id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
13600 | https://www.practicalclinicalskills.com/ultrasound/87/m-mode-for-pneumothorax | M-Mode for Pneumothorax
Pneumothorax -> M (Motion) Mode
Narration
One way that we could represent the motion from lung sliding in a still image is to use M-mode or motion mode, and this is where you take a one dimensional line put it across the pleural line and look at it over time and when you do that you can see that the portion of the image on the left has an area where there is motion represented in the inferior half of the field, this is called the “seashore sign” it looks sort of like waves coming into the beach and its nice and happy and normal.
Pneumothorax -> M (Motion) Mode
Narration
Now if you’re looking at the pleural line and there is no motion we’ll get what we call the barcode sign also sometimes called the stratosphere sign and it basically shows there is no motion and it will look the same throughout the field.
PTX - M (Motion) Mode – Seashore Vs. Bar Code
Normal lung sliding
PTX - no sliding
Narration
So here’s a comparison labeled the seashore sign which is using m-mode showing motion on the left and that’s normal and then the barcode or stratosphere sign on the right where you don’t have motion at the pleural line so it looks the same from front to back on the left portion of the image.
Lessons
| | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- |
| | Approach to Thoracic Ultrasound. ||
| | The Inferior Thoracic Space. ||
| | Pleural Effusions. ||
| | More Pleural Effusions. ||
| | The Anterior Pleural Line. ||
| | Ultrasound for Pneumothorax. ||
| ✓ | M-Mode for Pneumothorax. ||
| | Lung Point Sign. ||
| | The ABZ Lines. ||
| | B lines and AIS. ||
| | Pneumonia. ||
| | Rib Fracture. ||
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Closed Intervals
Solution Counting
Trigonometric Expressions
Solving Equations
Mathematical Equations
Mathematical Solutions
Trigonometric Functions
Functions and Equations
5
How do I find the number of solutions of a trigonometric equation in any given interval?
Mario-César Suárez Arriaga
PhD in Applied Mathematics & Geothermal Energy, National Autonomous University Of Mexico (UNAM) (Graduated 2000)
·
Author has 84 answers and 219.4K answer views
·
7y
There is no general formula or theorem for that. I strongly suggest you to make first a plot of the trigonometric equation in the given fixed interval. You can easily see how important is the visualization in this apparently simple example of Sin(2
x
) + Cos(5
x
^2):
There is no general formula or theorem for that. I strongly suggest you to make first a plot of the trigonometric equation in the given fixed interval. You can easily see how important is the visualization in this apparently simple example of Sin(2
x
) + Cos(5
x
^2):
Dasun Medagoda
Studied Mathematics & Trigonometry (mathematics) at Ananda College
·
5y
Actually, there are infinite solution to an any given trigonometric equation. But if you limit the values you may get limited number of results.
I order to find the solutions you can derive a general formula for each trigonometric ratio.
General formula for sine:
(n180)+(-1)^n{The principle value} where n is any integer value
General formula for cosine:
(n360)(+/-){The principle value} where n is any integer value
General formula for tangent:
n180+{The principle value} where n is any integer value
you can substitute any integer for n and get an angle as a result until the answers are out of the ran
Actually, there are infinite solution to an any given trigonometric equation. But if you limit the values you may get limited number of results.
I order to find the solutions you can derive a general formula for each trigonometric ratio.
General formula for sine:
(n180)+(-1)^n{The principle value} where n is any integer value
General formula for cosine:
(n360)(+/-){The principle value} where n is any integer value
General formula for tangent:
n180+{The principle value} where n is any integer value
you can substitute any integer for n and get an angle as a result until the answers are out of the range.
Also please note that these equations may give two identical results. So take only one.
Gregory Allen
MS in Mathematics, University of Florida (Graduated 1987)
·
Author has 2.3K answers and 3.4M answer views
·
7y
If you’re looking for a general formula then I don’t believe there is one. You can’t even rely on knowing the period and the length of the interval. For example, the intervals [0, pi] and [.001, pi + .001] both have the same length but the function sin(x) = 0 has two solutions on the first interval but just one solution on the second.
Related questions
Can someone give an easy method to find the number of solutions of a trigonometric equation?
What is the number of solutions for the trigonometric equation 5sin x = x on the interval x in [0,2 pi]?
How do I solve a trigonometric equation in an interval?
Can one trigonometric equation have two general solutions?
Can you give me some examples of trigonometric equations and their solutions?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.5M answer views
·
4y
Related
What is the number of solutions for the trigonometric equation 5sin x = x on the interval x in [0,2 pi]?
Just draw a graph and have a look!
As you can see there are THREE intersections in total but you only asked for intersections in [ 0, 2π ] so there are only TWO.
Incidentally the actual solutions are x = 0 and x = ± 2.596
Just draw a graph and have a look!
As you can see there are THREE intersections in total but you only asked for intersections in [ 0, 2π ] so there are only TWO.
Incidentally the actual solutions are x = 0 and x = ± 2.596
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1y
To find the number of solutions of a trigonometric equation in a given interval, you can follow these steps:
Identify the Equation and Interval
Start with the specific trigonometric equation (e.g., [math]\sin(x) = a[/math], [math]\cos(x) = b[/math], or [math]\tan(x) = c[/math]).
Determine the interval [math][a, b][/math] within which you want to find the solutions.
Analyze the Trigonometric Function
Understand the behavior of the trigonometric function over its period:
Sine and Cosine have a period of [math]2\pi[/math].
Tangent has a period of [math]\pi[/math].
Find the General Solutions
For equations like [math]\sin(x) = a[/math]:
The general solutions can be expressed as:
[math]x = \arcsi[/math]
To find the number of solutions of a trigonometric equation in a given interval, you can follow these steps:
Identify the Equation and Interval
Start with the specific trigonometric equation (e.g., [math]\sin(x) = a[/math], [math]\cos(x) = b[/math], or [math]\tan(x) = c[/math]).
Determine the interval [math][a, b][/math] within which you want to find the solutions.
Analyze the Trigonometric Function
Understand the behavior of the trigonometric function over its period:
Sine and Cosine have a period of [math]2\pi[/math].
