fibrosarcoma48, liposarcoma49, leiomyosarcoma50, chondrosarcoma51, osteosarcoma52,53 and malignant schwannoma54. It is likely that most thyroid neoplasms resembling sarcomas are examples of undifferentiated (sarcoma-toid) carcinomas. In addition, cartilage and bone pro-duction in association with benign and malignant thy-roid lesions have been observed39,55. However, the dis-tinction between true thyroid sarcomas and sarcoma-like undifferentiated carcinomas is of little importance since the natural history and response to therapy of both entities do not differ significantly. They occur in elderly patients, are rapidly growing and are uniform-ly fatal9. The sarcoma that apparently does arise in the thyroid is angiosarcoma. It occurs predominately in Alpine regions of central Europe where it typically arises in a gland with multinodular goiter56. Although several investigators have suggested that this entity actually represents a vascular variant of undifferenti-ated carcinoma, some of these tumors exhibit anasto-mosing vascular channels, Wiebel-Palade bodies ul-trastructurally, immunoreactivity for factor-VIII re-lated antigen and other factors consistent with endo-thelial differentiation57.
A.5. MALIGNANT LYMPHOMAS Primary non-Hodgkin lymphomas of the thyroid are now considered to be tumors of mucosa-associat-ed lymphoid tissue (MALT)15. They constitute about 8% of all thyroid malignancies2. Primary thyroid lym-phomas have a B cell phenotype and are highly asso-ciated with lymphocytic or Hashimotos thyroiditis2,15.
Thyroid malignant lymphomas are most common in adult or elderly women, clinically presented in the form of an enlarged thyroid, leading to symptoms of tra-cheal or laryngeal compression when extended out-side the gland2,9. Most patients are euthyroid58. Since MALT lymphomas characteristically metastasize to sites that contain similar tissue, it is not uncommon to see a lymphoma of the thyroid with concomitant involvement of the gastrointestinal tract30,59.
Other primary lymphoid tumors include plasma-cytoma and Hodgkin lymphoma2.
A.6. SECONDARY TUMORS OF THE THYROID Although any malignant tumor can metastasize to the thyroid gland, the latter is an infrequent site of tumor metastases. Direct extension into the thyroid may occur in carcinomas of the pharynx, larynx, tra-chea and esophagus2,9. Most of these neoplasms are of squamous cell type. Retrograde lymphatic spread into the thyroid, although unusual, has been report-ed, with breast carcinoma being the most frequent9.
Hematogenous metastases to the thyroid, particular-ly of malignant melanoma, lung, gastrointestinal, breast and renal cell carcinomas are commonly en-countered at autopsy series2,9. Rare sources of prima-ry tumors, such as choriocarcinoma, malignant phyl-loides tumor and osteosarcoma, have also been re-ported60. Some metastases are found in preexisting thyroid lesions, such as breast carcinoma into papil-lary carcinoma and lung and renal cell carcinoma into follicular adenoma2,61-63. Particularly diagnostically challenging are metastases of unknown origin. Of these the most common are renal cell carcinoma, large bowel adenocarcinoma and malignant melanoma9. Metastatic disease to the thyroid may present a diagnostic problem for the following reasons: a) Many cases are asymptomatic and too small in size to be detected clinically; b) The primary site is difficult to identify histologically, in small biopsy samples (FNA), since many metastatic lesions are poorly or moder-107 Histopathology of Thyroid Tumors ately differentiated (and sometimes undifferentiated) carcinomas; c) Metastases could be manifested long after the detection of a primary tumor (as long as 26 years); d) Many clinically detected lesions are solitary rather than multiple60,64.
B. MOLECULAR MARKERS OF THYROID TUMORS Several types of oncogene alterations have been described as possible mechanism(s) for thyroid tumorogenesis. Whether these alterations can serve as markers, predicting the biological behavior of thy-roid tumors or confirming unclear diagnoses is a mat-ter of debate. For a comprehensive review on molec-ular abnormalities involved in thyroid tumorogenesis see Sagev et al65.
B.1. RAS Oncogene RAS oncogenes (K-ras, H-ras, N-ras) activation has been identified in tumors originating from the follic-ular epithelium of the thyroid gland. Point mutations in ras oncogenes are more common in follicular ade-noma and carcinoma than in papillary carcinoma as well as in tumors from iodide-deficient areas66,67. Some studies suggest that ras activation is an early event in thyroid carcinogenesis while others indicate an asso-ciation between ras mutations and tumor progres-sion68,69.
B.2. RET Oncogene The RET (rearranged during transfection) onco-gene encodes two isoforms of a transmembrane ty-rosine-kinase receptor, which is involved in the devel-opment of the neural crest and kidney70. A common genetic alteration in thyroid tumors is the rearrange-ments of the RET oncogene leading to the so-called RET/PTC (for papillary thyroid carcinoma) onco-gene71. Experimental studies have indicated that the RET/PTC oncogene is specifically involved in the pathogenesis of thyroid tumors with morhpologic fea-tures of papillary carcinomas, while a high prevalence of RET/PTC rearrangements have been reported in radiation-induced papillary carcinoma, especially in children affected in the Chernobyl reactor accident70-73. It is of interest that spontaneous RET mutations are also associated with familial and sporadic medul-lary carcinomas65.
Recent studies showed that RET/PTC was more frequently expressed in papillary microcarcinomas than in clinically manifest tumors, and poorly differ-entiated thyroid cancer has a low prevalence in RET activation36,74. It therefore appears that rearrangement of RET/PTC is an early event in papillary thyroid car-cinoma development and is less important in tumor progression.
B.3. p53 The p53 tumor suppressor gene is the most fre-quent mutated gene in human cancer. With regard to the thyroid, p53 gene mutations are rarely observed in differentiated tumors and are more commonly found in poorly differentiated and anaplastic carcino-mas75,76. The high p53 protein expression in undiffer-entiated carcinomas, compared to papillary carcino-mas, and the absence of mutations in the residual pap-illary component suggest that p53 genetic alterations are late events in the sequence of thyroid carcinogen-esis and could be linked to their reported worse prog-nosis77,78.
B.4. MET c-MET oncogene encodes a tyrosine kinase acting as the receptor for hepatocyte growth factor/scatter factor (HGF/SF), a powerful mitogen for epithelial cells, including thyroid follicular cells65,79. MET over-expression is associated with the papillary thyroid car-cinoma phenotype, particularly its aggressive forms, being negative in medullary carcinomas and low or absent in poorly differentiated tumors79,80. Data on the significance of MET overexpression in thyroid tumors are inconsistent. Some studies found MET overeac-tivity to be associated with advanced tumor stage and histologic variants with poor prognosis, while others showed a relationship between negative/low expres-sion and vascular invasion and distant metastases79,81.
C. CONCLUSIONS Thyroid tumors can originate from follicular epi-thelium, from parafollicular or C cells or from nonep-ithelial stromal components. The traditional classifi-cation of thyroid cancer as well differentiated carci-nomas, characterized by relatively good prognosis, or poorly differentiated carcinomas associated with ag-gressive behavior, metastases and death, is no longer applicable since certain morphologic variants of pap-illary carcinoma are associated with poor prognosis.
108 C.D. SCOPA The hyalinizing trabecular adenoma of the thyroid remains a controversial entity, while the concept of poorly differentiated carcinoma still constitutes a com-plicated issue. The practical importance of the vari-ous cytoplasmic changes seen in thyroid lesions re-sides on differential diagnosis rather than their bio-logical behavior. Several types of molecular alterations occurring in thyroid tumors are under investigation in order to enhance our understanding of the possible mechanisms of thyroid tumorogenesis. The clinical significance of these studies remains to be defined.
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17336 | https://artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality?srsltid=AfmBOoqyjG8d74xfaoAP9q9kieJjIgtL7e32quurphYSB3LfOBgC8Z52 | Art of Problem Solving
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Power Mean Inequality
The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality.
Inequality
For positive real numbers and positive real weights with sum , the power mean with exponent , where , is defined by
( is the weighted geometric mean.)
The Power Mean Inequality states that for all real numbers and , if . In particular, for nonzero and , and equal weights (i.e. ), if , then
Considering the limiting behavior, we also have , and .
The Power Mean Inequality follows from Jensen's Inequality.
Proof
We prove by cases:
for
for with
Case 1:
Note that As is concave, by Jensen's Inequality, the last inequality is true, proving . By replacing by , the last inequality implies as the inequality signs are flipped after multiplication by .
Case 2:
For , As the function is concave for all , by Jensen's Inequality, For , becomes convex as , so the inequality sign when applying Jensen's Inequality is flipped. Thus, the inequality sign in is flipped, but as , is a decreasing function, the inequality sign is flipped again after applying , resulting in as desired.
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17337 | https://faculty.ecnu.edu.cn/picture/article/1284/f9/a9/3934337642009863578c8b93a6c0/7d6ac3dc-364a-4b47-bb8c-9839d3ed43cc.pdf | ２０１９年第 ３期 故学敉学 ３ － １ ＨＰＭ视角下的人口增长问题学习进阶设计 ＃ 姜浩哲 汪晓勤 （华东师范大学教师教育学院 ，上海 ２０００６２） １ 问题的提出 人口增长问题是中 、小学数学教材中常见 的数学问题．在小学数学 “ 认识百分数 ” 相关章 节中 ，学生就有接触到 “ 人口增长率 ” 的概念 ， 并开始简单预测未来人口增长趋势 ；在初中学 习了函数的相关概念后 ，学生能通过函数的图 像寻找人口数量随时间增长的规律 ；而高中等 比数列的有关内容则与人口指数增长模型有 着密切的联系 ． 在我国 ，尽管人口增长问题学生各学段的 数学学习中均有涉及 ，但通常教学却往往只重 视数学知识本身，忽视了课程内容的联系 、学 生思维的发展和建模能力的培养．近年来 ，关 于学习进阶 （Ｌｅａｒｎｉｎｇ Ｐｒｏｇｒｅｓｓｉｏｎｓ ）①的相关研 究和应用方兴未艾 ［ １］ ．吴颖康等认为 ， “ 学习进 阶揭示了学生在学习和探索某 一主题时 ，对该 主题的思考 、理解与实践活动在相当长的 一段 时间内是如何从简单到复杂、从低水平到高水 平、从新手到专家逐步发展的． ” ［３ ］倘若基于学 习进阶加以设计 ，使人口增长问题教学能围绕 主线 ，不断深人贯穿于不同学段的相关课程内 容之中 ，不失为系统帮助学生认识模型本质和 提升建模能力的良策 ． 在数学史上 ，人口增长问题同样引发了数 学家们的广泛兴趣 ？早在公元前４５００年，巴比 伦王国就举办了全国性调査 ，按族登记人 口 ［４］ ． １７世纪 ，英国学者格兰特 （Ｊ ． Ｇｒ ａｕｎｔ ， １６２０ － １６７４）编制了世界上第 一张生命表 ，并正 式开始人口数量规律的科学研究 ［５ ］ ．著名数学 家欧拉（Ｌ ． Ｅｕｌｅｒ ， １７０７ － １７８３ ）在《无穷分析引 论》中介绍指数和对数时引人了相关人口指数 增长的实例 ［６ ］ ，马尔萨斯 （Ｔ． Ｍａｌｔｈｕｓ ， １７６６ － １８３４）在此基础上对人口指数增长模型进行了 完善 ． 一方面，人口增长问题的发展历史为建立 假设的学习进阶 （Ｈｙｐｏｔｈｅｔ ｉｃａ ｌ Ｌｅａｒｎｉｎｇ Ｔｒａ ｊｅｃｔｏｒ ｉｅｓ ）提供了参考 ［７ ］ ；另 一方面 ，从ＨＰＭ 视角看 ，丰富的历史材料在选择和加工后融入 课堂教学同样具有较高的价值和意义． ２ 教学设计 ２ ．１ 小学阶段人口增长问題的教学设计 ２ ．１ ． １ 新课引入 教师向学生介绍新课历史背景 ．人口增长 问题由来已久 ，我国先秦时期管子、孔子、商 鞅、韩非子等人就在著作中进行了相关论述 ， 甚至提出了人口和土地之间应有 一个理想比 例的思想．古希腊柏拉图 （Ｐｌａｔｏ ，公元前４２７ － 公元前３４７ ）和亚里士多德 （Ａｒ ｉｓｔｏｔｌｅ ，公元前 ３８４ －公元前 ３２２）从城邦国家的防务 、安全和 行政管理角度研究了人口 “ 适度 ” 问题． １７世纪 后 ，许多数学家、统计学家或都探讨过人口增 长问题 ．当然 ，研究人口增长问题 ，必然离不开 人口统计数据 ．早在公元前４５００年 ，巴比伦王 国就举办了全国性调查 ，按族登记人口 ．我国 是世界上唯 一有长期不间断人口资料记录的 国家 ，据《后汉书》记载 ，公元前２２００年 ，大禹 就曾经 “ 平水土 ，分九州 ，数万民 ” ，所谓 “ 数万 民 ” 就是统计人口 ． ２ ．１ ．２ 例题解析 例 １ 我国 （不含港澳台 ） １９７０年、１９８０年 和 １９９０年人口总数分别为８． ３０亿 、９． ８７亿和 ＊上海高校 “ 立德树人 ” 人文社会科学重点研究基地之数学教育教学研究基地研究项 目 “ 数学课程与教学中落实立 德树人根本任务的研究 ” 之系列论文之一 ． ① Ｌｅａｒ ｎｉｎ ｇ Ｐｒ ｏｇｒｅｓ ｓｉｏｎｓ与 Ｌｅａｒ ｎｉｎｇ ７Ｖａ ） ｅｃｔ ｏｒ ｉｅｓ本质上具有 一致性 ，因而在此可不做区分 （参阅文献［２］和［３ ］ ） ． ３ －２ 故爹狄学 ２０１９年第３期 １１． ４３亿① ，试比较 １９７０年至 １９８０年间 、 １９８０ 年至 １９９０年间人口增长速度 ． 教师指出 ， １９７０年至 １９８０年间、１９８０年至 １９９０年间人口数量分别增长了 １． ５７亿 、１． ５６ 亿 ．然而 ，从历史上看 ，我国自 １９８２年起推行计 划生育政策 ， １９８０年至 １９９０年间人口增长速 度呈放缓趋势 ．教师继而指出 ：在人口基数不 同的情况下 ，不能通过人口增长的绝对数量比 较人口增长速度 ． 教师联系百分数、比例的有关内容 ，启发学 生寻找能表示人口增长速度的 “ 相对量 ” ，并给 出学生人口增长率的数学公式 ：设某地区Ｘ年 人口总数为Ｓ ｉ ， Ｘ＋ｉＶ年人口总数为Ｓ ２ ，则该地 区在Ｚ年至Ｚ＋Ｗ年间人口总增长率为 ｘ １００％ ．教师要求学生运用人口增长率的概念和公 式作答例 １ ，发现结论与历史相符 ． 教师指出 ，学生可以将数学模型理解为数 字、字母或其他数学符号组成的 ，描述现实对 象数量规律的数学公式、图形或算法 ，虽然真 正实际问题的数学模型通常要比例 １复杂得 多 ，但是建立数学模型的基本步骤已经包含在 了解决例 １的过程中了 ：对实际人口总数进行 近似以简化问题 ；用字母表示数 ；根据实际问 题列出数学公式 ；求出数学解答 ；用得到的数 学答案解释原问题 ；最后还用历史背景知识对 上述结果进行了简单检验 ． ２ ．１ ．３ 思维拓展 思考题 １ 若已知２０１０年Ａ、Ｂ两地区人口 总数均为 １０万人 ，２０１０年至２０１５年间 ，Ａ地区人 口总增长率为３〇％ ， Ｂ地区每年平均人口增长率 为６％ ，试比较２０１５年Ａ 、Ｂ两地区的人口总数 ？ 小学生在百分数应用的学习过程中常有 这样的思维误区 ，即＃年间人口总增长率等于 年平均人口增长率与Ｗ的乘积 ，教师通过思考 题 １引导学生思考和发现 ５年后Ａ 、 Ｂ两地区 人口总数并不相等 ，既帮助学生巩固了所学新 知 ，也为后续学习埋下伏笔 ． 在课堂最后 ，教师为学生简单拓展人口自 然增长率与出生率 、死亡率之间的关系 ，并补 充介绍格兰特生命表的有关历史内容 ． ２ ． １ ．４ 课后活动 教师布置课后活动 ？ ，要求学生通过网络 、 图书等收集世界人口历史数据 ，绘制简单的世 界人口调查统计表、折线统计图等 ，估算不同 时期世界人口增长率并对未来世界人口进行 合理预测 ，搜集资料了解人口发展与资源环境 承载能力之间的关系 ． 我国著名教育家张奠宙教授指出 ： “ 解决 数学应用问题的本质是数学建模 ． ” 在小学阶 段 ，学生运用数学公式解答应用问题的过程 ， 本身也是 一种典型 、简约 、形象的数学建模过 程 ［９ ］ ．教师往往也会直接为学生建立模型 ，学 生 一般只需要找出模型中数学符号在具体问 题情境中的含义 ，代人相关数值后即可求解得 到答案 ．教学设计既包含有中国传统式的课堂 教学 ，也将美国探究实践式的课后活动融人其 中 ．教师通过建立 “人口增长率 ” 模型帮助学生 学习百分数的应用 ，也使学生体会了数学建模 的基本步骤 ，从而对数学模型产生初步的认识 ． ２ ．２ 初中阶段人口增长问题的教学设计 ２ ．２ ．１ 新知传授 教师教授学生描点法作函数图像的 一般 步骤 ：列表 、描点、连线 ，并详细说明和举例 ＿ ２ ．２ ．２ 例题解析 教师引导学生从函数的观点分析人口增 长问题并给出例题 ． 例２ 表 １给出了近两个世纪的美国人口 统计数据 ，运用描点法作出人口数量随时间 （年份）变化的函数的图像 ？ 学生作答后 ，教师要求学生回忆小学阶段 学习的 “人口增长率 ” 概念并进行相关计算 ，随 后 ，教师引导学生发现 ：在人口增长率较高的时 期 ，函数图像上升 “ 陡峭 ” ，而在人口增长率较低 的时期 ，函数图像上升 “ 平缓 ” ，并指出通过函数 图像观察人口增长趋势更具直观性？同时 ，教师 补充了 １９３０年至 １９６０年经济萧条和二次大战 的历史背景 ，以解释其间人口增长缓慢的原因 ． ① 数据来源 ：中华人民共和国国家统计局 ． 中国统计年鉴２０１４［ Ｍ］ ．北京 ：中国统计出版社 ，２０１４． ② 本课后活动教学设计参考了全美数学教师协会 （ＮＣＴＭ）的官网资源和文献［８ ］ ． ２０１９年第３期 获学教 ■学 ３ — ３ 表 １ 美国历史人口数据 ［ １１］ 年 １７９０ １８００ １８１０ １８２０ １８３０ １８４０ １８５０ １８６０ １８７０ １８８０ １８９０ 人口 ／百万 ３ ． ９ ５． ３ ７ ． ２ ９ ． ６ １２． ９ １７ ． １ ２３ ．２ ３ １ ．４ ３８ ． ６ ５０ ．２ ６２． ９ 年 １９００ １９１０ １９２０ １９３０ １９４０ １９５０ １９６０ １９７０ １９８０ １９９０ ２０００ 人口 ／百万 ７６．０ ９２ ．０ １０６ ．５ １２３ ．２ １３１ ．７ １５０ ．７ １７９ ．３ ２０３ ． ２ ２２６．５ ２４８ ． ７ ２８１．４ 从简单运用已建立的模型到根据 一般步 骤自主建立模型 ，这 一阶段的学习让学生对数 学建模有了更进 一步的尝试．再看数学模型 ， 小学时 ，学生了解的数学模型仅仅是用来解决 实际问题的 ，它只是具有应用价值 ，但如今 ，教 师引导学生发现数学模型还是对现实情境的 一种简洁、清晰的表达 ， 一段函数图像 ，却很好 地阐明了近两个世纪的人口增长过程 ，这也很 好地解释了人们常说的 “ 数学是 一种语言 ” ，从 这个意义上说 ，学生在对数学模型本质的认识 上也完成了 一次进阶． ２ ．３ 高中阶段人口增长问题的教学设计 ２ ．３ ． １ 新课引入 首先 ，教师引导学生回忆小学时的思考题 １ ，要求学生猜想 ｉＶ年间人口总增长率与年平均 人口增长率之间的关系 ．同时 ，教师帮助学生回 顾初中阶段学习的例 ２ ，要求学生计算 １７９０年 至 １８６０年期间各相邻 １０年人口数量的比值 ，发 现近似相等 ．教师由此引导学生建立等比数列 模型 ：从 １９７０年起到 １８６０年美国每 １０年的人 口数量近似地构成 一个等比数列｛？？｝ ，若每 １０ 年人口总增长率为 贝 ！１ＫＪ的公比９ ＝ １ ＋＊． ２ ．３ ．２ 历史展示 教师向学生展示数学家欧拉研究人口增 长问题的历史． １７４８年 ，欧拉在《无穷分析引 论》中介绍指数和对数时就引人和讨论了相关 人口增长问题．欧拉认为 ，第 ／ｔ年的人口数量 满足圪＋１ ＝ （ １ ＋幻圪 ，其中ｎ为正整数４代 表人口增长率 为正实数 ．如果考虑初始人口 数量Ｐ 。 ，则广 ＝ （ １ ＋幻 这被称作是人口几 何级数增长模型或人口指数增长模型 ［ １° ］ ．其 后 ，人口学家马尔萨斯在调查了英国 １〇〇多年 的人口统计资料 ，对指数增长模型进行了完善 并补充了人口增长率不变的假设 ［１ １ ］ ．教师随后 要求学生建立模型解答下述选自《无穷分析引 论》中的例题？ 例 ３ 某地现有人口 １０万 ，年增长率为 ￥ ，求百年后该地人口数 ． 教师介绍 ，这道问题欧拉是根据 １７４７年柏 林人口调查数据提出的 ， １７４７年柏林人口数量 约为 １０． ７２２ ４万 ．欧拉通过人口指数增长模型 说明自然情况下任何地区在 一个世纪中人口 数量将增加 １０倍以上 ，而这 一事实也在伦敦被 观察证实 ［ １〇］ ． ２ ．３ ．３ 思维拓展 教师指出 ，在欧拉《无穷分析引论》指数与 对数的相关章节还讨论了贷款利率问题 ，与人 口指数增长模型十分类似 ，思考题２对原书问 题进行了适当改编． 思考题２ 某人以 ５％的年复利率借款４０ 万弗罗林，商定 年后 一次性还款，问ｎ年后连 本带息共应还款多少？ 教师引导学生运用类比的思想方法建立 等比数列模型解答 ，将两类问题进行适当比较 并要求学生联系小学阶段课后活动中人口发展 与资源环境承载能力关系的资料思考 ：在贷款 利率问题中 ，如果 ／１— ００ ，贝 ！１还款金额必定也会 趋向于无穷大 ；但是在人口增长问题中 ，如果时 间接近无限长 ，人口数量会趋向于无穷大吗？ 如果说小学阶段通过 “ 增长率 ” 研究人口 增长问题较为繁琐且缺少了直观性 ，初中阶段 通过函数的图像研究人口增长问题不能很好 满足定量分析的要求 ，那么在高中阶段 ，借助 等比数列相关知识得到的人口指数增长模型 则同时弥补了以上两种方法的缺陷．通过数列 通项公式 ，便可以简单明了计算和预测某 一地 区较长时间内人口数量 ．高中阶段 ，教师从小 学思考题和初中例题的内容再出发 ，引导学生 在事先没有步骤和方法可循的情况下自主建 立等比数列模型 ．思维拓展环节 ，教师结合《无 穷分析引论》的相关内容设置思考题，引导学 ８ －４ 故擎枚学 ２０１９年第 ３期 生通过类比的思想方法使数学模型应用到更 多新的领域 ，数学模型再也不是预制好的、 一 成不变的 ，而是可以根据具体问题不断调整、 灵活多变的 ；也不是为特定对象所独有的 ，而 是可以转移到所有合适领域中的．课堂最后 ， 教师再次结合小学课后活动内容启发学生进 行深度思考 ，帮助学生了解人口指数增长模型 也有不完善的地方 ，因为地球的资源和空间是 有限的 ，人口不可能无限制地增长 ，并引导学 生树立正确的人口观 ． ３ 数学史与学习进阶融合的方式 表 ２ 数学史与学习进阶融合的方式 学段 数学史运 用方式 教学 环节 本教学环节数 学史相关内容 回顾的其他学 段或环节的内容 回顾相关教学内容的目的 高中 整体性 重构 复制式 例题 解析 《无穷分析引论》中 人口增长问题 顺应式 新课 引人 运用指数增长模型 分析美国 １８世纪上 半叶人口数据 况年间人口总增长 率与年平均人口增 长率之间的关系 引导学生联系等比数列的有关内容建 立模型 思维 拓展 《无穷分析引论》中 贷款利率问题 《无穷分析引论》中 人口增长问题、人口 发展与资源环境承 载能力之间的关系 引导学生通过类比的思想方法建立和 应用数学模型 ，并对两类问题进行比 较 ．启发学生树立正确的人口观 初中 顺应式 例题 解析 通过函数的图像分 析美国历史人口 数据 通过增长率分析中 国历史人口数据 补充说明人口增长率在函数图像中的 体现，对比说明通过函数图像观察人口 增长趋势更具直观性 小学 附加式 新课 引人 介绍人口思想和人 口普査的历史 思维 拓展 介绍格兰特生命表 课后 活动 査阅资料了解世界 历史人口数据、人口 发展与资源环境承 载能力之间的关系 顺应式 例题 解析 通过增长率分析中 国历史人口数据 纵观从小学至髙中的人口增长问题教学 设计 ，数学史在整体性重构后与学习进阶相融 合的 ，史料和问题在教学中环环相扣、层层铺 垫 ，如表２所示 ．教学设计中 ，数学史料和问题 不断回顾、补充、对比 ，意图促进学生的认知和 元认知水平 ［ １２ ］ ． ４ 数学史与学习进阶融合的价值 在人口增长问题的教学设计中 ，不仅由 学习进阶联结而成的数学内容 、思想的体系 和网络呈现了 “ 方法之美 ” ，而且融合了的数 学史与学习进阶还共同体现了 “ 知识之谐 ” 、 “ 探究之乐 ” 、 “ 能力之助 ” 、 “ 文化之魅 ” 和 “ 德 育之效 ４ ． １ 方法之美 许多研究发现学习进阶有助于学生形成 更丰富的知识网络或概念序列 ［ ３ ’ １３ ］ ．通过学习 进阶 ，百分数、函数、数列等多样的数学方法 ， 类比、数形结合等精妙的数学思想在人口增长 问题教学中联结成系统全面、有机统 一的网络 体系 ，呈现出 “ 方法之美 ４ ．２ 知识之谐 一方面 ，基于历史人口统计数据或历史上 ２０１９年第 ３期 故学故学 ３ －５ 数学家们的认知引导学生发现和解决问题，使 得学生理解和学习模型有关知识的过程变得 自然而然、水到渠成 ？ ，另 一方面 ，基于学习进阶 的教学是依据学生的认知和理解而设计的 ，各 学段教学内容环环相扣、层次铺垫 ，学生也能 明白有关模型或知识不是 “ 降落伞 ” 从天而降 ， 体现了 “ 知识之谐 ４ ．３ 探究之乐 一方面 ，历史人口数据为数学建模提供了 丰富的资源依托 ，学生在探究如何更好地描述 历史人口增长规律的过程中既能被数学家们的 智慧吸引 ，也能因自身数学活动经验的积累而 收获乐趣．另 一方面 ，在学习进阶中 ，人口增长问 题环环相扣 、层层深人 ，面对愈发复杂的问题情 境 ，学生更能在好奇心的驱使下揭开问题的层 层面纱 ，刨根究底之中 “探究之乐 ” 也愈发浓烈 ． ４ ．４ 能力之助 一方面 ，历史问题有助于提升学生数学建 模等核心素养 ［ １４ ］ ；另 一方面 ，学习进阶为循序 渐进地引导学生发展建模能力提供了科学依 据 ．从开始学会运用已经建立的模型 ，到自主 建立和求解模型 ，学生数学建模能力不是 一蹴 而就的 ，基于学习进阶的教学也无疑成为 “ 能 力之助 ” ？ ４ ．５ 文化之魅 一方面 ，人口增长问题历史悠久 ，既展现 了数学与现实生活的联系 ，教学设计也充分体 现了 Ｍ ？克莱因 （Ｍ ． Ｋｌｉｎｅ ， １９０８ － １９９２）的 “数学文化原理 ” ；另 一方面 ，学习进阶过程中 ， 教学设计同样为学生呈现出多元的学习文 化 ［ １５ ］ ：既有中国传统文化注重的循序渐进的课 堂讲授 ，也引入了西方文化所提倡的旨在引导 学生实践 、探索 、体验的课后活动 ？ ４ ．６ 德育之效 一方面 ，教学设计落实了 “立德树人 ” 这 一 今日教育的根本任务 ，通过数学史的融人引导 学生更好地认识数学活动的本质 ，理解数学是 不断演进的 ，数学家们勤奋、执着、严谨的优秀 品质更向学生传递了数学背后的人文精神 ；另 一方面 ，基于学习进阶的教学也促进了 “ 人口 观 ” 的发展 ，教师引导学生逐渐理解了人口增 长应与资源环境相适应 ，潜移默化的 “ 德育之 效 ”也由此产生 ． ５ 结语 从 １９８９ － ２０１６年核心期刊文献的统计分 析可以发现 ，中小学生数学建模能力的提高是 数学教育研究人员关注的焦点 ，优秀的数学建 模案例还略显不足 ［ １６ ］ ． — 方面 ，数学史可以作 为假设的学习进阶建立的依据 ，而学习进阶又 能科学地、循序渐进地发展学生建模能力 ；另 一方面 ，数学史又为数学建模案例的开发提供 了丰富 、真实的背景资源 ．与此同时 ，数学史和 学习进阶融合后还体现了多个维度的价值（如 图 １所示 ） ．当然 ，考虑到学习进阶本身具有假 设性的特点 ，我们在学习进阶设计的过程中尚 未开展实证研究 ，因而还存在部分值得完善的 地方 ．但是 ，我们有理由相信 ，从ＨＰＭ视角下 设计数学建模的学习进阶无论是对于当前的 教学实践、案例开发 ，还是对今后教科书的修 订和编写 ，都具有重要的参考意义． 数学建模 图 １ ＨＰＭ视角下的数学建模学习进阶设计 知识之谐 方法之美 探究之乐 能力之助 文化之魅 德育之效 ３ － ６ 参考文献 ［ １ ］ Ｎａｔｉｏｎａｌ Ｒｅｓｅａｒｃｈ Ｃｏｕｎｃｉｌ （ ＮＲＣ ） ． Ｔａｋｉｎｇ Ｓｃｉｅｎｃｅ ｔｏ Ｓｃｈｏｏｌ ： Ｌ ｅａｒｎｉｎｇ ａｎｄ Ｔｅａｃｈｉｎｇ Ｓｃｉｅｎｃｅ ｉｎ Ｇｒａｄｅｓ Ｋ － ８ ［Ｍ ］ ， Ｗａｓｈｉｎｇ ｔｏｎ ： Ｎａｔｉｏｎａｌ Ａｃａｄｅｍｉｅｓ Ｐｒｅｓｓ ， ２００７ ：２１１ ＿ ２５１ ． ［ ２ ］ Ｄａ ｒｏ Ｐ ， Ｍｏｓｈｅｒ Ｆ Ａ ， Ｃｏｒｃｏｒ ａｎ Ｔ ． Ｌｅａｒｎｉｎｇ Ｔｒａ ｊｅｃｔｏｒｉｅｓ ｉｎ Ｍａ ｔｈｅｍａｔｉｃｓ ： Ａ Ｆｏｕｎｄａ ？ ｔｉｏｎ ｆｏｒ Ｓｔａｎｄａｒｄｓ ， Ｃｕｒｒ ｉｃｕｌｕｍ ， Ａｓｓｅｓｓｍｅｎｔ ， ａｎｄ Ｉｎｓｔｒｕｃｔｉｏｎ ［ＥＢ／ＯＬ］ ． （２０１５ －〇６ －〇３） ［２０１６ － １１ 一 ２０ ］ ． Ｃｏｎｓｏｒｔｉｕｍ ｆ ｏｒ Ｐｏｌｉｃｙ Ｒｅｓｅａｒ ｃｈ ｉｎ Ｅｄｕｃａ ｔｉｏｎ ， ２０１１ ． ｈｔｐ ：／／ｗｗｗ． ｃｐｒｅ． ｏｉ＾／ｓｉｔｅｓ／ｄｅ ｆ ａｕｌｔ／ ｆｉｌｅｓ／ｒｅｓｅａｒｃｈｒｅｐｏｒ ｔ／１２２０＿ ｌｅａｍｉｎｇｔｒ ａ ｊｅｃｔｏｒｉｅ ｓｉｎｍａｔｈｃｃ ｉｉｒｅｐｏｒ ｔ ． ｐｄｆ． ＤＯＩ ；１０． １２６９８／ｃｐ ｒｅ． ２０１１ ． ｒｒ６８． ［ ３ ］ 吴颖康 ，邓少博 ，杨洁 ．数学教育中 学习进阶的研究进展及启示 ［Ｊ］ ．数学教育学 报 ，２０１７ ，２６（６ ） ：４０ ＿４６． ［ ４ ］ 孙競新 ？人口普査的历史 ［ Ｊ］ ？江苏 统计，２０００（ｓｌ ） ：３８ －４１ ． ［ ５ ］ （美）Ｐｏｌｌａｒｄ ＪＨ著 ．人口增长的数 学模型［Ｍ ］ ．姚志坚 ，译 ．成都 ：四川大学出版 社 ， １９８８ ： １ － ２． ［ ６ ］ Ｅｕｌｅｒ Ｌ． Ｉｎｔｒ ｏｄｕｃｔ ｉｏｎ ｔｏ Ａｎａｌｙｓ ｉｓ ｏｆ ｔｈｅ Ｉｎｆ ｉｎｉｔｅ ［Ｍ ］ ， Ｎｅｗ Ｙｏｒｋ ： ＳＰＲＩＮＧＥＲ － ＶＥＲＬＡＧ ， １９８８ ：７５ － ９２． ［ ７ ］ Ｂａｋｋｅｒ Ａ ． Ｄｅｓｉｇｎ Ｒｅｓｅａｒｃｈ ｉｎ Ｓｔａｔｉｓｔｉｃｓ Ｅｄｕｃａｔｉｏｎ — ｏｎ Ｓｙｍｂｏｌｉｚｉｎｇ ａｎｄ Ｃｏｍｐｕｔｅｒ Ｔｏｏｌｓ ［Ｄ ］ ． Ｔｈｅ Ｆｒｅｕｄｅｎｔｈａｌ Ｉｎｓｔｉｔｕ ｔｅ ， Ｕｔ ｒｅｃｈｔ ， ２００４ ： ５１ ，８７ ． ［ ８ ］ 蓝非 ．掌握方法是解决数学问题的 ２０１９年第 ３期 关键 —美国小学数学拓展课《人口增长中的 数学》案例 ［Ｊ］ ？现代教学 ，２００５（ １ １） ：６０ ＿ ６１ ． ［ ９ ］ 张秋爽 ？在问题教学中建构数学模 型 一－有感于吴正宪《行走中的数学问题》的 教学［Ｊ］ ．教育视界 ，２０１５ （８） ：２５ －２６． ［ １０］ Ｂａｃａ＾ｒ Ｎ． Ａ Ｓｈｏｒ ｔ Ｈｉｓｔｏｒｙ ｏｆ Ｍａｔｈｅ ？ ｍａ ｔｉｃａ ｌ Ｐｏｐｕｌａ ｔｉｏｎ Ｄｙｎａｍｉｃｓ ［ Ｍ ］ ． Ｌｏｎｄｏｎ ： Ｓｐｒ ｉｎｇｅｒ Ｌｏｎｄｏｎ ， ２０１１ ： １３ － ３９． ［ １１ ］ 姜启源 ，谢金星 ，叶俊 ．数学模型 （第 ４版 ） ［Ｍ ］ ．北京 ：高等教育出版社 ， ２０１１ ： １６３ － １７３ ． ［ １２ ］ Ｆ ａｕｖｅｌ Ｊ ， Ｍａａｎｅｎ Ｊ Ｖ． Ｈｉｓｔｏｒｙ ｉｎ Ｍａｔｈｅｍａｔｉｃｓ Ｅｄｕｃａ ｔｉｏｎ ［ Ｍ ］ ． Ｄｏｒｄｒｅｃｈｔ ： Ｋｌｕｗｅｒ Ａｃａｄｅｍｉｃ Ｐｕｂｌｉｓｈｅｒ ， ２０００：２８９ － ２９０． ［ １３ ］ Ｎｏ ｒｄｉｎｅ Ｊ ， Ｋｒａ ｊｃｉｋ Ｊ ， Ｆｏｒｔｕｓ Ｄ ． Ｔｒａｎｓｆｏｒｍｉｎｇ Ｅｎｅｒ ｇｙ Ｉｎｓｔｒｕｃｔｉｏｎ ｉｎ Ｍｉｄｄｌｅ Ｓｃｈｏｏｌ ｔｏ Ｓｕｐｐｏｒｔ Ｉｎｔｅｇｒ ａｔｅｄ Ｕｎｄｅｒ ｓｔａｎｄｉｎｇ ａｎｄ Ｆｕｔｕｒｅ Ｌｅａｒｎｉｎｇ ［ Ｊ］ ． Ｓｃｉｅｎｃｅ Ｅｄｕｃａｔｉｏｎ ， ２０１１ ，９５（４ ） ： ６７０ －６９９． ［ １４ ］ Ｗａｎｇ Ｘ Ｑ ， Ｑ ｉ Ｃ Ｙ ， Ｗａ ｎｇ Ｋ ． Ａ Ｃａｔｅｇｏｒ ｉｚａｔｉｏｎ Ｍｏｄｅｌ ｆｏｒ Ｅｄｕｃａｔｉｏｎａ ｌ Ｖａｌｕｅｓ ｏｆ ｔｈｅ Ｈｉｓｔｏｒｙ ｏｆ Ｍａｔｈｅｍａ ｔｉｃｓ ： Ａｎ Ｅｍｐ ｉｒ ｉｃａｌ Ｓｔｕｄｙ ［Ｊ］ ． Ｓｃｉｅｎｃｅ ＆ Ｅｄｕｃａｔｉｏｎ ， ２０１７ ，２６（２） ； １０２９ ＿ １０５２． ［ １５］ 张三花 ，黄甫全．学习文化研究 ：价 值、进展与走向 ［ Ｊ ］ ．江苏高教 ， ２０１０ （６ ） ： １５ － １８ ． ［ １６］ 牛伟强 ，张倜 ，熊斌．中国中小学数 学建模研究的回顾与反思 —基于 １９８９ — ２０１６年核心期刊文献的统计分析［ Ｊ］ ．数学教 育学报 ，２０１７ ，２６（５） ：６６ － ７０． 故擎敉学 |
17338 | https://www.studocu.com/en-us/messages/question/5049790/what-are-the-possible-values-of-ml-for-each-of-the-following-values-of-l-0-1-2-3 | [Solved] What are the possible values of ml for each of the - General Chemistry I (CHEM 1411 ) - Studocu
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What are the possible values of ml for each of the following values of l?0123General Chemistry I (CHEM 1411 )
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St. Philip's College
General Chemistry I
Question
What are the possible values of ml for each of the
St. Philip's College
General Chemistry I
Question
Anonymous Student
1 year ago
What are the possible values of ml for each of the following values of l?
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123069396 Chem 11 Exam Review with Answers General Chemistry I Lecture notes None
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1 year ago
Answer
The magnetic quantum number, often denoted as ml, determines the number of orbitals and their orientation within a subshell. The possible values of ml can range from -l to +l, including zero. This means that for each value of the azimuthal quantum number l, there are 2l + 1 possible values of ml.
Let's break down the possible values of ml for each given value of l:
l = 0
For l = 0, the possible value of ml is:
0
l = 1
For l = 1, the possible values of ml are:
-1
0
+1
l = 2
For l = 2, the possible values of ml are:
-2
-1
0
+1
+2
l = 3
For l = 3, the possible values of ml are:
-3
-2
-1
0
+1
+2
+3
In summary, the possible values of ml increase as the value of l increases, ranging from -l to +l.
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17339 | https://spinningnumbers.org/a/thevenin-howto.html | Spinning Numbers
Thévenin's theorem how to
We proved Thévenin’s and Norton’s theorems in the previous article. This article covers the practical steps to create a Thévenin or Norton equivalent for a linear circuit.
Thévenin’s theorem
Any network of resistors and sources, when viewed from a port, can be simplified down to one voltage source in series with one resistor.
Norton’s theorem
Any network of resistors and sources, when viewed from a port, can be simplified down to one current source in parallel with one resistor.
The proof and the practical design method are separate ideas. You often find them mixed together in many texts.
Written by Willy McAllister.
Contents
Using Thévenin’s theorem
When is Thévenin’s theorem useful?
Strategy
Example 1
Example 1 simulation models
Example 2
Where we’re headed
To create a Thévenin or Norton equivalent,
Pick two nodes to be the port of the circuit you want to transform.
Remove the external component(s) connected to the port.
Find any two of these three: R,voc,isc — whichever two are the easiest,
R — Suppress the internal sources and simplify the resulting resistor network down to a single resistor.
voc — Restore internal sources, leave the port open, find the open-circuit voltage.
isc — Restore internal sources, short across the port, find the short-circuit current.
The third variable comes from the other two,
R=voc/isc
voc=iscR
isc=voc/R
The Thévenin equivalent is RT in series with VT,
VT=voc
RT=R
The Norton equivalent is RN in parallel with IN,
IN=isc
RN=R
One of the coolest ideas from linear circuit theory is the concept of an equivalent circuit. Say you have a circuit and you “look into” a port using a voltmeter and ammeter. You observe some sort of i-v behavior. If you have another circuit that displays the same i-v behavior then those two circuits are equivalent from the viewpoint of the ports.
In an earlier article on how to simplify a resistor network, we learned how to turn any resistor network into an equivalent single resistor. Thévenin’s and Norton’s theorems are the next step. They teach us how to simplify networks of resistors and sources. If you have a complicated linear circuit the theorems provide the instructions for how to construct a very simple equivalent circuit.
Using Thévenin’s theorem
Thévenin’s theorem looks like this in schematic form,
The top circuit is made of any number of resistors, voltage sources, and current sources. We select two internal nodes and mark them with little circles to define a port we care about. We draw the port so it pokes out the side of the circuit. We create a Thévenin equivalent “from the viewpoint” of this port.
When is Thévenin’s theorem useful?
Think about using Thévenin’s theorem when you want to focus on a specific part of a circuit and push the details of the rest into the background. For example, suppose you care about what an amplifier does at its output port. Thévenin’s theorem creates a simple equivalent version of the complicated amplifier with the exact same i-v behavior at the output.
Sometimes you can apply Thévenin’s theorem as an alternative to Kirchhoff’s Current Law or Voltage Law. It is another circuit design tool to have in your toolbox.
Strategy
We use aspects of the proof to come up with practical steps to find an equivalent circuit.
A Thévenin equivalent has a voltage source VT in series with a resistor R.
The Norton equivalent has a current source IN in parallel with the same resistor R.
To create a Thévenin equivalent,
Pick two nodes to be the port of the circuit you want to transform.
Remove any external component(s) connected to the port.
Transform what remains to its Thévenin equivalent.
The resistor—you can call it R or RT or RN—is the simplified equivalent resistance of the resistor network from the original circuit when all internal sources are suppressed.
To suppress a voltage source, replace it with a short.
To suppress a current source, replace it with an open.
The Thévenin voltage VT is voc, the voltage at the port when all internal sources are turned on and the port is left open.
Alternatively, you can find the Norton current IN. It is isc, the current flowing from the port when all internal sources are turned on and a short is placed across the port.
If you know isc and R, compute voc=VT=iscR for the Thévenin equivalent.
If you know voc and R, compute isc=IN=voc/R for the Norton equivalent.
When you find voc or isc you have to do a circuit analysis. Choose either one based on how hard you think the analysis task will be. If you can’t tell, flip a coin.
Example 1
We are asked to find the output voltage for several different values of the load resistor (the initial value of the load resistor is 2kΩ shown on the right),
We could solve the whole circuit for each value of the load. Or, we can find the Thévenin equivalent of everything to the left of the port (the two little circles) and solve a much simpler circuit for each value of the load.
Define the port
First, decide what part of the system you want to reduce to a Thévenin equivalent. Select two nodes you care about and define them as the port. Then remove any components external to the circuit.
We are interested in what’s happening at the 2kΩ resistor on the far right, so we identify the port by drawing two little circles. Our goal is to simplify everything to the left of the port by finding its Thévenin equivalent. The first step is to take away the 2kΩ resistor,
Find the Thévenin components
The next task is to identify the Thévenin voltage and the Thévenin resistor.
Thévenin resistance
To find the Thévenin resistance, suppress internal sources and compute a single equivalent R. We use this same resistance in both the Thévenin and Norton equivalent circuits, so we can call it RT or RN or just R.
Here is the circuit with the two internal sources suppressed—the voltage source is replaced with a short, the current short is replace with an open,
With all the sources gone, what’s left? A resistor network. So let’s reduce the network to a single equivalent resistance (∥ is a shorthand notation for “in parallel with”),
R=500+(1000∥1000)=500+1000+10001000⋅1000
R=1000Ω
Thévenin voltage
To find VT we have to do a circuit analysis. If it looks like it’s going to be difficult, think about solving for IN instead. Pick the easiest route.
To find the Thévenin voltage restore the internal sources and leave the port open. Then find the open circuit voltage, voc. voc is the voltage the circuit presents to the world when nothing is connected to its port. The Thévenin voltage is equal to voc.
Have a go at finding voc yourself. (It is not that easy.) Hint: When a circuit has multiple sources one of your choices is to solve by superposition.
The answer is tucked away here—try yourself before peeking,
Find voc
The circuit has multiple sources. When you see multiple sources, think superposition!.
Break the circuit into two sub-circuits, one for each source. Then work out voc for each one. The final voc is the superposition of the two sub-circuits, voc=voc1+voc2.
Sub-circuit #1 has the voltage source, with the current source suppressed,
Use the voltage divider formula to find voc1,
voc1=5V1000+10001000=2.5V
Sub-circuit #2 has the current source restored and the voltage source suppressed,
The 2mA current flows through the three resistors,
voc2=2mA(500+1000∥1000)=2mA(500+500)
voc2=2V
Now superimpose the two sub-circuits,
voc=voc1+voc2=2.5+2
voc=4.5V
If you are not already in love with superposition, try to solve this circuit with the Node Voltage Method. The difference in effort is striking.
Norton current
The third variable you might want to find is the Norton current, IN. You have to do a circuit analysis for this, too, but it may turn out to be easier than finding voc. Let’s see if that’s the case.
To find the Norton current restore all the internal sources and place a short circuit across the port. Then find the current in the shorting wire, isc. The Norton current is equal to isc. It is the current the circuit would generate if you forced the output voltage to zero. (Please be very careful if you do this to a real circuit. It might not be designed to drive a short.)
The solution for isc is tucked away here—see if you can do it yourself before peeking,
Find isc
This problem has multiple sources. Whenever I see multiple sources my first thought is superposition! Superposition turns one nasty circuit into multiple simple circuits.
Solve for isc twice, and add the results, isc=isc1+isc2
Sub-circuit #1 includes the voltage source, with the current source suppressed,
Find isc1,
5V=i1(1000+1000∥500)
(1K∥500)=1000+5001000⋅500=333Ω
i1=1000+3335V=3.75mA
Find vx with the voltage divider formula,
vx=51000+333333=5⋅0.250=1.25V
isc1 is the current flowing in the 500Ω resistor,
isc1=500vx=5001.25
isc1=2.5mA
Sub-circuit #2 has the current source restored and the voltage source suppressed. This one is particularly easy to solve.
The wire across the port directly shorts across the current source. That means the voltage across the current source and the resistor network is zero. Therefore, the entire 2mA current flows in the shorting wire,
isc2=2mA
Superimpose the two components of isc,
isc=isc1+isc2=2.5mA+2mA
isc=4.5mA
For this example the effort to find isc compared to voc was similar, with voc probably a little easier overall. No harm done, we got to see both. Hooray for superposition!
Find VT from isc and R
We can use isc and R to find the Thévenin voltage.
VT=iscR
VT=4.5mA1000Ω
VT=4.5V
The same answer is the same as when we analyzed the circuit for voc.
Assemble the equivalent circuits
Put it together. The Thévenin equivalent is R in series with the open-circuit voltage source,
And pretty much for free we get the Norton equivalent. The Norton equivalent is the same R in parallel with the short-circuit current source,
Example 1 simulation models
Open this simulation model.
Click on DC to run a DC analysis.
Look at the current and voltage for the 2kΩ load resistor of the first circuit. Compare that to the current and voltage of the Thévenin and Norton equivalents. Are they the same?
Explore: Change the load resistor to some other value in all three by double-clicking on the resistor symbol.
Run DC analysis again.
The the current and voltage will be different, but all three should still match because all three circuits have the same i-v equation. Pretty cool, eh?
If you want to simulate the superposition solutions, suppress the sources one at a time in the two versions in the lower left and run DC again.
Example 2
Here’s a practical application. This circuit shows a common way to set up a bipolar junction transistor (BJT) as an amplifier. The BJT is the symbol in the center right, with reference designator Q1. The power supply for the amplifier is provided by the 15V source.
The 100kΩ and 50kΩ resistors set the voltage of Q1’s base terminal to an intermediate value between the power supply and ground. Together these resistors are called the biasing network. We are going to convert the biasing network into its Thévenin equivalent.
Identify the port and isolate the biasing network by removing the external components.
Isolated biasing network
Find the two Thévenin components, a voltage source and resistance.
Thévenin voltage and resistance
The Thévenin voltage is the voltage on the port when we leave it open, voc. The circuit is a voltage divider so we’ll use that formula to find voc,
VT=voc=15V100k+50k50k=15V⋅31
VT=5V
To get the Thévenin resistance we suppress the voltage source by replacing it with a short circuit. The Thévenin resistance is what’s left,
The symbol on the right is an eyeball “looking into” the port to find RT.
RT=50k∥100k=50k+100k50k⋅100k=150k5000M
RT=33.3kΩ
Assemble the two components to get the Thévenin equivalent,
Thévenin equivalent
Embed the Thévenin equivalent back into the amplifier circuit.
Thévenin equivalent embedded into amplifier
Summary
Thévenin’s theorem is another design tool to put in your toolbox. Use it is an alternative to Kirchhoff’s Current Law or Voltage Law.
Thévenin’s theorem
A circuit made of any combination of resistors and sources can be simplified down to a single voltage source in series with a single resistor.
Norton’s theorem
A circuit made of any combination of resistors and sources can be simplified down to a single current source in parallel with a single resistor.
Norton and Thévenin forms are interchangeable because of what we learned in the article on source transformation.
To create a Thévenin or Norton equivalent,
Identify the port and remove the external component(s).
Find any two of these three: R,voc,isc, whichever two are the easiest,
R — Suppress the internal sources and simplify the resulting resistor network.
voc — Restore internal sources, leave the port open, find the open-circuit voltage.
isc — Restore internal sources, short across the port, find the short-circuit current.
The third variable can be derived from the other two,
R=voc/isc
voc=iscR
isc=voc/R
The Thévenin equivalent is R in series with voc.
The Norton equivalent is R in parallel with isc.
Questions
William Weigeshoff,
31 March 2021
This is an amazing explanation and example. I found it really useful and helpful while studying for an upcoming test, do you think you could do an example with a dependent source in the future?
↪︎ Reply to William Weigeshoff
William Weigeshoff,
31 March 2021
This is an amazing explination and example. I found it really useful and helpful while studying for an upcoming test, do you think you could do an example with a dependent source in he future?
↪︎ Reply to William Weigeshoff
Loyad Joseph Losan,
23 May 2020
Thank you for this bereft piece of knowledge. The subject was daunting me for quite a long period, but now I am relieved……Thank you Will
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17340 | https://math.stackexchange.com/questions/3691633/can-an-average-of-an-overall-set-be-equal-to-the-average-of-a-subset | statistics - Can an average of an overall set be equal to the average of a subset? - Mathematics Stack Exchange
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Can an average of an overall set be equal to the average of a subset?
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I read in a publication that the average salary for lawyers in America is ˉ x=$163,595 x¯=$163,595. Of these salaries, the average for men is ˉ x m=$183,687 x¯m=$183,687, and for women, it is ˉ x w=$163,595 x¯w=$163,595. I'm thinking, how is it possible for the average of women's salary be equal the average of the entire set?
Note that ˉ x w=ˉ x<ˉ x m x¯w=x¯0 x¯w,x¯,x¯m>0.
Can we prove that this is possible/impossible? Can we find 1 simple example where ˉ x w=ˉ x x¯w=x¯ (given ˉ x w,ˉ x,ˉ x m>0 x¯w,x¯,x¯m>0, and ˉ x m>ˉ x x¯m>x¯) is possible?
statistics
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edited May 26, 2020 at 14:33
StubbornAtom
18k 4 4 gold badges 42 42 silver badges 116 116 bronze badges
asked May 25, 2020 at 22:46
GileadGilead
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2 No it's not possible unless there are 0 0 men. The overall average will be a positive weighted sum of the two gender averages ( weighted by the portion of population of lawyers in each gender) which will be properly between the two subset averages. That is (overall average) = (men ave.)(men % of pop.) + (woman ave.)(woman % of pop.). I'm assuming these are actual means, not sample means or some other sort of estimate.Ned –Ned 2020-05-25 22:56:04 +00:00 Commented May 25, 2020 at 22:56
@twosigma This would not work if the remaining elements have an average that is higher.Gilead –Gilead 2020-05-25 22:56:22 +00:00 Commented May 25, 2020 at 22:56
1 @ThomasWinckelman: the subsets would need to be mutually exclusive. In your example, the first subset is (1,3,5) with a mean of 3, but the second subset is (1,5) also with a mean of 3. In my example, the second (higher) subset needs to be strictly greater than the mean of the entire set, which it's not in this case.Gilead –Gilead 2020-05-25 23:02:50 +00:00 Commented May 25, 2020 at 23:02
2 @ThomasWinckelman The men and women subsets are disjoint and exhaustive (i.e. a partition) so that sort of example doesn't apply here.Ned –Ned 2020-05-25 23:03:18 +00:00 Commented May 25, 2020 at 23:03
2 I was able to find those numbers at datausa.io/profile/soc/… but with the women's average at 132,637. This seems to be from some governmental source but I can't figure out which. I suspect your source comes from the same ultimate source as this one but with an error.Michael Lugo –Michael Lugo 2020-05-26 15:23:36 +00:00 Commented May 26, 2020 at 15:23
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This is the calculation behind Ned's comment.
Let m m be the number of men, w w the number of women and a m a m, a w a w respective averages. Then, the overall average is
a=m a m+w a w m+w.
a=m a m+w a w m+w.
If a=a w a=a w, from the above we get m a m+w a w=(m+w)a w m a m+w a w=(m+w)a w, i.e. m(a m−a w)=0 m(a m−a w)=0. Thus, either m=0 m=0 or a m=a w a m=a w.
Moreover, if a m≥a w a m≥a w, then
a w=m a w+w a w m+w≤m a m+w a w m+w≤m a m+w a m m+w=a m.
a w=m a w+w a w m+w≤m a m+w a w m+w≤m a m+w a m m+w=a m.
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edited May 25, 2020 at 23:22
answered May 25, 2020 at 23:04
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17341 | https://oeis.org/A102187 | A102187 - OEIS
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A102187
Arithmetic means of divisors of arithmetic numbers (arithmetic numbers, A003601, are those for which the average of the divisors is an integer).
10
1, 2, 3, 3, 4, 6, 7, 6, 6, 9, 10, 7, 8, 9, 12, 10, 15, 9, 16, 12, 12, 19, 15, 14, 21, 12, 22, 14, 13, 18, 24, 19, 18, 27, 15, 18, 15, 20, 30, 14, 31, 24, 21, 18, 34, 21, 24, 18, 36, 37, 24, 21, 40, 42, 27, 33, 30, 45, 28, 28, 32, 36, 30, 21, 49, 26, 51, 27, 52, 24, 54, 55, 27, 38
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Values of sigma(n)/tau(n) on the terms of A003601, where tau(n) (A000005) is the number of divisors of n and sigma(n) (A000203) is the sum of the divisors of n.
LINKS
Zak Seidov, Table of n, a(n) for n = 1..10000
O. Ore, On the averages of the divisors of a number, Amer. Math. Monthly, 55 (1948), 615-619.
FORMULA
a(n) = sigma(A003601(n))/tau(A003601(n)).
EXAMPLE
The first four terms are 1,2,3,and 3, being the averages of the divisors of the first four arithmetic numbers, 1,3,5 and 6, respectively. Indeed, 1/1=1, (1+3)/2=2, (1+5)/2=3 and (1+2+3+6)/4=3.
MAPLE
with(numtheory): p:=proc(n) if type(sigma(n)/tau(n), integer)=true then sigma(n)/tau(n) else fi end: seq(p(n), n=1..130);
MATHEMATICA
a003601[n_Integer] :=
Select[Range[n], IntegerQ[DivisorSigma[1, #]/DivisorSigma[0, #]] &]; a102187[n_Integer] :=
Map[DivisorSigma[1, #]/DivisorSigma[0, #] &, a003601[n]]; a102187 ( Michael De Vlieger, Aug 05 2014 )
CROSSREFS
Cf. A003601, A000005, A000203.
Sequence in context: A162627A023158A120882 A133610A256211A265109
Adjacent sequences: A102184A102185A102186 A102188A102189A102190
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Feb 16 2005
STATUS
approved
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17342 | https://www.homeworkforyou.com/static_media/uploadedfiles/Pie%20charts%20and%20Bar%20graphs.pdf | Statistical Reasoning Name: Unit 2 – Pie Charts and Bar Graphs – Day 1 Practice 1. MARITAL STATUS In the Statistical Abstract of the United States we find these data on the marital status of adult American women as of 2007: a. Make a bar graph to show the distribution of marital status (include title, labels and consistent scale) b. Calculate the degrees and percentages needed to construct a pie chart (do not create the pie chart. (round % to nearest percent, round degrees to nearest degree, double check your totals) MARITAL STATUS COUNT (thousands) Percent Degree Never Married 25,262 Married 65,128 Widowed 11,208 Divorced 13,210 Total 114,807 2. COLLEGE FRESHMEN A survey of college freshmen asked what field they planned to study. The results: 25.2% arts and humanities, 19.3% business, 7.1% education, 16.6% engineering and science, 7.8% professional and 15.3% social science. a. What percent plan to study fields other than those listed? b. Make a bar graph that compares the percents of college freshmen planning to study various fields. (be sure to include the title, labels and consistent scale) MARITAL STATUS COUNT (thousands) Never Married 25,262 Married 65,128 Widowed 11,208 Divorced 13,210 TOTAL 114,807 c. Calculate the degrees and percentages needed to construct a pie chart (do not create the pie chart. (round % to nearest tenth, one decimal, round degrees to nearest degree, double check your totals) FIELD OF STUDY Percent Degree Total 3. GIRLS EXCEL Is it true that girls perform better than boys in the study of languages and so-called soft sciences? Here are several Advanced Placement subjects and the percent of examinations taken by female candidates in 2007: English Language/Composition, 63%; French Language, 70%; Spanish Language, 64%; and Psychology, 65%. a. Explain clearly why we cannot use a pie chart to display these data, even if we know the percent of exams taken by girls for every subject. b. Make a bar graph of the data. Order the bars from tallest to shortest; this will make comparison easier (be sure to include the title, labels and consistent scale). |
17343 | https://ecampusontario.pressbooks.pub/uvicmicroeconomics/chapter/4-2-elasticity/ | 4.1 Calculating Elasticity – Principles of Microeconomics
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Contents
Topic 1: Introductory Concepts and Models
Case Study – Beer or Cancer?
Solutions: Case Study – Beer or Cancer?
Introduction to Microeconomics
1.1 What Is Economics, and Why Is It Important?
The Problem of Scarcity
The Division of and Specialization of Labor
Why the Division of Labor Increases Production
Trade and Markets
Why Study Economics?
Summary
Topic 1 Solutions
Topic 1 References
Topic 1 Multiple Choice Questions
1.2 Opportunity Costs & Sunk Costs
Sunk Costs
1.3 Marginal Analysis
Topic 2: Specialization and Trade
Case Study – Brexit
Solutions: Case Study – Brexit
Topic 2 References
Topic 2 Multiple Choice Questions
2.2 Production Possibility Frontier
Model of Production
Production Possibility Frontier
Shifting a PPF
Glossary
2.3 Trade
Production Possibilities with Trade
Linear PPF’s
Conclusion
Glossary
2.1 Economic Efficiency
Topic 2 Solutions
Introduction to Specialization & Trade
Topic 3: Supply, Demand, and Equilibrium
Case Study – The Housing Market
Solutions: Case Study – The Housing Market
Topic 3 References
Topic 3 Multiple Choice Questions
3.6 Equilibrium and Market Surplus
Equilibrium
Market Surplus
Topic 3 Solutions
3.5 Other Determinants of Supply
3.4 Building Supply and Producer Surplus
Introduction to Supply and Demand
3.1 The Competitive Market Model
An Economic Model of Demand and Supply
3.2 Building Demand and Consumer Surplus
The Law of Demand
Consumer Surplus
Completing the Demand Curve
Glossary
3.3 Other Determinants of Demand
1.Income
Market Demand
Topic 4 Part 1: Elasticity
4.1 Calculating Elasticity
4.3 Relative Elasticity
Factors That Influence Relative Elasticity
4.2 Elasticity and Revenue
Topic 4 Part 2: Applications of Supply and Demand
Topic 4 Solutions
Solutions: Case Study – Automation in Fast Food
Case Study – Automation in Fast Food
Topic 4 Multiple Choice Questions
Topic 4 References
4.4 Introduction to Government Policy
A Five Year Plan
Policy Implements
4.7 Taxes and Subsidies
Transfer and Deadweight Loss
Subsidy
4.9 Tariffs
Tariffs
Conclusion
4.5 Price Controls
Price Ceiling
Price Floor
4.6 Quantity Controls
Market Surplus
4.8 Elasticity and Policy
Maxwell Nicholson
Topic 5: Externalities
Case Study – Sulpher Dioxide
5.3 Directly Targeting Pollution
Pollution Standards
Cap and Trade
Pollution Tax
Key Concepts and Summary
Solutions: Case Study – Sulpher Dioxide
Topic 5 Solutions
5.2 Indirectly Correcting Externalities
A Source of Market Failure
Correcting or ‘Internalizing’ an Externality
Zero Externality isn’t the Answer
Topic 5 References
5.1 Externalities
Externalities
Enriching Our Model
A Negative Externality
Pareto Improvements and Potential Pareto Improvement
Potential Pareto Improvements to Externalities
Positive Externalities
Key Concepts and Summary
Introduction to Environmental Protection and Negative Externalities
Topic 6: Consumer Theory
Solutions: Case Study – The Liberal Gas Tax
Case Study – The Liberal Gas Tax
Topic 6 Solutions
Topic 6 References
6.3 Understanding Consumer Theory
6.4 Building Demand
The Foundations of Demand Curves
Introduction to Consumer Choices
Our Demand Model So Far
6.1 The Budget Line
The Budget Line
When Income Changes
When Price Changes
When Price and Income Change
Conclusion
6.2 The Indifference Curve
Conclusion
Topic 7: Producer Theory
Case Study – Oil Markets
Solution: Case Study – Oil Markets
Topic 7 Solutions
Topic 7 References
7.4 The Structure of Costs in the Long Run
Choice of Production Technology
Economies of Scale
Shapes of Long-Run Average Cost Curves
Shifting Patterns of Long-Run Average Cost
Key Concepts and Summary
7.1 Building Producer Theory
7.3 Producer Theory in the Long Run
7.2 Understanding Producer Theory
The Competitive Firm
Introduction to Cost and Industry Structure
Topic 8: Imperfect Competition
Case Study – Diamond’s Demise
Solutions: Diamond’s Demise
Topic 8 References
Topic 8 Solutions
Introduction to Imperfect Competition
8.2 Fixing Monopoly
8.3 Why Monopolies Persist
Natural Monopoly
Control of a Physical Resource
Legal Monopoly
Promoting Innovation
Intimidating Potential Competition
Summing Up Barriers to Entry
Key Concepts and Summary
8.4 Monopolistic Competition
Differentiated Products
Perceived Demand for a Monopolistic Competitor
How a Monopolistic Competitor Chooses Price and Quantity
Monopolistic Competitors and Entry
8.1 Monopoly
Market Surplus
Principles of Microeconomics
4.1 Calculating Elasticity
Learning Objectives
By the end of this section, you will be able to:
Calculate the price elasticity of demand
Calculate the price elasticity of supply
Calculate the income elasticity of demand and the cross-price elasticity of demand
Apply concepts of price elasticity to real-world situations
(Credit: Melo McC/ Flickr/ CC BY-NC-ND 2.0)
That Will Be How Much?
Imagine going to your favorite coffee shop and having the waiter inform you the pricing has changed. Instead of $3 for a cup of coffee with cream and sweetener, you will now be charged $2 for a black coffee, $1 for creamer, and $1 for your choice of sweetener. If you want to pay your usual $3 for a cup of coffee, you must choose between creamer and sweetener. If you want both, you now face an extra charge of $1. Sound absurd? Well, that is the situation Netflix customers found themselves in 2011 – a 60% price hike to retain the same service.
In early 2011, Netflix consumers paid about $10 a month for a package consisting of streaming video and DVD rentals. In July 2011, the company announced a packaging change. Customers wishing to retain both streaming video and DVD rental would be charged $15.98 per month – a price increase of about 60%. In 2014, Netflix also raised its streaming video subscription price from $7.99 to $8.99 per month for new U.S. customers. The company also changed its policy of 4K streaming content from $9.00 to $12.00 per month that year.
How did customers of the 18-year-old firm react? Did they abandon Netflix? How much will this price change affect the demand for Netflix’s products? The answers to those questions will be explored in this chapter with a concept economists call elasticity.
Click to read the rest of the Netflix story
Anyone who has studied economics knows the law of demand: a higher price will lead to a lower quantity demanded. What you may not know is how much lower the quantity demanded will be. Similarly, the law of supply shows that a higher price will lead to a higher quantity supplied. The question is: How much higher? This topic will explain how to answer these questions and why they are critically important in the real world.
To find answers to these questions, we need to understand the concept of elasticity.Elasticityis an economics concept that measures the responsiveness of one variable to changes in another variable. Suppose you drop two items from a second-floor balcony. The first item is a tennis ball, and the second item is a brick. Which will bounce higher? Obviously, the tennis ball. We would say that the tennis ball has greater elasticity.
But how is this degree of responsiveness seen in our models?Both the demand and supply curve show the relationship between price and quantity, and elasticity can improve our understanding of this relationship.
Theown price elasticity of demandis the percentage change in the quantity demanded of a good or service divided by the percentage change in the price. This shows the responsiveness of the quantity demanded to a change in price.
Theown price elasticity of supplyis the percentage change in quantity supplied divided by the percentage change in price. This shows the responsiveness of quantity supplied to a change in price.
Our formula for elasticity,%Δ Q u a n t i t y%Δ P r i c e, can be used for most elasticity problems, we just use different prices and quantities for different situations.
Why percentages are counter-intuitive
Recall that the simplified formula for percentage change is N e w V a l u e−O l d V a l u e O l d V a l u e, also written as Δ V a l u e O l d V a l u e.
Suppose there is an increase in quantity demanded from 4 coffees to 6 coffees. Calculating percentage change ((6−4)4) there has been a 50% increase in quantity demanded. Using the same numbers, consider what happens when quantity demanded decreases from 6 coffees to 4 coffees, ((4−6)6) this change results in a 33% decrease in quantity demanded.
Right away, this should raise a red flag about calculating the elasticity between at two points, if percentage change is dependant on the direction (A to B or B to A) then how can we ensure a consistent elasticity value?
Figure 4.1a
Let’s calculate elasticity from both perspectives:
Moving from A to B:
%ΔPrice:The coffee price falls from $4.50 to $3.00, meaning the percentage change is (3.00−4.50)4.50 = -33%. Price has fallen by 33%.
%ΔQuantity: The quantity of coffee sold increases from 4 to 6, meaning the percentage change is(6−4)4 = 50%. Quantity has risen by 50%
Elasticity:%Δ Q u a n t i t y%Δ P r i c e=−50%33% = 1.5
Moving from B to A:
%ΔPrice: The coffee price rises from $3.00 to $4.50, meaning the percentage change is (4.50−3.00)3.00 = 50%. Price has risen by 50%.
%ΔQuantity:
The quantity of coffee sold falls from 6 to 4, meaning the percentage change is (4−6)6= -33%. Quantity has fallen by 33%
Elasticity:%Δ Q u a n t i t y%Δ P r i c e=33%50% = 0.67
These two calculations give us different numbers. This type of analysis would make elasticity subject to direction which adds unnecessary complication. To avoid this, we will instead rely on averages.
Note that elasticity is an absolute value, meaning it is not affected by positive of negative values.
Mid-point Method
To calculate elasticity, instead of using simple percentage changes in quantity and price, economists use the average percent change. This is called the mid-point method for elasticity, and is represented in the following equations:
%c h a n g e i n q u a n t i t y Q 2−Q 1(Q 2+Q 1)/2×100%c h a n g e i n p r i c e P 2−P 1(P 2+P 1)/2×100
The advantage of the mid-point method is that one obtains the same elasticity between two price points whether there is a price increase or decrease. This is because the denominator is an average rather than the old value.
Using the mid-point method to calculate the elasticity between Point A and Point B:
%c h a n g e i n q u a n t i t y 6−4(6+4)/2×100 2 5×100 40%%c h a n g e i n p r i c e 3.00−4.50(3.00+4.50)/2×100−1.50 3.75×100−40%P r i c e E l a s t i c i t y o f D e m a n d 40%40%1
This method gives us a sort of average elasticity of demand over two points on our curve. Notice that our elasticity of 1 falls in-between the elasticities of 0.67 and 1.52 that we calculated in the previous example.
Point-Slope Formula
In Figure 4.1a we were given two points and looked at elasticity as movements along a curve. As we will see in Topic 4.3, it is often useful to view elasticity at a single point.To calculate this, we have to derive a new equation.
%Δ Q u a n t i t y%Δ P r i c e=E l a s t i c i t y
Since we know that a percentage change in price can be rewritten as
Δ P r i c e P r i c e
and a percentage change in quantity to
Δ Q u a n t i t y Q u a n t i t y
we can rearrange the original equation as
Δ Q u a n t i t y Q u a n t i t y Δ P r i c e P r i c e
which is the same as saying
Δ Q u a n t i t y⋅P r i c e Δ P r i c e⋅Q u a n t i t y=Δ Q Δ P⋅P Q
This gives us our point-slope formula. How do we use it to calculate the elasticity at Point A? The P/Q portion of our equation corresponds to the values at the point, which are $4.5 and 4. The ΔQ/ΔP corresponds to theinverse slope of the curve.Recall slope is calculated as rise/run.
In Figure 4.1, the slope is 3−4.5 6−4 = 0.75, which means the inverse is 1/0.75 = 1.33. Plugging this information into our equation, we get:
Δ Q Δ P⋅P Q
1.33⋅4.5 4 = 1.5
This analysis gives us elasticity as a single point. Notice that this gives us the same number as calculating elasticity from Point A to B. This is not a coincidence. When we are calculating from Point A to Point B, we are actually just calculating the elasticity at Point A, since we are using the values on Point A as the denominator for our percentage change. Likewise from Point B to Point A, we are calculating the elasticity at Point B. When we use the mid-point method, we are just taking an average of the two points. This solidifies the fact that there is a different elasticity at every point on our line, a concept that will be important when we discuss revenue.
Not Really So Different
Even though mid-point and Point-Slope appear to be fairly different formulas, mid-point can be rewritten to show how similar the two really are.
Δ Q(Q 1+Q 2)/2 Δ P(P 1+P 2)/2 =Δ Q Q 1+Q 2 Δ P P 1+P 2
Remember that when a fraction is divided by a fraction, you can rearrange it to a fraction multiplied by the inverse of the denominator fraction.
= Δ Q Δ P⋅(P 1+P 2)(Q 1+Q 2)
Notice that compared to point-slope:Δ Q Δ P⋅P Q, the only difference is that point-slope is the inverse of the slope multiplied by a single point, whereas mid-point is the inverse of the slope multiplied by multiple points. This reinforces the conclusion that mid-point represents an average.
Other Elasticities
Remember, elasticity is the responsiveness of one variable to changes in another variable. This means it can be applied to more that just the price-quantity relationship of our market model. In Topic 3 we discussed how goods can be inferior/normal or substitutes/complements. We will examine this even further when we introduce consumer theory, but for now we can develop our understanding by applying what we know about elasticities.
Own-price elasticity of supply(e P S)
Our analysis of elasticity has been centred around demand, but the same principles apply to the supply curve. Whereas elasticity of demand measures responsiveness of quantity demanded to a price change, own-price elasticity of supplymeasures the responsiveness of quantity supplied. The more elastic a firm, the more it can increase production when prices are rising, and decrease its production when prices are falling.Our equation is as follows:
%Δ Q S u p p l i e d%Δ P
Own-price elasticity of supply can be calculated using mid-point and point-slope formula in the same way as for e P D.
Cross-price elasticity of demand (e XP D)
Whereas the own-price elasticity of demand measures the responsiveness of quantity to a goods own price, cross-price elasticity of demand shows us how quantity demand responds to changes in the price of related goods. Whereas before we could ignore positives and negatives with elasticities, with cross-price, this matters. Our equation is as follows:
%Δ Q G o o d A%Δ P G o o d B
Consider our discussion of complements and substitutes in Topic 3.3. We defined complements as goods that individuals prefer to consume with another good, and substitutes as goods individuals prefer to consume instead of another good. If the price of a complement rises our demand will fall, if the price of a substitute rises our demand will rise.For cross-price elasticity this means:
A complement will have a negative cross-price elasticity, since if the % change in price is positive, the % change in quantity will be negative and vice-versa.
A substitutewill have a positive cross-price elasticity, since if the % change in price is positive, the % change in quantity will be positive and vice-versa.
This adds another dimension to our discussion of complements/substitutes. Now we can comment on the strength of the relationship between two goods. For example, a cross-price elasticity of -4 suggests an individual strongly prefers to consume two goods together, compared to a cross-price elasticity of -0.5. This could represent the cross-price elasticity of a consumer for a hot dog, with respect to ketchup and relish. The consumer might strongly prefer to consume hot dogs with ketchup, and loosely prefers relish.
Income elasticity of demand (e N D)
In Topic 3 we also explained how goods can be normal or inferior depending on how a consumer responds to a change in income. This responsiveness can also be measured with elasticity by the income elasticity of demand.Our equation is as follows:
%Δ Q%Δ I n c o m e
As with cross-price elasticity, whether our elasticity is positive or negative provides valuable information about how the consumer views the good:
A normal goodwill have a positive income elasticity, since if the % change in income is positive, the % change in quantity will be positive and vice-versa.
A inferior goodwill have a negative income elasticity, since if the % change in income is positive, the % change in quantity will be negative and vice-versa.
The value of our elasticity will indicate how responsive a good is to a change in income. A good with an income elasticity of 0.05, while technically a normal good (since demand increases after an increase in income) is not nearly as responsive as one with an income elasticity of demand of 5.
Summary
Elasticity is a measure of responsiveness, calculated by the percentage change in one variable divided by the percentage change in another.
Both mid-point and point-slope formulas are important for calculating elasticity in different situations. Mid-point gives an average of elasticities between two points, whereas point-slope gives the elasticity at a certain point. These can be calculated with the following formulas:
Base Formula Mid-Point Formula Point-Slope Formula
%Δ Q u a n t i t y%Δ P r i c e Δ Q Δ P⋅(P 1+P 2)(Q 1+Q 2)Δ Q Δ P⋅P Q
Since elasticity measures responsiveness, it can also be used to measure the own-price elasticity of supply, the cross-price elasticity of demand, and the income elasticity of demand.These can be calculated with the following formulas:
Own-Price Elasticity of Supply Cross-Price Elasticity of Demand Income Elasticity of Demand
%Δ Q S u p p l i e d%Δ P%Δ Q G o o d A%Δ P G o o d B%Δ Q%Δ I n c o m e
Glossary
Cross-price elasticity of demandthe percentage change in the quantity demanded of good A as a result of a percentage change in price of good BElasticityan economics concept that measures responsiveness of one variable to changes in another variableIncome elasticity of demandthe percentage change in quantity demanded of a good or service as a result of a percentage change in incomeOwn-price elasticity of demandpercentage change in the quantity demanded of a good or service divided the percentage change in priceMid-point MethodInvolves multiplying the inverse of the slope by the values of a single point.Own-price elasticity of supplypercentage change in the quantity supplied divided by the percentage change in pricePoint Slope MethodA method of calculating elasticity between two points.Involves calculating the percentage change of price and quantity with respect to an average of the two points.
Exercises 4.1
1. Use the demand curve diagram below to answer the following question.
What is the own-price elasticity of demand as price increases from $2 per unit to $4 per unit? Use the mid-point formula in your calculation.
a) 1/3.
b) 6/10.
c) 2/3.
d) None of the above.
2. Suppose that a 2% increase in price results in a 6% decrease in quantity demanded. Own-price elasticity of demand is equal to:
a) 1/3.
b) 6.
c) 2
d) 3.
3. If own-price elasticity of demand equals 0.3 in absolute value, then what percentage change in price will result in a 6% decrease in quantity demanded?
a) 3%
b) 6%
c) 20%.
d) 50%.
4. Suppose you are told that the own-price elasticity of supply equal 0.5. Which of the following is the correct interpretation of this number?
a) A 1% increase in price will result in a 50% increase in quantity supplied.
b) A 1% increase in price will result in a 5% increase in quantity supplied.
c) A 1% increase in price will result in a 2% increase in quantity supplied.
d) A 1% increase in price will result in a 0.5% increase in quantity supplied.
5. Suppose that a 10 increase in price results in a 50 percent decrease in quantity demanded. What does (the absolute value of) own price elasticity of demand equal?
a) 0.5.
b) 0.2.
c) 5.
d) 10.
6. If goods X and Y are SUBSTITUTES, then which of the following could be the value of the cross price elasticity of demand for good Y?
a) -1.
b) -2.
c) Neither a) nor b).
d) Both a) and b).
7. If pizza is a normal good, then which of the following could be the value of income elasticity of demand?
a) 0.2.
b) 0.8.
c) 1.4
d) All of the above.
8. If goods X and Y are COMPLEMENTS, the which of the following could be the value of cross price elasticity of demand?
a) 0.
b) 1.
c) -1.
d) All of the above could be the value of cross price elasticity of demand.
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Next: 4.3 Relative Elasticity
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17344 | https://math.stackexchange.com/questions/825821/what-is-the-theorem-called-that-states-that-equal-angles-gives-equal-sides | Stack Exchange Network
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What is the theorem called that states that equal angles gives equal sides?
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We have an isosceles triangle, what is the theorem called that states that the sides opposite it's congruent angles will have congruent lengths? Could someone also explain why this is.
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$\begingroup$ Note that you are asking about a sort of converse to the fifth Proposition of Euclid's Book 1, the Pons Asinorum, which states the congruence of angles opposite sides of equal lengths. $\endgroup$
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2014-06-08 12:45:58 +00:00
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The theorem you want is:
If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.
This works for any triangle, not just an isoceles one (obviously, the theorem implies that the triangle in question is isoceles, but you don't need to know that in advance).
This is an early theorem in Euclid's Elements. It's a simple consequence of the converse, that if a triangle has two equal sides then the opposite angles must also be equal, which is also proven in Euclid's elements.
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answered Jun 8, 2014 at 12:48
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$\begingroup$ Just what I needed, thanks. $\endgroup$
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17345 | https://brainly.com/question/35592004 | [FREE] "The positive integer N has exactly 8 positive divisors, including 1 and N . Two of these divisors are - brainly.com
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"The positive integer N has exactly 8 positive divisors, including 1 and N. Two of these divisors are 14 and 21. What is the sum of all 8 positive divisors of N?
(A) 35
(B) 47
(C) 53
(D) 96
(E) 103"
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Final answer:
The sum of all positive divisors of a number N is found to be 96.
Explanation:
Since 14 and 21 are divisors of N, and both numbers have common factors of 7, we can deduce that 7 is a prime factor of N.
Furthermore, since 14 is divisible by 2, N must also include 2 as a prime factor.
The number 14 can be factored as 2 7 and 21 as 3 7.
This information sheds light on the prime factorization of N, suggesting it is composed of a square of a prime number and another distinct prime number.
Specifically, we can infer that N = 2 7^2 3.
Calculating the sum of the divisors of N involves identifying all the divisors and summing them up.
The divisors are as follows: 1, 2, 3, 6, 7, 14, 21, 42.
Adding these divisors together yields
1 + 2 + 3 + 6 + 7 + 14 + 21 + 42 = 96.
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The positive integer N with exactly 8 positive divisors, including 14 and 21 as its divisors, is 42. The sum of all its divisors is 96. Hence, the answer is (D) 96.
Explanation
To find the positive integer N that has exactly 8 positive divisors, where two divisors are 14 and 21, we need to examine the prime factorization of N.
Since 14 can be factored into 2×7 and 21 can be factored into 3×7, we can see that the prime factors involved are 2, 3, and 7.
To have exactly 8 positive divisors, N must be in one of the following forms:
p 7 (one prime raised to the 7th power)
p 3×q (one prime raised to the 3rd power times another distinct prime)
p 1×q 1×r 1 (three distinct primes)
Given our prime factors (2, 3, 7), the factorization that fits is N=2 1×3 1×7 1=2×3×7=42.
Next, we find the divisors of N=42. They are:
1
2
3
6
7
14
21
42
Now we sum all these divisors:
1+2+3+6+7+14+21+42=96
Thus, the sum of all 8 positive divisors of N is 96.
Examples & Evidence
For example, if we had a number like 60, its divisors would be 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60, totaling 12 divisors. However, in this case, 42 only has 8 distinct divisors which fit our requirements.
The calculation of divisors derives from the fundamental theorem of arithmetic, which states that every positive integer can be uniquely represented as a product of prime factors. This method of counting divisors is proven through the use of their exponents in prime factorizations.
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The sum of all 8 positive divisors of N is 96.
Explanation
To find the sum of all 8 positive divisors of N, we need to find the prime factorization of N. Since two of the divisors are 14 and 21, we know that N must be a multiple of both 14 and 21.
The prime factorization of 14 is 2 7, and the prime factorization of 21 is 3 7. Therefore, N must have at least one factor of 2, one factor of 3, and one factor of 7.
Since N has exactly 8 positive divisors, including 1 and N, it must have two additional prime factors. Let's assume these factors are a and b.
Therefore, the prime factorization of N can be written as N = 2 3 7 a b.
Since N has exactly 8 positive divisors, the exponents of the prime factors must be 1. So, a and b must be 1.
Therefore, N = 2 3 7 1 1 = 42.
The sum of all 8 positive divisors of N can be found using the formula: (2^0 + 2^1) (3^0 + 3^1) (7^0 + 7^1) = (1 + 2) (1 + 3) (1 + 7) = 3 4 8 = 96.
Learn more about sum of divisors here:
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17346 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Applied_Discrete_Structures_(Doerr_and_Levasseur)/01%3A_Set_Theory/1.01%3A_Set_Notation_and_Relations | Skip to main content
1.1: Set Notation and Relations
Last updated
: Aug 16, 2021
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1: Set Theory
1.2: Basic Set Operations
Page ID
: 80495
Al Doerr & Ken Levasseur
University of Massachusetts Lowell
( \newcommand{\kernel}{\mathrm{null}\,})
Notion of a Set
Theterm set is intuitively understood by most people to mean a collection of objects that are called elements (of the set). This concept is the starting point on which we will build more complex ideas, much as in geometry where the concepts of point and line are left undefined. Because a set is such a simple notion, you may be surprised to learn that it is one of the most difficult concepts for mathematicians to define to their own liking. For example, the description above is not a proper definition because it requires the definition of a collection. (How would you define “collection”?) Even deeper problems arise when you consider the possibility that a set could contain itself. Although these problems are of real concern to some mathematicians, they will not be of any concern to us. Our first concern will be how to describe a set; that is, how do we most conveniently describe a set and the elements that are in it? If we are going to discuss a set for any length of time, we usually give it a name in the form of a capital letter (or occasionally some other symbol). In discussing set A, if x is an element of A, then we will write x∈A. On the other hand, if x is not an element of A, we write x∉A. The most convenient way of describing the elements of a set will vary depending on the specific set.
Enumeration. When the elements of a set are enumerated (or listed) it is traditional to enclose them in braces. For example, the set of binary digits is {0,1} and the set of decimal digits is {0,1,2,3,4,5,6,7,8,9}. The choice of a name for these sets would be arbitrary; but it would be “logical” to call them B and D, respectively. The choice of a set name is much like the choice of an identifier name in programming. Some large sets can be enumerated without actually listing all the elements. For example, the letters of the alphabet and the integers from 1 to 100 could be described as A={a,b,c,…,x,y,z}, and G={1,2,…,99,100}. The three consecutive “dots” are called an ellipsis. We use them when it is clear what elements are included but not listed. An ellipsis is used in two other situations. To enumerate the positive integers, we would write {1,2,3,…}, indicating that the list goes on infinitely. If we want to list a more general set such as the integers between 1 and n, where n is some undetermined positive integer, we might write {1,…,n}.
Standard Symbols. Sets that are frequently encountered are usually given symbols that are reserved for them alone. For example, since we will be referring to the positive integers throughout this book, we will use the symbol P instead of writing {1,2,3,…}. A few of the other sets of numbers that we will use frequently are:
P:the positive integers, {1,2,3,4,…}
N:the natural numbers, {0,1,2,3,…}
Z:the integers, {…,−3,−2,−1,0,1,2,3,…}
Q:the rational numbers
R:the real numbers
C:the complex numbers
Caution: Some people (roughly half of the world?) call the set {1,2,3,4,…} the natural numbers. We are not among them. We take the pythonic approach that assumes that starting with zero is more natural than starting at one.