Tangent has a period of [math]\pi[/math].
Find the General Solutions
For equations like [math]\sin(x) = a[/math]:
The general solutions can be expressed as:
[math]x = \arcsin(a) + 2k\pi[/math] and [math]x = \pi - \arcsin(a) + 2k\pi[/math] for any integer [math]k[/math].
For [math]\cos(x) = b[/math]:
The general solutions are:
[math]x = \arccos(b) + 2k\pi[/math] and [math]x = -\arccos(b) + 2k\pi[/math].
For [math]\tan(x) = c[/math]:
The general solution is:
[math]x = \arctan(c) + k\pi[/math].
Determine the Relevant Solutions in the Interval
Substitute integer values for [math]k[/math] to find solutions within the specified interval [math][a, b][/math]:
For each expression derived from the general solutions, increment [math]k[/math] starting from the lowest integer that keeps the solution within the interval until the highest integer that does the same.
Count the Solutions
Count all unique solutions that fall within the interval. Be mindful of solutions that may overlap or fall outside the interval.
Example
Consider the equation [math]\sin(x) = \frac{1}{2}[/math] and the interval [math][0, 2\pi][/math]:
General Solutions:
[math]x = \frac{\pi}{6} + 2k\pi[/math] and [math]x = \frac{5\pi}{6} + 2k\pi[/math].
Finding Solutions in [math][0, 2\pi][/math]:
For [math]k = 0[/math]:
[math]x = \frac{\pi}{6}[/math] (valid)
[math]x = \frac{5\pi}{6}[/math] (valid)
For [math]k = 1[/math]:
[math]x = \frac{\pi}{6} + 2\pi[/math] (out of range)
[math]x = \frac{5\pi}{6} + 2\pi[/math] (out of range)
Count:
There are 2 solutions in the interval [math][0, 2\pi][/math]: [math]\frac{\pi}{6}[/math] and [math]\frac{5\pi}{6}[/math].
Conclusion
By following these steps, you can systematically find the number of solutions of a trigonometric equation within any specified interval.
Dave Clark
Studied Mathematics at Queens' College, Cambridge
·
Author has 1.8K answers and 7.1M answer views
·
Updated 7y
Related
How do I find all the solutions for the trigonometric equation sin3x=-1/root2, given that -180 degrees 180 degrees?
I would solve this in two steps:
Find all the solutions for the equation;
Select those solutions that fall within the required range.
First we find all solutions for [math]\sin{3x} = -\frac{1}{\sqrt{2}}[/math].
The principal solution for [math]\sin{\theta} = -\frac{1}{\sqrt{2}}[/math] is [math]\theta = -45^\circ[/math].
The [math]\sin[/math] function is even around [math]-90^\circ[/math] so this gives us a second solution of [math]\theta = -135^\circ[/math], and these are the only two solutions in the range [math]- 180^\circ < \theta \le 180^\circ[/math].
The [math]\sin[/math] function is periodic with period [math]360^\circ[/math], so the full set of solutions is [math]\theta = -45^\circ + 360n^\circ[/math] or [math]-135^\circ + 36[/math]
I would solve this in two steps:
Find all the solutions for the equation;
Select those solutions that fall within the required range.
First we find all solutions for [math]\sin{3x} = -\frac{1}{\sqrt{2}}[/math].
The principal solution for [math]\sin{\theta} = -\frac{1}{\sqrt{2}}[/math] is [math]\theta = -45^\circ[/math].
The [math]\sin[/math] function is even around [math]-90^\circ[/math] so this gives us a second solution of [math]\theta = -135^\circ[/math], and these are the only two solutions in the range [math]- 180^\circ < \theta \le 180^\circ[/math].
The [math]\sin[/math] function is periodic with period [math]360^\circ[/math], so the full set of solutions is [math]\theta = -45^\circ + 360n^\circ[/math] or [math]-135^\circ + 360n^\circ[/math] where [math]n \in \mathbb{Z}[/math].
Setting [math]\theta = 3x[/math] we obtain the full set of solutions for [math]x[/math]:
[math]x = -15^\circ + 120n^\circ[/math] or [math]-45^\circ + 120n^\circ[/math]
Now we select the solutions that fall within the required range [math]-180^\circ < x < 180^\circ[/math].
When [math]n = 0[/math] we have potential solutions [math]x = -15^\circ[/math] or [math]-45^\circ[/math].
When [math]n = 1[/math] we have potential solutions [math]x = 105^\circ[/math] or [math]75^\circ[/math].
When [math]n = 2[/math] we have potential solutions [math]x = 225^\circ[/math] or [math]195^\circ[/math].
When [math]n = -1[/math] we have potential solutions [math]x = -135^\circ[/math] or [math]-165^\circ[/math].
When [math]n = -2[/math] we have potential solutions [math]x = -255^\circ[/math] or [math]-285^\circ[/math].
Values of [math]n > 2[/math] or [math]n < -2[/math] give solutions clearly outside the required range.
This gives the following solutions: [math]\boxed{ x = -165^\circ, -135^\circ, -45^\circ, -15^\circ, 75^\circ, 105^\circ }[/math]
Related questions
What is an example of a trigonometric equation that has two solutions?
How do you solve an equation on trigonometric functions?
How would I solve this type of trigonometry equation?
How do I solve the trigonometric equation tg (180 -2x) =-1?
How do you solve a trigonometric equation graphically?
Reuven Harmelin
Lecturer at Technion - Israel Institute of Technology (1982–present)
·
Author has 2.3K answers and 1.9M answer views
·
4y
Related
What is the number of solutions for the trigonometric equation 5sin x = x on the interval x in [0,2 pi]?
You may ask the question, how many times the graph of
intersects the horizontal line y=1, for a=5 and x in the interval [0,2pi], and add to that number of intersections the trivial solution x=0 to your equation.
First, notice that f(0)=5>1, and also f(pi)=0<1, and for pi<x<2pi we have
f(x)<0<1, and therefore, the graph of f(x) intersects the line y=1 only inside the interval [0,pi]. In order to prove that the number of theses intersections is exactly 1, we just need to show that f(x) is strictly decreasing in that interval [0,pi]. Indeed
and since cos(x) and x-tan(x) are of opposite signs in (0.pi
You may ask the question, how many times the graph of
intersects the horizontal line y=1, for a=5 and x in the interval [0,2pi], and add to that number of intersections the trivial solution x=0 to your equation.
First, notice that f(0)=5>1, and also f(pi)=0<1, and for pi<x<2pi we have
f(x)<0<1, and therefore, the graph of f(x) intersects the line y=1 only inside the interval [0,pi]. In order to prove that the number of theses intersections is exactly 1, we just need to show that f(x) is strictly decreasing in that interval [0,pi]. Indeed
and since cos(x) and x-tan(x) are of opposite signs in (0.pi/2) and in (pi/2,pi), we deduce the derivative f’(x) is negative in (0,pi), which completes our discussion. Hence, the answer to your question is 2.
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With today’s modern day tools there can be an overwhelming amount of tools to choose from to build your own website. It’s important to keep in mind these considerations when deciding on which is the right fit for you including ease of use, SEO controls, high performance hosting, flexible content management tools and scalability. Webflow allows you to build with the power of code — without writing any.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.5M answer views
·
7y
Related
How can I find the periodicity of a trigonometric function?
I like the following method for understanding “period” of trig functions.
This applies to both sine graphs and cosine graphs so I will just focus on sine graphs.
The basic y = sin(x) graph has one complete wave or period from x = 0 to 360 degrees.
I prefer to work in degrees because they are more meaningful and familiar to students.
___________________________________________________________________________________
Consider y = sin(2x)
What we say to ourselves is “When is 2x equal to 360 ?”
If 2x = 360 then x = 180 so one complete period is from x = 0 to 180.
__________________________________________
I like the following method for understanding “period” of trig functions.
This applies to both sine graphs and cosine graphs so I will just focus on sine graphs.
The basic y = sin(x) graph has one complete wave or period from x = 0 to 360 degrees.
I prefer to work in degrees because they are more meaningful and familiar to students.
___________________________________________________________________________________
Consider y = sin(2x)
What we say to ourselves is “When is 2x equal to 360 ?”
If 2x = 360 then x = 180 so one complete period is from x = 0 to 180.
___________________________________________________________________________________
Consider y = sin(3x)
We say to ourselves is “When is 3x equal to 360?”
If 3x = 360 then x = 120 so one complete period is from x = 0 to 120
______________________________________________________________________________
Similarly if y = sin(6x) we say “When is 6x = 360?” and we get x = 60
And if y = sin( ½ x) we say “When is ½ x = 360?” and we get x = 720
_____________________________________________________________________________
Final note: The only thing affecting the period of y = a + b sin (nx + p)
is the variable “n”.
If you want to work in radians, to find the period of a function like:
y = a + b sin (nx + p)
________________________________________________________________________
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.5M answer views
·
3y
Related
How do I find all the solutions of sin(x) =0.4?
I order to really understand this and not just apply a general formula just follow this explanation…
These are the best DEFINITIONS of sine, cosine and tangent.
In the diagram below we imagine that OP can rotate about the origin.
This is called a “unit circle” because OP = 1 unit.
Angles are measured from the positive x axis in an anti-clockwise direction.
(Negative angles are measured in a clockwise direction.)
We don’t need to refer to SOH CAH TOA for this work.
The definitions of sin (θ), cos (θ) and tan (θ) are clearly shown on this diagram.
Putting all this together, if sin (x) = 0.4, the very ba
I order to really understand this and not just apply a general formula just follow this explanation…
These are the best DEFINITIONS of sine, cosine and tangent.
In the diagram below we imagine that OP can rotate about the origin.
This is called a “unit circle” because OP = 1 unit.
Angles are measured from the positive x axis in an anti-clockwise direction.
(Negative angles are measured in a clockwise direction.)
We don’t need to refer to SOH CAH TOA for this work.
The definitions of sin (θ), cos (θ) and tan (θ) are clearly shown on this diagram.
Putting all this together, if sin (x) = 0.4, the very basic angle is 23.6 degrees but to get the one in the second quadrant we find 180 – 23.6 = 156.4 degrees
But we can ADD or subtract 360 degrees to either of these any number of times and the angles will be in the same positions with the same RED sine value as in the diagrams.
So the general solution is x = 23.6 +360n or 156.6 + 360n
where n is any integer.
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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Aaron Briseno
B.S in Mathematics & Teaching, University of California, Los Angeles (Graduated 2010)
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Author has 1.3K answers and 3.1M answer views
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How do I solve a trigonometric equation in an interval?
Step 1: Obtain:
f(x)g(x)=0
OR
f(x)=trig_value (well-known)
Where f(x) and g(x) involve a single type of trig expression.
Step 2: Think to yourself
“Which quadrants does my given trig function exist in the manner described?”
Step 3: Think to yourself
“Which reference angle deals with the trig value mentioned in step 2/the equation.”
Step 4: Set the inside = to references above don’t forget to apply period rules to the trig functions.
Your example
[math]\sin(2x)=-\dfrac{\sqrt{3}}{2}[/math]
Step 1 is done.
Step 2: Sine is negative in QIII and QIV
Step 3: sine is square-root 3 over 2 at all the “tall triangles” IE reference
Step 1: Obtain:
f(x)g(x)=0
OR
f(x)=trig_value (well-known)
Where f(x) and g(x) involve a single type of trig expression.
Step 2: Think to yourself
“Which quadrants does my given trig function exist in the manner described?”
Step 3: Think to yourself
“Which reference angle deals with the trig value mentioned in step 2/the equation.”
Step 4: Set the inside = to references above don’t forget to apply period rules to the trig functions.
Your example
[math]\sin(2x)=-\dfrac{\sqrt{3}}{2}[/math]
Step 1 is done.