Set-Builder Notation. Another way of describing sets is to use set-builder notation. For example, we could define the rational numbers as
Q={a/b∣a,b∈Z,b≠0}.
Note that in the set-builder description for the rational numbers:
a/b indicates that a typical element of the set is a “fraction.”
The vertical line, ∣, is read “such that” or “where,” and is used interchangeably with a colon.
a,b∈Z is an abbreviated way of saying a and b are integers.
Commas in mathematics are read as “and.”
The important fact to keep in mind in set notation, or in any mathematical notation, is that it is meant to be a help, not a hindrance. We hope that notation will assist us in a more complete understanding of the collection of objects under consideration and will enable us to describe it in a concise manner. However, brevity of notation is not the aim of sets. If you prefer to write a∈Z and b∈Z instead of a,b∈Z, you should do so. Also, there are frequently many different, and equally good, ways of describing sets. For example, {x∈R∣x2−5x+6=0} and {x∣x∈R,x2−5x+6=0} both describe the solution set {2,3}.
A proper definition of the real numbers is beyond the scope of this text. It is sufficient to think of the real numbers as the set of points on a number line. The complex numbers can be defined using set-builder notation as C={a+bi:a,b∈R}, where i2=−1.
In the following definition we will leave the word “finite” undefined.
Definition 1.1.1: Finite Set
A set is a finite set if it has a finite number of elements. Any set that is not finite is an infinite set.
Definition1.1.2: Cardinality
Let A be a finite set. The number of different elements in A is called its cardinality. The cardinality of a finite set A is denoted |A|.
As we will see later, there are different infinite cardinalities. We can't make this distinction until Chapter 7, so we will restrict cardinality to finite sets for now.
Subsets
Definition1.1.3: Subset
Let A and B be sets. We say that A is a subset of B if and only if every element of A is an element of B. We write A⊆B to denote the fact that A is a subset of B.
Example 1.1.1: Some Subsets
If A={3,5,8} and B={5,8,3,2,6}, then A⊆B.
N⊆Z⊆Q⊆R⊆C
If S={3,5,8} and T={5,3,8}, then S⊆T and T⊆S.
Definition 1.1.4: Set Equality
Let A and B be sets. We say that A is equal to B (notation A=B) if and only if every element of A is an element of B and conversely every element of B is an element of A; that is, A⊆B and B⊆A.
Example 1.1.2: Examples Illustrating Set Equality
In Example 1.1.1, S=T. Note that the ordering of the elements is unimportant.
The number of times that an element appears in an enumeration doesn't affect a set. For example, if A={1,5,3,5} and B={1,5,3}, then A=B. Warning to readers of other texts: Some books introduce the concept of a multiset, in which the number of occurrences of an element matters.
A few comments are in order about the expression “if and only if” as used in our definitions. This expression means “is equivalent to saying,” or more exactly, that the word (or concept) being defined can at any time be replaced by the defining expression. Conversely, the expression that defines the word (or concept) can be replaced by the word.
Occasionallythere is need to discuss the set that contains no elements, namely the empty set, which is denoted by ∅. This set is also called the null set.
It is clear, we hope, from the definition of a subset, that given any set A we have A⊆A and ∅⊆A. If A is nonempty, then A is called an improper subset of A. All other subsets of A, including the empty set, are called proper subsets of A. The empty set is an improper subset of itself.
Note 1.1.1
Not everyone is in agreement on whether the empty set is a proper subset of any set. In fact earlier editions of this book sided with those who considered the empty set an improper subset. However, we bow to the emerging consensus at this time.
Exercises
Exercise 1.1.1
List four elements of each of the following sets:
{k∈P∣k−1 is a multiple of 7}
{x∣x is a fruit and its skin is normally eaten}
{x∈Q∣1x∈Z}
{2n∣n∈Z,n<0}
{s∣s=1+2+⋯+n for some n∈P}
Answer
: These answers are not unique.
1. 8,15,22,29
2. apple, pear, peach, plum
3. 1/2,1/3,1/4,1/5
4. −8,−6,−4,−2
5. 6,10,15,21
Exercise 1.1.2
List all elements of the following sets:
{1n∣n∈{3,4,5,6}}
{α∈ the alphabet ∣α precedes F}
{x∈Z∣x=x+1}
{n2∣n=−2,−1,0,1,2}
{n∈P∣n is a factor of 24 }
Exercise 1.1.3
Describe the following sets using set-builder notation.
{5,7,9,…,77,79}
the rational numbers that are strictly between −1 and 1
the even integers
{−18,−9,0,9,18,27,…}
Answer
: 1. {2k+1∣k∈Z,2⩽k⩽39}
2. {x∈Q∣−1<x<1}
3. {2n∣n∈Z}
4. {9n∣n∈Z,−2≤n}
Exercise 1.1.4
Use set-builder notation to describe the following sets:
{1,2,3,4,5,6,7}
{1,10,100,1000,10000}
{1,1/2,1/3,1/4,1/5,...}
{0}
Exercise 1.1.5
Let A={0,2,3}, B={2,3}, and C={1,5,9}. Determine which of the following statements are true. Give reasons for your answers.
3∈A
{3}∈A
{3}⊆A
B⊆A
A⊆B
∅⊆C
∅∈A
A⊆A
Answer
: 1. True
2. False
3. True
4. True
5. False
6. True
7. False
8. True
Exercise 1.1.6
One reason that we left the definition of a set vague is Russell's Paradox. Many mathematics and logic books contain an account of this paradox. Two references are and . Find one such reference and read it.
1: Set Theory
1.2: Basic Set Operations |
17347 | https://www.quora.com/How-do-I-find-the-base-and-height-of-a-triangle | How to find the base and height of a triangle - Quora
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How do I find the base and height of a triangle?
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11mo
To find the base and height of a triangle, you can follow these steps:
Identify the Type of Triangle:
Right Triangle: The base and height are the two sides that form the right angle.
Equilateral Triangle: All sides are equal, and you can drop a perpendicular from one vertex to the opposite side to find the height.
Scalene or Isosceles Triangle: You may need additional information, like the coordinates of the vertices or the lengths of the sides.
Using the Area Formula:
The area of a triangle can be calculated using the formula:
Area=1 2×base×height Area=1 2×base×height
If
Continue Reading
To find the base and height of a triangle, you can follow these steps:
Identify the Type of Triangle:
Right Triangle: The base and height are the two sides that form the right angle.
Equilateral Triangle: All sides are equal, and you can drop a perpendicular from one vertex to the opposite side to find the height.
Scalene or Isosceles Triangle: You may need additional information, like the coordinates of the vertices or the lengths of the sides.
Using the Area Formula:
The area of a triangle can be calculated using the formula:
Area=1 2×base×height Area=1 2×base×height
If you know the area and either the base or height, you can rearrange the formula to find the unknown:
If you know the area and the base:
height=2×Area base height=2×Area base
If you know the area and the height:
base=2×Area height base=2×Area height
Finding the Height in Different Types of Triangles:
For a Right Triangle: The height is one of the legs of the triangle.
For an Equilateral Triangle: Use the formula for height h h:
h=√3 2×side length h=3 2×side length
For a Scalene or Isosceles Triangle: Use Heron's formula to find the area first, and then use the area to find the height.
Using Coordinates:
If the triangle's vertices are given by their coordinates (x 1,y 1)(x 1,y 1), (x 2,y 2)(x 2,y 2), and (x 3,y 3)(x 3,y 3), you can find the area using the formula:
Area=1 2|x 1(y 2−y 3)+x 2(y 3−y 1)+x 3(y 1−y 2)|Area=1 2|x 1(y 2−y 3)+x 2(y 3−y 1)+x 3(y 1−y 2)|
From the area, you can then find the height relative to a chosen base.
Conclusion
To determine the base and height, identify the type of triangle and use the appropriate method based on the information you have. If you have specific measurements or a type of triangle in mind, I can help you further!
Upvote ·
Henry Burek
Former Optometrist · Author has 2.2K answers and 3.2M answer views
·4y
Originally Answered: What is the base and height of a triangle? ·
What is the base and height of a triangle?
The base of a triangle can be any side of the triangle. The height (also known as the altitude) is the perpendicular distance from the base to the vertex opposite. A triangle can thus have three different base/height pairs but the product of base and height is a constant for any given triangle (and equal to twice the triangle’s area).
Continue Reading
What is the base and height of a triangle?
The base of a triangle can be any side of the triangle. The height (also known as the altitude) is the perpendicular distance from the base to the vertex opposite. A triangle can thus have three different base/height pairs but the product of base and height is a constant for any given triangle (and equal to twice the triangle’s area).
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Kumar Saurav
Math geek. · Author has 574 answers and 2.1M answer views
·8y
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How can I identify base and height of right angle triangle?
You can call either of the catheti( the perpendicular sides of a right triangle, and a very fancy word with which to impress your friends), the base or the height.
For the purposes of Trigonometry, it depends on the reference angle.
For example, in your figure if you take ∠B A C∠B A C to be the reference angle, then:
The side opposite to the angle will be the height. In this case, B C B C is the height.
The side adjacent to the angle will be your base. In this case, A B A B is the base.
So if you want to find tan∠B A C tan∠B A C,
tan∠B A C=P B=B C A B tan∠B A C=P B=B C A B
For finding area, it doesn't matter
Continue Reading
You can call either of the catheti( the perpendicular sides of a right triangle, and a very fancy word with which to impress your friends), the base or the height.
For the purposes of Trigonometry, it depends on the reference angle.
For example, in your figure if you take ∠B A C∠B A C to be the reference angle, then:
The side opposite to the angle will be the height. In this case, B C B C is the height.
The side adjacent to the angle will be your base. In this case, A B A B is the base.
So if you want to find tan∠B A C tan∠B A C,
tan∠B A C=P B=B C A B tan∠B A C=P B=B C A B
For finding area, it doesn't matter because
1 2(base)(height)=1 2(height)(base)1 2(base)(height)=1 2(height)(base)
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How do I find the height of a triangle given the area and base?
How can I identify base and height of right angle triangle?
How do I find the base of a triangle with the height and area given?
How do I find the height of a triangle given 3 sides?
How do I find the height of a triangle and its side?
Neal Schermerhorn
Lives in Massachusetts (1967–present) · Author has 6.6K answers and 1.7M answer views
·1y
If you are given a side of a triangle, it’s often best to consider that the base. If you are given no information about the lengths of any part of the construction, or insufficient information to determine one side of the triangle, then this will be impossible.
If you know all the sides, you can use Heron’s formula to find the area, then use the bh/2 formula to find the height from a given base.
If you know one side and two angles, or two sides and one angle, you can usually use the law of cosines or the law of sines to find the other side(s) and then use Heron’s formula. This relies on having a
Continue Reading
If you are given a side of a triangle, it’s often best to consider that the base. If you are given no information about the lengths of any part of the construction, or insufficient information to determine one side of the triangle, then this will be impossible.
If you know all the sides, you can use Heron’s formula to find the area, then use the bh/2 formula to find the height from a given base.
If you know one side and two angles, or two sides and one angle, you can usually use the law of cosines or the law of sines to find the other side(s) and then use Heron’s formula. This relies on having a workable combination of sides and angles—in some cases multiple triangles could have the same configuration.
Upvote ·
Terry Darwin
Self-employed Sole Trader (1991–present) · Author has 10.5K answers and 6.7M answer views
·5y
Originally Answered: How do you find the base and height of an isosceles triangle? ·
How do you find the base and height of an isosceles triangle?
By knowing at least two other facts about it: one length and one angle would be sufficient. The method employed would involve Pythagoras’ Theorem and/or trigometry, depending on what other facts you are given and personal preference.
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Klaus Ole Kristiansen
, M.Sc. Mathematics, University of Copenhagen (1992) · Author has 24.8K answers and 9.5M answer views
·4y
Originally Answered: A triangle's area is 25cm, the base and height are the same. What are the base and height? ·
A triangle's area is 25cm², the base and height are the same. What are the base and height?
a=1/2bh let b=h
a=1/2h²
25cm² = 1/2h²
50cm²=h²
h=5√2cm
The base and height are both 5√2cm long.
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How do I find a triangle’s height and area if all the sides are given?
How can I find the height of a triangle?
How do you find the height of a triangle when you know the area?
How do I find the base and height of a triangle with only the hypotenuse?
How do you find the height of a triangle without an area?
Danny
Studied at The High School Experience
·3y
Originally Answered: How do I find the base of a triangle with the height and area given? ·
Since we know that the formula for the area of a triangle is: A = 0.5bh, you simply substitute (plugging them into the formula) the value of the area and height into the formula, and solve for the base.
For instance, if we know that the area of a triangle is 3 cm^2, and the height is 2 cm, we substitute them into the formula, where we are given: 3 = 0.5(2) x b.
We then make the subject of the formula: b = 3/1 = 3. This means that the base of the triangle is 3 cm.
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·4y
Related
How do I find the height of a triangle given the area and base?
It’s easy!
The formula for area of a triangle is 1/2 x base x height
1/2 x base x height = area
If you’re given the area and base to find the height, substitute the height with x, h, k, or any letter of your choice. Then solve it. The answer is the height.
For example, find the height of a triangle when area = 20cm^2 and base = 5cm.
Let the height be h
1/2 x 5 x h = 20cm^2
5/2 x h = 20
Divide both sides by 5/2
h = 20 ÷ 5/2
h = 20 x 2/5
h = 4 x 2
h = 8
Height of the triangle = 8cm
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Steve Scott
Practicing Architect · Author has 148 answers and 129.5K answer views
·5y
First you need to define the triangle. To do this you need the all the values for one of the following combinations:
3 sides
2 sides & 1 angle
1 side & 2 angles
Once you have this by using trigonometry (which you didn’t ask about) you will have the base (which can be any side) and the other sides. The height will then be:
(side 2 or 3)squared - (1/2 the base) squared= square root of the height.
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Raymond Beck
Former Infantry Sergeant · Author has 42.4K answers and 10.1M answer views
·9mo
Originally Answered: What is the method for finding the base and height of a triangle with only two known angles? ·
if you know 2 angles, A and B, then you know the 3rd angle C=180-A-B
know any side and use the law of sines to find the other 2 sides
a/SinA = b/sinB = c/SinC
when you know the sides you can make any side the base, then find the height given A=bh/2. find A with Huron’s formula
if all you know is 2 angles, you can find the 3rd angle but you can’t find the sides. all you can find is the ratio of the sides. and the ratio of height to base
Upvote ·
Thomas Bell
Ph.D. in Mathematics, University of Oregon (Graduated 2013) · Author has 3.9K answers and 648.6K answer views
·2y
I suggest first looking to see if your book tells you. Otherwise maybe take out a ruler. If you know the coordinates, use the distance formula for the base and you might potentially need some trig for the height. Or maybe sonar.
Upvote ·
Phlyn Heubach
BS in Mechanical Engineering, California Polytechnic State University, San Luis Obispo (Graduated 1997) · Author has 478 answers and 877.4K answer views
·5y
Related
How do you find the base height of an isosceles triangle?
Question: How do you find the base height of an isosceles triangle?
Given an isosceles triangle the height constructed to the base, will bisect the base, and form two congruent right triangles. What is known will dictate how you solve for the height (red dashed lines). If you know the lengths of the sides of the isosceles triangle, then use Pythagorean Theorem.
h 1=√w 2−x 2 h 1=w 2−x 2 or h 2=√y 2−z 2 h 2=y 2−z 2
If an angle and one side length of the isosceles triangle is given, then trigonometry would be used. Examples:
h 1=w⋅s i n(α)h 1=w⋅s i n(α) or h 2=z⋅t a n(β)h 2=z⋅t a n(β)
Continue Reading
Question: How do you find the base height of an isosceles triangle?
Given an isosceles triangle the height constructed to the base, will bisect the base, and form two congruent right triangles. What is known will dictate how you solve for the height (red dashed lines). If you know the lengths of the sides of the isosceles triangle, then use Pythagorean Theorem.
h 1=√w 2−x 2 h 1=w 2−x 2 or h 2=√y 2−z 2 h 2=y 2−z 2
If an angle and one side length of the isosceles triangle is given, then trigonometry would be used. Examples:
h 1=w⋅s i n(α)h 1=w⋅s i n(α) or h 2=z⋅t a n(β)h 2=z⋅t a n(β)
Upvote ·
9 4
Louis M. Rappeport
B.S. from University of California, Berkeley · Author has 7.6K answers and 6.7M answer views
·4y
Originally Answered: A triangle's area is 25cm, the base and height are the same. What are the base and height? ·
The area of a triangle is 1/2 times the base by the height.
So, if b is both the base and the height:
1/2 b²=25
b²=50
b=√50 or 5√2
The base and the height are 5√2 cms in length ………………..
Upvote ·
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A Trigonometric Observation in Right Triangle
Problem
Solution
Consider a right triangle with the side lengths a,b,c:
We have:
π=arctan r r+arctan a−r r+arctan b−r r=1+arctan(a r−1)+arctan(b r−1)=1+arctan(a+b+c b−1)+arctan(a+b+c a−1)(for r=a b a+b+c)=1+arctan a+c b+arctan a+c a.
Hence,
arctan(b+c a)+arctan(a+c b)=3 π 4.
On the other hand, it's known that r=a+b−c 2. Thus a−r=a−b+c 2 and b−r=b−a+c 2. We conclude that
π=arctan r r+arctan a−r r+arctan b−r r=1+arctan a−b+c a+b−c+arctan b−a+c a+b−c.
From which
arctan(a−b+c a+b−c)+arctan(b−a+c a+b−c)=3 π 4.
Algebraic view
More generally, assume a b,b c,c a>0. Since, for x,y>0,0<arctan x+arctan y<π,
tan(arctan(a+c b)+arctan(b+c a))=a+c b+b+c a 1−a+c b⋅b+c a=−1,
because of a 2+b 2=c 2.
We deduce thus that π 2<tan(arctan(a+c b)+arctan(b+c a))<π and, since tangent is an injective function on (π 2,π), then
tan(arctan(a+c b)+arctan(b+c a))=tan 3 π 4,
and the conclusion follows.
Another geometric view
∠A 1 O B 1=∠A 2 O B 2=α;∠B 1 A 1 O=∠B 2 A 2 O=90∘−α.
In Δ A 1 O A 2, by construction, A 1 O=A 2 0,∠A 1 O A 2=180∘−α, implying
∠O A 1 A 2=∠O A 2 A 1=180∘−(180∘−α)2=α 2=arctan b+c a.
Similarly, arctan a+c b=90∘−β 2. In conclusion,
arctan b+c a+arctan a+c b=90∘−α 2+90∘−β 2=180∘−α+β 2=180∘−45∘=135∘=3 π 4.
Acknowledgment
John Molokach came up with the first formula after watching Trigonometry by Watching. After he informed me of his discovery, in an attempt to prove it, I stumbled on the second expression. Leo Giugiuc gave another (algebraic) proof after observing that the three numbers may as well be negative. Artyom Sedykh gave another purely geometric derivation.
Trigonometry
What Is Trigonometry?
Addition and Subtraction Formulas for Sine and Cosine
Sine of a Sum Formula
Addition and Subtraction Formulas for Sine and Cosine II
Addition and Subtraction Formulas for Sine and Cosine III
Addition and Subtraction Formulas for Sine and Cosine IV
Addition and Subtraction Formulas
The Law of Cosines (Cosine Rule)
Cosine of 36 degrees
Tangent of 22.5 o - Proof Wthout Words
Sine and Cosine of 15 Degrees Angle
Sine, Cosine, and Ptolemy's Theorem
arctan(1) + arctan(2) + arctan(3) = π
Trigonometry by Watching
arctan(1/2) + arctan(1/3) = arctan(1)
Morley's Miracle
Napoleon's Theorem
A Trigonometric Solution to a Difficult Sangaku Problem
Trigonometric Form of Complex Numbers
Derivatives of Sine and Cosine
ΔABC is right iff sin²A + sin²B + sin²C = 2
Advanced Identities
Hunting Right Angles
Point on Bisector in Right Angle
Trigonometric Identities with Arctangents
The Concurrency of the Altitudes in a Triangle - Trigonometric Proof
Butterfly Trigonometry
Binet's Formula with Cosines
Another Face and Proof of a Trigonometric Identity
cos/sin inequality
On the Intersection of kx and |sin(x)|
Cevians And Semicircles
Double and Half Angle Formulas
A Nice Trig Formula
Another Golden Ratio in Semicircle
Leo Giugiuc's Trigonometric Lemma
Another Property of Points on Incircle
Much from Little
The Law of Cosines and the Law of Sines Are Equivalent
Wonderful Trigonometry In Equilateral Triangle
A Trigonometric Observation in Right Triangle
A Quick Proof of cos(pi/7)cos(2.pi/7)cos(3.pi/7)=1/8
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17349 | https://chapelhillmathcircle.org/wp-content/uploads/2024/04/adv20240427egyptianfractions.pdf | Egyptian Fractions Chapel Hill Math Circle: Advanced Group Matt Crawford April 27, 2024 1 Introduction An Egyptian Fraction is a sum of positive unit fractions (fractions with 1 in the numerator, and a positive integer in the denominator). The Egyptians first did many calculations and kept records using these types of fractions. They also had special symbols for 2 3 and 3 4, but we will only consider the unit Egyptian fractions.
Figure 1: The Rhind Papyrus, purchased by Alexander Rhind in 1858 is dated to around 1650 BCE [RP1] We will explore Egyptian Fraction representations of the number 1 (often called “unity”), counting such representations, how to construct representations, and more!
2 Building Blocks Definition 2.1. A Unit Fraction is a fraction with 1 in the numerator, and a positive integer in the denominator.
Question 2.1. List 3 different unit fractions.
1 Definition 2.2. An Egyptian Fraction Representation of One is any sum of unit fractions that adds up to 1.
Question 2.2. List 3 different Egyptian Fraction Representations of One.
Notice that in our definition of an Egyptian Fraction Representation of 1, we don’t actually put any restriction on how many fractions we can use. We could use infinitely many if we wanted! One of the most famous infinite Egyptian Fraction Representations of 1 is 1 = 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + · · · Question 2.3. (Tricky) Can you think of another infinite Egyptian Fraction Representation of 1?
Let’s turn our attention back to finite sums.
Definition 2.3. An Egyptian Fraction Representation of 1 that uses exactly n fractions is called an Egyptian Fraction Representation of 1 in n terms.
Question 2.4. Can you list the only (1) Egyptian Fraction Representation of 1 using 1 term?
Question 2.5. Can you list the only (1) Egyptian Fraction Representation of 1 using 2 terms?
In order to count the number of Egyptian Fraction Representations of 1 with more than 2 terms, we need to talk about the order of the terms.
Definition 2.4. An Ordered Egyptian Fraction Representation of 1 in n terms is an Egyptian Fraction representation of 1 in n terms, with the denominators are listed in increasing order.
If the denominators are not listed in increasing order, we call this an Unordered Egyptian Fraction Representation of 1 in n terms.
2 Example 2.1. The Egyptian Fraction Representation of 1 on the left is ordered, while the representation on the right is unordered.
1 = 1 2 + 1 4 + 1 4 1 = 1 4 + 1 2 + 1 4 Question 2.6. Can you list the 3 Ordered Egyptian Fraction Representation of 1 using 3 terms? How many Unordered Egyptian Fraction Representations of 1 in 3 terms must there be?
I think we can all agree, it’s getting really old writing all of these fractions. So instead of writing the fractions over and over again, let’s get some new notation to make things easier to write down.
Definition 2.5. Let 1 = 1 x1 + 1 x2 + 1 x3 + · · · + 1 xn be an Egyptian Fraction Representation of 1 in n terms. We will now use the notation (x1, x2, x3, . . . , xn) to mean the same thing, that is, instead of listing out the fractions and adding them together, we will just list out the denominators in the fractions as a list.
Example 2.2. The 3 Ordered Egyptian Fraction Representations of 1 in 3 terms are (3, 3, 3), (2, 4, 4) and (2, 3, 6) in our new notation.
3 Splittings Let us look and see if there are any relationships between the Egyptian Fraction Represen-tations of 1 in 2 terms and those in 3 terms: The only Egyptian Fraction Representation of 1 in 2 terms is (2, 2) and the 3 Ordered Representations of 1 in 3 terms are (3, 3, 3), (2, 4, 4) and (2, 3, 6).
Question 3.1. Can you think of a way to “split” (2, 2) to get any of the other 3 represen-tations in 3 terms?
It may be easier to see the splittings in the fraction form: 3 Example 3.1. Let’s start with (2, 2) and manipulate it so that we can one of the represen-tations in 3 terms.
1 = 1 2 + 1 2 = 1 2 + 1 2(1) = 1 2 + 1 2 1 2 + 1 2 = 1 2 + 1 2 · 1 2 + 1 2 · 1 2 = 1 2 + 1 4 + 1 4 We were able to split one of the copies of 1 2 into two copies of 1 4!
Question 3.2. (Tricky) After seeing the above example, can you think of a way to split (2, 2) into (2, 3, 6)?
In order to split (2, 2) into (2, 3, 6), notice how we split (2, 2) into (2, 4, 4). We split one copy of 1 2 into two pieces, each piece had size one-half of the whole 1 2. Is there a way we could have split one of the copies of 1 2 into two pieces that were different sizes?
Example 3.2. Yes! Take (2, 2) again: 1 = 1 2 + 1 2 = 1 2 + 1 2(1) = 1 2 + 1 2 2 3 + 1 3 = 1 2 + 1 2 · 2 3 + 1 2 · 1 3 = 1 2 + 1 3 + 1 6 Question 3.3. (Super Tricky) After seeing the above examples, can you think of a way to split (2, 2) into (3, 3, 3)?
4 So how did we know to split up the copy of 1 2 the two different ways that we did? Notice that in the first example, we multiplied by 1, which we split up as 1 2 + 1 2. In the second example, we multiplied by 1, which we split up as 2 3 + 1 3. The trick to notice here is that (unbelievably) 1 + 1 = 2 and 2 + 1 = 3! That is, in the first example, the numerators of how we split up the 1 were 1 and 1, and the denominator of both pieces was 1 + 1 = 2. In the second example, the numerators of how we split up the 1 were 2 and 1, and the denominator of both pieces was 2 + 1 = 3.
Question 3.4. Do you have any ideas where we got the numerators 1 and 1 in the first example?
Do you have any ideas where we got the numerators 2 and 1 in the second example? Hint: What fraction are we trying to split?
This leads us to a Theorem!
Theorem 3.1. Let p q be a fraction in simplest terms. Then we can write p q = 1 a + 1 b if and only if we can find two other natural numbers c and d that are divisors of q and p divides c + d. The correspondence is given by a = q(c + d) pd b = q(c + d) pc Let’s see the theorem in action and how to use it.
Example 3.3. Suppose we want to write 5 12 as the sum of two unit fractions. Then our theorem above says that if we can find two numbers c and d that divide 12 (the denominator), and such that 5 (the numerator) divides c + d, then we know we can set a = 12(c + d) 5d and b = 12(c + d) 5c , and then we will have that 5 12 = 1 a + 1 b.
So let’s start looking for c and d.
• Now c and d both have to be divisors of 12, so let’s look at the divisors of 12. The divisors of 12 are {1, 2, 3, 4, 6, 12}. So c and d are going to both be numbers from this list.
• We also need 5 to divide c + d. So let’s just look through all the pairs of divisors and see whether their sums are divisible by 5!
• But do we really need to do ALL of the pairs of divisors? NO! If we look at the pair c = 4 and d = 6, that will give us the values a = 4 and b = 6... but if we look at the pair c = 6 and d = 4, that will give us the values a = 6 and b = 4! These are the same things just switched! So we only need to worry about pairs with c ≤d. Complete the table below.
5 c d c + d 5|(c + d)?
1 1 2 NO 1 2 3 NO 1 3 4 NO 1 4 5 YES 1 6 1 12 2 2 2 3 2 4 2 6 2 12 3 3 3 4 3 6 3 12 4 4 4 6 4 12 6 6 6 12 12 12 Thus, we see that our pairs of divisors of 12 that work are: c d 1 4 2 3 3 12 4 6 • Now for each pair of choices for c and d, we can compute the corresponding a and b from above (complete the table below): c d a = 12(c+d) 5d b = 12(c+d) 5c 1 4 3 12 2 3 3 12 4 6 • Do you notice anything about the values of a and b we got for the different values of c and d? Think about why the values c = 1 and d = 4 gives the same values of a and b as for the choices of c = 3 and d = 12.
6 • Because c = 3 and d = 12 are not relatively prime, we know that there are other values of c and d that are smaller that give the same a and b values, like c = 1 and d = 4!
Definition 3.1. Two numbers x and y are said to be relatively prime if gcd(x, y) = 1.
Remark 3.1.1. Note that two numbers being relatively prime does not mean that either of them are actually prime. For example x = 14 and y = 15 are relatively prime, but neither of them are prime numbers.
So we have learned that we can write 5 12 as the sum of two unit fractions in the following two ways: 5 12 = 1 3 + 1 12 5 12 = 1 4 + 1 6 Moving back to splittings, we can now use this theorem to split all of representations of 1 in 3 terms into representations of 1 into 4 terms!
Question 3.5. How many Ordered Egyptian Fraction Representations of 1 in 4 terms are there? What are they?
7 Question 3.6. (Challenge) How many Ordered Egyptian Fraction Representations of 1 in 5 terms are there? What are they?
8 4 Greedy Algorithm Question 4.1. Does every positive rational number (fraction) have an Egyptian Fraction Representation?
Example 4.1.
25 14 = 1 1 + 1 2 + 1 4 + 1 28 Question 4.2. In this example, why did we start with 1 1?
Note that once we “use” 1 1 we have: 25 14 = 1 1 + 11 14 So now we’ve reduced the question of whether 25 14 has an Egyptian Fraction Representation to whether 11 14 has an Egyptian Fraction Representation!
Question 4.3. Continuing with the same expansion as above, why did we use 1 2 as the next fraction? Did we have any choice for our next fraction after using 1 2?
A useful function when talking about Egyptian Fraction Representations is the floor function.
Definition 4.1. Let x be any real number. We define the Floor of x, denoted ⌊x⌋, to be the largest integer that is less than or equal to x.
Example 4.2.
• ⌊2.7⌋= 2.
• ⌊8⌋= 8.
• √ 2 = 1.
Question 4.4. What are is the value of ⌊x⌋for the following values of x?
• x = 0 9 • x = −π • x = 11 10 Another very useful theorem will lead to our Greedy Algorithm: Theorem 4.1. Every positive rational number p q can be written as p q = 1 x + r, where x is one more than q p rounded down to the nearest integer, and r is a rational number strictly smaller than p q. That is x = j q p k + 1 and r < p q.
Let’s see this theorem in action: Example 4.3. Write 5 7 using an Egyptian Fraction Representation.
• First, we note that 7 5 + 1 = 2.
• Next, we subtract 1 2 from 5 7: 5 7 −1 2 = 3 14 • Now we repeat the process. First we note that 14 3 + 1 = 5.
• Nex we subtract 1 5 from 3 14: 3 14 −1 5 = 1 70 • Since the last difference we computed was itself a unit fraction, we’re done! Now we can put the pieces back together and we get: 5 7 = 1 2 + 1 5 + 1 70 Definition 4.2. A Greedy Algorithm is any algorithm (set of instructions/process) that follows the pattern of making the biggest/most-optimal/greediest choice at any step.
Question 4.5. Why is the theorem above basically a Greedy Algorithm?
Remark 4.1.1. Notice that every time in the process we used above, we took the largest fraction we could away from the initial fraction we started with.
10 Algorithm 1 Greedy Egyptian Fraction Representation Algorithm 1: Greedy(p,q,n,ℓ) Input: p - the numerator of p q in simplest form, q - the denominator of p q in simplest form, n the number of terms to use, ℓ- the denominator of the last used fraction.
Output: Count - the number of Egyptian Fraction Representations of p q in n terms, with every denominator at east as big as ℓ.
2: Count ←0 3: if n is 1 then 4: if q is not divisible by p OR q is less than p × ℓthen 5: Count ←Count + 0 (no representations) 6: else 7: Count ←Count + 1 (one representation) 8: else 9: bdenom ← j nq p k 10: if q is less than p × ℓthen 11: adenom ←ℓ 12: else 13: adenom ← j q p k + 1 14: for adenom ≤i ≤bdenom do 15: Count ←Count + Greedy(ip −q, iq, n −1, i) 16: return Count Remark 4.1.2. The algorithm above can be modified to also keep track of and store the actual representations, not just the count of how many representations there are.
Question 4.6. In what order does the Greedy Algorithm find the 3 different Ordered Egyp-tian Fraction Representations of 1 in 3 terms?
Question 4.7. In what order does the Greedy Algorithm find the 14 different Ordered Egyptian Fraction Representations of 1 in 4 terms?
11 5 Distinct Terms Question 5.1. Given a positive rational number r, does r have a representation as an Egyptian Fraction with the denominators in the the representation all being distinct?
Question 5.2. How can we use our greedy algorithm to find an Egyptian Fraction Repre-sentation of 1 for any number of terms?
Question 5.3. Look at the the Egyptian Fraction Representations of 1 that have the largest denominators for each number of terms. For n = 1 the only representation is (1). For n = 2 the only representation is (2, 2). For n = 3 the representation with the largest denominator is (2, 3, 6). For n = 4 the only representation is (2, 3, 7, 42). Can you find the solution with the largest denominator for 5 terms?
Definition 5.1. Let n be a natural number. We define the Ordered Egyptian Fraction Rep-resentation of 1 in n terms with the largest denominator to be the Sylvester Representation of 1 in n terms.
Question 5.4. Let n ≥3 be a natural number. Can you find a pattern for every denominator in the Sylvester Representation of 1 in n terms (except for the last denominator), in terms of the previous terms in the Sylvester Representation of 1 in n terms?
Question 5.5. Let n ≥3 be a natural number. Can you find a pattern for the last denom-inator in the Sylvester Representation of 1 in n terms, in terms of the previous terms in the 12 Sylvester Representation of 1 in n terms?
Question 5.6. Same question as above, but can you find a recursive pattern that depends only on the next-to-last term?
Remark 5.0.3. The last denominators in each of the Sylvester Representations forms a se-quence: {1, 2, 6, 42, 1806, 3263442, . . .}.
It turns out [A076393] that there is a a number E ≈1.26408473530530 . . . such that the nth term in the above sequence, which we will denote sn, is given by sn = E2n+1 + 1 2 −1.
This shows us that sn grows even faster than an exponential function, it grows DOUBLY-Exponentially!
References [A076393] OEIS Foundation Inc. (2019), The On-Line Encyclopedia of Integer Sequences,
[RP1] Rhind Papyrus Image Wikipedia - Paul James Cowie (Pjamescowie) - Papyrus.jpg 13 |
17350 | https://cctbio.ece.umn.edu/wiki/images/archive/a/ab/20101025212721!Qian_Riedel_Rosenberg_Uniform_Approximation_and_Bernstein_Polynomials_with_Coefficients_in_the_Unit_Interval.pdf | Uniform Approximation and Bernstein Polynomials with Coefficients in the Unit Interval∗ Weikang Qian and Marc D. Riedel Electrical and Computer Engineering, University of Minnesota 200 Union St. S.E.
Minneapolis, MN 55455, USA {qianx030, mriedel}@umn.edu Ivo Rosenberg Mathematics and Statistics University of Montreal C.P. 6128, Succ. Centre-Ville Montreal, Quebec, Canada H3C 3J7 rosenb@DMS.UMontreal.CA Abstract This paper presents two main results. The first result pertains to uniform approximation with Bernstein polynomials. We show that, given a power-form polynomial g, we can obtain a Bernstein polynomial of degree m with coefficients that are as close as desired to the corre-sponding values of g evaluated at the points 0, 1 m, . . . , 1, provided that m is sufficiently large.
The second result pertains to a subset of Bernstein polynomials: those with coefficients that are all in the unit interval. We show that polynomials in this subset map the open interval (0, 1) into the open interval (0, 1) and map the points 0 and 1 into the closed interval [0, 1]. The motivation for this work is our research on probabilistic computation with digital circuits. Our design methodology, called stochastic logic, is based on Bernstein polynomials with coefficients that correspond to probability values; accordingly, the coefficients must be values in the unit interval. The mathematics presented here provide a necessary and sufficient test for deciding whether polynomial operations can be implemented with stochastic logic.
1 Introduction The Weierstrass approximation theorem is a famous theorem in mathematical analysis. It as-serts that every continuous function defined on a closed interval can be uniformly approximated as closely as desired by a polynomial function .
The Weierstrass Approximation Theorem: Let f be a continuous function defined on the closed interval [a, b]. For any ϵ > 0, there exists a polynomial function p such that for all x in [a, b], we have |f(x) −p(x)| < ϵ.
□ The theorem can be proved by a transformation with Bernstein polynomials . By a linear substitution, the interval [a, b] can be transformed into the unit interval [0, 1]. Thus, the original statement of the theorem holds if and only if the theorem holds for every continuous function f defined on the interval [0, 1].
∗This work is supported by a grant from the Semiconductor Research Corporation’s Focus Center Research Program on Functional Engineered Nano-Architectonics, contract No. 2003-NT-1107.
1 A Bernstein polynomial of degree n is a polynomial expressed in the following form : n X k=0 βk,nbk,n(x), (1) where each βk,n, k = 0, 1, . . . , n, is a real number and bk,n(x) = n k xk(1 −x)n−k.
(2) The coefficients βk,n are called Bernstein coefficients and the polynomials b0,n(x), b1,n(x), . . . , bn,n(x) are called Bernstein basis polynomials of degree n. Define the n-th Bernstein polynomial for f to be Bn(f; x) = n X k=0 f k n bk,n(x).
In 1912, Bernstein showed the following result [4,5]: The Bernstein Theorem: Let f be a continuous function defined on the closed interval [0, 1].
For any ϵ > 0, there exists a positive integer M such that for all x in [0, 1] and integer m ≥M, we have |f(x) −Bm(f; x)| < ϵ.
□ Note that the function Bm(f; x) is a polynomial on x. Thus, based on the Bernstein Theorem, the Weierstrass Approximation Theorem holds. Given a power-form polynomial g of degree n, it is well known that for any m ≥n, g can be uniquely converted into a Bernstein polynomial of degree m . Combining this fact with the Bernstein Theorem, we have the following corollary.
Corollary 1 Let g be a polynomial of degree n. For any ϵ > 0, there exists a positive integer M ≥n such that for all x in [0, 1] and integer m ≥M, we have m X k=0 βk,m −g k m bk,m(x) < ϵ, where β0,m, β1,m, . . . , βm,m satisfy g(x) = m X k=0 βk,mbk,m(x). □ In the first part of the paper, we prove a stronger result than this: Theorem 1 Let g be a polynomial of degree n ≥0. For any ϵ > 0, there exists a positive integer M ≥n such that for all integers m ≥M and k = 0, 1, . . . , m, we have βk,m −g k m < ϵ, where β0,m, β1,m, . . . , βm,m satisfy g(x) = m X k=0 βk,mbk,m(x). □ 2 (Combining Theorem 1 with the fact that m X k=0 bk,m(x) = 1, we can easily prove Corollary 1.) In the second part of the paper, we consider a subset of Bernstein polynomials: those with coefficients that are all in the unit interval [0, 1].
Definition 1 Define U to be the set of Bernstein polynomials with coefficients that are all in the unit interval [0, 1]: U = ( p(x) | ∃n ≥1, 0 ≤β0,n, β1,n, . . . , βn,n ≤1, such that p(x) = n X k=0 βk,nbk,n(x) ) .
□ The question we ask is: which polynomials can be converted into Bernstein polynomials in U?
Definition 2 Define the set V to be the set of polynomials which are either identically equal to 0 or equal to 1, or map the open interval (0, 1) into (0, 1) and the points 0 and 1 into the closed interval [0, 1], i.e., V = {p(x) | p(x) ≡0, or p(x) ≡1, or 0 < p(x) < 1, ∀x ∈(0, 1) and 0 ≤p(0), p(1) ≤1} .
□ We prove that the two sets are equivalent: Theorem 2 V = U.
□ In what follows, we will refer to a Bernstein polynomial of degree n converted from a polynomial g as “the Bernstein polynomial of degree n of g”. When we say that a polynomial is of degree n, we mean that the power-form of the polynomial is of degree n.
Example 1 Consider the polynomial g(x) = 3x −8x2 + 6x3. It maps the open interval (0, 1) into (0, 1) with g(0) = 0, g(1) = 1. Thus, g is in the set V . Based on Theorem 2, we have that g is in the set U.
We verify this by considering Bernstein polynomials of increasing degree.
• The Bernstein polynomial of degree 3 of g is g(x) = b1,3(x) −2 3b2,3(x) + b3,3(x).
Note that here the coefficient β2,3 = −2 3 < 0.
• The Bernstein polynomial of degree 4 of g is g(x) = 3 4b1,4(x) + 1 6b2,4(x) −1 4b3,4(x) + b4,4(x).
Note that here the coefficient β3,4 = −1 4 < 0.
• The Bernstein polynomial of degree 5 of g(x) is g(x) = 3 5b1,5(x) + 2 5b2,5(x) + b5,5(x).
Note that here all the coefficients are in [0, 1].
3 Since the Bernstein polynomial of degree 5 of g satisfies Definition 1, we conclude that g is in the set U. □ Example 2 Consider the polynomial g(x) = 1 4 −x + x2. Since g(0.5) = 0, thus g is not in the set V . Based on Theorem 2, we have that g is not in the set U. We verify this. By contraposition, suppose that there exist n ≥1 and 0 ≤β0,n, β1,n, . . . , βn,n ≤1 such that g(x) = n X k=0 βk,nbk,n(x).
Since g(0.5) = 0, therefore, n X k=0 βk,nbk,n(0.5) = 0. Note that for all k = 0, 1, . . . , n, bk,n(0.5) > 0.
Thus, we have that for all k = 0, 1, . . . , n, βk,n = 0. Therefore, g(x) ≡0, which contradicts the original assumption about g. Thus, g is not in the set U. □ The remainder of the paper is organized as follows. In Section 2, we present some mathematical preliminaries pertaining to Bernstein polynomials. In Section 3, we prove Theorem 1. Based on this theorem, in Section 4, we prove Theorem 2. Finally, we conclude the paper with a discussion on applications of these theorems to our research in probabilistic computation with digital circuits.
2 Properties of Bernstein Polynomials We list some pertinent properties of Bernstein polynomials.
(a) The positivity property: For all k = 0, 1, . . . , n and all x in [0, 1], we have bk,n(x) ≥0.
(3) (b) The partition of unity property: The binomial expansion of the left-hand side of the equality (x + (1 −x))n = 1 shows that the sum of all Bernstein basis polynomials of degree n is the constant 1, i.e., n X k=0 bk,n(x) = 1.
(4) (c) Converting power-form coefficients to Bernstein coefficients: The set of Bernstein basis polynomials b0,n(x), b1,n(x), . . . , bn,n(x) forms a basis of the vector space of polynomials of real coefficients and degree no more than n .
Each power basis function xj can be uniquely expressed as a linear combination of the n + 1 Bernstein basis polynomials: xj = n X k=0 σjkbk,n(x), (5) for j = 0, 1, . . . , n. To determine the elements of the transformation matrix σ, we write xj = xj(x + (1 −x))n−j 4 and perform a binomial expansion on the right hand side. This gives xj = n X k=j k j n j bk,n(x), for j = 0, 1, . . . , n. Therefore, we have σjk = ( σjk = k j n j −1, for j ≤k 0, for j > k.
(6) Suppose that a power-form polynomial of degree no more than n is g(x) = n X k=0 ak,nxk (7) and the Bernstein polynomial of degree n of g is g(x) = n X k=0 βk,nbk,n(x).
(8) Substituting Equations (5) and (6) into Equation (7) and comparing the Bernstein coefficients, we have βk,n = n X j=0 aj,nσjk = k X j=0 k j n j −1 aj,n.
(9) Equation (9) provide a means for obtaining Bernstein coefficients from power-form coefficients.
(d) Degree elevation: Based on Equation (2), we have that for all k = 0, 1, . . . , m, 1 m+1 k bk,m+1(x) + 1 m+1 k+1 bk+1,m+1(x) = xk(1 −x)m+1−k + xk+1(1 −x)m−k =xk(1 −x)m−k = 1 m k bk,m(x), or bk,m(x) = m k m+1 k bk,m+1(x) + m k m+1 k+1 bk+1,m+1(x) = m + 1 −k m + 1 bk,m+1(x) + k + 1 m + 1bk+1,m+1(x).
(10) Given a power-form polynomial g of degree n, for any m ≥n, g can be uniquely converted into a Bernstein polynomial of degree m. Suppose that the Bernstein polynomials of degree m and degree m + 1 of g are m X k=0 βk,mbk,m(x) and m+1 X k=0 βk,m+1bk,m+1(x), respectively. We have m X k=0 βk,mbk,m(x) = m+1 X k=0 βk,m+1bk,m+1(x).
(11) 5 Substituting Equation (10) into the left-hand side of Equation (11) and comparing the Bernstein coefficients, we have βk,m+1 = β0,m, for k = 0 k m+1βk−1,m + 1 − k m+1 βk,m, for 1 ≤k ≤m βm,m, for k = m + 1.
(12) Equation (12) provides a means for obtaining the coefficients of the Bernstein polynomial of degree m + 1 of g from the coefficients of the Bernstein polynomial of degree m of g. We will call this procedure degree elevation.
For convenience, given a Bernstein polynomial g(x) = Pn k=0 βk,nbk,n(x), we can also express it as g(x) = n X k=0 ck,nxk(1 −x)n−k, (13) where ck,n = n k βk,n, (14) for k = 0, 1, . . . , n. Substituting Equation (14) into Equation (12), we have ck,m+1 = c0,m, for k = 0 ck−1,m + ck,m, for 1 ≤k ≤m cm,m, for k = m + 1.
(15) 3 A Proof of Theorem 1 Suppose that the polynomial g is of degree n.
Applying Equation (15) recursively, we can express ck,m as a linear combination of c0,n, c1,n, . . . , cn,n.
Lemma 1 Let g be a polynomial of degree n. For any m ≥n, suppose that the Bernstein polynomial of degree m of g is g(x) = m X k=0 ck,mxk(1 −x)m−k. Let ck,m = 0 for all k < 0 and all k > m. Then for all k = 0, 1, . . . , m, we have ck,m = m−n X i=0 m −n i ck−m+n+i,n.
□ (16) Proof: We prove the lemma by induction on m −n.
Base case: For m −n = 0, the right-hand side of Equation (16) reduces to 0 0 ck,n = ck,m, so the equation holds.
Inductive step: Suppose that Equation (16) holds for some m ≥n and all k = 0, 1, . . . , m.
Consider m + 1. Since we assume that c−1,m = cm+1,m = 0, Equation (15) can be written as ck,m+1 = ck−1,m + ck,m, (17) 6 for all k = 0, . . . , m + 1. With our convention that ci,n = 0 for all i < 0 and i > n, it is easily seen that c−1,m = 0 = m−n X i=0 m −n i c−1−m+n+i,n, cm+1,m = 0 = m−n X i=0 m −n i cm+1−m+n+i,n.
Together with the induction hypothesis, we conclude that for all k = −1, 0, . . . , m, m + 1 ck,m = m−n X i=0 m −n i ck−m+n+i,n.
(18) Based on Equations (17) and (18), for all k = 0, 1, . . . , m + 1, we have ck,m+1 = m−n X i=0 m −n i ck−1−m+n+i,n + m−n X j=0 m −n j ck−m+n+j,n.
In the first sum, we change the summation index to j = i −1. We obtain ck,m+1 = m−n−1 X j=−1 m −n j + 1 ck−m+n+j,n + m−n X j=0 m −n j ck−m+n+j,n = m −n 0 ck−m+n−1,n + m−n−1 X j=0 m −n j + 1 + m −n j ck−m+n+j,n + m −n m −n ck,n.
Applying the basic formula r q = r −1 q −1 + r −1 q , we obtain ck,m+1 = ck−m+n−1,n + m−n−1 X j=0 m + 1 −n j + 1 ck−m+n+j,n + ck,n = m+1−n X i=0 m + 1 −n i ck−m−1+n+i,n.
Thus Equation (16) holds for m + 1. By induction, it holds for all m ≥k. □ Remark: Equation (16) can be formulated as ck,m = min{k,n} X i=max{0,k−m+n} m −n k −i ci,n, (19) for all m ≥n and k = 0, 1, . . . , m. Indeed, in Equation (16), first use the basic formula r q = r r −q and then change the summation index to j = k −m + n + i to obtain ck,m = m−n X i=0 m −n m −n −i ck−m+n+i,n = k X j=k−m+n m −n k −j cj,n.
Note that cj,n ̸= 0 implies 0 ≤j ≤n. This yields Equation (19). □ 7 Lemma 2 Let n be a positive integer. For all integer m, k and i such that m > n, 0 ≤k ≤m, max{0, k −m + n} ≤i ≤min{k, n}, (20) we have k m i 1 −k m n−i − m−n k−i m k ≤n2 m .
□ (21) Proof: For simplicity, we define δ = k m i 1 −k m n−i − m−n k−i m k . Now m−n k−i m k = (m −n)!
(k −i)!(m −n −k + i)! · k!(m −k)!
m!
= k(k −1) · · · (k −i + 1)(m −k)(m −k −1) · · · (m −n −k + i + 1) m(m −1) · · · (m −n + 1) = i−1 Y j=0 k −j m −j · n−i−1 Y j=0 m −k −j m −i −j = i−1 Y j=0 1 −m −k m −j · n−i−1 Y j=0 1 − k −i m −i −j .
(22) We obtain an upper bound for m−n k−i m k by replacing j in Equation (22) with its least value, 0: m−n k−i m k ≤ i−1 Y j=0 1 −m −k m · n−i−1 Y j=0 1 −k −i m −i = k m i m −k m −i n−i .
We need the following simple inequality: for real numbers 0 ≤x ≤y ≤1 and a non-negative integer l, yl −xl = (y −x) l−1 X j=0 yjxl−1−j ≤(y −x)l.
(23) From Equation (20), we obtain 0 ≤i ≤min{k, n} ≤k ≤m and so we can use Equation (23) for 0 ≤x = m −k m ≤m −k m −i = y ≤1, l = n −i ≥0.
We obtain δ = k m i 1 −k m n−i − m−n k−i m k ≥ k m i m −k m n−i − m −k m −i n−i!
= − k m i m −k m −i n−i − m −k m n−i!
≥− k m i m −k m −i −m −k m (n −i) = − k m i (m −k)i(n −i) (m −i)m .
Since 0 ≤k m ≤1, 0 ≤m −k m −i ≤1, and 0 ≤i ≤n, we obtain − k m i (m −k)i(n −i) (m −i)m ≥−i(n −i) m > −n2 m .
8 Therefore, δ = k m i 1 −k m n−i − m−n k−i m k > −n2 m .
(24) Similarly, we obtain a lower bound for m−n k−i m k by replacing the index j in Equation (22) with i in the first product and with n −i in the second product, obtaining m−n k−i m k = i−1 Y j=0 1 −m −k m −j · n−i−1 Y j=0 1 − k −i m −i −j ≥ i−1 Y j=0 1 −m −k m −i · n−i−1 Y j=0 1 −k −i m −n = k −i m −i i m −n −k + i m −n n−i ≥ k −i m −i i m −n −k + i m −n + i n−i .
Thus, proceeding as above, we have δ = k m i 1 −k m n−i − m−n k−i m k ≤ k m i m −k m n−i − k −i m −i i m −n −k + i m −n + i n−i = " k m i − k −i m −i i# m −k m n−i + "m −k m n−i − m −n −k + i m −n + i n−i# k −i m −i i .
Due to Equation (20), we have 0 ≤k −i m −i ≤k m ≤1, 0 ≤m −n −k + i m −n + i ≤m −k m ≤1, and so we obtain δ ≤ k m i − k −i m −i i + m −k m n−i − m −n −k + i m −n + i n−i .
(25) Applying Equation (23) twice to the right-hand side of Equation (25), we obtain δ ≤i k m −k −i m −i + (n −i) m −k m −m −n −k + i m −n + i = i2 m · m −k m −i + (n −i)2 m · k m −n + i.
From Equation (20), we have 0 ≤m −k m −i ≤1, 0 ≤ k m −n + i ≤1.
Therefore, δ = k m i 1 −k m n−i − m−n k−i m k ≤i2 + (n −i)2 m ≤ni + n(n −i) m = n2 m .
(26) Equations (24) and (26) together yield Equation (21). □ Now we give a proof of Theorem 1.
9 Theorem 1 Let g be a polynomial of degree n ≥0. For any ϵ > 0, there exists a positive integer M ≥n such that for all integer m ≥M and k = 0, 1, . . . , m, we have βk,m −g k m < ϵ, where β0,m, β1,m, . . . , βm,m satisfy that g(x) = m X k=0 βk,mbk,m(x). □ Proof: For n = 0, g is a constant polynomial. Suppose that g(x) = y, where y is a constant value. We select M = 1. Then, for all integers m ≥M and all integers k = 0, 1, . . . , m, we have βk,m = y = g k m . Thus, the theorem holds.
For n > 0, we select M such that M > max ( n2 ϵ n X i=0 |ci,n|, 2n ) , where the real numbers c0,n, c1,n, . . . , cn,n satisfy g(x) = n X i=0 ci,nxi(1 −x)n−i.
(27) Now consider any m ≥M. Since 2n ≤max ( n2 ϵ n X i=0 |ci,n|, 2n ) < M ≤m, we have m −n > n. Consider the following three cases for k.
1. The case where n ≤k ≤m −n. Here max{0, k −m + n} = 0 and min{k, n} = n. Thus, the summation indices in Equation (19) range from 0 to n. Therefore, βk,m = ck,m m k = n X i=0 m−n k−i m k ci,n.
(28) Substituting x with k m in Equation (27), we have g k m = n X i=0 ci,n k m i 1 −k m n−i .
(29) By Lemma 2, since 0 < n < m and 0 ≤k ≤m, Equation (21) holds for all 0 = max{0, k −m + n} ≤i ≤min{k, n} = n. Thus, by Equations (21), (28), (29) and the well-known inequality | P xi| ≤P |xi|, we have βk,m −g k m = n X i=0 " m−n k−i m k − k m i 1 −k m n−i# ci,n ≤ n X i=0 m−n k−i m k − k m i 1 −k m n−i |ci,n| ≤n2 m n X i=0 |ci,n|.
10 Since n2 ϵ n X i=0 |ci,n| < M ≤m, we have n2 m n X i=0 |ci,n| < ϵ.
(30) Therefore, for all n ≤k ≤m −n, we have βk,m −g k m < ϵ.
2. The case where 0 ≤k < n.
Since m > 2n, we have k −m + n < k −n < 0.
Thus, max{0, k −m + n} = 0 and min{k, n} = k. Thus, the summation indices in Equation (19) range from 0 to k. Therefore, βk,m = ck,m m k = k X i=0 m−n k−i m k ci,n.
(31) When k + 1 ≤i ≤n, we have that 1 ≤k + 1 ≤i and so k m i 1 −k m n−i = k m k m i−1 1 −k m n−i ≤k m < n m ≤n2 m .
(32) By Lemma 2, since 0 < n < m and 0 ≤k ≤m, Equation (21) holds for all 0 = max{0, k −m + n} ≤i ≤min{k, n} = k. Thus, by Equations (21), (29), (30), (31), (32) and the inequality | P xi| ≤P |xi|, we have βk,m −g k m = k X i=0 m−n k−i m k ci,n − n X i=0 k m i 1 −k m n−i ci,n = k X i=0 " m−n k−i m k − k m i 1 −k m n−i# ci,n − n X i=k+1 k m i 1 −k m n−i ci,n ≤ k X i=0 m−n k−i m k − k m i 1 −k m n−i |ci,n| + n X i=k+1 k m i 1 −k m n−i |ci,n| ≤n2 m n X i=0 |ci,n| < ϵ.
3. The case where m −n < k ≤m. Since m > 2n, we have n < m −n < k. Thus, max{0, k −m + n} = k −m + n and min{k, n} = n.
Now, the summation indices in Equation (19) range from k −m + n to n. Therefore, βk,m = ck,m m k = n X i=k−m+n m−n k−i m k ci,n.
(33) When 0 ≤i ≤k −m + n −1, we have that 1 ≤m + 1 −k ≤n −i. Thus, k m i 1 −k m n−i = 1 −k m k m i 1 −k m n−i−1 ≤m −k m < n m ≤n2 m .
(34) 11 By Lemma 2, since 0 < n < m and 0 ≤k ≤m, Equation (21) holds for all k −m + n = max{0, k −m + n} ≤i ≤min{k, n} = n. Thus, by Equations (21), (29), (30), (33), (34) and the inequality | P xi| ≤P |xi|, we have βk,m −g k m = n X i=k−m+n m−n k−i m k ci,n − n X i=0 k m i 1 −k m n−i ci,n = n X i=k−m+n " m−n k−i m k − k m i 1 −k m n−i# ci,n − k−m+n−1 X i=0 k m i 1 −k m n−i ci,n ≤ n X i=k−m+n m−n k−i m k − k m i 1 −k m n−i |ci,n| + k−m+n−1 X i=0 k m i 1 −k m n−i |ci,n| ≤n2 m n X i=0 |ci,n| < ϵ.
In conclusion, if m ≥M, then for all k = 0, 1, . . . , m, we have βk,m −g k m < ϵ.
□ 4 A Proof of Theorem 2 We demonstrate that the sets U and V defined in the introduction – see Definitions 1 and 2 – are one and the same. We demonstrate that U ⊆V and V ⊆U separately. First, we prove the former – the easier one. Then we use Theorem 1 to prove the latter.
Theorem 3 U ⊆V.
□ Proof: Let n ≥1 and βk,n = 0, for all 0 ≤k ≤n. Then the polynomial p(x) = n X k=0 βk,nbk,n(x) = 0.
Let n ≥1 and βk,n = 1, for all 0 ≤k ≤n. Then, by Equation (4), the polynomial p(x) = n X k=0 βk,nbk,n(x) = 1.
Thus 0 ∈U and 1 ∈U. From the definition of V , 0 ∈V and 1 ∈V .
Now consider any polynomial p ∈U such that p ̸≡0 and p ̸≡1.
There exist n ≥1 and 0 ≤β0,n, β1,n, . . . , βn,n ≤1 such that p(x) = n X k=0 βk,nbk,n(x).
From Equations (3), (4) and the fact that 0 ≤β0,n, β1,n, . . . , βn,n ≤1, for all x in [0, 1], we have 0 ≤p(x) = n X k=0 βk,nbk,n(x) ≤ n X k=0 bk,n(x) = 1.
We further claim that for all x in (0, 1), we must have 0 < p(x) < 1. By contraposition, we assume that there exists a 0 < x0 < 1, such that p(x0) ≤0 or p(x0) ≥1. Since for 0 < x0 < 1, we have 0 ≤p(x0) ≤1, thus p(x0) = 0 or 1.
12 We first consider the case that p(x0) = 0. Since 0 < x0 < 1, it is not hard to see that for all k = 0, 1, . . . , n, bk,n(x0) > 0. Thus, p(x0) = 0 implies that for all k = 0, 1, . . . , n, βk,n = 0. In this case, for any real number x, p(x) = Pn k=0 βk,nbk,n(x) = 0, which contradicts the assumption that p(x) ̸≡0.
Similarly, in the case that p(x0) = 1, we can show that p(x) ≡1, which contradicts the assumption that p(x) ̸≡1. In both cases, we get a contradiction; this proves the claim that for all x in (0, 1), 0 < p(x) < 1.
Therefore, for any polynomial p ∈U such that p ̸≡0 and p ̸≡1, we have p ∈V . Since we showed at the outset that 0 ∈U, 1 ∈U, 0 ∈V and 1 ∈V , thus, for any polynomial p ∈U, we have p ∈V . Therefore, U ⊆V . □ Next we prove the claim that V ⊆U. We will first show that each of four possible different categories of polynomials in the set V are in the set U. The different categories are tackled in Theorems 4 and 5 and Corollaries 2 and 3.
Theorem 4 Let g be a polynomial of degree n mapping the open interval (0, 1) into (0, 1) with 0 ≤g(0), g(1) < 1.
Then g ∈U. □ Proof: Since g is continuous on the closed interval [0, 1], it attains its maximum value Mg on [0, 1].
Since g(x) < 1, for all x ∈[0, 1], we have Mg < 1.
Let ϵ1 = 1 −Mg > 0. By Theorem 1, there exists a positive integer M1 ≥n such that for all integers m ≥M1 and k = 0, 1, . . . , m, we have βk,m −g k m < ϵ1, where β0,m, β1,m, . . . , βm,m satisfy that g(x) = m X k=0 βk,mbk,m(x). Thus, for all m ≥M1 and all k = 0, 1, . . . , m, βk,m < g k m + ϵ1 ≤Mg + 1 −Mg = 1.
(35) Denote by r the multiplicity of 0 as a root of g(x) (where r = 0 if g(0) ̸= 0) and by s the multiplicity of 1 as a root of g(x) (where s = 0 if g(1) ̸= 0). We can factorize g(x) as g(x) = xr(1 −x)sh(x), (36) where h(x) is a polynomial, satisfying that h(0) ̸= 0 and h(1) ̸= 0.
We show that h(0) > 0. By the way of contraposition, suppose that h(0) ≤0. Since h(0) ̸= 0, we have h(0) < 0. By the continuity of the polynomial h(x), there exists some 0 < x∗< 1, such that h(x∗) < 0. Thus, g(x∗) = x∗r(1 −x∗)sh(x∗) < 0. However, g(x) > 0, for all x ∈(0, 1).
Therefore, h(0) > 0. Similarly, we have h(1) > 0.
Since g(x) > 0 for all x in (0, 1), we have h(x) = g(x) xr(1 −x)s > 0 for all x in (0, 1). In view of the fact that h(0) > 0 and h(1) > 0, we have h(x) > 0, for all x in [0, 1]. Since h(x) is continuous on the closed interval [0, 1], it attains its minimum value mh on [0, 1]. Clearly, mh > 0.
Let ϵ2 = mh > 0. By Theorem 1, there exists a positive integer M2 ≥n−r −s, such that for all integers d ≥M2 and k = 0, 1, . . . , d, we have γk,d −h k d < ϵ2, where γ0,d, γ1,d, . . . , γd,d satisfy that h(x) = d X k=0 γk,dbk,d(x).
(37) 13 Thus, for all d ≥M2 and all k = 0, 1, . . . , d, γk,d > h k d −ϵ2 ≥mh −mh = 0.
Combining Equations (36) and(37), we have g(x) = xr(1 −x)sh(x) = xr(1 −x)s d X k=0 γk,dbk,d(x) = xr(1 −x)s d X k=0 γk,d d k xk(1 −x)d−k = d X k=0 γk,d d k d+r+s k+r d + r + s k + r xk+r(1 −x)d+s−k = d+r X k=r γk−r,d d k−r d+r+s k bk,d+r+s(x) = d+r+s X k=0 βk,d+r+sbk,d+r+s(x), where βk,d+r+s are the coefficients of the Bernstein polynomial of degree d + r + s of g and βk,d+r+s = 0, for 0 ≤k < r and d + r < k ≤d + r + s γk−r,d( d k−r) (d+r+s k ) > 0, for r ≤k ≤d + r.
Thus, when m = d + r + s ≥M2 + r + s, we have for all k = 0, 1, . . . , m, βk,m ≥0.
(38) According to Equations (35) and (38), if we select an m0 ≥max{M1, M2 + r + s}, then g(x) can be expressed as a Bernstein polynomial of degree m0: g(x) = m0 X k=0 βk,m0bk,m0(x), with 0 ≤βk,m0 ≤1, for all k = 0, 1, . . . , m0. Therefore, g ∈U. □ Theorem 5 Let g be a polynomial of degree n mapping the open interval (0, 1) into (0, 1) with g(0) = 0 and g(1) = 1. Then g ∈U. □ Proof: Denote by r the multiplicity of 0 as a root of g(x). We can factorize g(x) as g(x) = xrh(x), (39) where h(x) is a polynomial satisfying h(0) ̸= 0. By a similar reasoning as in the proof of Theorem 4, we obtain h(0) > 0. Since for all x in (0, 1], h(x) = g(x) xr > 0, we have for all x in [0, 1], h(x) > 0.
Since h(x) is continuous on the closed interval [0, 1], it attains its minimum value mh on [0, 1].
Clearly, mh > 0.
Let ϵ1 = mh > 0. By Theorem 1, there exists a positive integer M1 ≥n −r such that for all integers d ≥M1 and k = 0, 1, . . . , d, we have γk,d −h k d < ϵ1, where γ0,d, γ1,d, . . . , γd,d satisfy h(x) = d X k=0 γk,dbk,d(x).
(40) 14 Thus, for all d ≥M1 and all k = 0, 1, . . . , d, γk,d > h k d −ϵ1 ≥mh −mh = 0.
Combining Equations (39) and (40), we have g(x) = xrh(x) = xr d X k=0 γk,dbk,d(x) = xr d X k=0 γk,d d k xk(1 −x)d−k = d X k=0 γk,d d k d+r k+r d + r k + r xk+r(1 −x)d−k = d+r X k=r γk−r,d d k−r d+r k bk,d+r(x) = d+r X k=0 βk,d+rbk,d+r(x), where βk,d+r are the coefficients of the Bernstein polynomial of degree d + r of g and βk,d+r = 0, for 0 ≤k < r γk−r,d( d k−r) (d+r k ) > 0, for r ≤k ≤d + r.
Thus, when m = d + r ≥M1 + r, we have for all k = 0, 1, . . . , m, βk,m ≥0.
(41) Let g∗(x) = 1 −g(x).
(42) Then g∗maps the open interval (0, 1) into (0, 1) with g∗(0) = 1, g∗(1) = 0. Denote by s the multiplicity of 1 as a root of g∗(x). Thus, we can factorize g∗(x) as g∗(x) = (1 −x)sh∗(x), (43) where h∗(x) is a polynomial satisfying that h∗(1) ̸= 0. As in the proof of Theorem 4, we obtain h∗(1) > 0. Since for all x in [0, 1), h∗(x) = g∗(x) (1 −x)s > 0, we have for all x ∈[0, 1], h∗(x) > 0. Since h∗(x) is continuous on the closed interval [0, 1], it attains its minimum value m∗ h on [0, 1]. Clearly, m∗ h > 0.
Let ϵ2 = m∗ h > 0. By Theorem 1, there exists a positive integer M2 ≥n −s such that for all integers q ≥M2 and k = 0, 1, . . . , q, we have γ∗ k,q −h∗ k q < ϵ2, where γ∗ 0,q, γ∗ 1,q, . . . , γ∗ q,q satisfy h∗(x) = q X k=0 γ∗ k,qbk,q(x).
(44) Thus, for all q ≥M2 and all k = 0, 1, . . . , q, γ∗ k,q > h∗ k q −ϵ2 ≥m∗ h −m∗ h = 0.
Combining Equations (42), (43) and (44), we have g(x) = 1 −g∗(x) = 1 −(1 −x)sh∗(x) = 1 −(1 −x)s q X k=0 γ∗ k,qbk,q(x) = 1 −(1 −x)s q X k=0 γ∗ k,q q k xk(1 −x)q−k = 1 − q X k=0 γ∗ k,q q k q+s k q + s k xk(1 −x)q+s−k.
15 Further using (4), we obtain g(x) = q+s X k=0 bk,q+s(x) − q X k=0 γ∗ k,q q k q+s k bk,q+s(x) = q+s X k=0 βk,q+sbk,q+s(x), where the βk,q+s’s are the coefficients of the Bernstein polynomial of degree q + s of g: βk,q+s = 1 − γ∗ k,q(q k) (q+s k ) < 1, for 0 ≤k ≤q 1, for q < k ≤q + s.
Thus, when m = q + s ≥M2 + s, we have for all k = 0, 1, . . . , m, βk,m ≤1.
(45) According to Equations (41) and (45), if we select an m0 ≥max{M1 + r, M2 + s}, then g(x) can be expressed as a Bernstein polynomial of degree m0: g(x) = m0 X k=0 βk,m0bk,m0(x), with 0 ≤βk,m0 ≤1, for all k = 0, 1, . . . , m0, Therefore, g ∈U. □ Lemma 3 If a polynomial p is in the set U, then the polynomial 1 −p is also in the set U. □ Proof: Since p is in the set U, there exist n ≥1 and 0 ≤β0,n, β1,n, . . . , βn,n ≤1 such that p(x) = n X k=0 βk,nbk,n(x).
By Equation (4), we have 1 −p(x) = n X k=0 bk,n(x) − n X k=0 βk,nbk,n(x) = n X k=0 (1 −βk,n)bk,n(x) = n X k=0 γk,nbk,n(x), where γk,n = 1−βk,n satisfying 0 ≤γk,n ≤1, for all k = 0, 1, . . . , n. Therefore, 1−p is in the set U. □ Corollary 2 Let g be a polynomial of degree n mapping the open interval (0, 1) into (0, 1) with 0 < g(0), g(1) ≤1.
Then g ∈U. □ Proof: Let polynomial h = 1 −g. Then h maps (0, 1) into (0, 1) with 0 ≤h(0), h(1) < 1. By Theorem 4, h ∈U. By Lemma 3, g = 1 −h is also in the set U. □ Corollary 3 Let g be a polynomial of degree n mapping the open interval (0, 1) into (0, 1) with g(0) = 1 and g(1) = 0. Then g ∈U. □ 16 Proof: Let the polynomial h = 1 −g. Then h maps (0, 1) into (0, 1) with h(0) = 0, h(1) = 1. By Theorem 5, h ∈U. By Lemma 3, g = 1 −h is also in the set U. □ Combining Theorem 4, Theorem 5, Corollary 2 and Corollary 3, we show that V ⊆U.
Theorem 6 V ⊆U.
□ Proof: Based on the definition of V , for any polynomial p ∈V , we have one of following five cases.
1. The case where p ≡0 or p ≡1. In the proof of Theorem 3, we have shown that 0 ∈U and 1 ∈U. Thus p ∈U.
2. The case where p maps the open interval (0, 1) into (0, 1) with 0 ≤p(0), p(1) < 1.
By Theorem 4, p ∈U.
3. The case where p maps the open interval (0, 1) into (0, 1) with 0 < p(0), p(1) ≤1.
By Corollary 2, p ∈U.
4. The case where p maps the open interval (0, 1) into (0, 1) with p(0) = 0 and p(1) = 1. By Theorem 5, p ∈U.
5. The case where p maps the open interval (0, 1) into (0, 1) with p(0) = 1 and p(1) = 0. By Corollary 3, p ∈U.
In summary, for any polynomial p ∈V , we have p ∈U. Thus, V ⊆U. □ Based on Theorems 3 and 6, we have proved Theorem 2: V = U.
□ 5 Discussion We are interested in Bernstein polynomials with coefficients in the unit interval because this concept has applications in the area of digital circuit design. Specifically, the concept is a math-ematical prerequisite for a design methodology that we have been advocating called stochastic logic [7–9]. We provide a brief overview of this application and point the reader to further sources.
Stochastic logic implements Boolean function with inputs that are random Boolean variables.
A Boolean function f on n variables x1, x2, . . . , xn is a mapping f : {0, 1}n →{0, 1}.
With stochastic logic, the variables x1, x2, . . . , xn are a set of independent random Boolean variables, i.e., for 1 ≤i ≤n, xi has a certain probability pi (0 ≤pi ≤1) of being one and a probability 1 −pi of being zero. With random Boolean variables as inputs, the output is also a random Boolean variable: the function f has a certain probability po of being one and a probability 1 −po of being zero.
If implemented by digital circuitry, stochastic logic can be viewed as computation that trans-forms input probabilities into output probabilities . Given an arbitrary Boolean function f and 17 a set of input probabilities p1, p2, . . . , pn that correspond to the probabilities of the input random Boolean variables being one, the output probability po is a function on p1, p2, . . . , pn. In fact, we have shown that the general form of the function is a multivariate polynomial on variables p1, . . . , pn with integer coefficients and with the degree of each variable no more than one .
Example 3 Consider stochastic logic based on the Boolean function f(x1, x2, x3) = (x1 ∧x2) ∨(¬x1 ∧x3), where ∧means logical AND (conjunction), ∨means logical OR (disjunction), and ¬ means logical negation.
The Boolean function f evaluates to one if and only if the 3-tuple (x1, x2, x3) takes values from the set {(0, 0, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1)}. The probability of the output being one is po = Pr(f = 1) = Pr (x1, x2, x3 : (x1, x2, x3) ∈{(0, 0, 1), (0, 1, 1), (1, 1, 0), (1, 1, 1)}) = Pr(x1 = 0, x2 = 0, x3 = 1) + Pr(x1 = 0, x2 = 1, x3 = 1) + Pr(x1 = 1, x2 = 1, x3 = 0) + Pr(x1 = 1, x2 = 1, x3 = 1).
If x1, x2, and x3 are independent random Boolean variables with probability p1, p2, and p3 of being one, respectively, then we obtain po = (1 −p1)(1 −p2)p3 + (1 −p1)p2p3 + p1p2(1 −p3) + p1p2p3 = (1 −p1)p3 + p1p2 = p1p2 + p3 −p1p3, (46) which confirms that the function computed by stochastic logic is a multivariate polynomial on arguments p1, p2, and p3 with integer coefficients and with the degree of each variable no more than 1. □ In design problems, we encounter univariate polynomials that have real coefficients and degree greater than 1. Sometimes it is possible to implement these by setting some of the probabilities pi to be a common variable x and the others to be constants. For example, if we set p1 = p3 = x and p2 = 0.75 in Equation (46), then we obtain the polynomial g(x) = 1.75x −x2. With different underlying Boolean functions and different assignments of probability values, we can implement many different univariate polynomials.
An interesting and yet practical question is: which univariate polynomials can be implemented by stochastic logic? Define the set W to be the set of (univariate) polynomials that can be imple-mented. We are interested in characterizing the set W.
In we showed that U ⊆W, i.e., if a polynomial can be expressed as a Bernstein polynomial with all coefficients in the unit interval, then the polynomial can be implemented by stochastic logic. In this paper, we proved that V = U. Thus, we have V ⊆W.
Further, in we showed that W ⊆V , i.e., if a polynomial can be implemented by stochastic logic, then it is either identically equal to 0 or equal to 1, or it maps the open interval (0, 1) into the open interval (0, 1) and maps the points 0 and 1 into the closed interval [0, 1]. Therefore, we conclude that W = V , i.e., a polynomial can be implemented by stochastic logic if and only if it is either identically equal to 0 or equal to 1, or it maps the open interval (0, 1) into the open interval (0, 1) and maps the points 0 and 1 into the closed interval [0, 1].
This necessary and sufficient conditions allows us to answer the question of whether any given polynomial can be implemented by stochastic logic. Based on the mathematics, we have proposed a constructive design method . An overview of the method and its applications in circuit design will appear in a forthcoming “Research Highlights” article in Communications of the ACM .
18 References P. J. Davis, Interpolation and Approximation.
Blaisdell Publishing Company, 1963.
G. Lorentz, Bernstein Polynomials.
University of Toronto Press, 1953.
J. Berchtold and A. Bowyer, “Robust arithmetic for multivariate Bernstein-form polynomials,” Computer-Aided Design, vol. 32, no. 11, pp. 681–689, 2000.
S. N. Bernstein, “D´ emonstration du th´ eor` eme de Weierstrass fond´ ee sur le calcul des proba-bilit´ es,” Communications of the Kharkov Mathematical Society, vol. 13, pp. 1–2, 1912.
K. M. Levasseur, “A probabilistic proof of the Weierstrass approximation theorem,” The Amer-ican Mathematical Monthly, vol. 91, no. 4, pp. 249–250, 1984.
R. Farouki and V. Rajan, “On the numerical condition of polynomials in Bernstein form,” Computer Aided Geometric Design, vol. 4, no. 3, pp. 191–216, 1987.
B. Gaines, “Stochastic computing systems,” in Advances in Information Systems Science.
Plenum, 1969, vol. 2, ch. 2, pp. 37–172.
B. Brown and H. Card, “Stochastic neural computation I: Computational elements,” IEEE Transactions on Computers, vol. 50, no. 9, pp. 891–905, 2001.
W. Qian and M. D. Riedel, “The synthesis of robust polynomial arithmetic with stochastic logic,” in Design Automation Conference, 2008, pp. 648–653.
——, “The synthesis of stochastic logic,” Communications of the ACM (to appear), 2010.
19 |
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Nutritional rickets: a review of disease burden, causes, diagnosis, prevention and treatment
6 November 2019
| Publication
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Overview
This document aims to provide a literature review of nutritional rickets among infants, children and adolescents. It is intended to provide stakeholders with a summary of the aspects surrounding rickets in public health, including the burden of rickets and its causes, diagnosis, prevention and treatment.
This document is not a World Health Organization (WHO) guideline. It is a literature review that also includes the history of rickets epidemiology, the pathophysiology of the condition, and issues related to its diagnosis and consequences
As most cases of nutritional rickets are caused by low vitamin D intake and sun exposure and/or low calcium intake, the document focuses on nutritional rickets and discusses the physiology, functions and epidemiology of vitamin D and calcium deficiency and food sources of these nutrients. The current WHO recommendations for calcium and vitamin D in different populations and settings are discussed. This publication supports the Comprehensive implementation plan on maternal, infant and young child nutrition, calling for an update of the evidence for nutrition actions, in line with The global strategy for women’s, children’s and adolescents’ health (2016–2030). The document provides a review of nutritional rickets in infants, children and adolescents, using the approach suggested by The WHO strategy on research for health. Applying this strategy, the document covers the following areas:
overview of the history and epidemiology of rickets;
the magnitude and distribution of nutritional rickets in the population, especially in infants, children and adolescents;
the causes or determinants of rickets, whether they are biological, behavioural, social or environmental factors;
potential interventions to prevent or mitigate nutritional rickets in infants, children and adolescents;
implementation or delivery of solutions through nutritional policies and programmes;
evaluation of the actions for prevention or treatment of nutritional rickets; and
current research gaps.
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ISBN: 978-92-4-151658-7 |
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17353 | https://en.wikipedia.org/wiki/Electromagnetic_induction | Jump to content
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Contents
(Top)
1 History
2 Theory
2.1 Faraday's law of induction and Lenz's law
2.2 Maxwell–Faraday equation
2.3 Faraday's law and relativity
3 Applications
3.1 Electrical generator
3.2 Electrical transformer
3.2.1 Current clamp
3.3 Magnetic flow meter
4 Eddy currents
4.1 Electromagnet laminations
4.2 Parasitic induction within conductors
5 See also
6 References
6.1 Notes
6.2 References
7 Further reading
8 External links
Electromagnetic induction
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From Wikipedia, the free encyclopedia
Production of voltage by a varying magnetic field
Not to be confused with Magnetic inductance.
Electromagnetic or magnetic induction is the production of an electromotive force (emf) across an electrical conductor in a changing magnetic field.
Michael Faraday is generally credited with the discovery of induction in 1831, and James Clerk Maxwell mathematically described it as Faraday's law of induction. Lenz's law describes the direction of the induced field. Faraday's law was later generalized to become the Maxwell–Faraday equation, one of the four Maxwell equations in his theory of electromagnetism.
Electromagnetic induction has found many applications, including electrical components such as inductors and transformers, and devices such as electric motors and generators.
History
Electromagnetic induction was discovered by Michael Faraday, published in 1831. It was discovered independently by Joseph Henry in 1832.
In Faraday's first experimental demonstration, on August 29, 1831, he wrapped two wires around opposite sides of an iron ring or "torus" (an arrangement similar to a modern toroidal transformer).[citation needed] Based on his understanding of electromagnets, he expected that, when current started to flow in one wire, a sort of wave would travel through the ring and cause some electrical effect on the opposite side. He plugged one wire into a galvanometer, and watched it as he connected the other wire to a battery. He saw a transient current, which he called a "wave of electricity", when he connected the wire to the battery and another when he disconnected it. This induction was due to the change in magnetic flux that occurred when the battery was connected and disconnected. Within two months, Faraday found several other manifestations of electromagnetic induction. For example, he saw transient currents when he quickly slid a bar magnet in and out of a coil of wires, and he generated a steady (DC) current by rotating a copper disk near the bar magnet with a sliding electrical lead ("Faraday's disk").
Faraday explained electromagnetic induction using a concept he called lines of force. However, scientists at the time widely rejected his theoretical ideas, mainly because they were not formulated mathematically. An exception was James Clerk Maxwell, who used Faraday's ideas as the basis of his quantitative electromagnetic theory. In Maxwell's model, the time varying aspect of electromagnetic induction is expressed as a differential equation, which Oliver Heaviside referred to as Faraday's law even though it is slightly different from Faraday's original formulation and does not describe motional emf. Heaviside's version (see Maxwell–Faraday equation below) is the form recognized today in the group of equations known as Maxwell's equations.
In 1834 Heinrich Lenz formulated the law named after him to describe the "flux through the circuit". Lenz's law gives the direction of the induced emf and current resulting from electromagnetic induction.
Theory
Faraday's law of induction and Lenz's law
Main article: Faraday's law of induction
Faraday's law of induction makes use of the magnetic flux ΦB through a region of space enclosed by a wire loop. The magnetic flux is defined by a surface integral: where dA is an element of the surface Σ enclosed by the wire loop, B is the magnetic field. The dot product B·dA corresponds to an infinitesimal amount of magnetic flux. In more visual terms, the magnetic flux through the wire loop is proportional to the number of magnetic field lines that pass through the loop.
When the flux through the surface changes, Faraday's law of induction says that the wire loop acquires an electromotive force (emf).[note 1] The most widespread version of this law states that the induced electromotive force in any closed circuit is equal to the rate of change of the magnetic flux enclosed by the circuit: where is the emf and ΦB is the magnetic flux. The direction of the electromotive force is given by Lenz's law which states that an induced current will flow in the direction that will oppose the change which produced it. This is due to the negative sign in the previous equation. To increase the generated emf, a common approach is to exploit flux linkage by creating a tightly wound coil of wire, composed of N identical turns, each with the same magnetic flux going through them. The resulting emf is then N times that of one single wire.
Generating an emf through a variation of the magnetic flux through the surface of a wire loop can be achieved in several ways:
the magnetic field B changes (e.g. an alternating magnetic field, or moving a wire loop towards a bar magnet where the B field is stronger),
the wire loop is deformed and the surface Σ changes,
the orientation of the surface dA changes (e.g. spinning a wire loop into a fixed magnetic field),
any combination of the above
Maxwell–Faraday equation
See also: Faraday's law of induction § Maxwell–Faraday equation
In general, the relation between the emf in a wire loop encircling a surface Σ, and the electric field E in the wire is given by where dℓ is an element of contour of the surface Σ, combining this with the definition of flux we can write the integral form of the Maxwell–Faraday equation
It is one of the four Maxwell's equations, and therefore plays a fundamental role in the theory of classical electromagnetism.
Faraday's law and relativity
Faraday's law describes two different phenomena: the motional emf generated by a magnetic force on a moving wire (see Lorentz force), and the transformer emf that is generated by an electric force due to a changing magnetic field (due to the differential form of the Maxwell–Faraday equation). James Clerk Maxwell drew attention to the separate physical phenomena in 1861. This is believed to be a unique example in physics of where such a fundamental law is invoked to explain two such different phenomena.
Albert Einstein noticed that the two situations both corresponded to a relative movement between a conductor and a magnet, and the outcome was unaffected by which one was moving. This was one of the principal paths that led him to develop special relativity.
Applications
The principles of electromagnetic induction are applied in many devices and systems, including:
Current clamp
Electric generators
Electromagnetic forming
Graphics tablet
Hall effect sensors
Induction cooking
Induction motors
Induction sealing
Induction welding
Inductive charging
Inductors
Magnetic flow meters
Mechanically powered flashlight
Near-field communications
Pickups
Rowland ring
Transcranial magnetic stimulation
Transformers
Wireless energy transfer
Electrical generator
Main article: Electric generator
The emf generated by Faraday's law of induction due to relative movement of a circuit and a magnetic field is the phenomenon underlying electrical generators. When a permanent magnet is moved relative to a conductor, or vice versa, an electromotive force is created. If the wire is connected through an electrical load, current will flow, and thus electrical energy is generated, converting the mechanical energy of motion to electrical energy. For example, the drum generator is based upon the figure to the bottom-right. A different implementation of this idea is the Faraday's disc, shown in simplified form on the right.
In the Faraday's disc example, the disc is rotated in a uniform magnetic field perpendicular to the disc, causing a current to flow in the radial arm due to the Lorentz force. Mechanical work is necessary to drive this current. When the generated current flows through the conducting rim, a magnetic field is generated by this current through Ampère's circuital law (labelled "induced B" in the figure). The rim thus becomes an electromagnet that resists rotation of the disc (an example of Lenz's law). On the far side of the figure, the return current flows from the rotating arm through the far side of the rim to the bottom brush. The B-field induced by this return current opposes the applied B-field, tending to decrease the flux through that side of the circuit, opposing the increase in flux due to rotation. On the near side of the figure, the return current flows from the rotating arm through the near side of the rim to the bottom brush. The induced B-field increases the flux on this side of the circuit, opposing the decrease in flux due to r the rotation. The energy required to keep the disc moving, despite this reactive force, is exactly equal to the electrical energy generated (plus energy wasted due to friction, Joule heating, and other inefficiencies). This behavior is common to all generators converting mechanical energy to electrical energy.
Electrical transformer
Main article: Transformer
When the electric current in a loop of wire changes, the changing current creates a changing magnetic field. A second wire in reach of this magnetic field will experience this change in magnetic field as a change in its coupled magnetic flux, . Therefore, an electromotive force is set up in the second loop called the induced emf or transformer emf. If the two ends of this loop are connected through an electrical load, current will flow.
Current clamp
Main article: Current clamp
A current clamp is a type of transformer with a split core which can be spread apart and clipped onto a wire or coil to either measure the current in it or, in reverse, to induce a voltage. Unlike conventional instruments the clamp does not make electrical contact with the conductor or require it to be disconnected during attachment of the clamp.