Step 2: Sine is negative in QIII and QIV
Step 3: sine is square-root 3 over 2 at all the “tall triangles” IE reference angle of [math]\pi/3[/math], with period of 2pi
Step 4: [math]\pi/3[/math] angle in QIII is [math]4\pi/3[/math] and in QIV we are talking about [math]5\pi/3[/math]
[math]2x=\dfrac{4\pi}{3}+2\pi n[/math]
[math]2x=\dfrac{5\pi}{3}+2\pi n[/math]
Solve for x
[math]x=\dfrac{2\pi}{3}+\pi n[/math]
[math]x=\dfrac{5\pi}{6}+\pi n[/math]
Since we are on the interval [math]0[/math] to [math]2\pi[/math] we want to make sure our answers’ numerators are not more than twice the denominator. And list the answers
[math]x=2\pi/3, 5\pi/3, 5\pi/6, 11\pi/6[/math]
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Upvoted by
Richard Shearer
, MSc (Biochemistry) Biochemistry & Mathematics, University of the Witwatersrand · Author has 6.8K answers and 52.5M answer views
·
Updated 1y
Related
How do you find the general solution of a trigonometric equation?
I always advise teachers to use what I call the “understanding method” for finding the general solutions as opposed to the “rote method” of just substituting in the following formulas:
Students can follow and mechanically substitute the relevant values but clearly there is no real understanding of WHY these equations give the “general solutions”.
_____________________________________________________
I always advise teachers to use what I call the “understanding method” for finding the general solutions as opposed to the “rote method” of just substituting in the following formulas:
Students can follow and mechanically substitute the relevant values but clearly there is no real understanding of WHY these equations give the “general solutions”.
____________________________________________________________________
All trig equations, after various manipulations, end up as simple cases such as:
sinθ = ½ or cosθ = ½ or tanθ = 1 (I only chose these constants for convenience!)
____________________________________________________________________
This is the UNDERSTANDING Method:
Consider the equation: sinθ = ½
The basic solution is 30 degrees but there are TWO places from 0 to 360 degrees where sinθ = ½
The student knows “sine” is positive in 1st and 2nd quadrants
and basic angle is θ =30 degrees.
The solutions in 0 ≤ θ ≤ 360 are 30 degrees and 150 degrees but we can add 360 to each of these and 30 + 360 is in the same position as 30 so sin390 = ½
Also 150 + 360 is in the same position as 150 so sin 510 = ½
We can add any number of 360 degrees and the sine of the angle stays at ½
So the General solution is:
At first, students “think” in degrees and I suspect most teachers do too but the equivalent general solution in radians is of course:
____________________________________________________________________
Similarly, let’s consider the equation cosθ = ½
The basic solution is 60 degrees but there are TWO places from 0 to 360 degrees
where cosθ = ½
The student knows “cos” is positive in 1st and 4th quadrants
and the basic angle is θ = 60 degrees and the other is 300 degrees.
The solutions in 0 ≤ θ ≤ 360 are 60 degrees and 300 degrees but we can add 360 to each of these and 60 + 360 is in the same position as 60 so cos 420 = ½
Also 300 + 360 is in the same position as 300 so sin660 = ½
So the General solution in degrees is:
and in radians, the general solution is:
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math
·
Author has 9K answers and 23.2M answer views
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7y
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How do I find all the solutions for the trigonometric equation sin3x=-1/root2, given that -180 degrees 180 degrees?
Solve: sin(3x) = [math]\frac{-1}{\sqrt{2}}[/math] for x in ( -180° , 180° )
First: Review what you should have already memorized:
One of the first things I did in Trig class was memorize the sines and cosines of 0°, 30°, 45°, 60° and 90°:
sin(00°) = [math]\frac{\sqrt{0}}{2}[/math] = cos(90°)
sin(30°) = [math]\frac{\sqrt{1}}{2}[/math] = cos(60°)
sin(45°) = [math]\frac{\sqrt{2}}{2}[/math] = cos(45°)
sin(60°) = [math]\frac{\sqrt{3}}{2}[/math] = cos(30°)
si
Solve: sin(3x) = [math]\frac{-1}{\sqrt{2}}[/math] for x in ( -180° , 180° )
First: Review what you should have already memorized:
One of the first things I did in Trig class was memorize the sines and cosines of 0°, 30°, 45°, 60° and 90°:
sin(00°) = [math]\frac{\sqrt{0}}{2}[/math] = cos(90°)
sin(30°) = [math]\frac{\sqrt{1}}{2}[/math] = cos(60°)
sin(45°) = [math]\frac{\sqrt{2}}{2}[/math] = cos(45°)
sin(60°) = [math]\frac{\sqrt{3}}{2}[/math] = cos(30°)
sin(90°) = [math]\frac{\sqrt{4}}{2}[/math] = cos(00°)
Do you see the pattern? 0, 1, 2, 3, 4
I also memorized the period of the various functions.
The period of the sine wave is 360°, so if sin(30) = ½, then so does sin(30+360°) and sin(30–360°)
The next thing I memorized was the standard pattern of the sine and cosine waves.
The sine wave pattern tells me if sin(45°) = [math]\frac{\sqrt{2}}{2}[/math], then I can change both numbers to negative and see that:
sin(-45°) = [math]\frac{-\sqrt{2}}{2}[/math]
and visualizing the curve tells me that sin(-135°) has the same sine value.
Second: Simplify the problem and solve the simpler problem:
Let: z = 3x
Solve the simpler equation: sin(z) = [math]\frac{-1}{\sqrt{2}}[/math]
From what we memorized and figured out above:
z = -45° and -135°
period of z = 360°
We said that z = 3x, so solve:
3x = -45° and -135° → → x = -15° and -45°
period of 3x = 360° → → period of x = 120°
What I did was divide both sides of each equation by 3
Third: Look again at the original equation:
sin(3x) = [math]\frac{-1}{\sqrt{...
Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
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Author has 3.9K answers and 4M answer views
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Related
What is the solutions of the trigonometric equation 2cos² x-3cos x+1 =0?