Magnetic flow meter
Main article: Magnetic flow meter
Faraday's law is used for measuring the flow of electrically conductive liquids and slurries. Such instruments are called magnetic flow meters. The induced voltage ε generated in the magnetic field B due to a conductive liquid moving at velocity v is thus given by:
where ℓ is the distance between electrodes in the magnetic flow meter.
Eddy currents
Main article: Eddy current
Electrical conductors moving through a steady magnetic field, or stationary conductors within a changing magnetic field, will have circular currents induced within them by induction, called eddy currents. Eddy currents flow in closed loops in planes perpendicular to the magnetic field. They have useful applications in eddy current brakes and induction heating systems. However eddy currents induced in the metal magnetic cores of transformers and AC motors and generators are undesirable since they dissipate energy (called core losses) as heat in the resistance of the metal. Cores for these devices use a number of methods to reduce eddy currents:
Cores of low frequency alternating current electromagnets and transformers, instead of being solid metal, are often made of stacks of metal sheets, called laminations, separated by nonconductive coatings. These thin plates reduce the undesirable parasitic eddy currents, as described below.
Inductors and transformers used at higher frequencies often have magnetic cores made of nonconductive magnetic materials such as ferrite or iron powder held together with a resin binder.
Electromagnet laminations
Eddy currents occur when a solid metallic mass is rotated in a magnetic field, because the outer portion of the metal cuts more magnetic lines of force than the inner portion; hence the induced electromotive force is not uniform; this tends to cause electric currents between the points of greatest and least potential. Eddy currents consume a considerable amount of energy and often cause a harmful rise in temperature.
Only five laminations or plates are shown in this example, so as to show the subdivision of the eddy currents. In practical use, the number of laminations or punchings ranges from 40 to 66 per inch (16 to 26 per centimetre), and brings the eddy current loss down to about one percent. While the plates can be separated by insulation, the voltage is so low that the natural rust/oxide coating of the plates is enough to prevent current flow across the laminations.
This is a rotor approximately 20 mm in diameter from a DC motor used in a CD player. Note the laminations of the electromagnet pole pieces, used to limit parasitic inductive losses.
Parasitic induction within conductors
In this illustration, a solid copper bar conductor on a rotating armature is just passing under the tip of the pole piece N of the field magnet. Note the uneven distribution of the lines of force across the copper bar. The magnetic field is more concentrated and thus stronger on the left edge of the copper bar (a,b) while the field is weaker on the right edge (c,d). Since the two edges of the bar move with the same velocity, this difference in field strength across the bar creates whorls or current eddies within the copper bar.
High current power-frequency devices, such as electric motors, generators and transformers, use multiple small conductors in parallel to break up the eddy flows that can form within large solid conductors. The same principle is applied to transformers used at higher than power frequency, for example, those used in switch-mode power supplies and the intermediate frequency coupling transformers of radio receivers.
See also
| Electromagnetism |
| Electricity Magnetism Optics History Computational Textbooks Phenomena |
| Electrostatics Charge density Conductor Coulomb law Electret Electric charge Electric dipole Electric field Electric flux Electric potential Electrostatic discharge Electrostatic induction Gauss's law Insulator Permittivity Polarization Potential energy Static electricity Triboelectricity |
| Magnetostatics Ampère's law Biot–Savart law Gauss's law for magnetism Magnetic dipole Magnetic field Magnetic flux Magnetic scalar potential Magnetic vector potential Magnetization Permeability Right-hand rule |
| Electrodynamics Bremsstrahlung Cyclotron radiation Displacement current Eddy current Electromagnetic field Electromagnetic induction Electromagnetic pulse Electromagnetic radiation Faraday's law Jefimenko equations Larmor formula Lenz's law Liénard–Wiechert potential London equations Lorentz force Maxwell's equations Maxwell tensor Poynting vector Synchrotron radiation |
| Electrical network Alternating current Capacitance Current density Direct current Electric current Electric power Electrolysis Electromotive force Impedance Inductance Joule heating Kirchhoff's laws Network analysis Ohm's law Parallel circuit Resistance Resonant cavities Series circuit Voltage Watt Waveguides |
| Magnetic circuit AC motor DC motor Electric machine Electric motor Gyrator–capacitor Induction motor Linear motor Magnetomotive force Permeance Reluctance (complex) Reluctance (real) Rotor Stator Transformer |
| Covariant formulation Electromagnetic tensor Electromagnetism and special relativity Four-current Four-potential Mathematical descriptions Maxwell equations in curved spacetime Relativistic electromagnetism Stress–energy tensor |
| Ampère Biot Coulomb Davy Einstein Faraday Fizeau Gauss Heaviside Helmholtz Henry Hertz Hopkinson Jefimenko Joule Kelvin Kirchhoff Larmor Lenz Liénard Lorentz Maxwell Neumann Ohm Ørsted Poisson Poynting Ritchie Savart Singer Steinmetz Tesla Thomson Volta Weber Wiechert |
| v t e |
Alternator – Device converting mechanical into electrical energy
Crosstalk – Signals in one channel affecting another
Faraday paradox – Apparent paradox with Faraday's law of induction
Fleming's right-hand rule – Mnemonic for the direction of induced current in a moving magnetic field
Hall effect – Electromagnetic effect in physics
Inductance
Moving magnet and conductor problem
^ The EMF is the voltage that would be measured by cutting the wire to create an open circuit, and attaching a voltmeter to the leads. Mathematically, is defined as the energy available from a unit charge that has traveled once around the wire loop.
References
^ Poyser, A. W. (1892). Magnetism and Electricity: A Manual for Students in Advanced Classes. London and New York: Longmans, Green, & Co. p. 285.
^ a b Giancoli, Douglas C. (1998). Physics: Principles with Applications (5th ed.). pp. 623–624.
^ Ulaby, Fawwaz (2007). Fundamentals of applied electromagnetics (5th ed.). Pearson: Prentice Hall. p. 255. ISBN 978-0-13-241326-8.{{cite book}}: CS1 maint: publisher location (link)
^ "Joseph Henry". Distinguished Members Gallery, National Academy of Sciences. Archived from the original on 2013-12-13. Retrieved 2006-11-30.
^ Errede, Steven (2007). "A Brief History of The Development of Classical Electrodynamics" (PDF).
^ "Electromagnetism". Smithsonian Institution Archives.
^ Faraday, Michael (1831-08-29). "Faraday's notebooks: Electromagnetic Induction" (PDF). The Royal Institution of Great Britain. Archived from the original (PDF) on 2021-08-30.
^ Michael Faraday, by L. Pearce Williams, pp. 182–183
^ Michael Faraday, by L. Pearce Williams, pp. 191–195
^ a b Michael Faraday, by L. Pearce Williams, p. 510
^ Maxwell, James Clerk (1904), A Treatise on Electricity and Magnetism, Vol. II, Third Edition. Oxford University Press, pp. 178–179 and 189.
^ "Archives Biographies: Michael Faraday", The Institution of Engineering and Technology.
^ Good, R. H. (1999). Classical Electromagnetism. Saunders College Publishing. p. 107. ISBN 0-03-022353-9.
^ Feynman, R. P.; Leighton, R. B.; Sands, M. L. (2006). The Feynman Lectures on Physics, Volume 2. Pearson/Addison-Wesley. p. 17-2. ISBN 0-8053-9049-9.
^ Griffiths, D. J. (1999). Introduction to Electrodynamics (3rd ed.). Prentice Hall. pp. 301–303. ISBN 0-13-805326-X.
^ Tipler, P. A.; Mosca, G. (2003). Physics for Scientists and Engineers (5th ed.). W.H. Freeman. p. 795. ISBN 978-0716708100.
^ Jordan, E.; Balmain, K. G. (1968). Electromagnetic Waves and Radiating Systems (2nd ed.). Prentice-Hall. p. 100. ISBN 978-0132499958.
^ Hayt, W. (1989). Engineering Electromagnetics (5th ed.). McGraw-Hill. p. 312. ISBN 0-07-027406-1.
^ Schmitt, R. (2002). Electromagnetics Explained. Newnes. p. 75. ISBN 978-0750674034.
^ Whelan, P. M.; Hodgeson, M. J. (1978). Essential Principles of Physics (2nd ed.). John Murray. ISBN 0-7195-3382-1.
^ Nave, C. R. "Faraday's Law". HyperPhysics. Georgia State University. Retrieved 2011-08-29.
^ Maxwell, J. C. (1861). "On physical lines of force". Philosophical Magazine. 90 (139): 11–23. doi:10.1080/14786446108643033.
^ Griffiths, D. J. (1999). Introduction to Electrodynamics (3rd ed.). Prentice Hall. pp. 301–303. ISBN 0-13-805326-X. Note that the law relating flux to EMF, which this article calls "Faraday's law", is referred to by Griffiths as the "universal flux rule". He uses the term "Faraday's law" to refer to what this article calls the "Maxwell–Faraday equation".
^ "The flux rule" is the terminology that Feynman uses to refer to the law relating magnetic flux to EMF. Feynman, R. P.; Leighton, R. B.; Sands, M. L. (2006). The Feynman Lectures on Physics, Volume II. Pearson/Addison-Wesley. p. 17-2. ISBN 0-8053-9049-9.[permanent dead link]
^ Einstein, A. (1905). "Zur Elektrodynamik bewegter Körper" (PDF). Annalen der Physik. 17 (10): 891–921. Bibcode:1905AnP...322..891E. doi:10.1002/andp.19053221004.
: Translated in Einstein, A. (1923). "On the Electrodynamics of Moving Bodies" (PDF). The Principle of Relativity. Jeffery, G.B.; Perret, W. (transl.). London: Methuen and Company.
26. ^ a b c Images and reference text are from the public domain book: Hawkins Electrical Guide, Volume 1, Chapter 19: Theory of the Armature, pp. 270–273, Copyright 1917 by Theo. Audel & Co., Printed in the United States
Further reading
Maxwell, James Clerk (1881), A treatise on electricity and magnetism, Vol. II, Chapter III, §530, p. 178. Oxford, UK: Clarendon Press. ISBN 0-486-60637-6.
External links
Media related to Electromagnetic induction at Wikimedia Commons
The Laws of Induction - The Feynman Lectures on Physics
A free java simulation on motional EMF
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17354 | https://engoo.com/app/words/word/shed/zga0ELstQmCjlQAAAAiSXg | shed (【Verb】to lose a covering, such as leaves, hair, or skin ) Meaning, Usage, and Readings | Engoo Words
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17355 | https://www.youtube.com/watch?v=awNCX-yK8QU | Basic Absolute Value Function Translations: y=|x-h|+k
Mathispower4u
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This video introduces translations of the absolute value function
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welcome to a video on graphing translations of the basic absolute value function in this video the equations will be in standard form with a equal to one standard form of an absolute value function is y equals or f of x equals a times the absolute value of the quantity x minus h plus k but again for this video a is equal to one and therefore the equations will be in the form of y equals the absolute value of the quantity x minus h plus k before we discuss how h and k affect the graph there are a few things we should notice about the graph of the basic or parent absolute value function shown here on the left remember if we're ever not sure how to make a graph we can always make a table of values to graph the function notice how the graph of the absolute value function is a v shape in this case it opens up the lowest point on the graph or the point where the graph changes direction is this point here called the vertex notice how the vertex is at the origin which has an ordered pair zero comma zero notice when x is greater than zero or to the right of the vertex the graph is linear with the slope of positive one meaning if we start at any point on the graph on the right side and go up one and right one we can determine another point on the graph and then when x is less than zero or to the left of the vertex the slope is negative one now going back to the form of the equation y equals the absolute value of the quantity x minus h plus k the value of h will shift the graph left or right and the value of k will shift the graph up or down when h is positive or greater than zero we would have the absolute value of x minus h so when we have subtraction here h is positive and the graph is shifted right h units and when h is less than zero or negative we would have the absolute value of the quantity x minus negative h which simplifies to x plus h so when we have addition here h is negative and the graph is shifted left the absolute value of h units and now for k when k is positive or greater than zero the graph is shifted up k units and when k is less than zero or negative the graph is shifted down the absolute value of k units and then finally the vertex is the ordered pair h comma k let's look at an animation to better understand how the values of h and k affect the graph of the basic absolute value function let's first see how h affects the graph of the function notice when h is positive the graph is shifted right h units and when h is negative the graph is shifted left the absolute value of h units so if we stop here for a moment where h is negative six let's determine the equation of this graph well if h is negative six and we know a is one and k is zero this would give us the equation y equals the absolute value of the quantity x minus negative six simplifying we have the absolute value of the quantity x plus six so again when we have addition inside the absolute value the graph is shifted left in this case six units and this should make sense because notice how when x is negative six the ordered pair here at the vertex is negative six comma zero and notice when x is negative six we do have a zero inside the absolute value the absolute value of zero is zero and therefore the output or y value is zero when x is negative six and when h is positive let's say when h is positive four we would substitute positive four in for h which gives us the equation y equals the absolute value of the quantity x minus four again when we have subtraction inside the absolute value the graph is shifted right in this case four units and now let's see how k affects the graph so we'll set h back to zero and now change the value of k notice when k is positive the graph is shifted up k units and when k is negative the graph is shifted down the absolute value of k units so here where k is equal to negative four because we have addition here we normally don't write plus negative four we normally just write minus four and we know a is one and h is zero and therefore the equation of this purple graph is y equals the absolute value of x and then minus four and let's take a look at an equation when k is positive let's change k to positive two notice how the graph is shifted up two units and because k is positive two the equation of this purple graph or this graph here is y equals the absolute value of x plus two let's take a look at a few more examples let's say we're asked to graph y equals or g of x equals the absolute value of x plus three again because we have addition here h is going to be negative three we need to be thinking that we can write x plus three as x minus negative three to determine the value of h again because in standard form we do have subtraction inside the absolute value so because h is negative three the graph is shifted left the absolute value of negative three or three units and so notice how the graph would be this orange graph here where we have the parent graph or basic absolute value function graphed here in blue we shift this left three units to form the graph of y equals the absolute value of the quantity x plus three and then to graph y equals the absolute value of x minus one because we have subtraction here the graph is shifted right one unit because h is positive one so again the blue graph is the basic absolute value function we would take this graph and shift it right one unit to graph the given function and now let's look at two more examples involving the value of k if we want to graph y equals the absolute value of x plus four k is positive four which means which means we shift the basic absolute value function up four units so in blue we have the basic absolute value function because k is four we shift the graph up four units to graph the given function and if we have y equals the absolute value of x minus two we shift the graph of the parent function or the basic absolute value function down two units to graph the given function so this is an overview of how h and k affect the graph of the basic absolute value function in the next video we will graph absolute value functions that have both the value of h and k in the equation i hope you found this helpful |
17356 | https://en.wikipedia.org/wiki/Percentage_point | Jump to content
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1 Differences between percentages and percentage points
2 Related units
3 See also
4 References
Percentage point
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From Wikipedia, the free encyclopedia
Unit for the arithmetic difference of two percentages
"Percent point" redirects here; not to be confused with Percent point function.
A percentage point or percent point is the description for the arithmetic difference between two percentages. For example, moving up from 40 percent to 44 percent is an increase of 4 percentage points (although it is a 10-percent increase in the quantity being measured, if the total amount remains the same). In written text, the unit (the percentage point) is usually either written out, or abbreviated as pp, p.p., or %pt. to avoid confusion with percentage increase or decrease in the actual quantity. After the first occurrence, some writers abbreviate by using just "point" or "points".
Differences between percentages and percentage points
[edit]
Consider the following hypothetical example: In 1980, 50 percent of the population smoked, and in 1990 only 40 percent of the population smoked. One can thus say that from 1980 to 1990, the prevalence of smoking decreased by 10 percentage points (or by 10 percent of the population) or by 20 percent when talking about smokers only – percentages indicate proportionate part of a total.
Percentage-point differences are one way to express a risk or probability. Consider a drug that cures a given disease in 70 percent of all cases, while without the drug, the disease heals spontaneously in only 50 percent of cases. The drug reduces absolute risk by 20 percentage points. Alternatives may be more meaningful to consumers of statistics, such as the reciprocal, also known as the number needed to treat (NNT). In this case, the reciprocal transform of the percentage-point difference would be 1/(20pp) = 1/0.20 = 5. Thus if 5 patients are treated with the drug, one could expect to cure one more patient than would have gotten well without receiving the treatment.
For measurements involving percentages as a unit, such as, growth, yield, or ejection fraction, statistical deviations and related descriptive statistics, including the standard deviation and root-mean-square error, the result should be expressed in units of percentage points instead of percentage. [citation needed] Mistakenly using percentage as the unit for the standard deviation is confusing, since percentage is also used as a unit for the relative standard deviation, i.e. standard deviation divided by average value (coefficient of variation).
Related units
[edit]
Percentage (%) 1 part in 100
Per mille (‰) 1 part in 1,000
Permyriad (‱) 1 part in 10,000
Basis point (bp) difference of 1 part in 10,000
Per cent mille (pcm) 1 part in 100,000
Parts-per notation parts per million (ppm) etc.
See also
[edit]
Baker percentage – Mathematical notation in cooking
Parts-per notation – Set of units to describe small values
Percent point function – Statistical function that defines the quantiles of a probability distribution
Per-unit system – In power systems, expression of system quantities as fractions
Relative change, also known as relative difference – Comparisons in quantitative sciences
References
[edit]
^ Brechner, Robert (2008). Contemporary Mathematics for Business and Consumers, Brief Edition. Cengage Learning. p. 190. ISBN 9781111805500. Archived from the original on 18 May 2015. Retrieved 7 May 2015.
^ Wickham, Kathleen (2003). Math Tools for Journalists. Cengage Learning. p. 30. ISBN 9780972993746. Archived from the original on 18 May 2015. Retrieved 7 May 2015.
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17357 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_3?srsltid=AfmBOooQqRJHjw4FKl2maAcweFySB4PRj2lPqeawQrItKDDNzfEPfTr7 | Art of Problem Solving
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1983 IMO Problems/Problem 3
Problem
Let , and be positive integers, no two of which have a common divisor greater than . Show that is the largest integer which cannot be expressed in the form , where , and are non-negative integers.
Solution 1
First off, I prove is un-achievable. Also, assume WLOG .
Assume , then take the equation to give us . By CRT and , we take this equation mod a and mod b to give us , and (and using all numbers are relatively prime pairwise). Substituting this back into the original equation gives us , contradiction (this is where the part comes in).
Now, let , and if a solution exists where then we are done because we can add to always.
Consider the set , where and where .
Therefore, we get . This is the same as after doing some simplifying. By CRT, this must hold mod a and mod b and because . Mod gives us , for which a value of obviously exists mod , which can be chosen from the set of values we have assigned for . Similar method shows a value of exists mod from the set we have given to .
Now, we already know that . Also, the LHS of the equation is at least , therefore and we are done.
This solution was posted and copyrighted by binomial-theorem. The original thread for this problem can be found here:
Solution 2
First we will prove is unattainable, as such: Suppose . Then, taking this mod , we have that , so , and . Similarly, , and , so , , and . Thus, , so , which is a contradiction.
Now we will prove all is attainable, as such: consider the integer such that and . Rearranging the equation yields , so set . We see that , so is a positive integer (obviously . Now, note that since , we have that , so . Thus, so , so by Chicken Mc-Nugget theorem, there exist that satisfy the equation and are now done.
This solution was posted and copyrighted by Stormersyle. The original thread for this problem can be found here:
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17358 | https://scicomp.stackexchange.com/questions/962/linear-programming-feasibility-problem-with-strict-positivity-constraints | optimization - Linear programming feasibility problem with strict positivity constraints - Computational Science Stack Exchange
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Linear programming feasibility problem with strict positivity constraints
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There is a system of linear constraints A x≤b A x≤b . I wish to find a strictly positive vector x>0 x>0 that satisfies these constraints. That means, x i>0 x i>0 is required for every component x i x i of x x. How can I use an LP solver to find such a strictly positive vector x x (or confirm that no x x exists)? I cannot simply introduce another system of constraints x i>0 x i>0, because equality must always be permitted in an LP—but I can use the LP solver several times, with changing objective functions. I think I should use the slack variable method, but I don't know how.
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edited Jan 24, 2012 at 11:54
David Ketcheson
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You can circumvent the problem of choosing a small ϵ>0 ϵ>0 by being a bit more ambitious: Try to find x x such that A x≤b A x≤b and that the smallest entry in x x is largest possible.
To that end, introduce a new variable
y=[x ϵ]∈R n+1 y=[x ϵ]∈R n+1
(if x x was in R n R n) and solve the following problem by an LP-solver
max y[0…0 1]⋅y s.t.[A 0]y≤b and 0≤⎡⎣⎢⎢⎢⎢⎢1 0 0 0 1⋱⋯⋯⋯0 0⋮1−1−1−1⎤⎦⎥⎥⎥⎥⎥y.max y[0…0 1]⋅y s.t.[A 0]y≤b and 0≤[1 0⋯0−1 0 1⋯0−1⋱⋮0⋯1−1]y.
This is a reformulation of the following problem:
max ϵ s.t A x≤b and x≥ϵ 1.max ϵ s.t A x≤b and x≥ϵ 1.
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edited Mar 9, 2019 at 20:50
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well done, this is equivalent to a trick a coauthor and I just used in a recent paper, and definitely superior to the approach I suggested.Aron Ahmadia –Aron Ahmadia 2012-01-24 12:26:15 +00:00 Commented Jan 24, 2012 at 12:26
Agreed. Well played, sir.Geoff Oxberry –Geoff Oxberry 2012-01-24 13:00:41 +00:00 Commented Jan 24, 2012 at 13:00
The reformulated problem may have an unbounded objective in cases where the answer to the original problem is trivial. For example, if the system of constraints is just x≥−1 x≥−1. That is fine as long as you check for feasible, optimal or unbounded in the return status of your lp solver, or explicitly bound the ϵ ϵ.David Nehme –David Nehme 2012-02-17 03:50:46 +00:00 Commented Feb 17, 2012 at 3:50
@DavidNehme: One can add the constraint y n+1≤1 y n+1≤1 to get a bounded objective.Arnold Neumaier –Arnold Neumaier 2012-06-22 19:51:48 +00:00 Commented Jun 22, 2012 at 19:51
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For an LP feasibility problem, I wouldn't use standard simplex. Standard primal (or dual) simplex algorithms will only visit the vertices of the feasible set of the primal (or dual) problems.
Let the feasible set of the problem you actually want to solve be F={x:A x≤b,x>0}F={x:A x≤b,x>0}, and suppose instead you were to solve the problem (F ε F ε):
s.t.min x 0 A x≤b x≥ε⋅1.min x 0 s.t.A x≤b x≥ε⋅1.
The closest approximant of the problem you want to solve is F 0 F 0, which admits slightly too many points. The problem is that the boundary of the positive orthant (i.e., the set B={x:x≥0,∃i:x i=0}B={x:x≥0,∃i:x i=0} could make up part of the boundary of the feasible set of F 0 F 0. We'd like to exclude those points. One way of doing that is to do what Aron suggested, which is to set ε ε to some small positive value, and then use any standard LP algorithm. This strategy is a good one, and will probably work in a wide variety of situations. However, it will fail if F ε F ε is infeasible. We know that F 0⊂F⊂F ε F 0⊂F⊂F ε for all ε>0 ε>0 (to abuse notation and refer to a feasible set by its corresponding problem), and it's possible that even if you pick small positive values of ε ε, the LP solver will indicate that your LP is infeasible.
For an LP solver, I'd use any interior point algorithm for LPs that starts with a feasible point and stays feasible, which is another way to exclude points in B B. You needn't have to supply a feasible point to these algorithms; standard solvers will do it for you. Methods like affine scaling, potential reduction, and barrier methods set up auxiliary LPs that will find feasible solutions, and the iterates for these algorithms traverse the interior of the feasible region. You only need to locate one point in your feasible region, so as long as the auxiliary problems used by the LP solvers locate a feasible point for your problem, and that feasible point is strictly positive, you should be all right. If solving F ε F ε fails for small positive values of ε ε, you might still be able to use these methods to locate a strictly positive feasible point within F 0 F 0.
Don't use simplex, though, because it will only explore the vertices of F ε F ε, which is exactly what you want to avoid doing.
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answered Jan 24, 2012 at 10:37
Geoff OxberryGeoff Oxberry
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Feasibility problems are a slightly trickier game than general linear problems, which you have noted. If you are solving approximately (by using a floating-point representation of the system of equations and constraints), it is legitimate to require x i>=ϵ x i>=ϵ, where ϵ ϵ is some very small numerical value, big enough to assure that x i x i actually lives in R+ℜ+, but small enough that a solution on the boundary is not considered.
You might have to adjust ϵ ϵ, and your solution will be qualified to "within a factor of ϵ ϵ", but this is sufficient for many situations.
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answered Jan 24, 2012 at 9:22
Aron AhmadiaAron Ahmadia
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The answer given by aeismail is to be read carefully, regard the lp
max(x 1+x 2)max(x 1+x 2)
s.t.
x 1+x 2≤1 x 1+x 2≤1
x 1,x 2≥0 x 1,x 2≥0
It has solutions (1,0)(1,0) and (0,1)(0,1) as well as others (degenerated). The generality of the question implys that these cases need to be treated as well.
Since you are able to choose you objective function, you could try to modify it iteratively. E.g. Start with all coefficients for all variables equal to one, check wether you gain an approprate solution. If one variable is zero, rise it's coefficient and start again...
Though I can not give a mathematical prove that this works (or a well defined procedure how to modify the objective function). I hope this helps :)
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edited Jan 24, 2012 at 18:46
Dan
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4
However, if you have a large number of degenerate solutions, how would you deal with this numerically? Won't pretty much any numerical solver throw up a warning (or worse) about solving this problem?aeismail –aeismail 2012-01-24 09:42:15 +00:00 Commented Jan 24, 2012 at 9:42
No, they won't; they'll just return the first optimal solution encountered. The way that you would continue to generate solutions is to add cutting planes (or other constraints) that exclude previously calculated optimal solutions. In this case, adding such cutting planes would enable you to return a discrete approximation of the infinite set of optimal solutions.Geoff Oxberry –Geoff Oxberry 2012-01-24 10:40:14 +00:00 Commented Jan 24, 2012 at 10:40
I would view that as a strange programming decision; why wouldn't you want to tell the user that the objective function was doing something weird in the neighborhood of the reported solution? For a nonlinear solver, I could see there being an issue with figuring out what's going on; but shouldn't that be easier to tell with a linear system?aeismail –aeismail 2012-01-24 13:19:54 +00:00 Commented Jan 24, 2012 at 13:19
I'd have to think about how one would detect degeneracy by actually constructing problems, but typically, users want an optimal solution, so the most important information for an LP is to return if the solution is optimal, feasible (but not optimal), infeasible, or unbounded. (These statuses are, in fact, what a solver like CPLEX would return.) Degeneracy is primarily a theoretical issue; the only reason it would be discussed in a numerical context is either in algorithm design or in practice, to note that degeneracy typically slows down a solver.Geoff Oxberry –Geoff Oxberry 2012-01-24 20:10:54 +00:00 Commented Jan 24, 2012 at 20:10
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17359 | https://chemistry.coach/general-chemistry-2/colligative-properties-of-solutions | Prev chapter
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Colligative Properties of Solutions | General Chemistry 2
Colligative properties of solutions are studied in this chapter: solution and solute, types of solutions, colloids, solution process and enthalpy of solution, concentration units, factors affecting solubility and Henry’s law, vapor-pressure lowering and Raoult’s law, boiling-point elevation, freezing-point depression, osmotic pressure, electrolyte solutions, van't Hoff factor.
Types of Solutions
The Solution Process
Concentration Units
Factors Affecting Solubility
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
Electrolyte Solutions
Types of Solutions
Solution and solute:
A solution is a homogeneous mixture where one or more substances (solutes) are uniformly dispersed in another substance (solvent). The solvent typically dictates the phase of the solution, and the particles in a solution are very small, typically less than 1 nanometer in size.
A solute is the substance that is dissolved in the solvent. The amount of solute relative to the solvent determines the concentration of the solution.
In a saltwater solution:
Salt (sodium chloride, NaCl) is the solute,
Water (H2O) is the solvent.
The salt dissolves completely in the water, creating a homogeneous mixture where the salt particles are uniformly distributed at the molecular level.
Saturated, unsaturated, and supersaturated solutions:
Saturated solution: A solution that contains the maximum amount of solute that can dissolve at a given temperature. Any additional solute added will not dissolve and will remain undissolved in the solution.
Unsaturated solution: A solution that contains less solute than the maximum amount that can be dissolved at a given temperature. More solute can still be dissolved in the solvent.
Supersaturated solution: A solution that contains more solute than the solvent can theoretically dissolve at a given temperature. This is an unstable state, and the excess solute can precipitate out of the solution if disturbed.
Colloids:
A colloid is a mixture of a solvent and suspended particles that are larger than those in a true solution but smaller than those in a suspension. The particles in a colloid range in size from 1 to 100 nanometers, making them larger than simple molecules but small enough to remain in suspension and not settle out.
Hydrophilic colloids: Colloids in which the dispersed particles have an affinity for water. They are usually stable in aqueous solutions.
Hydrophobic colloids: Colloids in which the dispersed particles do not have an affinity for water. They tend to aggregate and separate from the dispersion medium unless stabilized by surfactants.
Hydrophilic colloids: gelatin, starch.
Hydrophobic colloids: oil droplets in water.
Tyndall effect: This is the scattering of light by colloidal particles. It is a characteristic feature of colloids, where a beam of light passing through the colloid becomes visible due to the scattering by the dispersed particles.
The Solution Process
Steps in the solution process:
Step 1 - Solute separation: The first step in the solution process involves breaking the intermolecular forces or ionic bonds within the solute. This step requires energy, known as the lattice energy in ionic compounds or the cohesive energy in molecular solutes. The energy required to separate the solute particles is always endothermic.
Step 2 - Solvent separation: The solvent molecules must also be separated to create space for the solute particles. This step also requires energy and is endothermic. The solvent molecules must overcome their intermolecular forces, such as hydrogen bonding in water, to make room for the solute.
Step 3 - Solvation (or hydration in water): The final step involves the interaction between solute particles and solvent molecules. Solvent molecules surround the solute particles, stabilizing them in the solution. This step releases energy, making it an exothermic process.
Enthalpy of solution:
The overall enthalpy change of the solution process (ΔHsolution) depends on the balance of energy changes in the three steps:
ΔHsolution = ΔH1 + ΔH2 + ΔH3
ΔHsolution = overall enthalpy change of the solution process
ΔH1 = enthalpy change of the solute separation
ΔH2 = enthalpy change of the solvent separation
ΔH3 = enthalpy change of the solvation
If ΔHsolution < 0: the dissolution process is exothermic, and the solution will release heat.
If ΔHsolution > 0: the dissolution process is endothermic, and the solution will absorb heat.
Entropy (S) of the solution process:
The dissolution process often leads to an increase in entropy because the solute particles become more dispersed in the solvent. This increase in disorder usually drives the solution process forward, even if the enthalpy change is positive (endothermic).
Concentration Units
Molarity (M):
Molarity is defined as the number of moles of solute dissolved in one liter of solution:
M =
M = molarity (in mol.L-1)
nsolute = moles of solute (in mol)
Vsolution = volume of solution (in L)
Molality (m):
Molality is defined as the number of moles of solute dissolved in one kilogram of solvent:
m =
m = molality (in mol.kg-1)
nsolute = moles of solute (in mol)
msolution = mass of solvent (in kg)
Mass percent (w/w%):
Mass percent is the mass of the solute divided by the total mass of the solution, multiplied by 100:
w/w% = x 100
w/w% = mass percent
msolute = mass of solute (in kg)
msolution = mass of solution (in kg)
msolution = msolute + msolvent
Volume percent (v/v%):
Volume percent is the volume of the solute divided by the total volume of the solution, multiplied by 100:
% v/v = x 100
v/v% = volume percent
Vsolute = volume of solute (in L)
Vsolution = volume of solution (in L)
Mole fraction (χ):
Mole fraction is the ratio of the number of moles of a component to the total number of moles of all components in the solution:
χA =
χA = mole fraction of component A
nA = moles of component A (in mol)
ntot = total moles of all components (in mol)
Parts per million (ppm) and parts per billion (ppb):
These units are used to express very dilute concentrations of substances. Ppm is the number of parts of solute per million parts of the solution, and ppb is the number of parts of solute per billion parts of the solution.
ppm = x 106
ppm = parts per million
msolute = mass of solute (in kg)
msolution = mass of solution (in kg)
Factors Affecting Solubility
Effect of temperature:
For most solid solutes, solubility increases as the temperature rises. The increased kinetic energy of the molecules at higher temperatures allows more solute particles to overcome the forces holding them in the solid state, thus dissolving in the solvent.
The solubility of sugar in water increases as the temperature is increased. This is why sugar dissolves more quickly in hot water compared to cold water.
The solubility of gases in liquids typically decreases with an increase in temperature. Higher temperatures provide gas molecules with more energy, allowing them to escape from the solvent into the gas phase more easily.
Carbonated beverages, such as soda, lose carbonation more rapidly when they are warm because the solubility of carbon dioxide decreases as the temperature increases.
Effect of pressure:
Pressure significantly affects the solubility of gases in liquids. According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid:
c = kH P
c = molar concentration or solubility (in mol.L-1)
kH = proportionality constant called the Henry's law constant
P = partial pressure of the gas above the liquid (in atm)
The Henry's Law constant varies depending on the gas and the solvent, and it typically decreases with increasing temperature. This constant quantifies how much gas will dissolve in the solvent under a given pressure.
Carbonated beverages are bottled under high pressure to increase the solubility of carbon dioxide in the liquid. When the bottle is opened, the pressure is released, reducing the solubility of CO2, which escapes as bubbles, causing the beverage to go flat.
Vapor-Pressure Lowering
Raoult's law:
Raoult's law states that the vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution. This law is applicable to solutions where the solute does not volatilize and does not react with the solvent.
Psolution = χsolvent Posolvent
Psolution = vapor pressure of the solvent in the solution (in atm)
χsolvent = mole fraction of the solvent in the solution
Posolvent = vapor pressure of the pure solvent (in atm)
Vapor-pressure lowering:
When a non-volatile solute is added to a solvent, the solute particles occupy space at the surface of the liquid, reducing the number of solvent molecules that can escape into the vapor phase. As a result, the vapor pressure of the solvent above the solution is lower than that of the pure solvent.
The vapor-pressure lowering (ΔP) can be calculated as:
ΔP = Posolvent - Psolution
= Posolvent (1 - χsolvent)
ΔP = vapor-pressure lowering (in atm)
Posolvent = vapor pressure of the pure solvent (in atm)
Psolution = vapor pressure of the solvent in the solution (in atm)
χsolvent = mole fraction of the solvent in the solution
Alternatively, since χsolute+ χsolvent = 1, we can express the vapor-pressure lowering as:
ΔP = Posolvent χsolute
ΔP = vapor-pressure lowering (in atm)
Posolvent = vapor pressure of the pure solvent (in atm)
χsolvent = mole fraction of the solute in the solution
Boiling-Point Elevation
Boiling-point elevation:
Boiling-point elevation (ΔTb) is the increase in the boiling point of a solvent upon the addition of a non-volatile solute. It is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent:
ΔTb = Tb - Tob
ΔTb = boiling-point elevation (in oC)
Tb = boiling point of the solution (in oC)
Tob = boiling point of the pure solvent (in oC)
The change in boiling point is directly proportional to the molal concentration of the solute in the solution:
ΔTb = Kb m
ΔTb = boiling-point elevation (in oC)
Kb = molal boiling-point elevation constant (in oC.kg.mol-1)
m = molality of the solute (in mol.kg-1)
Mechanism:
Lowered vapor pressure: The addition of a non-volatile solute decreases the solvent's vapor pressure because fewer solvent molecules can escape into the vapor phase. As a result, a higher temperature is required for the vapor pressure to equal atmospheric pressure, leading to a higher boiling point.
Colligative property: Boiling-point elevation is a colligative property, meaning it depends on the number of solute particles in the solution, not the identity of the solute. This property is particularly useful for determining the molar mass of unknown solutes.
Freezing-Point Depression
Freezing-point depression:
Freezing-point depression (ΔTf) is the decrease in the freezing point of a solvent when a solute is dissolved in it. It is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
ΔTf = Tof - Tf
ΔTf = freezing-point depression (in oC)
Tof = freezing point of the pure solvent (in oC)
Tf = freezing point of the solution (in oC)
This effect is proportional to the number of solute particles present in the solution, regardless of their identity:
ΔTf = - Kf x m
ΔTf = freezing-point depression (in oC)
Kf = molal freezing-point depression constant (in oC.mol.kg-1)
m = solution molality (in mol.kg-1)
Mechanism:
Disruption of freezing: The presence of solute particles interferes with the ability of solvent molecules to arrange themselves into a solid crystalline structure. As a result, the temperature must be lowered further to reach the point where the solvent can solidify.
Colligative property: Freezing-point depression is a colligative property, meaning it depends only on the number of solute particles in the solution and not on their chemical identity.
Osmotic Pressure
Semipermeable membrane:
A semipermeable membrane allows only certain molecules, such as solvent molecules, to pass through it while blocking solute particles. This differential permeability creates a natural movement of solvent from the region of lower solute concentration to the region of higher solute concentration.
Osmosis:
Osmosis is the process by which solvent molecules move across a semipermeable membrane from a dilute solution (or pure solvent) into a more concentrated solution. This movement continues until the concentrations on both sides of the membrane are equal, or until osmotic pressure is applied to stop the flow.
Osmotic pressure:
Osmotic pressure (Π) is the pressure that must be applied to a solution to prevent the inward flow of water (or another solvent) across a semipermeable membrane. It is directly proportional to the concentration of solute particles in the solution:
Π = MR T
Π = osmotic pressure (in atm)
M = molarity of the solution (in mol.L-1)
R = ideal gas constant = 0.0821 (in L.atm.K-1.mol-1)
T = absolute temperature (in K)
Osmotic pressure is a colligative property, meaning it depends on the number of solute particles in the solution, not their identity. The more solute particles present, the higher the osmotic pressure.
Electrolyte Solutions
Electrolytes:
Electrolytes are substances that, when dissolved in water or other solvents, dissociate into ions. These ions are responsible for conducting electricity in the solution. There are 2 types of electrolytes:
Strong electrolytes: Completely dissociate into ions in solution, resulting in high electrical conductivity.
Weak electrolytes: Partially dissociate into ions in solution, resulting in lower electrical conductivity.
Sodium chloride (NaCl), potassium hydroxide (KOH), hydrochloric acid (HCl) are strong electrolytes.
Acetic acid (CH₃COOH), ammonia (NH₃), weak organic acids, and bases are weak electrolytes.
Van 't Hoff factor (i):
The van 't Hoff factor represents the number of particles into which a solute dissociates in solution. It can be expressed as:
i =
i = van 't Hoff factor
nparticles = moles of particles in solution
nsolute = moles of solute dissolved
Thus, i is 1 for all nonelectrolytes, and greater than 1 for strong electrolytes.
i(Sucrose): 1; i(NaCl): 2; i(KBr): 2; i(CaCl2): 3; i(FeCl3): 4.
Colligative properties of electrolyte solutions:
Electrolyte solutions exhibit colligative properties such as boiling-point elevation, freezing-point depression, and osmotic pressure. The presence of more ions (as in strong electrolytes) amplifies these effects compared to nonelectrolyte solutions. The equations for colligative properties are modified as follows:
ΔTb = iKb m
ΔTb = boiling-point elevation (in oC)
i = van't Hoff factor
Kb = molal boiling-point elevation constant (in oC.kg.mol-1)
m = molality of the solute (in mol.kg-1)
ΔTf = - iKf x m
ΔTf = freezing-point depression (in oC)
i = van't Hoff factor
Kf = molal freezing-point depression constant (in oC.mol.kg-1)
m = solution molality (in mol.kg-1)
Π = i MR T
Π = osmotic pressure (in atm)
i = van't Hoff factor
M = molarity of the solution (in mol.L-1)
R = ideal gas constant = 0.0821 (in L.atm.K-1.mol-1)
T = absolute temperature (in K)
Check your knowledge about this Chapter
#### What are colligative properties and how do they depend on the number of solute particles?
Colligative properties are physical properties of solutions that depend on the number of dissolved solute particles, rather than the type
of particles. These properties include vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure.
The impact of solute particles on these properties is directly related to the concentration of the particles; for example, adding more solute
particles will further lower the vapor pressure and increase the boiling point and osmotic pressure of the solution, while also causing a greater
decrease in the freezing point. These effects occur because the presence of solute particles disrupts the orderly process of solvent molecules
entering the gas phase or forming a solid, thus altering the physical behavior of the solvent.
#### How does the nature of the solute and solvent affect the solubility of a substance?
The solubility of a substance is significantly influenced by the chemical nature of both the solute (the substance being dissolved) and the solvent (the substance doing the dissolving).
According to the principle "like dissolves like," polar solvents, such as water, are more effective at dissolving polar solutes or ionic compounds, because the solvent can effectively surround the solute particles and separate them into solution due to similar intermolecular forces. Nonpolar solvents, on the other hand, are better at dissolving nonpolar solutes, because the intermolecular forces between solute and solvent are similarly weak, allowing the solute to blend into the solvent with ease.
Additionally, specific solute-solvent interactions, such as hydrogen bonding and dipole-dipole interactions, can either enhance or reduce solubility depending on compatibility.
#### What is the role of intermolecular forces in the solution process?
The role of intermolecular forces in the solution process is crucial for determining whether a substance will dissolve in a solvent. During solvation, solute particles must overcome their intermolecular forces to separate and disperse among the solvent molecules. In turn, the solvent molecules must overcome their intermolecular forces to accommodate the solute particles. Successful dissolution occurs when the solute-solvent interactions are strong enough to offset the energy required to break apart the solute and solvent's intermolecular attractions. Consequently, similar intermolecular forces between solute and solvent, such as hydrogen bonding or dipole-dipole interactions, often lead to better solubility.
#### How do you calculate the molality of a solution, and in what situations is it preferred over molarity?
To calculate the molality of a solution, you divide the number of moles of solute by the mass of the solvent in kilograms. The formula is:
m =
m = molality (in mol.kg-1)
nsolute = moles of solute (in mol)
msolution = mass of solvent (in kg)
Molality is preferred over molarity when dealing with temperature-dependent studies because molality is not affected by changes in temperature; it relies on mass, which does not change with temperature like volume. This distinction makes molality especially useful for colligative property calculations which are influenced by the amount of solute particles, not their concentration in volume.
#### What is Raoult's Law and how does it relate to vapor-pressure lowering in solutions?
Raoult's Law states that the vapor pressure of a volatile component in a solution is directly proportional to its mole fraction in the solution. This law indicates that the presence of a non-volatile solute lowers the vapor pressure of the solvent in a solution because the solute molecules occupy some of the surface area, reducing the number of solvent molecules that can escape into the vapor phase. As a result, the solution has a lower vapor pressure compared to the pure solvent, which is a phenomenon known as vapor-pressure lowering.
#### Why do ionic and polar solutes lower the vapor pressure of solvent more effectively than nonpolar solutes?
Ionic and polar solutes have a significant impact on vapor pressure because they dissociate in solution to form ions or
interact strongly with the solvent molecules through dipole-dipole interactions, respectively. This strong solute-solvent
interaction makes it more difficult for the solvent molecules to escape into the gas phase, which reduces the rate of
evaporation. Consequently, the solution's vapor pressure, which depends on the number of solvent molecules in the gas
phase, is lower than that of the pure solvent. Nonpolar solutes do not form such strong interactions with the solvent
molecules and therefore have a less pronounced effect on the vapor pressure.
#### What is the boiling-point elevation, and how is it quantitatively determined?
Boiling-point elevation occurs when a solute is dissolved in a solvent, causing the boiling point of the solution to be higher than that of the pure solvent. This phenomenon is due to the fact that the presence of solute particles lowers the solvent's vapor pressure, requiring a higher temperature to reach the vapor pressure necessary for boiling.
The quantitative determination of boiling-point elevation is described by the formula:
ΔTb = Kb m
ΔTb = boiling-point elevation (in oC)
Kb = molal boiling-point elevation constant (in oC.kg.mol-1)
m = molality of the solute (in mol.kg-1)
Freezing-point depression occurs when a solute is dissolved in a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. This phenomenon happens because the presence of solute particles interferes with the formation of the solid lattice structure that is typical of the frozen solvent. As a result, additional energy (lower temperature) is required to solidify the solution.
Freezing-point depression is calculated using the formula:
ΔTf = - Kf x m
ΔTf = freezing-point depression (in oC)
Kf = molal freezing-point depression constant (in oC.mol.kg-1)
m = solution molality (in mol.kg-1)
Osmotic pressure is the pressure required to prevent the flow of solvent molecules through a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration. It is a colligative property, meaning it is influenced by the number of solute particles in a solution rather than their identity.
The osmotic pressure is directly proportional to the molarity of the solute particles in the solution, according to the formula:
Π = MR T
Π = osmotic pressure (in atm)
M = molarity of the solution (in mol.L-1)
R = ideal gas constant = 0.0821 (in L.atm.K-1.mol-1)
T = absolute temperature (in K)
Therefore, increasing the concentration of solute particles increases the osmotic pressure.
The van't Hoff factor (i) quantifies the number of particles an electrolyte produces when dissolved in water. For colligative
properties, which depend on the number of particles in solution (not their identity), the van't Hoff factor is crucial. Electrolytes
that ionize completely have a van't Hoff factor greater than 1, amplifying colligative effects like boiling-point elevation and
freezing-point depression. In contrast, nonelectrolytes have a factor of 1, as they do not produce additional particles.
The dissociation or association of solute molecules in solution can significantly affect colligative properties because these properties depend on the number of particles in solution, not their identity. When a solute dissociates into ions, the number of particles in the solution increases, which intensifies colligative effects such as boiling-point elevation and freezing-point depression. Conversely, when solute molecules associate, the number of particles decreases, diminishing the effect on colligative properties. So, the calculated effect on the colligative properties using the van't Hoff factor, which accounts for the actual number of particles formed from the solute, can offer a more precise understanding of these changes. |
17360 | https://www.oxfordlearnersdictionaries.com/definition/english/immutable | Definition of immutable adjective from the Oxford Advanced Learner's Dictionary
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17361 | https://www.math.lsu.edu/~rich/4038_Downloads/Bessel_Series_example.pdf | Math 4038 Fourier-Bessel Series Example Supplement Example 0.1. In class we have calculated the coefficients of the Fourier-Bessel expansion of the function f(x) = k if 0 ≤x ≤a 0 if a < x ≤R . We determined that if αmn is the mth zero of the function Jn on the positive x-axis, then f(x) = ∞ m=1 cmJn αmnx R , where cm = 2 R2J2 n+1(αmn) R 0 f(x)Jn αmnx R xdx.
The convergence takes place in the square norm corresponding to the given scalar product with the weight function p(x) = x.
In order to use Mathematica to compute the graphs of some partial sums of this Fourier-Bessel series, I specify in the following that k = 1 = a and R = 2. I chose to use n = 0. In case you use Mathematica, I will include the Mathematica input code that I employed for this example, together with the output.
Example 0.2. Here is the Mathematica code with the Output. First we calculate the first 50 zeros of J0: zeros = N[BesselJZero[0, Range]] output = 2.40483, 5.52008, 8.65373, 11.7915, 14.9309, 18.0711, 21.2116, 24.3525, 27.4935, 30.6346, 33.7758, 36.9171, 40.0584, 43.1998, 46.3412, 49.4826, 52.6241, 55.7655, 58.907, 62.0485, 65.19, 68.3315, 71.473, 74.6145, 77.756, 80.8976, 84.0391, 87.1806, 90.3222, 93.4637, 96.6053, 99.7468, 102.888, 106.03, 109.171, 112.313, 115.455, 118.596, 121.738, 124.879, 128.021, 131.162, 134.304, 137.446, 140.587, 143.729, 146.87, 150.012, 153.153, 156.295 Then we write the sum of the first ten terms of the Bessel series: Sum[N[BesselJ[1, zeros/2]]/(zerosN[BesselJ[1, zeros]]2)BesselJ[0, zeros∗x/2], {k, 1, 10}] output = 0.769756 BesselJ[0, 1.20241 x] + 0.661472 BesselJ[0, 2.76004 x] - 0.282963 BesselJ[0, 4.32686 x] - 0.464336 BesselJ[0, 5.89577 x] + 0.198712 BesselJ[0, 7.46546 x] + 0.378402 BesselJ[0, 9.03553 x] - 0.160955 BesselJ[0, 10.6058 x] - 0.327418 BesselJ[0, 12.1762 x] + 0.138625 BesselJ[0, 13.7467 x] + 0.292704 BesselJ[0, 15.3173 x] Then we plot the sum of the first 10 terms of the Bessel series in Fig. 1, after which we perform the final two operations above, changing to the sum of the first 50 terms in Fig. 2: Plot[{%, 1 - HeavisideTheta[x - 1]}, {x, 0, 2}] 1 Math 4038 Fourier-Bessel Series Example Supplement 0.5 1.0 1.5 2.0 0.2 0.4 0.6 0.8 1.0 1.2 Figure 1: 10th partial sum, with f(x).
0.5 1.0 1.5 2.0 0.2 0.4 0.6 0.8 1.0 Figure 2: 50th partial sum S50, with f(x).
Notice how little area there is between the graph of y = f(x) and the graph of the 50th partial sum of the Fourier-Bessel series. In fact, we know that ∥f −Sn∥2 →0, as n →∞.
In Fig. 3 we have shown the square of (S50 −f(x))2. Note that there is a narrow peak at x = 1 because the continuous partial sum S50(x) is about half-way between one and zero at x = 1.
This is necessary since the jump must be bridged. In fact, ∥S50 −f∥2 = 0.00400453 according to Mathematica.
Example 0.3. You may have noticed that it took a rather long partial summation of the Fourier-Bessel series to give a good approximation to a function that has a jump discontinuity at the 2 Math 4038 Fourier-Bessel Series Example Supplement 0.5 1.0 1.5 2.0 0.05 0.10 0.15 0.20 Figure 3: (S50 −f(x))2, with step function f(x).
midpoint, x = 1. Here is a different example, without a discontinuity but with points of nondif-ferentiability. Define the tent function as follows: f(x) = 1 −|x −1|, 0 ≤x ≤2.
After tabulating the zeros of J0 as in the first two examples, we use the following method of finding the tenth partial sum of the Fourier-Bessel series: Sum[2/(22 ∗N[BesselJ[1, zeros]]2)Integrate[(1−Abs[x−1])∗BesselJ[0, (zeros∗x/2)]∗ x, {x, 0, 2}] ∗BesselJ[0, zeros ∗x/2], {k, 1, 10}] Output= 1.01488 BesselJ[0, 1.20241 x] - 0.644837 BesselJ[0, 2.76004 x] - 0.302905 BesselJ[0, 4.32686 x] + 0.0365356 BesselJ[0, 5.89577 x] + 0.08648 BesselJ[0, 7.46546 x] - 0.0668066 BesselJ[0, 9.03553 x] - 0.0765271 BesselJ[0, 10.6058 x] + 0.0143768 BesselJ[0, 12.1762 x] + 0.0372709 BesselJ[0, 13.7467 x] - 0.0259684 BesselJ[0, 15.3173 x] Notice that this time we have instructed Mathematica to perform the weighted integral to find the needed scalar products for the coefficients.
Plot[{%, 1 - Abs[x - 1]}, {x, 0, 2}] See Fig. 4 In this example ∥f −S10∥2 = 0.000119022, ∥f −S10∥= 0.0109 although S10 is the sum of only the first ten terms of the Fourier Bessel series for the tent function shown. In Fig. 5 there is a graph of (f −S10)2.
Note the exaggerated scale on the vertical axis.
3 Math 4038 Fourier-Bessel Series Example Supplement 0.5 1.0 1.5 2.0 0.2 0.4 0.6 0.8 1.0 Figure 4: 10th partial S10, with f(x) = 1 −|x −1|.
0.5 1.0 1.5 2.0 0.001 0.002 0.003 0.004 0.005 Figure 5: (f(x) −S10(x))2, with f(x) = 1 −|x −1|.
4 |
17362 | https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOoo0YENUl2LOYvrjsHt1UPcm73jL1R1eibD0ExrmrmYu37Z0NAE0 | Art of Problem Solving
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
[hide]
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
Main article: Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
Main article: Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
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17363 | https://www.statology.org/residual-plot-ti-84/ | How to Create a Residual Plot on a TI-84 Calculator
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How to Create a Residual Plot on a TI-84 Calculator
by Zach BobbittPosted onLast updated on April 22, 2021
A residual plot is used to assess whether or not the residuals in a regression analysis are normally distributed and whether or not they exhibit heteroscedasticity.
This tutorial provides a step-by-step example of how to create a residual plot for the following dataset on a TI-84 calculator:
Step 1: Enter the Data
First, we will enter the data values.
Press Stat, then press EDIT. Then enter the x-values of the dataset in column L1 and the y-values in column L2:
Step 2: Perform Linear Regression
Next, we will fit a linear regression model to the dataset.
Press Stat, then scroll over to CALC. Then scroll down to LinReg(ax+b) and press ENTER.
Press ENTER once again to perform linear regression:
The fitted regression model is: y = 7.397 + 1.389x
Step 3: Create the Residual Plot
Next, press 2nd and then press Y=. In the new screen that appears, press ENTER on the first plot option.
Hover over the “On” option and press press ENTER. Then scroll down to YList and press 2nd and then press STAT. Then press “7” to choose the residuals:
The term “RESID” will then appear next to Ylist:
Lastly, press ZOOM and then scroll down to ZoomStat and press ENTER.
The residual plot will appear:
The x-axis displays the x values from the dataset and the y-axis displays the residuals from the regression model.
To see the actual values of the residuals, press 2nd and then press STAT. Then press “7” to choose the residuals:
Press ENTER once more to display the residuals.
Scroll to the right to see the values for each of the residuals.
Additional Resources
How to Perform Linear Regression on a TI-84 Calculator
How to Perform Quadratic Regression on a TI-84 Calculator
How to Calculate a Correlation Coefficient on a TI-84 Calculator
Posted in Programming
Zach Bobbitt
Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.
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2 Replies to “How to Create a Residual Plot on a TI-84 Calculator”
danesays: October 19, 2022 at 4:56 pm thank you! Reply
Billsays: May 3, 2023 at 10:57 pm This was a big help! I always forget to do the Linreg function before trying to make the residual plot. Reply
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Name :___ Date : ___ 1) F i n d t h e a r e a o f t h e s h a d e d r e g i o n 2) F i n d t h e a r e a o f t h e s h a d e d r e g i o n.if the big diameter is 12 in, and the small diameter is 8 in.3) F i n d t h e a r e a o f t h e s h a d e d r e g i o n:4) F i n d t h e a r e a o f t h e s h a d e d r e g i o n 5) Find the area of the shaded region if the 6) Find the area of the shaded region radis of the !ir!le is " in,
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Name :___ Date : ___#) Find the area of the shaded region 8) Find the area of the shaded region if the radis of the big !ir!le is 12 !m.") F i n d t h e a r e a o f t h e s h a d e d r e g i o n.1$) %r i a n g l e &'( i s ei l a t e r a l. %h e a r! i s a r t o f a !ir!le +ith !enter ' and radis '&. hat is the area of the shaded region-6 !m 11) Find the area of the shaded area in this 12) %his roblem is +i!ed hard/// 0f o dont +ant figre in +hi!h ea!h side of the right to do it, dont : 0f o +ant a !hallenge, gie it a triangle has a semi!ir!le !reated sing tr. Find the area of the shaded region. 0 left some the sid e as the dia met er. %his is one of tho se som e oi nts in the re to get o sta rte d.o hae to loo for the se!ret ones:)6 !m
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Name :___ Date : ___
Mensuration
1.Let h be t he h eig ht an d a be the sid e of an e qui lat era l tri ang le, the n h = (3/2) a. 2.
C i r c u m f e r e n c e a n d a r e a o f a c i r c l e.
I f r i s t h e r a d i u s o f a c i r c l e, t h e n(i) t h e c i r c u m f e r e n c e o f t h e c i r c l e = 2 r(ii) t he a rea of the circle = r² 3.
A r e a o f a c i r c u l a r r i n g.
I f a n d r a r e t h e r a d i i o f t h e b i g g e r a n d s m a l l e r (c o n c e n t r i c) c i r c l e s, t h e n area of the ring = (² !r²). ".
C i r c u m f e r e n c e a n d a r e a o f a s e c t o r o f a c i r c l e.
If r is t he ra di u s of t he ci r cl e an d th e ar c su bt en d s an a ng le of n# a t t he ce nt er, t he n(i) t h e l e n g t h o f t h e a r c = (n/3$%).2 r = n r/1&%(ii) t he are a of t he sec tor = (n/3$%).r² '.
Cir cum fer enc e and are a of cir cum scr ibe d and ins cri bed cir cle s of an equ ila ter al tri ang le.
If a nd r ar e t he ra di i o f t he ci r cu m sc ri be d a nd in sc r ib ed ci r cl es of th e tr ia ng le, th en(i) = (2/3)h a n d r = (1/3)h(i i) t h e c i r c u m f e r e n c e o f t h e c i r c u m s c r i b e d c i r c l e = 2 = ("/3)h(i i i) t h e a r e a o f t h e c i r c u m s c r i b e d c i r c l e = ² = ("/)h ²(i) t h e c i r c u m f e r e n c e o f t h e i n s c r i b e d c i r c l e = 2 r = (2/3)h() the area of the inscribed circle = r² = (1/) h² $.
Ci rc um fe re nce an d ar ea of ci rc ums cr ib ed an d in sc ri bed ci rc le s of a re gul ar he xa gon.
Let a be the side of a regular heagon and , r be the radii of the circumscribed and inscribed circles r e s+e c t ie l o f t h e h ea g o n, t h e n(i) = a a n d r = (3/2) a(i i) t h e c i r c u m f e r e n c e o f t h e c i r c u m s c r i b e d c i r c l e = 2 = 2 a(i i i) t h e a r e a o f t h e c i r c u m s c r i b e d c i r c l e = ² = a ²(i) t h e c i r c u m f e r e n c e o f t h e i n s c r i b e d c i r c l e = 2 r = 3 a(i) the area of the inscr ibed circle = r² = (3/")a².
Surface area and volume (of solids)
1.
S o l i d C y l i n d e r.
L e t r b e t h e r a d i u s a n d h b e h e i g h t o f a s o l i d cl i n d e r, t h e n(i) c u re d (l a t e r a l) s u r f a c e a r e a = 2 r h(i i) t o t a l s u r f a c e a r e a = 2 r(h -r)(iii) olume = r²h 2.
H o l l o w c y l i n d e r.
Le t an d r be th e ete rn al an d in te rn al ra di i, an d h be th e he ig ht of a ho llo cli nde r, the n(i) et e r n a l c u re d s u r f a c e a r e a = 2h(i i) i n t e r n a l c u re d s u r f a c e a r e a = 2 r h(i i i) t o t a l s u r f a c e a r e a = 2(h -r h -² !r ²)(i) olume of material = (² !r²)h 3.
Cone.
L e t r, h a n d l b e t h e r a d i u s, h e i g h t a n d s l a n t h e i g h t r e s+e c t ie l o f a c o n e, t h e n(i) s l a n t h e i g h t = r ² -h ²(i i) c u re d (l a t e r a l) s u r f a c e a r e a = r l(i i i) t o t a l s u r f a c e a r e a = r(l -r)(i) olume = (1/3) r²h ".
S o l i d sh e r e.
L e t r b e t h e r a d i u s o f a s o l i d s+h e r e, t h e n
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Name :___ Date : ___
(i) s u r f a c e a r e a = "r ²(ii) olume = ("/3) r '.
Sh e r i c a l s h e l l.
L e t a n d r b e t h e r a d i i o f t h e o u t e r a n d i n n e r s+h e r e s, t h e n(i) t h i c 0 n e s s o f t h e s h e l l = !r(ii) o lume o f m aterial = ("/3) ( !r) $.
S o l i d h e m i sh e r e.
L e t r b e t h e r a d i u s o f a h e m i s+h e r e, t h e n(i) c u re d (l a t e r a l) s u r f a c e a r e a = 2 r ²(i i) t o t a l s u r f a c e a r e a = 3 r ²(iii) olume = (2/3) r .
H e m i sh e r i c a l s h e l l.
L e t a n d r b e t h e r a d i i o f t h e o u t e r a n d i n n e r h e m i s+h e r e s, t h e n(i) t h e t h i c 0 n e s s o f t h e s h e l l = !r(i i) et e r n a l c u re d s u r f a c e a r e a = 2 ²(i i i) i n t e r n a l c u re d s u r f a c e a r e a = 2 r ²(i) t o t a l s u r f a c e a r e a = (3² -r ²)() o lume o f mat erial = (2/3)( !r)
!xercise
1.ind the length of the d iamet er of a ci rcle h ose circ umfe rence is "" cm. 2.ind the circu mfere nce of a circl e hose area is 1"" cm². 3.o man t imes i ll the hee l of a car rotat e in a 4ourne of && 0m if it is 0n on tha t the diam eter of th eheel is '$ cm5 ".(a) 6alc ulate th e radius of a ccle heel in cm hich ma0 es % reol ution s in moing a dis tanc e of 1'"m.(b) 7 buc0et is raised from a ell b means of a ro+e hich is ound round a heel of diameter cm.8ien that the buc0et ascends in 1 minute 2& seconds ith a uniform s+eed of 191 m/s, calculate the number of com+lete reolutions the heel ma0es in raisin g the buc0et. :a 0e to be 22/. '.7 roa d hich is m id e surro und s a cir cul ar +ar0 h ose cir cum fer enc e is 3'2 m. in d the cost of+aing the road at s 2% +er m². $.:h e su m of rad ii of to cir cl es is cm and dif fe re nc e of the ir circ um fe re nc es is &cm. i nd the circumferences of the to circles. .:o circle s touc h et ernall. :h e sum o f thei r area s is '&cm² an d dista nce be teen t heir c ente rs is 1%cm. ind the radii of the to circles. &.:he diame ters of the to circle s are in the rati o 3 ; " and the sum of the are as of the circ les is equa l to the area of a circle hose diameter is 3$ cm. ind the radii of the circles. .7 c o++er i re h en ben t in th e for m of an e qui lat era l tri ang le ha s are a 121 3 cm². If th e sam e ir e is bent into the form of a circle, find the area enclosed b the ire. 1%.(a) In the figure (i) gien belo, 7<6 is a square inscribed in a circle of radius cm. 6alculate the area o f t h e c i r c l e a n d t h e a r e a o f t h e s h a d e d r e g i o n.(b) In the figure (ii) gien belo, 7<6 is a rectangle ith sides 7< = "2 cm and <6 = 2& cm. :o qu a rt er ci rc l es a re d ran a s sh on in f ig u re. 6a lc u la t e t he a re a of t he sh ad ed +ar t.(i) (ii) 11.7 rect angle ith one side " cm is inscri bed in a circle of radius 29' cm. ind the area of the rect angle.
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17365 | https://brainly.com/question/46745110?source=archive | [FREE] Out of 21 days marked with the numbers from 1 to 21, three are drawn at random. What is the probability - brainly.com
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Out of 21 days marked with the numbers from 1 to 21, three are drawn at random. What is the probability that the numbers form an Arithmetic Progression (AP)?
A) 1/14
B) 1/21
C) 1/7
D) 3/21
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The probability of selecting three numbers in arithmetic progression out of 21 days is 1/14, calculated by dividing the number of favorable outcomes (90 AP sequences) by the total number of ways to pick three days (1330 combinations).
Explanation
The student is asking about the probability of selecting three days out of 21 where the numbers form an arithmetic progression (AP). The total number of possible combinations of three days from 21 is calculated using the combination formula C(21, 3), which is 1330.
For the three numbers to be in an AP, the middle number must be the average of the other two. Considering this condition, we can see that starting with the day marked '1', there are 9 possibilities (1, 3, 5, ..., 17, 19). For the day marked '2', there are similarly 9 possibilities (2, 4, 6, ..., 18, 20), and so on until the day marked '10', which also has 9 possibilities. For the days marked '11' onwards, the number of possibilities begins to decrease again, mirroring the first half. Therefore, the total number of AP sequences possible is 9 + 9 + ... + 9 (10 times) which gives us 90.
To find the probability, we divide the number of favorable outcomes by the total number of outcomes: Probability = 90/1330, which simplifies to 1/14. Hence, the correct answer is a) 1/14.
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The probability of randomly choosing three numbers from 21 that form an arithmetic progression is 14 1. This is calculated by finding the total combinations of three numbers (1330) and the number of favorable outcomes (90 AP sequences). Therefore, the answer is (A) 1/14.
Explanation
To determine the probability that three randomly drawn numbers from a set of 21 form an Arithmetic Progression (AP), we need to follow these steps:
Calculate the total number of combinations of selecting 3 numbers from 21. This can be done using the combination formula, which is given by:
C(n,k)=k!(n−k)!n!
For our case, n=21 and k=3, therefore:
C(21,3)=3!(21−3)!21!=3×2×1 21×20×19=1330
So, there are 1,330 ways to choose any 3 numbers from the set of 21.
Count the number of favorable outcomes where the 3 numbers form an AP. For three numbers to be in an AP, they must be equidistant from each other, which means if we denote the three numbers as a−d,a, and a+d, the middle number must be the average of the other two.
We can observe the sequences:
For any chosen middle number (from 2 to 20), we can create valid AP sequences by controlling the distance d such that the first number is at least 1 and the last one is at most 21.
Starting with the middle number as 2, the sequences would be: (1, 2, 3), (2, 4, 6), (2, 5, 8), ..., up to (2, 10, 18).
Continuing this thought process up through 20, we find that there are valid combinations for each middle number that satisfies being within the range while maintaining equal spacing.
After calculating for each middle number (2 to 20), we find there are a total of 90 distinct AP sequences.
Calculate the probability by dividing the number of favorable outcomes by the total outcomes:
P=Total combinations Number of AP sequences=1330 90=14 1
Thus, the probability that a randomly chosen set of 3 numbers will form an Arithmetic Progression is 14 1.
Therefore, the correct answer is (A) 1/14.
Examples & Evidence
For example, if the three drawn numbers are 4, 6, and 8, they form an AP with a common difference of 2. Another set like 10, 14, and 18 also forms an AP with a common difference of 4.
The combination formula and the properties of Arithmetic Progressions used in this explanation are well-established mathematical principles, helping in constructing an accurate solution.
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17367 | https://empendium.com/mcmtextbook/table/031_9142 | Selected drugs causing a disulfiram-like reaction - McMaster Textbook of Internal Medicine
Table 19.3-1. Selected drugs causing a disulfiram-like reaction– Abacavir
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– Chloramphenicol
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– Phenylbutazone
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17368 | https://www.oxfordlearnersdictionaries.com/us/definition/english/liaison | Definition of liaison noun from the Oxford Advanced Learner's Dictionary
liaison
noun
/liˈeɪzn/
/ˈliːəzɑːn/,/liˈeɪzɑːn/
jump to other results
[uncountable, singular] a relationship between people or organizations involving the exchange of information or ideas
We are hoping to establish better customer liaison.
liaison between A and B Our role is to ensure liaison between schools and parents.
liaison with somebody We work in close liaison with the police.
Extra Examples
She is responsible for liaison with researchers at other universities.
We are hoping to establish better liaison with these groups.
We maintained a close liaison with the trade union.
a customer liaison manager
good liaison between management and staff
Oxford Collocations Dictionaryadjective
close
effective
good
…verb + liaison
maintain
establish
improve
…liaison + noun
committee
group
team
…preposition
in liaison with
liaison between
liaison with
…See full entry
Definitions on the go
Look up any word in the dictionary offline, anytime, anywhere with the Oxford Advanced Learner’s Dictionary app.
2. [countable] liaison (to/with somebody/something) a person whose job is to make sure there is a good relationship between two groups or organizations
the White House liaison to organized labor
We hired someone as customer liaison.
3. [countable] liaison (with somebody) a secret sexual relationship, especially if one or both partners are already in a relationship with somebody else synonym affair
He finally admitted to several sexual liaisons.
She was having a romantic liaison with her husband's best friend.Topics Family and relationshipsc2
Oxford Collocations Dictionaryadjective
romantic
sexual
adulterous
…verb + liaison
havepreposition
liaison withSee full entry
Word Originmid 17th cent. (as a cookery term): from French, from lier ‘to bind’.
See liaison in the Oxford Advanced American Dictionary
Check pronunciation:
liaison
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: liaison officer noun
liaison officers
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liable adjective
liaise verb
liaison noun
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Liam
28 September 2025
conclude
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17369 | http://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1098316.html | SOLUTION: given the sum S10=910 of an arithmetic sequence & A20=95, find A1?
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| | | | | | | | | | | --- --- --- --- | | | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Algebra: Sequences of numbers, series and how to sum themSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | | | | | | | --- | | --- | | Click here to see ALL problems on Sequences-and-series --- Question 1098316: given the sum S10=910 of an arithmetic sequence & A20=95, find A1? Answer by greenestamps(13200) About Me (Show Source): You can put this solution on YOUR website! Let a be the first term and d be the common difference. If the sum of the first 10 terms of an arithmetic sequence is 910, then that sum of 910 is the sum of 5 pairs of numbers, each with a sum of 910/5 = 182. One of those pairs is the sum of the first and tenth numbers. The first number is a; the 10th number is a+9d: so a+%2B+a%2B9d+=+182 2a%2B9d+=+182 The 20th term, 95, is a+19d: a%2B19d+=+95 That gives you two equations in a and d which you can solve to find the answer to the problem. But I'm wondering if you have shown the right numbers, because with the numbers you show, the terms of the sequence turn out to be ugly fractions. So I'm not going to show the ugly arithmetic; you can solve the pair of equations on your own. | | | | |
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| | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Algebra: Sequences of numbers, series and how to sum themSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | | |
| | | | | | | |
--- --- ---
| | | | | | | | --- --- --- | | Algebra: Sequences of numbers, series and how to sum themSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | | |
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| Algebra: Sequences of numbers, series and how to sum themSection | SolversSolvers | LessonsLessons | Answers archiveAnswers | | |
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| | --- | | Click here to see ALL problems on Sequences-and-series --- Question 1098316: given the sum S10=910 of an arithmetic sequence & A20=95, find A1? Answer by greenestamps(13200) About Me (Show Source): You can put this solution on YOUR website! Let a be the first term and d be the common difference. If the sum of the first 10 terms of an arithmetic sequence is 910, then that sum of 910 is the sum of 5 pairs of numbers, each with a sum of 910/5 = 182. One of those pairs is the sum of the first and tenth numbers. The first number is a; the 10th number is a+9d: so a+%2B+a%2B9d+=+182 2a%2B9d+=+182 The 20th term, 95, is a+19d: a%2B19d+=+95 That gives you two equations in a and d which you can solve to find the answer to the problem. But I'm wondering if you have shown the right numbers, because with the numbers you show, the terms of the sequence turn out to be ugly fractions. So I'm not going to show the ugly arithmetic; you can solve the pair of equations on your own. | | |
| | |
--- |
| Click here to see ALL problems on Sequences-and-series --- Question 1098316: given the sum S10=910 of an arithmetic sequence & A20=95, find A1? Answer by greenestamps(13200) About Me (Show Source): You can put this solution on YOUR website! Let a be the first term and d be the common difference. If the sum of the first 10 terms of an arithmetic sequence is 910, then that sum of 910 is the sum of 5 pairs of numbers, each with a sum of 910/5 = 182. One of those pairs is the sum of the first and tenth numbers. The first number is a; the 10th number is a+9d: so a+%2B+a%2B9d+=+182 2a%2B9d+=+182 The 20th term, 95, is a+19d: a%2B19d+=+95 That gives you two equations in a and d which you can solve to find the answer to the problem. But I'm wondering if you have shown the right numbers, because with the numbers you show, the terms of the sequence turn out to be ugly fractions. So I'm not going to show the ugly arithmetic; you can solve the pair of equations on your own. | | |
17370 | https://www.sciencedirect.com/science/article/pii/S2768276524008320 | Unicameral Bone Cysts: Treatment Rationale and Approach - ScienceDirect
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Current Concept Review
Unicameral Bone Cysts: Treatment Rationale and Approach
Author links open overlay panel Soroush Baghdadi MD 1, Alexandre Arkader MD 1 2
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Abstract:
Unicameral bone cyst (UBC) is a benign cystic lesion most commonly diagnosed in the proximal aspects of the humerus and femur of growing children. Medullary venous obstruction is the leading pathogenesis theory, resulting in fluid accumulation, bone resorption, and cortical thinning. Most UBCs are asymptomatic and likely go undiagnosed while the most common presentation is a pathologic fracture. Younger children tend to present with active lesions, which are uniloculated, abut the physis, and have a higher recurrence or persistence rate after treatment. Latent lesions in older children have migrated away from the growth plate and might become multiloculated. Radiographs are often diagnostic although advanced imaging is also helpful in some cases. Most UBCs do not need treatment; when indicated, management ranges from aspiration and steroid injection to decompression, curettage and grafting, and internal fixation. Percutaneous or open surgical approaches are acceptable and often yield good results although high recurrence rates should be expected, especially in younger children with active lesions.
Key Concepts:
•UBCs are not a true neoplasm and do not have malignant potential. However, they may cause complications such as pathologic fractures and growth disturbance.
•Treatment should be personalized to each patient's demographics, functional needs, desires of the family, and lesion characteristics.
•Patients/families should be aware of the high recurrence rate and the possible need for repeat intervention.
•A percutaneous or open surgical approach, with decompression, curettage of the cyst, grafting, and internal fixation if necessary, is the mainstay of management.
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Introduction
Unicameral bone cyst (UBC) is a common benign, cystic lesion of the skeleton, also known as solitary or simple bone cyst. While it does not represent a true neoplasm, expansion and weakening of the bone might lead to symptoms and/or pathologic fractures. Virchow is often credited with the first report of UBC, although his description was published in the pre-Roentgen era. Several authors have expanded our knowledge in the twentieth century, including Jaffe and Lichtenstein, who defined the pathological characteristics of the lesion, as well as the cyst lining and fluid features.
Although UBC is currently favored in the literature over solitary bone cyst, the term “unicameral” is not always accurate. Some UBC have one cavity in the initial stages, yet many develop a multiloculated structure, especially after fractures or treatment. UBC typically arises in the metaphysis of tubular bones adjacent to the growth plate and is almost exclusively diagnosed in the immature skeleton (although it can persist into adulthood), with reported mean age of 11 years at diagnosis. UBC represents up to 3% of all reported bone lesions, but similar to other benign pathologies, most cases may go undiagnosed; hence its true incidence is likely higher.
While UBCs have no malignant potential and have a relatively favorable long-term prognosis regardless of the treatment, controversy still exists regarding the natural history, indications for treatment, and preferred technique. In this review, we focus on the different therapeutic approaches based on the available literature and institutional experience.
Etiology and Pathogenesis
The exact etiology and pathogenesis of UBC remain unclear. The most favored pathogenesis theory is a blockage in the intramedullary venous drainage postulated by Cohen.1., 2. In brief, an insult to the bone causes venous obstruction, which in turn leads to fluid accumulation, increased pressure, and bone resorption.1., 2., 3. Several therapeutic approaches have been established based on this mechanism, including decompression or fenestration of the cyst.
On gross pathologic examination, UBC is usually a single fluid-filled cavity. The pathognomonic feature is clear, straw-colored fluid on aspiration. However, the fluid might have a bloody tinge following a pathologic fracture. Histologic analysis of the cyst lining reveals a thin fibrous membrane with occasional multinucleated giant cells. Secondary changes, including the presence of hemosiderin, macrophages, and reactive woven bone, are also common in the presence of a fracture.
At the cellular level, osteoclasts are believed to play a role in the pathogenesis of UBCs. Osteoclasts are present in the cyst lining and fluid, and the messengers and products of the continued osteoclast-mediated bone destruction have been confirmed in the cyst,3 including prostaglandin-E2 (PGE2),4lysosomal enzymes,5 and acid phosphatase levels.6 However, a complex signaling pathway exists between cells of different lineages, and osteoblasts are required for activation of osteoclastic bone degradation. Additionally, increased osteoblastic activity and concentration of RANKL (receptor activator of nuclear factor kappa-B ligand) which activates osteoclasts, has been found in UBC lining and fluid.3
Clinical Presentation
Similar to other benign bone lesions of childhood, most UBCs are asymptomatic and likely go undiagnosed unless there is an incidental finding or pathologic fracture. Symptomatic patients typically present with either an acute pathologic fracture or an insufficiency/stress fracture. In the latter group, the patient might present with mechanical symptoms, painful range of motion, or limping. In a systematic review, more than three-quarters of the published UBC cases were diagnosed after an acute fracture, 15% with an insufficiency fracture, and the rest as an incidental finding.7
Although UBCs may occur in any site, the metaphysis and diaphysis of long bones are the most common location. More than half of the lesions are diagnosed in the proximal humerus, followed by the proximal femur and proximal tibia.7 The calcaneus is also a relatively common site for UBC which tends to occur in the slightly older population. Boys are twice as likely to be diagnosed with a UBC, although it is unclear whether this represents a higher incidence or is the result of higher fracture risk in boys.
Imaging
UBCs appear as centrally located, well-defined, lucent metaphyseal lesions on plain radiographs, with a narrow zone of transition, and often abutting the physis (active phase). With longitudinal growth, the lesion “migrates” away from the physis and becomes diaphyseal (latent phase). Larger lesions may lead to bone expansion and result in cortical thinning. Periosteal reaction or cortical destruction are absent unless a fracture occurs. Soft tissue extension is not a feature. The “fallen fragment sign” was first described by Reynolds to help distinguish UBC from other radiolucent but solid intramedullary lesions.8., 9. Following a fracture, the fluid within the UBC has no resistance to the bone fragments falling into the cavity, thus creating the distinctive intracavitary displaced bone fragment. Although the fallen fragment sign is helpful in some cases, it has a low sensitivity.10
Advanced imaging may be valuable in differentiating a UBC from similar lesions in equivocal cases. Computed tomography (CT) scan shows cortical thinning, fluid density inside the lesion, and occasionally the fallen fragment sign.11 On magnetic resonance imaging (MRI), UBC is characterized by a hypointense signal on T1-weighted images and hyperintense on T2-weighted and fluid-sensitive sequences. Septations, if present, are also visible on MRI. On post-contrast images, peripheral enhancement is appreciated. When a fracture is present, T1-weighted images will be hypersignal due to blood, while marrow edema, periosteal reaction, and surrounding edema are hyperintense on T2-weighted and fluid-sensitive sequences. Fluid-fluid levels may be present, especially in the presence of a pathologic fracture. MRI is particularly useful for the evaluation of a suspected insufficiency fracture and physeal involvement/disruption by the cyst.
Differential Diagnosis
All primarily cystic bone lesions are included in the differential diagnosis list, including aneurysmal bone cyst (ABC), non-ossifying fibroma (NOF), and fibrous dysplasia. In addition to the imaging findings, the specific demographics and pertinent clinical symptoms of each lesion are helpful in distinguishing the diagnosis.
ABC is the primary differential diagnosis of UBC due to similarities in demographics and imaging characteristics. ABC is an expansile, lytic lesion that affects the metaphysis of long bones, similar to UBC, but in an eccentric location. ABC can be locally aggressive with cortical expansion and disruption. On MRI, multiple septations and fluid-fluid levels are present due to the bloody contents of the cavities. The signal intensity is heterogeneous, with individual lobules having different signal characteristics.12 An intact UBC is easier to distinguish from ABC while a fractured or partially healed UBC can be challenging to differentiate. During surgery, aspiration of clear straw-colored fluid is diagnostic for UBC. A bloody fluid, although not diagnostic, is more suggestive of ABC.
NOFs have a lytic, lobular appearance but are cortical based, rather than central, and are classically surrounded by a sharp sclerotic border. Mild cortical expansion and pathologic fracture can be present. MRI demonstrates the fibrous content of the lesion.
Monostotic fibrous dysplasia is in the differential diagnosis of latent, diaphyseal UBCs. The presence of a ground-glass appearance instead of a fluid-filled cavity helps differentiate the two lesions on advanced imaging. Furthermore, fibrous dysplasia lesions may undergo cystic degeneration which increases the similarities with UBC and ABC.13
Natural History
The literature is inconclusive regarding the true natural history of UBCs. Most lesions are believed to heal by the time the patient reaches skeletal maturity although some studies have shown that the cyst persists, but the risk of pathologic fracture decreases. Several risk factors for complications (i.e., pathologic fracture, recurrence/persistence after treatment) have been identified. One of the earliest factors recognized is the distance of the lesion from the physis.14Active cysts abutting the growth plate have a higher risk of progression, growth, fracture, and recurrence after treatment, while latent cysts distant from the growth plate are more likely to resolve with or without treatment. In one study, healing was achieved in cysts that were > 2 cm away from the growth plate, but not in those closer to the physis.15 Age is also an important predictor for healing and recurrence after surgery. Patients older than 10 years have a lower rate of persistence or recurrence.16., 17. The risk of pathologic fracture is related to the lesion size and location. UBCs of the lower limb (weight bearing bones) have a higher risk of fracture as do larger lesions. The exact threshold that will increase fracture risk is uncertain, although evidence suggests that lesions wider than 50-80% of the bone width are at a high risk for fracture.18., 19. Overall, around 10-15% of cysts are expected to heal after a fracture,16., 20., 21., 22. especially for complete or displaced fractures.
Healing of the UBCs can be assessed with the classification system developed by Neer,23., 24. who defined a healed cyst as one filled by new bone formation with radiolucent areas < 1 cm in size. “Healing with defect” is a lesion with radiolucent areas < 50% of the diameter of the bone with enough cortical thickness to prevent fracture. A persistent cyst has radiolucent areas > 50% of the bone diameter with a thin cortical rim. Finally, a recurrent cyst is one that has reappeared in a previously healed area, or more commonly, a radiolucent area that has increased in size.
Management
Over the years, several management strategies and surgical techniques have been introduced for UBC. Nevertheless, there is significant controversy on the optimal management strategy, indications for treatment, and the surgical procedure of choice. This controversy is, in part, rooted in the benign, nonaggressive nature of UBCs. The risk of complications, including pathologic fracture and growth arrest, should be weighed against the risk of surgical treatment and recurrence, as well as the potential for spontaneous healing over time. Treatment should be personalized based on each patient's demographics and unique lesion characteristics. Age, patient's activity level, location, size of the lesion, and family preferences are important factors in tailoring a treatment plan for UBC (Figure 1).
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Figure 1. A sample algorithm for management of unicameral bone cysts. Refer to text for details.
Asymptomatic patients
Most asymptomatic, incidentally found UBCs are amenable to observation alone, with the notable exception of lesions around the hip and those with very high risk of pathologic fracture, and juxta physeal lesions with a risk of potential growth disturbance. Asymptomatic upper extremity lesions can almost universally be treated conservatively. It is important that the patient and family are reassured about the benign nature of the cyst and educated about the risk of pathologic fracture as well as surgical risks and the potential for recurrence. Active patients with a high risk for fracture or symptoms may elect to undergo prophylactic treatment for large, active lesions. Large (> 50% bone width) femoral lesions, especially around the hip, have a high risk of fracture, and prophylactic treatment is most often recommended.
Symptomatic patients
A thorough clinical and imaging workup is warranted to evaluate the cause of the patient's symptoms. In general, the culprit is an impending or insufficiency/stress fracture evident by surrounding marrow edema (MRI) and cortical disruption (CT or MRI). Although a trial of conservative treatment can initially be pursued, these patients often require surgery, especially for lower extremity lesions.
Pathologic fractures
Pathologic fractures may lead to decompression of the cyst and stimulate partial or complete healing in 10-15% of the cases. Therefore, conservative treatment is usually the initial preferred method, particularly for nondisplaced pathologic fractures and fractures of the upper extremities. Recurrent fractures may be an indication for surgical treatment. Proximal femoral UBC is a special case in which surgical treatment of an impending or overt fracture is indicated at presentation due to the morbidity associated with a displaced fracture around the hip.
Treatment Options
Steroid Injection
The use of intralesional steroid injection, most often methylprednisolone acetate (MPA), was first attempted by Scaglietti in the 70s,25 and later adapted by several other authors.26., 27., 28., 29., 30., 31., 32. The increased concentrations of prostaglandins in the cyst fluid motivated its use. The mechanism of action of steroids is still unknown, and some believe that the cyst decompression, rather than the steroids itself, is responsible for healing. The results vary widely and the recurrence or persistence rate ranges from 15-80% after one injection. For the most part, MPA injection alone yields unpredictable results and complete healing after the first injection is uncommon.33 In an attempt to increase healing rates, autologous bone marrow (ABM) and/or demineralized bone matrix (DBM) have been used in combination with MPA, betting on their osteogenic potential to increase the chances of healing.
The literature is inconclusive on which injection material (or combination of) leads to a higher healing rate26., 27., 28., 30., 32., 34. but the addition of bone marrow derivatives does not seem to add a significant benefit and may add donorsite morbidity in the case of ABM.26 Overall, one injection as a standalone procedure tends to be unsuccessful and most patients eventually need multiple injections. In a systematic review, MPA injection had a 71.4% healing rate, while ABM and DBM injection had a 77.9-98% healing rate,7 although many patients required multiple injections.
Surgical Treatment
While indications for surgical treatment are not standardized, surgical intervention can be considered for symptomatic lesions, stress or recurrent fractures, displaced pathologic fractures of lower extremities, lesions posing risk to the physis, or lesions with a high risk for complications such as those around the hip. There is a wide array of techniques available, and the treating surgeon can personalize the treatment plan based on the several available techniques, patient's characteristics, availability, and personal preference. The range of all the treatment modalities are described below and include aspiration and decompression, curettage, grafting, and internal fixation (Table 1).
Table 1. Principles of Treatment of UBC
| Technique | Options | Details | Recommendation |
--- --- |
| Decompression | Aspiration Needle Jamshidi trocar Curette Flexible nail | Decompression is believed to promote healing Cyst fluid confirms diagnosis Pathologic examination possible | Jamshidi or a similar large needle is preferred Perform for all cysts |
| Curettage | Curettes Pituitary rongeurs | Cyst content and lining are removed and sent for pathology Promotes healing | A cystography is performed prior to curettage to ensure complete removal of the septations and lining |
| Graft Materials | Autograft Allograft Bone substitutes Calcium phosphate/sulfate pellets or cement | Provides osteoconductive and osteoinductive material for bone regeneration No graft option is superior to others | Pellets or cement is preferred due to ease of use and unlimited supply |
| Internal fixation | Flexible nail Kirshner wire Screws Side plate + screw | Provides stability to the weakened bone The classic hardware for proximal femoral lesions include screws or side plate + screws depending on age and location of the lesion | Humerus generally does not need fixation unless unstable Side plate + screw is recommended for all proximal femoral lesions |
Aspiration and Decompression
UBC is a closed cavity with fluid under pressure. Therefore, drainage and decompression of the lesion may promote healing. Cyst decompression has historically been performed from an outside-in (extramedullary decompression) technique utilizing large-bore needles (e.g., Jamshidi or similar needle), curettes, or Kirshner wires. Cannulated screws and absorbable pins have also been utilized to provide “continuous decompression” of the cyst35 although there is no evidence that the decompression is actually continuous or that this method leads to a higher healing rate. Intra-medullary decompression with flexible intramedullary nails is an alternative technique.36., 37., 38., 39., 40., 41., 42. The intramedullary hardware violates the cyst membrane and connects the UBC to the normal medullary cavity, which may provide bone marrow as an autologous bone graft. Intramedullary nailing with or without concomitant curettage and bone grafting has been reported to have a high healing rate.7., 36., 37., 38., 39., 40., 41., 42. Additionally, nails may be retained to provide structural support.
Curettage
Curettage of the cyst can be done percutaneously, endoscopic, or open. Curettage not only decompresses the cyst, but also connects the cyst with the adjacent normal medullary canal and removes the cyst lining which may decrease the risk of recurrence and provides pathologic confirmation of the diagnosis. Fluoroscopic guidance, especially when combined with intralesional injection of contrast material for a “cystography,” provides visual confirmation of the nature and extent of the cyst. It also helps to determine if septations are present in the cyst. The minimally invasive technique of percutaneous curettage with the aid of fluoroscopy and bone grafting was introduced by Dormans et al. and is one of the most successful treatment strategies in UBC with excellent short-term results.22 However, a follow-up study of the same patient cohort showed a persistence/recurrence rate of 20%,18 which is similar to other techniques.
Grafting
The use of bone grafting following curettage and decompression provides structural support and may also help to promote a healing response through osteoconductive/osteoinductive properties. While the bone graft and/or bone substitute easily incorporates in the UBC, recurrence or persistence may still occur. The use of autograft, allograft (cancellous and cortical strut), and synthetic bone substitutes have all been reported in the literature with similar results. There is no clear advantage to a particular graft type although bone substitute is more attractive due to their unlimited supply, absence of donor site morbidity, and no risk of disease transmission. Furthermore, several bone substitute formulations can provide immediate structural support to the lesion. The literature abounds with studies on curettage and grafting of UBCs, and although most studies are case series or otherwise low-level evidence, virtually all graft types have been successful (to a certain degree) in promoting healing of the lesion with a lower recurrence rate than decompression and curettage alone.7., 18., 22., 28., 29., 36., 43., 44., 45., 46., 47., 48., 49.
Internal Fixation
Internal fixation is not routinely required as part of the surgical management of UBCs, especially for upper extremity lesions where functional bracing is usually preferred. Intramedullary flexible nails provide sufficient stability for most upper extremity lesions and some lower extremity cysts, but lesions around the hip often require a more stable, rigid fixation for adequate healing and early mobilization. Surgical management of UBCs does not generally require internal fixation.
Surgical Technique
Lesion and patient characteristics dictate the preferred surgical approach. Upper extremity lesions, those in younger children, diaphyseal lesions, and UBCs of the tibia are generally amenable to a percutaneous approach. Most proximal femoral lesions will require a formal open approach, as well as other locations, if surgical exposure is to be performed for internal fixation.
Percutaneous Approach
As previously described,18., 22. the cyst is located with fluoroscopy, a 1 cm incision is made, and a Jamshidi or similar large-bore needle is used to perforate the cyst. The intra-lesional fluid is aspirated to confirm the clear serous or serous-sanguineous appearance. Radiopaque contrast is injected into the cyst to delineate the extent of the cyst. The cystogram also provides visual confirmation of the nature of the cyst, as UBCs lack or have minimal septations. Under fluoroscopic guidance, curettes and pituitary rongeurs can be used to remove the cyst lining and contents, which are sent to pathology. Care is taken not to miss the proximal and distal extents of the cyst. Next, a flexible nail is entered through the same incision and advanced distally (and proximally if possible) to decompress the cyst into the normal medullary canal. The nail is then removed, and a bone graft of choice is used to fill the cyst cavity. After fluoroscopic confirmation, the incision is sutured and temporary immobilization, in the form of a functional brace (e.g., Sarmiento), is recommended (Figure 2).
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Figure 2. Percutaneous treatment of UBC: Lesion is located and perforated (A-B), cystogram and a thorough curettage is performed (C-D), and a flexible nail is used to decompress the lesion (E-F). Bone graft (in this case calcium sulfate pellets) is packed in the cavity (G), and a functional brace is fitted (H). Figure reproduced with permission from CHOP Orthopaedics, Children’s Hospital of Philadelphia, Philadelphia, PA
Open Approach
Similar to the percutaneous approach, fluoroscopy is used to locate the lesion. A cortical window is made over the cyst, and the cavity is curetted thoroughly. While some authors recommend screw or wire fixation for femoral neck lesions, we recommend utilizing fixation with a lateral buttress (plates) for all cysts around the hip. Lesions that extend to the subtrochanteric region may be stabilized with loadsharing devices (intramedullary nails). If the lesion extends to the proximal femoral physis, epiphyseal fixation is recommended, especially in skeletally mature patients. In skeletally immature patients, fixation is preferably stopped short of the physis, or if possible, the physis is crossed with a nonthreaded fixation that would allow continued growth (Figure 3).
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Figure 3. Open approach to UBC: Lesion is located, an oval window is created, and a thorough curettage is performed (A-B). Adjuvant cauterization is optional to achieve extended curettage (C-D). The cavity is filled with bone graft (in this case, calcium cement), and internal fixation is performed. Growth-friendly options with partially threaded screws are also available to prevent growth arrest (F-G). Figure reproduced with permission from CHOP Orthopaedics, Children’s Hospital of Philadelphia, Philadelphia, PA
While surgical management of UBCs may have a higher success rate than observation (natural history) or injection-only, both percutaneous and open approaches still have a recurrence rate of up to 25%. The patient/family should be aware of the chance of persistence or recurrence even with the most comprehensive treatment strategies and that repeated treatment might be required.7 Younger children with active lesions (abutting the physis) have a substantially higher rate of recurrence compared to older patients with latent diaphyseal cysts. The approach to a recurrent lesion is generally similar to an untreated lesion and should be personalized to each patient, considering their function and expectations as well as lesion characteristics. However, the use of internal fixation at the initial surgery, especially in the form of screw/side plate constructs in the proximal femur, may allow the surgeon to be more conservative if a recurrence occurs.
Other Complications and Pitfalls
While the most common, a fracture is not the only complication of UBCs. Recurrence is a major concern that occurs in up to 25% of cases7., 21., 31., 50. and is generally treated similarly to the primary lesion. Active lesions abutting the physis have the potential to grow into the epiphysis through the growth plate. With the “physis at risk,” there is the potential for physeal injury and growth arrest: therefore, surgical treatment may be indicated for such lesions. Depending on the location and extent of physeal injury, angular deformity and/or longitudinal growth disturbance may occur (Figure 4).
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Figure 4. Complications of UBC include pathologic fractures (A), recurrence (B), physeal injury (C), as seen on coronal fat-saturated MRI), and growth disturbance (D). Figure reproduced with permission from CHOP Orthopaedics, Children's Hospital of Philadelphia, Philadelphia, PA
Summary
In conclusion, UBCs have a benign, likely improving or resolving natural history. Most patients will fare well with observation alone. The decision to surgically treat lesions should be made after an in-depth discussion with the family/patient. The surgical approach should similarly be individualized to the patient's demographics, functional needs, and lesion characteristics.
A comprehensive percutaneous or open surgical treatment including decompression, curettage, and grafting, plus internal fixation in select cases, has the highest success rate, although a 20-25% recurrence rate should be expected. The current literature on the topic is generally low-level evidence, which future large, multicenter studies might be able to tackle.
Additional Links
•How I do it and why? Unicameral Bone Cysts Alexandre Arkader, MD; IPOS.
•Pathologic fractures of the Proximal Femur Alexandre Arkader, MD; IPOS
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17371 | https://pubs.acs.org/doi/abs/10.1021/ie200541q | Single-Event MicroKinetics of Aromatics Hydrogenation on Pt/H-ZSM22 | Industrial & Engineering Chemistry Research
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Single-Event MicroKinetics of Aromatics Hydrogenation on Pt/H-ZSM22
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Download Hi-Res ImageDownload to MS-PowerPointCite This:Ind. Eng. Chem. Res. 2011, 50, 23, 12933-12945
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Article May 19, 2011
Single-Event MicroKinetics of Aromatics Hydrogenation on Pt/H-ZSM22
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Tapan Bera†
Joris W. Thybaut†
Guy B. Marin†
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†Laboratory for Chemical Technology, Ghent University, Krijgslaan 281 − S5, B-9000 Ghent, Belgium
Fax: +32 (0)9 264 49 99. E-mail: Joris.Thybaut@UGent.be
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Industrial & Engineering Chemistry Research
Cite this: Ind. Eng. Chem. Res. 2011, 50, 23, 12933–12945
Click to copy citation Citation copied!
Published May 19, 2011
Publication History
Received
17 March 2011
Accepted
19 May 2011
Revised
5 May 2011
Published
online 13 June 2011
Published
in issue 7 December 2011
research-article
Copyright © 2011 American Chemical Society
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Abstract
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A fundamental Single-Event MicroKinetic (SEMK) model for the hydrogenation of aromatic components on a Pt catalyst has been developed. It is based on the Horiuti-Polanyi mechanism considering atomic hydrogen addition steps to the (partially hydrogenated) aromatic species on the catalyst surface. The reaction network used accounts for the position at which the hydrogen atoms are added to the ring. In accordance with a quantum chemical assessment of the reaction pathway it was assumed that the kinetic parameters only depend on the saturation degree of the nearest neighbor carbon atoms and the branching degree of the carbon atom involved in the hydrogen atom addition. Six reactions families, of which three occur in the reaction network for benzene, are considered. The total number of 18 model parameters was reduced to 7 by calculation of the pre-exponential factors and by accounting for thermodynamic constraints. Experimental benzene hydrogenation data measured at temperatures in the range from 423 to 498 K, benzene inlet partial pressures in the range from 10 to 60 kPa, and hydrogen inlet partial pressures from 100 to 600 kPa on Pt catalyst have been regressed. In accordance with quantum chemical, statistical, and thermodynamic calculations, the selected version of the model gives the best description of the data with an F value of 4150. According to this selected SEMK model, the activation energies for the hydrogen addition to a carbon atom between two unsaturated or two saturated carbon atoms are identical and lower than the activation energy for hydrogen addition to a carbon atom between an unsaturated and saturated hydrogen atom. The estimated chemisorption enthalpy of hydrogen amounts to −59.4 kJ mol–1 and corresponds with an average surface coverage of 30%. A value of −56.0 kJ mol–1 for the chemisorption enthalpy of benzene is obtained. The total surface coverage by hydrocarbon species amounts to 60% under typical reaction conditions, without a pronounced Most Abundant Surface Intermediate (MASI).
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Adsorption
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Industrial & Engineering Chemistry Research
Cite this: Ind. Eng. Chem. Res. 2011, 50, 23, 12933–12945
Click to copy citation Citation copied!
Published May 19, 2011
Publication History
Received
17 March 2011
Accepted
19 May 2011
Revised
5 May 2011
Published
online 13 June 2011
Published
in issue 7 December 2011
Copyright © 2011 American Chemical Society
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Article Views
1072
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Citations
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Citations are the number of other articles citing this article, calculated by Crossref and updated daily.Find more information about Crossref citation counts.
The Altmetric Attention Score is a quantitative measure of the attention that a research article has received online. Clicking on the donut icon will load a page at altmetric.com with additional details about the score and the social media presence for the given article. Find more information onthe Altmetric Attention Score and how the score is calculated.
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17372 | https://www.geogebra.org/m/k2j4gcvd | Line Symmetry – GeoGebra
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In this worksheet we will investigate the line symmetry of the letter H.
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17373 | https://www.quora.com/Why-it-is-necessary-when-using-the-gas-laws-to-express-the-temperature-on-the-Kelvin-temperature-scale | Something went wrong. Wait a moment and try again.
Temperature Scale
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Why it is necessary, when using the gas laws, to express the temperature on the Kelvin temperature scale?
·
Using the Kelvin temperature scale when applying gas laws is essential for several reasons:
Absolute Zero
: The Kelvin scale is an absolute temperature scale, with 0 K representing absolute zero, the point at which molecular motion stops. This is critical in gas laws, as many equations rely on temperature being proportional to the average kinetic energy of gas molecules.
Direct Proportionality
: Gas laws, such as the Ideal Gas Law (PV = nRT), require temperature to be in Kelvin because the relationships between pressure, volume, and temperature are linear only when temperature is expressed in Kelv
Using the Kelvin temperature scale when applying gas laws is essential for several reasons:
Absolute Zero
: The Kelvin scale is an absolute temperature scale, with 0 K representing absolute zero, the point at which molecular motion stops. This is critical in gas laws, as many equations rely on temperature being proportional to the average kinetic energy of gas molecules.
Direct Proportionality
: Gas laws, such as the Ideal Gas Law (PV = nRT), require temperature to be in Kelvin because the relationships between pressure, volume, and temperature are linear only when temperature is expressed in Kelvin. In Celsius or Fahrenheit, the zero points are arbitrary and do not reflect the absence of thermal energy.
Mathematical Consistency
: When performing calculations involving gas laws, using Kelvin ensures that the equations yield consistent and accurate results. For example, if temperature is given in Celsius, it can lead to negative values when substituted into equations, which are not physically meaningful in the context of gas behavior.
Standardization
: The scientific community universally uses the Kelvin scale for temperature in thermodynamic equations, providing a standard reference that facilitates communication and comparison of results across different studies and applications.
In summary, expressing temperature in Kelvin is necessary to maintain the physical correctness of gas law calculations and ensure that the laws accurately describe the behavior of gases under varying conditions.
Related questions
Why is Kelvin scale better than Celsius scale temperature?
When making calculations using the ideal gas law: PV=nRT, do you express the temperature in Celsius or in Kelvin?
When solving gas law problems, it is essential that temperatures be converted to the Kelvin scale. Why is the Celsius scale not appropriate for solving gas-law problems?
Why does the temperature in the ideal gas law have to be expressed in Kelvin?
What does the temperature scale really represent? Is 300 kelvin really “twice” the temperature of 150 kelvin? Or is the proper scaling nonlinear (e.g. logarithmic, though I doubt that)? What is being measured? Is it a certain amount of energy?
Rick O’Malley
Author has 4.2K answers and 4.1M answer views
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8y
Because the equations only work with an absolute scale. And because the equations are meant to model observations.
In the ideal gas law, the quantity PV/T is constant, so if we double the temp we must double the product of pressure x volume.
Try using a relative temp scale (either Fahrenheit or Celsius, it doesn't matter).
P = 1 (atm), V = 10 (liters), T = 10 (deg), so PV/T = 1.
What is PV at T = -10? At T = 0?
Both non-sensical answers. (PV can't be negative or infinite.)
No, we use Kelvin (or Rankine) because the equations are all about molecular activity, which is only zero at absolute zero.
Celsi
Because the equations only work with an absolute scale. And because the equations are meant to model observations.
In the ideal gas law, the quantity PV/T is constant, so if we double the temp we must double the product of pressure x volume.
Try using a relative temp scale (either Fahrenheit or Celsius, it doesn't matter).
P = 1 (atm), V = 10 (liters), T = 10 (deg), so PV/T = 1.
What is PV at T = -10? At T = 0?
Both non-sensical answers. (PV can't be negative or infinite.)
No, we use Kelvin (or Rankine) because the equations are all about molecular activity, which is only zero at absolute zero.
Celsius and Fahrenheit are merely scales of convenience.
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Jake Dickerman
HS Chemistry and Physics Teacher
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Author has 1.4K answers and 2.3M answer views
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7y
Let’s start out with a counter example. Why can you use both liters and gallons in a gas law problem? Let’s say I’ve got 4 gallons. That’s 15.14 litters. Let’s say I’ve got 2 gallons. That’s 7.57 litters. What’s the same about both of those? They’re both a 1 : 3.59 ratio. So if I wind up doing a gas laws problem with gallons, it’s exactly like doing a gas laws problem with some number of liters times 3.59. It’s a proportional relationship.
Let’s say now that I compare Celsius and Kelvin in the same way. I’ve got water at boiling temp (100 °C, 373 K). Now let’s double the temperature, except doe
Let’s start out with a counter example. Why can you use both liters and gallons in a gas law problem? Let’s say I’ve got 4 gallons. That’s 15.14 litters. Let’s say I’ve got 2 gallons. That’s 7.57 litters. What’s the same about both of those? They’re both a 1 : 3.59 ratio. So if I wind up doing a gas laws problem with gallons, it’s exactly like doing a gas laws problem with some number of liters times 3.59. It’s a proportional relationship.
Let’s say now that I compare Celsius and Kelvin in the same way. I’ve got water at boiling temp (100 °C, 373 K). Now let’s double the temperature, except does that mean I go to 200 °C, which is 474 K or does it mean I go to 746 K, which is 473 °C?
Gas laws are proportional relationships, but there is no proportional relationship between Celsius and Kelvin. What we have instead is a 1:1 correspondence. This is also why there is no proportional relationship between Celsius and Farenheit. The conversion between the two includes an addition or subtraction of 32, so they correspond, but there is not a proportionality between the two. Going back to your question, this tells us that between Celsius and Kelvin, there is no way that both can work in a gas laws problem.
So if we can’t use both, how do we know that the one we should use is Kelvin and not °Celsius? To understand that, let’s think about what zero means in our different units. What is zero liters? It’s no volume. What is zero mol? It’s no amount. What is zero °C. It’s… the temperature where water freezes. That’s sort of a random thing.
Zero K on the other hand, that’s a zero energy state. It’s not possible, but it is clear. Kelvin is an absolute scale. You could also, if you really wanted to, use the Rankine scale. Those two scales should have proportional relationships to each other and both have the same zero.
Walden Bjørn-Yoshimoto
Postdoc at University of Copenhagen
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Author has 890 answers and 2.7M answer views
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8y
It is not.
But if you use any other temperature, you might need to adjust everything else accordingly.
Let me exemplify.
The molar gas constant is 8.314472 J/(moK). That is the same as 8.314472 J/(mol°C), because this constant is per unit, it doesn’t assume a zero point, and the difference between 1 and 2 degrees K is the same as the difference between 1 and 2 degrees C. But if you use Fahrenheit, then all of a sudden, a change of 1°F is equal to a change of 5/9 degrees K (or degrees C). So now the molar gas constant, expressed in°F is (8.3144725/9) = 4.619151 J/(mol°F).
However, it is importa
It is not.
But if you use any other temperature, you might need to adjust everything else accordingly.
Let me exemplify.
The molar gas constant is 8.314472 J/(moK). That is the same as 8.314472 J/(mol°C), because this constant is per unit, it doesn’t assume a zero point, and the difference between 1 and 2 degrees K is the same as the difference between 1 and 2 degrees C. But if you use Fahrenheit, then all of a sudden, a change of 1°F is equal to a change of 5/9 degrees K (or degrees C). So now the molar gas constant, expressed in°F is (8.3144725/9) = 4.619151 J/(mol°F).
However, it is important to use a scale that starts at zero. Not at some randomly defined zero (such as Fahrenheit or Celsius), but actually at zero. The temperature in the gas laws (let’s talk about the ideal gas law, for simplicity) relate temperature to pressure and volume. The pressure/volume is related to how fast the molecules move (temperature), so the only zero point that can make sense to use as zero, is absolute zero (where the molecules don’t move). Everything else is arbitrary, and would need adjustments in the formulas to accommodate (and those adjustments would be to ‘relate’ to absolute zero anyway, so it wouldn’t make sense. E.g., if using °C, you would need to ‘adjust’ by expressing “°C - 273.15°C”; for Fahrenheit you would need to use “°F - 459.67 °F”).
But using the modified gas constant we calculated above, we can use the fact that a change of 1 degree Fahrenheit equals a change of 1 degree on the (horrible) Rankine scale. So there is nothing at all wrong with using:
pV = nRT,
where T is expressed in Rankine, and R is 4.619151 J/(mol°R).
And you can do the same with any scale you define, as long as it has 0K (absolute zero) as zero.
Related questions
Why isn't it possible to have temperature less than -273 Kelvin?
How do I convert temperature to the Kelvin scale?
Why is the Kelvin scale called a thermodynamic temperature scale?
What are the highest and lowest possible temperatures on the Kelvin scale?
Why do we write small t (temperature) for the Kelvin scale?
Gary Ahlers
Lapidary, HVAC Design, Astrophotographer, Constr'n (1975–present)
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Author has 8.1K answers and 6.5M answer views
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3y
Originally Answered: When solving gas law problems, it is essential that temperatures be converted to the Kelvin scale. Why is the Celsius scale not appropriate for solving gas-law problems?
·
The Kelvin scale is used in gas law problems because the pressure and volume of a gas depend on the kinetic energy or motion of the particles. The Kelvin scale is proportional to the KE of the particles, that is, 0 K (absolute zero) means 0 kinetic energy.
For all gas law problems it is necessary to work in the Kelvin scale because temperature is in the denominator in the combined gas laws (P/T, V/T and PV/T) and can be derived in the ideal gas law to the denominator (PV/RT) . If we measured temperature in celsius we could have a value of zero degrees celsius and this would solve as no solution
The Kelvin scale is used in gas law problems because the pressure and volume of a gas depend on the kinetic energy or motion of the particles. The Kelvin scale is proportional to the KE of the particles, that is, 0 K (absolute zero) means 0 kinetic energy.
For all gas law problems it is necessary to work in the Kelvin scale because temperature is in the denominator in the combined gas laws (P/T, V/T and PV/T) and can be derived in the ideal gas law to the denominator (PV/RT) . If we measured temperature in celsius we could have a value of zero degrees celsius and this would solve as no solution, as you cannot have zero in the denominator.
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Robert Mueller
Reader,writer at This Site Not Employed in the QPP
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Author has 2.8K answers and 3.3M answer views
·
3y
Originally Answered: When solving gas law problems, it is essential that temperatures be converted to the Kelvin scale. Why is the Celsius scale not appropriate for solving gas-law problems?
·
The Kelvin scale places the absolute zero temperature at 0 K, where is the imaginary case of a truly ideal gas the volume would shrink to zero. That does not happen at the temperature indicated as 0 °C on our household thermometers. If you are happier you could conceptually use Celsius scale temperatures but you would have to drag around and properly handle that 273.15 K difference by adding it onto all Celsius temperatures in the gas law. I believe you will quickly decide doing the math on the Kelvin scale and converting only before calculating and again after is more convenient.
Akash Jaiswal
Studying science
·
9y
Originally Answered: Why is Kelvin scale so important to be used while dealing with ideal gases?
·
Celcius scale
When first celcius scale was marked as standard measurement for temperature (that time only way to measure temperature). It was done so by taking water. The temperature at which water freezers was taken to be 0° celcius and temperature at which it boiled was marked 100° celcius. Now the the mean length was divided into 100 equal parts.
Fahrenheit Scale
The problem with the celcius scale was it had large temperature difference even between 1°C which made it less accurate. So a new standard was to be mentioned. Then BRINE (Equal amt of ice and salt) was taken. Its freezing point
Celcius scale
When first celcius scale was marked as standard measurement for temperature (that time only way to measure temperature). It was done so by taking water. The temperature at which water freezers was taken to be 0° celcius and temperature at which it boiled was marked 100° celcius. Now the the mean length was divided into 100 equal parts.
Fahrenheit Scale
The problem with the celcius scale was it had large temperature difference even between 1°C which made it less accurate. So a new standard was to be mentioned. Then BRINE (Equal amt of ice and salt) was taken. Its freezing point and boiling points were taken. Now in that scale freezing point of water was 32°F and 212°F was its boiling point. So now the difference between freezing and boiling point was divided into 180 parts. This gave more accuracy in temperature measurement and revolutionized the body temperature measurement method to find fever.
Kelvin Scale
All went good till Lord Kelvin decided to find relation between different parameters of bulk of molecules. He found that even at 0°C the matter had some Kinetic energy and as Kinetic energy is a property of temperature. There must be some temperature when the molecules have random movement. So to find that he made the volume constant and it was observed that on one degree rise in temperature in celcius scale. The increment in pressure was 1/273.17 atm. So according to his hypothesis if the temperature would be decreased to 273.17 then pressure shall be decrease by (273.17/273.17=1) 1.i.e now the pressure shall be zero. Which means no Kinetic energy in the molecules. Hence all the earlier equations face the experimental value putting this absolute temperature.
HENCE, IT WAS A MAJOR LEEP IN PHYSICAL CHEMISTRY, AND THERMODYNAMICS WOULD NOT HAVE EXISTED WITHOUT THIS.
Regarding your question whatever process you take it will work only in absolute temperature. Due to the above mentioned reason.
Sorry for being that long. :p
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Randall Marks
Former Teacher (1965–2007)
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Author has 86 answers and 58.4K answer views
·
4y
Originally Answered: Why should the Kelvin scale be used for gas law calculations?
·
The Kelvin scale is directly related to the energy of the moving particles whereas celcius and fahrenheit are not. When a subatance is at 0K the energy is theoretically zero. In the other 2 scales there is plenty of energy in the substances at 0C and 0F. If you double the Kelvin temp of a gas you will double the average kinetic energy of it’s molecules that will not be the case with C or F. Example if you heat a gas from 5C to 10C you do not double the energy- in using Kelvin you will actually be heating from 278K to 283K which is not even close to doubling. Remember to convert from celcius to
The Kelvin scale is directly related to the energy of the moving particles whereas celcius and fahrenheit are not. When a subatance is at 0K the energy is theoretically zero. In the other 2 scales there is plenty of energy in the substances at 0C and 0F. If you double the Kelvin temp of a gas you will double the average kinetic energy of it’s molecules that will not be the case with C or F. Example if you heat a gas from 5C to 10C you do not double the energy- in using Kelvin you will actually be heating from 278K to 283K which is not even close to doubling. Remember to convert from celcius to Kelvin just add 273
Pranav Gupta
Wanna visit Aliens, the Star Trek way.
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Author has 139 answers and 105.9K answer views
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9y
Originally Answered: Why is Kelvin scale so important to be used while dealing with ideal gases?
·
Earleir (before the introduction of Kelvin scale), degree Celcius and degree Farenhiet were the common scales. It was noticed that at a constant volume, Decreasing temprature decreased the pressure of the gas.But it was observed that even at zero degree Celcius, there was some existing pressure of the gas. So a relation was formed i.e. P(atm) ∝ T (degree c)+ k(some constant) at constant volume. After some practical experiments, the value of this constant(k) was found to be 273.17. This was found be William Thomson a.k.a Lord Kelvin. Hence the Kelvin scale was introduced as it could be used as
Earleir (before the introduction of Kelvin scale), degree Celcius and degree Farenhiet were the common scales. It was noticed that at a constant volume, Decreasing temprature decreased the pressure of the gas.But it was observed that even at zero degree Celcius, there was some existing pressure of the gas. So a relation was formed i.e. P(atm) ∝ T (degree c)+ k(some constant) at constant volume. After some practical experiments, the value of this constant(k) was found to be 273.17. This was found be William Thomson a.k.a Lord Kelvin. Hence the Kelvin scale was introduced as it could be used as it is in the equations. Hope this helped.
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Dennis Leppin
B.E.and M.E. in Chemical Engineering, City College of New York
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Author has 2.2K answers and 1.6M answer views
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Aug 18
Originally Answered: When solving gas law problems, it is essential that temperatures be converted to the Kelvin scale. Why is the Celsius scale not appropriate for solving gas-law problems?
·
How would that work at negative temperatures like minus 50 C. That would require either negative pressures or negative volumes. Both of those are physically meaningless. One can actually prove the absolute scale of temperature is the correct one to use. You would need to study a thermodynamics textbook and given that your question is rather naive, I’ll assume that level of effort might be beyond your ken.
Geoffrey Widdison
Been learning science since I was six, still getting started.
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Author has 14.4K answers and 156.9M answer views
·
12y
Because negative temperatures don't make sense in science.
Celsius and Fahrenheit scales both have zero set at some arbitrary point, so you have negative and positive numbers which have meaning only by convention.The Kelvin scale, on the other hand, is an absolute scale, where zero means that no thermal energy is present. Thus, doubling the temperature means you have twice as much thermal energy. Temperature has to be calculated in an absolute scale for gas laws to work.
On a side note, there are other absolute scales you can use (such as the Rankine scale) which would work for the gas law
Because negative temperatures don't make sense in science.
Celsius and Fahrenheit scales both have zero set at some arbitrary point, so you have negative and positive numbers which have meaning only by convention.The Kelvin scale, on the other hand, is an absolute scale, where zero means that no thermal energy is present. Thus, doubling the temperature means you have twice as much thermal energy. Temperature has to be calculated in an absolute scale for gas laws to work.
On a side note, there are other absolute scales you can use (such as the Rankine scale) which would work for the gas laws, but those are so rarely used that most people just stick with Kelvin.
César Tomé López
Studied at University of Granada
·
12y
Originally Answered: Why does the application of Charles's law and the combined gas law require the use of kelvin?
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1 It is not a convention as such. Only the size of the unit (kelvin) is conventional.
2 Charles' law and, consequently, the CGL are fulfilled more precisely in the limit of pressure 0.
3 In this limit the deviations of Charles' Law are the same for different gases. Every gas shows the same behaviour temperature vs. volume at constant pressure in this limit.
4 If you graph volume vs. temperature (it doesn't matter the temperature scale now) at very low pressures and extrapolate till you find that the curve cuts the temperature axis, you'll find that all gases cut at exactly the same point, namely
1 It is not a convention as such. Only the size of the unit (kelvin) is conventional.
2 Charles' law and, consequently, the CGL are fulfilled more precisely in the limit of pressure 0.
3 In this limit the deviations of Charles' Law are the same for different gases. Every gas shows the same behaviour temperature vs. volume at constant pressure in this limit.
4 If you graph volume vs. temperature (it doesn't matter the temperature scale now) at very low pressures and extrapolate till you find that the curve cuts the temperature axis, you'll find that all gases cut at exactly the same point, namely, using degrees Celsius now, -273,15 ºC.
5 So it makes sense to define a temperature scale which is a) independent of the substance you use and b) fulfills completely Charles' Law at very low pressures. This is the abolute temperature scale and it uses kelvins.
6 Henceforth all gas laws, actually all thermodynamics, use this reference in its formulae and constants.
7 This does not mean you can not use Celsius or Fahrenheit or whatever, you only need to be aware of the a) algebra b) value of the constants c) system limitations that doing it would imply.
Ajay Prakash
M.Phil. in Physics & Atmospheric Science, Indian Institute of Technology, Kanpur (IITK) (Graduated 1980)
·
3y
Originally Answered: When solving gas law problems, it is essential that temperatures be converted to the Kelvin scale. Why is the Celsius scale not appropriate for solving gas-law problems?
·
Depends on what units you using. Check the Gas Constant units. Normally defined as —
8.31446261815324 kg⋅m2⋅s−2⋅K−1⋅mol−
Convert this to Centigrade then use it in an equation where temperatures are in centigrade.
Related questions
Why is Kelvin scale better than Celsius scale temperature?
When making calculations using the ideal gas law: PV=nRT, do you express the temperature in Celsius or in Kelvin?
When solving gas law problems, it is essential that temperatures be converted to the Kelvin scale. Why is the Celsius scale not appropriate for solving gas-law problems?
Why does the temperature in the ideal gas law have to be expressed in Kelvin?
What does the temperature scale really represent? Is 300 kelvin really “twice” the temperature of 150 kelvin? Or is the proper scaling nonlinear (e.g. logarithmic, though I doubt that)? What is being measured? Is it a certain amount of energy?
Why isn't it possible to have temperature less than -273 Kelvin?
How do I convert temperature to the Kelvin scale?
Why is the Kelvin scale called a thermodynamic temperature scale?
What are the highest and lowest possible temperatures on the Kelvin scale?
Why do we write small t (temperature) for the Kelvin scale?
Which is a better temperature scale, Kelvin or Rankine?
How do you convert the temperature of 25 degrees to the Kelvin scale?
Why is the temperature in Kelvin 273 degrees?
What is Kelvin temperature scale normally used for?
Why is temperature used to measure gas?
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17374 | https://arxiv.org/pdf/1506.02555 | EIGENVALUES FOR MAXWELL’S EQUATIONS WITH DISSIPATIVE BOUNDARY CONDITIONS
FERRUCCIO COLOMBINI, VESSELIN PETKOV † AND JEFFREY RAUCH
Abstract. Let V (t) = etG b , t ≥ 0, be the semigroup generated by Maxwell’s equations in an exterior domain Ω ⊂ R3 with dissipative boundary condition
Etan − γ(x)( ν ∧ Btan ) = 0 , γ (x) > 0, ∀x ∈ Γ = ∂Ω. We prove that if γ(x) is nowhere equal to 1, then for every 0 < 1 and every N ∈ N the eigenvalues of Gb lie in the region Λ ∪ R N , where Λ = {z ∈ C : | Re z| ≤ C(| Im z| 12 + +1) , Re z < 0}, RN = {z ∈ C : | Im z| ≤ CN (| Re z| + 1) −N , Re z < 0}.
Introduction
Suppose that K ⊂ { x ∈ R3 : |x| ≤ a} is an open connected domain and Ω := R3 \ ¯K is an open connected domain with C∞ smooth boundary Γ. Consider the boundary problem
∂tE = curl B, ∂tB = −curl E in R+
t
× Ω,Etan − γ(x)( ν ∧ Btan ) = 0 on R+
t
× Γ,E(0 , x ) = e0(x), B(0 , x ) = b0(x).
(1.1) with initial data f = ( e0, b 0) ∈ (L2(Ω)) 6 = H. Here ν(x) denotes the unit outward normal to ∂Ω at x ∈ Γ pointing into Ω, 〈, 〉 denotes the scalar product in C3, utan :=
u − 〈 u, ν 〉ν, and γ(x) ∈ C∞(Γ) satisfies γ(x) > 0 for all x ∈ Γ. The solution of the problem (1.1) is given by a contraction semigroup ( E, B ) = V (t)f = etG b f, t ≥ 0, where the generator Gb has domain D(Gb) that is the closure in the graph norm of functions u = ( v, w ) ∈ (C∞
(0)
(R3)) 3 × (C∞
(0)
(R3)) 3 satisfying the boundary condition
vtan − γ(ν ∧ wtan ) = 0 on Γ .
In an earlier paper we proved that the spectrum of Gb in Re z < 0 consists of isolated eigenvalues with finite multiplicity. If Gbf = λf with Re λ < 0, the solution
u(t, x ) = V (t)f = eλt f (x) of (1.1) has exponentially decreasing global energy. Such solutions are called asymptotically disappearing and they are invisible for inverse scattering problems. It was proved that if there is at least one eigenvalue
λ of Gb with Re λ < 0, then the wave operators W± are not complete, that is Ran W − 6 = R an W +. Hence we cannot define the scattering operator S related to the Cauchy problem for the Maxwell system and (1.1) by the product W −1+ W−.For the perfect conductor boundary conditions for Maxwell’s equations, the energy is conserved in time and the unperturbed and perturbed problems are associated to unitary groups. The corresponding scattering operator S(z) : ( L2(S2)) 2 →
(L2(S2)) 2 satisfies the identity
S−1(z) = S∗(¯ z), z ∈ C (1.2)
2000 Mathematics Subject Classification. Primary 35P20, Secondary 47A40, 35Q61.
†
The author was partially supported by the ANR project Nosevol BS01019 01 .
1
arXiv:1506.02555v4 [math.AP] 4 Aug 2016 2F. COLOMBINI, V. PETKOV AND J. RAUCH x
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Figure 1. Eigenvalues of Gb
if S(z) is invertible at z. The scattering operator S(z) defined in is such that
S(z) and S∗(z) are analytic in the ”physical” half plane {z ∈ C : Im z < 0} and the above relation for conservative boundary conditions implies that S(z) is invertible for Im z > 0. For dissipative boundary conditions the relation (1.2) in general is not true and S(z0) may have a non trivial kernel for some z0, Im z0 > 0. Lax and Phillips proved that this implies that iz0 is an eigenvalue of Gb. The analysis of the location of the eigenvalues of Gb is important for the location of the points where the kernel of S(z) is not trivial. The main result of this paper is the following (see Figure 1)
Theorem 1.1. Assume that for all x ∈ Γ, γ(x) 6 = 1 . Then for every 0 < 1
and every N ∈ N there are constants C > 0 and CN > 0 such that the eigenvalues of Gb lie in the region Λ ∪ R N , where
Λ = {z ∈ C : | Re z| ≤ C(| Im z|1/2+ + 1) , Re z < 0},
RN = {z ∈ C : | Im z| ≤ CN (| Re z| + 1) −N , Re z < 0}.
If Re λ < 0 and Gb(E, B ) = λ(E, B ) 6 = 0, then
λE = curl B on Ω,λB = −curl E on Ω,
div E = div B = 0 , on Ω,Etan − γ(ν ∧ Btan ) = 0 on Γ.
(1.3) This implies that u := ( E, B ) satisfies ∆u − λ2u = 0 , on Ω.
The eigenvalues of Gb are symmetric with respect to the real axis, so it is sufficient to examine the location of the eigenvalues whose imaginary part is nonnegative. The mapping z 7 → z2 maps the positive quadrant {z ∈ C : Re z > 0 , Im z > 0}
bijectively to the upper half space. Denote by √z the inverse map. The part of the MAXWELL EIGENVALUES 3
hδ
Z1
Z2
Z3
Figure 2. Contours Z1, Z 2, Z 3, δ = 1 /2 −
spectral domain {λ ∈ C : Re λ < 0 , Im λ > 0} is mapped by λ = i√z to the upper half plane {z ∈ C : Im z > 0}. In {z ∈ C : Im z ≥ 0} introduce the sets
Z1 := {z ∈ C : Re z = 1 , hδ ≤ Im z ≤ 1}, 0 < h 1, 0 < δ < 1/2,Z2 := {z ∈ C : Re z = −1, 0 ≤ Im z ≤ 1},Z3 := {z ∈ C : | Re z| ≤ 1, Im z = 1 }.
Set λ = i√z/h, z ∈ Z1 ∪ Z2 ∪ Z3. To study the eigenvalues λ, |λ| > R 0, it is sufficient to consider 0 < h 1. As z runs over the rectangle in Figure 2, with 0 < h 1, λ sweeps out the large values in the intersection of left and upper half planes. The values of z ∈ Z2 near the lower left hand corner, z = −1, of the rectangle go the spectral values near the negative real axis. The spectral analysis near these values in Z2 for dissipative Maxwell’s equations does not have clear analogue with the spectral problems for the wave equation with dissipative boundary conditions. In fact, for the wave equation if 0 < γ (x) < 1, ∀x ∈ Γ,
the eigenvalues of the generator of the corresponding semigroup are located in the domain Λ (see Section 3, and ). For Maxwell’s equations the eigenvalues of
Gb lie in the domain Λ ∪ R N and for 0 < γ (x) < 1 and γ(x) > 1 we have the same location (see Appendix for the case K = {x ∈ R3 : ‖x| ≤ 1}). Equation (1.3) implies that on Ω each eigenfunction u = ( E, B ) of Gb satisfies
√z E = h
i curl B , √z B = − h
i curl E , (1.4) and therefore ( −h2∆ − z)E = ( −h2∆ − z)B = 0 . For eigenfunctions ( E, B ) 6 = 0, we derive a pseudodifferential system on the boundary involving Etan = E − 〈 E, ν 〉ν
and Enor = 〈E, ν 〉. A semi-classical analysis shows that for z ∈ Z1 ∪ Z3 this system implies that for h small enough we have E|Γ = 0 which yields E = B = 0. By scaling one concludes that the eigenvalues λ = i√zh of Gb lie in the region Λ ∪ M ,where
M = {z ∈ C : | arg z − π| ≤ π/ 4, |z| ≥ R0 > 0, Re z < 0}.
The strategy for the analysis of the case z ∈ Z1 ∪ Z3 is similar to that ex-ploited in and . In these papers the semi-classical Dirichlet-to-Neumann map N (z, h ) plays a crucial role and the problem is reduced to the proof that some 4 F. COLOMBINI, V. PETKOV AND J. RAUCH
h−pseudodifferantial operators is elliptic in a suitable class. For the Maxwell sys-tem the pseudodifferential equation on the boundary is more complicated. Using the equation div E = 0 , yields a pseudodifferential system for Etan and Enor . We show that if ( E, B ) 6 = 0 is an eigenfunction of Gb, then ‖Enor ‖H1
h(Γ)
is bounded by Ch ‖Etan ‖H1
h(Γ)
. The term involving Enor then plays the role of a negligible perturbation in the pseudodifferentrial system on the boundary and this reduces the analysis to one involving only Etan . The system concerning Etan has a diagonal leading term and we may apply the same arguments as those of to conclude that
Etan = 0 and hence Enor = 0 .
The analysis of the case z ∈ Z2 is more difficult since the principal symbol g
of the pseudodifferential system for Etan need not be elliptic at some points (see Section 3). Even where g is elliptic, if | Im z| ≤ h1/2 it is difficult to estimate the norm of the difference Op h(g)Op h(g−1) − I. To show that the eigenvalues of Gb
lying in M are in fact confined to the region RN for every N ∈ N, we analyze the real part of the following scalar product in L2(Γ)
Q(E0) := Re 〈(N (z, h ) − √zγ )E0, E 0〉L2(Γ) , E0 := E|Γ.
We follow the approach in , based on a Taylor expansion of Q(E0) at z = −1and the fact that for z = −1 we have Q(E0) = O(hN ), ∀N ∈ N. In the Appendix we treat the case when K = {x ∈ R3 : |x| ≤ 1} is a ball and γ = const. We prove that for γ ≡ 1 the operator G has no eigenvalues in {Re z < 0}, while for every
γ ∈ R+ \ { 1} we have infinite number of real eigenvalues. 2. Pseudodifferential equation on the boundary
Introduce geodesic normal coordinates ( y1, y ′) ∈ R3 on a neighborhood of a point
x0 ∈ Γ as follows. For a point x, y′(x) is the closest point in Γ and y1 = dist ( x, Γ). Define ν(x) to be the unit normal in the direction of increasing y1 to the surface
y1 = constant through x. Thus ν(x) is an extension of the unit normal vector to a unit vector field. The boundary Γ is mapped to y1 = 0 and
x = α(y1, y ′) = β(y′) + y1ν(y′).
We have
∂∂x k
= νk(y′) ∂∂y 1
+
3
∑
j=2
∂y j
∂x k
∂∂y j
, k = 1 , 2, 3.
Moreover,
3
∑
k=1
νk(y′) ∂y j
∂x k
(y1, y ′) = 〈ν, ∂y j
∂x 〉 = 0 , j = 1 , 2, 3, and
3
∑
k=1
νk(x)∂xk f (x) = ∂y1 (f (α(y1, y ′)) .
Since ‖ν(x)‖ = 1, 〈ν, ∂ xj ν〉 = 0 , j = 1 , 2, 3.
A straight forward computation yields
ν(x) ∧ h
i curl u(x) = ih∂ ν utan +
(
〈Dx1 u, ν 〉, 〈Dx2 u, ν 〉, 〈Dx3 u, ν 〉
)∣ ∣∣tan
= ih∂ ν utan +
(
grad h〈u, ν 〉
)∣ ∣∣tan
− ihg 0(utan ), x ∈ Γ,MAXWELL EIGENVALUES 5
where
Dxj = −ih∂ xj , j = 1 , 2, 3, grad hf = {Dxj f }j=1 ,2,3,g0(utan ) = {〈 utan , ∂ xj ν〉} j=1 ,2,3.
Setting Enor = 〈E, ν 〉, from (1.3) one deduces
ν ∧ B = − 1
√z ν ∧ h
i curl E = 1
√z Dν Etan − 1
√z
[(
grad hEnor
)∣ ∣∣tan
− ihg 0(Etan )
]
,
where Dν = −ih∂ ν and the boundary condition in (1.3) becomes
(
Dν − 1
γ
√z
)
Etan −
(
grad hEnor
)∣ ∣∣tan
ihg 0(Etan ) = 0 , x ∈ Γ. (2.1) Next grad hf (x)|tan =
{ 3∑
j=2
∂y j
∂xk
Dyj f (α(y1, y ′))
}
k=1 ,2,3
and for u = ( u1, u 2, u 3) ∈ C3,
h
i div u(α(y1, y ′)) = 〈Dy1 u(α(y1, y ′)) , ν (y′)〉 +
3
∑
k=1 3
∑
j=2
∂y j
∂xk
Dyj uk(α(y1, y ′)) = Dy1
(
unor (y1, y ′)
)
+
3
∑
j=2
Dyj
〈
utan (α(y1, y ′)) , ∂y j
∂x
〉
h〈utan , Z 〉,
where 〈u(α(y1, y ′)) , ν (y′)〉 := unor (y1, y ′) and Z depends on the second derivatives of yj , j = 2 .3. Apply the operator Dy1 − √zγ(y′) to div E(α(y1, y ′)) = 0 to find (D2
y1
−√zγ(y′) Dy1 )Enor (y1, y ′) +
3
∑
j=2
Dyj
〈
(Dy1 −√zγ(y′) )Etan (α(y1, y ′)) , ∂y j
∂x
〉
= h〈(Dy1 −√zγ )Etan , Z 〉 + h〈Etan , Z 1〉,
where γ(y′) := γ(β(y′)) .
Taking the trace y1 = 0 and applying the boundary condition (2.1), yields
(
D2
y1
+
3
∑
j,μ =2 3
∑
k=1
∂y j
∂x k
∂y μ
∂x k
D2
yj,y μ
)
Enor (0 , y ′) −√zγ(y′) Dy1 Enor (0 , y ′)= h
〈(
grad hEnor
)∣∣tan (0 , y ′), Z
〉
hQ 1(Etan (0 , y ′)) , (2.2) with
‖Q1(Etan (0 , y ′)) ‖L2(R2) ≤ C2‖Etan (0 , y ′)‖H1
h(R2)
.
Here Hsh(Γ) , s ∈ R, denotes the semi-classical Sobolev spaces with norm ‖〈 h∂ x〉su‖L2(Γ) ,
〈h∂ x〉 = (1 + ‖h∂ x‖2)1/2. In the exposition below we use the spaces ( L2(Γ)) 3 and (Hsh(Γ)) 3 of vector-valued functions but we will omit this in the notations writing simply L2(Γ) and Hsh(Γ). The operator −h2∆x − z in the coordinates ( y1, y ′) has the form
P(z, h ) = D2
y1
r(y, D y′ ) + q1(y, D y ) + h2 ˜q − z6 F. COLOMBINI, V. PETKOV AND J. RAUCH
with r(y, η ′) = 〈R(y)η′, η ′〉, q 1(y, η ) = 〈q1(y), η )〉. Here
R(y) =
{ 3∑
k=1
∂y j
∂x k
∂y μ
∂x k
}3
j,μ =2
=
{〈 ∂y j
∂x , ∂y μ
∂x
〉} 3
j,μ =2
is a symmetric (2 × 2) matrix and r(0 , y ′, η ′) = r0(y′, η ′), where r0(y′, η ′) is the principal symbol of the Laplace-Beltrami operator −h2∆Γ on Γ equipped with the Riemannian metric induced by the Euclidean one in R3. We have
(
P(z, h )Enor
)
(0 , y ′) = 〈P (z, h )E, ν 〉(0 , y ′) + hQ 2(E(0 , y ′)) ,
where
‖Q2(E(0 , y ′)) ‖L2(R2) ≤ C2 ‖E(0 , y ′)‖H1
h(R2)
.
Since P(z, h )E = 0, this lets us replace the terms with all second derivatives of
Enor in (2.4) by zE nor (0 , y ′) modulo terms having a factor h and containing first order derivatives of Enor . This follows from the form of the matrix R(y) given above. After a multiplication by − γ(y′)
√z
the equation (2.2) yields (Dy1 − γ(y′)√z)Enor (0 , y ′) = hQ 3(E(0 , y ′)) , (2.3) where Q3(E(0 , y ′)) has the same properties as Q2(E(0 , y ′)) .
Let ψ(x) ∈ C∞
0
(R3) be a cut-off function with support in small neighborhood of
x0 ∈ Γ. Replace E, B by Eψ = Eψ, B ψ = Bψ. The above analysis works for Eψ
and Bψ with lower order terms depending on ψ. We obtain
〈(Dν − γ(x)√z)E|Γψ(x), ν (x)〉 = h Q 3,ψ (E|Γ).
Taking a partition of unity in a neighborhood of Γ, yields
〈(Dν − γ(x)√z)E|Γ, ν 〉 = hQ 4(E|Γ), ‖Q4(E|Γ)‖L2(Γ) ≤ C‖E|Γ‖H1
h(Γ)
. (2.4) For z ∈ Z1 ∪ Z2 ∪ Z3 let ρ(x′, ξ ′, z ) = √z − r0(x′, ξ ′) ∈ C∞(T ∗Γ) be the root of the equation
ρ2 + r0(x′, ξ ′) − z = 0 with Im ρ(x′, ξ ′, z ) > 0. For large |ξ′|,
ρ(x′, ξ ′, z ) ∼ |ξ′|, Im ρ(x′, ξ ′, z ) ∼ |ξ′|,
while for bounded |ξ′|,Im ρ(x′, ξ ′, z ) ≥ hδ
C .
We recall some basic facts about h-pseudodifferential operators that the reader can find in . Let X be a C∞ smooth compact manifold without boundary with dimension d ≥ 2. Let ( x, ξ ) be the coordinates in T ∗(X) and let a(x, ξ, h ) ∈
C∞(T ∗(X)) . Given m ∈ R, l ∈ R, δ > 0 and a function c(h) > 0, one denotes by
Sl,m δ (c(h)) the set of symbols so that
|∂αx ∂βξ a(x, ξ, h )| ≤ Cα,β (c(h)) −l−δ(|α|+|β|)(1 + |ξ|)m−| β|, ∀α, ∀β, (x, ξ ) ∈ T ∗(X).
If c(h) = h, we denote Sl,m δ (c(h)) simply by Sl,m δ . Symbols restricted to a domain where |ξ| ≤ C will be denoted by a ∈ Slδ (c(h)) . The h−pseudodifferential operator MAXWELL EIGENVALUES 7
with symbol a(x, ξ, h ) acts by (Op h(a)f )( x) := (2 πh )−d+1
∫
T∗X
e−i〈x−y,ξ 〉/h a(x, ξ, h )f (y)dydξ.
For matrix valued symbols we use the same definition. This means that every element of a matrix symbol is in the class Sl,m δ (c(h)) .
Now suppose that a(x, ξ, h ) satisfies the estimates
|∂αx a(x, ξ, h )| ≤ c0(h)h−| α|/2, (x, ξ ) ∈ T ∗(X) (2.5) for |α| ≤ d − 1, where c0(h) > 0 is a parameter. Then there exists a constant C > 0independent of h such that
‖Op h(a)‖L2(X)→L2(X) ≤ C c 0(h). (2.6) For 0 ≤ δ < 1/2 products of h-pseudodifferential operators are well behaved. If
a ∈ Sl1,m 1
δ
, b ∈ Sl2,m 2
δ
and s ∈ R, then
‖Op h(a)Op h(b) − Op h(ab )‖Hs(X)→Hs−m1−m2+1 (X) ≤ Ch −l1−l2−2δ+1 . (2.7) Let u ∈ C3 be the solution of the Dirichlet problem (−h2∆ − z)u = 0 on Ω, u = F on Γ. (2.8) Introduce the semi-classical Dirichlet-to-Neumann map
N (z, h ) : Hsh(Γ) 3 F −→ Dν u|Γ ∈ Hs−1
h
(Γ) .
G. Vodev established for bounded domains K ⊂ Rd, d ≥ 2, with C∞ boundary the following approximation of the interior Dirichlet-to-Neumann map Nint (z, h )related to (2.8), where the equation ( −h2∆ − z)u = 0 is satisfied in K.
Theorem 2.1 () . For every 0 < 1 there exists 0 < h 0() 1 such that for
z ∈ Z1, := {z ∈ Z1, | Im z| ≥ h 12 −} and 0 < h ≤ h0() we have
‖N int (z, h )( F ) − Op h(ρ + hb )F ‖H1
h(Γ)
≤ Ch
√| Im z| ‖F ‖L2(Γ) , (2.9)
where b ∈ S00,1(Γ) does not depend on h and z. Moreover, (2 .9) holds for z ∈ Z2 ∪Z3
with | Im z| replaced by 1.
With small modifications (2.9) holds for the Dirichlet-to-Neumann map N (z, h )related to (2.8) (see ). Applying (2.9) with N (z, h ) and F = E0 = E|Γ, we obtain
∥∥∥〈N (z, h )E0, ν 〉 − 〈 Op h(ρ)E0, ν 〉
∥∥∥L2(Γ)
≤ Ch
√| Im z| ‖E0‖L2(Γ) . (2.10) Therefore (2.4) yields
∥∥∥〈Op h(ρ) − γ√z)E0, ν 〉 − hQ 4(E0)
∥∥∥L2(Γ)
≤ Ch
√| Im z| ‖E0‖L2(Γ) . (2.11) The commutator [ Op h(ρ), ν (x)] is a pseudodifferential operator with symbol in
h1−δ S0,0
δ
and so
‖[Op h(ρ), ν k(x)] Enor ‖Hjh(Γ) ≤ C2h1−δ ‖Enor ‖Hjh(Γ) , k = 1 , 2, 3, j = 0 , 1.
The last estimate combined with (2.11) implies
∥∥∥(Op h(ρ) − γ√z)Enor − hQ 4(E0)
∥∥∥L2(Γ)
≤ C3
( h
√| Im z| + h1−δ )
‖E0‖L2(Γ) . (2.12) 8 F. COLOMBINI, V. PETKOV AND J. RAUCH
Eigenvalues-free regions
For z ∈ Z1, we have ρ ∈ S0,1
δ
with 0 < δ = 1 /2− < 1/2, while for z ∈ Z2 ∪Z3 we have ρ ∈ S0,10 (see ). Since Γ is connected one has either γ(x) > 1 or 0 < γ (z) < 1. We present the analysis in the case where 0 < γ (x) < 1, ∀x ∈ Γ. The case 1 < γ (x)is reduced to this case at the end of the section. Clearly, there exists 0 > 0 such that
0 ≤ γ(x) ≤ 1 − 0, ∀x ∈ Γ.
Combing (2.4) and (2.9), yields
‖〈 (Op h(ρ) − γ(x)√z)E0, ν (x)〉‖ L2(Γ) ≤ C h
√| Im z| ‖E0‖L2(Γ) + C1h‖E0‖H1
h(Γ)
,
where for z ∈ Z2 ∪ Z3 we can replace | Im z| by 1. This estimate for E0 and the estimate for the commutator [ Op h(ρ), ν k(x)] imply
‖(Op h(ρ) − γ(x)√z)Enor ‖L2(Γ) ≤ C3 h
√| Im z| ‖E0‖L2(Γ) + C4h1−δ ‖E0‖H1
h(Γ)
. (3.1) Let ( x′, ξ ′) be coordinates on T ∗(Γ). Consider the symbol
c(x′, ξ ′, z ) := ρ(x′, ξ ′, z ) − γ(x′)√z, x′ ∈ Γ.
Following the analysis in Section 3, , we know that c is elliptic in the case 0 <γ(x′) < 1 and if z ∈ Z1 we have c ∈ S0,1
δ
, | Im z|c−1 ∈ S0,−1
δ
, while if z ∈ Z2 ∪ Z3
one gets c ∈ S0,10 , c −1 ∈ S0,−10 . This implies
‖Op h(c−1)Op h(c)Enor ‖H1
h(Γ)
≤ C
| Im z| ‖Op h(c)Enor ‖L2(Γ) .
On the other hand, according to Section 7 in , the symbol of the operator
Op h(c−1)Op h(c) − I is given by
N
∑
j=1
(ih )j
j!
∑
|α|=j
Dαξ′ (c−1)( x′, ξ ′)Dαy′ c(y′, η ′)∣∣x′=y′,ξ ′=η′ + ˜bN (x′, ξ ′):= bN (x′, ξ ′) + ˜bN (x′, ξ ′),
where
|∂αx′ ˜bN (x′, ξ ′)| ≤ CαhN (1 −2δ)−sd−| α|/2.
Taking into account the estimates for c−1 and c, and applying (2.5), and (2.6) yields
∥∥∥(
Op h(c−1)Op h(c) − I
)
Enor
∥∥∥Hjh(Γ)
≤ C5
h
| Im z|2 ‖Enor ‖Hjh(Γ) , j = 0 , 1.
Repeating the argument in Section 3 in concerning the case 0 < γ (x′) < 1, for
z ∈ Z1 and 0 < δ < 1/2, one finds
‖Enor ‖H1
h(Γ)
≤
∥∥∥(
Op h(c−1)Op h(c) − I
)
Enor
∥∥∥H1
h(Γ)
+
∥∥∥Op h(c−1)Op h(c)Enor
∥∥∥H1
h(Γ)
≤ C6h1−2δ ‖E0‖L2(Γ) + C5h1−2δ ‖Enor ‖H1
h(Γ)
C7h1−δ ‖E0‖H1
h(Γ)
.
(3.2) Clearly,
‖E0‖Hkh (Γ) ≤ ‖ Etan ‖Hkh (Γ) + Bk‖Enor ‖Hkh (Γ) , k ∈ NMAXWELL EIGENVALUES 9
with Bk independent of h. Hence we can absorb the terms involving the norms of
Enor in the right hand side of (3.2) choosing h small enough, and we get
‖Enor ‖H1
h(Γ)
≤ Ch 1−2δ ‖Etan ‖H1
h(Γ)
. (3.3) The analysis of the case z ∈ Z2 ∪ Z3 is simpler since in the estimates above we have no coefficient | Im z|−1 and we obtain the same result with a factor h on the right hand side of (3.3). With a similar argument it is easy to show that
‖Enor ‖L2(Γ) ≤ C′h1−2δ ‖Etan ‖L2(Γ) . (3.4) In fact from (2.12) one obtains
∥∥∥Op h(c−1)
[
( Op h(ρ)−γ√z)Enor −hQ 4(E0)
]∥ ∥∥L2(Γ)
≤ C8
| Im z|
( h
√| Im z| +h1−δ ‖E0‖L2(Γ)
)
and
‖Op h(c−1)Q4(E0)‖L2(Γ) ≤ C9
| Im z| ‖E0‖L2(Γ) .
Combining these estimates with the estimate of ‖Op h(c−1)Op h(c) − I‖L2(Γ) →L2(Γ)
yields (3.4). Going back to the equation (2.1), we have
(
Dν − 1
γ
√z
)
E =
(
Dν − γ√z
)
Enor ν − ( 1
γ − γ)√zE nor ν
+ihg 0(Etan ) +
(
grad h(Enor )
)∣∣tan , x ∈ Γ. (3.5) Notice that for the first term on the right hand side of (3.5) we can apply the equality (2.4), while for Enor and
(
grad h(Enor )
)∣∣tan we have a control by the estimate (3.3). Consequently, setting E0 = E|Γ, the right hand side of (3.5) is bounded by Ch 1−2δ ‖E0‖H1
h(Γ)
. Next 1 < 11 − 0
≤ 1
γ(x) ≤ 1
0
, ∀x ∈ Γ.
This corresponds to the case ( B) examined in Section 4 of . The approximation of the operator N (z, h ) given by (2.9) yields the estimate
‖(Op h(ρ) − 1
γ
√z)E0‖L2(Γ) ≤ C
( h
√| Im z| ‖E0‖L2(Γ) + h1−2δ ‖E0‖H1
h(Γ)
)
. (3.6) For z ∈ Z1 ∪ Z3 the symbol
d(x′, ξ ′, z ) := ρ(x′, ξ ′, z ) − 1
γ(x′)
√z
is elliptic (see Section 4, ) and d ∈ S0,1
δ
, d −1 ∈ S0,−1
δ
. Then from (3.6) we esti-mate ‖E0‖H1
h(Γ)
and we obtain E0 = 0 for h small enough. This implies E = B = 0. Now recall that we have Re λ = − Im √zh , Im λ = Re √zh .
Suppose that z ∈ Z1. Then 10 F. COLOMBINI, V. PETKOV AND J. RAUCH
| Re λ| ≥ C(h−1)1−δ , | Im λ| ≤ C1h−1 ≤ C2| Re λ| 11−δ .
So if
| Re λ| ≥ C3| Im λ|1−δ , Re λ ≤ − C4 < 0,
there are no eigenvalues λ = i√zh of Gb. In the same way we handle the case z ∈ Z3
and we conclude that if z ∈ Z1 ∪ Z3 for every > 0 the eigenvalues λ = i√zh of Gb
lie in the domain Λ ∪ M , where
M = {z ∈ C : | arg z − π| ≤ π/ 4, |z| ≥ R0 > 0, Re z < 0},
Λ being the domain introduced in Theorem 1.1. Of course, if we consider the domain
Z3,δ 0 = {z ∈ C : | Re z| ≤ 1, Im z = δ0 > 0},
instead of Z3, we obtain an eigenvalue-free region with M replaced by
Mδ0 = {z ∈ C : | arg z − π| ≤ arctg δ 0, |z| ≥ R0(δ0) > 0, Re z < 0}.
The investigation of the case z ∈ Z2 is more complicated since the symbol d may vanish for Im z = 0 and ( x′
0
, ξ ′
0
) ∈ T ∗(Γ) satisfying the equation
√
1 + r0(x′
0
, ξ ′
0
) − 1
γ(x′
0
) = 0.
To cover this case and to prove that the eigenvalues λ = i√zh with z ∈ Z2
are confined in the domain RN , ∀N ∈ N, we follow the arguments in and . For z ∈ Z2 we introduce an operator T (z, h ) that yields a better approximation of N (z, h ). In fact, T (z, h ) is defined by the construction of the semi-classical parametrix in Section 3, for the problem (2.8) with F = E0. We refer to for the precise definition of T (z, h ) and more details. For our exposition we need the next proposition. Since (∆ − z)E = 0, as in , we obtain
Proposition 3.1. For z ∈ Z2 and every N ∈ N we have the estimate
‖N (z, h )E0 − T (z, h )E0‖H1
h(Γ)
≤ CN h−s0 hN ‖E0‖L2(Γ) (3.7)
with constants CN , s 0 > 0, independent of E0, h and z, and s0 independent of N .Proof of Theorem 1.1 in the case z ∈ Z2. Consider the system
(
Dν − 1
γ
√z
)
Etan −
(
grad hEnor
)∣ ∣∣tan
ihg 0(Etan ) = 0 , x ∈ Γ,
div hEtan + div h
(
Enor ν
)
= 0 , x ∈ Γ, (3.8) where div hF = ∑3
k=1
Dxk Fk.
Take the scalar product 〈, 〉L2(Γ) in L2(Γ) of the first equation of (3.8) and Etan .Applying Green formula, it easy to see that
− Re 〈grad hEnor
∣∣∣tan
, E tan 〉L2(Γ) = − Re 〈div hEtan , E nor 〉L2(Γ) . (3.9) We claim that Im 〈g0(Etan ), E tan 〉L2(Γ) = 0 . (3.10) Let Etan = ( w1, w 2, w 3). Then
〈g0(Etan ), E tan 〉C3 =
3
∑
k,j =1
wk
∂ν k
∂x j
wj = 1
q
3
∑
k,j =1
wk
∂V k
∂x j
wj = 1
q 〈Sw, w 〉C3 ,MAXWELL EIGENVALUES 11
where S := { ∂V k
∂x j
}3
k,j =1
with V (x) = q(x)ν(x), q (x) > 0 because ∑3
k=1
(∂xj q)wkνk =0. Thus if the boundary is given locally by x3 = G(x1, x 2), we choose V (x) = (−∂x1 G, −∂x2 G, 1) and it is obvious that S is symmetric. Therefore Im 〈Sw, w 〉C3 =0 and this proves the claim. Hence (3.10) implies Re[ ih〈g0(Etan ), E tan 〉L2(Γ) ] = 0 . (3.11) From the L2(Γ) scalar product of the second equation in (3.8) with Enor , we obtain Re 〈div hEtan , E nor 〉L2(Γ) + Re 〈Dν Enor , E nor 〉L2(Γ) = 0 . (3.12) In fact, div h(Enor ν) = Dν Enor − ihE nor div ν
and Im
(
div ν|Enor |2
)
= 0 .
Taking together (3.9), (3.11) and (3.12), we conclude that Re
[
〈(Dν −√zγ )Etan , E tan 〉L2(Γ) + 〈Dν Enor ν, E nor ν〉L2(Γ)
]
= Re
〈
Dν E, E
〉
L2(Γ)
− Re 〈√zγ Etan , E tan 〉L2(Γ) = 0 .
Here we have used the fact that
〈Dν Etan , E nor ν〉C3 = Dν
(
〈Etan , E nor ν〉C3
)
= 0 .
Applying Proposition 3.1 with E|Γ = E0, yields
∣∣∣Re
〈
T (z, h )E0, E 0
〉
L2(Γ)
− Re
〈 √zγ Etan , E tan
〉
L2(Γ)
∣∣∣ ≤ CN h−s0 hN ‖E0‖L2(Γ) .
(3.13) For z = −1, as in Lemma 3.9 in and Lemma 4.1 in , we have
| Re 〈T (−1, h )E0, E 0〉L2(Γ) | ≤ CN h−s0+N ‖E0‖2
L2(Γ)
= 0 .
Consequently, by using Taylor formula for the real-valued function Re
[〈
T (z, h )E0, E 0
〉
L2(Γ)
−
〈 √zγ Etan , E tan
〉
L2(Γ)
]
,
we get for every N ∈ N the estimate
∣∣∣Im
[〈
( ∂T ∂z (zt, h )) E0, E 0
〉
L2(Γ)
−
〈 γ1
2√zt
Etan , E tan
〉
L2(Γ)
]∣ ∣∣
≤ CN
h−s0+N
| Im z| ‖E0‖2
L2(Γ)
, (3.14) where zt = −1 + it Im z, 0 < t < 1, γ 1 = γ−1.
According to Lemma 3.9 in , in (3.14) we can replace ∂T ∂z (zt, h ) by Op h( ∂ρ ∂z (zt)) and this yields an error term bounded by Ch ‖E0‖2
H−1
h(Γ)
. On the other hand,
∣∣∣〈
Op h( ∂ρ ∂z (zt)) Etan , E nor ν
〉
L2(Γ)
+
〈
Op h( ∂ρ ∂z (zt)) Enor , E tan ν
〉
L2(Γ)
∣∣∣
≤ Ch ‖E0‖2
L2(Γ)
since the estimate (3.4) holds for z ∈ Z2 with factor h and ∂ρ ∂z (zt) ∈ S0,−10 .12 F. COLOMBINI, V. PETKOV AND J. RAUCH
Thus the problem is reduced to a lower bound of
J :=
∣∣∣Im
[
〈
(
Op h( ∂ρ ∂z (zt)) − γ1
2√z
)
Etan , E tan 〉L2(Γ) +〈Op h( ∂ρ ∂z (zt)) Enor ν, E nor ν〉L2(Γ)
]∣ ∣∣
≥
∣∣∣Im 〈
(
Op h( ∂ρ ∂z (zt)) − γ1
2√z
)
Etan , E tan 〉L2(Γ)
∣∣∣ − C1‖Enor ‖2
L2(Γ)
.
Since γ1(x) > 1, ∀x ∈ Γ, applying the analysis of Section 4 in for the scalar product involving Etan , one deduces
∣∣∣Im
〈(
Op h( ∂ρ ∂z (zt)) − γ1
2√z
)
Etan , E tan
〉
L2(Γ)
∣∣∣ ≥ η1‖Etan ‖2
L2(Γ)
, η1 > 0.
By using once more the estimate (3.4), for h small enough we obtain
J ≥ η1
(
‖Etan ‖2
L2(Γ)
+‖Enor ‖2
L2(Γ)
)
−B0h‖Etan ‖2
L2(Γ)
≥ η2‖E0‖2
L2(Γ)
, 0 < η 2 < η 1.
Consequently, (3.14) yields (η2 − B1h)‖E0‖2
L2(Γ)
≤ CN
h−s0+N
| Im z| ‖E0‖2
L2(Γ)
and for small h we conclude that for z ∈ Z2 the eigenvalues λ = i√zh of Gb lie in the region RN . This completes the analysis of the case 0 < γ (x) < 1, ∀x ∈ Γ.
To study the case γ(x) > 1, ∀x ∈ Γ, we write the boundary condition in (1.1) as 1
γ(x) (ν ∧ Etan ) − (ν ∧ (ν ∧ Btan )) = 1
γ(x) (ν ∧ Etan ) + Btan = 0 .
Next
ν ∧ E = 1
√z ν ∧ h
i curl B = − 1
√z Dν Btan + 1
√z
[(
grad hBnor
)∣ ∣∣tan
− ihg 0(Btan )
]
and one obtains
(
Dν − γ(x)√z
)
Btan −
(
grad hBnor
)∣ ∣∣tan
ihg 0(Btan ) = 0 , x ∈ Γ (3.15) which is the same as (2.1) with Etan , Enor replaced respectively by Btan , B nor
and 1
γ(x)
replaced by γ(x) > 1. We apply the operator Dy1 − γ√z to the equation div B = 0 and repeat without any change the above analysis concerning Etan , E nor .
Thus the proof of Theorem 1.1 is complete.
Remark 3.2 . The result of Theorem 1.1 holds for obstacles K = ∪Jj=1 Kj , where
Kj , j = 1 , ..., J are open connected domains with C∞ boundary and Ki ∩ Kj =
∅, i 6 = j. Let Γ j = ∂K j , j = 1 , ..., J. In this case we may have γ(x) < 1 for some obstacles Γ j and γ(x) > 1 for other ones. The proof extends with only minor modifications. The construction of the semi-classical parametrix in is local and for the Dirichlet-to-Neumann map Nj (z, h ) related to Γ j we get the estimate
‖N j (z, h )( F ) − Op h(ρ + hb )F ‖H1
h(Γ j)
≤ Ch
√| Im z| ‖F ‖L2(Γ j ).
The boundary condition in (1.1) is local and we can reduce the analysis to a fixed obstacle Kj . If ( E, B ) 6 = 0 is an eigenfunction of Gb, our argument implies Etan = 0 for x ∈ Γj if 0 < γ (x) < 1 on Γ j and Btan = 0 for x ∈ Γj in the case γ(x) > 1 on Γj . By the boundary condition we get Etan = 0 on Γ and this yields E = B = 0 MAXWELL EIGENVALUES 13
since the Maxwell system with boundary condition Etan = 0 has no eigenvalues in
{z ∈ C : Re z < 0}.
Appendix
In this Appendix, assume that γ > 0 is constant. Our purpose is to study the eigenvalues of Gb in case the obstacle is equal to the ball B3 = {x ∈ R3 : |x| ≤ 1}.Setting λ = iμ, Im μ > 0, an eigenfunction ( E, B ) 6 = 0 of Gb satisfies curl E = −iμB, curl B = iμE. (4.1) Replacing B by H = −B yields for ( E, H ) ∈ (H2(|x| ≤ 1)) 6,
{
curl E = iμH, curl H = −iμE, for x ∈ B3,Etan + γ(ν ∧ Htan ) = 0 , for x ∈ S2. (4.2) Expand E(x), H (x) in the spherical functions Y mn (ω), n = 0 , 1, 2, ..., |m| ≤ n, ω ∈ S2
and the modified Hankel functions h(1)
n
(z) of first kind. An application of Theorem 2.50 in (in the notation of it is necessary to replace ω by μ ∈ C \ { 0}) says that the solution of the system (4.2) for |x| = r = 1 has the form
Etan (ω) =
∞
∑
n=1
∑
|m|≤ n
[
αmn
(
h(1)
n
(μ) + ddr h(1)
n
(μr )|r=1
)
U mn (ω) + βmn h(1)
n
(μ)V mn (ω)
]
,Htan (ω) = − 1
iμ
∞
∑
n=1
∑
|m|≤ n
[
βmn
(
h(1)
n
(μ)+ ddr h(1)
n
(μr )|r=1
)
U mn (ω)+ μ2αmn h(1)
n
(μ)V mn (ω)
]
.
Here U mn (ω) = 1√n(n+1) grad S2 Y mn (ω) and V mn (ω) = ν ∧ U mn (ω) for n ∈ N, −n ≤
m ≤ n form a complete orthonormal basis in
L2
t
(S2) = {u ∈ (L2(S2)) 3 : 〈ν, u 〉 = 0 on S2}.
To find a representation of ν ∧ Htan , observe that ν ∧ (ν ∧ U mn ) = −U mn , so (ν∧Htan )( ω) = − 1
iμ
∞
∑
n=1
∑
|m|≤ n
[
βmn
(
h(1)
n
(μ)+ ddr h(1)
n
(μr )|r=1
)
V mn (ω)−μ2αmn h(1)
n
(μ)U mn (ω)
]
and the boundary condition in (4.2) is satisfied if
αmn
[
h(1)
n
(μ) + ddr (h(1)
n
(μr )) |r=1 − γiμh (1)
n
(μ)
]
= 0 , ∀n ∈ N, |m| ≤ n, (4.3)
− βmn γ
iμ
[
h(1)
n
(μ) + ddr (h(1)
n
(μr )) |r=1 − iμγ h(1)
n
(μ)
]
= 0 , ∀n ∈ N, |m| ≤ n. (4.4) For γ ≡ 1, there are no eigenvalues.
Proposition 4.1. For γ ≡ 1 the operator Gb has no eigenvalues in {Re z < 0}.
Proof. The functions h(1)
n
(z) have the form (see for example )
h(1)
n
(x) = (−i)n+1 eix
x
n
∑
m=0
im
m!(2 x)m
(n + m)! (n − m)! = (−i)n+1 eix
x Rn
( i
2x
)
with
Rn(z) :=
n
∑
m=0
zm
m!(n + m)! (n − m)! =
n
∑
m=0
amzm.14 F. COLOMBINI, V. PETKOV AND J. RAUCH
Therefore the term in the brackets [ ... ] in (4.3) becomes (1 − γ)iμR n
( i
2μ
)
−
n
∑
m=0
amm
( i
2μ
)m
.
Setting w = i
2μ
, we must study for Re w > 0 the roots of the equation
gn(w) := 1 − γ
2w Rn(w) + wR ′
n
(w) = 0 . (4.5) For γ = 1 one obtains R′
n
(w) = 0 . A result of Macdonald says that the zeros of the function h(1)
n
(z) lie in the half plane Im z < 0 (see Theorem 8.2 in ), hence
Rn(w) 6 = 0 for Re w ≥ 0. By the theorem of Gauss-Lucas we deduce that the roots of
R′
n
(w) = 0 lie in the convex hull of the set of the roots of Rn(w) = 0 , so R′
n
(w) 6 = 0 for Re w > 0. Consequently, (4.3) and (4.4) are satisfied only for αmn = βmn = 0 and
Etan = 0 . This implies E = H = 0 .
For the case γ 6 = 1, there are an infinite number of real eigenvalues.
Proposition 4.2. Assume that γ ∈ R+ { 1} is a constant. Then Gb has an infinite number of real eigenvalues. Let γ0 = max {γ, 1
γ
}. Then all real eigenvalues λ with exception of the eigenvalue
λ1 = − 2(γ0 − 1)
(
1 +
√
1 + 4
γ0−1
) . (4.6)
satisfy the estimate
λ ≤ − 1max {(γ0 − 1) , √γ0 − 1} . (4.7)
Proof. Assume first that γ > 1. Then qn(w) = wg n(w) = 0 has at least one real root w0 > 0. Indeed, qn(0) = 1−γ
2
< 0, q n(w) → +∞ as w → +∞. Choosing
αm0
n
6 = 0 for an integer m0, |m0| ≤ n and taking all other coefficients αmn , β mn equal to 0, yields Etan 6 = 0 and Gb has an eigenfunction with eigenvalue λ = − 12w0 < 0.
It is not excluded that gn(w) and gm(w) for n 6 = m have the same real positive root. If we assume that for Re w > 0 the sequence of functions {gn(w)}∞
n=1
has only a finite number of real roots w1, ..., w N , wj ∈ R+, then there exists an infinite number of functions gnj (w) having the same root which implies that we have an eigenvalue of Gb with infinite multiplicity. This is a contradiction, and the number of real eigenvalues of Gb is infinite. It remains to establish the bound on the real eigenvalues. First, consider the case n = 1. Then one obtains the equation 2w2
2w + 1 = γ − 12which has a positive root w0 = 14
(
γ − 1 + √(γ − 1) 2 + 4( γ − 1)
)
. This yields the
λ1 from (4.6) Next examine the case n ≥ 2. For a root w0 ∈ R+ one has
w0
(
w0
R′
n
(w0)
Rn(w0)
)
= γ − 12 .MAXWELL EIGENVALUES 15
Case 1. w0 ≥ 12√3 . Then the inequality
∑nm=2 ma mwm
0
a1w0
∑nm=2 amwm
0
a1w0 + 1 ≥ 2 ∑nm=2 amwm
0
a1w0
∑nm=2 amwm
0
a1w0 + 1 ≥ 1is satisfied since a2 = 12 (n + 2)( n + 1) n(n − 1) ≥ 12 . Consequently, 2 w0 ≤ γ − 1 and this implies that the eigenvalue λ = − 12w0 satisfies
λ < − 1
γ − 1 . (4.8)
Case 2. 0 < w 0 ≤ 12√3 . Apply the inequality
∑nm=2 ma mwm−10 + a1
w0
∑nm=2 amwm−10 + a1w0 + 1 ≥ 2 ∑nm=2 amwm−10 + a1
w0
∑nm=2 amwm−10 + a1w0 + 1 ≥ 2that is equivalent to 2
[
(1 − w0)S0 − a1w0
]
a1 ≥ 2with S0 = ∑nm=2 amwm−10 . This inequality holds because (1 − w0)
n
∑
m=2
amwm−10 − a1w0 ≥ ( 12 a2 − a1)w0 , a1 = ( n + 1) n ≥ 2,
and, 12 a2 − a1 = 14 (n + 2)( n + 1) n(n − 1) − (n + 1) n = n(n + 1)
[ 14 (n + 2)( n − 1) − 1
]
≥ 0.
Therefore, 2w20 ≤ w20
∑nm=1 ma mwm−10
∑nm=1 amwm
0
1 = γ − 12 .
This easily yields
λ ≤ − 1
√(γ − 1) . (4.9) In the case 0 < γ < 1 one has 1 /γ > 1 and we apply the above analysis to the equation (4.4). Setting γ0 = max {γ, 1
γ
} and taking into account (4.8) and (4.9), we obtain the result. This completes the proof.
Remark 4.3 . Proposition 4.2 yields a more precise result than that in since we prove the existence of an infinite number of real eigenvalues Gb for every γ ∈
R+ \ { 1}. In the case γ = 11+ , > 0 the eigenvalue λ1 has the form
λ1 = 12
(
1 −
√
1 + 4
)
and this result for small > 0 has been obtained in . Clearly, as γ → 1 the real eigenvalues of Gb go to −∞ . .It is easy to see that for γ > 1 the equation gn(w) = 0 has no complex roots. Denote by
zj , Re zj < 0, j = 1 , ..., n, n ≥ 116 F. COLOMBINI, V. PETKOV AND J. RAUCH
the roots of Rn(w) = 0. Suppose that gn(w0) = 0 , n ≥ 1 with Re w0 > 0, Im w0 6 = 0 .
Then Im
[ 1 − γ
2w0
w0
n
∑
j=1
1
w0 − zj
]
= 0 and
− (1 − γ) Im w0
2|w0|2 + Re w0
[
−
n
∑
j=1
Im w0
|w0 − zj |2 +
n
∑
j=1
Im zj
|w0 − zj |2
]
Im w0
n
∑
j=1
Re w0 − Re zj
|w0 − zj |2 = 0 . (4.10) On the other hand, if zj with Im zj 6 = 0 is a root of Rn(w) = 0, then ¯ zj is also a root and Im zj
|w0 − zj |2 − Im zj
|w0 − ¯zj |2 = Im zj
|w0 − zj |2|w0 − ¯zj |2
(
|w0 − ¯zj |2 − | w0 − zj |2)
= 4 Im w0(Im zj )2
|w0 − zj |2|w0 − ¯zj |2 .
Equation (4.10) becomes Im w0
[ γ − 12|w0|2 −
n
∑
j=1
Re zj
|w0 − zj |2 + ∑
Im zj>0
4 Re w0(Im zj )2
|w0 − zj |2|w0 − ¯zj |2
]
= 0 . (4.11) The term in the brackets [ ... ] is positive, and one concludes that Im w0 = 0 .
Repeating the argument of the Appendix in , one can show that for 0 < γ < 1
the complex eigenvalues of Gb lie in the region
{
z ∈ C : |arg z − π| > π/ 4, Re z < 0
}
.
Remark 4.4 . We do not know if there exist non real eigenvalues for B3.
Acknowledgment. We thank Georgi Vodev for many useful discussions and remarks concerning an earlier version of the paper.
References
F. Colombini, V. Petkov and J. Rauch, Incoming and disappearing solutions of Maxwell’s equations , Proc. AMS, 139 (2011), 2163-2173. F. Colombini, V. Petkov and J. Rauch, Spectral problems for non elliptic symmetric systems with dissipative boundary conditions , J. Funct. Anal. 267 (2014), 1637-1661. M. Dimassi and J. Sj¨ ostrand, Spectral asymptotics in semi-classical limits , London Mathe-matical Society, Lecture Notes Series, 268 , Cambridge University Press, 1999. A. Kirsch and F. Hettlich, The Mathematical Theory of Time-Harmonic Maxwells Equations ,vol. 190 of Applied Mathematical Sciences, Springer, Switzerland, 2015. P. Lax and R. Phillips, Scattering theory for dissipative systems , J. Funct. Anal. 14 (1973), 172-235. A. Majda, The location of the spectrum for the dissipative acoustic operator , Indiana Univ. Math. J. 25 (1976), 973-987. F. Olver, Asymptotics and Special Functions , Academic Press,New York, London, 1974. V. Petkov, Location of the eigenvalues of the wave equation with dissipative boundary con-ditions , Inverse Problems and Imaging, to appear, (arXiv: math.AP. 1504.06408v4). MAXWELL EIGENVALUES 17
G. Vodev, Transmission eigenvalue-free regions . Commun. Math. Phys. 336 (2015), 1141-1166.
Dipartimento di Matematica, Universit` a di Pisa, Italia
E-mail address : colombini@dm.unipi.it
Institut de Math´ ematiques de Bordeaux, 351, Cours de la Lib´ eration, 33405 Talence, France
E-mail address : petkov@math.u-bordeaux.fr
Department of Mathematics, University of Michigan, USA
E-mail address : rauch@umich.edu |
17375 | https://www.sciencing.com/calculate-phase-shift-5157754/ | How To Calculate The Phase Shift
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How To Calculate The Phase Shift
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How To Calculate The Phase Shift
By John Papiewski Updated Aug 30, 2022
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Phase shift is a small difference between two waves; in math and electronics, it is a delay between two waves that have the same period or frequency. Typically, phase shift is expressed in terms of angle, which can be measured in degrees or radians, and the angle can be positive or negative. For example, a +90 degree phase shift is one quarter of a full cycle; in this case, the second wave leads the first by 90 degrees. You can calculate phase shift using the frequency of the waves and the time delay between them.
Sine Wave Function and Phase
Sine Wave Function and Phase
In math, the trigonometric sine function produces a smooth wave-shaped graph that cycles between a maximum and a minimum value, repeating every 360 degrees or 2 pi radians. At zero degrees, the function has a value of zero. At 90 degrees, it reaches its maximum positive value. At 180 degrees, it curves back down toward zero. At 270 degrees, the function is at its maximum negative value, and at 360, it returns to zero, completing one full cycle. Angles greater than 360 simply repeat the previous cycle. A sine wave with a phase shift begins and ends at a value other than zero, although it resembles a "standard" sine wave in every other respect.
Choosing the Wave Order
Choosing the Wave Order
Calculating phase shift involves comparing two waves, and part of that comparison is choosing which wave is "first" and which is "second." In electronics, the second wave is typically the output of an amplifier or other device, and the first wave is the input. In math, the first wave may be an original function and the second a subsequent or secondary function. For example, the first function may be y = sin(x), and the second function may be y = cos(x). The order of the waves does not affect the absolute value of the phase shift, but it does determine if the shift is positive or negative.
Comparing the Waves
Comparing the Waves
When comparing the two waves, arrange them such that they read left to right using the same x-axis angle or time units. For example, the graph for both may start at 0 seconds. Find a peak on the second wave, and find the corresponding peak on the first. When looking for a corresponding peak, stay within one full cycle, otherwise the phase difference result will be incorrect. Note the x-axis values for both peaks, then subtract them to find the difference. For example, if the second wave peaks at 0.002 seconds and the first peaks at 0.001 seconds, then the difference is 0.001 – 0.002 = -0.001 seconds.
Calculating Phase Shift
Calculating Phase Shift
To calculate the phase shift, you need the frequency and period of the waves. For example, an electronic oscillator may produce sine waves at a frequency of 100 Hz. Dividing the frequency into 1 gives the period, or duration of each cycle, so 1/100 gives a period of 0.01 seconds. The phase shift equation is ps = 360 td / p, where ps is the phase shift in degrees, td is the time difference between waves and p is the wave period. Continuing the example, 360 -0.001 / 0.01 gives a phase shift of -36 degrees. Because the result is a negative number, the phase shift is also negative; the second wave lags behind the first by 36 degrees. For a phase difference in radians, use 2 pi td / p; in our example, this would be 6.28 -.001 / .01 or -.628 radians.