Let [math]y=\cos{(x)}[/math], then:
Therefore, we see that either [math]y=\frac{1}{2}[/math] or [math]y=1[/math]. Since [math]y=\cos{(x)}[/math], we have:
[math]x=\frac{\pi}{3}+2\pi n[/math]
[math]x=\frac{5\pi}{3}+2\pi m[/math]
[math]x=2\pi q[/math]
where [math]n[/math], [math]m[/math] and [math]q[/math] are integers.
These are the roots of the function:
which looks like this:
Let [math]y=\cos{(x)}[/math], then:
[math]2y^2-3y+1=0[/math]
math(y-1)=0[/math]
Therefore, we see that either [math]y=\frac{1}{2}[/math] or [math]y=1[/math]. Since [math]y=\cos{(x)}[/math], we have:
[math]\cos{(x)}=\frac{1}{2}[/math]
[math]x=\frac{\pi}{3}+2\pi n[/math]
[math]x=\frac{5\pi}{3}+2\pi m[/math]
or
[math]\cos{(x)}=1[/math]
[math]x=2\pi q[/math]
where [math]n[/math], [math]m[/math] and [math]q[/math] are integers.
These are the roots of the function:
[math]f(x)=2\cos^2{(x)}-3\cos{(x)}+1[/math]
which looks like this:
Related questions
Can someone give an easy method to find the number of solutions of a trigonometric equation?
What is the number of solutions for the trigonometric equation 5sin x = x on the interval x in [0,2 pi]?
How do I solve a trigonometric equation in an interval?
Can one trigonometric equation have two general solutions?
Can you give me some examples of trigonometric equations and their solutions?
What is an example of a trigonometric equation that has two solutions?
How do you solve an equation on trigonometric functions?
How would I solve this type of trigonometry equation?
How do I solve the trigonometric equation tg (180 -2x) =-1?
How do you solve a trigonometric equation graphically?
How do I find all the solutions for the trigonometric equation sin3x=-1/root2, given that -180 degrees 180 degrees?
How do you find all trigonometric values of a trigonometry equation (trigonometry, math)?
How do you solve tan2x=4tanxsolve the trigonometric equation above?
How do you solve the trigonometric equation 2sinxsin3x = 1?
What exactly am I solving for what I attempt to solve a trigonometric equation?
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Common Ion Effect - Definition, Explanation, Examples
Last Updated : 23 Jul, 2025
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Common ion effect is a phenomenon in chemistry that describes the decrease in solubility of a compound when it is dissolved in a solution that already contains one of its constituent ions. This effect occurs due to the principle of chemical equilibrium and Le Chatelier's principle.
In this article, we will learn in detail about the common ion effect, its impact on solubility, and its validity in the case of acids, bases, and salts. We will also learn its limitations and FAQs based on the common ion effect.
What is Common Ion Effect?
Common ion effect is a chemical phenomenon that causes a decrease in the solubility of an ionic precipitate by the addition of a soluble compound with an ion in common with the precipitate. It is based on the principle of chemical equilibrium, which states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. The common ion effect is a result of Le Chatelier's principle.
Explanation of Common Ion Effect
To understand the common ion effect, let us first understand the common ion. Common ion is when addition of an ion in the solution that comes from two different sources. Imagine there are two solutions, NaCl and AgCl, where Cl is common. Hence, it becomes a common ion. So the common ion effect is when the addition of an ion, already present in the solution disrupts the equilibrium of the solution.
We know that when AgCl dissolves in water, it forms Ag+ (silver ions) and Cl- (chloride ions).
Now imagine in the NaCl solution we put some undissolved AgCl.
As the concentration of the common ion Cl- increases in the solution, according to Le Chatelier's principle, the equilibrium will shift to manage these increasing Cl- ions. As a result, AgCl will precipitate out of the solution and reduce the solubility of AgCl in solution due to presence of dissolved ions of NaCl.
Impact of Common Ion Effect on Solubility
If you observe in the above example, solubility of AgCl was affected due to common ion effect. The other impacts of common ion effect on solubility is as described below,
Decreased Solubility due to Common Ions
As shown in NaCl and AgCl example, the equilibrium shifts due to excess chloride ions and reduces the solubility of AgCl. According to Le Chatelier's principle, the solution precipitate out AgCl and reduces its solubility in NaCl solution due to the presence of common chloride ions.
Common Ion Effect on pH
Common ions may result in effects of disassociation of weak acid and bases in order to maintain equilibrium in the solution. It Influences the degree of dissociation of weak acids and bases by shifting the equilibrium in a particular direction that reduces the concentration of common ion and hence disrupts the pH level of solution. Let us see Common Ion Effect in both strong and weak acids and bases.
Common Ion Effect in Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Common Ion Effect in Weak Bases
Let us consider one weak base ammonia(NH 3), we will add ammonium chloride which will decrease the dissociation of ammonia and disrupts the impact of common ion on ammonia, which will reduce basicity i.e.pHdecreases of the solution.
Dissociation of ammonia in water
NH3+ H2O → NH4+ OH-
When we add ammonium chloride (NH 4 Cl)
NH4Cl → NH4+ Cl-
When Ammonium Chloride is added to Dissolved ammonia solution, the extra NH 4 and the common ion effect results in a decrease of OH- and hence reduces basicity of Ammonia in water.
Common Ion Effect in Weak Acids
When acetic acid(weak acid) andsodium acetate(strong electrolyte) generate acetate ions, where sodium acetate is dissolved completely but acetic acid only partly ionizes, since common ion effect results in reduced dissociation of acetic acid, pH increases and will be less acidic.
Dissociation of acetic acid in water
CH3COOH(acetic acid) + H2O(water) → CH3COO-(acetate ion) + H3O+(Hydronium ion)
When we add sodium acetate
CH3COONa → CH3COO-+ Na+
which decreases the concentration of H 3 O+ and makes the solution less acidic.
Common Ion Effect in Strong Acids and Bases
Solutions Containing Strong Acids and Bases, its common ion effect is minimal or negligible as these elements completely dissociate in water. For example, take one strong acid like HCl which dissociates in H+ and Cl-. When we add NaCl on this, it will have negligible effect as H+ ions is already completely ionized.