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Dead Time
In subject area:Mathematics
Dead time is defined as the period following a bolus injection during which initial measurements may be inaccurate due to the prominence of high activity, necessitating adequate correction to ensure accurate data collection.
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Signal processing
2015, Physics and Engineering of Radiation Detection (Second Edition)Syed Naeem Ahmed
A.2Dead time
The total time it takes the ADC to acquire a signal, complete the conversion, and become available for the next acquisition is called dead time because during this time it cannot accept a new signal. The dead time typically consists of:
•
: signal acquisition time,
•
: conversion time,
•
: data transfer to buffers,
•
: reset time.
In well-designed systems, the transfer of converted signal to the memory is performed during the reset time. How that memory is subsequently read out and handled is not part of ADC performance. However, since such a memory is generally of limited capacity (such as 1 kB FIFOs), it must be read out continuously to avoid overflow, which leads to loss of information.
Additionally, operations on ADCs are controlled through digital signals, which have their own response times and uncertainties. Such uncertainties are referred to as time jitters and should be given proper consideration in high-resolution systems.
If the event rate is such that pulses arrive during the dead time of an ADC, the information gets lost unless the charge is dynamically integrated on some capacitor for later acquisition by the ADC. Another possibility is to determine the average dead time of the system and then correct for it in the final analysis. This strategy works well for random signals where all events experience the same dead time. Determination of the dead time is a straightforward process in which the ADC is fed with a known stream of pulses and the output is recorded. The comparison of output to input gives a quantitative measure of the dead time.
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Single-Photon Generation and Detection
2013, Experimental Methods in the Physical SciencesMartin J. Stevens
2.3.6Dead Time, Reset Time, and Recovery Time
The dead time, , is the duration of time, beginning at the start of a detection event, during which a detection system is incapable of producing an output electrical signal in response to additional incident photons. During the dead time, the detection efficiency is zero, as illustrated in Fig. 2.18. The dead time may be caused by intrinsic processes in the photosensitive system or it may be induced by external control systems in order to produce a particular performance characteristic.
The reset time, , is the time over which the detection efficiency increases from zero back to its initial value. If the detection efficiency approaches this initial value very slowly, it may be necessary to specify as the elapsed time after which the detection efficiency changes by less than some small percentage of its value.
The total time required for the detection efficiency to recover to its steady-state value after a detection event is the recovery time, . In detectors or systems with a very short reset time, , and the terms can be used interchangeably.
The operation of the detector during its reset requires some special consideration. The fact that the detection efficiency is in transition during its reset can strongly affect measurements at high count rates. In addition, the reset action in some types of detectors (notably SPADs) can affect the ability of the electronics to sense a detection event that may have occurred during the reset. This is the origin of the so-called twilight events in some actively quenched SPAD detectors, and is discussed in Chapter 8.
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Book series2013, Experimental Methods in the Physical SciencesMartin J. Stevens
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Gas-filled detectors
2015, Physics and Engineering of Radiation Detection (Second Edition)Syed Naeem Ahmed
3.6.BDead time
The physical process of a Geiger breakdown takes some time to subside in a GM counter. The output pulse, therefore, is not only large but also fairly long. The problem is that from the initiation of the breakdown until it has died and the pulse has been recorded, the counter remains dead for subsequent ionization events. This time is called the dead time of the GM counter. Since it is not possible to eliminate dead time, one must make corrections in the recorded count rate to account for the missed events. If Nc and Nt are the recorded and true count rates then, for a GM counter, they are related by
(3.6.1)
where η can be thought to represent the efficiency of the counter. Efficiency depends on many factors, such as dead time due to the discharge process, dead time due to electronics, efficiency in discriminating the good events from noise, etc. Let us now see if we can derive a simple relation for efficiency using intuitive arguments. Suppose there is an average dead time τ during which the detector becomes unable to record any new ionizing event. We can assume this since under constant operating conditions the time it will take the avalanche to spread throughout the detector, cause breakdown, and then subside should not vary from pulse to pulse. Now, if C is the total number of counts recorded by the detector in a time t, then the recorded count rate will be
(3.6.2)
Since τ is the dead time of the detector, the rate Nlost at which the true events are not recorded is given by
(3.6.3)
where Nt is the true count rate, or the rate at which the detector would record pulses if it had no dead time. It is simply the sum of the recorded count rate and the lost count rate:
(3.6.4)
Using the above two equations, we can write
(3.6.5)
Hence the efficiency of a GM tube can be written as
(3.6.6)
Typical GM tubes have a dead time of the order of 100 µs. It is obvious from the above relation that such a detector will have an efficiency of 50% if operated in a radiation field of 10 kHz. In other words, on average, it will detect one particle out of two incident particles.
Example:
A GM detector having an efficiency of 67% is placed in a radiation field. On average, it reads a count rate of 1.53×104 per second. Find the true rate of incident radiation and the dead time of the detector.
Solution:
The rate of incident radiation is the true count rate of Eq. (3.6.5). Hence we have
For the dead time we use Eq. (3.6.6) as follows:
Dead time for GM tubes is generally determined experimentally in laboratories by using the so-called two-source method. This involves recording the count rates from two sources independently and then combining them together. According to Eq. (3.6.5), the true count rates Nt,1, Nt,2, and Nt,12 of the two sources independently and combined are given by
where Nc,x with x=1, 2, 12 represent the recorded count rates in the three respective configurations. Now, since the atoms in the two sources decay independently of each other, their true rates should add up; that is,
(3.6.7)
Substituting the true count rate expressions in this equation gives
(3.6.8)
Note that, in order to determine the dead time using this method, one does not require knowledge of the true count rates of either of the two sources. Hence any source with arbitrary strength can be chosen for the purpose. However, one must make certain that the decay rate is neither very low nor very high to ensure that the efficiency of the detector does not fall too low.
Example:
In an attempt to find the dead time of a GM counter, three measurements are taken. The first two with separate radiation sources give average count rates of 124 and 78 s−1. With both the sources in front of the detector, the count rate is found to be 197 s−1 Estimate the dead time of the counter.
Solution:
We can use Eq. (3.6.8) to estimate the dead time as follows:
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Chapter
Efficient Experimental Design for fMRI
2007, Statistical Parametric MappingR. Henson
Keep the subject as busy as possible
This refers to the idea that ‘dead-time’ – time during which the subject is not engaged in the task of interest – should be minimized. Again, of course, there may be psychological limits to the subject's performance (e.g. they may need rests), but apart from this, factors such as the SOA should be kept as short as possible (even within blocks of trials). The only situation where you might want longer SOAs (or blocks of rest) is if you want to measure ‘baseline’. From a cognitive perspective though, baseline is rarely meaningful, since it is rarely under strong experimental control (see below).
Only stop the scanner – i.e. break your experiment into sessions – if it is strictly necessary. Breaks in scanning disrupt the spin equilibrium (i.e. require extra dummy scans), reduce the efficiency of any temporal filtering (since the data no longer constitute a single time-series), and introduce other potential ‘session’ effects (McGonigle et al., 2000).
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Single-Photon Generation and Detection
2013, Experimental Methods in the Physical SciencesIvo Pietro Degiovanni, ... Franco N.C. Wong
7.2.2.3Dead Time Reduction
As we have seen, spatial multiplexing of detectors can not only enable PNR capabilities, but can also reduce the effective dead time, allowing for higher detection rates. Particularly, we saw that the effective dead time of a multiplexed system is reduced relative to a single detector by a factor of , the number of detectors in the passive “detector tree” arrangement.
There exists a method to improve detection rates in a more efficient way than just randomly sending the photons toward the elements of the click/no-click detector array. This method requires a means to actively monitor the state of each detector to check if it is ready to register a photon or it is dead, and an optical switch to route subsequent incoming photons to a detector that is known to be ready [37–40], as shown in Fig. 7.4. We analyze this strategy analytically and numerically, and show that this scheme allows the -detector system to be operated at a detection rate significantly higher than -times the detection rate of an individual detector, while reducing the overall dead time.
The system’s switching operation could be sequential, with each detector firing in order (the control system only switches to the next detector if the previous has fired), or it could be set up to switch the input to any live detector regardless of whether the previous detector had just fired. This latter implementation may allow for optimum use of an array of detectors where each detector may have a different dead time or when the switching time of the system is not negligible. In the simple model discussed here, we assume that all the detectors have the same dead time and switch transition time. The switch transition time includes any latency or other possible delays. While we include switch latency as part of the switch transition time, rather than as a separate parameter, we point out that latency may affect the choice of what firing order is used depending on what detection characteristic is most important for the particular application. For example, an application might have a repetitive pulsed photon source where the time to sense a detection is longer than the pulse period while the switch time by itself is less than the pulse period. In that case the detection system might benefit from operating in a mode where the input is immediately switched to another live detector regardless of whether the previous detector fired. This would reduce the effect of the long latency, although at a cost of a decreased likelihood of having at least one live detector available.
The switching strategy for the optical switch discussed here consists of simply re-routing photons to the next detector in the sequence of -detectors after the previous detector fires. This is the simplest implementation and is all that is required when the optical switching time is not a large fraction of each individual click/no-click detector’s dead time. Even when the switching time is non-negligible, our assessment shows an advantage from using this scheme versus passive detector-tree schemes.
The relevant figure of merit in this context is dead time fraction (DTF), defined as the ratio of missed- to incident-events. A good device-independent benchmark for comparing different detection systems at high photon detection rates is the rate of incoming photons that results in a DTF of 10%, . This is a practical limit for detector operation in many real-world applications.
We analytically estimate the DTF from the mean total count rate of the overall detector pool and effective dead time for each detector (which depends on their position in the switching system). We consider a Poissonian source, as described above, and a pool of identical detectors (with both equal detection efficiencies and equal non-paralyzable dead times ). From Eq. (7.17) we find the DTF for a detector tree to be
(7.18)
Analogously, for a detection system with an array of the same -detectors, and active multiplexing, an overall or “effective” dead time can be introduced, treating the whole system as a single detection unit:
(7.19)
Therefore, the task reduces to calculating the effective dead time of the system. Because the optical switch only switches photons to a new detector after a count is registered, the effective dead time is given by the statistical contribution of the switching time, and the single-detector dead time, , governed by the two cases illustrated in Fig. 7.5: (a) events are counted in a time interval longer than , or (b) they occur in a time interval shorter than . In the latter case, a photon is switched back to a detector that is still dead, and the entire detector assembly saturates. Due to this saturation, the assembly dead time is longer than by the remaining dead time of an individual detector. We write the effective dead time for an -detector assembly as:
(7.20)
where
(7.21)
and
(7.22)
are the probabilities that case (a) or (b) occurs for , which is the probability density distribution of the time interval , between a count and the preceding one. Note that the dependence of the above probabilities on requires solving an integral equation to obtain . The mean time interval between a count and a preceding count when case (b) occurs is given by:
(7.23)
Note that Eq. (7.12) allows writing an expression for the probability density distribution , assuming a Poissonian input and introducing a constant effective dead time :
(7.24)
An analytical formula for exists only for detectors:
(7.25)
where is the principal value of the Lambert -function . For more detectors we use numerical methods to determine .
Interestingly, while neglecting the dynamic nature of the dead time and introducing a constant effective dead time seems to be a restrictive assumption, the results obtained with this approach are in excellent agreement with experimental results, as well as Monte-Carlo simulations in all the regimes considered . This is not surprising, because such a treatment merely swaps the order of integration when computing the averages.
Figure 7.6a shows the dead time fraction versus the incoming photon rate for systems with detectors with single-detector dead times of and switching times equal to 1% and 10% of the single-detector dead time. For the effect of switching time on the system is negligible, while for , the multiplexed scheme shows much less increase of the points with increasing detector number. Figure 7.6b compares the analytic theory with the Monte-Carlo results, showing good agreement with the simulation for all switching times . Figure 7.6c compares the active multiplexed scheme just described with a passive scheme (detector/beam-splitter tree configuration) for . As judged by the points, the active multiplexed scheme surpasses the passive arrangement for relatively few detectors, .
Figure 7.7a shows the mean effective dead time for up to 5, versus the mean incident photon rate (), for and . The effective dead time clearly satisfies the condition . We see that the maximum effective dead time of the multiplexed scheme coincides with the detector-tree dead time. This means that for an optical switch with , the active scheme surpasses what is possible with a passive scheme. Figure 7.7b shows the ratio of the mean count rate for the multiplexed scheme to the count rate of a single detector, versus the mean incoming photon rate. We see that, as expected for high count rates, the maximum gain is -times the rate that would be obtained by a single detector.
Figure 7.8 shows versus the number of detectors for the active switching system at several switching times. The result differs little from the case when the switching time is neglected. Up to the results show significant advantage of the active-switch scheme over a passive beam-splitter tree for all numbers of detectors shown. Above the advantage of the active system is significantly reduced until ultimately its figure of merit falls below that of the passive scheme for just a few detectors.
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Chapter
MEASUREMENT OF SOURCE STRENGTH
1968, Alpha-, Beta- and Gamma-Ray SpectroscopyR.A. Allen
3.1.6Short half-life materials
If the half life of the nuclide being measured is so short that the disintegration rate reduces significantly during the measurements, allowance must be made for the decay. Suppose the disintegration rate at arbitrary time zero (the start of the measurement) is N0 and at the end of the measurement, after time t, it is Nt. We know that Nt = N0 exp(—λt) and the number of disintegrations N in time t is N0 [1—exp(—λt)]/λ; the ratio of the total counts in time t, nβt nγt/nct is equal to N and hence N0 can be calculated. The situation is complicated if the count-rate dependent corrections are significant since the magnitude of these corrections is variable during the determination.
We must now consider the corrections which arise from the effects of the finite resolving time of the coincidence unit and the finite dead times of the detectors (including subsequent amplifiers, discriminators etc.).
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Book1968, Alpha-, Beta- and Gamma-Ray SpectroscopyR.A. Allen
Chapter
MEASUREMENT OF SOURCE STRENGTH
1968, Alpha-, Beta- and Gamma-Ray SpectroscopyR.A. Allen
1.2COUNTING STATISTICS AND LOSSES
The decay process is strictly random; the probability of a number of disintegrations – and hence the number of counts recorded – in a given time is described by Poisson's law which, for large numbers, approximates to the Gaussian distribution. Thus, the error associated with the observation of a large number of events, n, is expressed by σ = √n where σ is the standard deviation. The standard deviation (σN) in the estimate of the count rate N, which is equal to n/t, t being the time of observation, is (√n)/t = = √/(N/t). The relationship between the often used phrase ‘confidence limit’ and standard deviation is simply the method of quoting the error of a measurement. If the error is quoted as ±σ, the chance that the true value lies within these limits is 0.68; called the 68% confidence limit. Other values are:
Limit of error quoted: ± 0.675σ, σ, 2σ, 3σ corresponds to confidence limit of 50%, 68%, 95% and 99.7% respectively. The standard deviation (σ) after the subtraction of background count rate (B) is given by
(1)
Associated with each counting channel is a ‘dead time’ (τD), the period after each detection of a pulse during which the channel is inoperative. For non-extending dead times, the loss of counts is expressed by the relation N = n/(1 — n τD) where N is the true detection rate and n the observed rate. One useful way of measuring the dead time is by the paired-source technique. Two approximately equal sources (1 and 2) are counted in the order 1, 1 + 2, 2; from the three counting rates and the background count rate (nB) the dead time can be calculated, for
(2)
alternatively, τD can be measured electronically using a double pulse generator or observing the pulses at very high counting rates on an oscilloscope screen.
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Single-Photon Generation and Detection
2013, Experimental Methods in the Physical SciencesSergey V. Polyakov
8.2Definitions
Key properties of detectors based on a range of underlying principles have been introduced in the preceding chapters. Here we establish a generalized list of parameters that are relevant to calibrations with accuracies down to the level. More accurate calibrations may require considering additional parameters.
Recovery Time, Dead Time, and Reset Time. Many detectors, such as SPADs and single SNSPDs, can detect only one photon of the input field at a time, after which they become insensitive to incoming radiation for a certain amount of time. This period of time is called the device’s dead time. Note also that returning to operation from this insensitive (dead) state may be a complex process, during which both the ability to detect photons and the device’s detection latency (i.e. the time between a photon entering the detector and a signal output, see Chapter 2) can differ significantly from that of normal operation. This mode of a detector is called reset time. The above two modes of a detector, when the detector’s operation is abnormal, are called the recovery time. Although the instantaneous detection efficiency at any point during this transient period may be a function of time elapsed since the detection event, one possible simplification useful for characterization with continuous wave sources is assuming a constant detection efficiency when the detector is alive and zero detection efficiency when the detector is dead, and adjusting the effective dead time accordingly.
Detection Efficiency. Detection efficiency is commonly defined as the probability of a detector to produce a successful detection given a single-photon input, when the detector is operating normally (far from its previous recovery), . Because of dead time (and assuming that both afterpulsing and dark count rates are zero), the actual rate of detection will be bound by , where is the rate of incoming photons. Obviously, just knowing is not enough to describe the operation of a detector with realistic input fields. Alternatively, detection efficiency can be defined as a proportionality coefficient relating the number of detection signals to the number of photons incident on a detector, the auto-correlation properties of the input field, etc.
The detection efficiency of a full PNR detector is defined in a similar fashion. This probability should scale with the number of incident photons. That is, if is the probability to detect a single photon given a single-photon input, the is the probability to detect a photon pair given two input photons, and in general is the probability to detect photons given an -photon state input. Any deviation of a real-life device from the full PNR assumption can significantly limit the accuracy of a precision measurement.
Dark Counts. Dark counts are output signals produced by the detector that do not correspond to actual photon detections. In most cases, nothing about the processes that cause dark counts makes them distinguishable from photon detection events. It is often assumed that dark counts obey Poissonian statistics (although at very high count rates that may no longer be true because of dead time effects). For an accurate calibration, dark count rates need to be characterized and included in the analysis.
Afterpulsing. Afterpulsing counts occur some time after a detection, sometimes even during the recovery of the detector. These counts originate from within the detector, and not from incoming photons. Particularly, in semiconductor avalanche detectors (see Chapter 4) afterpulses are triggered by carriers that were trapped and subsequently released from localized trap sites in the avalanche region after a previous avalanche. In a calibration the afterpulsing probability needs to be characterized as well.
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Single-Photon Generation and Detection
2013, Experimental Methods in the Physical SciencesIvo Pietro Degiovanni, ... Franco N.C. Wong
7.2.3.3Dead Time Fraction Reduction via Active Detector Multiplexing
As shown from theory, active multiplexing of the detectors can boost detection rates while maintaining low saturation of the assembly by a factor exceeding the number of detectors in the assembly. This was verified in a proof-of-principle experiment with an assembly of just two detectors and a fast optical switch [39,40].
The experimental setup, presented in Fig. 7.9a, is built around a parametric down-conversion crystal that produces photon pairs at two different frequencies. The photon at 810 nm is detected by a silicon SPAD (with a dead time of 50 ns, that is negligible compared to the dead time of the infrared detectors in the system under test). The detection of an 810 nm photon heralds a 1550 nm photon in the signal arm, where different detector arrangements were tested. Two InGaAs detectors () connected through a fast optical switch were used to implement the multiplexing arrangements. The real-time logic for the active detector arrangements was built using a Field Programmable Gate Array (FPGA) (for some background on this, see for example ). For dead time reduction as discussed earlier in this chapter, the design for two detectors was built around an asynchronous Set-Reset Flip-Flop (RS-Trigger), see Fig. 7.9b. Note that even though FPGAs usually operate synchronously, asynchronous codes of low complexity can be implemented, although extensive testing of the asynchronous gate solutions is required. In particular, devices built using asynchronous gates may exhibit dead time. Fortunately, when SPAD outputs are used, single-photon detector dead time is usually longer than that of the FPGA logic. Another complication is the need to carefully consider timing (latency) and duration of the single-photon detector’s output, which may differ from one detector to the other (even if the detectors are from the same batch) and be ready to modify the logical circuit accordingly.
Several detector configurations were compared: (i) a single detector, (ii) a detector-tree arrangement, and (iii) a multiplexed arrangement that was designed to reduce detection dead time, as discussed above (Fig. 7.10). The count rate for configuration (iii) is the highest for all values of DTF. In particular, for our chosen threshold of DTF = 10% we see a heralding count rate 2.3 times higher for the actively multiplexed scheme (iii) as compared to a single detector. Note that this improvement factor is achieved with just two detectors in the assembly.
Another interesting feature of active multiplexing is that other properties of detectors can be improved along with the dead time reduction. Because the output of only one of the detectors is monitored at any point of the protocol, the overall dark-count rate is just that of a single detector instead of scaling linearly with the number of detectors, as happens with passive multiplexing arrangements . Another benefit of active multiplexing is that the afterpulse probability of an active arrangement will be always lower than that of a single detector or a detector tree and will depend on count rate .
While analyzing the performance limitations of active multiplexing, it was found that the maximum count rate increase is limited by a feature often found in gated InGaAs detectors that is not commonly appreciated. In addition to detection dead time, these detectors have a dead time of its gate input (i.e., in the case when a detector is gated on but no photon is detected, the gate circuitry requires some time to be ready to respond to a subsequent gate), thus a strategy for suppressing the impact of this gate dead time was developed, requiring only a modification of the FPGA firmware. A measurement using this modification with a two-detector multiplexed assembly in configuration (iii) resulted in a dead time reduction factor of nearly 5 when compared with a single InGaAs detector (Fig. 7.10). The details of this modification are beyond the scope of this chapter, but can be found in .
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MEASUREMENT OF SOURCE STRENGTH
1968, Alpha-, Beta- and Gamma-Ray SpectroscopyR.A. Allen
3.1.7Count-rate dependent corrections
Much has been written on the subject of dead time and resolving time corrections in coincidence equations, particularly in consideration of high efficiency detectors46,47,48. Comparison of the various derivations is hampered by authors’ different terminology for the various effects – such that, for example, the term ‘accidental coincidence’ does not necessarily encompass the same events in all papers, and final expressions differ from one another; to a first approximation, the variation in results obtained by using the various formulae is insignificant.
For the individual single channel count rates (β-channel and γ-channel) the correction is straightforward. If the dead time in each channel is constant and non-extending, and is denoted respectively by τβ and τγ, then the detection rates Nβ and Nγ are given by the relations
For the coincidence mixer, two effects must be considered. Coincidences are lost because the input pulses are lost during the ‘dead times’ of the input channels (τβ and τγ above); also, coincidences are gained because the resolution time of the coincidence unit is finite. Input pulses to the coincidence mixer are usually shaped rectangular, of duration time (τR)β and (τR)γ. For two completely independent series of pulses, of arrival rates nβ and nγ, the random (chance, accidental) count rate nr is given by
(14)
where 2τR=(τR)β + (τR)γ. Indeed, this formulation is commonly used to correct for random coincidences in coincidence measurements of activity when low efficiency detectors are used. The correction is not strictly accurate however, since the rate is dependent upon the efficiency of the detectors; neglecting background radiations and dead times, the random coincidence rate is zero if either efficiency is 100% and the decay scheme is simple.
Now the random rate may be described by the expression
(15)
ignoring dead time losses, the factor N0εβ (1 − εγ) is simply the count rate of β-particles whose corresponding coincident γ-ray is not detected, with a similar explanation for N0εγ (1 − εβ).
An alternative approach is given by Campion46; if n'β, n'γ and n'c are observed counting rates, including backgrounds, the portion of the β-count rate which did not give rise to a true coincidence is simply (n'β — nc); the random coincidence rate due to these pulses is (n'β − nc)n'γτR. The expression (n'γ — nc)n'βτR is the number of random coincidences associated with the γ-channel count rate corrected for true coincidences. The random rate is given by nr = n'c — nc and hence
(16)
which is in terms of observables. Genuine background coincidences are included in nc and must be subtracted after the resolving time correction.
The coincidence rate is further modified by dead time losses. Consider the coincidence mixer, resolving time τR, fed by a series of pulses which have had imposed upon them a dead time τD. In practice τD is usually greater than τR. Following the approach made by Hay ward48 we can derive an expression for the number of probable coincidences lost in a time τD after an event has been registered. If a coincidence event is registered, the mixer is dead for a time τD following the event; if a single channel event is registered, the mixer is dead for a time (τD — τR) since it remains live for an interval τR after the event. Remembering that the true coincidence rate is given by N0 εβεγ, the number occurring in a time τD, and hence the number lost in the coincidence mixer after each coincidence is recorded, is N0 εβεγ τD. The rate of loss is therefore nc N0 εβεγ τD. The β-rate is N0 εβ, the number of non-coincident γ-rays occurring in time (τD — τR) after each event is N0εγ (1 − εβ)(τD − τR), with a similar expression if the roles of the channels are reversed: N0εβ (1 − εγ)(τD − τR). Thus the number of lost coincidences is given by
(17)
We can equate the two sums: (theoretical true coincidence rate plus theoretical random rate) equals (observed coincidence rate plus lost coincidences) i.e.
Rearrangement and simplification to the first order of approximation (i.e. setting nc /nβ = εγ and nc /nγ = εβ) gives an expression for N0.
(18)
If the background count rate is significant, equation (18) is modified only in the first factor which becomes (nβ − nBGβ)(nγ − nBGγ)/(nc − nBGc) where nBGβ etc. is the background count rate. The factor inside the curly bracket is unaltered.
Although the derivation of these and similar correction formulae may be complex, their application is not difficult since, in general, the corrections are finally expressed in terms of observables and, apart from measurement of τD and τR, no subsidiary determinations of constants has to be undertaken.
It has been suggested49,50 that the true disintegration rate N0 may be expressed in the form N0 = 1 + a1 N + a2 N2 + … where ai is a constant and N = (nβnγ/nc)f(ε), the apparent disintegration rate allowing for decay scheme corrections but not for time dependent corrections. If efficiencies and geometries can be kept strictly constant, the ai's may be calculated by determination of N for several sources whose relative strengths are known. This may be effected either by repeated measurements on a decaying source or by measurements on several sources prepared in an identical manner and whose only variable is the activity.
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17377 | https://pubmed.ncbi.nlm.nih.gov/10897162/ | Refractory idiopathic absence status epilepticus: A probable paradoxical effect of phenytoin and carbamazepine - PubMed
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. 2000 Jul;41(7):887-94.
doi: 10.1111/j.1528-1157.2000.tb00258.x.
Refractory idiopathic absence status epilepticus: A probable paradoxical effect of phenytoin and carbamazepine
I Osorio1,R C Reed,J N Peltzer
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1 Comprehensive Epilepsy Center, University of Kansas Medical Center, Kansas City, Kansas 66160, USA. iosorio@kumc.edu
PMID: 10897162
DOI: 10.1111/j.1528-1157.2000.tb00258.x
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Refractory idiopathic absence status epilepticus: A probable paradoxical effect of phenytoin and carbamazepine
I Osorio et al. Epilepsia.2000 Jul.
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. 2000 Jul;41(7):887-94.
doi: 10.1111/j.1528-1157.2000.tb00258.x.
Authors
I Osorio1,R C Reed,J N Peltzer
Affiliation
1 Comprehensive Epilepsy Center, University of Kansas Medical Center, Kansas City, Kansas 66160, USA. iosorio@kumc.edu
PMID: 10897162
DOI: 10.1111/j.1528-1157.2000.tb00258.x
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Purpose: To compare the frequency of seizures and status epilepticus and their response to first-line drugs in patients with idiopathic generalized epilepsies receiving carbamazepine or phenytoin to those receiving other drugs or no treatment.
Methods: We performed a retrospective chart review of all cases of idiopathic generalized epilepsies treated by the authors between 1985 and 1994. We compared seizure frequency and mean intravenous benzodiazepine dose required to control absence status epilepticus, intraindividually in subjects on carbamazepine or phenytoin before and after discontinuation of these compounds, and interindividually to subjects without treatment or receiving other drugs.
Results: Bouts of absence or tonic-clonic status epilepticus and seizures in subjects treated with phenytoin or carbamazepine at therapeutic concentrations were considerably more frequent and proved intractable to treatment with valproic acid or benzodiazepines, compared with a cohort of subjects also with idiopathic generalized epilepsies, but naive to, or receiving subtherapeutic or therapeutic doses of other agents.
Conclusions: Our observations strongly suggest that therapeutic concentrations of phenytoin and carbamazepine exacerbate idiopathic generalized epilepsies. Subjects in whom absence is one of the seizure types seem at a particularly high risk for responding paradoxically. These findings underscore the value of accurate classification of seizures and particularly the syndromic approach to diagnosis and point to the potential for iatrogenic complications with indiscriminate use of antiseizure drugs.
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Refractory idiopathic status epilepticus.Osorio I.Osorio I.Epilepsia. 2001 Feb;42(2):288-9. doi: 10.1046/j.1528-1157.2001.29300-2.x.Epilepsia. 2001.PMID: 11240606 No abstract available.
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17378 | https://elifesciences.org/articles/48767 | Skip to Content
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Induction of Sertoli-like cells from human fibroblasts by NR5A1 and GATA4
Jianlin Liang
Nan Wang
Jing He
Jian Du
Yahui Guo
Lin Li
Wenbo Wu
Chencheng Yao
Zheng Li
Kehkooi Kee
Tsinghua University, China;
National Institute of Biological Sciences, China;
Shanghai Jiao Tong University School of Medicine, Shanghai General Hospital, China;
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Jianlin Liang
Nan Wang
Jing He
Jian Du
Yahui Guo
Lin Li
Wenbo Wu
Chencheng Yao
Zheng Li
Kehkooi Kee
(2019)
Induction of Sertoli-like cells from human fibroblasts by NR5A1 and GATA4
eLife 8:e48767.
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Abstract
Sertoli cells are essential nurse cells in the testis that regulate the process of spermatogenesis and establish the immune-privileged environment of the blood-testis-barrier (BTB). Here, we report the in vitro reprogramming of fibroblasts to human induced Sertoli-like cells (hiSCs). Initially, five transcriptional factors and a gene reporter carrying the AMH promoter were utilized to obtain the hiSCs. We further reduce the number of reprogramming factors to two, NR5A1 and GATA4, and show that these hiSCs have transcriptome profiles and cellular properties that are similar to those of primary human Sertoli cells. Moreover, hiSCs can sustain the viability of spermatogonia cells harvested from mouse seminiferous tubules. hiSCs suppress the proliferation of human T lymphocytes and protect xenotransplanted human cells in mice with normal immune systems. hiSCs also allow us to determine a gene associated with Sertoli cell only syndrome (SCO), CX43, is indeed important in regulating the maturation of Sertoli cells.
Introduction
Sertoli cells are the first somatic cell type to differentiate in the testis and the only somatic cell type inside the seminiferous tubules. Sertoli cells play a critical role in directing testis morphogenesis and the creation of an immune-privileged microenvironment, which is required for male germ cell development. During early gonad development, male somatic cells express the male sex-determining gene SRY, which directs the sex-specific vascular development and seminiferous cord formation (Bott et al., 2006; Brennan et al., 2003; Koopman et al., 1990) via the initiation of a cascade of genes, including SOX9, FGF9, AMH and PGD2 (Barrionuevo et al., 2009; Moniot et al., 2009). NR5A1 (or SF1), GATA4, and WT1 are major transcriptional factors that direct somatic cells to become fetal Sertoli cells (Rotgers et al., 2018). Five transcriptional factors have been demonstrated to successfully reprogram mouse fibroblasts to Sertoli cells (Buganim et al., 2012). The expanding fetal Sertoli cells and another type of testicular somatic cell (i.e., peritubular cells) regulate the final organization and morphogenesis of the developing gonad into a testis (Griswold, 1998; McLaren, 2000).
Sertoli cells are the pivotal somatic cell regulators inside the seminiferous cord. Sertoli cells embed male germ cells during all differentiating stages and provide immunological, nutritional and structural support for germ cell development (Oatley and Brinster, 2012). Sertoli cells secrete the growth factors and cytokines needed for proper spermatogenesis, including the maintenance of spermatogonial stem cells, meiosis initiation of spermatocytes, and maturation of spermatozoa (Hai et al., 2014). Furthermore, Sertoli cells have the unique ability to modulate immunoreactions that protect the developing germ cells from immunological attacks. The immune-privileged potential of Sertoli cells has been utilized in many allo- and xeno-grafts to reduce the immune response in the field of cell transplantation (Kaur et al., 2015; Mital et al., 2010; Valdés-González et al., 2005). Preclinical studies have transplanted Sertoli cells with various other cell types for the treatment of diabetes, neurodegenerative diseases, Duchenne muscular dystrophy, skin allografts and other diseases (Luca et al., 2018).
Recently, co-cultures of differentiated rodent primordial germ cells and neonatal testicular somatic cells have successfully enabled meiosis completion and round spermatid formation in vitro (Zhou et al., 2016), highlighting the potential use of testicular somatic cells in the field of reproductive medicine although more experimental validations and improvements are needed. Human pluripotent stem cells have been differentiated to spermatid-like cells (Easley et al., 2012; Kee et al., 2009), but the co-culturing of stem cells with Sertoli cells could enhance the efficiencies of obtaining functional male gametes. However, the procurement of human Sertoli cells is not feasible because of biological and ethical constraints. The availability of donated Sertoli cells is limited, and expanding the limited number of human Sertoli cells in vitro remains a challenge (Chaudhary et al., 2005; Kulibin and Malolina, 2016). Therefore, the generation of Sertoli cells from fibroblasts could alleviate these issues and fulfill the basic research and clinical demands.
Direct lineage reprogramming has been considered a promising strategy for obtaining functional cell types with lower teratoma risks than directed differentiation of pluripotent stem cells (Cherry and Daley, 2012; Xu et al., 2015). The induction of cell type conversion between divergent lineages has been achieved using combinations of lineage-specific transcription factors (Hendry et al., 2013; Huang et al., 2014; Nam et al., 2013; Yamanaka and Blau, 2010). Fibroblasts are common cells in animal connective tissues that can be conveniently obtained from patients. Therefore, fibroblasts are often used as initiating cells in many lineage reprogramming experiments. The direct reprogramming of Sertoli cells from fibroblasts has been demonstrated in mouse (Buganim et al., 2012), but the direct lineage conversion of human Sertoli cells from fibroblasts has not been described. Here, we report the efficient induction of human Sertoli cells (hiSCs) from both primary human fibroblasts and fibroblasts derived from human embryonic stem cells (hESCs). These hiSCs exhibit an epithelial morphology, lipid droplet accumulation, and transcriptomes similar to those of primary Sertoli cells; sustain the growth of mouse spermatogonia cells; and perform immune-privileged function during transplantation experiments.
Connexin 43 (CX43) is a predominant gap junction protein expressed in BTBs that affects the maturation of Sertoli cells and spermatogenesis (Brehm et al., 2007; Gerber et al., 2016; Sridharan et al., 2007; Weider et al., 2011). The deletion of CX43 in Sertoli cells, but not germ cells, causes infertility in mice (Brehm et al., 2007; Günther et al., 2013). The absence of CX43 expression in human Sertoli cells is associated with Sertoli cell-only syndrome (SCO) and impaired spermatogenesis in male patients (Brehm et al., 2002; Defamie et al., 2003), but whether the deletion of CX43 directly affects the characteristics of human Sertoli cells has not been demonstrated. Utilizing our in vitro hiSC model, we demonstrate that the deletion of CX43 affects the transcriptome profile and maturation of hiSCs.
Results
Testing the reprogramming capability of five putative transcriptional factors
Based on the reprogramming capability of the transcriptional factors reported in a mouse study (Buganim et al., 2012), we first tested the reprogramming capabilities of the human homologs of the five transcriptional factors (5TFs: NR5A1, GATA4, WT1, SOX9 and DMRT1) to convert human fibroblasts to hiSCs. All five human homologs were correctly cloned into lentiviral vectors and expressed at high levels as verified by immunofluorescent staining of HPFs and transcriptional level in human embryonic kidney cells (Figure 1A and Figure 1—figure supplement 1A). After the lentiviral transduction with all five factors and culturing in selective medium for 5 days (Figure 1B), many HPFs started to transform from the typical elongated morphology of fibroblasts into the squamous morphology that typically appears in epithelial cells (Figure 1—figure supplement 1B). The analysis of the transcriptional expression showed that genes enriched in mesenchymal-to-epithelial transition (Buganim et al., 2012; Li et al., 2011; Samavarchi-Tehrani et al., 2010), including MUC1, CLDN1, CLDN7, CLDN11 and TJP1, exhibited increased expression in this mixed population of transformed HPFs (Figure 1—figure supplement 1C). In addition, the transcriptional expression of genes enriched in Sertoli cells, such as AR, KRT18, CLU, PTGDS, SCF, BMP4 and INHA, exhibited increased expression in this mixed population of transformed HPFs (Figure 1—figure supplement 1D). To determine whether these 5TFs were able to reprogram other fibroblast sources, we derived human fibroblast-like cells from hESC line H1 (dH1) and reprogrammed these cells as described for the HPFs. The dH1 morphology resembled that of fibroblasts, and no detectable expression of pluripotent markers was observed, but the expression of many markers of fibroblasts was observed (Figure 1—figure supplement 2A, B and C). After the transduction of the 5TFs, dH1 underwent a fibroblast to epithelial transformation similar to that observed in the HPFs (Figure 1—figure supplement 2D), suggesting that the 5TFs can transform both types of fibroblasts into epithelial-like cells and increase the expression of Sertoli cell markers.
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To isolate and enrich the hiSCs in the mixed population of reprogrammed fibroblasts, we constructed a gene reporter system utilizing the 1.6 kb promoter region of AMH, which is a gene specifically expressed in Sertoli cells (Franke et al., 2004), connected to EGFP (Figure 1C). Recent report of single-cell transcriptome analysis also confirms that AMH is specifically expressed in Sertoli cells but not other somatic cells of adult human testis (Wang et al., 2018). We constructed a gene reporter carrying AMH promoter fusing to EGFP gene and transduced the reporter into adult human Sertoli cells to confirm EGFP expression (Figure 1—figure supplement 3). When the lentivector carrying the reporter, AMH:EGFP, was stably integrated into HPFs and dH1, none of the cells would express EGFP without transcriptional induction. After the transduction and selection of the fibroblasts, some reprogrammed cells were expected to express EGFP to allow us to isolate them by fluorescence-activated cell sorting (FACS). We found that both HPF and dH1 reproducibly yielded a clear AMH:EGFP-positive population after 10 days of transduction with the 5TFs (Figure 1D). Intriguingly, the AMH:EGFP+ population in the dH1 group (~15%) was much higher than that in the HPF group (~6%), indicating that the conversion of the dH1 cells was more efficient. The AMH:EGFP+ cells were isolated by FACS and adhered to culture dishes and exhibited an epithelial morphology (Figure 1E). We verified that the endogenous AMH expression was activated because the expression level of the AMH gene was significantly upregulated in the AMH:EGFP+ cell population compared to that in the control dH1 cells (CTRL) and AMH:EGFP- cells (Figure 1F). The specific expression of endogenous AMH in AMH:EGFP+ cells validated the function of the reporter to isolate the reprogrammed Sertoli cells from the mixed populations which might contain other cell types. Moreover, an epithelial marker expressed in Sertoli cells, that is, KRT18, was used to validate the transformation of the fibroblasts to Sertoli cells as one kind of epithelial cells. The cytoskeleton pattern of KRT18 expression was observed in the AMH:EGFP+ cells but not the control dH1 cells (Figure 1G).
Whole-genome transcriptional profiling of hiSCs resembling adult Sertoli cells
To determine whether hiSCs reprogrammed with the 5TFs (5F-hiSCs) are similar to human Sertoli cells, we compared the transcriptomes of the AMH:EGFP+ 5F-hiSCs, dH1 cells infected with p2k7 empty virus (dH1-2K7) as negative controls, and primary adult Sertoli cells (aSCs) from human biopsy samples (Wen et al., 2017). We focused our analysis on the differentially expressed genes (DEGs,>2 fold change, p-value<0.05) between 5F-hiSCs and dH1-2K7 and between aSCs and dH1-2K7. In total, 7533 genes were differentially expressed between 5F-hiSCs and dH1-2K7, including 4528 upregulated genes and 3005 downregulated genes (Figure 2A and B). Additionally, 5377 genes were differentially expressed between aSCs and dH1-2K7, including 3343 upregulated genes and 2034 down-regulated genes. The Venn analysis showed that 3626 genes were shared among the DEGs in both hiSCs and aSCs, accounting for approximately 67% of the DEGs in aSCs. Among this shared group of DEGs (CO-DEGs), 1973 genes were upregulated, while 1314 genes were down-regulated in both the hiSCs and aSCs (Figure 2C), indicating that the trends in transcriptional expression between the hiSCs and aSCs were the same in these genes. The cluster analysis of dH1-2K7, hiSCs and aSCs also showed that the CO-DEGs had a similar expression pattern between the hiSCs and aSCs, and consistency was observed between duplicate samples (Figure 2D). The Gene Ontology (GO) analysis of the CO-DEGs showed that among the 1973 upregulated genes, many genes were involved in the regulation of cell communication, regulation of immune response processes, response to hormones, and lipid metabolic process, whereas among the 1314 down-regulated genes, many genes were involved in the mitotic cell cycle and microtubule-based processes (Figure 2E). These changes in gene expression indicated that the hiSCs acquired unique cellular characteristics that were distinct from the original fibroblasts. To further confirm that the AMH:EGFP+ 5F-hiSCs have the signature of Sertoli cells, we examined the expression of several Sertoli cell markers, including CLU, NCAM2, DHH, ERBB4, INHB, INHA, SHBG, GATA6, CDKN1B, TGFα and LMMA3, by quantitative PCR (qPCR). Compared to the control cells, all Sertoli cell markers were highly enriched in the AMH:EGFP+ 5F-hiSCs (Figure 2F). Taken together, the transcriptional profile of the AMH:EGFP+ 5F-hiSCs resembled that of the aSCs, and many Sertoli cell markers were expressed.
Figure 2
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NR5A1 and GATA4 are sufficient to reprogram fibroblasts to hiSCs as 5F-hiSCs
Although the 5TFs yielded the AMH:EGFP+ hiSCs, the combination of all 5TFs may not be necessary to reprogram fibroblasts to hiSCs. Therefore, we used fewer transcription factors to generate hiSCs and compared the percentage of AMH:EGFP+ cells in all 31 combinations of NR5A1, GATA4, SOX9, WT1 and DMRT1. The FACS results indicated that 16 combinations of transcriptional factors yielded varying levels of AMH:EGFP+ cells after 10 days of reprogramming (Figure 3A). NR5A1 was the only common factor found in all 16 combinations in AMH:EGFP+ cells. This transcriptional factor alone generated approximately 3.79% of AMH:EGFP+ cells, and the combination of all 5TFs produced the highest percentage of AMH:EGFP+ cells. Surprisingly, the combinations with NR5A1 and GATA4 generated as many AMH:EGFP+ cells as all 5TF combined. Moreover, all combinations containing NR5A1 and GATA4 resulted in a similar level higher than the combinations with NR5A1 (Figure 3B). The AMH:EGFP+ cells generated by 2TFs and 5TFs showed similar morphologies, including a large cell body with an epithelial morphology, and expressed KRT18 (Figures 1G and 3C).
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Then, we analyzed and compared the transcriptome of the 2F-hiSCs and with the transcriptome of the aSCs and 5F-hiSCs. We identified the common DEGs among the dH1/aSCs, dH1/5F-hiSCs (dH1), dH1/2F-hiSCs (dH1) and dH1/5F-hiSCs (HPF) and performed hierarchical clustering using their FPKM values. The analysis revealed that the transcriptome profiles of the 2F-hiSCs were more similar to the profiles of the 5F-hiSCs and adult Sertoli cells (aSCs) than to the dH1 and HPF profiles (data not shown). To identify the putative signature genes similar among the 2F-hiSCs, 5F-hiSCs and aSCs, we generated a heat map of the 1638 CO-DEGs and carried out gene correlation clustering. Notably, the differentially expressed genes were grouped into three groups of 512, 689 and 437 genes (Figure 3D). The Gene Ontology analysis showed that many of the 512 highly expressed genes mostly shared by the 2F-hiSCs, 5F-hiSCs and aSCs were involved in reproductive structure development, immune effector processes and response to hormones. These genes, including KRT18, PTGDS and SOX9, are used as markers of Sertoli cells or are highly expressed in Sertoli cells (Buganim et al., 2012; Sharpe et al., 2003). We further compared the expression of 59 genes that are highly expressed in Sertoli cells (Bouma et al., 2010; Boyer et al., 2004; Mincheva et al., 2018) among the 2F-hiSCs, 5F-hiSCs and aSCs. All three groups exhibited similar expression of many markers that are expressed in more mature Sertoli cells, including CDKN1B (or p27kip1) and CLU (Wang et al., 2016). However, some markers in the aSCs, including NCAM2, INHA and KRT18 (Kanatsu-Shinohara et al., 2012; Wang et al., 2016), were expressed at lower levels (Figure 3—figure supplement 1) than those in the other two hiSCs. The Sertoli cell marker expression was very similar between the 2F-hiSCs and 5F-hiSCs. Principal component analysis (PCA) indicated that aSCs, 2F-hiSCs (dH1), 5F-hiSCs (dH1) clustered closely, confirming their co-expressed gene profiles are similar (Figure 3—figure supplement 2). SOX9 expressed at similar level in aSCs, dH1 and 2F-hiSCs but not in undifferentiated hESCs, showing that SOX9 expression increased after hESCs became fibroblasts, and maintained similar expression during 2F-hiSC formation (Figure 3—figure supplement 3). Therefore, we focused on the 2F-hiSCs for the subsequent more thorough characterization.
Although we used AMH:EGFP reporter to isolate Sertoli-like cells, it was still possible that the isolated cells were Leydig cells. Hence, we examined whether the cells expressed any Leydig cell marker 3β-HSD (Zhang et al., 2015). We confirmed that neither 2F-hiSCs nor aSCs expressed any 3β-HSD (Figure 3—figure supplement 4A). Based on the RNA-sequencing results, we also confirmed other gene markers of Leydig cells were not detected or expressed at low level (Figure 3—figure supplement 4B).
We also validated that NR5A1 and GATA4 was able to reprogram primary human skin fibroblasts (HSF) and yielded AMH:EGFP+ hiSCs (Figure 3—figure supplement 5A, B and C). 12.9% of AMH:EGFP+ cells were induced after overexpression of the 2TFs. KRT18 were only detected in the AMH:EGFP+ cells after induction along with other genes enriched in Sertoli cells (Figure 3—figure supplement 5D and E). These results further supported that 2TFs can reprogram other primary human cell types.
2F-hiSCs attract human endothelial cells and accumulate lipid droplets
Sertoli cells mediate the migration of endothelial cells to seminiferous tubules during testicular cord formation (Brennan et al., 2003; Cools et al., 2011). We investigated whether the hiSCs attracted human umbilical vein endothelial cells (HUVECs). The migration assay showed the number of HUVECs that were attracted and passed through the membrane was ~1.5 fold higher in both 2F-hiSCs and aSCs compared to the controls (Figure 4A,B). These results indicate that significantly more endothelial cells were attracted by the conditioned medium collected from the 2F-hiSCs than by that collected from the dH1-2K7 cells.
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Another unique feature of Sertoli cells in humans and other mammalian species is the accumulation of lipid droplets in the cytoplasm (Gorga et al., 2017; Wang et al., 2006). The Gene Ontology analysis showed that genes that participate in lipid metabolism processes were upregulated in the 2F-hiSCs, 5F-hiSCs and aSCs (Figure 3D), supporting the presence of high numbers of lipid droplets in the hiSCs. We used BODIPY to stain and count the number of cells with high numbers of lipid droplets to determine whether lipid droplets appeared in the 2F-hiSCs. The average percentage of cells that exhibited strong BODIPY positivity was approximately 15%, 40% and 60% in the control cells, 2F-hiSCs, and aSCs respectively (Figure 4C,D). Therefore, the percentage of cells containing high quantities of lipid droplets was higher in both 2F-hiSCs and aSCs than that in the dH1-2K7 cells. Oil-red O staining also confirmed the same result showing lipid droplet enrichment in 2F-hiSCs and aSCs (Figure 4—figure supplement 1).
2F-hiSCs sustain in vitro culturing of mouse spermatogonia cells
Mouse spermatogonia cells were isolated from seminiferous tubules of 6 dpp mice according to established protocols and co-cultured with dH1 or 2F-hiSCs to examine whether 2F-hiSCs sustain the growth of male germ cells (Figure 5A, Figure 5—figure supplement 1A, B, C, D, E and F). The observations that more mouse germ cells attached and survived on the 2F-hiSCs than dH1 cells began at approximately 12 hr of co-culturing (Figure 5—figure supplement 1G). The morphology of the round cells attached to the hiSCs appeared alive and resembled spermatogonia cells but the cells attached to dH1 appeared apoptotic and degenerated. After 48 hr of co-culturing, the samples were fixed and immunostained to further confirm that the 2F-hiSCs formed attachments with the spermatogonia cells. The immunostaining of the germ cell-specific marker DAZL (to identify mouse spermatogonia cells) and human-nuclear specific marker NuMA (to identify hiSCs and dH1) indicated that significantly more DAZL-positive spermatogonia cells attached to the hiSCs, but almost no DAZL-positive cells attached to the dH1 cells despite the similar numbers of plated hiSCs and dH1 cells (Figure 5B and C). Similar immunostaining patterns were observed when aSCs were co-cultured with mouse spermatogonia cells (Figure 5—figure supplement 2). Sertoli cells directly contact male germ cells in seminiferous tubules in vivo, and we hypothesized that the hiSCs would directly contact the spermatogonia cells. We immunostained the co-cultured cells with DAZL and KRT18 and found that many DAZL-positive cells localized to areas occupied by hiSCs, typically at the edge of the cell bodies (Figure 5D).
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2F-hiSCs suppress T cell proliferation, IL-2 production and protect transplanted human cells
A specialized property of Sertoli cells is their functional role in creating an immune-privileged environment in seminiferous tubules to protect germ cells from immunological attacks. Previous studies have shown this unique function, which has been exploited in therapeutic transplantation for the protection of many other cell types (Kaur et al., 2015). We first investigated whether the medium from the 2F-hiSCs cells exhibited any suppressive effect on the proliferation of Jurkat E6 cells (human T lymphocytes) to examine whether the 2F-hiSCs could suppress the immunoreaction of immunological cells. The suppressive effect was evaluated using an assay commonly used to determine the active metabolism of a proliferating cell, that is, WST-1 (Chui et al., 2011; Figure 6A). The Jurkat E6 cells were treated with various concentrations of 2F-hiSCs-conditioned medium or aSCs-conditioned medium and exhibited a significant dose-responsive decrease in cell proliferation compared to that in cells treated with dH1-2K7-conditioned medium (Figure 6B). The proliferation level was ~35% lower in the Jurkat lymphocytes exposed to the highest concentration of the 2F-hiSC-conditioned medium or aSCs-conditioned medium. We collected the Jurkat cells and analyzed the protein levels of interleukin-2 (IL-2), which plays an essential role in the immune system. The ELISA indicated that the IL-2 levels were significantly lower in the cells cultured with the 2F-hiSCs-conditioned or aSCs-condictioned medium than those in the cells cultured with the dH1-2K7-conditioned medium (Figure 6C, figure 6-figure supplement 1C). 22 genes previously known to participate in immune-modulation (Kaur et al., 2014; Mital et al., 2010) or categorized as immune effector gene by gene ontology were activated after reprogramming of dH1 (Figure 6D), further supported that hiSCs acquired immunosuppressive function.
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Approximately, 1.3 × 106 human 293FT cells stably integrated with a luciferase-expressing vector were co-transplanted with 2.5 × 105 dH1-2K7, 2F-hiSCs or aSCs into mice with normal immune systems via hypodermic injection to determine whether the 2F-hiSCs could protect xenotransplanted cells. The transplantation experiment was performed to investigate the immunosuppressive effects at different locations (foreleg, hindleg, left and right sides of the animals as indicated on the figures) in different animals, and the transplanted sites were monitored for up to 10 days. D-luciferin was injected into the animals to follow the surviving transplanted 293FT cells, and the signal was monitored using live imaging 15 min post-injection beginning 3 days after transplantation. The transplanted 293FT cells gradually diminished in the immunocompetent mouse from day 3 to day 10 (mouse #1, #2) as indicated by the reduced luciferase activity at the transplanted sites (Figure 6E, Figure 6—figure supplement 1A). All 293FT cells co-transplanted with hiSCs exhibited higher luciferase activity, which ranged from 1.7- to 3.9-fold, 3 days after transplantation. Three of the four groups of transplanted cells survived until day 10, and two of the three groups of 293FT cells with hiSCs survived at least 10 days after transplantation with strong luciferase activity. In contrast, their counterpart control cells exhibited less than 40-fold or no detectable luciferase activity (mouse #1 foreleg group and mouse #2 hindleg group). Similar level of protection was observed when the same number of 293FT cells were transplanted with aSCs into mouse #3 and #4 (Figure 6—figure supplement 1B). The transplants at day 10 were collected and dissected to examine the remaining cell types. NuMA staining indicated that there were many human cells survived in the transplants of 2F-hiSCs and some cells clustered together as cell aggregates (Figure 6—figure supplement 2A). These human cells could be 293FT cells or 2F-hiSCs. Positive immunostainings of NR5A1 confirmed the existence of 2F-hiSCs in the transplant but not the control transplant or the surrounding mouse tissue (Figure 6—figure supplement 2B).
hiSCs forms cell aggregates and do not actively proliferate
Mouse Sertoli cells forms cell aggregates during in vitro culturing (Buganim et al., 2012). 2F-hiSCs also exhibited similar morphology of spherical cellular aggregates when cultured in 10% FBS on matrigel (Figure 6—figure supplement 3A). Different from the mouse Sertoli cells, we observed formation of partial ring-like structures when 2F-hiSCs were cultured in 2% FBS medium (Figure 6—figure supplement 3B). Adult human Sertoli cells tend to be more quiescent and less proliferative (Sharpe et al., 2003). We examined the proliferation of 2F-hiSCs by EDU staining and found that these cells were not proliferative (Figure 6—figure supplement 4A and B).
CX43 deletion disrupts gap junctions and alters the expression profile of hiSCs
We investigated whether the deletion of the gap junction protein CX43 could affect hiSCs formation and determined whether hiSCs exhibit the same genetic requirements for development as Sertoli cells in vivo. We created a homozygous deletion of CX43 hESC line, derived fibroblasts from this line and compared the reprogramming efficiency of the hiSCs (Figure 7A). We successfully generated three targeted mutations at CX43, that is, one single allele mutation (6#) and two double allele mutations (21# and 26#) at Exon 2 of the CX43 gene, using a CRISPR-CAS9 gene editing system (Figure 7B). The protein expression was completely disrupted in the CX43-/- (21#) and CX43-/- (26#) cell lines, but the heterozygous protein level remained similar to the wild-type CX43 level based on the Western blot analyses (Figure 7—figure supplement 1A). The immunostaining of CX43 (Figure 7C) and photo bleaching assay (Figure 7D and E) of ES-derived fibroblasts (dH1) both showed that the expression of CX43 and gap junctions between neighboring cells were disrupted because there was no detectable CX43 staining or diffusing fluorescent dye recovered after photo bleaching in the CX43-/- (26#) cell line.
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We compared the reprogramming efficiency of the 2F-hiSCs between the wild-type dH1 and CX43KO (CX43-/- (#26)) cell lines. The time course experiments showed that the percentage of AMH:EGFP+ in the WT hiSCs peaked at ~13.8% on day 15 and decreased to 3.3% on day 25 (Figure 7F and G; Figure 7—figure supplement 1B). Remarkably, the percentage of AMH:EGFP+ in the CX43-/- (#26) cell line was 23.9% on day 15 and decreased to 18.5% on day 25, but the overexpression of CX43 in the deletion cell line revealed a much lower percentage of AMH:EGFP+ from days 4 to 25. Therefore, the expression level of CX43 in the cells was indirectly proportional to the percentage of AMH:EGFP+ cells. This result is consistent with the higher AMH expression in CX43 knockout mice reported in a previous study (Weider et al., 2011) and suggests that the effect of the CX43 deletion leads to the dedifferentiation of Sertoli cells to a less mature state.
The high percentage of AMH:EGFP+ cells in the CX43KO cells may be due to their more immature status than that of WT AMH:EGFP+ hiSCs. Therefore, we compared the transcriptional profiles of these two populations to examine whether CX43KO affected gene expression or any cellular processes. The volcano analysis revealed that 2736 genes were differentially expressed (p-value less than 0.01) between the CX43KO 2F-hiSCs and WT 2F-hiSCs (Figure 7—figure supplement 2A). We found 754 genes with a difference in the transcript level greater than two-fold; 512 genes were down-regulated and 242 genes were upregulated in the CX43KO 2F-hiSCs compared to those in the WT 2F-hiSCs (Figure 7—figure supplement 2B). We further analyzed the 754 genes using a heat map and GO analysis to identify specific genes or processes affected by CX43KO. The genes were classified into eight gene sets according to the expression patterns among the WT dH1, WT 2F-hiSCs, CX43KO dH1 and CX43KO 2F-hiSCs (Figure 7H, Figure 7—figure supplement 2C). The genes in Groups 1 and 4 exhibited lower expression levels in the WT 2F-hiSCs and CX43 2F-hiSCs than in the other two groups of control cells, suggesting that these genes were affected by the reprogramming process. The genes in Groups 5 and 6 exhibited similar patterns between the WT cells and CX43KO cells, suggesting that these two groups were affected by CX43KO. In contrast, the expression profiles of the genes in Groups 2, 3, 7, and eight in the WT and CX43KO hiSCs differed from those in their counterpart controls and between the WT and CX43KO hiSCs. These genes may reflect the cellular maturation status of these two reprogrammed cell populations. The GO analysis of enriched terms in these four groups revealed that Group two was enriched with genes involved in catabolic processes, including nucleobase-containing compound catabolic process, and Group three contained genes involved in steroid metabolic or lipid biosynthetic processes. Group seven included many genes that participated in the cytoskeleton, and Group eight contained genes related to gamete generation.
Ten marker genes previously reported to be more highly expressed in mature or immature Sertoli cells (Sharpe et al., 2003; Wang et al., 2016) were chosen for the expression level comparisons between WT and CX43KO hiSCs. The CX43KO hiSCs exhibited a higher expression of some immature markers and lower expression of mature markers, suggesting that the CX43KO cells were more immature than the WT hiSCs.
Discussion
This study reported that human fibroblasts can be reprogrammed to Sertoli-like cells using 5TFs (NR5A1, GATA4, WT1, DMRT1 and SOX9) or 2TFs (NR5A1 and GATA4). Fibroblasts from two sources, that is, human pulmonary fibroblasts and fibroblasts derived from human embryonic stem cells, were reprogrammed to Sertoli-like cells and exhibited transcriptomes similar to those of primary adult Sertoli cells. The present study generated a Sertoli cell-specific gene reporter, that is, AMH:EGFP, to enhance the efficiency of isolating a relatively pure population of hiSCs from a mixed population of cells at different reprogramming stages (Figure 1). AMH has been used in many conditioned knockout studies to specify the expression of Cre recombinase in mouse Sertoli cells (Lécureuil et al., 2002). Recent single-cell transcriptome study of adult human testicular cells indeed confirmed that AMH is specifically expressed in Sertoli cells but not other somatic cells or germ cells (Wang et al., 2018). We cloned the human promoter of AMH and fused it to EGFP and showed that the cell population expressing EGFP appeared only after reprogramming, and these cells expressed many Sertoli cell marker genes. This isolation step eliminates the possibility that the reprogrammed cells we studied were other somatic cells such as Leydig cells which share some gene markers. The success of reprogramming human fibroblasts with the same factors used in a previous mouse study (Buganim et al., 2012) confirms that the reprogramming capability of these 5TFs is conserved between humans and mice.
Another achievement of this study was the reduction in the reprogramming factors from 5TFs to 2TFs. NR5A1, which is often called SF1, was the most essential of the five factors because almost no AMH:EGFP+ cells appeared in the reprogramming experiments without NR5A1, even if GATA4, WT1, DMRT1 and SOX9 were overexpressed (Figure 3A and B). Notably, only the addition of GATA4, but not the addition of the three other TFs, to the reprogramming combination increased the percentage of AMH:EGFP+ cells. Several TFs, including NR5A1 and GATA4, are key regulators establishing the mouse Sertoli cell identity (Rotgers et al., 2018). Therefore, unsurprisingly, NR5A1 and GATA4 were sufficient to reprogram human fibroblasts to Sertoli-like cells. WT1 and SOX9 are also determining factors in mouse Sertoli cells, but their reprogramming abilities were lower than those of NR5A1 and GATA4. SOX9 interacts with NR5A1 to trigger the specific onset of AMH expression (Rotgers et al., 2018), but SOX9 overexpression in our reprogramming experiments only slightly increased the percentage of AMH:EGFP+ cells in a few combinations of TFs (Figure 3A and B).
The cellular characterizations of the 2F-hiSCs showed that these cells carried many known properties of Sertoli cells. Lipid metabolism in Sertoli cells is important for providing nutritional and energy supplies to germ cells (Gorga et al., 2017). Previous studies have shown that the high metabolic activity requirements for lipid and steroid synthesis are associated with the differentiated status of Sertoli cells (Johnston et al., 2008). Notably, the hiSCs in the present study exhibited a high expression of lipid and steroid related genes (Figure 3D), and the WT hiSCs exhibited a higher expression than the CX43KO hiSCs (Figure 7H). These results suggest that the hiSCs were similar to terminally differentiated Sertoli cells in vivo. The lipid droplet staining assays clearly showed the presence of high numbers of lipid droplets in the hiSCs but not the fibroblasts used for reprogramming (Figure 4C).
Sustaining the viability and differentiation of spermatogonia cells is the most essential function of Sertoli cells. We examined the ability of hiSCs to sustain mouse spermatogonia cells because of the inaccessible procurement of human spermatogonia cells from hospitals. Previous studies performing xenotransplantation experiments transplanting primate spermatogonia stem cells into mouse testis demonstrated that primate germ cells colonized in the seminiferous tubules of the recipient mice, but the primate germ cells did not differentiate because of the evolutionary differences between these species (Nagano et al., 2001). The isolated mouse spermatogonia cells in our study attached and survived on the hiSCs, but no differentiated spermatocytes were detected (Figure 5B; data not shown). These results suggest that the reprogrammed Sertoli cells were only capable of providing a niche for the survival of male germ cells in vitro. Recent studies have reported in vitro derivations of immature human gametes from hESCs (Easley et al., 2012; Jung et al., 2017; Kee et al., 2009). The male and female gametes in these studies were immature likely due to the absence of fully competent somatic cells in the differentiating culture. Therefore, adding induced Sertoli cells or granulosa cells to in vitro differentiating cultures may be a key step in obtaining fully functional gametes in vitro.
Another prominent role of Sertoli cells is the creation of an immune-privileged environment to protect germ cells from immune attacks of lymphocytes. Many studies have documented that Sertoli cells secrete factors that suppress the proliferation of T cells, B cells and NK cells (Kaur et al., 2014; Luca et al., 2018), including the suppression of IL-2 from T cells (Selawry et al., 1991). Our results indicate that culture medium incubated with hiSCs suppress the proliferation of Jurkat cells (the immortalized cell line of human T lymphocytes) and reduce IL-2 production in Jurkat cells treated with hiSC-conditioned medium (Figure 6B and C). Many genes previously known to participate in immune effector processes were upregulated in the hiSCs compared to those in fibroblasts (Figure 6D), suggesting that hiSCs modulate immune suppression via multiple immunological pathways. Notably, all 293FT cells co-transplanted with hiSCs into two different immunocompetent mice survived longer than their counterpart control cells co-transplanted with fibroblasts (Figure 6E and Figure 6—figure supplement 1A,B). These results suggest that the hiSCs protected the xenotransplanted 293FT cells from host immune cells. Previous studies have reported that Sertoli cells protect many cell types in allogenic and xenogenic transplantation, and several studies used immunosuppressive drugs or immune-deficient mice in long-term follow-up experiments (Kaur et al., 2015; Mital et al., 2010). One study xenotransplanted porcine Sertoli cells with rat islet cells into the kidney capsule of immunocompetent rats (Yin et al., 2009). The islet allografts survived 8 to 9 days when 1.5 × 106 porcine Sertoli cells were co-transplanted. In our experiments, 2.5 × 105 hiSCs were co-transplanted with 293FT cells and survived at least 10 days in two mice. Therefore, our xenotransplanted 293FT cells survived a similar or longer duration with lower numbers of reprogrammed human Sertoli cells in immunocompetent mice. Long-term transplantations of hiSCs with clinically relevant cell types, including pancreatic islets and skin grafts, should be investigated for potential clinical applications, such as the treatment of diabetes and skin burns. Our hiSCs have the advantage of being of human origin, which alleviates the issue of xenotransplantation and potential of animal virus infection.
Reprogrammed Sertoli cells may be used as a model for examining the cellular and genetic mechanisms of human Sertoli cell biology. A previous study reported an association between Sertoli-only syndrome (SCO) and the lower mRNA expression of CX43 (Defamie et al., 2003) and suggested that the absence of CX43 rendered the Sertoli cells more immature. However, the precise mechanism by which the absence of CX43 causes infertility is unclear. Our studies revealed that multiple pathways, including lipid metabolism and nucleobase catabolism, were more highly expressed in the WT than in the CX43KO hiSCs, indicating that the absence of CX43 disrupts these molecular pathways in Sertoli cells. The expression of several markers of immature Sertoli cells was higher and the expression of markers of mature Sertoli cells was lower in CX43KO than in the WT hiSCs. Taken together, our results suggest that the deletion of CX43 disrupts multiple molecular pathways and delays the maturation of Sertoli cells. Infertility in men may be caused by genetic mutations in germ cells or Sertoli cells. Genetic or cellular defects in Sertoli cells have not been well studied to understand how they cause infertility due to a lack of in vitro model of human Sertoli cells. Our in vitro induced system now provides a platform for basic research and potential treatment of male infertility caused by Sertoli cells in the future.
Materials and methods
Key resources table
| Reagent type (species) or resource | Designation | Source or reference | Identifiers | Additional information |
--- ---
| Mouse | C57BL/6 | Vital River Laboratory Animal Technology |
| Cell line (Homo-sapiens) | 293FT cells | Thermo Fisher Scientific | Cat# R70007 |
| Cell line (Homo-sapiens) | H1 ES cells | WiCell Research Institute | Cat# WA01 |
| Cell line (Homo-sapiens) | Human Pulmonary Fibroblasts | National Infrastructure of Cell Line Resource | Cat# CCC-HPF-1 (PUMC, Beijing) |
| Transfected construct (Homo-sapiens) | p2k7-EF1alpha-NR5A1 | This paper | Lentiviral construct to express target gene |
| Transfected construct (Homo-sapiens) | p2k7-EF1alpha-GATA4 | This paper | Lentiviral construct to express target gene |
| Transfected construct (Homo-sapiens) | p2k7-EF1alpha-SOX9 | This paper | Lentiviral construct to express target gene |
| Transfected construct (Homo-sapiens) | p2k7-EF1alpha-WT1 | This paper | Lentiviral construct to express target gene |
| Transfected construct (Homo-sapiens) | p2k7-EF1alpha-DMRT1 | This paper | Lentiviral construct to express target gene |
| Transfected construct | p2k7-EF1alpha-luciferin | This paper | Lentiviral construct to express target gene |
| Transfected construct (Homo-sapiens) | p2k7-EF1alpha-CX43 | This paper | Lentiviral construct to express target gene |
| Recombinant DNA reagent | p2k7-AMH-GFP | This paper | Lentiviral construct for AMH reporter |
| Recombinant DNA reagent | pX335-U6-Chimeric_BB-CBh-hSpCas9n(D10A) | Addgene #42335 | Expressing Cas9 and gRNA (Cong et al., 2013) |
| Biological sample (Homo-sapiens) | Primary adult human Sertoli cells | Shanghai Jiao Tong University | Freshly isolated |
| Antibody | anti-DAZL(Mouse polyclonal) | AbD Serotec | Cat#: MCA2336; RRID: AB_2292585 | IF(1:50) |
| Antibody | anti-NuMA (Rabbit polyclonal) | Abcam | Cat#: Ab97585; RRID: AB_GR27454-16 | IF(1:100) |
| Antibody | anti-human KRT18 (Rabbit polyclonal) | Proteintech | 10830–1-AP | IF(1:100) |
| Antibody | anti-SOX9 (Rabbit monoclonal) | Abcam | Cat#: Ab170660; RRID: AB_GR155689-1 | IF(1:100) |
| Antibody | anti-human CX43 (Rabbit polyclonal) | Cell Sigaling | Cat#: 3512S | IF(1:100) |
| Antibody | anti-human DMRT1 (Rabbit polyclonal) | Abcam | Cat#: Ab1786 | IF(1:100) |
| Antibody | anti-human WT1 (Rabbit polyclonal) | Proteintech | Cat#: 12609–1-AP | IF(1:100) |
| Antibody | anti-human GATA4 (Rabbit polyclonal) | Proteintech | Cat#: 19530–1-AP | IF(1:100) |
| Antibody | anti-human NR5A1 (Rabbit polyclonal) | Proteintech | Cat#: 18658–1-AP | IF(1:100) |
| Antibody | donkey anti-Mouse IgG (H+L) Highly Cross-Adsorbed Secondary Antibody, Alexa Fluor 488 | Invitrogen | Cat#: A-21202; RRID: AB_141607 | IF(1:1000) |
| Antibody | donkey anti-Mouse IgG (H+L) Highly Cross-Adsorbed Secondary Antibody, Alexa Fluor 555 | Invitrogen | Cat#: A-31572; RRID: AB_1567203 | IF(1:1000) |
| Strain, strain background (Escherichia coli) | TransStbl3 Chemically Competent Cell | TransGen Biotech | Cat#: CD521 |
| Peptide, recombinant protein | bFGF, Recombinant Human FGF basic Protein | R and D Systems | Cat#: 233-FB-CF |
| Chemical compound, drug | G418, Geneticin | Thermo Fisher Scientific | Cat#: 10131035 |
| Chemical compound, drug | Blasticidin | Thermo Fisher Scientific | Cat#: R21002 |
| Software, algorithm, website | ImageJ | NIH | |
| Software, algorithm, website | FlowJo (v10.3) | BD | |
| Software, algorithm, website | Prism7 (v7.0 a) | Graphpad Software | |
| Software, algorithm | DAVID Bioinformatics Resources (v6.8; GO) | |
| Software, algorithm, website | Tophat2/cufflinks | (Kim et al., 2013) | |
| Software, algorithm, website | R (v3.4.1; PCA, cluster and DEG) | |
| Other | DAPI stain | Invitrogen | D1306 | (1 µg/mL) |
Experimental model and subject details
Animal Care and Use
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C57BL/6 mice were purchased from Vital River Laboratory Animal Technology Co., Ltd (Beijing, China). All animal maintenance and experimental procedures were performed according to the guidelines of the Institutional Animal Care and Use Committee (IACUC) of Tsinghua University, Beijing, China.
Cell Lines
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Human ES cell line used in this study were H1 (XY), purchased from WiCell, Inc Undifferentiated H1 were maintained on MEF feeder cells as previous described (Jung et al., 2017). All cells were cultured at 37°C in a humidified incubator supplied with 5% CO2. ES medium were standard knockout serum replacer (KSR) consisted of 20% knockout serum replacer, 0.1 mM nonessential amino acids, 1 mM L-glutamine, 0.1 mM -mercaptoethanol, and 4 ng/ml recombinant human basic fibroblast growth factor (bFGF, R and D systems).
To obtain adherent fibroblast differentiation, human ESC clones were first transferred to 1% Matrigel coated plates by colony picking using a glass needle and cultured in ES cell conditioned medium (ES medium incubated overnight on irradiated MEFs) for ~5 days to remove all the residual MEF cells. Differentiation of hESCs to fibroblasts (dH1) began after aspirating of conditioned medium, washing with PBS without Ca2+ and Mg2+ twice, and replacing with differentiation medium (knockout DMEM with 20% fetal bovine serum, 0.1 mM nonessential amino acids, 1 mM L-glutamine). Differentiation medium was changed every 3 days. When the cell reached 90% confluency (~7 days), the resulting dH1 were collected and kept in liquid nitrogen tank for long time preservation or passaged by the ratio 1:3 if more cells were needed.
Primary human pulmonary fibroblasts (HPF) were purchased from National Infrastructure of Cell Line Resource. Primary human skin fibroblast (HSF) was a gift from Professor Ting Chen at National Institute of Biological Sciences, Beijing, and prepared according to the published protocol (Qian et al., 2018). Human adult Sertoli cells were a generous gift from Professor Zheng Li at Shanghai Jiao Tong University and prepared according to the established protocol (Wen et al., 2017). The HPF and HSF cells were maintained at 75 ml culture flask in 15 ml DMEM culture medium (Corning) containing 10% FBS (Gibco), 0.1 mM nonessential amino acids and 1 mM L-glutamine. Primary adult Sertoli cells were cultured as previously described in DF12 medium consisting of DMEM/F12% and 10% FBS (Gibco) and passaged every 5 days when cell confluence reached 80%.
Human umbilical vein endothelial cells (HUVEC) were a gift from Professor Jie Na at Tsinghua University. HUVECs were cultured in ECM medium (Sciencell) containing 10% FBS (Gibco) and passaged every 3 days when cell confluence reached 80%.
All of the cells above were cultured in a 37°C humidified incubator supplied with 5% CO2.
Method details
Overexpression vector construction and lentivirus production
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All overexpression vectors carrying EF1α promoter and desired gene were constructed using the Gateway system (Invitrogen) as previously described (Jung et al., 2017). Briefly, the candidate cDNA was first introduced into pENTR/1A or pENTR/D-topo donor vectors and transferred to 2K7 destination vectors with EF1α promoter in pENTR/5’ topo by LR recombination (Suter et al., 2006). Modified destination plasmids containing the cDNA were then introduced into 293FT cells together with the helper plasmids vsvg and ∆8.9 by LIPO3000 (Invitrogen) transfection to produce virus. Approximately, a total of 37 ml of virus supernatant was harvested on day1 and day3 after transfection and filtered with a 0.45 μm filter. At the time of virus infection, 8 μg/ml of polybrene was supplemented to increase infection efficiency.
AMH:EGFP reporter and creating reporter cell line
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1.6 kb of human AMH promoter upstream of the transcriptional start site was PCR amplified from genomic DNA of 293FT cells and cloned to pENTR5’-TOPO. Cloned plasmids were then recombined with pENTR/D-TOPO that carried the EGFP cDNA to create p2K7-AMH:EGFP recombinant plasmid and generated lentiviral supernatant as described above in overexpression vector construction section. Fibroblast HPF or dH1 in early passage (with 50% confluence) were transduced overnight on plate in fibroblast medium and recovered for one day after removal of virus. Subsequent drug selection by blasticidin (10 μg/ml) required another 3 days. Selected human fibroblasts were passaged for two times to expand the cell number for further experiments or frozen in liquid nitrogen.
hiSC reprogramming and enrichment by FACS
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Sertoli-like cell reprogramming was carried out in a T75 flask coated with Matrigel (1%). In brief, 1.5 × 105 human fibroblast cells carrying AMH:GFP reporter were seeded into T75 flask 1 day prior to overnight transduction with lentivirus NR5A1, GATA4, DMRT1, SOX9 and WT1 (for 5F-hiSCs) or NR5A1 and GATA4 (for 2F-hiSCs), recovered for 24 hr, following by drug selection with geneticin (1 mg/ml) for 5 days in MEF medium. After drug selection, transduced cells were cultured in DF12 medium, maintained for the indicated reprogramming duration, and harvested for FACS by digestion with TrypLE Express (Invitrogen). The single cell suspension for FACS was prepared with MACS medium (10% FBS in PBS with 0.0125 mM EDTA) and filtrated through a BD cell sorter. Cell sorting was proceeded on a high-speed cell sorter (Influx, BD) and was sorted to collecting tube containing DF12 medium.
Experiment aiming at examining different combinations of reprogramming factors was performed in six well plates and all possible combinations of NR5A1, GATA4, DMRT1, SOX9 and WT1 were tested. At the time of virus infection, 200 ul of each virus was added per well, and brought to the same final volume of 1 ml with p2k7 empty virus.
Quantitative PCR and statistical analysis
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Quantitative PCR was conducted as previously described (Jung et al., 2017). Briefly, total RNA was collected according to the instructions provided by QIAGEN RNeasy kit (QIAGEN) or TRIZOL (Invitrogen). CDNA was generated by EasyScript One-Step gDNA Removal and cDNA Synthesis SuperMix kit (TRANSGEN) according to manufacturer’s protocols using up to 500 ng RNA for each sample. 20 μl reaction (for Bio-rad 96-well System) were prepared and conducted with TransStart Green qPCR SuperMix kit (TRANSGEN). Gene expression was calculated using Bio-Rad CFX Manager program for relative expression formulation (dC(t)) and normalized to housekeeping genes (ACTB or GAPDH). Then, the gene expression of different samples were again normalized to the expression of control cells infected by p2k7 empty virus (CTRL) and reported as fold change (Vandesompele et al., 2002). Statistical analysis was carried out using Student’s t-test or one-way ANOVA by Prism 6.0 software.
RNA sequencing
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~2.5 × 105 cells were collected by FACS and total RNA was extracted by TRIZOL (Invitrogen). The quality and integrity of the purified RNAs were checked by Agilent 2100 bioanalyzer. Qualified RNA from the following samples was used for RNA Sequencing analysis: (1) HPF or dH1 carrying AMH:EGFP reporter, transduced with empty virus p2k7 and followed the same reprogramming procedure as described above. (2) Day10 hiSCs generated with five factors (5F-hiSCs) or generated with two factors (2F-hiSCs). (3) Human primary adult Sertoli cells (aSCs) cultured in vitro in DMEMF12 medium + 10% FBS. Sequencing libraries preparation and sequencing operations were carried out by ANNOROAD, a company providing RNA sequencing service, complied with the whole set of processes from Illumina.
Immunofluorescence of cultured cells
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Dissociated cells from FACS enrichment were collected onto a slide by Cytospin (800 r.p.m. for 5 min) or replated onto a 6-well plate. After that, cells were washed one time with PBS, fixed in 4% paraformaldehyde for 10 min and treated with 2.5% Triton X-100 for 15 min. For antibodies staining, slides were first blocked in 2.5% donkey serum for 1 hr, then incubated overnight in 4°C with primary antibody (1:200 for KRT18, 1:100 for NuMA, 1:200 for CX43, 1:100 for VASA, all rabbit-derived, Abcam; 1:50 for DAZL, mouse-derived, AbD Serotec; 1:100 for NR5A1, GATA4, WT1, SOX9, DMRT1, all rabbit-derived, Proteintech). Slides were then washed five times (each 3 min) with PBST (0.1% Tween-20/PBS), followed by anti-rabbit secondary antibodies (anti-rabbit-555 or anti-rabbit-488, Invitrogen) incubating for 1 hr at room temperature and washing for another five times. All sections were then mounted with Prolong Diamond Anti-Fade Mounting Reagent (ThermoFisher) and cover slip.
Effect of 2F-hiSCs conditioned medium on Jurkat cell proliferation and IL-2 production
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1 ml conditioned medium were collected from 3.5 × 104 of 2F-hiSCs or dH1-2K7 fibroblast (control) cultured at 50% confluence 48 hr after plating. For each assay, 2 × 104 human T lymphocytes (Jurkat E6-1 cells, gift from Professor Hai Qi, Tsinghua University) were seeded in 1 well of 96-well plate with 120 μl 1064 medium with 10% FBS, 0.1 mM nonessential amino acids, 1 mM L-glutamine, 0.1 mM–mercaptoethanol and added with indicated amount of conditioned medium from either 2F-hiSCs or dH1-2K7 (as control). The Jurkat cells were then cultured in a 37°C incubator with 5% CO2 for 3 days. Metabolism of WST-1 was used to determine the proliferative ability of lymphocytes in each well according to manufacturer’s instructions (Beyotime, Co., Ltd.). Three hours prior to analysis, 12 μl of WST-1 was added to each well and the final absorbance was measured by a microplate reader at 450 nm (SPECTRA max PLUS, Molecular Devices Inc). 1064 medium alone with WST-1 was used as a control to subtract background absorbance. For IL-2 quantification, 2 × 105 of Jurkat cells were seeded in one 6-well plate containing either 50% 2F-hiSCs or dH1-2K7 conditioned medium and cultured for 3 days. Then, cells were collected and lysed with 20 μl RIPA buffer. The concentration of IL-2 was determined by ELISA kit according to the manufacturer’s instructions (Cusabio Biotech, Co., Ltd.).
Cell migration assays
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Confluent P8 HUVEC cells were cultured with fresh ECM medium overnight before experiment. Then cells were incubated with medium containing 2.5 μM Calcein-AM fluorescent dye for one hour in incubator supplied with 37°C, 5% CO2. Cells were then trypsinized, counted and suspended in migration medium (50% ECM+50% MEF medium). Migration assays were carried out in Corning FluoroBlok 24-multiwell insert plate with 8.0 μm pores (Cat. No. 351157. Corning). Prior to seeding HUVEC (150 k, in 100 μl volume) into the insert, 50% conditioned medium pre-incubated with hiSCs or dH1 was mixed with 50% ECM and added to the basal chamber (600 μl in total). Following incubation of HUVECs at 37°C, 5% CO2 needed for 20 hr. The migration cells passed through the membrane were monitored by Calcein-AM with the help of an inverted microscope (Leica) at 485/535 nm (Ex/Em). Images were captured using Leica-Pro software provided by the microscope company. The number of migration cell was measured by calculating of the Calcein-AM green signal of each image.
Mouse spermatogonia isolation and coculture with 2F-hiSCs
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Mouse spermatogonia cells were isolated from ~24 testis of C57BL/6 mice at day six after birth according to previous protocol with minor modifications (Kanatsu-Shinohara et al., 2003; Wang et al., 2015). Briefly, the decapsulated testis and seminiferous tubules were detached with the help of tweezers. After washing with DPBS for three times, the seminiferous tubules were transferred to a new 15 ml tube and subjected to enzymatic digestion. Firstly, the seminiferous tubules were digested with 1 mg/ml collagenase IV for 5 min at 37°C in a water bath with shaking, then centrifuged at 50 × g to collect the segregated tubules and washed three times with DMEM/F12 medium. The small fragments were resuspended in 2 ml culture medium composed of DMEM/F12 medium with 10% FBS and seeded on 2 wells of 6-well plates (Corning) and incubated at incubator supplied with 37°C, 5% CO2. After attachment for 24 hr, the somatic cells were tightly attached to the dish and formed patches, while spermatogonia were just loosely attached on the somatic cells. The spermatogonia were then transferred to a new plate coated with 1% matrigel and cultured for two passages in the medium (DMEM/F12 medium, 20% KSR, 100 × Glutamax, 100 × NEAA, 100 × Pen/Strep, human GDNF 20 ng/mL, human bFGF 5 ng/mL). 24 hr prior to co-culturing experiments,~1.5 × 105 dH1 or 2F-hiSCs were plated to 1 well of the 48 well plate. The spermatogonia cells were collected by gently washing with fresh medium and transferred to the well plated with the dH1 or 2F-hiSCs in DMEM/F12 medium with 10% FBS.
Co-transplantation of hiSCs with 293FT cells
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All animal experiments were approved by the Institutional Animal Care and Use Committees at Tsinghua University. ~ 1.3 × 106 number of 293FT cells stably transduced with a luciferase reporter were mixed with 2.5 × 105 cells of either dH1 or 2F-hiSCs in 100 μl Matrigel. The suspension was transplanted into C57BL/6 (Purchased from Vital River Laboratory Animal Technology) mice with normal immune system by subcutaneous injection. Live imaging of transplanted mice was performed at the indicative day after transplantation. 15 min prior to live imaging, 100 μl of D-luciferin (15 mg/ml, 10 μL/g mouse weight) was injected to the mice by intraperitoneal injection and luciferase activity was measured in the IVIS Spectrum machine (PerkinElmer Health Sciences, Inc).
BODIPY and Oil Red O staining of lipid droplets
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Cells were washed with PBS and fixed with 4% PFA for 15 min. For BODIPY staining, cells were stained with BODIPY (Invitrogen, working solution1 μg/ml) at room temperature for 15 min and washed in PBS for three times. Then, mounted cells with Prolong gold anti-fade reagent (Invitrogen).
For Oil Red O staining, 0.5% Oil Red O Stock Solution (in isopropanol) was used. Staining solution contains 6 parts of stock Oil Red O and 4 parts of distilled water, filtered with Whitman paper before used. After cell fixation, stained the cells with 1 ml staining solution for 15 min, then clear background using 60% isopropanol. Finally, wash with distilled water.
Cell lines
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All primary cells and cell lines have been tested to confirm no mycoplasma contamination before they were used in the experiments. Information about three established cell lines 293FT, H1 and CCC-HPF-1 were listed in the key resource table. None of the cell line is found in the list of misidentified cell lines maintained by the International Cell Line Authentication Committee.
Quantification and statistical analysis
Statistical details of analysis including statistical test used, value of n and statistical significance were all described in the figure legends.
RNA-seq data processing
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All RNA-seq data were mapped to human genome build hg19 (UCSC) by TopHat (version 2.1.1), reads from PCR duplicates were dropped. The gene expression level was calculated by Cufflinks (version 2.2.1) using the protein coding genes GTF file extracted from Ensembl database (Homo_sapiens.GRCh37.75.gtf). Read counts were obtained using HTSeq (version 0.9.1). Differentially expressed genes (DEGs) were analyzed using R package DESeq2 and selected using p-value<0.01 as a threshold. Heat map was plotted using heatmap.2 function of R, and gene expressions were scaled to FPKM or Z-scores. We used R function cor to do sample correlation clustering and principal component analysis (PCA). K-means clustering was performed using Cluster 3.0 package (K = 3, Spearman Correlation, Complete-linkage) and clustered heat maps were produced by TreeView.
Data availability
All data generated or analysed during this study are included in the manuscript and supporting files.
The following data sets were generated
Liang J(2018) NCBI Gene Expression Omnibus
ID GSE133757. HiSC RNA sequence data.
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Article and author information
Author details
Jianlin Liang
Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
Tsinghua-Peking Center for Life Sciences, School of Life Sciences, Tsinghua University, Beijing, China
##### Contribution
Conceptualization, Formal analysis, Validation, Investigation, Writing—original draft
##### Competing interests
No competing interests declared
"This ORCID iD identifies the author of this article:" 0000-0003-3027-2421
2. #### Nan Wang
Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
##### Contribution
Formal analysis, Validation, Investigation
##### Competing interests
No competing interests declared
3. #### Jing He
Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
##### Contribution
Data curation, Formal analysis, Validation
##### Competing interests
No competing interests declared
4. #### Jian Du
Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
##### Contribution
Formal analysis, Validation, Methodology
##### Competing interests
No competing interests declared
5. #### Yahui Guo
Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
##### Contribution
Formal analysis, Validation, Methodology
##### Competing interests
No competing interests declared
6. #### Lin Li
Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
##### Contribution
Supervision, Validation, Methodology
##### Competing interests
No competing interests declared
7. #### Wenbo Wu
National Institute of Biological Sciences, Beijing, China
##### Contribution
Validation, Investigation, Methodology
##### Competing interests
No competing interests declared
8. #### Chencheng Yao
Department of Andrology, the Center for Men's Health, Urologic Medical Center, Shanghai Key Laboratory of Reproductive Medicine, Shanghai Jiao Tong University School of Medicine, Shanghai General Hospital, Shanghai, China
##### Contribution
Validation, Investigation, Methodology
##### Competing interests
No competing interests declared
9. #### Zheng Li
Department of Andrology, the Center for Men's Health, Urologic Medical Center, Shanghai Key Laboratory of Reproductive Medicine, Shanghai Jiao Tong University School of Medicine, Shanghai General Hospital, Shanghai, China
##### Contribution
Resources, Validation
##### Competing interests
No competing interests declared
10. #### Kehkooi Kee
1. Center for Stem Cell Biology and Regenerative Medicine, Department of Basic Medical Sciences, School of Medicine, Tsinghua University, Beijing, China
2. Tsinghua-Peking Center for Life Sciences, School of Life Sciences, Tsinghua University, Beijing, China
3. Beijing Advanced Innovation Center for Structural Biology, School of Life Sciences, Tsinghua University, Beijing, China
##### Contribution
Conceptualization, Resources, Data curation, Formal analysis, Supervision, Funding acquisition, Validation, Investigation, Visualization, Methodology, Writing—original draft, Project administration, Writing—review and editing
##### For correspondence
kkee@tsinghua.edu.cn
##### Competing interests
No competing interests declared
"This ORCID iD identifies the author of this article:" 0000-0001-6926-7203
Funding
Ministry of Science and Technology of the People's Republic of China (2018YFA0107703)
Kehkooi Kee
Ministry of Science and Technology of the People's Republic of China (2017YFC1001601)
Kehkooi Kee
The funders had no role in study design, data collection and interpretation, or the decision to submit the work for publication.
Acknowledgements
We are indebted to Prof. Zuping He helping us to obtain primary human Sertoli cells, Prof. Ting Chen for providing us the primary human keratinocytes, and Prof. Jianbin Wang for helping us to analyze CNV of Sertoli cells. Research funding is provided by the Ministry of Science and Technology of China [2018YFA0107703, 2017YFC1001601] and Cross-discipline Foundation of Tsinghua University.
Ethics
Animal experimentation: C57BL/6 mice were purchased from Vital River Laboratory Animal Technology Co., Ltd (Beijing, China). All animal maintenance and experimental procedures were performed according to the guidelines of the Institutional Animal Care and Use Committee (IACUC) of Tsinghua University, Beijing, China (Approval number: 17-JJK1).
Copyright
© 2019, Liang et al.
This article is distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use and redistribution provided that the original author and source are credited.
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Jianlin Liang
Nan Wang
Jing He
Jian Du
Yahui Guo
Lin Li
Wenbo Wu
Chencheng Yao
Zheng Li
Kehkooi Kee
(2019)
Induction of Sertoli-like cells from human fibroblasts by NR5A1 and GATA4
eLife 8:e48767.
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Equation of tangent perpendicular to line
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$\begingroup$
I've got the homework question which I cannot solve.
Find the equation of the tangents to $4x^2+y^2=72$ that are perpendicular to the line $2y+x+3=0$.
What I have done so far:
I have found the gradient of the line which is $m_1 = -\frac12$.
Which means that the equation perpendicular to the tangent of $4x^2+y^2=72$ is $m_2 = 2$, since $m_1m_2 = -1$.
I've found the derivative of $4x^2+y^2=72$ which is $\dfrac{dy}{dx} = -\dfrac{4x}{y}$.
The next thing I have done is set $m_2=\dfrac{dy}{dx}$.
Now what?
calculus
ordinary-differential-equations
derivatives
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edited May 2, 2014 at 20:10
Matthew Conroy
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asked May 2, 2014 at 19:43
Luke TaylorLuke Taylor
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3
$\begingroup$ Looks right so far, so we get $-4x/y=2$. Substitute in the equation of the ellipse. To diagnose your answer, one would have to know what it is. $\endgroup$
André Nicolas
– André Nicolas
2014-05-02 19:47:44 +00:00
Commented May 2, 2014 at 19:47
$\begingroup$ What do you mean by substitute in the equation of the ellipse? $\endgroup$
Luke Taylor
– Luke Taylor
2014-05-02 19:53:19 +00:00
Commented May 2, 2014 at 19:53
$\begingroup$ I wrote out a solution. $\endgroup$
André Nicolas
– André Nicolas
2014-05-02 20:05:20 +00:00
Commented May 2, 2014 at 20:05
Add a comment |
3 Answers 3
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$\begingroup$
$$2y+x+3=0$$ The important thing is that you know what being perpendicular means. In terms of slopes, one line has to have a slope that's equal to the negative reciprocal of the other. This is a property you should know from your Algebra class. For this problem, we want to find the slope of the line that's perpendicular to the line we have above. To do that, we will first need to find the slope of the line above. Let's rearrange stuff to find it: $$2y=-x-3 \ y=-\frac12x-1.5$$
You can get $\frac{dy}{dx}$ of this, but really it's in the form of $y=mx+b$ and we know that the slope of this line is $-\frac12$. Thus, the slope of the line perpendicular to this one is $2$. $$4x^2+y^2=72$$ We've got the slope, but not the line yet. We want to find a line that will be tangent to the ellipse that was given. How do we find that? Start with finding the point at which it is tangent. You probably already know how to do this. First take the derivative. For this problem, implicit differentiation is not a really good idea. Thus, isolate for $y$ first and then solve it:
$$y=\sqrt{72-4x^2}$$
$$\frac{dy}{dx}=\frac12 \left(72-4x^2\right)^{-\frac12}(-8x)$$
Now that you have the derivative, find the value for $x$ where the slope of the ellipse is the same as the perpendicular line. In other words, $\frac{dy}{dx}=2$:
$$2=\frac12 \left(72-4x^2\right)^{-\frac12}(-8x)$$
A little messy, but you'll find that $x=\pm3$. Therefore the point on the ellipse where the tangent will have this slope is $(\pm3, \pm6)$ (plugged in to the original to get the $y-$value). From this point, you can easily get the line now. You've got $y=mx+b$. $m$ is $2$, and now to find $b$ we'll plug in $x/y$:
$$6=(2)(-3)+b$$
$$b=12$$
Or for $x=3$, $b=-12$.
Our final answer ends up being:
$$\therefore y=2x\pm12$$
Pictorial proof:
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edited May 2, 2014 at 20:10
answered May 2, 2014 at 20:02
ShaharShahar
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$\begingroup$ Thanks for the detailed answer. :) $\endgroup$
Luke Taylor
– Luke Taylor
2014-05-02 20:14:35 +00:00
Commented May 2, 2014 at 20:14
Add a comment |
1
$\begingroup$
Let the point of tangency be the point $(x,y)$ on our curve. Your calculation shows that $\frac{-4x}{y}=2$, that is, $y=-2x$.
Also, we have $4x^2+y^2=72$. Substituting $-2x$ for $y$ in this equation, we get $8x^2=72$, and therefore $x=\pm 3$. That gives the two points $(3,-6)$ and $(-3,6)$.
The line with slope $2$ through $(3,-6)$ has equation $y-(-6)=2(x-3)$, which simplifies to $y=2x-12$. The other line has equation $y=2x+12$.
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edited May 2, 2014 at 20:12
answered May 2, 2014 at 20:04
André NicolasAndré Nicolas
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$\endgroup$
1
$\begingroup$ You are welcome. For no very good reason, I would have chosen $(a,b)$ as the coordinates of the point on the curve, and solved for $a$ and $b$. Same calculation, just a change of name. $\endgroup$
André Nicolas
– André Nicolas
2014-05-02 20:10:19 +00:00
Commented May 2, 2014 at 20:10
Add a comment |
1
$\begingroup$
HINT :
You found $m_1=-\dfrac{1}{2}$ and $m_2=2$. This is correct. Then $$ \begin{align} 4x^2+y^2&=72\ y^2&=72-4x^2\ y&=\sqrt{72-4x^2}\ \frac{dy}{dx}&=\frac{d}{dx}(72-4x^2)^{\Large\frac12}\ m_2&=\frac12(72-4x^2)^{-\Large\frac12}(-8x)\ 2&=-\frac{4x}{\sqrt{72-4x^2}}\ \sqrt{72-4x^2}&=-2x\ 72-4x^2&=(-2x)^2\ 72-4x^2&=4x^2\ x^2&=9\ x&=\pm3 \end{align} $$
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answered May 2, 2014 at 20:02
Tunk-FeyTunk-Fey
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17380 | https://www.quora.com/How-can-I-find-the-point-of-intersection-of-three-lines-simultaneously | Something went wrong. Wait a moment and try again.
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5
How can I find the point of intersection of three lines simultaneously?
David Marilley
Former Financial Engineer
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Author has 2.9K answers and 1.9M answer views
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5y
Three distinct lines present several cases. In one case all three are parallel and this can be easily seen if slopes all match. Obviously there is no point of intersection, even if two lines are parallel. If there is no parallel pair, then find where any two cross. If that point works in the third equation, you found a common point for three lines. Otherwise there is no point shared by all three lines. Generally, I would graph all three lines, and then look and see the regions they all bound. Usually you get a triangle and six unbounded regions. When all three lines share a point, you expect t
Three distinct lines present several cases. In one case all three are parallel and this can be easily seen if slopes all match. Obviously there is no point of intersection, even if two lines are parallel. If there is no parallel pair, then find where any two cross. If that point works in the third equation, you found a common point for three lines. Otherwise there is no point shared by all three lines. Generally, I would graph all three lines, and then look and see the regions they all bound. Usually you get a triangle and six unbounded regions. When all three lines share a point, you expect to see just the six unbounded regions, since the triangle has collapsed to a point.
Mark Barton
Researches suspensions for gravity wave detectors
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Author has 18.2K answers and 24.5M answer views
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10y
Three lines in a plane don't normally intersect at a single point. You can make three pairs of lines from three lines (1-2, 2-3, 3-1), and each of the pairs will either intersect at a single point or be parallel. You first need to check each of those pairs separately. If any of them are parallel, there's no intersection point of all three lines. If you do get three intersection points, you then need to check whether they're the same.
And if you're doing this on a computer, you probably need to do the check for parallelism and comparison of points with some tolerance ('are they the same to the
Three lines in a plane don't normally intersect at a single point. You can make three pairs of lines from three lines (1-2, 2-3, 3-1), and each of the pairs will either intersect at a single point or be parallel. You first need to check each of those pairs separately. If any of them are parallel, there's no intersection point of all three lines. If you do get three intersection points, you then need to check whether they're the same.
And if you're doing this on a computer, you probably need to do the check for parallelism and comparison of points with some tolerance ('are they the same to the nearest pixel/0.001"?' etc) because with rounding errors, even pairs of lines that you tried to make intersect at the same point will be ever so slightly off.
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To find the point of intersection of three lines simultaneously, you need to determine whether the three lines intersect at a single point. Here’s how to do it step-by-step:
Set Up the Equations
Each line can be represented by a linear equation in the form:
- Line 1: [math]a_1x + b_1y = c_1[/math]
- Line 2: [math]a_2x + b_2y = c_2[/math]
- Line 3: [math]a_3x + b_3y = c_3[/math]
Formulate the System of Equations
You now have a system of three equations with two variables (x and y). To find a unique solution, the three lines must be concurrent (i.e., they intersect at a single point).
Use Two Equations to Solve for x and y
First,
To find the point of intersection of three lines simultaneously, you need to determine whether the three lines intersect at a single point. Here’s how to do it step-by-step:
Set Up the Equations
Each line can be represented by a linear equation in the form:
- Line 1: [math]a_1x + b_1y = c_1[/math]
- Line 2: [math]a_2x + b_2y = c_2[/math]
- Line 3: [math]a_3x + b_3y = c_3[/math]
Formulate the System of Equations
You now have a system of three equations with two variables (x and y). To find a unique solution, the three lines must be concurrent (i.e., they intersect at a single point).
Use Two Equations to Solve for x and y
First, solve any two of the equations simultaneously. For example, use the first two lines:
From Line 1 and Line 2, you can express [math]y[/math] in terms of [math]x[/math] or vice versa.
Substitute the expression for [math]y[/math] (or [math]x[/math]) from one equation into the other to find the values of [math]x[/math] and [math]y[/math].
Substitute into the Third Equation
Once you have [math]x[/math] and [math]y[/math] from the first two lines, substitute these values into the third line's equation. If the equation holds true, then the three lines intersect at that point. If not, the lines do not intersect at a single point.
Example
Consider the lines:
- Line 1: [math]2x + 3y = 6[/math]
- Line 2: [math]x - y = 1[/math]
- Line 3: [math]3x + 4y = 10[/math]
Step 1: Solve the first two equations:
1. From Line 2: [math]y = x - 1[/math]
2. Substitute [math]y[/math] into Line 1:
[math]2x + 3(x - 1) = 6 \ 2x + 3x - 3 = 6 \ 5x = 9 \ x = \frac{9}{5}[/math]
3. Substitute [math]x[/math] back to find [math]y[/math]:
[math]y = \frac{9}{5} - 1 = \frac{9}{5} - \frac{5}{5} = \frac{4}{5}[/math]
Step 2: Check with Line 3:
[math]3\left(\frac{9}{5}\right) + 4\left(\frac{4}{5}\right) = \frac{27}{5} + \frac{16}{5} = \frac{43}{5} \neq 10[/math]
Since Line 3 does not hold true, the three lines do not intersect at a single point.
Conclusion
To summarize, you can use substitution or elimination methods to solve the equations of any two lines and then check the third line. If all three equations are satisfied at the same point, then the lines intersect at that point; otherwise, they do not.
Edward Ponderer
CareerTechn.Ed., Engr./arch, & Energy/env./util.
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Author has 681 answers and 1.1M answer views
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Originally Answered: How can I find the point of intersection of three lines simultaneously ?
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It may not exist. Three lines in a plane will always meet in a triangle unless tow of them or all three are parallel. The triple intersection is a special case where the sides of this triangle go to zero. [Not that this isn’t an important case. In mechanical engineering, “drop lines” will always pass through the “centroid” of a cross-sectional area. Two drop lines are used to find that centroid, and a third verifies it (or indicates the measurement area).]
So what you do is equate the y = mx + b for any two of the three possible pairs of lines. If there is one solution (not zero or infinitely m
It may not exist. Three lines in a plane will always meet in a triangle unless tow of them or all three are parallel. The triple intersection is a special case where the sides of this triangle go to zero. [Not that this isn’t an important case. In mechanical engineering, “drop lines” will always pass through the “centroid” of a cross-sectional area. Two drop lines are used to find that centroid, and a third verifies it (or indicates the measurement area).]
So what you do is equate the y = mx + b for any two of the three possible pairs of lines. If there is one solution (not zero or infinitely many), for them both, then all three will give the same point—the triple intersection.
Lorene Hollingsworth
Studied Mathematics (Graduated 1967)
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Author has 303 answers and 416.8K answer views
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5y
When you have 3 lines, you do not know if they will all share exactly one point. However, if you set up a system of 2 equations and solve them, that would give you their intersection point, then take the third equation and using one of the other two equations solve the same kind of system. If the solution is the same as the first two equations, then they all meet at one point. Lines do not have to intersect in the same point and usually do not. They could be parallel, or the same line. The rare case would be if all meet at the same point.
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Related questions
What is the point at which the given lines intersect?
What are the three lines on a map that intersect at the same point?
If 2 lines intersect, is their intersection point always unique?
Can you provide an example where knowing that two lines intersect would be useful, even if the exact point of intersection is unknown?
How can we determine if three points on a line intersect at one point?
Mr AB
If you need an answer, you will get an answer
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Author has 4K answers and 2.5M answer views
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2y
Related
How do you find the intersection of three lines in geometry?
First we need the equation of the three lines.
Three straight lines are said to be concurrent if they passes through a point i.e., they meet at a point.
The point at which concurrent lines intersect is called point of concurrence.
We can find the point of intersection by simultaneously equating the equation of two lines.
We will get the point of intersection.
Now we need to check ✔️, whether the third line is concurrent or not.
An image will clarify this step:
To conclude if the above three lines are concurrent, the following condition shown below as a determinant should be evaluated to 0.
There are o
First we need the equation of the three lines.
Three straight lines are said to be concurrent if they passes through a point i.e., they meet at a point.
The point at which concurrent lines intersect is called point of concurrence.
We can find the point of intersection by simultaneously equating the equation of two lines.
We will get the point of intersection.
Now we need to check ✔️, whether the third line is concurrent or not.
An image will clarify this step:
To conclude if the above three lines are concurrent, the following condition shown below as a determinant should be evaluated to 0.
There are other important methods too, for checking, but if I give that, the answer would go on to a long description.
Roland Aldridge
Former Became enraptured with sound at Oxford in 1969 (1973–1979)
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Author has 1K answers and 1.9M answer views
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10y
Originally Answered: How can I find the point of intersection of three lines simultaneously ?
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If you know that they all three intersect at a single point (and normally they won't), then you can just calculate the intersection of the easiest two lines. The third will have to go through that point too.
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Bhim Mu
Former Retd Gov Servant (1971–2007)
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Author has 3.4K answers and 498.2K answer views
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1y
Related
How do I find the intersection of two lines?
Each line has an equation in 2 variable
solve 2 equations in 2 variables
The value you get for 2 variable say x,y are the intersection point co-ordinates
say
x + y = 1 ————[A]
2 x + 3 y = 3——-[B]
2 x+3 y+3
2 x+2 y=2 ←—2[A]
0+y=1 ←- subtracting
y =1
substitute y=1 in [A]
x+1 =1
x=1–1 =0 intersection point is (0,-1)
Each line has an equation in 2 variable
solve 2 equations in 2 variables
The value you get for 2 variable say x,y are the intersection point co-ordinates
say
x + y = 1 ————[A]
2 x + 3 y = 3——-[B]
2 x+3 y+3
2 x+2 y=2 ←—2[A]
0+y=1 ←- subtracting
y =1
substitute y=1 in [A]
x+1 =1
x=1–1 =0 intersection point is (0,-1)
Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994)
·
Author has 10.6K answers and 23.5M answer views
·
5y
Related
How do I get the point of intersection of two lines using a cross product if I know two points of each line?
The meet of two lines and the join of two points are both handled by the cross product in a projective setting. Let’s work it out.
We use capital letters math[/math] for coordinates in 3 space and small letters math[/math] for our 2D Cartesian plane, which we elevate to [math]Z=1[/math].
A line through the origin and point math[/math] in 3 space has parametric representation [math]t(X,Y,Z).[/math] Only the ratio of the coordinates matter, so let’s abbreviate this line math.[/math] It intersects our Cartesian plane, the [math]Z=1[/math] plane, at math[/math], i.e.
[math]x=\dfrac X Z, \qquad y = \dfrac Y Z[/math]
So math[/math] is a projective way to refer
The meet of two lines and the join of two points are both handled by the cross product in a projective setting. Let’s work it out.
We use capital letters math[/math] for coordinates in 3 space and small letters math[/math] for our 2D Cartesian plane, which we elevate to [math]Z=1[/math].
A line through the origin and point math[/math] in 3 space has parametric representation [math]t(X,Y,Z).[/math] Only the ratio of the coordinates matter, so let’s abbreviate this line math.[/math] It intersects our Cartesian plane, the [math]Z=1[/math] plane, at math[/math], i.e.
[math]x=\dfrac X Z, \qquad y = \dfrac Y Z[/math]
So math[/math] is a projective way to refer to a point in the Cartesian plane, namely the point math\;[/math]. The Cartesian point math[/math] corresponds to the line in three space through the origin math.[/math]
What about the Cartesian line [math]ax+by+c=0?[/math] When we include the origin in three space, we get a plane whose equation is [math]aX+bY+cZ=0.[/math] We abbreviate the plane and its corresponding line in the Cartesian plane as [math][a:b:c]\;[/math] as the values only matter in ratio.
The join of two points math[/math] and math[/math], i.e the line between two points, which we’ll write as the product math(d:e:f),\;[/math] is given by the cross product.
math(d:e:f) = [ bf-ce : cd-af : ae-bd ][/math]
The meet of two lines [math][a:b:c]\;[/math] and [math][d:e:f][/math] is given identically by the cross product:
[math][a:b:c][d:e:f] = ( bf-ce : cd-af : ae-bd )[/math]
Let’s work out an example where we know the answer. Line one points math, (1,1)[/math], line two points math,(3,1)[/math] so a final answer of math[/math].
The line between math[/math] and math[/math] is
math(1:1:1)=[0(1)-1(1) :1(1)-0(1) : 0(1)-0(1)]=[-1 :1: 0 ][/math]
which is [math]-x + y +0 = 0,[/math] aka [math]y=x \quad\checkmark[/math]
The line between math[/math] and math[/math] is
math(3:1:1)=[3(1)-1(1) :1(3)-1(1) : 1(1)-3(3)][/math]
[math]=[2:2:-8]=[1:1:-4] [/math]
which is [math] x + y + -4= 0,[/math] aka [math]x+y=4 \quad\checkmark[/math]
The meet we seek is
math ( (1:3:1)(3:1:1) )[/math]
[math]= [-1 :1: 0][1:1:-4] = (1(-4)-0(1) : 0(1)-(-1)(-4) : -1(1)-1(1) )[/math]
[math]= (-4 : -4 : -2 ) = (2 : 2 : 1) = (2,2) \quad\checkmark[/math]
Math works!
Homework: Determine the relation between the point math[/math] and the line [math][a:b:c][/math] in the Cartesian plane. Hint: The polarity of Apollonius.
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Related questions
What is the value of slope in finding the intersection point between two lines?
What is the relation between the intersection of two lines and the midpoint of those lines?
If 2 finite lines meet at a point, such as the corner of a square, is it called an intersection? Otherwise, what is it called?
How do you find the intersection points when two lines have different slopes?
If the intersection of two lines is a point and the intersection of two planes is a line, what is the intersection of two cubes?
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What is the point at which the given lines intersect?
What are the three lines on a map that intersect at the same point?
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How can we determine if three points on a line intersect at one point?
What is the value of slope in finding the intersection point between two lines?
What is the relation between the intersection of two lines and the midpoint of those lines?
If 2 finite lines meet at a point, such as the corner of a square, is it called an intersection? Otherwise, what is it called?
How do you find the intersection points when two lines have different slopes?
If the intersection of two lines is a point and the intersection of two planes is a line, what is the intersection of two cubes?
How can three coplanar lines intersect?
How do I find an equation of a line that passes through intersection of two lines without finding the point of intersection?
What is the common point of two intersecting straight lines?
How many points of intersection will be found between two parallel planes?
What is the maximum and the minimum number of points of intersection when three lines intersect?
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Topographic Mapping of Isolated Thalamic Infarcts Using Vascular and Novel Probabilistic Functional Thalamic Landmarks
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Abstract
Purpose
We aimed to re-evaluate the relationship between thalamic infarct (TI) localization and clinical symptoms using a vascular (VTM) and a novel functional territorial thalamic map (FTM).
Methods
Magnetic resonance imaging (MRI) and clinical data of 65 patients with isolated TI were evaluated (female n = 23, male n = 42, right n = 23, left n = 42). A VTM depicted the known seven thalamic vascular territories (VT: inferolateral, anterolateral, inferomedial, posterior, central, anteromedian, posterolateral). An FTM was generated from a probabilistic thalamic nuclei atlas to determine six functionally defined territories (FT: anterior: memory/emotions; ventral: motor/somatosensory/language; medial: behavior/emotions/nociception, oculomotor; intralaminar: arousal/pain; lateral: visuospatial/somatosensory/conceptual and analytic thinking; posterior: audiovisual/somatosensory). Four neuroradiologists independently assigned diffusion-weighted imaging (DWI) lesions to the territories mapped by the VTM and FTM. Findings were correlated with clinical features.
Results
The most frequent symptom was a hemisensory syndrome (58%), which was not specific for any territory. A co-occurrence of hemisensory syndrome and hemiparesis had positive predictive values (PPV) of 76% and 82% for the involvement of the inferolateral VT and ventral FT, respectively. Thalamic aphasia had a PPV of 63% each for involvement of the anterolateral VT and ventral FT. Neglect was associated with involvement of the inferolateral VT/ventral FT. Interrater reliability for the assignment of DWI lesions to the VTM was fair (κ = 0.36), but good (κ = 0.73) for the FTM.
Conclusion
The FTM revealed a greater reproducibility for the topographical assignment of TI than the VTM. Sensorimotor hemiparesis and neglect are predictive for a TI in the inferolateral VT/ventral FT. The hemisensory syndrome alone does not allow any topographical assignment.
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Introduction
Thalamic infarcts (TI) account for approximately 3.1–4.4% of all ischemic strokes and despite their small size they are often associated with major neurological deficits [1,2,3]. Due to its strategic localization, the thalamus is deeply embedded in functional brain connectivity and plays a key mediating role in motor, sensory, coordinative, memory, cognitive and behavioral functions [1, 4,5,6,7]. The clinical diagnosis of TI may be challenging, since patients may present with a wide variety of symptoms depending on infarct location, volume and lateralization [1, 7,8,9].
However, the reported symptoms and syndromes associated with the involvement of certain vascular thalamic territories (VT: anterolateral—tuberothalamic artery; inferomedial—paramedian arteries; inferolateral—thalamogeniculate artery; posterior—posterior medial and lateral choroidal arteries; see Table 1) are not consistent in the literature [1, 6,7,8, 10]. Magnetic resonance imaging (MRI), in particular diffusion-weighted imaging (DWI), may provide valuable information on the topographic allocation of TI, as location and extent of infarction guide further diagnostic work-up and treatment and could have impact on prognosis [1, 3, 4, 11]. Previous studies investigating the relationship between infarct location and symptoms mostly relied on computed tomography (CT) alone [1, 12], were based on both CT and MRI [4, 10, 13] or focused on specific aspects of TI [5, 14,15,16]. The aim of our study was to (re-)evaluate the attribution of symptoms with respect to vascular thalamic territories. Furthermore, we aimed to create a novel territorial model based on functional thalamic properties.
Material and Methods
Data Acquisition
Ethics approval for this retrospective study was obtained from the local authority. We reviewed our institutional radiological database for stroke patients who underwent MRI from 2010 to 2020, scanning a total of 6289 patients with ischemic stroke that presented to our stroke unit. Search terms applied in this collective included “MRI” and “thalamic stroke”, “thalamic infarction” or “infarction of the thalamus”. Only cases with both clinically and radiologically confirmed diagnosis of TI were included (n = 165). Patients with additional acute ischemia outside the thalamus (n = 47), or with underlying basilar artery occlusion (n = 48) as well as patients with bilateral TI (n = 5) were excluded. Data of the remaining 65 patients with isolated TI were used for further analysis.
Imaging
MRI examinations were performed on two scanners, and the undermentioned sequences were used for further analysis. Philips Achieva 1.5 T (Philips Health Systems, Eindhoven, The Netherlands):
T1 weighted, time repetition (TR) 3.8 ms, time echo (TE) 1.7 ms, flip angle 8°, section thickness 2.2 mm, matrix 320 × 320, field of view (FOV) 350 mm2
Diffusion weighted imaging (DWI) axial, TR 3240 ms, time echo 75 ms, slice thickness 5 mm, gap 0.5 mm, matrix 256 × 256, FOV 220 mm2
Siemens Skyra 3.0T (Siemens Healthineers, Erlangen, Germany):
T1 weighted, TR 3.1 ms, TE 1.4 ms, flip angle 8°, section thickness 2.2 mm, matrix 320 × 320, FOV 350 mm2
DWI axial, TR 3800 ms, TE 95 ms, slice thickness 5 mm, gap 0.5 mm, matrix 384 × 384, FOV 230 mm2
Thalamic Maps
A vascular thalamic map (VTM) including the four traditional and three variant type VTs as described by Carrera et al. was manually created based on literature ([8, 10, 12, 17]; Fig. 1) and projected onto the Montreal Neurosciences Institutes (MNI) 152 standard template with 1 mm isotropic resolution [18:S169–S88.
.")].
Vascular thalamic map: sectional topography of vascular thalamic territories in axial (upper row) and coronal sections (lower row) projected onto the MNI-152 standard template
For creation of the VTM, we used the four traditional vascular thalamic territories (see Table 1) and, in addition, variant territories based on variable arterial supply and border zone ischemia :
The anterolateral territory—supplied by the tuberothalamic artery that originates from the posterior communicating artery.
The inferomedial territory—supplied by the paramedian arteries arising from the pre-communicating (P1) segment of the posterior cerebral artery (PCA).
The inferolateral territory—supplied by the thalamogeniculate artery that arises from the post-communicating P2 segment of the PCA.
The posterior territory—supplied by the posterior medial and lateral choroidal arteries that usually originate from the post-communicating P2-segment of the PCA.
The central territory—defined by the central thalamic part comprising parts of adjacent vascular territories.
The anteromedian territory—including the anterior part of the paramedian territory and the posterior part of the anterior territory.
The posterolateral territory—defined by infarcts involving the classical inferolateral and posterior territories, combining the posterior part of the inferolateral territory and the anterior part of the posterior territory.
Using the open source FreeSurfer software suite ( RRID:SCR_001847), a novel thalamic map including six functionally defined territories (FTM; Fig. 2) was generated from a freely available probabilistic thalamic nuclei atlas by Iglesias et al. . This atlas was created by using ex vivo MRI and histology and consists of 26 thalamic subregions that are based on the topography of the thalamic nuclei.
Functional thalamic map: sectional topography of functionally defined thalamic territories in axial (upper row) and coronal sections (lower row) projected onto the MNI-152 standard template
To ensure an accurate correlation of neurological symptoms and functional territory (FT), the 26 subregions were grouped into 6 territories, considering both the topography and the following fundamental thalamic functions:
anterior: memory, emotions
ventral: motor, somatosensory, language (left)
medial: behavior, emotions, nociception, oculomotor
intralaminar: arousal, pain
lateral: visuospatial, somatosensory, conceptual/analytic thinking
posterior: audiovisual, somatosensory
Territorial maps were then transferred from FreeSurfer space onto the MNI-152 standard template using MATLAB (MathWorks, Natick, MA, USA; version R2013b; RRID:SCR_001622) and converted to the Neuroimaging Informatics Technology Initiative (nifti) file format. In the next step, subregions were selected and combined according to the six predefined territories. The functional thalamic map (Fig. 2) was created to facilitate evaluation of MRIs.
Data Analysis
MRIs were reviewed by four experienced neuroradiologists and diffusion-weighted imaging (DWI) lesion location was assessed at the same time using templates of both the VTM and FTM (Fig. 1 and 2). The raters were aware of the patients’ clinical symptoms.
If more than one territory appeared to be affected, the raters were required to opt for the territory that was predominantly affected by the DWI lesions. For discrepancies, final locations of lesions were determined in consensus with a fifth senior neuroradiologist.
Clinical data included baseline demographic characteristics, type and severity of symptoms as assessed by a neurological examination and NIHSS score at admission. Data were then analyzed by two neurologists and correlated with imaging findings.
To generate lesion overlay maps, DWI (b = 1000 s/mm2) images were aligned via co-registered three-dimensional T1-weighted data onto the MNI-152 standard space template. For this purpose, T1-weighted data were brain extracted and tissues segmented using the software tools BET and FAST from the FMRIB software library (FSL, version 5.0.7, RRID:SCR_001847) toolbox. DWI (b = 0 s/mm2) images were aligned with the T1-weighted dataset via a boundary-based registration according to the segmented white matter. The T1-weighted data set was aligned to the MNI-152 template using a combination of linear and nonlinear registration. By combining the first (DWI to T1-weighted) and second (T1-weighted to MNI-152) transformation matrices, the co-registration was then applied on the DWI (b = 1000 s/mm2) images. TI masks were manually marked based on DWI in MNI-152 standard space and an overlap was used to generate corresponding lesion overlay maps.
Statistical Analysis
Statistical analysis was performed using Microsoft Excel version 16.49 (Microsoft, Redmond, WA, USA) and GraphPad Prism version 6 (GraphPad Software, San Diego, CA, USA). Values are reported as the mean and standard deviation unless otherwise specified. Interrater agreement was evaluated using Fleiss’ kappa. Normal data distribution was ascertained using the D’Agostino-Pearson omnibus normality test.
Results
A total of 65 patients (female n = 23, male n = 42) with isolated TI were included in the analysis. Mean age was 64 years (range 10–93 years). The majority of 42 patients (66.7%) presented with left sided TI, whereas only 23 patients (33.3%) had right sided TI. Clinical findings with respect to VTs and FTs are shown in Tables 2 and 3, respectively.
Interrater reliability for the neuroradiological assignment of infarcts to the VTM was fair (κ = 0.36), but good (κ = 0.73) for the FTM. Consistency between all raters was best for the anterolateral VT (6/10; 60%) and for the ventral FT (31/41; 75.6%). The lowest interrater agreement was found for the central VT (2/12; 16.2%) and for the intralaminar FT (2/5; 50%). The most frequently affected territories were the inferolateral VT (26/65; 40%) and the ventral FT (41/65; 63.1%).
Evaluation of lesion overlay maps (Fig. 3) showed an approximate geometrical congruence of affected areas for motor symptoms (hemiparesis, facial paresis, dysarthria). All patients with aphasia (n = 8) had TI located in the rostral portions of the left thalamus (anterolateral VT, inferomedial VT, central VT, inferomedial VT) (Fig. 3). Patients with ataxia had lesion location in the lateral portions of the thalamus in which the inferolateral VT was affected in 60%.
Thalamic infarction lesion overlay maps in axial orientation concerning most frequently observed symptoms. The overlay images are projected onto the MNI-152 space. a Hemisensory, b hemiparesis, c dysarthria, d aphasia, e facial paresis, f ataxia, g neuropsychological, h vigilance
Based on the overlays, concordance between actual clinical symptoms and those expected based on the anatomic location of TI was found in 46/65 patients (70.1%) when applying the VTM and in 46/65 (78.5%) patients when applying the FTM. Concordance between the manually assigned TI localizations (as per consensus) and the corresponding locations resulting from the overlays was 62/65 (95.4%) for the VTM and 63/65 (97.0%) for the FTM.
Discussion
We evaluated the topographic mapping of stroke symptoms in patients with isolated TI using two different territorial classifications of the thalamus: (1) the common separation based on the vascular supply (VTM) and a (2) functional subdivision of the thalamus based on a functional probabilistic map (FTM). We depicted the acute TI on DWI and four radiologists performed the topographic mapping based on the VTM and FTM. The interrater reliability in mapping DWI lesions to the VTM was rather low. Clinical symptoms were known to the raters, therefore, this knowledge might have influenced the choice of VTs. Due to the variability of VTs, defined clinical symptoms and syndromes sometimes cannot be clearly assigned to a particular VT, resulting in dichotomous allocation. The inconsistency of VTs may be based on thalamus-supplying vascular variants, which are common [20,21,22,23,24], but not considered in the traditional vascular-based approach. Consecutively, our study showed a better interrater reliability when using the FTM compared to the VTM.
Moreover, from a neuroradiological point of view, thalamus-supplying vessels are of such a small caliber that they usually are not seen in clinical routine MR or CT angiography. Therefore, attribution of TI to a particular vessel occlusion and a single VT is often not feasible [7, 14].
In contrast, a specific clinical symptomatology can usually be functionally anatomically well assigned to thalamic nuclei groups. We therefore created a new territorial map based on functional thalamic properties. There was a good interrater agreement in mapping infarcts on DWI (b = 1000 s/mm2) to this newly created FTM. Only the intralaminar FT (corresponding to the central VT) showed a low interrater agreement most likely due to the smallest volume of all thalamic territories.
Therefore, the mapping of stroke symptoms in isolated TI seems to be more valid using the FTM. This was to be expected, since here the topographical assignment of the thalamic regions is based on their respective function. Conversely, however, our results show that these classifications, which are mostly based on empiricism, seem to be valid because the assignment was successful. Nevertheless, this encompasses not all neurological symptoms. The topographic mapping for the sensorimotor hemiparesis and the rare neglect was reliable, while the sensory hemisyndromes and aphasia could not be assigned with certainty even on the FTM. Hemisensory disturbance occurred as a predominant symptom in 57% of patients without a preponderance for a certain territory. Regarding its assignment to territories, it is a non-specific symptom and occurred as the most frequent symptom in four of seven VTs and three of the six FTs.
However, a study investigating pure sensory syndromes in TI showed that sensory dysfunction and disturbance were more common in TI involving the ventrocaudal nucleus and the ventro-oral nucleus intermedius . These nuclei are part of the ventral FT, which was most commonly affected in our patient group, accordingly.
Moreover, the co-occurrence of hemisensory disturbance and hemiparesis showed a positive predictive value (PPV) of 76% and 82% for the involvement of the inferolateral VT and ventral FT, respectively. The appearance of thalamic aphasia had a PPV of 63% each for involvement for the anterolateral VT and the ventral FT, nevertheless, no assignment to a specific territory was possible. All eight patients with aphasia had TI located on the left side affecting the rostral portions of the thalamus, which is consistent with recent reports suggesting that appropriate TIs disrupt input from left cortical areas to anterior thalamic nuclei, leading to aphasic symptoms . Ataxia was predominantly associated with involvement of the lateral portions of the thalamus, including the inferolateral VT in 60%, and the ventral FT in 80%. The finding may be explained by affection of the ventral lateral nucleus (VL) that is located in these territories and forms part of the motor functional division [12, 25]. Dysarthria was predominantly associated with involvement of the inferolateral VT and ventral FT, with a PPV of 64% for the involvement of the ventral FT.
As mentioned above, interrater agreement was lowest for the small central VT, as its location may impede a distinct delineation from adjacent territories. Carrera et al. found a wide variety of neurological and neuropsychological syndromes as a result of involvement of adjacent thalamic structures . They determined hypesthesia in all of these patients as the leading symptom. Although the involvement of the central VT occurred more often in our patients (18.5%), the main symptom of hemisensory disturbance likewise occurred in 75% of patients.
However, the assignment of stroke symptoms to the vascular supply of the thalamus may be more relevant than functional mapping because it does not answer the most important questions in acute stroke, whether it is caused by microangiopathy or macroangiopathy or whether there is an embolic source in the anterior or posterior circulation. Anyhow, nowadays this should be answered by CT angiography or MR angiography performed in the context of stroke evaluation. Moreover, the overlap of the thalamic blood supply of the different territories is too pronounced to be adequately defined.
Previous studies pointed out that involvement of particular vascular territories was not significantly associated with a specific cause for infarction, although it was hypothesized that affection of the inferolateral vascular territory results from small vessel hypertensive arteriolopathy in the thalamogeniculate artery . In contrast, cardioembolism was found more often in infarction of the inferomedial territory which is supplied by the paramedian arteries [1, 7, 15, 26]. Therefore, the information on stroke etiology given by TI localization based on a vascular map may be limited. Our results underline this overlap of symptoms and blood supply, so that functional thalamic mapping should be more appropriate for clinical use. A potential criticism of our study is the fact that we assigned the infarcts on DWI (b = 1000 s/mm2) to a single territory for the purpose of further analysis. In lesions overlapping two or more territories, raters were instructed to define the majorly affected territory. However, a recent study showed that in isolated TI, 97% of ischemic lesions were confined to one vascular territory, whereas in posterior cerebral artery territory infarct, top of the basilar artery syndrome and extended posterior circulation infarctions, ischemic lesions tended to involve multiple thalamic territories .
Another criticism may be that our novel functional thalamic map includes six vascular territories in contrast to the seven traditional vascular territories typically mentioned in literature [7, 8]. However, our intention was to establish a straightforward model with high discriminatory power that represents basic thalamic functions.
In conclusion, the presented results underline that the topographic mapping of stroke symptoms in patients with isolated TI based on vascular territories is often not possible. Therefore, we propose a new more reliable classification system for TIs based on functional anatomy that demonstrated a better reproducibility compared to the traditional vascular territory model.
In daily clinical practice, radiological description of TI using the traditional vascular territory model may leave stroke physicians in uncertainty of the affected anatomy and expected associated clinical syndrome. The possible benefit of specifying the occluded (perforator) artery often bears no immediate clinical relevance and may remain of merely academic value. In contrast, a functional territory model enables the treating neurologist to link TI lesions more reliably to the observed clinical findings. When the affected functional territories satisfactorily match the observed neurological deficits, therapeutic and rehabilitative measures may be arranged promptly. However, if the respective TI lesion is not fully explanatory for the observed clinical syndrome (e.g., impaired consciousness, personality changes, aphasia), clinicians can be confidently encouraged to initiate further diagnostic work-up, such as additional cerebrospinal fluid analysis, electroencephalography, or laboratory assessment. In summary, the proposed clinical radiologic model may help to more consistently assign TI to specific neurological symptoms, and therefore improve the interaction between radiologists and stroke physicians towards a more focused patient care.
Effectiveness and user acceptance of the novel territorial model may be evaluated in follow-up studies using larger patient collectives. The new model may also be beneficial for further research investigating the role of thalamic lesions including correspondent research in movement disorders , dementia , neuropsychology or psychiatry [4, 29].
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Authors and Affiliations
Institute for Neuroradiology, University Hospital, Johann Wolfgang Goethe-University, Theodor Stern Kai 7, 60590, Frankfurt am Main, Germany
Maximilian Rauch, Jan-Rüdiger Schüre, Fee Keil, Eike Steidl, Se-jong You, Elke Hattingen & Stefan Weidauer
Department of Neurology, University Hospital, Johann Wolfgang Goethe-University, Theodor Stern Kai 7, 60590, Frankfurt am Main, Germany
Franziska Lieschke, Christian Foerch & Martin A. Schaller-Paule
Department of Psychiatry and Psychotherapy, University Medical Center of the Johannes Gutenberg-University, Untere Zahlbacher Str. 8, 55131, Mainz, Germany
Martin A. Schaller-Paule
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M. Rauch, J.-R. Schüre, F. Lieschke, F. Keil, E. Steidl, S.-j. You, C. Foerch, E. Hattingen, S. Weidauer and M.A. Schaller-Paule declare that they have no competing interests.
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Rauch, M., Schüre, JR., Lieschke, F. et al. Topographic Mapping of Isolated Thalamic Infarcts Using Vascular and Novel Probabilistic Functional Thalamic Landmarks.
Clin Neuroradiol 33, 435–444 (2023).
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Salir: Imperfect Tense
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Using the chart below you can learn how to conjugate the Spanish verb salir in Imperfect tense.
Definition
to leave, go out
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| Is it irregular? | Is it reflexive? | What is the gerund? | What is the past participle? |
--- --- |
| Yes | No | saliendo | salido |
Remember: these verb charts are only a tool to use while one is learning the language. In other words, one must eventually forget the verb chart and it must become second nature.
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Spanish tense name: Imperfecto
Mode: Indicative
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--- |
| Yo | salía |
| Tu | salías |
| El/Ella | salía |
| Nosotros | salíamos |
| Vosotros | salíais |
| Ellos/Ellas | salían |
Practice Salir (Imperfect Tense) Conjugations
All Tenses For The Verb Salir
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基礎 講義
信望愛 文教 基金會 ‧數學種子教師團隊 1
圓與直線關係
圓的方程式
平面上定義圓為給定一點 O 以及固定長度 r，在座標平面上所有與 O 點距離為 r 之點集合稱之為圓，其中這個 O 就是我們所熟悉的圓心， r 即為半徑。
關於圓的方程式我們存在以下幾種表示方法：
圓的標準式 ：
座標平面上給定圓心 (a,b)，以及半徑長 r。
則圓 C 之方程式為
C:( x−a) 2 +( y−b)2 = r2
圓的一般式 ：
將圓的標準式 (x−a) 2 +( y−b)2 = r2 展開化簡成二元二次方程式的形式，
x2 +y2+dx +ey +f= 0
此方程式即稱為圓的一般式
圓的直徑式 ：
座標平面上給定一圓的直徑兩端點 (x1 ,y1 ),( x2 ,y2)，
則圓 C的方程式為：
C:( x−x1 )( x−x2 )+( y−y1 )( y−y2 )= 0
點與圓關係
假設已知一圓 C: ( x−h) 2 +( y−k) 2 =r2，以及點 P(a,b)則P點與圓之關係可分成下列幾種方法
點在圓外
(I) 判別方式：點到圓心距離 >半徑
�(𝑎 − ℎ )2 + ( 𝑏 − 𝑘 )2>r
(II) 將圓化為一般式 x2 +y2 +dx +ey +f=0 之後
將 P(a,b) 之座標代入左式，
若代入之結果 a2 +b2 +dx +ey >0 ，則 P 點在圓外 2
1122
ax by cd a b
+= +
點在圓上
(I) 判別方式：點到圓心距離 = 半徑
�(𝑎 − ℎ )2 + ( 𝑏 − 𝑘 )2=r
(II) 將圓化為一般式 x2 +y2 +dx +ey +f=0 之後
將 P(a,b) 之座標代入左式，
若代入之結果 a2 +b2 +dx +ey =0 ，則 P 點在圓上。
點在圓內
(I) 判別方式：點到圓心距離
�(𝑎 − ℎ )2 + ( 𝑏 − 𝑘 )2<r
(II) 將圓化為一般式 x2+y2 +dx +ey +f=0 之後
將 P(a,b) 之座標代入左式，
若代入之結果 a2 +b2 +dx +ey <0 ，則 P 點在圓內
點與直線距離
已知座標平面上一直線 L:ax +by +c=0 以及點 P(x1 ,y1) 則 P 到 L 的距離
圓與直線關係
平面上圓與直線之間可能存在三種關係：
圓與直線不相交（相離）
圓與直線交於 1點（相切）
圓與直線交於兩點（相割）
要判斷圓與直線關係有兩種方法，第一種是藉助點與直線距離公式計算圓心到直線的 距離，再將此長度與半徑做比較。第二種為代數判斷法，尋找在圓方程式以及直線方程式 之間是否存在聯立解，接下來將一一詳述之。 3
I. 設圓
C 與直線
L 的方程式
圓心 (h,k)到直線 L的距離為 d = |𝑎ℎ+𝑏𝑏+𝑐 |
√𝑎 2+𝑏 2
d > r圓與直線不相交（相離）
d = r圓與直線交於 1點（相切）
d < r圓與直線交於兩點（相割）
II. 代數判定法
設圓 C與直線 L的方程式如下，並解其聯立方程式
利用 ax +by +c=0 將其中一個變數消去
移項後得到 y= −𝑎𝑎−𝑐 𝑏 將這個 y代入圓方程式（將 x代換掉也可以得到一樣的效果）
得到 (X−h) 2+( −𝑎𝑎−𝑐 𝑏 −k) 2 = r2
最後將方程式移項化簡成 Ax 2 +Bx +c = 0
令其判別式為 Δ
Δ < 0 則圓與直線不相交（相離）
Δ = 0 則圓與直線交於 1點（相切）
Δ > 0 則圓與直線交於兩點（相割）
從代數上的意義來看這種判定法的中心思想就是，如果 Δ>0 代表存在兩組相異的點座標
(x,y)可以同時滿足兩個式子，即代表圓跟直線會交於兩點； Δ=0 的話就是只有一個 (x,y)可以 滿足，代表只有一個交點也就是切點； Δ<0 代表這個方程式無解，也就是說在座標平面上找 不到點可以同時位於這兩個圖形上，也就代表圓與直線會相離。
圓的切線
切線的定義：與圓恰有一個交點的直線稱為切線，該交點稱為切點。
切線有一個很重要的性質是切線必垂直於經過該切點的半徑，在求切線的過程中江蕙 會大量利用到這項性質來幫助我們計算。
平面座標上切線的求法依照已知的條件不同，做法主要可以分成以下三種：
給定通過的切點，求切線。
給定切線斜率，求切線。
給定圓外一點，求切線。 4
給定通過切點算切線
假設圓心為 Q點、切點為 P點，利用 QP 必垂直切線的性質推出切線斜率。最後利用點 斜式結合已知 P點座標跟切線斜率得到切線方程式
例題 已知圓方程式 C: ( x+1) 2+( y−3) 2 = 25 ，求過切點 (2,7) 的切線方程式
ANS
利用點 (2,7) 以及圓心 (-1,3) 求出過切點的半徑方程式斜率為 4
3
切線需與 QP ���� 垂直故切線斜率必為 −3 4 ，接下來利用點斜式解過點 (2,7) 斜率
−3 4
的直線方程式為 y − 7 = −3 4 (x − 2)
給定切線斜率，求切線
座標平面上給定圓 C以及斜率 m，可以求得兩條切線方程式，且這兩條切線將對稱於斜 率同樣為 m的直徑。
假設已知圓 C: ( x−h) 2 +( y−k) 2 = r2並給定斜率 m，則切圓 C且斜率為 m之切線方程式為：
如果是不喜歡記公式的同學，這裡提供另外一個相當直覺的算法。
給定斜率 m就可以假設切線方程式為 y = mx + b，接下來利用圓心到切線距離需等於半徑列 出關係式 |𝑚ℎ− 𝑏 +𝑏 |
√𝑚 2+1
= r，在這個等式裡除了 b以外全部都是已知常數，
故可輕鬆求得兩切線方程式。我們用一個簡單例子來示範這個作法。
21)( mrhxmky +±−=−5
例題 圓C：x2 +y2 = 1
求與圓 C相切，且斜率 =−3
4
的切線
ANS
設切線方程式為 y= −3
4
x + 𝑏 ，則圓心 (0,0) 到切線的距離為 |0−0+𝑏 |
�(−3
4)2+1
=1
b=± 54，所以與圓 C相切，且斜率 = −3 4 的切線將會有兩條分別為 y =−3 4 𝑎 + 54以及 y = −3
4
𝑎 − 5
4
給定圓外一點，求切線
圓外任意一點對圓必可做出兩條切線，給定圓外一點 P(x0 ,y0 ) 則可假設過此點的方程 式為 y−y0 =m(x−x0 )。接下來我們有兩種方法可以求出 m，第一種利用圓心到直線距離等於 半徑；第二種利用圓與直線關係中的判別式 Δ=0 ，兩種作法完全等價。值得注意的一點是， 已知過圓外一點必可做出兩條切線，但若是最後只算出一個 m，千萬不要以為就是只有一條 切線，而是另外一條是斜率無法表達的垂直切線，過 (x0, y0 ) 的垂直切線方程式即為 x=x0。
例題 圓C：(x−4) 2+( y+1) 2 =25 點P(9,9)
求過 P點與 C相切之切線方程式
ANS
首先必須先確定 P點與圓之相對位置
P代入 C→(9 −4) 2 +(9+1) 2 =125 >25 ，確定 P在圓 C，將會有兩條切線
圓心 (4, −1) r=5
過P點之方程式 y−9= m(x−9)
mx −y+9 −9m=0
d=|4𝑚+1+9−9𝑚 |
√𝑚 2+1
=5
→ |10 − 5𝑚 |2 = 25 m2 +25
→100 m=75
m=34，只求出一個 m代表另一個斜率為無法表達的垂直切線的斜率
故切線方程式為 y−9= 3
4
(𝑎 − 9)以及 𝑎 = 96
小試身手
例題 1 設方程式 x2+y2+2( m+1) x−2my +3 m2−2=0 之圖形為一圓，
試求 m 之範圍，以及當 m=？時此圓之半徑有最大值 =？
例題 2 已知一圓之圓心位於直線 x+y−1=0 上，圓之半徑 r = 5，且過點 (6 ,0) ，
試求此圓之方程式
例題 3 座標平面上一點 A(6,3) ，點 P 為圓 C：x2 +y2−4x+2 y+3=0 上之一動點，
試求 PA 之最大、最小值，以及此時 P 點之座標
例題 4 坐標平面上圓 C：(x−7) 2+( y−8) 2 =9 圖形上有幾個點與圓點的距離正好為一整數
例題 5 試求坐標平面上過點 (1,2) 且與圓 x2+y2−8x−6y+20=0 相些之切線方程式為何
例題 6 求斜率為 2 且與圓 x2 +y2−4x+6 y−3=0 相切的直線方程式為何
解答與解析
例題 1：將方程式配方後可得 (x+( m+1)) 2 +( y-m) 2 =-m2 +2 m+3
圖形為一圓
∴-m2 +2 m+3>0
⇒-1< m<3
圓之半徑為 √−𝑚 2 + 2 𝑚 + 3 = �− (𝑚 − 1) 2 + 4
當 m = 1 時圓之半徑 r 有最大值 = 2
例題 2：利用直線參數式設圓心坐標為 (t,1 −t)，以及已知過點 (6.0) 半徑為 5
⇒�(𝑡 − 6) 2 + (1 − 𝑡 )2=5
⇒ (t−6) 2 +(1 −t)2 =25
⇒t2−7t+6=0
t=6 或 1
圓心為 (6,-5) (1,0)
故圓之方程式為 (x−6) 2 +( y+5) 2 = 25 或 (x−1) 2 +y2 = 25 7
例題 3：62 +3 2−4 × 6 + 2 × 3 + 3 = 30 > 0
⇒A(6,3) 在圓 C 之外部
x2 +y2−4x+2 y+3=0 ⇒ (x−2) 2 +( y+1) 2 = 2
故圓心為 𝑂 (2,-1) 半徑為 √2
O𝐴 ���� = 4 √2 ⇒ PA �� ��之最小值 = 4√2 − √ 2 = 3 √2，此時 OP ���� ∶ PA ���� = 1: 3
由分點公式可知 P 點坐標為 (3×2+1 ×61+3 , 3×( −1 )+1 ×31+3 ) = (3,0)
PA ���� 之最大值 = 4 √2 + √2 = 5 √2
O(2,-1) 為(x,y)及(3,0) 之中點， ⇒ (𝑎 , 𝑦 ) = (1,-2)
例題 4：圓C: ( x−7) 2 +( y−8) 2 = 9 ⇒ 圓心 (7,8) 半徑 = 3
設圓上 P 點滿足到原點之距離為整數
∵ OS ���� ≤ OP ���� ≤ OR ����
⟹ O𝐴 ���� − r ≤ O𝑃 ���� ≤ O𝐴 ���� + r
⟹ √ 72 + 8 2 − 3 ≤ OP ���� ≤ √ 72 + 8 2 + 3
⟹ 7. ⋯ ≤ OP ���� ≤ 13. ⋯
13 ,12 ,11 ,10 ,9,8=OP ，其中每個滿足距離為 8,9,10,11,12,13 的點在左右半圓都各有一 個點，故總共有 2 × 6 = 12 個
例題 5：點(1,2) 在圓外
設切線為 y−2 = m(𝑎 −1)
⇒ mx −y−m+2 = 0
又圓 x2 +y2−8x−6y+20=0 ⇒ (x−4) 2 +( y−3) 2 = 5
圓心 (4,3) 半徑 √5
圓心到切線距離 = 半徑
∴|4𝑚−3−𝑚+2 |
√𝑚 2+1
= √5 ⇒ |3 𝑚 − 1| = √5𝑚 2 + 5
左右平方化簡後可得 2m2−3m-2=0 ⟹ (m−2)(2 m+1)=0
m = 2 或 −1 2
故切線方程式為 2x−y=0 與 x+2 y−5 = 0
例題 6：設切線方程式為 y=2 x+k
圓 x2 +y2−4x+6 y−3=0 ⟹ (x−2) 2 +( y+3) 2 = 16
圓心 (2,-3) 半徑 4
圓心至切線距離等於半徑 ⇒ |2×2− (−3 )+𝑏 |
�2 2+(−1 )2
= 4
|7+ 𝑏 |
√5
= 4
7+ k= ±4 √5
k=−7 ± 4 √5
故切線方程式為 2x−y−7±4 √5=0 |
17385 | https://www.bubblyprimes.com/cross-cancellation/ | Cross-cancellation
What is cross-cancellation?
Cross-cancellation is a shortcut that you can use to make multiplying fractions easier. Sometimes you have to simplify fractions after doing arithmetic with them. Cross-cancellation is a simplification that can be done before. That’s great, because simplifying before means when you do multiply, you’ll have smaller numbers, and smaller numbers are easier to work with.
Did you catch that? Cross-cancellation makes fractions easier.
But what operation can you use cross-cancellation on? Multiplication of fractions.
In mathematics, we use the word operation for something basic that you do to a number. The most common examples of operations are adding, subtracting, multiplying, and dividing (there’s others, but let’s not get into that right now).
Cross-cancellation and Division
In addition to multiplication, cross-cancellation can be used to make math easier when fractions are divided. That’s because you can always convert a fraction division problem to a fraction multiplication problem. Just take the reciprocal of the divisor. Taking the reciprocal of a fraction just means swapping the numerator and the denominator (put the top on the bottom, and the bottom on the top). The divisor is the number that you are dividing by.
How do you cross-cancel?
You can cross-cancel systematically, but there’s also some shortcuts. Here’s the systematic way to cross-cancel:
If a division problem is being converted into multiplication, first rewrite it converting the divisor into its reciprocal.
Find the prime factorization of both the numerator and denominator of both fractions.
If either factor was not in simplest form, perform the simplification now.
Cancel any factors found both in a numerator and the opposite denominator. In fact, it’s because these factors are located diagonally from each other, that we call it cross-cancellation. If multiple copies of any factor are present, deal with them by canceling only as many copies as are present in the opposite location (the same as when simplifying fractions).
Repeat the previous step on the other diagonal.
Multiply all factors on the top of both fractions to form the numerator of the answer, and multiply all factors on the bottom of both fractions to form the denominator of the answer.
Step-by-step cross-cancellation example
Do you always have to do all that? No, not always. Did you notice where we used the word systematic to describe that step-by-step way of doing cross-cancellation? Some people probably ignored that word, and some other people might have wondered what it meant, and had a sinking feeling that we’re using fancy words, and things might soon stop making sense. Please don’t worry. What we meant by systematic was that anyone who can follow those steps will get the right answer at the end. When you’re familiar with the process, you can usually combine or skip steps, use alternative techniques, or take short cuts.
Cross-cancellation with the GCF Shortcut
You can save work if the numbers are relatively small and you can see some common factors. In such a case you don’t have to find the complete prime factorization. Instead, we usually cross out numbers across a diagonal and write replacement numbers next to them with the Greatest Common Factors removed. Here’s an example of what this technique looks like.
Example of using GCFs to simplify a fraction
Cross-cancellation and Fraction Simplification
Cross-cancellation is really a special version of simplifying fractions. You can only take advantage of it when multiplying or dividing fractions. It’s worth practicing simplifying fractions first in order to get a larger perspective, and to get an appreciation for how it’s useful. In fact, the advantage of cross-cancellation over multiplying and then simplifying is that before the numbers are multiplied they are smaller and easier to work with. You get the same answer if you multiply first, and then simplify, but it can be a lot more work. Cross-cancellation is just a shortcut, but a good one.
How do you Get Good at Cross-cancelling?
Getting good at most things takes practice. We have created a game, Bubbly Primes, with the purpose of giving students lots of practice factoring so they gain the ability to see common factors like in the GCF technique. It’s a fun way to cultivate skills that traditionally come from spending lots of time on worksheets and quizzes. We encourage you to spend some time playing the game.
What’s the difference between cross-cancellation and cross-multiplication?
Sometimes students confuse cross-cancellation with cross-multiplication. It’s easy to do, because the names are similar, both have to do with fractions, and furthermore, both have to do with multiplication. However:
Use cross multiplication to algebraically manipulate an equality involving fractions on either side of the equal sign, usually to solve for a variable.
Use Cross-cancellation as a simplification technique when two fractions are being multiplied.
Usually the two techniques are for entirely different kinds of problems. Here’s an example of each.
Next Topic: Kinds of Numbers
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17386 | https://www.researchgate.net/publication/273509036_Mobius_transformations_of_matrix_polynomials | Möbius transformations of matrix polynomials | Request PDF
Article
Möbius transformations of matrix polynomials
April 2015
Linear Algebra and its Applications 470(4):120-184
DOI:10.1016/j.laa.2014.05.013
Authors:
D. Steven Mackey
Western Michigan University
Niloufer Mackey
Western Michigan University
Christian Mehl
Technische Universität Berlin
Volker Mehrmann
Technische Universität Berlin
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Citations (66)References (70)
Abstract
We discuss Möbius transformations for general matrix polynomials over arbitrary fields, analyzing their influence on regularity, rank, determinant, constructs such as compound matrices, and on structural features including sparsity and symmetry. Results on the preservation of spectral information contained in elementary divisors, partial multiplicity sequences, invariant pairs, and minimal indices are presented. The effect on canonical forms such as Smith forms and local Smith forms, on relationships of strict equivalence and spectral equivalence, and on the property of being a linearization or quadratification are investigated. We show that many important transformations are special instances of Möbius transformations, and analyze a Möbius connection between alternating and palindromic matrix polynomials. Finally, the use of Möbius transformations in solving polynomial inverse eigenproblems is illustrated.
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... To achieve it, we use an argument based on Möbius transformations (see, [3,Section 6] and ). A Möbius transformation is applied to the prescribed data to remove infinity as an eigenvalue. ...
... By [26,Theorem 7.4], K(s) and L(s) T are minimal bases with column degrees k 1 , . . . , k r and ℓ 1 , . . . ...
... By taking α = 0, β = 1, γ = 1, and δ = −a in[3, Proposition 6.16] (see also[26, Theorem 5.3]) we deduce:(i) For any c ∈ F \ {0}, if (s − c) t 1 , . . . , (s − c)tr are the local invariant rational functions at s − c of B(s) then s − a − 1 c t 1 , . . . ...
Polynomial and rational matrices with the invariant rational functions and the four sequences of minimal indices prescribed
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May 2025
Itziar Baragaña
A. Roca
Froilan Dopico
S. Marcaida
The complete eigenstructure, or structural data, of a rational matrix R(s) is comprised by its invariant rational functions, both finite and at infinity, which in turn determine its finite and infinite pole and zero structures, respectively, and by the minimal indices of its left and right null spaces. These quantities arise in many applications and have been thoroughly studied in numerous references. However, other two fundamental subspaces of R(s) in contrast have received much less attention: the column and row spaces, which also have their associated minimal indices. This work solves the problems of finding necessary and sufficient conditions for the existence of rational matrices in two scenarios: (a) when the invariant rational functions and the minimal indices of the column and row spaces are prescribed, and (b) when the complete eigenstructure together with the minimal indices of the column and row spaces are prescribed. The particular, but extremely important, cases of these problems for polynomial matrices are solved first and are the main tool for solving the general problems. The results in this work complete and non-trivially extend the necessary and sufficient conditions recently obtained for the existence of polynomial and rational matrices when only the complete eigenstructure is prescribed.
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... In Section 4 this realization result is extended to include elementary divisors at ∞ in the given spectral data. In order to achieve this extension, we use the well-known tool of Möbius transformations [1,4,12,17,21]. In particular, some results on Möbius transformations of matrix polynomials over arbitrary fields developed in will be very useful for our purposes. At this point the signature result of the paper -the quasi-triangularization of any regular matrix polynomial over an arbitrary field in Theorem 4.5 -now follows easily. ...
... Next, we introduce the Index Sum Theorem [5, Theorem 6.5] for regular matrix polynomials. Theorem 2.9 can be seen to be a consequence of Theorem 5.2 in , together with the use of Möbius transformations . Theorem 2.9 (Index Sum Theorem for Regular Matrix Polynomials). ...
... An important concept, used in [1,4,6,9,19,21] for handling regular matrix polynomials that have nontrivial infinite spectral structure, is that of Möbius transformations of matrix polynomials. These transformations and their properties were studied in [1,12,17,21], and will be important for the results in Section 4. ...
Quasi-triangularization of matrix polynomials over arbitrary fields
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Feb 2023
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Luis Miguel Anguas
Froilan Dopico
Richard Hollister
D. Steven Mackey
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... Proof. First we prove the bounds in (21). By setting W 1 = 0 and W 2 to be any nonsingular matrix in (19) and (20) respectively and using Theorem 12, we have |λ| u 1 . ...
... Lower bounds (21) 0.8452 (26) 0.6664 (30) 0.8980 (31) 0.6053 The numerical experiments show that the bounds u and l 1 in Theorem 15 can be much tighter than those from (31). Observe that these two sets of bounds involve comparable computational effort, while those involving U 0 , U 1 , L 1 and L 2 are costlier to compute. ...
... For a detailed study of the effect of such transforms on matrix polynomials, see . ...
Locating eigenvalues of quadratic matrix polynomials
Article
May 2022
LINEAR ALGEBRA APPL
Nandita Roy
Shreemayee Bora
The location of the roots of a quadratic scalar polynomial may be identified from its coefficients. This paper shows that when the coefficients of the polynomial are square matrices, then appropriate generalizations of some of these statements hold for the eigenvalues of the resulting quadratic matrix polynomial. The locations of the eigenvalues are described with respect to the imaginary axis, the unit circle or the real line. The results lead to upper bounds on some important distances associated with quadratic matrix polynomials. The principal tool used is an eigenvalue localization technique using block Geršgorin sets applied to certain linearizations of these polynomials that come from well known vector spaces. New bounds on the eigenvalues of the matrix polynomial arising from these localizations are also presented.
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... In Section 4 we extend this realization result to include elementary divisors at ∞ in the given spectral data. In order to achieve this extension, we use the well-known tool of Möbius transformations , although once again we will need to do a significant amount of work to adapt them to work smoothly for matrix polynomials over arbitrary fields. At this point the signature result of the paper -the quasi-triangularization of any regular matrix polynomial over an arbitrary field in Theorem 4.9 -now follows easily. ...
... An important concept, used in and for handling regular matrix polynomials that have nontrivial infinite spectral structure, is that of Möbius transformations of matrix polynomials. These transformations and their properties were studied in , and will be important for the results in Section 4. Definition 2.13 (Möbius transformations of matrix polynomials). Let P (λ) be a matrix polynomial of grade g over the field F, and suppose A = a b c d ∈ F 2×2 is nonsingular. ...
... For conceptual clarity, it is important to keep in mind that for any fixed field F and grade g, the formula in (2.10) actually defines a whole family of transformations, one for each matrix size m × n. Once one has fixed the underlying field F, grade g, input matrix size m × n, and matrix A, then it is known that the Möbius transformation M A defines a bijection on the set of all m × n, grade g matrix polynomials over F. However, even by staying within the confines of one of these bijections, the transformation M A (P ) does not in general preserve the degree of the input P ; degree may increase, decrease, or remain unchanged . This lack of degree preservation happens even for the simplest size matrix polynomials, i.e., for scalar (1 × 1) polynomials. ...
Quasi-Triangularization of Matrix Polynomials over Arbitrary Fields
Preprint
Full-text available
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Luis Miguel Anguas
Froilan Dopico
Richard Hollister
D. Steven Mackey
In , Taslaman, Tisseur, and Zaballa show that any regular matrix polynomial P(λ)P(\lambda) over an algebraically closed field is spectrally equivalent to a triangular matrix polynomial of the same degree. When P(λ)P(\lambda) is real and regular, they also show that there is a real quasi-triangular matrix polynomial of the same degree that is spectrally equivalent to P(λ)P(\lambda), in which the diagonal blocks are of size at most 2×2 2 \times 2. This paper generalizes these results to regular matrix polynomials P(λ)P(\lambda) over arbitrary fields F\mathbb{F}, showing that any such P(λ)P(\lambda) can be quasi-triangularized to a spectrally equivalent matrix polynomial over F\mathbb{F} of the same degree, in which the largest diagonal block size is bounded by the highest degree appearing among all of the F\mathbb{F}-irreducible factors in the Smith form for P(λ)P(\lambda).
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... by[36, Prop. 3.29], the elementary divisors of P (λ) are (λ − µ 1 ), . . . ...
... 3.29], the elementary divisors of P (λ) are (λ − µ 1 ), . . . , (λ − µ s ) and t infinite elementary divisors of degree 1 by[36, Thm. 5.3], and the minimal indices of P (λ) and Q(λ) are identical by[36, Thm. ...
... , (λ − µ s ) and t infinite elementary divisors of degree 1 by[36, Thm. 5.3], and the minimal indices of P (λ) and Q(λ) are identical by[36, Thm. 7.5]. ...
Generic Symmetric Matrix Polynomials with Bounded Rank and Fixed Odd Grade
Article
Full-text available
Jul 2020
SIAM J MATRIX ANAL A
Fernando De Terán
Andrii Dmytryshyn
Froilan Dopico
We determine the generic complete eigenstructures for n×n n \times n complex symmetric matrix polynomials of odd grade d and rank at most r. More precisely, we show that the set of n×n n \times n complex symmetric matrix polynomials of odd grade d, i.e., of degree at most d, and rank at most r is the union of the closures of the ⌊r d/2⌋+1\lfloor rd/2\rfloor+1 sets of symmetric matrix polynomials having certain, explicitly described, complete eigenstructures. Then we prove that these sets are open in the set of n×n n \times n complex symmetric matrix polynomials of odd grade d and rank at most r. In order to prove the previous results, we need to derive necessary and sufficient conditions for the existence of symmetric matrix polynomials with prescribed grade, rank, and complete eigenstructure in the case where all their elementary divisors are different from each other and of degree 1. An important remark on the results of this paper is that the generic eigenstructures identified in this work are completely different from the ones identified in previous works for unstructured and skew-symmetric matrix polynomials with bounded rank and fixed grade larger than 1, because the symmetric ones include eigenvalues while the others not. This difference requires using new techniques.
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... When A k = 0, the degree and grade are equal, otherwise grade is strictly larger than the degree. Even though the classical references on matrix polynomials only consider the notion of degree, several recent papers [11,37,41] show multiple advantages of working with the grade of a matrix polynomial instead of its degree. ...
... Before we continue, it is worth mentioning that some references have also included under the name "T -palindromic polynomials" those P (λ) satisfying the condition (rev j P )(λ) = −P T (λ) . More recently, such matrix polynomials are referred to as T -anti-palindromic , and are not studied in this paper. ...
... Note that one can prove the above statements either directly, or by observing that Q(λ) is just obtained from P (λ) by a special Möbius transformation . In the latter case, statements (a)-(b) follow from [37,Thm. ...
Quadratic realizability of palindromic matrix polynomials
Article
Full-text available
Apr 2019
LINEAR ALGEBRA APPL
Fernando De Terán
Froilan Dopico
D. Steven Mackey
Vasilije Perović
Let L=(L1,L2) be a list consisting of a sublist L1 of powers of irreducible (monic) scalar polynomials over an algebraically closed field F, and a sublist L2 of nonnegative integers. For an arbitrary such list L, we give easily verifiable necessary and sufficient conditions for L to be the list of elementary divisors and minimal indices of some T-palindromic quadratic matrix polynomial with entries in the field F. For L satisfying these conditions, we show how to explicitly construct a T-palindromic quadratic matrix polynomial having L as its structural data; that is, we provide a T-palindromic quadratic realization of L. Our construction of T-palindromic realizations is accomplished by taking a direct sum of low bandwidth T-palindromic blocks, closely resembling the Kronecker canonical form of matrix pencils. An immediate consequence of our in-depth study of the structure of T-palindromic quadratic polynomials is that all even grade T-palindromic matrix polynomials have a T-palindromic strong quadratification. Finally, using a particular Möbius transformation, we show how all of our results can be easily extended to quadratic matrix polynomials with T-even structure.
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... The use of Möbius transformations of matrix polynomials can be traced back to at least [24,25], where they are defined for general rational matrices which are not necessarily polynomials. Since Möbius transformations change the eigenvalues of a matrix polynomial in a simple way and preserve most of the properties of the polynomial , they have often been used to transform a matrix polynomial with infinite eigenvalues into another polynomial with only finite eigenvalues and for which a certain problem can be solved more easily. Recent examples of this theoretical use can be found, for instance, in [12,33]. ...
... Such results were extended to Hamiltonian and symplectic matrix pencils, i.e., matrix polynomials of degree one, in [26,27] (with the goal of relating discrete and continuous control problems) and generalized to several classes of structured matrix polynomials of degree larger than one in . A thorough treatment of the properties of Möbius transformations of matrix polynomials is presented in a unified way in . ...
... Assume also that a backward stable structured algorithm is available for a certain type of structured pencils and that L can be transformed into a pencil with such structure through a Möbius transformation, M A . By [23,Corollary 8.6], M A (L) is a (strong) linearization of M A (P ). However, even if the structured algorithm guarantees that the PEP associated with M A (P ) is solved in a backward stable way , it is not guaranteed that it solves the PEP associated with P in a backward stable way as well. ...
Conditioning and backward errors of eigenvalues of homogeneous matrix polynomials under M\"{o}bius transformations
Preprint
Full-text available
Oct 2018
Luis Miguel Anguas
Maribel Bueno
Froilan Dopico
M\"{o}bius transformations have been used in numerical algorithms for computing eigenvalues and invariant subspaces of structured generalized and polynomial eigenvalue problems (PEPs). These transformations convert problems with certain structures arising in applications into problems with other structures and whose eigenvalues and invariant subspaces are easily related to the ones of the original problem. Thus, an algorithm that is efficient and stable for some particular structure can be used for solving efficiently another type of structured problem via an adequate M\"{o}bius transformation. A key question in this context is whether these transformations may change significantly the conditioning of the problem and the backward errors of the computed solutions, since, in that case, their use may lead to unreliable results. We present the first general study on the effect of M\"{o}bius transformations on the eigenvalue condition numbers and backward errors of approximate eigenpairs of PEPs. By using the homogeneous formulation of PEPs, we are able to obtain two clear and simple results. First, we show that, if the matrix inducing the M\"{o}bius transformation is well conditioned, then such transformation approximately preserves the eigenvalue condition numbers and backward errors when they are defined with respect to perturbations of the matrix polynomial which are small relative to the norm of the polynomial. However, if the perturbations in each coefficient of the matrix polynomial are small relative to the norm of that coefficient, then the corresponding eigenvalue condition numbers and backward errors are preserved approximately by the M\"{o}bius transformations induced by well-conditioned matrices only if a penalty factor, depending on those coefficients, is moderate. It is important to note that these simple results are no longer true if a non-homogeneous formulation is used.
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... The use of Möbius transformations of matrix polynomials can be traced back to at least [24,25], where they are defined for general rational matrices which are not necessarily polynomials. Since Möbius transformations change the eigenvalues of a matrix polynomial in a simple way and preserve most of the properties of the polynomial , they have often been used to transform a matrix polynomial with infinite eigenvalues into another polynomial with only finite eigenvalues and for which a certain problem can be solved more easily. Recent examples of this theoretical use can be found, for instance, in [12,33]. ...
... Such results were extended to Hamiltonian and symplectic matrix pencils, i.e., matrix polynomials of degree one, in [26,27] (with the goal of relating discrete and continuous control problems) and generalized to several classes of structured matrix polynomials of degree larger than one in . A thorough treatment of the properties of Möbius transformations of matrix polynomials is presented in a unified way in . ...
... Assume also that a backward stable structured algorithm is available for a certain type of structured pencils and that L can be transformed into a pencil with such structure through a Möbius transformation, M A . By [23,Corollary 8.6], M A (L) is a (strong) linearization of M A (P ). However, even if the structured algorithm guarantees that the PEP associated with M A (P ) is solved in a backward stable way , it is not guaranteed that it solves the PEP associated with P in a backward stable way as well. ...
Conditioning and backward errors of eigenvalues of homogeneous matrix polynomials under Mobius transformations.
Preprint
Oct 2018
Maribel Bueno
Luis Miguel Anguas
Froilan Dopico
Möbius transformations have been used in numerical algorithms for computing eigen-values and invariant subspaces of structured generalized and polynomial eigenvalue problems (PEPs). These transformations convert problems with certain structures arising in applications into problems with other structures and whose eigenvalues and invariant subspaces are easily related to the ones of the original problem. Thus, an algorithm that is efficient and stable for some particular structure can be used for solving efficiently another type of structured problem via an adequate Möbius transformation. A key question in this context is whether these transformations may change significantly the conditioning of the problem and the backward errors of the computed solutions, since, in that case, their use might lead to unreliable results. We present the first general study on the effect of Möbius transformations on the eigenvalue condition numbers and backward errors of approximate eigenpairs of PEPs. By using the homogeneous formulation of PEPs, we are able to obtain two clear and simple results. First, we show that, if the matrix inducing the Möbius transformation is well conditioned, then such transformation approximately preserves the eigenvalue condition numbers and backward errors when they are defined with respect to perturbations of the matrix polynomial which are small relative to the norm of the whole polynomial. However, if the perturbations in each coefficient of the matrix polynomial are small relative to the norm of that coefficient, then the corresponding eigenvalue condition numbers and backward errors are preserved approximately by the Möbius transformations induced by well-conditioned matrices only if a penalty factor, depending on the norms of those matrix coefficients, is moderate. It is important to note that these simple results are no longer true if a non-homogeneous formulation of the PEP is used.
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... For the sake of simplification of proofs in the remainder of this paper, we will use Möbius transformations to reduce the case of perturbations of the form (1.2) to the special case e = 0 which has the effect that just one of the coefficient matrices of the pencil is perturbed. Although the effect of Möbius transformations on most invariants of matrix pencils or matrix polynomials is well understood (see, e.g., ), it seems that the change of the sign characteristic of real eigenvalues of Hermitian or real symmetric pencils has not yet been considered in the literature. Therefore, we will fill this gap and investigate how the sign characteristic of a Hermitian or real symmetric pencil is related to the one that is obtained under a Möbius transformation that keeps the Hermitian or real symmetric structure of the pencil invariant. ...
... (Here and throughout the remainder of the paper, we interpret a fraction d 0 with d ∈ C \ {0} as ∞.) Following , we extend this definition onto linear pencils as ...
... Thus, we may assume without loss of generality that the pencil under consideration is already in one of the canonical forms from section 2, and it is sufficient to investigate the effect of the Möbius transformation on each diagonal block independently. It was shown in that the eigenvalues of M b,c P (λ) are the Möbius transforms of the eigenvalues of P (λ) and the Kronecker structure is preserved. In particular, Theorem 5.3 of implies the following two lemmas displaying the effect of Möbius transformations on the blocks in the structured canonical forms. ...
Parameter-Dependent Rank-One Perturbations of Singular Hermitian Or Symmetric Pencils
Article
Full-text available
Jan 2017
SIAM J MATRIX ANAL A
Christian Mehl
Michał Wojtylak
Volker Mehrmann
Structure-preserving generic low-rank perturbations are studied for classes of structured matrix pencils, including real symmetric, complex symmetric, and complex Hermitian pencils. For singular pencils it is analyzed which characteristic quantities stay invariant in the perturbed canonical form, and it is shown that the regular part of a structured matrix pencil is not affected by generic perturbations of rank one. When the rank-one perturbations involve a scaling parameter, the behavior of the canonical forms dependent on this parameter is analyzed as well.
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... At first sight this looks artificial, but under some circumstances, and especially for structured matrix polynomials, it is a very useful concept, see [27,28]. In particular, and this is the main reason for using the grade instead of the degree, it has been shown in [27,31] that Möbius transformations, which play an important role in our analysis, are grade-preserving, but in general not degree-preserving. ...
... At first sight this looks artificial, but under some circumstances, and especially for structured matrix polynomials, it is a very useful concept, see [27,28]. In particular, and this is the main reason for using the grade instead of the degree, it has been shown in [27,31] that Möbius transformations, which play an important role in our analysis, are grade-preserving, but in general not degree-preserving. Once the grade g of a matrix polynomial P (x) = g j=0 P j x j is fixed, the reversal of P (x) is given by ...
... Proof. The conservation of the partial multiplicities follows immediately from [27,Theorem 5.3] or [31,Theorem 4.1]. Thus, it suffices to prove the statements on the sign features for which we apply Theorems 3.4 and 3.5, or equivalently Example 3.6. ...
On the sign characteristics of Hermitian matrix polynomials
Article
Full-text available
Sep 2016
LINEAR ALGEBRA APPL
Volker Mehrmann
Vanni Noferini
Françoise Tisseur
Hongguo Xu
The sign characteristics of Hermitian matrix polynomials are discussed, and in particular an appropriate definition of the sign characteristics associated with the eigenvalue infinity. The concept of sign characteristic arises in different forms in many scientific fields, and is essential for the stability analysis in Hamiltonian systems or the perturbation behavior of eigenvalues under structured perturbations. We extend classical results by Gohberg, Lancaster, and Rodman to the case of infinite eigenvalues. We derive a systematic approach, studying how sign characteristics behave after an analytic change of variables, including the important special case of Möbius transformations, and we prove a signature constraint theorem. We also show that the sign characteristic at infinity stays invariant in a neighborhood under perturbations for even degree Hermitian matrix polynomials, while it may change for odd degree matrix polynomials. We argue that the non-uniformity can be resolved by introducing an extra zero leading matrix coefficient.
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... does not have the eigenvalue \infty . This is possible, because the considered M\" obius transformation does not change the Kronecker structure of the pencil except for rotating the spectrum around the origin with a specific angle; see . Furthermore, the transformed pencil is still Hermitian, as \zeta 1 and \zeta 2 are taken to be real. ...
... These transformations map (anti-)palindromic pencils to even or odd pencils and vice versa; see, e.g., . Since Cayley transformations are special M\" obius transformations, it follows from the results in that the eigenvalues change in a predictable bijective way while the eigenvectors remain invariant. Also the change in the sign characteristic under a M\" obius transformation is well understood . ...
Solving Singular Generalized Eigenvalue Problems. Part III: Structure Preservation
Article
Full-text available
Aug 2025
Michiel Hochstenbach
Christian Mehl
Bor Plestenjak
In Parts I and II of this series of papers, three new methods for the computation of eigenvalues of singular pencils were developed: rank-completing perturbations, rank-projections, and augmentation. It was observed that a straightforward structure-preserving adaption for symmetric pencils was not possible and it was left as an open question how to address this challenge. In this Part III, it is shown how the observed issue can be circumvented by using Hermitian perturbations. This leads to structure-preserving analogs of the three techniques from Parts I and II for Hermitian pencils (including real symmetric pencils) as well as for skew-Hermitian, ∗\ast-even, ∗\ast-odd, and ∗\ast-(anti-)palindromic pencils. It is an important feature of these methods that the sign characteristic of the given pencil is preserved.
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... does not have the eigenvalue ∞. This is possible, because the considered Möbius transformation does not change the Kronecker structure of the pencil except for rotating the spectrum around the origin with a specific angle; see . Furthermore, the transformed pencil is still Hermitian, as ζ 1 and ζ 2 are taken to be real. ...
... These transformations map (anti-)palindromic pencils to even or odd pencils and vice versa; see, e.g., . Since Cayley transformations are special Möbius transformations, it follows from the results in that the eigenvalues change in a predictable bijective way while the eigenvectors remain invariant. Also the change in the sign characteristic under a Möbius transformation is well understood . ...
Solving singular generalized eigenvalue problems. Part III: structure preservation
Preprint
Full-text available
Jun 2024
Michiel Hochstenbach
Christian Mehl
Bor Plestenjak
In Parts I and II of this series of papers, three new methods for the computation of eigenvalues of singular pencils were developed: rank-completing perturbations, rank-projections, and augmentation. It was observed that a straightforward structure-preserving adaption for symmetric pencils was not possible and it was left as an open question how to address this challenge. In this Part III, it is shown how the observed issue can be circumvented by using Hermitian perturbations. This leads to structure-preserving analogues of the three techniques from Parts I and II for Hermitian pencils (including real symmetric pencils) as well as for related structures. It is an important feature of these methods that the sign characteristic of the given pencil is preserved. As an application, it is shown that the resulting methods can be used to solve systems of bivariate polynomials.
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... where λ, a, b, c, d ∈ C. Over arbitrary fields, the Möbius transformation preserves a number of spectral features of matrix polynomials, such as regularity, rank, minimal indicies, the location of zero entries, symmetry, and skew-symmetry . In particular, every Möbius transformation preserves the relation of spectral equivalence . ...
... Over arbitrary fields, the Möbius transformation preserves a number of spectral features of matrix polynomials, such as regularity, rank, minimal indicies, the location of zero entries, symmetry, and skew-symmetry . In particular, every Möbius transformation preserves the relation of spectral equivalence . The computational eigenvalue problem, after introducing a PML, truncating the domain, and applying finite elements, can be written as ...
Nonlinear eigenvalue problems for coupled Helmholtz equations modeling gradient-index graphene waveguides
Preprint
Mar 2020
Jung Heon Song
Matthias Maier
Mitchell Luskin
We discuss a quartic eigenvalue problem arising in the context of an optical waveguiding problem involving atomically thick 2D materials. The waveguide configuration we consider consists of a gradient-index (spatially dependent) dielectric equipped with conducting interior interfaces. This leads to a quartic eigenvalue problem with mixed transverse electric and transverse magnetic modes, and strongly coupled electric and magnetic fields. We derive a weak formulation of the quartic eigenvalue problem and introduce a numerical solver based on a quadratification approach in which the quartic eigenvalue problem is transformed to a spectrally equivalent linear eigenvalue problem. We verify our numerical framework against analytical solutions for prototypical geometries. As a practical example, we demonstrate how an improved quality factor (defined by the ratio of the real and the imaginary part of the computed eigenvalues) can be obtained for a family of gradient-index host materials with internal conducting interfaces. We outline how this result lays the groundwork for solving related shape optimization problems.
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... and t infinite elementary divisors of degree 1 by[36, Thm. 5.3], and the minimal indices of P (λ) and Q(λ) are identical by[36, Thm. ...
... and t infinite elementary divisors of degree 1 by[36, Thm. 5.3], and the minimal indices of P (λ) and Q(λ) are identical by[36, Thm. 7.5]. ...