Common Ion Effect in Salts
Common Ion Effect in Salts leads to reduced solubility of salts when it already contains one of the ions. The effect is seen in the form of solubility and precipitation.
Common Ion Effect in Solubility of Salts
Solubility of salts reduces in presence of common ion. Consider NaCl, when it is dissolved in water it dissociates in sodium and chloride ions and when some more chloride ions from any other source like AgCl , reduces the solubility of the salt and may result in precipitation.
Common Ion Effect in Precipitation Reactions
In Precipitation Reactions, the common ion effect reduces the solubility and results in precipitation. The increase in common ion shifts the equilibrium and decreases the solubility.
Applications of Common Ion Effect
The applications of common ion effect is mentioned below:
Control of Precipitation: The common ion effect is utilized to control the precipitation of sparingly soluble salts. For Example, Common Ion Effect is used to extract calcium carbonate out of drinking water by adding sodium carbonate. Sodium Carbonate is highly soluble while calcium carbonate is not(sparingly soluble).This helps in treatment of water and reduce calcium carbonate in water.
Buffer Solution:The common ion effect is essential in the functioning of buffer solutions. Buffer solutions are aqueous solutions that resist changes in pH when small amounts of acid or base are added to them.
Industrial Process: Common Ion effect can also be observed in salting out process of soap manufacture. The soaps are precipitated out by adding sodium chloride to the soap solution in order to reduce its solubility.
Limitations of Common Ion Effect
Common ion effect is useful in understanding equilibrium in solutions but its affects can depend on various factors like temperature and other external factors. Some of the limitations of Common ion effect are listed below,
Common ion effect only works with strong electrolytes(where complete dissociation occurs) When weak electrolytes, affects are negligible.
Common ion effect needs relatively high concentration of ions, as a result in dilute solutions, the changes are negligible.
Sometimes, In Precipitation reactions, if the ratio of common ion and solution is not appropriate, precipitation will occur without common ion effect.
Changes in temperature affect ionization and solubility equilibrium.
Conclusion
Common ion effect gives an appropriate idea of equilibrium in any solution which we have to handle carefully(take care of temperature, the presence of ions and strong electrolytes). The effect on its solubility, common ion effects in salt and common ion effects in acid or bases is discussed thoroughly in the article.
| Related Articles |
| Ionic Equilibrium | Solubility Equilibria |
| Solubility Formula | What do All Bases Have in Common |
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13603 | https://bio723-class.github.io/Bio723-book/simulating-sampling-distributions.html | A Simulating sampling distributions | Biology 723: Statistical Computing for Biologists
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1 Introduction
1.1 Accessing older versions of the course notes
1.2 How to use these lecture notes
2 Getting Started with R
2.1 What is R?
2.2 What is RStudio?
2.3 Entering commands in the console
2.4 Comments
2.5 Using R as a Calculator
2.5.1 Common mathematical functions
2.6 Variable assignment
2.6.1 Valid variable names
2.7 Data types
2.7.1 Numeric data types
2.7.2 Logical values
2.7.3 Character strings
2.8 Packages
2.8.1 Installing packages from the console
2.8.2 Install the tidyverse package
2.8.3 Installing packages from the RStudio dialog
2.8.4 Loading packages with the library() function
2.9 The R Help System
2.9.1 Getting help from the console
3 R Markdown and R Notebooks
3.1 R Notebooks
3.2 Creating an R Notebook
3.3 The default R Notebook template
3.4 Code and Non-code blocks
3.4.1 Non-code blocks
3.4.2 Code blocks
3.5 Running a code chunk
3.6 Running all code chunks above
3.7 “Knitting” R Markdown to HTML
3.8 Sharing your reproducible R Notebook
4 Data structures
4.1 Vectors
4.1.1 Vector Arithmetic
4.1.2 Vector recycling
4.1.3 Simple statistical functions for numeric vectors
4.1.4 Indexing Vectors
4.1.5 Comparison operators applied to vectors
4.1.6 Combining Indexing and Comparison of Vectors
4.1.7 Vector manipulation
4.1.8 Vectors from regular sequences
4.1.9 Additional functions for working with vectors
4.2 Lists
4.2.1 Length and type of lists
4.2.2 Indexing lists
4.2.3 Naming list elements
4.2.4 The $ operator
4.2.5 Changing and adding lists items
4.2.6 Combining lists
4.2.7 Converting lists to vectors
4.3 Data frames
4.3.1 Creating a data frame
4.3.2 Type and class for data frames
4.3.3 Length and dimension for data frames
4.3.4 Indexing and accessing data frames
4.3.5 Logical indexing of data frames
4.3.6 Adding columns to a data frame
5 Functions and control flow statements
5.1 Writing your own functions
5.1.1 Function arguments
5.1.2 Writing functions with optional arguments
5.1.3 Putting R functions in Scripts
5.2 Control flow statements
5.2.1if and if-else statements
5.2.2 for loops
5.2.3break statement
5.2.4repeat loops
5.2.5next statement
5.2.6 while statements
5.2.7ifelse
5.3map and related tools
5.3.1 basic map
5.3.2map_if and map_at
5.3.3 mapping in parallel using map2
5.3.4map variants that return vectors
6 Introduction to ggplot2
6.1 Loading ggplot2
6.2 Example data set: Anderson’s Iris Data
6.3 Template for single layer plots in ggplot2
6.4 An aside about function arguments
6.5 Strip plots
6.5.1 Jittering data
6.5.2 Adding categorical information
6.5.3 Rotating plot coordinates
6.6 Histograms
6.6.1 Variations on histograms when considering categorical data
6.7 Faceting to depict categorical information
6.8 Density plots
6.9 Violin or Beanplot
6.10 Boxplots
6.11 Building complex visualizations with layers
6.12 Useful combination plots
6.12.1 Boxplot plus strip plot
6.12.2 Setting shared aesthetics
6.13 ggplot layers can be assigned to variables
6.13.1 Violin plot plus strip plot
6.14 Adding titles and tweaking axis labels
6.15 ggplot2 themes
6.15.1 Further customization with ggplot2::theme
6.16 Other aspects of ggplots can be assigned to variables
6.17 Bivariate plots
6.17.1 Scatter plots
6.17.2 Adding a trend line to a scatter plot
6.18 Bivariate density plots
6.19 Combining Scatter Plots and Density Plots with Categorical Information
6.20 Density plots with fill
6.21 2D bin and hex plots
6.22 The cowplot package
7 Introduction to dplyr
7.1 Libraries
7.2 Reading data with the readr package
7.2.1 Reading Excel files
7.2.2 Example data: NC Births
7.3 A note on “tibbles”
7.4 Data filtering and transformation with dplyr
7.5 dplyr’s “verbs”