Generic symmetric matrix polynomials with bounded rank and fixed odd grade
Preprint
Full-text available
Nov 2019
Andrii Dmytryshyn
Fernando De Terán
Froilan Dopico
We determine the generic complete eigenstructures for n×n n \times n complex symmetric matrix polynomials of odd grade d and rank at most r. More precisely, we show that the set of n×n n \times n complex symmetric matrix polynomials of odd grade d, i.e., of degree at most d, and rank at most r is the union of the closures of the ⌊r d/2⌋+1\lfloor rd/2\rfloor+1 sets of symmetric matrix polynomials having certain, explicitly described, complete eigenstructures. Then, we prove that these sets are open in the set of n×n n \times n complex symmetric matrix polynomials of odd grade d and rank at most r. In order to prove the previous results, we need to derive necessary and sufficient conditions for the existence of symmetric matrix polynomials with prescribed grade, rank, and complete eigenstructure, in the case where all their elementary divisors are different from each other and of degree 1. An important remark on the results of this paper is that the generic eigenstructures identified in this work are completely different from the ones identified in previous works for unstructured and skew-symmetric matrix polynomials with bounded rank and fixed grade larger than one, because the symmetric ones include eigenvalues while the others not. This difference requires to use new techniques.
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... At the cost of passing to the algebraic closure F, one could instead list all the finite poles and zeros of R(λ), together with their structural index sequences. This description may be more natural in the important case F = C. [9,18,30] are called the partial multiplicity sequences at the finite eigenvalues. ...
... As we have seen so far in this section, it is possible to coherently define the structure at infinity of a polynomial matrix P (λ) by viewing it as a rational matrix, and then using the structural indices at infinity of this (special) rational matrix. However, this has not been the typical practice in the literature on polynomial matrices [9,11,18,29,30], or even for matrix pencils [17,38,43]. Instead, the standard way to define the structure at infinity of polynomial matrices has been via the reversal polynomial (2.2), as in the following definition. ...
Van Dooren's Index Sum Theorem and Rational Matrices with Prescribed Structural Data
Article
Full-text available
Jun 2019
SIAM J MATRIX ANAL A
Richard Hollister
Luis Miguel Anguas
Froilan Dopico
D. Steven Mackey
The structural data of any rational matrix R(λ)R(\lambda), i.e., the structural indices of its poles and zeros together with the minimal indices of its left and right nullspaces, is known to satisfy a simple condition involving certain sums of these indices. This fundamental constraint was first proved by Van Dooren in 1978; here we refer to this result as the rational index sum theorem. An analogous result for polynomial matrices has been independently discovered (and rediscovered) several times in the past three decades. In this paper we clarify the connection between these two seemingly different index sum theorems, describe a little bit of the history of their development, and discuss their curious apparent unawareness of each other. Finally, we use the connection between these results to solve a fundamental inverse problem for rational matrices---for which lists L\cal L of prescribed structural data does there exist some rational matrix R(λ)R(\lambda) that realizes exactly the list L\cal L? We show that Van Dooren's condition is the only constraint on rational realizability; that is, a list L\cal L is the structural data of some rational matrix R(λ)R(\lambda) if and only if L\cal L satisfies the rational index sum condition.
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... Theorem A.3.4 in is another example, closer to the ideas to be developed here, where the local structure is considered. On the other hand, the change of the finite and infinite elementary divisors by Möbius transformations on a matrix polynomial has been studied in [16,27] using a "global" approach in contrast with the local approach to be used here, which allows to extend the study to matrices of rational functions. ...
... We deal now with the problem of the relationship between the finite and infinite elementary divisors of two matrix polynomials obtained from each other by a Möbius transformation. The following result was proved in for nonsingular polynomial matrices using a completely different technique (see also [16,19]). 7. Equivalence that preserves the finite and infinite elementary divisors. In the previous section, and for notational convenience, we have included polynomials equal to 1 in the list of infinite elementary divisors. ...
Finite and infinite structures of rational matrices: A local approach
Article
Full-text available
Jun 2015
A. Amparan
S. Marcaida
Ion Zaballa
The structure of a rational matrix is given by its Smith-McMillan invariants. Some properties of the Smith-McMillan invariants of rational matrices with elements in different principal ideal domains are presented: In the ring of polynomials in one indeterminate (global structure), in the local ring at an irreducible polynomial (local structure), and in the ring of proper rational functions (infinite structure). Furthermore, the change of the finite (global and local) and infinite structures is studied when performing a Möbius transformation on a rational matrix. The results are applied to define an equivalence relation in the set of polynomial matrices, with no restriction on size, for which a complete system of invariants are the finite and infinite elementary divisors.
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... We finally consider the problem of constructing symmetric or skew-symmetric strong linearizations for P (λ) when its trailing coefficient is nonsingular. The key tool is the following lemma, which is a particular case of [35,Corollary 8.6] where the authors study the interaction between linearizations and Möbius transformations of matrix polynomials 2 . ...
... the rev(·) operation is a particular case of a Möbius transformation of a matrix polynomial. ...
Constructing symmetric structure-preserving strong linearizations
Article
Feb 2017
Heike Fassbender
Javier Perez Alvaro
Nikta Shayanfar
Polynomials eigenvalue problems with structured matrix polynomials arise in many applications. The standard way to solve polynomial eigenvalue problems is through the classical Frobenius companion linearizations, which may not retain the structure of the matrix polynomial. Particularly, the structure of the symmetric matrix polynomials can be lost, while from the computational point of view, it is advisable to construct a linearization which preserves the symmetry structure. Recently, new families of block-Kronecker pencils have been introduced in . Applying block-Kronecker pencils, we present structure-preserving strong linearizations for symmetric matrix polynomials. When the matrix polynomial has an odd degree, these linearizations are strong regardless of whether the matrix polynomial is regular or singular. Additionally, we construct structure-preserving strong linearizations for regular symmetric matrix polynomials of even degree under some simple nonsingularity conditions.
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... We emphasize that such condition includes a high level of redundancy while (17) does not, because unimodular matrices are very particular instances of matrices invertible in F λ (λ). The equivalence of these two conditions when (16) holds is a consequence of the effect of Möbius transformations on the elementary divisors of polynomial matrices [4,23,30]. ...
... Definition 7.1 R 1 (λ) and R 2 (λ) as in (23) are said to be strictly system equivalent at infinity if there exist biproper matrices B 1 (λ), B 2 (λ) ∈ F pr (λ) n×n and proper matrices W (λ) ∈ F pr (λ) p×n , Z(λ) ∈ F pr (λ) n×m such that ...
Strong Linearizations of Rational Matrices
Article
Full-text available
Nov 2018
SIAM J MATRIX ANAL A
A. Amparan
S. Marcaida
Ion Zaballa
Froilan Dopico
This paper defines for the first time strong linearizations of arbitrary rational matrices, studies in depth properties and characterizations of such linear matrix pencils, and develops infinitely many examples of strong linearizations that can be explicitly and easily constructed from a minimal state-space realization of the strictly proper part of the considered rational matrix and the coefficients of the polynomial part. As a consequence, the results in this paper establish a rigorous foundation for the numerical computation of the complete structure of zeros and poles, both finite and at infinity, of any rational matrix by applying any well-known backward stable algorithm for generalized eigenvalue problems to any of the strong linearizations constructed in this work.
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... Thus, the degree of P(λ) is fixed while its grade d is a choice that must satisfy d ≥ deg(P). The concept of grade has been used previously in [17,56] and is convenient when the degree of a polynomial is not known in advance. Throughout this paper when the grade of P(λ) is not explicitly stated, we consider its grade equal to its degree. ...
... These are the minimal bases of interest in this work. The proof of Theorem 2.7 is omitted since it follows from results on row-wise reversals of minimal bases [13,56]. For a simpler proof based on Theorem 2.2, see the extended version of this paper . ...
Block Kronecker Linearizations of Matrix Polynomials and their Backward Errors
Article
Full-text available
Oct 2018
NUMER MATH
Froilan Dopico
Piers W. Lawrence
Javier Pérez
Paul Michel Van Dooren
We introduce a new family of strong linearizations of matrix polynomials---which we call "block Kronecker pencils"---and perform a backward stability analysis of complete polynomial eigenproblems. These problems are solved by applying any backward stable algorithm to a block Kronecker pencil, such as the staircase algorithm for singular pencils or the QZ algorithm for regular pencils. This stability analysis allows us to identify those block Kronecker pencils that yield a computed complete eigenstructure which is exactly that of a slightly perturbed matrix polynomial. The global backward error analysis in this work presents for the first time the following key properties: it is a rigurous analysis valid for finite perturbations (i.e., it is not a first order analysis), it provides precise bounds, it is valid simultaneously for a large class of linearizations, and it establishes a framework that may be generalized to other classes of linearizations. These features are related to the fact that block Kronecker pencils are a particular case of the new family of "strong block minimal bases pencils", which are robust under certain perturbations and, so, include certain perturbations of block Kronecker pencils. We hope that this robustness property will allow us to extend the results in this paper to other contexts.
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... We finally consider the problem of constructing symmetric or skew-symmetric strong linearizations for P (λ) when its trailing coefficient is nonsingular. The key tool is the following lemma, which is a particular case of [35,Corollary 8.6] where the authors study the interaction between linearizations and Möbius transformations of matrix polynomials 2 . lemma:sym_even Lemma 5.15. ...
... the rev(·) operation is a particular case of a Möbius transformation of a matrix polynomial. ...
Symmetric and skew-symmetric block-Kronecker linearizations
Article
Jun 2016
Heike Fassbender
Javier Perez Alvaro
Nikta Shayanfar
Many applications give rise to structured matrix polynomials. The problem of constructing structure-preserving strong linearizations of structured matrix polynomials is revisited in this work and in the forthcoming ones \cite{PartII,PartIII}. With the purpose of providing a much simpler framework for structure-preserving linearizations for symmetric and skew-symmetric matrix polynomial than the one based on Fiedler pencils with repetition, we introduce in this work the families of (modified) symmetric and skew-symmetric block Kronecker pencils. These families provide a large arena of structure-preserving strong linearizations of symmetric and skew-symmetric matrix polynomials. When the matrix polynomial has degree odd, these linearizations are strong regardless of whether the matrix polynomial is regular or singular, and many of them give rise to structure-preserving companion forms. When some generic nonsingularity conditions are satisfied, they are also strong linearizations for even-degree regular matrix polynomials. Many examples of structure-preserving linearizations obtained from Fiedler pencils with repetitions found in the literature are shown to belong (modulo permutations) to these families of linearizations. In particular, this is shown to be true for the well-known block-tridiagonal symmetric and skew-symmetric companion forms. Since the families of symmetric and skew-symmetric block Kronecker pencils belong to the recently introduced set of minimal bases pencils \cite{Fiedler-like}, they inherit all its desirable properties for numerical applications. In particular, it is shown that eigenvectors, minimal indices, and minimal bases of matrix polynomials are easily recovered from those of any of the linearizations constructed in this work.
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... Substituting these equations in equations (4.5)-(4.10), we obtain, Differentiating equation (4.25) concerning § in the Caputo sense, we obtain, [52,53] used in combinatorial mathematics to study partially ordered sets (posets) and incidence algebras and Abel polynomial [54,55] used in combinatorial mathematics to study partitions and permutations. All the computations are carried in MATHEMATICA 13.0 in the 11th Gen Intel(R) Core(TM) i3-1115G4 @ 3.00GHz, 2901 Mhz, 2 Core(s), 4 Logical Processor. ...
A potential rook polynomial integration approach for seventh-order time frame fractional KdV models
Article
Full-text available
May 2025
PHYS SCRIPTA
Nirmala A.N.
Kumbinarasaiah S
Nonlinear evolution equations are an intriguing and challenging field of mathematics and physics. The Korteweg–de Vries (KdV) model is a prominent nonlinear evolution paradigm extensively explored in computational physics owing to its significance in replicating shallow water waves. Higher-order KdV equations, such as the Sawada-Kotera-Ito, Kaup-Kuperschimdt, and Lax simulations, have advanced wave research applications encompassing hydrodynamics, plasma physics, and nonlinear optical phenomena. Time fractional higher-order KdV models contribute to more effective problem-solving in the actual world. Seventh-order time fractional KdV models are incredibly complex, with higher-order and fractional time derivatives. Traditional numerical approaches may be insufficient to describe these equations’ behavior accurately. Advanced numerical techniques use fewer processing resources and less time. As a result, we developed a novel, efficient numerical strategy from the backdrop of algebraic graph theory to solve these models. This numerical algorithm includes an operational integration matrix based on the Rook polynomials to approximate the KdV models. The Caputo fractional derivatives are incorporated to deal with the time fractional derivatives in the seventh-order KdV models. With the standard collocation points, the emerging equation is turned into an ensemble of nonlinear algebraic equations and solved by Newton’s method to provide an alternative Rook polynomial collocation mechanism (RCM) solution. Numerical examples demonstrate the RCM’s efficacy. Absolute and residual errors are computed and compared with the recent, efficient methods to confirm the viability of the RCM algorithm. The comparative tabular results and graphical analysis witnessed the dynamic behaviors in the fractional domain and the RCM’s compatibility with the accurate results. The study demonstrates that the proposed methodology is an effective and convenient solution for fractional KdV models and generates scope for solving other variable order KdV models.
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... In this section we consider the following Möbius transform [16,17] T and its inverse T −1 : ...
Para-Hermitian Rational Matrices
Article
Full-text available
Dec 2024
Froilan Dopico
Vanni Noferini
María C. Quintana
Paul Michel Van Dooren
In this paper, we study para-Hermitian rational matrices and the associated structured rational eigenvalue problem (REP). Para-Hermitian rational matrices are square rational matrices that are Hermitian for all z on the unit circle that are not poles. REPs are often solved via linearization, that is, using matrix pencils associated to the corresponding rational matrix that preserve the spectral structure. Yet, nonconstant polynomial matrices cannot be para-Hermitian. Therefore, given a para-Hermitian rational matrix R(z), we instead construct a ∗-palindromic linearization for (1+z)R(z), whose eigenvalues that are not on the unit circle preserve the symmetries of the zeros and poles of R(z). This task is achieved via Möbius transformations. We also give a constructive method that is based on an additive decomposition into the stable and antistable parts of R(z). Analogous results are presented for para-skew-Hermitian rational matrices, i.e., rational matrices that are skew-Hermitian upon evaluation on those points of the unit circle that are not poles.
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... We will show that the Möbius group action reduces the dimension of the minimization problem for the tube-volume criterion. For a recent paper in which the Möbius transformation acts on polynomials, see Mackey, et al (2015). ...
Optimal experimental design that minimizes the width of simultaneous confidence bands
Preprint
Apr 2017
Henry P. Wynn
Satoshi Kuriki
We propose an optimal experimental design for a curvilinear regression model that minimizes the band-width of simultaneous confidence bands. Simultaneous confidence bands for curvilinear regression are constructed by evaluating the volume of a tube about a curve that is defined as a trajectory of a regression basis vector (Naiman, 1986). The proposed criterion is constructed based on the volume of a tube, and the corresponding optimal design that minimizes the volume of tube is referred to as the tube-volume optimal (TV-optimal) design. For Fourier and weighted polynomial regressions, the problem is formalized as one of minimization over the cone of Hankel positive definite matrices, and the criterion to minimize is expressed as an elliptic integral. We show that the M\"obius group keeps our problem invariant, and hence, minimization can be conducted over cross-sections of orbits. We demonstrate that for the weighted polynomial regression and the Fourier regression with three bases, the tube-volume optimal design forms an orbit of the M\"obius group containing D-optimal designs as representative elements.
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... In Proposition 3.4, we state without proofs those that will be relevant in this work. For a thorough study of the properties of Möbius transformations of matrix polynomials we refer the reader to . ...
Structured backward error analysis of linearized structured polynomial eigenvalue problems
Preprint
Full-text available
Dec 2016
Froilan Dopico
Javier Pérez
Paul Michel Van Dooren
We introduce a new class of structured matrix polynomials, namely, the class of M_A-structured matrix polynomials, to provide a common framework for many classes of structured matrix polynomials that are important in applications: the classes of (skew-)symmetric, (anti-)palindromic, and alternating matrix polynomials. Then, we introduce the families of M_A-structured strong block minimal bases pencils and of M_A-structured block Kronecker pencils,, and show that any M_A-structured odd-degree matrix polynomial can be strongly linearized via an M_A-structured block Kronecker pencil. Finally, for the classes of (skew-)symmetric, (anti-)palindromic, and alternating odd-degree matrix polynomials, the M_A-structured framework allows us to perform a global and structured backward stability analysis of complete structured polynomial eigenproblems, regular or singular, solved by applying to an M_A-structured block Kronecker pencil a structurally backward stable algorithm that computes its complete eigenstructure, like the palindromic-QR algorithm or the structured versions of the staircase algorithm. This analysis allows us to identify those M_A-structured block Kronecker pencils that yield a computed complete eigenstructure which is the exact one of a slightly perturbed structured matrix polynomial.These pencils include (modulo permutations) the well-known block-tridiagonal and block-antitridiagonal structure-preserving linearizations. Our analysis incorporates structure to the recent (unstructured) backward error analysis performed for block Kronecker linearizations by Dopico, Lawrence, P\'erez and Van Dooren, and share with it its key features, namely, it is a rigorous analysis valid for finite perturbations, i.e., it is not a first order analysis, it provides precise bounds, and it is valid simultaneously for a large class of structure-preserving strong linearizations.
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... In this section we consider the following Möbius transform [16,17] T and its inverse T −1 : ...
Para-Hermitian rational matrices
Preprint
Full-text available
Jul 2024
Froilan Dopico
Vanni Noferini
María C. Quintana
Paul Michel Van Dooren
In this paper we study para-Hermitian rational matrices and the associated structured rational eigenvalue problem (REP). Para-Hermitian rational matrices are square rational matrices that are Hermitian for all z on the unit circle that are not poles. REPs are often solved via linearization, that is, using matrix pencils associated to the corresponding rational matrix that preserve the spectral structure. Yet, non-constant polynomial matrices cannot be para-Hermitian. Therefore, given a para-Hermitian rational matrix R(z), we instead construct a ∗-palindromic linearization for (1+z)R(z), whose eigenvalues that are not on the unit circle preserve the symmetries of the zeros and poles of R(z). This task is achieved via M\"{o}bius transformations. We also give a constructive method that is based on an additive decomposition into the stable and anti-stable parts of R(z). Analogous results are presented for para-skew-Hermitian rational matrices, i.e., rational matrices that are skew-Hermitian upon evaluation on those points of the unit circle that are not poles.
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... The elliptical and hyperbolic quaternion concepts are defined with the help of generalized scalar products. For the generalized scalar product spaces, and for skew symmetric and orthogonal matrices defined by the generalized scalar product, see [16,30,31,34,45,46]. In Table 1, we give a summary of different types of quaternions briefly. ...
Generalized elliptical quaternions with some applications
Article
Full-text available
Jan 2023
Harun Barış Çolakoğlu
Mustafa Özdemir
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... The final solution comes from changing the variable λ to λ := d λ λ and from combining this with the multiplication by the constant d R and the diagonal scaling T discussed above. Note that the change of variable transforms the zeros and the poles of R(λ) in a very simple way, preserving their partial multiplicities, and that does not change at all its minimal indices [15,20]. The combination of all these scalings yields a new transfer function ...
Structural backward stability in rational eigenvalue problems solved via block Kronecker linearizations
Article
Full-text available
Jan 2023
Froilan Dopico
María C. Quintana
Paul Michel Van Dooren
In this paper we study the backward stability of running a backward stable eigenstructure solver on a pencil S(λ)S(\lambda ) that is a strong linearization of a rational matrix R(λ)R(\lambda ) expressed in the form R(λ)=D(λ)+C(λ I ℓ−A)−1 B R(\lambda )=D(\lambda )+ C(\lambda I_\ell -A)^{-1}B, where D(λ)D(\lambda ) is a polynomial matrix and C(λ I ℓ−A)−1 B C(\lambda I_\ell -A)^{-1}B is a minimal state-space realization. We consider the family of block Kronecker linearizations of R(λ)R(\lambda ), which have the following structure S(λ):=[M(λ)K^2 T C K 2 T(λ)B K^1 A−λ I ℓ 0 K 1(λ)0 0],\begin{aligned} S(\lambda ):=\left[ \begin{array}{ccc} M(\lambda ) &{} {\widehat{K}}2^T C &{} K_2^T(\lambda ) \ B {\widehat{K}}_1 &{} A- \lambda I\ell &{} 0\ K_1(\lambda ) &{} 0 &{} 0 \end{array}\right] , \end{aligned}where the blocks have some specific structures. Backward stable eigenstructure solvers, such as the QZ or the staircase algorithms, applied to S(λ)S(\lambda ) will compute the exact eigenstructure of a perturbed pencil S^(λ):=S(λ)+Δ S(λ)\widehat{S}(\lambda ):=S(\lambda )+\varDelta S(\lambda ) and the special structure of S(λ)S(\lambda ) will be lost, including the zero blocks below the anti-diagonal. In order to link this perturbed pencil with a nearby rational matrix, we construct in this paper a strictly equivalent pencil S~(λ)=(I−X)S^(λ)(I−Y)\widetilde{S}(\lambda )=(I-X)\widehat{S}(\lambda )(I-Y) that restores the original structure, and hence is a block Kronecker linearization of a perturbed rational matrix R~(λ)=D~(λ)+C~(λ I ℓ−A~)−1 B~{{\widetilde{R}}}(\lambda ) = {{\widetilde{D}}}(\lambda )+ {{\widetilde{C}}}(\lambda I\ell - {{\widetilde{A}}})^{-1} {{\widetilde{B}}}, where D~(λ){{\widetilde{D}}}(\lambda ) is a polynomial matrix with the same degree as D(λ)D(\lambda ). Moreover, we bound appropriate norms of D~(λ)−D(λ){{\widetilde{D}}}(\lambda )- D(\lambda ), C~−C{{\widetilde{C}}} - C, A~−A{{\widetilde{A}}} - A and B~−B{{\widetilde{B}}} - B in terms of an appropriate norm of Δ S(λ)\varDelta _S(\lambda ). These bounds may be, in general, inadmissibly large, but we also introduce a scaling that allows us to make them satisfactorily tiny, by making the matrices appearing in both S(λ)S(\lambda ) and R(λ)R(\lambda ) have norms bounded by 1. Thus, for this scaled representation, we prove that the staircase and the QZ algorithms compute the exact eigenstructure of a rational matrix R~(λ){{\widetilde{R}}}(\lambda ) that can be expressed in exactly the same form as R(λ)R(\lambda ) with the parameters defining the representation very near to those of R(λ)R(\lambda ). This shows that this approach is backward stable in a structured sense. Several numerical experiments confirm the obtained backward stability results.
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... Now, let us assume that μ = ∞. It is straightforward to see (but we refer otherwise to [27,Remark 4.3]) that, for a given pencil L(λ) and μ ∈ C, the identity W (μ, L) = W (μ −1 , revL) holds. It is also immediate to see that rank (revL) = rank L. ...
On bundles of matrix pencils under strict equivalence
Article
Nov 2022
LINEAR ALGEBRA APPL
Fernando De Terán
Froilan Dopico
Bundles of matrix pencils (under strict equivalence) are sets of pencils having the same Kronecker canonical form, up to the eigenvalues (namely, they are an infinite union of orbits under strict equivalence). The notion of bundle for matrix pencils was introduced in the 1990's, following the same notion for matrices under similarity, introduced by Arnold in 1971, and it has been extensively used since then. Despite the amount of literature devoted to describing the topology of bundles of matrix pencils, some relevant questions remain still open in this context. For example, the following two: (a) provide a characterization for the inclusion relation between the closures (in the standard topology) of bundles; and (b) are the bundles open in their closure? The main goal of this paper is providing an explicit answer to these two questions. In order to get this answer, we also review and/or formalize some notions and results already existing in the literature. We also prove that bundles of matrices under similarity, as well as bundles of matrix polynomials (defined as the set of m×n matrix polynomials of the same grade having the same spectral information, up to the eigenvalues) are open in their closure.
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... Detailed information on the Cayley transform can be found in many articles [31,. For the relationships between the Cayley transform and the M öbius transformation, see the article . We know that these properties are valid also in n-dimensional space. ...
Non-parabolic conical rotations
Article
Aug 2022
J COMPUT APPL MATH
Harun Barış Çolakoğlu
İskender Öztürk
Mustafa Özdemir
In this paper, we determine non-parabolic conical motions that occur on any given ellipse or hyperbola without using affine transformations. To achieve this aim, first, we define a generalized inner product whose circle is the given ellipse or hyperbola, and then determine elliptical and hyperbolic versions of skew-symmetric and orthogonal matrices using the associated inner product. Finally, we generate elliptical and hyperbolic versions of rotation and reflection matrices using the famous Rodrigues, Cayley, and Householder transformations. For each of the generalized formulas, we give numerical examples.
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... Transformasi Möbius menarik untuk dibahas dan dipelajari, salah satunya karena memiiki sifat-sifat yang memiliki koneksi kuat dengan geometri non-euclid, ilmu terapan, dan fisika . ...
Sifat Preservasi Lingkaran dan Garis Pada Transformasi Möbius
Article
Full-text available
Jun 2022
Guntur Maulana Muhammad
Iden Rainal Ihsan
Roni Priyanda
This article discusses Möbius transformation from the point of view of algebra to describe one of its geometric properties, i.e. preserving circles and lines in complex planes. In simple terms, this preservation means that Möbius transformation maps a collection of circles and lines (back) into a collection of circles and lines. In general, the discussion begins with an explanation of the definition of the Möbius transformation in the complex plane. The discussion continues on defining the basic mapping and direct affine transformation. These two concepts are used to prove the existence of the preservation properties of circles and lines in the Möbius transformation. It can be shown that the Möbius transformation can be expressed as a composition of the direct affine transform and the inverse. It can also be shown that the direct affine transform and the inverse both have the property of preserving circles and lines in the complex plane. Thus, it can be concluded that in this study the Möbius transformation has the property of preserving circles and lines in the complex plane.
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... Now, let us assume that µ " 8. It is straightforward to see (but we refer otherwise to [27,Remark 4.3]) that, for a given pencil Lpλq and µ P C, the identity W pµ, Lq " W pµ´1, revLq holds. It is also immediate to see that rank prevLq " rank L. ...
On bundles of matrix pencils under strict equivalence
Preprint
Full-text available
Apr 2022
Fernando De Terán
Froilan Dopico
Bundles of matrix pencils (under strict equivalence) are sets of pencils having the same Kronecker canonical form, up to the eigenvalues (namely, they are an infinite union of orbits under strict equivalence). The notion of bundle for matrix pencils was introduced in the 1990's, following the same notion for matrices under similarity, introduced by Arnold in 1971, and it has been extensively used since then. Despite the amount of literature devoted to describing the topology of bundles of matrix pencils, some relevant questions remain still open in this context. For example, the following two: (a) provide a characterization for the inclusion relation between the closures (in the standard topology) of bundles; and (b) are the bundles open in their closure? The main goal of this paper is providing an explicit answer to these two questions. In order to get this answer, we also review and/or formalize some notions and results already existing in the literature. We also prove that bundles of matrices under similarity, as well as bundles of matrix polynomials (defined as the set of m×n m\times n matrix polynomials of the same grade having the same spectral information, up to the eigenvalues) are open in their closure.
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... In this paper, we consider the analogous approach for invariant pairs of regular structured matrix polynomials that can be employed to preserve desired spectral properties in the perturbed polynomials. We mention that Mackey et al. have investigated preserving invariant pairs under Möbius transformation of a matrix polynomial . Now we describe the structured matrix polynomials which we consider in this paper. ...
Eigenvalue embedding problem for quadratic regular matrix polynomials with symmetry structures
Article
May 2022
LINEAR ALGEBRA APPL
Tinku Ganai
Bibhas Adhikari
In this paper, we consider structure-preserving eigenvalue embedding problem (SEEP) for quadratic regular matrix polynomials with symmetry structures. First, we determine perturbations of a quadratic matrix polynomial, unstructured or structured, such that the perturbed polynomials reproduce a desired invariant pair while maintaining the invariance of another invariant pair of the unperturbed polynomial. If the latter is unknown, it is referred to as no spillover perturbation. Then we use these results for solving the SEEP for structured quadratic matrix polynomials that include: symmetric, Hermitian, ⋆-even and ⋆-odd quadratic matrix polynomials. Finally, we show that the obtained analytical expressions of perturbations can realize existing results for structured polynomials that arise in real-world applications, as special cases.
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... In this paper, we consider the analogous approach for invariant pairs of regular structured matrix polynomials that can be employed to preserve desired spectral properties in the perturbed polynomials. We mention that Mackey et al. have investigated preserving invariant pairs under Möbius transformation of a matrix polynomial . Now we describe the structured matrix polynomials which we consider in this paper. ...
Eigenvalue embedding problem for quadratic regular matrix polynomials with symmetry structures
Preprint
Full-text available
Apr 2021
Tinku Ganai
Bibhas Adhikari
In this paper, we propose a unified approach for solving structure-preserving eigenvalue embedding problem (SEEP) for quadratic regular matrix polynomials with symmetry structures. First, we determine perturbations of a quadratic matrix polynomial, unstructured or structured, such that the perturbed polynomials reproduce a desired invariant pair while maintaining the invariance of another invariant pair of the unperturbed polynomial. If the latter is unknown, it is referred to as no spillover perturbation. Then we use these results for solving the SEEP for structured quadratic matrix polynomials that include: symmetric, Hermitian, ⋆\star-even and ⋆\star-odd quadratic matrix polynomials. Finally, we show that the obtained analytical expressions of perturbations can realize existing results for structured polynomials that arise in real-world applications, as special cases. The obtained results are supported with numerical examples.
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... Beberapa artikel yang membahas terkait pengembangan transformasi bilinear antara lain artikel Niamsup dan Jing yang membahas karakteristik transformasi bilinear serta artikel Mackey dkk menjelaskan transformasi Mobius dari matriks polinomial. Selain itu, Demirel memberikan penjelasan terkait transformasi bilinear dalam kaitannya dengan geometri hiperbolik dan Kaur menjelaskan terkait sifat-sifat dari transformasi bilinear dan Ozgur menjelaskan sifat grup Mobius transformasi. ...
TITIK TETAP INVERS TRANSFORMASI BILINEAR DAN TRANSFORMASI BILINEAR KONJUGAT
Article
Full-text available
Dec 2020
Ahmad Ansar
Bilinear transformation is one of the basic transformations studied in complex function analysis and it has many applications in various fields. This article discusses about the fixed point of bilinear transformation and its properties. The results obtained are used to investigating the fixed point invers bilinear transformation. This article also explains the fixed point of the conjugate bilinear transformation.
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... The final solution comes from changing the variable λ to λ := d λ λ and from combining this with the multiplication by the constant d R and the diagonal scaling T discussed above. Note that the change of variable transforms the zeros and the poles of R(λ) in a very simple way, preserving their partial multiplicities, and that does not change at all its minimal indices [14,19]. The combination of all these scalings yields a new transfer function F ), such that A has norm smaller than or equal to 1. ...
Structural backward stability in rational eigenvalue problems solved via block Kronecker linearizations
Preprint
Full-text available
Mar 2021
María C. Quintana
Froilan Dopico
Paul Michel Van Dooren
We study the backward stability of running a backward stable eigenstructure solver on a pencil S(λ)S(\lambda) that is a strong linearization of a rational matrix R(λ)R(\lambda) expressed in the form R(λ)=D(λ)+C(λ I ℓ−A)−1 B R(\lambda)=D(\lambda)+ C(\lambda I_\ell-A)^{-1}B, where D(λ)D(\lambda) is a polynomial matrix and C(λ I ℓ−A)−1 B C(\lambda I_\ell-A)^{-1}B is a minimal state-space realization. We consider the family of block Kronecker linearizations of R(λ)R(\lambda), which are highly structured pencils. Backward stable eigenstructure solvers applied to S(λ)S(\lambda) will compute the exact eigenstructure of a perturbed pencil S^(λ):=S(λ)+Δ S(λ)\widehat S(\lambda):=S(\lambda)+\Delta_S(\lambda) and the special structure of S(λ)S(\lambda) will be lost. In order to link this perturbed pencil with a nearby rational matrix, we construct a strictly equivalent pencil S~(λ)\widetilde S(\lambda) to S^(λ)\widehat S(\lambda) that restores the original structure, and hence is a block Kronecker linearization of a perturbed rational matrix R~(λ)=D~(λ)+C~(λ I ℓ−A~)−1 B~\widetilde R(\lambda) = \widetilde D(\lambda)+ \widetilde C(\lambda I_\ell- \widetilde A)^{-1} \widetilde B, where D~(λ)\widetilde D(\lambda) is a polynomial matrix with the same degree as D(λ)D(\lambda). Moreover, we bound appropriate norms of D~(λ)−D(λ)\widetilde D(\lambda)- D(\lambda), C~−C\widetilde C - C, A~−A\widetilde A - A and B~−B\widetilde B - B in terms of an appropriate norm of Δ S(λ)\Delta_S(\lambda). These bounds may be inadmissibly large, but we also introduce a scaling that allows us to make them satisfactorily tiny. Thus, for this scaled representation, we prove that the staircase and the QZ algorithms compute the exact eigenstructure of a rational matrix R~(λ)\widetilde R(\lambda) that can be expressed in exactly the same form as R(λ)R(\lambda) with the parameters defining the representation very near to those of R(λ)R(\lambda). This shows that this approach is backward stable in a structured sense.
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... More recently, in the authors showed that Möbius transformations for matrix polynomials allow to provide a common framework for the most frequent classes of structured matrix polynomials, such as (skew-)symmetric, (skew-)Hermitian, (anti-)palindromic, and alternating polynomials. A thorough study on the influence of Möbius transformations on relevant properties of general matrix polynomials over arbitrary fields has been carried out in . ...
Structured strong ℓ\boldsymbol{\ell}-ifications for structured matrix polynomials in the monomial basis
Preprint
Full-text available
Oct 2020
Carla Hernando
Javier Perez Alvaro
Fernando De Terán
In the framework of Polynomial Eigenvalue Problems, most of the matrix polynomials arising in applications are structured polynomials (namely (skew-)symmetric, (skew-)Hermitian, (anti-)palindromic, or alternating). The standard way to solve Polynomial Eigenvalue Problems is by means of linearizations. The most frequently used linearizations belong to general constructions, valid for all matrix polynomials of a fixed degree, known as {\em companion linearizations}. It is well known, however, that is not possible to construct companion linearizations that preserve any of the previous structures for matrix polynomials of even degree. This motivates the search for more general companion forms, in particular {\em companion ℓ\ell-ifications}. In this paper, we present, for the first time, a family of (generalized) companion ℓ\ell-ifications that preserve any of these structures, for matrix polynomials of degree k=(2 d+1)ℓ k=(2d+1)\ell. We also show how to construct sparse ℓ\ell-ifications within this family. Finally, we prove that there are no structured companion quadratifications for quartic matrix polynomials.
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... where α, β ∈ R are such that α 2 + β 2 = 1 and such that M α,β (A − λB) has neither the eigenvalues 0 nor ∞. Note that this Möbius transformation just has the effect of "rotating" eigenvalues on the extended real line R ∪ {∞}, but it leaves eigenvectors and the Jordan structure invariant; see, e.g., . The result then follows by applying the already proved parts of the theorem on M α,β (A − λB) followed by applying the inverse Möbius transformation ...
Solving Singular Generalized Eigenvalue Problems by a Rank-Completing Perturbation
Article
Full-text available
Sep 2019
SIAM J MATRIX ANAL A
Michiel Hochstenbach
Christian Mehl
Bor Plestenjak
Generalized eigenvalue problems involving a singular pencil are very challenging to solve, with respect to both accuracy and efficiency. The existing package Guptri is very elegant but may be time-demanding, even for small and medium-sized matrices. We propose a simple method to compute the eigenvalues of singular pencils, based on one perturbation of the original problem of a certain specific rank. For many problems, the method is both fast and robust. This approach may be seen as a welcome alternative to staircase methods.
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... We shall further show that λ is the unique eigenvalue of M 0 + λM 1 in the interval D = (λ 1 , λ 2 ). To simplify the analysis, we apply a Möbius transformation to the matrix pencil (15). Specifically, recalling thatλ ≥ 0 is known such that A +λB 0, we defineM := M 0 +λM 1 and consider the eigenvalues of ...
Eigenvalue-based algorithm and analysis for nonconvex QCQP with one constraint
Article
Full-text available
Nov 2017
Satoru Adachi
Yuji Nakatsukasa
A nonconvex quadratically constrained quadratic programming (QCQP) with one constraint is usually solved via a dual SDP problem, or Moré’s algorithm based on iteratively solving linear systems. In this work we introduce an algorithm for QCQP that requires finding just one eigenpair of a generalized eigenvalue problem, and involves no outer iterations other than the (usually black-box) iterations for computing the eigenpair. Numerical experiments illustrate the efficiency and accuracy of our algorithm. We also analyze the QCQP solution extensively, including difficult cases, and show that the canonical form of a matrix pair gives a complete classification of the QCQP in terms of boundedness and attainability, and explain how to obtain a global solution whenever it exists.
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... In this subsection, we introduce the Möbius group (transformation) acting on the set of design measures and the set of information matrices in polynomial design. For a recent paper in which the Möbius transformation acts on polynomials, see Mackey, et al (2015). ...
Optimal experimental design that minimizes the width of simultaneous confidence bands
Article
Apr 2017
Satoshi Kuriki
Henry P Wynn
We propose an optimal experimental design for a curvilinear regression model that minimizes the band-width of simultaneous confidence bands. Simultaneous confidence bands for nonlinear regression are constructed by evaluating the volume of a tube about a curve that is defined as a trajectory of a regression basis vector (Naiman, 1986). The proposed criterion is constructed based on the volume of a tube, and the corresponding optimal design is referred to as the minimum-volume optimal design. For Fourier and weighted polynomial regressions, the problem is formalized as one of minimization over the cone of Hankel positive definite matrices, and the criterion to minimize is expressed as an elliptic integral. We show that the M\"obius group keeps our problem invariant, and hence, minimization can be conducted over cross-sections of orbits. We demonstrate that for the weighted polynomial regression and the Fourier regression with three bases, the minimum-volume optimal design forms an orbit of the M\"obius group containing D-optimal designs as representative elements.
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... In Proposition 3.4, we state without proofs those that will be relevant in this work. For a thorough study of the properties of Möbius transformations of matrix polynomials we refer the reader to . ...
Structured backward error analysis of linearized structured polynomial eigenvalue problems
Article
Full-text available
Dec 2016
Froilan Dopico
Javier Pérez
Paul Michel Van Dooren
We introduce a new class of structured matrix polynomials, namely, the class of M_A-structured matrix polynomials, to provide a common framework for many classes of structured matrix polynomials that are important in applications: the classes of (skew-)symmetric, (anti-)palindromic, and alternating matrix polynomials. Then, we introduce the families of M_A-structured strong block minimal bases pencils and of M_A-structured block Kronecker pencils,, and show that any M_A-structured odd-degree matrix polynomial can be strongly linearized via an M_A-structured block Kronecker pencil. Finally, for the classes of (skew-)symmetric, (anti-)palindromic, and alternating odd-degree matrix polynomials, the M_A-structured framework allows us to perform a global and structured backward stability analysis of complete structured polynomial eigenproblems, regular or singular, solved by applying to an M_A-structured block Kronecker pencil a structurally backward stable algorithm that computes its complete eigenstructure, like the palindromic-QR algorithm or the structured versions of the staircase algorithm. This analysis allows us to identify those M_A-structured block Kronecker pencils that yield a computed complete eigenstructure which is the exact one of a slightly perturbed structured matrix polynomial.These pencils include (modulo permutations) the well-known block-tridiagonal and block-antitridiagonal structure-preserving linearizations. Our analysis incorporates structure to the recent (unstructured) backward error analysis performed for block Kronecker linearizations by Dopico, Lawrence, P\'erez and Van Dooren, and share with it its key features, namely, it is a rigorous analysis valid for finite perturbations, i.e., it is not a first order analysis, it provides precise bounds, and it is valid simultaneously for a large class of structure-preserving strong linearizations.
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Minimal rank factorizations of polynomial matrices
Article
Jan 2025
LINEAR ALGEBRA APPL
Andrii Dmytryshyn
Froilan Dopico
Paul Michel Van Dooren
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The DL ( P ) vector space of pencils for singular matrix polynomials
Article
Full-text available
Aug 2023
LINEAR ALGEBRA APPL
Froilan Dopico
Vanni Noferini
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A load balancing parallel algorithm for solving large-scale tridiagonal linear systems
Conference Paper
Dec 2021
Min Tian
Shan Qiao
Junjie Wang
Wei Du
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Orthogonal polynomials and Möbius transformations
Article
Sep 2021
R. S. Vieira
Vanessa Avansini Botta Pirani
Given an orthogonal polynomial sequence on the real line, another sequence of polynomials can be found by composing them with a Möbius transformation. In this work, we study the properties of such Möbius-transformed polynomials in a systematically way. We show that these polynomials are orthogonal on a given curve of the complex plane with respect to a particular kind of varying measure, and that they enjoy several properties common to the orthogonal polynomials on the real line. Moreover, many properties of the orthogonal polynomials can be easier derived from this approach, for example, we can show that the Hermite, Laguerre, Jacobi, Bessel and Romanovski polynomials are all related with each other by suitable Möbius transformations; also, the orthogonality relations for Bessel and Romanovski polynomials on the complex plane easily follows.
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Nonlinear eigenvalue problems for coupled Helmholtz equations modeling gradient-index graphene waveguides
Article
Dec 2020
Jung Heon Song
Matthias Maier
Mitchell Luskin
We discuss a quartic eigenvalue problem arising in the context of an optical waveguiding problem involving atomically thick 2D materials. The waveguide configuration we consider consists of a gradient-index (spatially dependent) dielectric equipped with conducting interior interfaces. This leads to a quartic eigenvalue problem with mixed transverse electric and transverse magnetic modes, and strongly coupled electric and magnetic fields. We derive a weak formulation of the quartic eigenvalue problem and introduce a numerical solver based on a quadratification approach in which the quartic eigenvalue problem is transformed to a spectrally equivalent companion problem. We verify our numerical framework against analytical solutions for prototypical geometries. As a practical example, we demonstrate how an improved quality factor (defined by the ratio of the real and the imaginary part of the computed eigenvalues) can be obtained for a family of gradient-index host materials with internal conducting interfaces. We outline how this result lays the groundwork for solving related shape optimization problems.
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On the computation of the Möbius transform
Article
Full-text available
Dec 2019
THEOR COMPUT SCI
Morgan Barbier
Hayat Cheballah
Jean-Marie Le Bars
The Möbius transform is a crucial transformation into the Boolean world; it allows to change the Boolean representation between the True Table and Algebraic Normal Form. In this work, we introduce a new algebraic point of view of this transformation based on the polynomial form of Boolean functions. It appears that we can perform a new notion: the Möbius computation variable by variable and new computation properties. As a consequence, we propose new algorithms which can produce a huge speed up of the Möbius computation for sub-families of Boolean function. Furthermore we compute directly the Möbius transformation of some particular Boolean functions. Finally, we show that for some of them the Hamming weight is directly related to the algebraic degree of specific factors.
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Root polynomials and their role in the theory of matrix polynomials
Article
Full-text available
Sep 2019
LINEAR ALGEBRA APPL
Froilan Dopico
Vanni Noferini
We develop a complete and rigorous theory of root polynomials of arbitrary matrix polynomials, i.e., either regular or singular, and study how these vector polynomials are related to the spectral properties of matrix polynomials. We pay particular attention to the so-called maximal sets of root polynomials and prove that they carry complete information about the eigenvalues (finite or infinite) of matrix polynomials and that they are related to the matrices that transform any matrix polynomial into its Smith form. In addition, we describe clearly, for the first time in the literature, the extremality properties of such maximal sets and identify some of them whose vectors have minimal grade. Once the main theoretical properties of root polynomials have been established, the interaction of root polynomials with three problems that have attracted considerable attention in the literature is analyzed. More precisely, we study the change of root polynomials under rational transformations, or reparametrizations, of matrix polynomials, the recovery of the root polynomials of a matrix polynomial from those of its most important linearizations, and the relationship between the root polynomials of two dual pencils. We emphasize that for the case of regular matrix polynomials all the results in this paper can be translated into the language of Jordan chains, as a consequence of the well known relationship between root polynomials and Jordan chains. Therefore, a number of open problems are also solved for Jordan chains of regular matrix polynomials. We also briefly discuss how root polynomials can be used to define eigenvectors and eigenspaces for singular matrix polynomials.
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Linearizations of Matrix Polynomials in Newton Bases
Article
Jun 2018
LINEAR ALGEBRA APPL
Vasilije Perović
D. Steven Mackey
We discuss matrix polynomials expressed in a Newton basis, and the associated polynomial eigenvalue problems. Properties of the generalized ansatz spaces associated with such polynomials are proved directly by utilizing a novel representation of pencils in these spaces. Also, we show how the family of Fiedler pencils can be adapted to matrix polynomials expressed in a Newton basis. These new Newton–Fiedler pencils are shown to be strong linearizations, and some computational aspects related to them are discussed.
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Inertia laws and localization of real eigenvalues for generalized indefinite eigenvalue problems
Article
Nov 2017
LINEAR ALGEBRA APPL
Yuji Nakatsukasa
Vanni Noferini
Sylvester's law of inertia states that the number of positive, negative and zero eigenvalues of Hermitian matrices is preserved under congruence transformations. The same is true of generalized Hermitian definite eigenvalue problems, in which the two matrices are allowed to undergo different congruence transformations, but not for the indefinite case. In this paper we investigate the possible change in inertia under congruence for generalized Hermitian indefinite eigenproblems, and derive sharp bounds that show the inertia of the two individual matrices often still provides useful information about the eigenvalues of the pencil, especially when one of the matrices is almost definite. A prominent application of the original Sylvester's law is in finding the number of eigenvalues in an interval. Our results can be used for estimating the number of real eigenvalues in an interval for generalized indefinite and nonlinear eigenvalue problems.
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Generic rank-two perturbations of structured regular matrix pencils
Article
Mar 2016
OPER MATRICES
Leonhard Batzke
The generic spectral behavior of classes of structured regular matrix pencils is examined under structure-preserving rank-2 perturbations, i.e., perturbations of normal rank two. For T -alternating, palindromic, and skew-symmetric matrix pencils we observe the following effects at each eigenvalue lambda under a generic, structure-preserving rank-2 perturbation: 1) The largest two Jordan blocks at lambda are destroyed. 2) If hereby the eigenvalue pairing imposed by the structure is violated, also the largest remaining Jordan block at lambda will grow in size by one. 3) If lambda is a single (double) eigenvalue of the perturbating pencil, one (two) new Jordan blocks of size one will be created at lambda.
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Finite and infinite structures of rational matrices: A local approach
Article
Full-text available
Jun 2015
A. Amparan
S. Marcaida
Ion Zaballa
The structure of a rational matrix is given by its Smith-McMillan invariants. Some properties of the Smith-McMillan invariants of rational matrices with elements in different principal ideal domains are presented: In the ring of polynomials in one indeterminate (global structure), in the local ring at an irreducible polynomial (local structure), and in the ring of proper rational functions (infinite structure). Furthermore, the change of the finite (global and local) and infinite structures is studied when performing a Möbius transformation on a rational matrix. The results are applied to define an equivalence relation in the set of polynomial matrices, with no restriction on size, for which a complete system of invariants are the finite and infinite elementary divisors.
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On the structure invariants of proper rational matrices with prescribed finite poles
Article
Full-text available
Jan 2013
A. Amparan
S. Marcaida
Ion Zaballa
The algebraic structure of matrices defined over arbitrary fields whose elements are rational functions with no poles at infinity and prescribed finite poles is studied. Under certain very general conditions, they are shown to be matrices over an Euclidean domain that can be classified according to the corresponding invariant factors. The relationship between these invariants and the local Wiener–Hopf factorization indices will be clarified. This result can be seen as an extension of the classical theorem on pole placement by Rosenbrock in control theory.
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Finite and Infinite Elementary Divisors of Matrix Polynomials: A Global Approach
Article
Full-text available
Ion Zaballa
Françoise Tisseur
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Skew-symmetric matrix polynomials and their Smith forms
Article
Full-text available
Jun 2013
LINEAR ALGEBRA APPL
D. Steven Mackey
Niloufer Mackey
Christian Mehl
Volker Mehrmann
Two canonical forms for skew-symmetric matrix polynomials over arbitrary fields are characterized—the Smith form, and its skew-symmetric variant obtained via unimodular congruences. Applications include the analysis of the eigenvalue and elementary divisor structure of products of two skew-symmetric matrices, the derivation of a Smith-McMillan-like canonical form for skew-symmetric rational matrices, and the construction of minimal symmetric factorizations of skew-symmetric rational matrices. A sufficient condition for the existence of solutions to matrix polynomial Sylvester equations, and results on the existence and construction of structured linearizations for regular and singular skew-symmetric matrix polynomials are also presented.
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Mathematical Systems Theory I: Modelling, State Space Analysis, Stabilityand Robustness
Article
Full-text available
Jan 2006
D. Hinrichsen
A. J. Pritchard
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Linearization of matrix polynomials expressed in polynomial bases
Article
Full-text available
Feb 2008
Amir Amiraslani
Robert M. Corless
Peter Lancaster
Companion matrices of matrix polynomials L( ) (with possibly singular leading co- ecient) are a familiar tool in matrix theory and numerical practice leading to so-called "linearizations" B A of the polynomials. Matrix polynomials as approximations to more general matrix functions lead to the study of matrix polynomials represented in a variety of classical systems of polynomials, including orthogonal systems and Lagrange polynomials, for example. For several such representations, it is shown how to con- struct (strong) linearizations via analogous companion matrix pencils. In case L( ) has Hermitian or alternatively complex symmetric coecients, the determination of linearizations B A with A and B Hermitian or complex symmetric is also discussed.
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Bilinear Transformation of Infinite-Dimensional State-Space Systems and Balanced Realizations of NonRational Transfer Functions
Article
Full-text available
Mar 1990
Raimund J Ober
Stephen Montgomery-Smith
The bilinear transform maps the open right half plane to the open unit disk and is therefore a suitable tool for carrying over results for continuous-time systems to discrete-time systems and vice versa. Corresponding state-space formulae are widely used and well understood for the case of finite-dimensional systems. In this paper infinite-dimensional generalizations of these formulae are studied for a general class of infinite-dimensional state-space systems. In particular, it is shown that reachability and observability are carried over and that the reachability and observability gramians are preserved under this transformation. N. J. Young [Operator theory and systems, Proc. Workshop, Amsterdam 1985, Oper. Theory, Adv. Appl. 19, 449-471 (1986; Zbl 0611.93005)] showed that a wide class of nonrational discrete-time transfer functions admit a balanced state-space representation. It is shown that this result carries over to the continuous-time situation via the bilinear transformation.
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The Algebraic Riccati Equation
Book
Full-text available
Jan 1995
Leiba Rodman
Peter Lancaster
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Structured Polynomial Eigenvalue Problems: Good Vibrations from Good Linearizations
Article
Full-text available
Dec 2006
SIAM J MATRIX ANAL A
D. Steven Mackey
Niloufer Mackey
Christian Mehl
Volker Mehrmann
Many applications give rise to nonlinear eigenvalue problems with an underlying structured matrix polynomial. In this paper several useful classes of structured polynomials (e.g., palindromic, even, odd) are identified and the relationships between them explored. A special class of linearizations which reflect the structure of these polynomials, and therefore preserve symmetries in their spectra, is introduced and investigated. We analyze the existence and uniqueness of such linearizations and show how they may be systematically constructed.
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Computer Controlled System Theory and Design
Book
Full-text available
Jan 1984
K.J. Åström
Bjorn Wittenmark
Practically all modern control systems are based upon microprocessors and complex microcontrollers that yield high performance and functionality. This volume focuses on the design of computer-controlled systems, featuring computational tools that can be applied directly and are explained with simple paper-and-pencil calculations. The use of computational tools is balanced by a strong emphasis on control system principles and ideas. Extensive pedagogical aids include worked examples, MATLAB macros, and a solutions manual (see inside for details). The initial chapter presents a broad outline of computer-controlled systems, followed by a computer-oriented view based on the behavior of the system at sampling instants. An introduction to the design of control systems leads to a process-related view and coverage of methods of translating analog designs to digital control. Concluding chapters explore implementation issues and advanced design methods.
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Sharp lower bounds for the dimension of linearizations of matrix polynomials
Article
Full-text available
Oct 2008
ELECTRON J LINEAR AL
Fernando De Terán
Froilan Dopico
A standard way of dealing with matrix polynomial eigenvalue problems is to use linearizations. Byers, Mehrmann and Xu have recently defined and studied linearizations of dimen-sions smaller than the classical ones. In this paper, lower bounds are provided for the dimensions of linearizations and strong linearizations of a given m × n matrix polynomial, and particular lineariza-tions are constructed for which these bounds are attained. It is also proven that strong linearizations of an n × n regular matrix polynomial of degree must have dimension nn × nn.
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Linearization of regular matrix polynomials
Article
Full-text available
Jan 2008
ELECTRON J LINEAR AL
Peter Lancaster
This note contains a short review of the notion of linearization of regular matrix polynomials. The objective is clarification of this notion when the polynomial has an "eigenvalue at infinity". The theory is extended to admit reduction by locally unimodular analytic matrix functions.
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Triangularizing Quadratic Matrix Polynomials
Article
Full-text available
Apr 2013
SIAM J MATRIX ANAL A
Ion Zaballa
Françoise Tisseur
We show that any regular quadratic matrix polynomial can be reduced to an upper triangular quadratic matrix polynomial over the complex numbers preserving the finite and infinite elementary divisors. We characterize the real quadratic matrix polynomials that are triangularizable over the real numbers and show that those that are not triangularizable are quasi-triangularizable with diagonal blocks of sizes 1×1 1\times 1 and 2×2 2 \times 2. We also derive complex and real Schur-like theorems for linearizations of quadratic matrix polynomials with nonsingular leading coefficients. In particular, we show that for any monic linearization λ I+A\lambda I+A of an n×n n\times n quadratic matrix polynomial there exists a nonsingular matrix defined in terms of n orthonormal vectors that transforms A to a companion linearization of a (quasi-)triangular quadratic matrix polynomial. This provides the foundation for designing numerical algorithms for the reduction of quadratic matrix polynomials to upper (quasi-)triangular form.
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General theory of regular matrix polynomials and band Toeplitz operators
Article
Full-text available
Jan 1988
I. Gohberg
M. A. Kaashoek
Peter Lancaster
La theorie generale des polynomes matriciels reguliers est etendue et developpee de telle facon que les resultats soient symetriques par rapport aux donnees spectrales a gauche et a droite. On donne des applications a la factorisation canonique et a l'inversion des operateurs de Toeplitz a bande bloc
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Continuation of Invariant Subspaces for Parameterized Quadratic Eigenvalue Problems
Article
Full-text available
Dec 2009
SIAM J MATRIX ANAL A
Wolf-Jürgen Beyn
Vera Thmmler
We consider quadratic eigenvalue problems with large and sparse matrices depending on a parameter. Problems of this type occur, for example, in the stability analysis of spatially discretized and parameterized nonlinear wave equations. The aim of the paper is to present and analyze a continuation method for invariant subspaces that belong to a group of eigenvalues, the number of which is much smaller than the dimension of the system. The continuation method is of predictor-corrector type, similar to the approach for the linear eigenvalue problem in [Beyn, Kleß, and Thümmler, Ergodic Theory, Analysis, and Efficient Simulation of Dynamical Systems, Springer, Berlin, 2001], but we avoid linearizing the problem, which will double the dimension and change the sparsity pattern. The matrix equations that occur in the predictor and the corrector step are solved by a bordered version of the Bartels–Stewart algorithm. Furthermore, we set up an update procedure that handles the transition from real to complex conjugate eigenvalues, which occurs when eigenvalues from inside the continued cluster collide with eigenvalues from outside. The method is demonstrated on several numerical examples: a homotopy between random matrices, a fluid conveying pipe problem, and a traveling wave of a damped wave equation.
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An Interpretation of Rosenbrock's Theorem via Local Rings
Chapter
Jul 2012
A. Amparan
S. Marcaida
Ion Zaballa
Consider a linear time invariant system ˙ x(t) = Ax(t) + Bu(t). (1) Rosenbrock's Theorem on pole assignment characterizes the possible invariant factors of the closed-loop system ˙ x(t) = (A + BF)x(t) + Bv(t) (2) when a feedback u(t) = F x(t) + v(t) is applied on the original system. This result is equivalent to the problem of the existence of a non-singular polynomial matrix (polynomial matrix representation of (2)) with prescribed invariant factors and left Wiener–Hopf factorization indices at infinity. Namely, the rational matrix G(s) = (sI −(A+ BF)) −1 B is the transfer function matrix of (2) and can be written as an irreducible fraction matrix description G(s) = N (s)P (s) −1 , where N (s) and P (s) are right coprime polynomial matrices. P (s) is called polynomial matrix representation of (2). On the one hand, the invariant factors form a complete system of invariants for the equivalence of polynomial matrices. We recall this relation for rational matrices. Two rational matrices G 1 (s), G 2 (s) are equivalent if there exist unimodular matrices U 1 (s), U 2 (s) such that G 2 (s) = U 1 (s)G 1 (s)U 2 (s). Any rational matrix G(s) is equivalent to a diagonal matrix Diag 1 (s) ψ 1 (s) ,. .. , m (s) ψ m (s) where i (s), ψ i (s) are monic and coprime polynomials such that 1 (s) | · · · | m (s) while ψ m (s) | · · · | ψ 1 (s) with | meaning divisibility. The irreducible rational functions i (s) ψ i (s) are the invariant rational functions of G(s) or the invariant factors when the matrix is polynomial. On the other hand, any rational matrix is left Wiener–Hopf equivalent at infinity to a diagonal matrix Diag(s k 1 ,. .. , s km), where the integers k 1 ,. .. , k m (we can assume in non-increasing order) form a complete system of invariants for the left Wiener–Hopf equivalence at infinity: Two rational matrices G 1 (s), G 2 (s) are said to be left Wiener–Hopf equivalent at infinity if there exist matrices U (s), unimodular, and B(s), biproper, such that 1
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Theorie der linearen Formen mit ganzen Coefficienten
Chapter
Jan 1968
Ferdinand Georg Frobenius
Jean-Pierre Serre
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Robust Multivariable Control
Article
Dec 2005
Oscar R. González
Atul G. Kelkar
This chapter presents fundamental tools in the analysis and design of linear, time-invariant, continuous time, robust, multivariable control systems. It introduces a mathematical model of the physical process that needs to be controlled for control system design and analysis. The analysis tools include basic measures of performance, frequency response, and stability theorems. Linear quadratic, H∞ and passivity-based control synthesis techniques are also introduced. The derivation of a mathematical model for a linear, time invariant system typically starts by writing differential equations relating the inputs to a standard set of variables, such as loop currents and the configuration variables. In some processes, it is possible to consider models with a single input and output (SISO) for which a vast amount of literature is available. In many other cases, it is necessary to consider models with multiple inputs and/or outputs (MIMO).
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MINIMAL BASES OF RATIONAL VECTOR SPACES, WITH APPLICATIONS TO MULTIVARIABLE LINEAR SYSTEMS.
Article
May 1975
G.David Forney Jr
A minimal basis of a vector space V of n-tuples of rational functions is defined as a polynomial basis such that the sum of the degrees of the basis n-tuples is minimum. Conditions for a matrix G to represent a minimal basis are derived. By imposing additional conditions on G, a minimal basis for V is arrived at that is unique. It is shown how minimal bases can be used to factor a transfer function matrix G in the form G equals ND minus 1, where N and D are polynomial matrices that display the controllability indices of G and its controller canonical realization. Transfer function matrices G solving equations of the form PG equals Q are also obtained by this method. Applications to the problem of finding minimal order inverse systems are given.
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A generalization of the matrix-sign-function solution for algebraic Riccati equations
Article
Sep 1986
JD GARDINER
AJ LAUB
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Numerical Geometric Integration
Article
Ernst Hairer
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Triangularizing matrix polynomials
Article
Oct 2013
LINEAR ALGEBRA APPL
Leo Taslaman
Françoise Tisseur
Ion Zaballa
For an algebraically closed field F, we show that any matrix polynomial P(lambda) is an element of Flambda, n <= m, can be reduced to triangular form, preserving the degree and the finite and infinite elementary divisors. We also characterize the real matrix polynomials that are triangularizable over the real numbers and show that those that are not triangularizable are quasi-triangularizable with diagonal blocks of sizes 1 x 1 and 2 x 2. The proofs we present solve the structured inverse problem of building up triangular matrix polynomials starting from lists of elementary divisors.
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A survey of modern algebra. Revised ed
Article
Garrett Birkhoff
Saunders MacLane
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Texts in Applied Mathematics
Article
Jan 2005
D. Hinrichsen
A. J. Pritchard
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Numerical solution of algebraic Riccati equations, volume 9 of Fundamentals of Algorithms
Article
Jan 2011
Dario A. Bini
Bruno Iannazzo
Beatrice Meini
This concise and comprehensive treatment of the basic theory of algebraic Riccati equations describes the classical as well as the more advanced algorithms for their solution in a manner that is accessible to both practitioners and scholars. It is the first book in which nonsymmetric algebraic Riccati equations are treated in a clear and systematic way. Some proofs of theoretical results have been simplified and a unified notation has been adopted. Readers will find a discussion of doubling algorithms, which are effective in solving algebraic Riccati equations, and a detailed description of all classical and advanced algorithms for solving algebraic Riccati equations, along with their MATLAB codes. This will help the reader gain understanding of the computational issues and provide ready-to-use implementation of the different solution techniques. Audience: This book is intended for researchers who work in the design and analysis of algorithms and for practitioners who are solving problems in applications and need to understand the available algorithms and software. It is also intended for students with no expertise in this area who wish to approach this subject from a theoretical or computational point of view. The book can be used in a semester course on algebraic Riccati equations or as a reference in a course on advanced numerical linear algebra and applications.
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The theory of complex functions
Article
R.A. Kortram
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Spectral equivalence of matrix polynomials and the Index Sum Theorem
Article
Oct 2014
LINEAR ALGEBRA APPL
Fernando De Terán
Froilan Dopico
D. Steven Mackey
The concept of linearization is fundamental for theory, applications, and spectral computations related to matrix polynomials. However, recent research on several important classes of structured matrix polynomials arising in applications has revealed that the strategy of using linearizations to develop structure-preserving numerical algorithms that compute the eigenvalues of structured matrix polynomials can be too restrictive, because some structured polynomials do not have any linearization with the same structure. This phenomenon strongly suggests that linearizations should sometimes be replaced by other low degree matrix polynomials in applied numerical computations. Motivated by this fact, we introduce equivalence relations that allow the possibility of matrix polynomials (with coefficients in an arbitrary field) to be equivalent, with the same spectral structure, but have different sizes and degrees. These equivalence relations are directly modeled on the notion of linearization, and consequently inherit the simplicity, applicability, and most relevant properties of linearizations; simultaneously, though, they are much more flexible in the possible degrees of equivalent polynomials. This flexibility allows us to define in a unified way the notions of quadratification and ℓ-ification, to introduce the concept of companion form of arbitrary degree, and to provide concrete and simple examples of these notions that generalize in a natural and smooth way the classical first and second Frobenius companion forms. The properties of ℓ-ifications are studied in depth; in this process a fundamental result on matrix polynomials, the “Index Sum Theorem”, is recovered and extended to arbitrary fields. Although this result is known in the systems theory literature for real matrix polynomials, it has remained unnoticed by many researchers. It establishes that the sum of the (finite and infinite) partial multiplicities, together with the (left and right) minimal indices of any matrix polynomial is equal to the rank times the degree of the polynomial. The “Index Sum Theorem” turns out to be a key tool for obtaining a number of significant results: on the possible sizes and degrees of ℓ-ifications and companion forms, on the minimal index preservation properties of companion forms of arbitrary degree, as well as on obstructions to the existence of structured companion forms for structured matrix polynomials of even degree. This paper presents many new results, blended together with results already known in the literature but extended here to the most general setting of matrix polynomials of arbitrary sizes and degrees over arbitrary fields. Therefore we have written the paper in an expository and self-contained style that makes it accessible to a wide variety of readers.
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A generalized structured doubling algorithm for the numerical solution of linear quadratic optimal control problems
Article
Jan 2013
Federico Poloni
Volker Mehrmann
We propose a generalization of the structured doubling algorithm to compute invariant subspaces of structured matrix pencils that arise in the context of solving linear quadratic optimal control problems. The new algorithm is designed to attain better accuracy when the classical Riccati equation approach for the solution of the optimal control problem is not well suited because the stable and unstable invariant subspaces are not well separated (because of eigenvalues near or on the imaginary axis) or in the case when the Riccati solution does not exist at all. We analyze the convergence of the method and compare the new method with the classical structured doubling algorithm as well as some structured QR methods. Copyright © 2012 John Wiley & Sons, Ltd.
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Book reviews: Algorithms for linear quadratic optimization
Article
Feb 2001
AUTOMATICA
Pablo A Iglesias
A scheme is proposed for control of multi-body, multi-input and multi-output nonlinear systems with joint backlash, flexibility and damping, represented by a gun turret-barrel model which consists of two subsystems: two motors driving two loads (turret ...
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Unified theorems on completions of matrix pencils
Article
Dec 1991
LINEAR ALGEBRA APPL
Isabel Cabral
Fernando C. Silva
We generalize and unify some theorems concerned with the existence of a pencil with a prescribed subpencil and prescribed strict equivalence class.
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Theorie der linearen Formen mit ganzen Coefficienten
Article
Jan 1879
J REINE ANGEW MATH
G. Frobenius
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Realization and Interpolation of Rational Matrix Functions
Article
Jan 1988
Joseph A. Ball
Israel Gohberg
Leiba Rodman
In this paper we generalize for matrix valued functions a number of well known interpolation problems for scalar rational functions and obtain explicit formulas for the solutions. The realization approach toward the study of rational matrix functions from systems theory serves here as the main tool. The main results recently appeared in the literature; here we give a more systematic and transparent exposition based exclusively on analysis in finite dimensional spaces.
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On the Sensitivity of the Eigenvalue Problem $Ax = \lambda Bx
Article
Dec 1972
G. W. Stewart
This paper considers the sensitivity of the eigenvalues and eigenvectors of the generalized matrix eigenvalue problem Ax equals lambda Bx to perturbations of A and B. The notion of a deflating subspace for the problem is introduced, and error bounds for approximate deflating subspaces obtained. The bounds also provide information about the eigenvalues of the problem. The resulting perturbation bounds estimate realistically the sensitivity of the eigenvalues, even when B is singular or nearly singular. The results are applied to the important special case where A is Hermitian and B is positive definite.
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Interpolation of Rational Matrix Functions Operator Theory: Advances and Applications
Article
Jan 1990
J. A. Ball
I. Gohberg
Leiba Rodman
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Discrete-Time Signal Processing
Article
Jan 1998
A.V. Oppenheim
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Problems and Theorems in Linear Algebra
Article
Jan 1994
Victor V. Prasolov
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Hamiltonian and Symplectic Algorithms for the Algebraic Riccati Equation
Article
Jan 1983
R. Byers
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A survey of matrix theory and matrix inequalities. (Obzor po teorii matric i matricnyh neravenstv.) Übersetzung aus dem Englischen. Herausgegeben von V. B. Lidskii
Article
Dec 1965
Marvin Marcus
Henryk Minc
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Stratification of full rank polynomial matrices
Article
Aug 2013
LINEAR ALGEBRA APPL
Stefan Johansson
Bo Kågström
Paul Michel Van Dooren
We show that perturbations of polynomial matrices of full normal-rank can be analyzed via the study of perturbations of companion form linearizations of such polynomial matrices. It is proved that a full normal-rank polynomial matrix has the same structural elements as its right (or left) linearization. Furthermore, the linearized pencil has a special structure that can be taken into account when studying its stratification. This yields constraints on the set of achievable eigenstructures. We explicitly show which these constraints are. These results allow us to derive necessary and sufficient conditions for cover relations between two orbits or bundles of the linearization of full normal-rank polynomial matrices. The stratification rules are applied to and illustrated on two artificial polynomial matrices and a half-car passive suspension system with four degrees of freedom.
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Algorithms for Linear-Quadratic Optimization
Book
Jan 1996
Vasile Sima
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Minimal Bases of Rational Vector Spaces, with Applications to Multivariable Linear Systems
Article
May 1975
SIAM J Contr
G.D.Jr. Forney
A minimal basis of a vector space V of n-tuples of rational functions is defined as a polynomial basis such that the sum of the degrees of the basis n-tuples is minimum. Conditions for a matrix G to represent a minimal basis are derived. By imposing additional conditions on G we arrive at a minimal basis for V that is unique. We show how minimal bases can be used to factor a transfer function matrix G in the form G=N D−1 G = ND^{ - 1} , where N and D are polynomial matrices that display the controllability indices of G and its controller canonical realization. Transfer function matrices G solving equations of the form P G=Q PG = Q are also obtained by this method; applications to the problem of finding minimal order inverse systems are given. Previous applications to convolutional coding theory are noted. This range of applications suggests that minimal basis ideas will be useful throughout the theory of multivariable linear systems. A restatement of these ideas in the language of valuation theory is given in an Appendix. A minimal basis of a vector space V of n-tuples of rational functions is defined as a polynomial basis such that the sum of the degrees of the basis n-tuples is minimum. Conditions for a matrix G to represent a minimal basis are derived. By imposing additional conditions on G we arrive at a minimal basis for V that is unique. We show how minimal bases can be used to factor a transfer function matrix G in the form G=N D−1 G = ND^{ - 1} , where N and D are polynomial matrices that display the controllability indices of G and its controller canonical realization. Transfer function matrices G solving equations of the form P G=Q PG = Q are also obtained by this method; applications to the problem of finding minimal order inverse systems are given. Previous applications to convolutional coding theory are noted. This range of applications suggests that minimal basis ideas will be useful throughout the theory of multivariable linear systems. A restatement of these ideas in the language of valuation theory is given in an Appendix.
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Fiedler companion linearizations for rectangular matrix polynomials
Article
Aug 2012
LINEAR ALGEBRA APPL
Fernando De Terán
Froilan Dopico
D. Steven Mackey
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17387 | https://www.quora.com/Can-this-be-true-for-some-x-ln-x-+-ln-2x-ln-3x | Can this be true for some [math]x[/math],[math]\ln(x) + \ln(2x) = \ln(3x)[/math]? - Quora
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5
Can this be true for some x x,ln(x)+ln(2 x)=ln(3 x)ln(x)+ln(2 x)=ln(3 x)?
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Anoni Meous
Studied Mathematics&Physics · Author has 184 answers and 345.2K answer views
·8y
It is true for x=1.5.
Look at the graphs of y=ln(x)+ln(2x) and y=ln(3x)
They meet at x=1.5, y=1.504, which implies x=1.5 is a solution
Continue Reading
It is true for x=1.5.
Look at the graphs of y=ln(x)+ln(2x) and y=ln(3x)
They meet at x=1.5, y=1.504, which implies x=1.5 is a solution
Upvote ·
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Bryan Harris
20 years of trying to make complex things simple · Author has 52 answers and 485.9K answer views
·8y
Again, lots of correct answers here, but here’s another way to think about it:
The question is asking, “Is there a left-to-right position on the graph where the lines are at the same place up-and-down?” or “Is there a number I can plug in for x where the y values are equal?” or “At what x value do the lines cross?” 1.5
Continue Reading
Again, lots of correct answers here, but here’s another way to think about it:
The question is asking, “Is there a left-to-right position on the graph where the lines are at the same place up-and-down?” or “Is there a number I can plug in for x where the y values are equal?” or “At what x value do the lines cross?” 1.5
Upvote ·
9 1
Tazwar
Author has 518 answers and 1.2M answer views
·Updated 8y
Originally Answered: Can this be true for some x ,㏑x + ㏑(2x) =㏑(3x)? ·
We have: ln(x)+ln(2 x)=ln(3 x)ln(x)+ln(2 x)=ln(3 x)
⇒ln(3 x)−ln(2 x)−ln(x)=0⇒ln(3 x)−ln(2 x)−ln(x)=0
Using the laws of logarithms:
⇒ln(3 x 2 x⋅x)=0⇒ln(3 x 2 x⋅x)=0
⇒ln(3 x 2 x 2)=0⇒ln(3 x 2 x 2)=0
⇒3 x 2 x 2=e 0⇒3 x 2 x 2=e 0
⇒3 x 2 x 2=1⇒3 x 2 x 2=1
⇒3 x=2 x 2⇒3 x=2 x 2
⇒2 x 2−3 x=0⇒2 x 2−3 x=0
⇒x(2 x−3)=0⇒x(2 x−3)=0
Using the null factor law:
⇒x=0⇒x=0
or
⇒2 x−3=0⇒2 x−3=0
⇒2 x=3⇒2 x=3
⇒x=3 2⇒x=3 2
However, the domain of ln(x)ln(x) is x>0 x>0.
Therefore, the equation is true for x=3 2 x=3 2.
Upvote ·
99 13
9 1
Related questions
More answers below
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Yassine Alouini
Studied Mathematics at École Centrale Paris (Graduated 2013) · Author has 1K answers and 5.1M answer views
·8y
You need to use the fact that: l n(a b)=l n(a)+l n(b)l n(a b)=l n(a)+l n(b) (for a,b a,b strictly positive numbers).
Then, l n(x)+l n(2)+l n(x)=l n(3)+l n(x)l n(x)+l n(2)+l n(x)=l n(3)+l n(x)
Simplifying, one gets: l n(x)=l n(3)−l n(2)l n(x)=l n(3)−l n(2).
Finally, applying the exponential
function, one finds that: x=3 2 x=3 2.
Let me know if you need more details!
Footnotes
Exponential function - Wikipedia
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To determine if the equation ln(x)+ln(2 x)=ln(3 x)ln(x)+ln(2 x)=ln(3 x) can be true for some x x, we can use properties of logarithms to simplify the equation.
Starting with the left-hand side:
ln(x)+ln(2 x)=ln(x)+ln(2)+ln(x)=2 ln(x)+ln(2)ln(x)+ln(2 x)=ln(x)+ln(2)+ln(x)=2 ln(x)+ln(2)
Now, rewrite the right-hand side:
ln(3 x)=ln(3)+ln(x)ln(3 x)=ln(3)+ln(x)
Now we can set the two sides equal to each other:
2 ln(x)+ln(2)=ln(3)+ln(x)2 ln(x)+ln(2)=ln(3)+ln(x)
Next, we can isolate ln(x)ln(x):
2 ln(x)−ln(x)+ln(2)=ln(3)2 ln(x)−ln(x)+ln(2)=ln(3)
This simplifies to:
ln(x)+ln(2)=ln(3)ln(x)+ln(2)=ln(3)
Now, we can isolate ln(x)ln(x):
ln(x)=ln(3)−ln(2)ln(x)=ln(3)−ln(2)
Using the properties of logarithms, we can combine the right-hand s
Continue Reading
To determine if the equation ln(x)+ln(2 x)=ln(3 x)ln(x)+ln(2 x)=ln(3 x) can be true for some x x, we can use properties of logarithms to simplify the equation.
Starting with the left-hand side:
ln(x)+ln(2 x)=ln(x)+ln(2)+ln(x)=2 ln(x)+ln(2)ln(x)+ln(2 x)=ln(x)+ln(2)+ln(x)=2 ln(x)+ln(2)
Now, rewrite the right-hand side:
ln(3 x)=ln(3)+ln(x)ln(3 x)=ln(3)+ln(x)
Now we can set the two sides equal to each other:
2 ln(x)+ln(2)=ln(3)+ln(x)2 ln(x)+ln(2)=ln(3)+ln(x)
Next, we can isolate ln(x)ln(x):
2 ln(x)−ln(x)+ln(2)=ln(3)2 ln(x)−ln(x)+ln(2)=ln(3)
This simplifies to:
ln(x)+ln(2)=ln(3)ln(x)+ln(2)=ln(3)
Now, we can isolate ln(x)ln(x):
ln(x)=ln(3)−ln(2)ln(x)=ln(3)−ln(2)
Using the properties of logarithms, we can combine the right-hand side:
ln(x)=ln(3 2)ln(x)=ln(3 2)
Exponentiating both sides gives:
x=3 2 x=3 2
So, yes, the equation ln(x)+ln(2 x)=ln(3 x)ln(x)+ln(2 x)=ln(3 x) can be true for x=3 2 x=3 2.
Upvote ·
Utkarsh Priyam
Studied Mathematics at Harker School (Graduated 2021) · Author has 262 answers and 147.5K answer views
·8y
ln(2x^2) = ln(3x) → ln(a) + ln(b) = ln(ab)
2x^2 = 3x
2x = 3 → x = 0 is a root
x = 3/2 or x = 0
Upvote ·
9 1
Related questions
More answers below
Which of the following is true? 5 44>4 53 5 44>4 53 or 2 100+3 100<4 100 2 100+3 100<4 100?
What is ∫x ln x d x∫x lnx d x?
Is ∫∞−∞x 2 p(x)d x≥1 2 π e 2 2∫∞−∞p(x)log 2 p(x)d x∫−∞∞x 2 p(x)d x≥1 2 π e 2 2∫−∞∞p(x)log 2p(x)d x true for all p(x)p(x)?
What is ∫1 0 ln(1−x)ln(1+x)x d x∫0 1 ln(1−x)ln(1+x)x d x?
Why is lim x→∞(ln(1+x)ln(x))x ln(x)=e lim x→∞(ln(1+x)ln(x))x ln(x)=e?
Peter v. Elbing
Algebra is favorite. Calculus is necessary. · Author has 718 answers and 943.1K answer views
·8y
There is a law for calculation of logarithms:∀a,b>0:ln(a b)=ln(a)+ln(b)∀a,b>0:ln(a b)=ln(a)+ln(b). So this equation reads
ln(x)+ln(2 x)=2 ln(x)+ln(2)=ln(3 x)=ln(x)+ln(3)⇒ln(x)=ln(3)−ln(2)=ln(3/2)⇒x=3/2 ln(x)+ln(2 x)=2 ln(x)+ln(2)=ln(3 x)=ln(x)+ln(3)⇒ln(x)=ln(3)−ln(2)=ln(3/2)⇒x=3/2.
Note that the logarithm is a strictly increasing function. So the solution is unique.
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Divyansh Tripathi
I like calculus. · Author has 168 answers and 312.7K answer views
·Updated 8y
Originally Answered: Can this be true for some x ,㏑x + ㏑(2x) =㏑(3x)? ·
ln x+ln 2 x=ln 3 x lnx+ln2 x=ln3 x
Recall that ln a b=ln a+ln b,∀a,b>0.Recall that lna b=lna+lnb,∀a,b>0.
⟹ln(x⋅2 x)=ln 3 x⟹ln(x⋅2 x)=ln3 x
⟹ln(2 x 2)=ln 3 x⟹ln(2 x 2)=ln3 x
⟹e ln 3 x=2 x 2⟹e ln3 x=2 x 2
⟹3 x=2 x 2()()⟹3 x=2 x 2
⟹2 x 3−3 x=0⟹2 x 3−3 x=0
⟹x(2 x−3)=0⟹⟹x(2 x−3)=0⟹
x=3 2 Recall that 0 is not an option here()()x=3 2 Recall that 0 is not an option here
Upvote ·
9 1
Faye Taylor
Studied Chemistry (college major) · Upvoted by
Ron Lessnick
, BSEE and MS Electrical Engineering & Mathematics, City College of New York (1965) · Author has 1K answers and 7M answer views
·8y
Originally Answered: Can this be true for some x ,㏑x + ㏑(2x) =㏑(3x)? ·
Log Rules: (As a refresher).
1) logb(mn) = logb(m) + logb(n)
2) logb(m/n) = logb(m) – logb(n)
3) logb(mn) = n · logb(m)
Using your log rule 1,
ln(x)+ln(2x)=ln(2x^2)
ln(2x^2)=ln(3x)
2x^2=3x
2x^2–3x=0
x(2x-3)=0 (2x=3, x=3/2)
So, x=0 and x=3/2
But, since log can only be defined when x>0, x=0 cannot be an answer. Therefore x=3/2.
Upvote ·
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Robert Szankowski
B.S. in Mathematics, Rutgers University (Graduated 2003) · Author has 671 answers and 1.4M answer views
·8y
l n(x)+l n(2 x)=l n(3 x)l n(x)+l n(2 x)=l n(3 x)
l n(2 x 2)=l n(3 x)l n(2 x 2)=l n(3 x)
e l n(2 x 2)=e l n(3 x)e l n(2 x 2)=e l n(3 x)
2 x 2=3 x 2 x 2=3 x
2 x 2−3 x=0 2 x 2−3 x=0
2 x(x−3 2)=0 2 x(x−3 2)=0
x(x−3 2)=0 x(x−3 2)=0
l n(0)l n(0) is undefined
therefore
x=3 2 x=3 2
Upvote ·
9 3
Jay Cashen
Author has 740 answers and 1.7M answer views
·8y
Easy because of the add the logs for multiplication rule:
Ln(2x^2) = ln(3x)
2x^2 = 3x
2x^2 - 3x = 0
x(2x - 3) = 0
x = 0, x = 3/2
ln goes to minus infinity as x ->0, so
x = 3/2
Upvote ·
Abhishek Kaushik
Childhood hobby, adulthood addiction · Author has 152 answers and 296.7K answer views
·8y
) logb(mn) = logb(m) + logb(n)
2) logb(m/n) = logb(m) – logb(n)
3) logb(mn) = n · logb(m)
Using your log rule 1,
ln(x)+ln(2x)=ln(2x^2)
ln(2x^2)=ln(3x)
2x^2=3x
2x^2–3x=0
x(2x-3)=0 (2x=3, x=3/2)
So, x=0 and x=3/2
But, since log can only be defined when x>0, x=0 cannot be an answer. Therefore x=3/2.
Upvote ·
Zane Heyl
Former Actuarial Analyst (2017–2022) · Author has 275 answers and 768.6K answer views
·8y
Originally Answered: Can this be true for some x ,㏑x + ㏑(2x) =㏑(3x)? ·
You should try solving this using log laws.
ln(x) + ln(2x) = ln(3x)
→ ln(2x^2) = ln(3x)
How would you proceed now?
Kind regards,
Zane Heyl
Upvote ·
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17388 | https://www.chemteam.info/Equilibrium/Writing-Ksp-expression.html | Ksp: Writing the solubility-product constant expression
| | | | | |
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| Solving Ksp Problems:Part One - s2 | Solving Ksp Problems:Part Two - 4s3 | Solving Ksp Problems:Part Three - 27s4 | Solving Ksp Problems:Part Four - 108s5 | Back to Equilibrium Menu |
This principle was first put forth by Walther Nernst in 1899. It has to do with solid substances usually considered insoluble in water. In each case, we will consider a saturated solution of the insoluble substance that is in contact with some undissolved solid. Important points to consider are:
1) Some of the solid does dissolve. Not very much, but enough. 2) The substance that dissolves will dissociate 100%. 3) There exists an equilibrium between the undissolved solid and ions in solution.
By the way, all of the examples discussed her and elsewhere are all occurring at standard temperature, which is 25.0 °C. Seldom is discussed a Ksp problem at anything other than standard temperature.
Since equilibrium principles can be used, that is where we start. Our first example is silver chloride, AgCl. When it dissolves, it dissociates like this:
AgCl(s) ⇌ Ag+(aq) + Cl¯(aq)
An equilibrium expression can be written:
Kc = ( [Ag+] [Cl¯] ) / [AgCl]
Now, we come to an important point. When the AgCl is enclosed in square brackets like this − [AgCl] − that means the "molar concentration" of solid AgCl. This value is a constant!! Why?
Answer: The "molar concentration" of a solid (it's not a useful chemistry idea, so it is seldom mentioned) can be directly related to the density, which is also a constant. Here is a short dimensional analysis which summarizes the relationship:
| | | | | | |
--- --- --- |
| g | | mole | | 1000 cm3 | |
| ––––––– | x | ––––––– | x | ––––––– | = mol/L <---molarity |
| cm3 | | g | | 1 L | |
| density | | molar mass | | convert cm3to liter | |
What we do is move the [AgCl] to the other side and incorporate it with the equilibrium constant. We can do this because [AgCl] is a constant.
Kc [AgCl] = [Ag+] [Cl¯]
Since Kc [AgCl] is a constant (because it's a constant times a constant which yields a constant), we replace it with a single symbol. Like this:
Ksp = [Ag+] [Cl¯]
(Just a side point: as you go on in chemistry, you'll get introduced to the concept of activity. The activity of a solid is defined as equal to the value of one. Since the activity of AgCl(s) = 1, it just drops out of the above expression. However, like I said, activity is for the future. Not right now.)
It turns out that the Ksp value can be either directly measured or calculated from other experimental data. Knowing the Ksp, we can calculate the solubility of the substance in a very straightforward fashion.
Here are three more examples of dissociation equations and their Ksp expressions:
| | |
--- |
| Sn(OH)2(s) ⇌ Sn2+(aq) + 2OH¯(aq) | Ksp = [Sn2+] [OH¯]2 |
| Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42¯(aq) | Ksp = [Ag+]2 [CrO42¯] |
| Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH¯(aq) | Ksp = [Fe3+] [OH¯]3 |
In order to write Ksp expressions properly, you must know how each ionic substance dissociates in water. That means you have to know your chemical nomenclature, polyatomic ions, and the charges associated with each ion.
Also, and this is important, so pardon the shouting:
EACH CONCENTRATION IN THE Ksp EXPRESSION IS RAISED TO THE POWER OF ITS COEFFICIENT IN THE BALANCED EQUATION.
Two more examples:
| | |
--- |
| Hg2Br2(s) ⇌ Hg22+(aq) + 2Br¯ (aq) | Ksp = [Hg22+] [Br¯]2 |
| Zn3(AsO4)2(s) ⇌ 3Zn2+(aq) + 2AsO43¯(aq) | Ksp = [Zn2+]3 [AsO43¯]2 |
Note how the mercury(I) ion is written. Hg22+ is correct. Do not write it as 2Hg+. Writing [Hg+]2 in the Ksp expression is wrong.
Here are ten chemical formulas. Write the chemical equation showing how the substance dissociates and write the Ksp expression.
| | | | | |
--- ---
| 1) AlPO4 | 2) BaSO4 | 3) CdS | 4) Cu3(PO4)2 | 5) CuSCN |
| 6) AgCN | 7) Mn(IO3)2 | 8) PbBr2 | 9) SrCO3 | 10) Bi2S3 |
Go to the answers
By the way, a word of warning. You may have done quite well at learning chemical nomenclature. However, in the study of Ksp, there may be some polyatomic ions used that you did not study in the nomenclature section. Thiocyanate (SCN¯) might be one example. Another could be the arsenate ion, (AsO43-). Note that arsenic is just below phosphorous in the periodic table. Compare arsenate with phosphate (PO43-).
| | | | | |
--- ---
| Solving Ksp Problems:Part One - s2 | Solving Ksp Problems:Part Two - 4s3 | Solving Ksp Problems:Part Three - 27s4 | Solving Ksp Problems:Part Four - 108s5 | Back to Equilibrium Menu | |
17389 | https://www.irmo.ie/7.Lemoine.pdf | Lemoine Point.
Figure 1: Two lines AS and AT through the vertex A of an an-gle are said to be isogonal if they are equally inclined to the arms of b A, or equivalently, to the bisector of b A (Figure 1).
The isogonals of the medians of a triangle are called symmedians.
We will show in a little while that the symmedians are concurrent and their point of concurrency is called the symmedian point. It is also called the Lemoine point.
Figure 2: As before, in a triangle ABC, the midpoint BC is denoted by A1, the intersection of BC and the bisector of b A is A3 and then the symmedian of AA1 will be AA′ 1 (Figure 2). Thus AA′ 1 = SymAA3(AA1).
Figure 3: Recall Steiner’s theorem which states that in a tri-angle ABC, if AA1 and AA2 are isogonal (Figure 3), then 1 |AB|2 |AC|2 = |BA1||BA2| |CA1||CA2|.
We now apply this to get the following.
Theorem 1 A line AA′ 1 in a triangle ABC (Figure 4) is a symmedian if and only if |BA′ 1| |CA′ 1| = |AB|2 |AC|2 = c2 b2 .
Figure 4: Proof The line AA′ 1 is a symmedian if AA1 is a median and AA′ 1 = SymAA3(AA1).
Then |BA1| = |CA1|, so on applying Steiner’s theorem, we get that AA′ 1 is a symmedian if and only if |AB|2 |AC|2 = |BA′ 1||BA1| |CA′ 1||CA1| = |BA′ 1| |CA′ 1|.
Remark It is well known that the bisector of an angle of a triangle divides the opposite side into the ratio of the sides about the angle. Then, be the above theorem, a symmedian does it in the ratio of the squares of the sides.
2 Figure 5: We can now apply the previous result to show that the symmedians are concurrent.
Theorem 2 Let AA′ 1, BB′ 1 and CC′ 1 be the symmedians of a triangle. Then these lines are concurrent at a point L called the Lemoine point (Figure 5).
Proof An easy application of Ceva’s the-orem and Theorem 1 above gives the result. We have |A′ 1B| |A′ 1C| = c2 b2, |B′ 1C| |B′ 1A = a2 c2 and |C′ 1A| |C′ 1B = b2 a2.
Then, by Ceva’s theorem, the symmedians are concurrent since the product of the ratios is 1.
Using van Aubel’s theorem we get the ratios in which L divides the sym-medians AA′ 1.
|LA| |LA′ 1| = |C′ 1A| |C′ 1B| + |B′ 1A| |B′ 1C| = b2 a2 + c2 a2 = b2 + c2 a2 .
Theorem 3 The tangents to the circumcircle C(ABC) of a triangle ABC at two of its vertices meet on the symmedian from the third vertex.
Figure 6: Proof Let the tangents to the circumcircles at the points B and C meet at the point K. Join A to K and let A′′ 1 be the point of intersection of BC and AK (Figure 6). We need to show that AA′′ 1 is the symmedian from the vertex A, i.e.
|BA′′ 1| |CA′′ 1| = c2 b2.
3 Consider |BA′′ 1| |CA′′ 1| = area(ABA′′ 1) area(ACA′′ 1) = area(BKA′′ 1) area(CKA′′ 1) = area(ABA′′ 1) + area(BKA′′ 1) area(ACA′′ 1) + area(CKA′′ 1) = area(ABK) area(ACK) = |AB||BK| sin(A b BK) |AC||CK| sin(A b CK) . . . (i).
Now make some observations. We have |KB| = |KC| since KB and KC are tangents from K to C(ABC). Furthermore, using the property that the angle between a tangent and a chord is equal to the angle in the segment on the opposite side of the chord, we have K b BC = K b CB = b A.
Thus A b BK = ( b A + b B), so sin(A b BK) = sin( b C), and A b CK = ( b A + b C), so sin(A b CK) = sin( b B).
Thus, from (i) we have |BA′′ 1| CA′′ 1| = |AB| sin( b C) |AC| sin( b B) = |AB|2 |AC|2 , by sine rule.
Then, by Theorem 1, AA′′ 1 is symmedian and so AK is the extension of a symmedian.
□ 4 Figure 7: In a triangle, the median is the locus of the mid-points of the line segments joining points on two sides and parallel to the third side (Figure 7). In the case of symmedians, we take line segments antiparallel to the third side. This is the next result.
Figure 8: Theorem 4 In a triangle ABC, if M and N are points on the sides AB and AC respectively, such that MN is antiparallel to BC, then the midpoint P of MN lies on the symmedian AA′ 1 (Figure 8).
Proof Let AA3 be the bisector of the angle b A. Points M ′ and N ′ on sides AC and AB, respectively are images of M and N under reflection in the line AA3 (Figure 9), Figure 9: M ′ = SymAA3(M), N ′ = SymAA3(N).
5 Then |AN| = |AN ′|, |AM| = |AM ′| and it follows that the triangles AN ′M ′ and ANM are congruent. Thus Ac N ′M ′ = A b NM = A b BC, since MN is antiparallel to BC. Thus N ′M ′ is parallel to BC. Then the midpoint of M ′N ′ lies on the median AA1 and so the midpoint of MN will lie on the symmedian AA′ 1 since mapping SymAA3( ) maps midpoints of segments to midpoints of images.
1 Properties of Lemoine Point.
Theorem 5 If X is a point on the symmedian from the vertex A of a triangle ABC, then the distances from X to the sides AB and AC are in the ratios of the lengths of these sides.
Figure 10: Proof Let AA′ 1 be the symmedian from A and let X be a point on AA′ 1.
Drop perpendiculars XX1 and XX2 to the sides AB and AC respectively.
Also, drop perpendiculars A′ 1f X1 and A′ 1f X2 to the sides AB and AC, respectively, from A′ 1 (Figure 10).
We claim that d(X, AB) |AB| = d(X, AC) |AC| , i.e. |XX1| |AB| = |XX2| |AC| .
Consider d(X, AB) d(X, AC) = d(A′ 1, AB) d(A′ 1, AC) = |BA′ 1| sin b B |CA′ 1| sin b C = |AB|2 sin b B |AC|2 sin b C = |AB| |AC|, as required.
Theorem 6 (Grebe’s first.) If L is the Lemoine point of a triangle ABC, then 6 d(L, BC) |BC| = d(L, AC) |AC| = d(L, AB) |AB| .
This follows immediately from Theorem 5 since L lies on all 3 symmedians.
Theorem 7 (Grebe’s second.) The point X in the plane of a triangle ABC which minimisses the quantity d2(X, BC) + d2(X, AC) + d2(X, AB) is the Lemoine point.
Proof In proving this we shall apply the Cauchy-Schwarz inequal-ity. Recall that if {a1, a2, . . . , an} and {b1, b2, . . . , bn} are sequences of real numbers then (Pn i=1 aibi)2 ≤(Pn i=1 a2 i )(Pn i=1 b2 i ) with equality if and only if a1 b1 = a2 b2 = · · · = an bn .
Figure 11: Let X be a point of the plane and drop per-pendiculars from X to the sides AB, BC and CA.
Let Xa, Xb and Xc be the feet of the perpendiculars (Figure 11).
Consider a|XXa| + b|XXb| + c|XXc| = 2 area(ABC), if X is inside ABC.
Then by the Cauchy-Schwarz inequality, 4[area(ABC)]2 ≤(a2 + b2 + c2){|XXa|2 + |XXb|2 + |XXc|2} Thus |XXa|2 + |XXb|2 + |XXc|2 ≥4(area(ABC))2 a2 + b2 + c2 , with equality if and only if |XXa| a = |XXb| b = |XXc c = constant, and this is true if and only if X = L, the Lemoine point.
Theorem 8 (Rigby?) The Lemoine point of a triangle is the centroid of its pedal triangle.
7 Figure 12: Proof Let the tangents to the circumcircle, C(ABC), of the triangle ABC at the points B and C meet at K.
From K drop perpendiculars Ka, Kb and Kc to the sides BC, AC(extended) and AB(extended) (Figure 11).
We claim that KKbKaKc is a parallelo-gram.
First extend the line segment KcK beyond K to a point L and extend KbK beyond K to a point M.
Since KKcAKb is a cyclic quadrilateral then since exte-rior angles are equal to interior opposites, we have L b KKb = b A and M b KKc = b A.
The quadrilateral KKcBKa is cyclic so Kc b BK = Kc c KaK (chord KcK) and L b KKa = Kc b BK (exterior equal to opposite interior).
The last equation can be written as L b KKb + Kb b KKa = Kc c KaK + K b BKa = Kc c KaK + b A (angle between tangent and chord).
But L b K = b A so we get Kb b KKa = Kc c KaK.
Thus lines KcKa is parallel to KKb.
Similarly, by considering the cyclic quadrilateral KKbCKa it can be shown that Kc b KKa = K c KaKb, and so lines KcK and KaKb are parallel.
Thus KKbKaKc is a parallelogram, as claimed.
It follows that KKa bisects KcKb and so KKa passes through the midpoint of KcKb.
8 Figure 13: Now drop perpendicular lines from the Lemoine point L to the sides BC, CA and AB.
Let La, Lb and Lc be the feet of these perpendiculars (Figure 13).
Clearly LLc is parallel to KKc and LLb is parallel to KKb.
The triangles ALcL and AKcK are similar so ALc AKc = AL AK .
The triangle ALLb and AKKb are similar so AL AK = ALb AKb .
Thus, combining both equalities, ALc AKc = ALb AKb .
Thus LcLb is parallel to KcKb.
So we have that the triangles LcLbL and KcKbK are similar.
Since KKa is parallel to LLa and KKa is a median of the triangle KKcKb, then LaL, when extended, is a median of the triangle LLcLb. Thus L lies on the median of the triangle LaLbLc from the vertex La. Similarly it can be shown that L also lies on the other medians of the triangle LaLbLc.
Result follows since LaLbLc is pedal triangle of the point L.
□ Figure 14: Recall that if ABC is a triangle X, Y, Z are points of the sides BC, CA and AB, then the perimeter of the triangle XY Z is a minimum if XY Z is the or-thic triangle. Now suppose we wish to minimise the quantity |XY |2 + |Y Z|2 + |ZX|2.
9 The next theorem tells us when that is done.
Theorem 9 If X, Y and Z are 3 points on the sides BC, CA and AB, respectively, then the quantity |XY |2 + |Y Z|2 + |ZX|2.
is a minimum when XY Z is the pedal triangle of the Lemoine point L.
Figure 15: Proof First we show that there is a unique set of points X0, Y0, Z0, on the sides such that |X0Y0|2 + |Y0Z0|2 + |Z0X0|2 is a minimum. Let x = |BX|, y = |CY | and z = |AZ| (Figure 15).
Now consider the function P(x, y, z) whose value is the quantity |ZY |2 + |Y X|2 + |XZ|2.
Then P(x, y, z) = z2 + (b −y)2 −2z(b −y) cos(A) = x2 + (c −z)2 −2x(c −z) cos(B) = y2 + (a −x)2 −2y(a −x) cos(C) = 2(x2 + y2 + z2) + (a2 + b2 + c2) −2by −2cz −2ax −2bz cos(A) −2cx cos(B) −2ay cos(C) +2yz cos(A) + 2yx cos(B) + 2xz cos(C) = 2(x2 + y2 + z2) + (a2 + b2c2) +2{2xy cos(C) + 2yz cos(A) + 2zx cos(B)} −2b(y + z cos(A)) −2c(z + x cos(B)) + 2a(x + y cos(C)).
Figure 16: Since P(x, y, z) represents a sphere or an ellip-soid then there exists a unique solution (x0, y0, z0) which minimises P(x, y, z). This gives correspond-ing points X0, Y0, Z0 on the sides of the triangle XY Z (Figure 16).
10 Now let X0Y0Z0 be the triangle which minimises |X0Y0|2 + |Y0Z0|2 + |Z0X0|2.
Let G′ be the centroid of X0Y0Z0 and X0X′ 0 be the median from the point X0.
By the median property of triangles |X0Z0|2 + |X0Y0| = 2|X0X′ 0|2 + 2|X′ 0Z0|2 (use cosine rule.) = 2|X0X′ 0|2 + |Z0Y0|2/2.
Thus |Z0Y0|2 + |Z0X0|2 + |X0Y0|2 = 2|X0X′ 0|2 + 3 2(|Z0Y0|2).
This is minimised if X0X′ 0 is perpendicular to the side BC. Similarly we need the other two medians of X0Y0Z0 to be perpendicular to the other two sides of the triangle.
Thus the centroid G′ of X0Y0Z0 has X0Y0Z0 as its pedal triangle.
It follows that G′ is the Lemoine point L of the triangle ABC. Result follows.
11 |
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Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
The Root-Mean Power-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (RMP-AM-GM-HM) or Exponential Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (EM-AM-GM-HM) or Quadratic Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (QM-AM-GM-HM), is an inequality of the root-mean power, arithmetic mean, geometric mean, and harmonic mean of a set of positivereal numbers that says:
, where , and is the .
The geometric mean is the theoretical existence if the root mean power equals 0, which we couldn't calculate using radicals because the 0th root of any number is undefined when the number's absolute value is greater than or equal to 1. This creates the indeterminate form of . Then, we can say that the limit as x goes to 0 is the geometric mean of the numbers.
The quadratic mean's root mean power is 2 and the arithmetic mean's root mean power is 1, as and the harmonic mean's root mean power is -1 as . Similarly, there is a root mean cube (or cubic mean), whose root mean power equals 3.
When the root mean power approaches , the mean approaches the highest number. When the root mean power reaches , the mean approaches the lowest number.
with equality if and only if . This inequality can be expanded to the power mean inequality, and is also known as the Mean Inequality Chain.
As a consequence, we can have the following inequality: If are positive reals, then with equality if and only if ; which follows directly by cross multiplication from the AM-HM inequality. This is extremely useful in problem-solving.
The Root Mean Power of 2 is also known as the quadratic mean, and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality.
Proof
The inequality is a direct consequence of the Cauchy-Schwarz Inequality; Alternatively, the RMS-AM can be proved using Jensen's inequality: Suppose we let (We know that is convex because and therefore ). We have: Factoring out the yields: Taking the square root to both sides (remember that both are positive):
The inequality is called the AM-GM inequality, and proofs can be found here.
The inequality is a direct consequence of AM-GM; , so , so .
Therefore, the original inequality is true.
Geometric Proofs
The inequality is clearly shown in this diagram for
Desmos SlidersDesmos Equation NOTE: The Desmos equation will not show the line when the numbers are negative.
(Note how the RMS is "sandwiched" between the minimum and the maximum)
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17391 | https://www.sciencedirect.com/topics/medicine-and-dentistry/retroperitoneal-hemorrhage | Retroperitoneal Hemorrhage - an overview | ScienceDirect Topics
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Retroperitoneal Hemorrhage
In subject area: Medicine and Dentistry
Retroperitoneal bleeding can result from visceral trauma or injury to the pelvic area or vertebral processes.
From: Critical Care Nursing Clinics of North America, 2006
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Ultrasound: Part 1
2014, Critical Care ClinicsSarah R. Williams MD, ... Laleh Gharahbaghian MD
Retroperitoneal Hemorrhage
Retroperitoneal hemorrhage is poorly visualized on ultrasonography.43,47 Retroperitoneal bleeding can result from a multitude of sources: pelvic fracture, injury to the great vessels (inferior vena cava [IVC] and aorta), and renal injuries. If the hemorrhage remains encapsulated in the retroperitoneal space and does not flow into the abdominal/pelvic compartments, the FAST can remain negative.
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Journal2014, Critical Care ClinicsSarah R. Williams MD, ... Laleh Gharahbaghian MD
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Access Complications
2022, Complications in Endovascular SurgeryMichael Yacoub MD, FACS, RPVI, Ali F. AbuRahma MD, FRCS, FACS, RVT, RPVI
Retroperitoneal Hemorrhage
Retroperitoneal hemorrhage is a result of a poor access technique that fails to identify the CFA, causing unsafe cannulation and sheath insertion. High punctures, cephalad to the lowest portion of the inferior epigastric artery, increase the risk of developing retroperitoneal hemorrhage. A puncture at this location is difficult to compress because there is no bony structure (i.e., the femoral head). In addition, the retroperitoneum is a large cavity and can accommodate large amounts of blood, which can result in hemodynamic instability in some patients. Retroperitoneal hemorrhage is rare, but can be life-threatening. High suspicion and early diagnosis are keys to decreasing morbidity and mortality.
Clinical Presentation and Diagnosis
This diagnosis is readily made on clinical presentation and manifests as unexplained hypotension or vagal reaction, either during the procedure or immediately after CFA sheath removal. If the patient complains of vague back pain or abdominal pain after a CFA access procedure, this should raise suspicions of retroperitoneal hemorrhage. A large retroperitoneal bleed may compress the femoral nerve, causing neuropathy. The physical exam is usually unremarkable; however, patients may present with abdominal fullness and flank ecchymosis (Grey Turner sign). Computed tomography (CT) is the diagnostic test of choice, and the use of contrast can aid in identifying the source of bleeding and guiding treatment. Patients with active retroperitoneal hemorrhage can present with continuous dropping of hematocrit and hemodynamic instability. These patients should be stabilized prior to obtaining a CT scan.
Management
Retroperitoneal hemorrhage can be managed conservatively in most cases by aggressive fluid resuscitation, correction of coagulopathy, and transfusion of packer red blood cells to maintain hematocrit. The patient should be placed on bed rest and undergo serial abdominal exams. Serial laboratory tests, including hemoglobin, platelets, and serum creatinine, should be obtained every 4–6 hours. Indications for surgical intervention include hemodynamic instability, continuous drop in hematocrit, and intractable pain and neuropathy. Patients with signs of hemodynamic instability, who respond appropriately to fluid resuscitation, should be evaluated by CT angiography to identify the source of bleeding. Hypotensive patients who are unresponsive to fluid resuscitation should undergo emergent surgical intervention. Both open and endovascular approaches have been described to manage patients with retroperitoneal hemorrhage. There are no randomized trials to guide the treatment strategies for retroperitoneal hemorrhage, and the evidence is based on small cohort series or isolated case reports. In the authors’ institution, most physicians prefer an endovascular approach, and open surgery is used only if the bleeding cannot be controlled. A contralateral CFA access is obtained and an angiogram of the iliac and femoral vessels is performed to identify the bleeding source. Hemostasis can be achieved by prolonged balloon inflation at the arteriotomy site or placement of a cover stent. If use of a covered stent is planned, care should be taken not to undersize the graft, as this will lead to continued extravasation. If extravasation from a branch vessel is identified, coil embolization of this vessel can be easily performed. If bleeding can’t be controlled using an endovascular approach, retroperitoneal exploration is performed to identify and control the bleeding source.
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Book2022, Complications in Endovascular SurgeryMichael Yacoub MD, FACS, RPVI, Ali F. AbuRahma MD, FRCS, FACS, RVT, RPVI
Chapter
Coagulopathic (Retroperitoneal) Hemorrhage
2016, Diagnostic Imaging: Genitourinary (Third Edition)
Definitions
•
Hemorrhage occurring in retroperitoneal compartments due to various etiologies
•
Term retroperitoneal hemorrhage may sometimes be a misnomer
○
Many cases occur primarily in posterior abdominal wall musculature
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Book2016, Diagnostic Imaging: Genitourinary (Third Edition)
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Retroperitoneal tumors and retroperitoneal fibrosis
2024, Penn Clinical Manual of Urology Third EditionDaniel Roberson MD, Daniel J. Lee MD
Retroperitoneal hemorrhage
Spontaneous retroperitoneal hemorrhage is rare and can occur in a wide variety of age groups depending on the etiology that can be highly varied (Table 24.3). When the bleeding is renal in origin, it has been classically referred to as Wunderlich syndrome. This is frequently the initial presentation of a renal angiomyolipoma, which is, in fact, the most common cause of spontaneous retroperitoneal hemorrhage. Patients may present in hypovolemic shock or complain of back, flank, abdominal, hip, or upper thigh pain. Physical examination may be significant for flank ecchymosis, hypotension, or gross hematuria. Cross-sectional imaging is most useful in uncovering the extent of bleed and the possibility of a malignant origin.
Patients should first be resuscitated and stabilized, anticoagulant medications should be discontinued, and any potential coagulopathies treated and reversed. Patients can be monitored with serial imaging to evaluate for resolution of the hematoma and bleeding. If conservative therapy fails and the patient becomes unstable, angiography with selective embolization of active bleeding vessels can be performed. Retroperitoneal exploration should be approached with caution as it often results in massive blood loss and, when bleeding is renal in origin, nephrectomy.
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Deaths: Trauma, Abdominal Cavity – Pathology
2016, Encyclopedia of Forensic and Legal Medicine (Second Edition)J.A. Prahlow
Retroperitoneum
Involvement of the retroperitoneum by trauma is typically manifest as retroperitoneal hemorrhage. Significant amounts of retroperitoneal hemorrhage (retroperitoneal hematomas) can result in shock and death. Relatively large amounts of retroperitoneal hemorrhage can remain hidden clinically. In addition, such bleeding can occur slowly, over time, before becoming symptomatic. Retroperitoneal hemorrhage frequently occurs with pelvic or spinal fractures, but may also occur with injuries of the pancreas, duodenum, and urinary tract, including the kidneys. External evidence of retroperitoneal hemorrhage is occasionally evident as ecchymosis of the flank; this is referred to as the ‘Grey-Turner sign’ (Prahlow, 2010; Figure 7).
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Figure 7. The so-called ‘Grey-Turner’ sign: hemorrhagic discoloration of the flank region, indicative of retroperitoneal hemorrhage.
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Reference work2016, Encyclopedia of Forensic and Legal Medicine (Second Edition)J.A. Prahlow
Review article
The Acute Abdomen
2015, Radiologic Clinics of North AmericaNicholas A. Bodmer MD, Kiran H. Thakrar MD
Epidemiology
Retroperitoneal hemorrhage may be spontaneous, traumatic, or iatrogenic (Box 7). Spontaneous retroperitoneal hemorrhage indicates the lack of a traumatic or iatrogenic cause and has a reported incidence of 0.6% to 6.6% in patients on therapeutic anticoagulation.107–110 In a series of 89 patients with spontaneous retroperitoneal hemorrhage, 66% were on anticoagulant medication, but only one-third had a supratherapeutic international normalized ratio (INR) or activated partial thromboplastin time.111
Box 7Spontaneous retroperitoneal and rectus sheath hemorrhage key points
PearlsFluid-fluid level or hematocrit sign occurs in patients with coagulopathy.Active extravasation may be arterial or venous and may be differentiated on multiphase contrast CT.Arcuate line divides the rectus sheath and affects the imaging appearance of rectus sheath hematomas.PitfallsOccult renal neoplasm may cause spontaneous perirenal retroperitoneal hemorrhage.Most occur in patients on anticoagulant medication, though most do not have supratherapeutic international normalized ratio or activated partial thromboplastin time.ControversiesAlthough nontraumatic by definition, spontaneous hemorrhage is likely caused by unrecognized minor trauma in susceptible patients
Rectus sheath hematoma also typically occurs in patients on systemic anticoagulation without a supratherapeutic INR.112
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Journal2015, Radiologic Clinics of North AmericaNicholas A. Bodmer MD, Kiran H. Thakrar MD
Chapter
Angiomyolipoma
2016, Diagnostic Pathology: Genitourinary (Second Edition)
Prognosis
•
Overwhelming majority with benign clinical behavior
•
Retroperitoneal hemorrhage is rare complication that can be fatal
○
Large tumor size important factor in this complication
•
Rare aggressive behavior, particularly those with predominant epithelioid and other atypical features
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Introduction to the Retroperitoneum
2016, Diagnostic Imaging: Genitourinary (Third Edition)
Posterior Pararenal Space
Disease originating within the posterior pararenal space is uncommon, essentially limited to hemorrhage and tumor.
“Retroperitoneal hemorrhage― is a misnomer since most spontaneous, coagulopathic hemorrhage originates within the abdominal wall, the iliopsoas compartment, or the rectus sheath. Only when hemorrhage extends beyond these fascial boundaries does it enter the retroperitoneum. Rectus sheath hematomas enter the extraperitoneal pelvic spaces through a defect in the caudal (infraumbilical) portion of the sheath. Iliopsoas hemorrhage often extends into any or all of the retroperitoneal compartments, predominantly along the main fascial planes. The hallmarks of coagulopathic hemorrhage are: Bleeding out of proportion to trauma, multiple sites of bleeding, and the presence of the “hematocrit― sign, a fluid-cellular debris level within the hematoma.
Retroperitoneal sarcomas, most commonly liposarcoma, often originate within 1 of the retroperitoneal compartments, and the site of origin can be determined by the relative mass effect on various organs and structures such as the kidneys, colon, and great vessels. Most liposarcomas have some identifiable fat within them and seem to be encapsulated, allowing for excision, although recurrent disease is common.
If retroperitoneal nodes are included in the discussion, the most common retroperitoneal tumor is non-Hodgkin lymphoma (NHL). NHL often results in massive lymphadenopathy. This characteristically involves the mesenteric and retroperitoneal nodes that are confluent and anteriorly displace the aorta and inferior vena cava from the spine. Retroperitoneal nodes are also frequently involved by malignancies originating in pelvic organs, such as the prostate, rectum, and cervix.
The other large, though uncommon group of primary retroperitoneal tumors are of neurogenic origin, including nerve sheath tumors, ganglioneuroma, neuroblastoma, and others. These often share the characteristics of appearing as well-defined, moderately enhancing masses that do not appear to arise from nodes nor abdominal viscera. Many, in fact, arise along the sympathetic nerve trunks, while others are part of a syndrome such as neurofibromatosis that may involve multiple nerves in a paraspinal or presacral distribution.
The great vessels, the aorta and inferior vena cava (IVC), are located in the retroperitoneum and are usually depicted as lying within the retromesenteric plane. Although primary disease of the IVC is rare, it may be the site of primary tumor (sarcoma) or the site of spread from a renal or adrenal carcinoma. More common are anomalies of the embryologic development of the IVC. Some 10% of the population have some anomaly of the embryologic sub- and supracardinal veins, usually at or below the level of the renal veins, resulting in variations such as duplicated IVC and retro- and circumaortic renal vein. While these are uncommonly of clinical significance (limited to affecting surgical and interventional procedures), they may be mistaken for pathologic conditions, most commonly enlarged retroperitoneal lymph nodes.
Abdominal aortic aneurysm is a major health concern, and rupture is usually fatal. Accurate diagnosis and precise mapping of the size and shape of an aneurysm allows effective, minimally invasive prophylactic treatment with endovascular stenting.
Retroperitoneal fibrosis is an inflammatory disorder that may be misinterpreted as a malignant process, as it envelops the aorta and IVC, often causing displacement and encasement of the ureters. It may occur as an isolated process or as part of a multisystem autoimmune disorder.
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Nontrauma Abdomen
2009, Emergency RadiologyStephan w. Anderson, ... Jorge A. Soto
Spontaneous Retroperitoneal Hemorrhage
In addition to rupture of the various visceral organ related aneurysms and pseudoaneurysms, a significant cause of spontaneous retroperitoneal hemorrhage is rupture of an abdominal aortic aneurysm (AAA). Although less commonly encountered than spontaneous hemorrhage from underlying hepatic and splenic pathology, spontaneous renal and pancreatic hemorrhage can be severe and life-threatening. The most common causes of spontaneous renal hemorrhage include benign and malignant lesions, commonly renal angiomyolipoma and renal cell carcinoma, respectively. Other causes of renal hemorrhage include vasculitides such as polyarteritis nodosa (PAN) and Wegener’s granulomatosis.
Similar to those with intraperitoneal hemorrhage, patients with spontaneous retroperitoneal hemorrhage often present with acute abdominal pain. When the hemorrhage is severe and massive, patients may present with hypotension and possibly even shock. In cases related to underlying renal pathology, hematuria may be associated.
Imaging Findings
Ultrasound may identify the retroperitoneal hemorrhage and possibly characterize the underlying etiology, as in the case of the renal tumors mentioned above; however, CT is often employed.
In cases of ruptured AAA, CT is often the first-line imaging modality (Fig. 9-28). CT protocols may include the use of intravenous contrast to evaluate for active extravasation, although unenhanced CT may readily identify cases of AAA rupture. In cases of rupture, CT demonstrates hematoma within the retroperitoneum, contiguous with the aorta. Often, when large, the hemorrhage is identified within multiple retroperitoneal compartments. CT findings that increase the specificity of the diagnosis of ruptured AAA include the “crescent” sign, which refers to hyperattenuating clefts within mural thrombus or the aneurysmal wall itself; a “draped” aorta, which refers to displacement of the aorta onto the spine with lateral “draping” over a vertebral body; discontinuity of the aortic wall; and frank active contrast extravasation. Ruptured AAA represents a surgical emergency that warrants rapid intervention. In cases of renal tumors, specific findings of angiomyolipoma may be identified, such as areas of macroscopic focal fat. However, these can become somewhat obscured by hemorrhage. Also, in cases of renal malignancy, the underlying mass may become obscured and follow-up imaging by CT and MRI may be necessary for identification and characterization of the mass lesion. The spatial resolution and multiplanar capabilities of the current generation of CT scanners offer the ability to accurately diagnose vascular abnormalities secondary to vasculitis, such as the multiple small aneurysms characteristic of PAN. The common causes of spontaneous hemorrhage specific to the pancreas include hemorrhage related to pseudocysts that cause pseudoaneurysm formation in the surrounding arteries secondary to exposure to the various pancreatic enzymes. Other causes include pancreatic malignancies, a rare etiology.
In cases of spontaneous hemorrhage within the abdomen, conventional angiography may demonstrate ongoing active arterial hemorrhage, and percutaneous transcatheter embolization can be used as definitive treatment of the hemorrhage. However, given the myriad of causes of hemorrhage, its underlying etiology often remains to be definitively managed.
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Book2009, Emergency RadiologyStephan w. Anderson, ... Jorge A. Soto
Chapter
Lumbosacral Plexopathy
2013, Electromyography and Neuromuscular Disorders (Third Edition)David C. Preston MD, Barbara E. Shapiro MD, PhD
Structural
Retroperitoneal hemorrhage (anticoagulation, hemophilia)
Pelvic or abdominal tumor
Aneurysm (common or internal iliac artery)
Endometriosis
Trauma
Nonstructural
Inflammatory (plexitis)
Infarction
Postpartum
Diabetes (diabetic amyotrophy)
Radiation
Postsurgical (retractor injury)
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Book2013, Electromyography and Neuromuscular Disorders (Third Edition)David C. Preston MD, Barbara E. Shapiro MD, PhD
Related terms:
Pulmonary Embolism
Hematocrit
Pseudoaneurysm
Neoplasm
Pelvis
Cerebral Hemorrhage
Hemoperitoneum
Anticoagulation
Computer Assisted Tomography
Artery
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Tools & Reference>Clinical Procedures
Focused Assessment With Sonography in Trauma (FAST)
Updated: Nov 07, 2022
Author: Timothy B Jang, MD, FAAEM, FRSM; Chief Editor: Mahan Mathur, MD more...;)
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Focused Assessment With Sonography in Trauma (FAST)
Sections Focused Assessment With Sonography in Trauma (FAST)
Practice Essentials
Technique
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Practice Essentials
Focused Assessment With Sonography in Trauma (FAST) is an ultrasound protocol developed to assess for hemoperitoneum and hemopericardium. The sensitivity of FAST has been measured between 85 and 96%, and specificity greater than 98%. In hypotensive trauma patients, sensitivity has approached 100%. FAST can be performed by experienced personnel in less than 5 minutes, and its use has been shown to decrease the time to surgical intervention, the patient length of stay, and the need for computed tomography (CT) and diagnostic peritoneal lavage (DPL). It is estimated that over 96% of level 1 trauma centers incorporate FAST into their trauma algorithms.
Blunt abdominal trauma (BAT) is a common reason for presentation to the emergency department (ED). Unfortunately, patient history and physical examination often lack the necessary sensitivity and specificity for accurate diagnosis of acute traumatic pathology. Diagnostic peritoneal lavage was historically used to determine which patients needed exploratory laparotomy, but DPL is difficult to perform in pregnant patients, it cannot be used for serial assessment, and it is overly sensitive, which leads to a high negative laparotomy rate. Abdominal CT has better specificity than DPL for intra-abdominal injury in BAT, but it can be difficult to perform in hemodynamically unstable patients, it is expensive, it requires removing patients from the clinical arena, and it may be relatively contraindicated in pregnant patients. FAST is an important and valuable diagnostic alternative to DPL and CT that can often facilitate timely diagnosis for patients with BAT. [4, 5, 6, 7, 8, 9]
Guidelines for FAST examination have been published by the American Institute of Ultrasound in Medicine (AIUM) and the American College of Emergency Physicians (ACEP). The primary FAST examination classically includes the subxiphoid window of the heart to denote pericardial fluid. Indications for FAST include evaluation of the torso for free fluid suggesting injury to the peritoneal, pericardial, and pleural cavities, particularly in cases of trauma. FAST examination may be used to evaluate the lungs for pneumothorax. [10, 11, 12]
Benefits of the FAST examination include the following:
Decreases time to diagnosis for acute abdominal injury in BAT
Helps accurately diagnose hemoperitoneum
Helps assess the degree of hemoperitoneum in BAT
Is noninvasive
Can be integrated into the primary or secondary survey and can be performed quickly, without removing patients from the clinical arena
Can be repeated for serial examinations
Is safe in pregnant patients and children, as it requires less radiation than CT
Leads to fewer DPLs; in the proper clinical setting, can lead to fewer CT scans (patients admitted to the trauma service and to receive serial abdominal examinations)
An extended version of the standard FAST examination (E-FAST) has been established and offers additional information. Along with images of the abdomen, the E-FAST examination includes views of bilateral hemithoraces to assess for hemothorax and views of bilateral upper anterior chest walls to assess for pneumothorax. [15, 16, 17, 18, 19, 20] For the remainder of this monograph, the FAST examination is referred to as the E-FAST examination, as appropriate.
The wide variation in the accuracy and reliability of FAST and E-FAST for children after blunt abdominal trauma reflects user expertise. FAST and E-FAST that are performed by expert personnel tend to be more accurate and reliable.
Several accepted indications have been identified for the FAST examination, including the following:
BAT
Stable penetrating trauma
Assessment of the degree of intraperitoneal free fluid
When emergency treatments such as intravenous (IV) fluids or blood transfusions are indicated, performance of a FAST examination should not delay initiation of these treatments.
The E-FAST allows clinicians to rapidly diagnose traumatic thoracoabdominal injuries at the bedside without use of ionizing radiation. It has high specificity and is extremely useful as an initial test to rule in dangerous diagnoses such as hemoperitoneum, pericardial effusion, hemothorax, and pneumothorax. Its moderate sensitivity means that it should not be used alone as a tool to rule out dangerous thoracoabdominal injuries. In patients with a mechanism or presentation of concern, additional imaging should be obtained despite a negative FAST examination.
Although ongoing resuscitation and a patient in extremis are not contraindications, the FAST examination can be difficult to perform in such situations.
Next:
Technique
Anesthesia generally is not necessary for sonographic evaluation. Analgesics may be required for pain control secondary to the particular trauma. Patients should be evaluated in the supine position but may be moved to the Trendelenburg or lateral decubitus position for improved visualization of particular views if there are no contraindications (eg, spinal precautions). Male patients should have the entire abdomen exposed for the examination. Take care with female patients to minimize exposure of sensitive areas. Typically, no complications are associated with this procedure, and no special efforts at complication prevention are required.
Focused assessment with sonography for trauma (FAST) should include views of (1) the hepatorenal recess (Morison pouch), (2) the perisplenic area, (3) the subxiphoid pericardial window, and (4) the suprapubic window (Douglas pouch). If an extended FAST (E-FAST) examination is performed, views of (1) bilateral hemithoraces and (2) the upper anterior chest wall should also be obtained.
(The videos below show demonstrations of FAST and E-FAST.)
Demonstration of focused assessment with sonography for trauma (FAST). Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
Focused assessment with sonography for trauma (FAST) cardiac extension for E-FAST. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
Blood tends to pool in dependent areas. The hepatorenal recess (Morison pouch) is the most dependent space in the supramesocolic region. The suprapubic view allows assessment of fluid in the most dependent area in the inframesocolic region. In women, this space (the rectouterine space) is known as the pouch of Douglas.
To visualize the Morison pouch, the transducer-probe should be placed in the right upper quadrant or laterally along the thoracoabdominal junction (see the images and videos below). This placement uses the liver as an acoustic window and avoids interference from air-filled bowel. The probe should be moved toward the inferior margin of the liver to obtain improved images of the right kidney.
Probe placement for right upper quadrant laterally.
View Media Gallery)
Right upper quadrant view.
View Media Gallery)
Focused assessment with sonography for trauma (FAST) that depicts fluid in the Morison pouch. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
Focused assessment with sonography for trauma (FAST) depicting normal right upper quadrant findings. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
In cases of acute hemoperitoneum, blood appears as an anechoic stripe in the recess (see the image below).
Free fluid in Morison pouch.
View Media Gallery)
To obtain the perisplenic view, the transducer-probe should be placed over the left flank, lateral to the spleen along the posterior axillary line (see the images below). When it is placed in this position, the handle of the probe should nearly touch the gurney. This placement allows the spleen to be used as an acoustic window and avoids interference from air-filled bowel. The probe should then be moved superiorly (toward the thoracoabdominal junction) and inferiorly to assess for the presence of free fluid above the spleen and along the spleen tip.
Probe placement for left upper quadrant laterally.
View Media Gallery)
Left upper quadrant view.
View Media Gallery)
Be sure to assess the hepatodiaphragmatic and splenodiaphragmatic spaces (see the image and videos below); blood often accumulates in these areas. A common pitfall is to scan only through the hepatorenal and splenorenal spaces.
Blood in the splenodiaphragmatic recess.
View Media Gallery)
Focused assessment with sonography for trauma (FAST) depicting fluid in the splenorenal space. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
Focused assessment with sonography for trauma (FAST) depicting a normal splenorenal space. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
To obtain the suprapubic view, place the probe just above the pubic symphysis and direct it inferiorly into the pelvis (see the images and video below). This view is easier to obtain when the bladder is full and before a Foley catheter is placed. Be sure to obtain both sagittal and transverse suprapubic views.
Suprapubic probe placement.
View Media Gallery)
Suprapubic view.
View Media Gallery)
Focused assessment with sonography for trauma (FAST) depicting fluid in the pelvis, sagittal view. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
For the subxiphoid view, the transducer-probe should be placed in the subxiphoid area and directed into the chest toward the left shoulder to provide a view of the diaphragm and the heart (see the images below). This view can be difficult to obtain if the patient is experiencing significant abdominal pain. It often requires pressing the probe into the abdomen and angling the probe so that it is nearly parallel to the skin. In such cases, it is helpful to place the palm over the top of the probe with the thumb on the indicator.
Subxiphoid probe placement.
View Media Gallery)
Subxiphoid view that demonstrates traumatic tamponade.
View Media Gallery)
If the patient is experiencing significant abdominal pain or is obese, consider switching to a parasternal long-axis view. The subxiphoid long-axis view is another view that can be used to assess for pericardial effusions. This view also allows the examiner to assess the size and collapsibility of the inferior vena cava (IVC).
If an E-FAST examination is being performed to rule out pneumothorax, place a high-frequency linear probe (8-12 MHz) with the indicator toward the patient’s head in a long-axis orientation. Place the probe high on the patient’s chest, just below the clavicles in the midclavicular line. Look for the pleural line sitting at the back of the ribs. The presence of sliding between the visceral and parietal pleura indicates absence of a pneumothorax in the area being scanned. The absence of sliding implies the presence of a pneumothorax.
(See the videos below.)
Extended focused assessment with sonography for trauma (E-FAST) that shows no pneumothorax. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
Extended focused assessment with sonography for trauma (E-FAST) that shows pneumothorax. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
View Media Gallery)
If an E-FAST examination is being performed to rule out hemothorax, place the transducer probe laterally on the lower thorax just above the diaphragm. This can be visualized by sliding the probe superiorly from the standard right and left upper quadrant views. Blood appears as an anechoic stripe in the thorax.
Additional Considerations
If rib shadowing is an obstacle, rotate the transducer probe by 30° to fit between the ribs. Consider switching to a probe with a smaller footprint (eg, a phased array probe) if such a device is available.
If the desired recesses are difficult to visualize, ask the patient to take a slow, deep breath and, if possible, to hold it. This may move the recess into view.
Be sure to fully interrogate each region by scanning through it in its entirety. A single negative view in each region does not constitute a negative E-FAST examination.
Intraperitoneal free fluid may not be hemoperitoneum. Consider ascites, urine, peritoneal dialysate, and other sources of intraperitoneal fluid. Be aware of false positives from fatty tissue, and attempt to determine precisely where the visualized fluid is located. In pregnant patients, the presence of free fluid after BAT may not be physiologic, especially if there is >2 mm to 4 mm and if the patient has no history of ovarian hyperstimulation syndrome.
Hemoperitoneum may take time to accumulate. Maintain a low threshold for repeating the E-FAST examination, especially if the patient’s vital signs or examination findings change. Serial E-FAST examinations increase the sensitivity with which intraperitoneal free fluid secondary to blunt abdominal trauma can be detected.
The E-FAST examination is an excellent initial imaging modality for identifying the presence of hemothorax or pneumothorax in the setting of trauma. Although it is quite specific, it is not sensitive enough to rule out all significant pathology.
Peer teaching can be applied to teach practical skills such as E-FAST without loss of clinical application skills. This relieves the burden of removing doctors from patient care situations while maintaining teaching standards.
Miniaturization of ultrasound has enabled helicopter emergency medical services (HEMS) to use point-of-care ultrasound to care for trauma patients on scene, allowing HEMS to detect life-threatening, time-sensitive conditions such as pneumothorax, pericardial effusion, and intraperitoneal hemorrhage. HEMS E-FAST has the potential to triage certain trauma patients directly to the operating room, bypassing the emergency department and saving crucial time.
Evaluation of the lung during E-FAST cannot be used in the trauma setting to identify patients with active COVID-19 infection or to stratify patients as having high or low risk of infection. This is likely due to differences in lung imaging techniques and the presence of concomitant thoracic injury.
Previous
References
Bloom BA, Gibbons RC. Focused assessment with sonography for trauma. StatPearls. [Internet]. Treasure Island, FL: StatPearls Publishing; 2022 Jul 25. [Full Text].
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Bahner D, Blaivas M, Cohen HL, Fox JC, Hoffenberg S, Kendall J, et al. AIUM practice guideline for the performance of the focused assessment with sonography for trauma (FAST) examination. J Ultrasound Med. 2008 Feb. 27(2):313-8. [QxMD MEDLINE Link].
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Grünherz L, Jensen KO, Neuhaus V, Mica L, Werner CML, Ciritsis B, et al. Early computed tomography or focused assessment with sonography in abdominal trauma: what are the leading opinions?. Eur J Trauma Emerg Surg. 2017 Jul 20. [QxMD MEDLINE Link].
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American Institute of Ultrasound in Medicine, American College of Emergency Physicians. AIUM practice guideline for the performance of the focused assessment with sonography for trauma (FAST) examination. J Ultrasound Med. 2014 Nov. 33 (11):2047-56. [QxMD MEDLINE Link].
Montoya J, Stawicki SP, Evans DC, Bahner DP, Sparks S, Sharpe RP, et al. From FAST to E-FAST: an overview of the evolution of ultrasound-based traumatic injury assessment. Eur J Trauma Emerg Surg. 2015 Mar 14. [QxMD MEDLINE Link].
Soult MC, Weireter LJ, Britt RC, Collins JN, Novosel TJ, Reed SF, et al. Can routine trauma bay chest x-ray be bypassed with an extended focused assessment with sonography for trauma examination?. Am Surg. 2015 Apr. 81 (4):336-40. [QxMD MEDLINE Link].
Calder BW, Vogel AM, Zhang J, Mauldin PD, et al. Focused assessment with sonography for trauma in children after blunt abdominal trauma: A multi-institutional analysis. J Trauma Acute Care Surg. 2017 Aug. 83 (2):218-224. [QxMD MEDLINE Link].
Helling TS, Wilson J, Augustosky K. The utility of focused abdominal ultrasound in blunt abdominal trauma: a reappraisal. Am J Surg. 2007 Dec. 194(6):728-32; discussion 732-3. [QxMD MEDLINE Link].
Kirkpatrick AW, Sirois M, Laupland KB, Liu D, Rowan K, Ball CG, et al. Hand-held thoracic sonography for detecting post-traumatic pneumothoraces: the Extended Focused Assessment with Sonography for Trauma (EFAST). J Trauma. 2004 Aug. 57(2):288-95. [QxMD MEDLINE Link].
Sheng AY, Dalziel P, Liteplo AS, Fagenholz P, Noble VE. Focused Assessment with Sonography in Trauma and Abdominal Computed Tomography Utilization in Adult Trauma Patients: Trends over the Last Decade. Emerg Med Int. 2013. 2013:678380. [QxMD MEDLINE Link]. [Full Text].
Fuchs PA, del Junco DJ, Fox EE, Holcomb JB, Rahbar MH, Wade CA, et al. Purposeful variable selection and stratification to impute missing Focused Assessment with Sonography for Trauma data in trauma research. J Trauma Acute Care Surg. 2013 Jul. 75(1 Suppl 1):S75-81. [QxMD MEDLINE Link]. [Full Text].
Barbosa RR, Rowell SE, Fox EE, Holcomb JB, Bulger EM, Phelan HA, et al. Increasing time to operation is associated with decreased survival in patients with a positive FAST examination requiring emergent laparotomy. J Trauma Acute Care Surg. 2013 Jul. 75(1 Suppl 1):S48-52. [QxMD MEDLINE Link]. [Full Text].
Bhoi S, Sinha TP, Ramchandani R, Kurrey L, Galwankar S. To determine the accuracy of focused assessment with sonography for trauma done by nonradiologists and its comparative analysis with radiologists in emergency department of a level 1 trauma center of India. J Emerg Trauma Shock. 2013 Jan. 6(1):42-6. [QxMD MEDLINE Link]. [Full Text].
Kärk Nielsen S, Ewertsen C, Svendsen LB, Hillingsø JG, Nielsen MB. Focused assessment with sonography for trauma in patients with confirmed liver lesions. Scand J Surg. 2012. 101(4):287-91. [QxMD MEDLINE Link].
Kornblith AE, Addo N, Plasencia M, et al. Development of a consensus-based definition of focused assessment with sonography for trauma in children. JAMA Netw Open. 2022 Mar 1. 5 (3):e222922. [QxMD MEDLINE Link].
Matsushima K, Khor D, Berona K, Antoku D, Dollbaum R, Khan M, et al. Double Jeopardy in Penetrating Trauma: Get FAST, Get It Right. World J Surg. 2017 Aug 4. [QxMD MEDLINE Link].
Thiessen MEW, Riscinti M. Application of focused assessment with sonography for trauma in the intensive care unit. Clin Chest Med. 2022 Sep. 43 (3):385-92. [QxMD MEDLINE Link].
Hussain ZJ, Figueroa R, Budorick NE. How much free fluid can a pregnant patient have? Assessment of pelvic free fluid in pregnant patients without antecedent trauma. J Trauma. 2011 Jun. 70(6):1420-3. [QxMD MEDLINE Link].
Szalai C, Shehada SE, Iancu S, et al. Does the clinical experience of a tutor influence how students learn extended focused assessment with sonography for trauma: a randomized controlled trial. Med Teach. 2022 Oct 18. 1-6. [QxMD MEDLINE Link].
Strony R, Slimmer K, Slimmer S, Corros P, Davis R, Zhu B, et al. Helicopter emergency medical services performed extended focused assessment with sonography: training, workflow, and sustainable quality. Air Med J. 2022 Mar-Apr. 41 (2):209-16. [QxMD MEDLINE Link].
Diaz-Miron J, Reppucci ML, Weinman J, et al. The use of ultrasound in establishing COVID-19 infection as part of a trauma evaluation. Emerg Radiol. 2022 Apr. 29 (2):227-34. [QxMD MEDLINE Link].
Media Gallery
Probe placement for right upper quadrant laterally.
Right upper quadrant view.
Free fluid in Morison pouch.
Probe placement for left upper quadrant laterally.
Left upper quadrant view.
Blood in the splenodiaphragmatic recess.
Suprapubic probe placement.
Suprapubic view.
Subxiphoid probe placement.
Subxiphoid view that demonstrates traumatic tamponade.
Focused assessment with sonography for trauma (FAST) cardiac extension for E-FAST. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Demonstration of focused assessment with sonography for trauma (FAST). Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Extended focused assessment with sonography for trauma (E-FAST) that shows no pneumothorax. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Extended focused assessment with sonography for trauma (E-FAST) that shows pneumothorax. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Focused assessment with sonography for trauma (FAST) that depicts fluid in the Morison pouch. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Focused assessment with sonography for trauma (FAST) depicting fluid in the pelvis, sagittal view. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Focused assessment with sonography for trauma (FAST) depicting fluid in the splenorenal space. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Focused assessment with sonography for trauma (FAST) depicting normal right upper quadrant findings. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
Focused assessment with sonography for trauma (FAST) depicting a normal splenorenal space. Video courtesy of Meghan Kelly Herbst, MD. Also courtesy of Yale School of Medicine, Emergency Medicine.
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Author
Timothy B Jang, MD, FAAEM, FRSM Professor of Clinical Emergency Medicine, University of California, Los Angeles, David Geffen School of Medicine; Director of Emergency Ultrasonography, Department of Emergency Medicine, Harbor-UCLA Medical Center
Timothy B Jang, MD, FAAEM, FRSM is a member of the following medical societies: American College of Emergency Physicians, American Institute of Ultrasound in Medicine, Christian Medical and Dental Associations, Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Coauthor(s)
Zahir Basrai, MD Fellow in Emergency Ultrasound, Division of Emergency Medicine, Harbor-UCLA Medical Center
Disclosure: Nothing to disclose.
Specialty Editor Board
Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Nothing to disclose.
Chief Editor
Mahan Mathur, MD Assistant Professor of Radiology and Biomedical Imaging, Yale University School of Medicine; Director, Medical Student Education, Associate Director, Diagnostic Radiology Residency Program, Yale-New Haven Hospital
Mahan Mathur, MD is a member of the following medical societies: American College of Radiology, American Roentgen Ray Society, Radiological Society of North America
Disclosure: Nothing to disclose.
Additional Contributors
James Quan-Yu Hwang, MD, RDMS, RDCS, FACEP Staff Physician, Emergency Department, Kaiser Permanente
James Quan-Yu Hwang, MD, RDMS, RDCS, FACEP is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, American Institute of Ultrasound in Medicine, Society for Academic Emergency Medicine
Disclosure: Received salary from 3rd Rock Ultrasound, LLC for speaking and teaching; Received consulting fee from Schlesinger Associates for consulting; Received consulting fee from Philips Ultrasound for consulting.
Gowthaman Gunabushanam, MD, FRCR Assistant Professor, Department of Diagnostic Radiology, Yale University School of Medicine
Gowthaman Gunabushanam, MD, FRCR is a member of the following medical societies: American Roentgen Ray Society, Connecticut State Medical Society
Disclosure: Nothing to disclose.
Acknowledgements
Christopher Angemi, DO Clinical Instructor, University of California, Los Angeles, David Geffen School of Medicine; Emergency Ultrasound Fellow, Department of Emergency Medicine, Harbor-UCLA Medical Center; Staff Physician, Department of Emergency Medicine, Bakersfield Memorial Hospital
Christopher Angemi, DO is a member of the following medical societies: American College of Emergency Physicians, American College of Osteopathic Emergency Physicians, American Osteopathic Association, and California Medical Association
Disclosure: Nothing to disclose.
Acknowledgments
Medscape Reference thanks Meghan Kelly Herbst, MD, Emergency Ultrasound Director, Department of Emergency Medicine, Hartford Hospital, for assistance with the video contribution to this article. Medscape Reference also thanks Yale School of Medicine, Emergency Medicine for assistance with the video contribution to this article.
encoded search term (Focused Assessment With Sonography in Trauma (FAST)) and Focused Assessment With Sonography in Trauma (FAST)
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View synonyms for sumptuous
sumptuous
[suhmp-choo-uhs]
adjective
entailing great expense, as from choice materials, fine work, etc.; costly.
a sumptuous residence.
2. luxuriously fine or large; lavish; splendid.
a sumptuous feast.
Synonyms: munificent, luxurious, magnificent
sumptuous
/ ˌsʌmptjʊˈɒsɪtɪ, ˈsʌmptjʊəs/
adjective
expensive or extravagant
sumptuous costumes
2. magnificent; splendid
a sumptuous scene
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Origin ofsumptuous1
1475–85; < Latin sūmptuōsus, equivalent to sūmptu ( s ) expense ( sumptuary ) + -ōsus -ous
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Origin ofsumptuous1
C16: from Old French somptueux, from Latin sumptuōsus costly, from sumptus; see sumptuary
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17394 | https://www.quora.com/What-does-increasing-the-pressure-on-a-gas-at-constant-temperature-do-to-the-volume-of-the-gas | What does increasing the pressure on a gas at constant temperature do to the volume of the gas? - Quora
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What does increasing the pressure on a gas at constant temperature do to the volume of the gas?
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According to Boyle's Law, when the temperature of a gas is held constant, increasing the pressure on the gas will result in a decrease in its volume. This relationship can be expressed mathematically as:
P 1 V 1=P 2 V 2 P 1 V 1=P 2 V 2
where:
P 1 P 1 and V 1 V 1 are the initial pressure and volume,
P 2 P 2 and V 2 V 2 are the final pressure and volume.
In simpler terms, if you increase the pressure (P 2>P 1 P 2>P 1), the volume must decrease (V 2<V 1 V 2<V 1) to maintain the equality, assuming the amount of gas and temperature remain constant. This inverse relationship indicates that pressure and volume are inversely proportional und
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According to Boyle's Law, when the temperature of a gas is held constant, increasing the pressure on the gas will result in a decrease in its volume. This relationship can be expressed mathematically as:
P 1 V 1=P 2 V 2 P 1 V 1=P 2 V 2
where:
P 1 P 1 and V 1 V 1 are the initial pressure and volume,
P 2 P 2 and V 2 V 2 are the final pressure and volume.
In simpler terms, if you increase the pressure (P 2>P 1 P 2>P 1), the volume must decrease (V 2<V 1 V 2<V 1) to maintain the equality, assuming the amount of gas and temperature remain constant. This inverse relationship indicates that pressure and volume are inversely proportional under constant temperature conditions.
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Ian Miller
PhD., FRSC, chemist, theoretician, author, · Author has 9K answers and 6.7M answer views
·8y
Increasing the pressure of a gas at constant temperature reduces the volume. If the gas is hydrogen or helium, it will come close to following the equation PV = nRT, where n is the number of moles and R is the molar gas constant.
Very few gases obey this exactly, although if the pressure change is small, deviations are not too bad for many of them. With very large changes of pressure, no gas follow
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Increasing the pressure of a gas at constant temperature reduces the volume. If the gas is hydrogen or helium, it will come close to following the equation PV = nRT, where n is the number of moles and R is the molar gas constant.
Very few gases obey this exactly, although if the pressure change is small, deviations are not too bad for many of them. With very large changes of pressure, no gas follows it exactly. The reason is, if you derive the relation theoretically, you assume the gases are points (i.e. they have no volume themselves) and they have no interactions between the...
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Gary Ahlers
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·4y
Originally Answered: At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container? ·
If temperature increases the pressure in a confined space - which is the volume (l x w x h) remains the same. If the temperature increases without volume constraint (like a balloon) the pressure remains constant and the volume increases. If temperature is constant, the pressure cannot increase. If you add gas to increase pressure in a constrained space the temperature will rise.
The Relationship between Pressure and Volume: Boyle's Law
The Relationship between Temperature and Volume: Charles's Law
The Relationship between Amount and Volume: Avogadro's Law
The volume of a gas is inversely proportio
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If temperature increases the pressure in a confined space - which is the volume (l x w x h) remains the same. If the temperature increases without volume constraint (like a balloon) the pressure remains constant and the volume increases. If temperature is constant, the pressure cannot increase. If you add gas to increase pressure in a constrained space the temperature will rise.
The Relationship between Pressure and Volume: Boyle's Law
The Relationship between Temperature and Volume: Charles's Law
The Relationship between Amount and Volume: Avogadro's Law
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature.
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What happens to the volume when the temperature of a gas is increased at a constant pressure?
At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container?
Why would the temperature of a gas increase if you increased the pressure at constant volume?
If the temperature of a gas is increased,will its volume increase or decrease?
At a constant temperature, how much pressure is increased of a gas when its volume decreases 10%?
Richard Routhier
MSc. in Chemistry, University of Montréal · Author has 3.2K answers and 4.6M answer views
·4y
Originally Answered: At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container? ·
At constant temperature, the product PV is a constant. P being the pressure and V the volume. If you increase the pressure, the volume will be reduced accordingly to keep the constant. To simplify if PV=1 then doubling the pressure means the volume will become 50% of the original.
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Jakub Adamik
BSC in Mechanical Engineering&Thermodynamics, University of Łódź (Graduated 2012) · Author has 55 answers and 54.3K answer views
·7y
Originally Answered: What will happen to the volume as temperature is increased at a constant pressure? ·
If U want to have constant pressure then You would have to enlarge volume of the container. Youve asked question from the right side, but still If the constant pressure is most important for You then You will have to work as a pressostat by yourself.
That is caused simply by the rule of the conservation of energy. If you increase temperature of gas, You increase its internal energy which in fact causes particles to move more rapidly, coliding with each other, causing for the volume the same result as if You increased the pressure.
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David Shires
BSc in Physics, University of Southampton (Graduated 1972) · Author has 155 answers and 90.1K answer views
·4y
Originally Answered: At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container? ·
If you pump gas into a container and allow it to cool to its original temperature, you will be left with more gas and more pressure in the container ( PV = nRT ). You are increasing the number of moles n, thus increasing the pressure P. This pressure increase will cause the container to increase in volume very slightly as the skin of the container is forced to stretch (i.e. put under tangential stress). Even a steel gas bottle will become a little bit larger - but not much!
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A container holds a gas and is held at a constant temperature. The pressure is increased. What happens to the volume of the gas?
What holds true for volume if the temperature of a given amount of gas increases and pressure remains constant?
If the temperature of gas stays the same and the pressure of gas increases, does the volume increase or decrease?
When we decrease the volume but keep the same amount of gas at constant temperature, what will happen to its pressure? Will it increase or decrease? Why?
If the temperature of a gas is decreased at constant pressure, what happened to its volume?
James
MSChE from Virginia Polytechnic Institute (VA TECH) (Graduated 1973) · Author has 10K answers and 5.1M answer views
·5y
the ideal gas law: PV = nRT
Since the right side of the equation remains constant, V must decrease as P increases in order to remain constant.
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Pradip Chandra Bhattacharyya
Former Retired · Author has 1.4K answers and 841.1K answer views
·4y
Originally Answered: At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container? ·
It will depend on whether a Safety Valve is installed on the container.
If not, indefinite increase in pressure shall lead to sudden break-up of the container, unless it is very strongly built, and likely causing a catastrophe!
Like a pressure cooker blowing up (although in this scenario the cooker is getting continuous heat input)!
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Glen Reese
Ph.D in Physics, Kansas State University (Graduated 1970) · Author has 5.4K answers and 5.1M answer views
·8y
Originally Answered: How does the volume of a fixed mass of gas increase when temperature increases under constant pressure? ·
Gas pressure in a container depends on the ferocity of the assault on its walls, depending on the mass, mass density, and velocity of the molecules clanging into them. Increase temperature, and you increase molecular velocity. That would increase the pressure, unless you allowed the volume to expand, decreasing density, to bring the pressure back in line.
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Robert Cook
BS-NE, MS-QA in Nuclear Energy&Physics, Texas A&M University (TAMU) (Graduated 1978) · Author has 12.2K answers and 7.8M answer views
·3y
Originally Answered: A container holds a gas and is held at a constant temperature. The pressure is increased. What happens to the volume of the gas? ·
Internal pressure is said to increase for some reason.
If the container cannot expand as pressure increases- and, in the real world every container does expand a little bit as stress increases - then the volume of the gas remains the same.
If the container can expand as pressure increases, then the volume increases.
Upvote ·
Rakib Ahmed
Studied Mechanical Engineering
·8y
Originally Answered: If the temperature of an ideal gas increases at a constant pressure, what happens to the volume of the gas? ·
The volume will increase according to the Charles law for ideal gas. The law states that if we hold the pressure constant, the volume will increase proportionally with it’s kelvin temperature. Thus the volume will tend to increase.
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9 1
Soumyaneel Mukherjee
Lives in Kolkata, West Bengal, India (2005–present) · Author has 2.2K answers and 1.7M answer views
·5y
When the temperature of a fixed mass of gas is kept constant if the pressure is increased symmetrically the volume of the.gas increases. The volume of the gas goes on decreasing as the pressure is increased if the temperature of the fixed mass of gas is kept constant.
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Mark Ross
Former Programmer (Retired) at Washington State Patrol (1986–2015) · Upvoted by
Kim Aaron
, Has PhD in fluid dynamics from Caltech · Author has 22.6K answers and 9.3M answer views
·7y
Originally Answered: What will happen to the volume as temperature is increased at a constant pressure? ·
If the pressure remains constant, the volume will increase. One exception to this is water at a near freezing temperature. I don’t know if other fluids do this.
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Related questions
What happens to the volume when the temperature of a gas is increased at a constant pressure?
At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container?
Why would the temperature of a gas increase if you increased the pressure at constant volume?
If the temperature of a gas is increased,will its volume increase or decrease?
At a constant temperature, how much pressure is increased of a gas when its volume decreases 10%?
A container holds a gas and is held at a constant temperature. The pressure is increased. What happens to the volume of the gas?
What holds true for volume if the temperature of a given amount of gas increases and pressure remains constant?
If the temperature of gas stays the same and the pressure of gas increases, does the volume increase or decrease?
When we decrease the volume but keep the same amount of gas at constant temperature, what will happen to its pressure? Will it increase or decrease? Why?
If the temperature of a gas is decreased at constant pressure, what happened to its volume?
What is the effect of increasing the temperature of a gas on its pressure or volume?
If the volume of a gas increases, what happens to the temperature if the pressure remains constant?
If the pressure of a gas is increased by 25%, at constant temperature, what is the % change in its volume?
What happens to the temperature of a gas if its pressure is increased or decreased while its volume is kept constant?
What happens to voltage/current when we decrease/increase pressure on a gas at constant temperature and volume?
Related questions
What happens to the volume when the temperature of a gas is increased at a constant pressure?
At a constant temperature, if the pressure of a gas increases, what will happen to the volume of the container?
Why would the temperature of a gas increase if you increased the pressure at constant volume?
If the temperature of a gas is increased,will its volume increase or decrease?
At a constant temperature, how much pressure is increased of a gas when its volume decreases 10%?
A container holds a gas and is held at a constant temperature. The pressure is increased. What happens to the volume of the gas?
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Show that the radiant heat transfer between two infinitely large parallel plates seprated by n shield is
| | |
--- |
| written 8.7 years ago by teamques10 ★ 70k | • modified 5.4 years ago |
$Q_(n-shields)$=$\frac{AÏ(T_1^4-T_2^4)}{((n+1)[\frac{2}{ϵ}-1])}$
heat transfer
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| written 8.7 years ago by teamques10 ★ 70k | • modified 8.7 years ago |
Radiation heat transfer between two large parallel plates of emissivitiesε_1 and ε_2 maintained at uniform temperatures T1 and T2 is given by
Q_(12,no shield)=(AÏ(T_1^4-T_2^4))/(1/ε_1 +1/ε_2 -1)
Now consider a radiation shield placed between these two plates, as shown in Figure
Fig: The radiation shield placed between two parallel plates and the radiation network associated with it.
Let the emissivities of the shield facing plates 1 and 2 be $ε_3,1$and $ε_3,2$, respectively. Note that the emissivity of different surfaces of the shield may be different. The radiation network of this geometry is constructed, as usual, by drawing a surface resistance associated with each surface and connecting these surface resistances with space resistances, as shown in the figure. The resistances are connected in series, and thus the rate of radiation heat transfer i
$Q_12,one shield$=$\frac{(E_b1-E_b2)}{(\frac{(1-ε_1)}{(A_1 ε_1 )}+\frac{1}{(A_1 F_12 )}+\frac{(1-ε_3,1)}{(A_3 ε_31 )}+\frac{(1-ε_3,2)}{(ε_3,2 A_3 )}+\frac{1}{(A_3 F_32 )}+\frac{(1-ε_2)}{(A_2 ε_2 )}+\frac{1}{(A_3 F_32 )}+\frac{(1-ε_2)}{(A_2 ε_2 )}))}$
Noting that F13 =F23 = 1 and A1 = A2 = A3 =A for infinite parallel plates,
$Q_12,one shield$ =$\frac{(AÏ(T_1^4-T_2^4))}{((\frac{1}{ε_1} +\frac{1}{ε_2} -1)+(\frac{1}{ε_3,1} +\frac{1}{ε_3,2} -1) )}$
where the terms in the second set of parentheses in the denominator represent the additional resistance to radiation introduced by the shield. The appearance of the equation above suggests that parallel plates involving multiple radiation shields can be handled by adding a group of terms like those in the second set of parentheses to the denominator for each radiation shield. Then the radiation heat transfer through large parallel plates separated by N radiation shields becomes,
$Q_12,N shields = \frac{A\sigma T_1^4-T_2^4}{\frac{1}{ε_1}+\frac{1}{ε_2}-1+\frac{1}{ε_3,1}+\frac{1}{ε_3,2}-1+....\frac{1}{ε_N,1}+\frac{1}{ε_N,2}-1}$
If the emissivities of all surfaces are equal,
$Q_12,N shields$=$\frac{AÏ(T_1^4-T_2^4)}{(N+1)(\frac{1}{ε}+\frac{1}{ε-1})}$=$\frac{(AÏ(T_1^4-T_2^4))}{((N+1)(\frac{2}{ε}-1) )}$
Hence proved.
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. 2016 Jan 16;8(1):e29967. doi: 10.5812/numonthly.29967
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Improving Compliance With a Single Post-Operative Dose of Intravesical Chemotherapy After Transurethral Resection of Bladder Tumour
Luke Stroman
Luke Stroman
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1,, Ben Tschobotko
Ben Tschobotko
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1, Hamid Abboudi
Hamid Abboudi
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1, David Ellis
David Ellis
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1, Elsie Mensah
Elsie Mensah
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1, Harikesh Kaneshayogan
Harikesh Kaneshayogan
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1, Evangelos Mazaris
Evangelos Mazaris
1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
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1
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1 Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, London, England
Corresponding author: Luke Stroman, Department of Urology, St Mary’s Hospital, Imperial College NHS Trust, Praed Street, W2 1NY, London, England. Tel: +44-2033126666, +44-7921369203, E-mail: stromanl@doctors.net.uk
Received 2015 Jun 10; Accepted 2015 Aug 3; Collection date 2016 Jan.
Copyright © 2016, Nephrology and Urology Research Center.
This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License ( which permits copy and redistribute the material just in noncommercial usages, provided the original work is properly cited.
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PMCID: PMC4779310 PMID: 26981495
Abstract
Background:
Post-operative single dose intravesical chemotherapy (PSDIVC) in patients with non-muscle invasive bladder cancer has been shown to reduce recurrence rates by up to 39%. However, some studies have suggested poor compliance with PSDIVC stating logistical issues and reluctance to give chemotherapy prior to histological confirmation as some of the reasons.
Objectives:
This study aims to analyse appropriate administration of PSDIVC practice in St. Mary’s Hospital against European Association of Urology guidelines and implement an intervention bundle to improve practice.
Patients and Methods:
All patients that underwent transurethral resection of bladder tumour (TURBT) between March 2012 and February 2013 were analysed retrospectively to review indication for post-operative chemotherapy, instillation rates and limiting factors preventing appropriate instillation. An intervention bundle including pre-operative delivery of mitomycin C (MMC) to the theatre suite, proforma placed in the operative notes and designated roles for PSDIVC induction was introduced to improve instillation and documentation rates. Prospective re-audit data was collected over six months between July 2013 and December 2013 following intervention.
Results:
Sixty-four patients in group A underwent TURBT prior to introduction of the intervention bundle. Fifty-four patients had non-muscle invasive bladder cancer (NMIBC), which would have been eligible for PSDIVC. Fifteen (28% of NMIBC) were administered PSDIVC. Twenty-three (36% of all patients) were either given PSDIVC or had a documented contraindication. Thirty-one patients in group B underwent TURBT following induction of intervention bundle. Twelve (50% of NMIBC) patients were given PSDIVC. Twenty-eight (90% of all patients) were either given PSDIVC or had a documented contraindication.
Conclusions:
The intervention bundle prompted increased administration of PSDIVC and documentation. Similar centres may benefit from an intervention to improve compliance.
Keywords: Mitomycin C, Bladder Cancer, Intravesical Instillation, Bladder Neoplasms
1. Background
A single instillation of mitomycin C (MMC) following transurethral resection of bladder tumour (TURBT) has been shown to reduce the recurrence of bladder cancer by up to 39% (1-3). Tumour recurrence following TURBT is thought to be due to cells from the resected tumour re-implanting into the urothelium following resection. Time from resection to recurrence is thought to be associated with tumour grade, lamina propria invasion and the presence of more than one tumour. The mechanism of action of intravesical chemotherapy involves destruction of circulating tumour cells by inhibiting DNA synthesis and ablation of the bladder cancer resection sites (4).
Sylvester et al. completed an initial meta-analysis in 2004 which showed an 11.7% reduction in recurrence and reduction of 39% in the odds of recurrence with PSDIVC (1). Mitomycin C, epirubicin, thiotepa, pirarubici and doxorubicin were all shown to have a beneficial effect (1).Two meta-analyses published in 2013 supported this evidence. Abern et al. found a 13% reduction in tumour recurrence using a single post-operative dose of IVC (2). Perlis et al. found a 12% reduction in early recurrence and 38% increase in the interval until recurrence but criticised 12 out of 13 studies for publication bias (3). These findings prompted endorsement by the European association of urology (EAU) and British association of urological surgeons to introduce guidelines suggesting that all patients with NMIBC undergoing TURBT should receive a post-operative instillation of intravesical chemotherapy within 24 hours (5, 6). Similarly the American Urological Association Guidelines recommend the use of PSDIVC for NMIBC perioperatively or postoperatively in an adjuvant fashion (7).
2. Objectives
We aim to audit the instillation of Mitomycin C at St Mary’s Hospital, London, against EAU guidelines, create an intervention bundle to overcome obstacles preventing administration and re-audit to show improvement in practice.
3. Patients and Methods
All patients that underwent TURBT over a 12-month period between March 2012 and February 2013 were allocated to group A and analysed to audit PSDIVC administration. Retrospective data was collected from operative notes, inpatient notes and post-operative medication charts. This data included indication for PSDIVC, whether instillation was given and documentation stating a contraindication if not given. Following discussion at a departmental meeting involving consultant urological surgeons, nurse practitioners and residents an intervention bundle was suggested. An intervention bundle including pre-operative delivery of MMC by the surgical resident to theatre suite, proforma placed pre-operatively in operation notes (Figure 1) and administration of MMC post-operatively by a surgical resident or nurse specialist was introduced, roles that were not previously clearly defined. Prospective re-audit data from group B was collected over a 6-month period between July 2013 and December 2013 in study two following intervention using the proforma.
Figure 1. Proforma Placed in Patient Notes Pre-Operatively in Group B.
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4. Results
Sixty-four patients in group A underwent TURBT prior to the introduction of the intervention bundle (Table 1). Fifty-four patients had NMIBC which would have been eligible for PSDIVC. Fifteen (28% of NMIBC) were treated with intravesical mitomycin C post-operatively within 24 hours. Reasons why PSDIVC was not administered were documented in eight cases (Table 2). Mitomycin C was documented to be not available for three patients, contraindicated in four patients and prescribed but not given to one patient. Forty-one patients had no documented reason why mitomycin was not given. Twenty-three (36% of all patients) were either given PSDIVC or had a documented contraindication in group A.
Table 1. Comparing Differences in Mitomycin C Administration Between Group A and Group B.
| | Group A | Group B |
---
| Total number of TURBT | 64 | 31 |
| Non-muscle invasive disease (CIS, Ta, T1) | 54 | 24 |
| Muscle invasive disease (T2 and above) | 10 | 7 |
| Given PSDIVC within 24 hours | 15 | 12 |
| Given PSDIVC after 24 hours | 0 | 0 |
| Contraindication documented | 8 | 16 |
| No documented reason why not given | 41 | 3 |
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Table 2. Documented Reasons Why PSDIVC Not Given.
| Reason Why PSDIVC Not Given | Group A | Group B |
---
| MMC not available | 3 | 0 |
| Palliative | 3 | 0 |
| Documented haematuria | 1 | 0 |
| Written in post-op notes but not given | 1 | 0 |
| Previously given adjuvant chemotherapy | 0 | 9 |
| Muscle invasive bladder cancer | 0 | 7 |
| No documented reason why PSDIVC not given | 41 | 3 |
| Total PSDIVC not given | 49 | 19 |
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Thirty-one patients in group B underwent TURBT following induction of intervention bundle. Twenty-four patients had NMIBC. Twelve (50% of NMIBC) patients were given PSDIVC. Contraindications were documented for 16 patients; three had no reason documented as to why treatment was not administered. Twenty-eight (90% of all patients) in group B were either given PSDIVC or had a documented contraindication.
5. Discussion
This is the first audit performed in our department that has looked at MMC administration rates. The importance of early postoperative instillation has repeatedly been shown. Disappointingly administration rates in group A were low. As was the number of patients who had contraindications documented. The patients in group B had increased administration rates of PSDIVC and documentation levels.
Administration of PSDIVC was increased in group B when compared to group A. One of the possible reasons for this could include the increased availability of mitomycin C post-operatively – there were no patients in group B that did not receive mitomycin C compared to three in the first study due to lack of availability. Following the initial audit a pathway to ensure availability of mitomycin was introduced. As mitomycin is a controlled medication and used infrequently, pharmacy and nursing staff were reluctant to keep mitomycin in the theatre suite. Surgical residents ordered mitomycin from pharmacy pre-operatively and took responsibility for delivery to the theatre suite on the day of surgery. It was the responsibility of surgical residents or surgical nurse specialists to administer and document post-operatively, roles not previously determined. Ideally responsibility of pharmacy logistics would be taken over by pharmacy staff, allowing the surgical team to focus on other responsibilities. A further improvement to the intervention bundle could include pharmacy staff being made aware of the required medications pre-operatively and ensuring medications are delivered to theatre suite.
It is also worth noting that meta-analyses by Abern et al. (2) and Perlis et al. (3) were both published after group A, however earlier EAU guidelines published in 2011 recommend PSDIVC for non-muscle invasive bladder cancer (8). To increase awareness of EAU guidelines a departmental meeting was held where the results of group A were presented and urological surgeons and trainees reminded of adjuvant chemotherapy guidelines. Proformas placed pre-operatively in the operative notes acted as a reminder for the surgeon to consider mitomycin C on the day of the operation. In addition proformas acted as a method of improving documentation in group B.
Absolute contraindications for intravesical chemotherapy include post-operative bleeding, hypersensitivity, bladder perforation, myelosuppression, and thrombocytopenia (9). In both groups of the 27 documented reasons why mitomycin was not given only one was due to excessive bleeding. Nine patients were not given mitomycin due to recurrent disease which would be more suitable for Bacillus Calmette-Guerin (BCG) treatment. Seven patients were deemed not suitable for PSDIVC due to muscle invasive disease.
In both groups administration of PSDIVC was lower than expected. There has been some suggestion that PSDIVC in other centres has also been modest (3). A national study completed in the United States of 1010 patients with NMIBC found that only 16.9% were given a post-operative dose of intravesical chemotherapy (10). A national snapshot audit study in the United Kingdom showed more promising results. Gan et al. asked all urological surgeons in the UK to send details of one patient with newly diagnosis bladder cancer, 192 consultants replied, of which 61% of patients were given mitomycin C post-operatively (11).
Attitudes of practice towards IVC have also been slow to progress. A national survey of 269 urologists in the United States found that 61% never use post-operative IVC and only 8% use IVC frequently or always (10). Another survey of 104 urologists determined 5 key reasons why they were reluctant to give IVC following TURBT. These included (a) reluctance to give chemotherapy until histological confirmation (b) uncertainty of tumour invasiveness, (c) pharmacy logistics, (d) suspected bladder perforation and (e) toxicity (10). Lack of evidence was also stated as a possible reason why IVC was not given. Factors such as pharmacy logistics and lack of knowledge of recent evidence are obstacles that may be overcome to improve installation rates of post-operative MMC and improve adherence to EAU guidelines for NMIBC.
The obstacles we encountered in administration of adjuvant chemotherapy are likely to be shared in similar centres. Such intervention bundles could therefore be introduced at other urological centres to tackle obstacles such as awareness of indications for adjuvant chemotherapy, pathways for delivery of controlled medications and poor documentation.
Improvements were seen in rates of administration and documentation of contraindications to improve compliance with EAU guidelines. A pathway including improvement in pharmacy logistics and awareness of latest research may help to overcome barriers preventing the administration of post-operative adjuvant chemotherapy.
Footnotes
Authors’ Contribution:Ben Tschobotko and Evangelos Mazaris designed the project. Luke Stroman, Ben Tschobotko, David Ellis, Elsie Mensah and Harikesh Kaneshayogan acquired data. Luke Stroman and Ben Tschobotko statistically analysed data. Luke Stroman drafted the manuscript. Luke Stroman, Hamid Abboudi and Evangelos Mazaris critically revised the manuscript for intellectual content. Evangelos Mazaris supervised the study.
References
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17399 | https://sixsigmadsi.com/glossary/type-i-error/ | Type I errors play a significant role in statistical hypothesis testing. It occurs when researchers mistakenly reject a null hypothesis that is actually true.
Understanding Type I errors is crucial in any field that relies on data analysis and hypothesis testing, such as medicine, psychology, and economics. This error can lead to incorrect conclusions and inappropriate decisions, so minimizing its occurrence is vital in research design.
Table of contents
What is a Type I Error?
Difference Between Type I and Type II errors
Why Does It Happen?
How Type I Error Occurs?
The Null Hypothesis
Types of Errors in Hypothesis Testing
The Rejection Region
How to Identify Type I errors?
Real-World Examples
Why We Care About Type I Errors?
Controlling Type I Errors
Final Words
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What is a Type I Error?
A Type I error, also known as a false positive, happens when a statistical test rejects the null hypothesis when it is actually true. Essentially, the researcher concludes there is an effect or difference when, in fact, there is none.
For instance, in a medical trial, a Type I error might occur if a new treatment is mistakenly considered effective when it has no real benefit.
It is the error of accepting the alternative hypothesis (the hypothesis we wish to test) when the difference or effect we observe is simply due to random variation. The probability of a Type I error is represented by the significance level, often denoted as α.
A Type I error happens when you say something is true, but it’s actually false. Imagine you’re testing if a new medicine works. You find it does, but really, it doesn’t. This is a Type I error. It’s a “false positive.” You reject the idea of “no effect” when “no effect” is right.
Difference Between Type I and Type II errors
| | | |
---
| Feature | Type I Error (False Positive) | Type II Error (False Negative) |
| Definition | Rejecting a true null hypothesis. Saying there is an effect when there isn’t. | Failing to reject a false null hypothesis. Saying there is no effect when there is. |
| Outcome | Finding a “false positive” result. | Missing a “true positive” result. |
| Probability | Represented by alpha (α), the significance level. | Represented by beta (β). |
| What it means | Incorrectly concluding that there is a significant effect. | Incorrectly concluding that there is no significant effect. |
| Real-world analogy | A medical test saying a healthy person has a disease. | A medical test saying a sick person is healthy. |
| Consequences | Risk in the legal system | This may lead to unnecessary actions or interventions. |
| Risk in legal system | Convicting an innocent person. | Acquitting a guilty person. |
| Relation to Null Hypothesis | Occurs when a true null hypothesis is rejected. | Occurs when a false null hypothesis is failed to be rejected. |
| How to decrease the chance | To decrease type 1 error risk, lowering the alpha value is used. | This may lead to missed opportunities for effective actions or interventions. |
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Why Does It Happen?
Statistics uses random samples. Samples might not perfectly show the whole group. Even if there’s no real difference, samples can show a difference by chance. This chance difference can trick us.
How Type I Error Occurs?
Type I error occurs in the context of hypothesis testing. In a typical experiment, a null hypothesis (H0) is tested against an alternative hypothesis (H1). The null hypothesis assumes that there is no effect or difference. Researchers conduct tests to see if the data provides sufficient evidence to reject this null hypothesis in favor of the alternative hypothesis.
When researchers reject the null hypothesis without sufficient evidence, they commit a Type I error. This error happens when the data shows a significant result, but the result is purely due to chance rather than a true effect.
The Null Hypothesis
We start with a “null hypothesis.” It says there’s no difference. For example, “The new medicine is no better than the old one.” We test if this is likely. If we reject it, we say there’s a difference.
Types of Errors in Hypothesis Testing
In hypothesis testing, there are two main types of errors: Type I and Type II.
Type I error (α error): Rejecting the null hypothesis when it is true.
Type II error (β error): Failing to reject the null hypothesis when it is false.
While Type I error involves detecting an effect when there is none, Type II error refers to failing to detect an effect when one actually exists. Both types of errors are inevitable in hypothesis testing, but researchers aim to minimize them.
The Rejection Region
The rejection region is a critical concept in hypothesis testing. It represents the area in the distribution of the test statistic where the null hypothesis is rejected. This region is determined by the significance level, α, which is typically set to 0.05 or 0.01.
If the test statistic falls within this rejection region, the null hypothesis is rejected, leading to a decision that there is a significant effect or difference. However, if the null hypothesis is actually true, there is still a small probability (α) that the test statistic could fall in the rejection region due to random variation, leading to a Type I error.
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How to Identify Type I errors?
Identifying a Type I error involves understanding the context of hypothesis testing and recognizing when a statistically significant result might be a false positive. Here’s a breakdown of how to approach this:
Understand the Null Hypothesis: Begin by clearly defining the null hypothesis. This is the statement of “no effect” or “no difference” that you’re testing. Example: “There is no difference in the effectiveness of Drug A and a placebo.”
Recognize Statistical Significance: A Type I error occurs when a statistical test leads you to reject the null hypothesis when it’s actually true. This rejection is typically based on a p-value being less than the chosen significance level (alpha, α). So, when you see a result declared “statistically significant,” it means there’s a possibility of a Type I error.
Consider the Significance Level (Alpha): The alpha level (e.g., 0.05) represents the probability of making a Type I error. If you set alpha at 0.05, you’re accepting a 5% risk of falsely rejecting the null hypothesis. Therefore, even with statistically significant results, there’s always that chance.
Evaluate the Context:
Multiple Comparisons: If you perform numerous statistical tests, the chance of a Type I error increases. Be wary of studies with many tests, especially if they only highlight a few “significant” results.
P-hacking: This refers to practices that manipulate data or analyses to achieve a statistically significant p-value. This greatly increases the risk of type 1 errors.
The nature of the study: Consider the likeliness of the alternative hypothesis being true. If a study claims to have found an extremely unlikely result, then it is more likely that a type 1 error has occurred.
Real-World Examples
Medical Tests: A test says you have a disease, but you don’t. This is a Type I error. It can cause stress and unnecessary treatment.
Court Cases: A jury finds someone guilty, but they’re innocent. This is a Type I error. It has serious consequences.
Factory Quality Control: A machine is said to be faulty, but it’s working fine. This is a Type I error. It can lead to wasted time and money.
Scientific Research: A study finds a new drug works, but it doesn’t. This is a Type I error. It can mislead future research.
Security Systems: An alarm goes off, but there’s no intruder. This is a Type I error. It can cause panic and wasted resources.
Why We Care About Type I Errors?
False Claims: Type I errors spread false information.
Wasted Resources: They can lead to wasted time, money, and effort.
Harm: In medicine or safety, they can cause harm.
Erosion of trust: Too many false positives erode trust in research.
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Controlling Type I Errors
Set a Lower Alpha: Use a smaller significance level (e.g., 0.01). This reduces the risk, but it also increases the risk of a Type II error (missing a real effect).
Bonferroni Correction: When doing many tests, adjust the alpha level. This helps control the overall Type I error rate.
Careful Study Design: Good planning and data collection reduce errors.
Replication: Repeat studies to check for consistent results.
Pre-registration: Publicly state your hypothesis and methods before starting the research. This prevents “p-hacking” (manipulating data to get a significant result).
Final Words
Type I error is a critical concept in hypothesis testing. It represents the erroneous rejection of a true null hypothesis, leading to false positives. While Type I errors cannot be completely avoided, researchers can minimize their likelihood by carefully selecting the significance level (α) and employing rigorous statistical methods.
Type I error has significant consequences, especially in fields like medicine, where false positives can lead to harmful decisions. By understanding Type I error and its implications, researchers can ensure that their findings are both valid and reliable, contributing to more accurate conclusions and better decision-making in various fields.
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