7.5.1select
7.5.2filter
7.5.3mutate
7.5.4arrange
7.5.5summarize
7.5.6group_by
7.5.7 Combining grouping and summarizing
7.5.8 Scoped variants of mutate and summarize
7.5.9 Combining summarize with grouping aesthetics in ggplot2
7.6 Pipes
7.6.1 Install and load magrittr
7.6.2 The basic pipe operator
7.6.3 An example without pipes
7.6.4 The same example using pipes
7.6.5 Assigning the output of a statement involving pipes to a variable
7.6.6 Compound assignment pipe operator
7.6.7 The dot operator with pipes
7.6.8 The exposition pipe operator
8 Data wrangling
8.1 Libraries
8.2 Data
8.3 Loading the data
8.4 Check your data!
8.5 Renaming data frame columms
8.6 Dropping unneeded columns
8.6.1 Finding all empty columns
8.6.2 Dropping columns by matching names
8.7 Merging data frames
8.8 Reshaping data with tidyr
8.8.1 Wide to long conversions using pivot_longer()
8.8.2 Extracting information from combined variables using extract()
8.8.3 Subsetting rows
8.8.4 Splitting columns
8.8.5 Combining data frames
8.8.6 Sorting data frame rows
8.9 Using your tidy data
8.9.1 Visualizing gene expression time series
8.9.2 Finding the most variable genes
8.9.3 Heat maps
8.10 Long-to-wide conversion using pivot_wider()
8.11 Exploring bivariate relationships using “wide” data
8.11.1 Large scale patterns of correlations
8.11.2 Adding new columns and combining filtered data frames
8.11.3 A heat mapped sorted by correlations
8.11.4 A “fancy” figure
9 Vector algebra
9.1 Libraries
9.2 Vector Mathematics in R
9.3 Simple statistics in vector form
9.3.1 Mean
9.3.2 Mean centering
9.3.3 Variance and standard deviation
9.3.4 Covariance and correlation
10 Matrices in R
10.1 Creating matrices in R
10.2 Matrix arithmetic operations in R
10.3 Descriptive statistics as matrix functions
10.3.1 Mean vector and matrix
10.3.2 Deviation matrix
10.3.3 Covariance matrix
10.3.4 Correlation matrix
10.4 Matrix Inverse
10.5 Solving sets of simultaneous equations
11 Linear Regression Models I
11.1 Linear functions
11.2 Linear regression
11.3 The optimality criterion for least-squares regression
11.4 Geometry of linear regression
11.5 Solution for the least-squares criterion
11.5.1 Bivariate regression, estimating coefficients
11.5.2 Multiple regression, estimating coefficients
11.6 Residuals
11.7 Regression as sum-of-squares decomposition
11.8 Variance “explained” by a regression model
11.9 Goodness of fit
11.10 Interpretting Regression
11.11 Illustrating linear regression with simulated data
11.12 Calculating the regression model using vector arithmetic
11.12.1 Calculating the predicted values and residuals
11.13 Plotting various aspects of a regression
11.13.1 Residual plots
11.14 Calculating the coefficient of determination
12 Linear Regression Models II
12.1 New Libraries to install
12.2 Standard libraries
12.3 Specifying Regression Models in R
12.3.1 Example data
12.3.2 Fitting the model using lm()
12.3.3 Interpretting summary output from lm()
12.4 Quick bivariate regression plots in ggplot
12.5 More about the data structure returned by lm()
12.5.1 Fitted values
12.5.2 Getting the model coefficients
12.6 Broom: a library for converting model results into data frames
12.6.1broom::tidy
12.6.2broom::augment
12.6.3broom::glance
12.7 qq-plots
12.8 Multiple regression
12.8.1 Exploration of the trees data set
12.9 3D Plots
12.9.1 scatterplot3d
12.10 Fitting the regression model
12.10.1 Visualizing the regression model in scatterplot3d
12.11 Interpretting the regression model
12.12 Exploring the Vector Geometry of the Regression Model
12.13 Exploring the Residuals from the Model Fit
12.14 An alternate model
12.15 Exploring the impact of nearly collinear predictors on regression
13 Principal Components Analysis
13.1 Libraries
13.2 Matrices as linear transformations
13.3 Eigenanalysis in R
13.4 Principal Components Analysis in R
13.4.1 Bioenv dataset
13.4.2 PCA of the Bioenv dataset
13.4.3 Calculating Factor Loadings
13.5 Drawing Figures to Represent PCA
13.5.1 PC Score Plots
13.5.2 Simultaneous Depiction of Observations and Variables in the PC Space
14 Canonical Variates Analysis
14.1 Libraries
14.2 Discriminant Analysis in R
14.2.1 Shorthand Formulae in R
14.2.2 Working with the output of lda()
14.3 Estimating confidence regions for group means in CVA
14.3.1 Drawing the CVA confidence regions
14.4 Calculating the Within and Between Group Covariance Matrices
14.4.1 Recapitulating the CVA analysis of lda()
15 Clustering in R
15.1 Libraries
15.2 Data set
15.3 Hierarchical Clustering in R
15.4 Manipulating hierarchical clusterings with dendextend
15.5 Plotting dendrograms in dendextend
15.6 Cutting dendrograms
15.7 Looking at clusters
15.8 Generating a heat map from a cluster
15.9 Working with sub-trees
15.10 Setting dendrogram parameters in dendextend
15.11 Combining heatmaps and dendrograms
15.12 K-means/K-medoids Clustering in R
15.12.1 Heat map from k-medoids cluster
15.13 Combining clusters and correlation matrix heatmaps
15.13.1 Hierarchical clustering, visualized on correlation matrix
15.13.2 K-medoids clustering visualized on the correlation matrix
15.14 Depicting the data within clusters
16 Non-linear regression models
16.1 LOESS regression
16.2 Logistic regression
16.2.1 A web app to explore the logistic regression equation
16.2.2 Titanic data set
16.2.3 Subsetting the data
16.2.4 Visualizing survival as a function of age
16.2.5 Fitting the logistic regression model
16.2.6 Visualizing the logistic regression
16.2.7 Quick and easy visualization
16.2.8 Impact of sex and passenger class on the models
16.2.9 Fitting multiple models based on groupings use dplyr::do
17 Randomization, Jackknife, and Bootstrap
17.1 Libraries
17.2 Randomization
17.3 Using randomization to test for a difference in means
17.4 Using randomization to test for equality of variances
17.5 Jackknifing in R
17.6 Bootstrapping in R
17.6.1 A more sophisticated application of the bootstrap
17.6.2 About logistic regression
17.6.3 Bumpus house sparrow data set
17.6.4 Predicting survival as a function of body weight
17.6.5 Fitting the logistic regression
17.6.6 Defining an appropriate function for boot
17.6.7 Calculating bootstrap confidence intervals
17.6.8 Visualizing bootstrap confidence intervals
Appendix
A Simulating sampling distributions
A.1 Sampling distributions
A.2 Simulating sampling from an underlying population
A.3 Seeding the pseudo-random number generator
A.4 Libraries
A.4.1 Properties of the underlying population
A.4.2 Other R functions related to the normal distribution
A.5 Random sampling from the simulated population
A.5.1 Another random sample
A.5.2 Simulating the generation of many random samples
A.5.3 A function to estimate statistics of interest in a random sample
A.5.4 Generating statistics for many random samples
A.6 Simulated sampling distribution of the mean
A.6.1 Differences between sampling distributions and sample/population distributions
A.6.2 Using sampling distributions to understand the behavior of statistics of interest
A.6.3 Sampling distributions for different sample sizes
A.6.4 Discussion of trends for sampling distributions of different sample sizes
A.7 Standard Error of the Mean
A.8 Sampling Distribution of the Standard Deviation
A.8.1 Standard error of standard deviations
A.9 What happens to the sampling distribution of the mean and standard deviation when our sample size is small?
A.9.1 For small samples, sample standard deviations systematically underestimate the population standard deviation
A.9.2 Underestimates of the standard deviation for small n n lead to understimates of the SE of the mean
A.9.3 The t-distribution is the appropriate distribution for describing the sampling distribution of the mean when n n is small
B Simulating confidence intervals
B.1 Confidence intervals
B.2 Generic formulation for confidence intervals
B.3 Example: Confidence intervals for the mean
B.3.1 Simulation of means
B.3.2 Distance between sample means and true means
B.3.3 Calculating a CI
B.4 A problem arises!
B.5 Confidence intervals under sample estimates of standard errors
B.5.1 The spread of sample estimates of the mean in units of true and estimates SEs
B.5.2 Quantifying deviations of sample estimates of the mean
B.6 t-distribution for confidence intervals of the mean
B.6.1 t-distribution vs standard normal distribution
B.6.2 Formula for confidence intervals of the mean based on sample estimates of the SE and the t-distribution
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Biology 723: Statistical Computing for Biologists
A Simulating sampling distributions
In these appendices we will explore how we can use R to carry out statistical simulations. Such simulations can be useful for a variety of purposes, such as understanding classical results in statistical inference, deriving novel results for unusual statistics or underlying distributions, or to help design experiments. More generally, simulation is a useful tool for honing our intuition or revealing key properties about complex systems.
In this chapter we will focus on how we can use simulation to derive well known and not so well known results regarding sampling distributions for a variety of statistics.
A.1 Sampling distributions
Usually when we collect biological data, it’s because we’re trying to learn about some underlying “population” of interest. Population here could refer to an actual population (e.g.all males over 20 in the United States; brushtail possums in the state of Victoria, Australia), an abstract population (e.g.corn plants grown from Monsanto “round up ready” seed; yeast cells with genotypes identical to the reference strain S288c), outcomes of a stochastic process we can observe and measure (e.g.meiotic recombination in flies; rainfall per square meter in the Brazilian amazon), etc.
It is often impractical or impossible to measure all objects/individuals/instances of a population of interest, so we take a sample (ideally a random sample) from the population and make measurements on the variables of interest in that sample. We do so with the hope that the various statistics we calculate on the variables of interest in that sample will be useful estimates of those same statistics in the underlying population.
However, we must always keep in mind that the statistics we calculate from our sample will almost never exactly match those of the underlying population. That is when we collect a sample, and measure a statistic (e.g.mean or standard deviation or skew or…) on variable X X in the sample, there is a degree of uncertainty about how well our estimate matches the true value of that statistic in the underlying population. We know that if we took another random sample from the same population, and re-calculated the statistic of interest, we’re likely to obtain an estimate that differs from that in our first sample. We can imagine repeating this procedure again and again a large number of times, recording our estimate each time which would yield a new distribution, which we’ll call a sampling distribution. A sampling distribution is the probability distribution of a given statistic for samples of a given size. Sampling distributions are a linchpin at the heart of statistical inference, how we quantify the uncertainty associated with statistics and use that information to test hypotheses and evaluate models.
Traditionally sampling distributions were derived analytically. In this class session we’ll see how to approximate sampling distributions for any statistic using computer simulation.
A.2 Simulating sampling from an underlying population
To illustrate the concept of sampling distributions, we’ll simulate drawing samples for an underlying population that we’re trying to estimate statistics about. This will allow us to compare the various statistics we calculate and their sampling distributions to their “true” values.
Let’s simulate a population consisting of 25,000 individuals with a single trait of interest – height (measured in centimeters). We will simulate this data set based on information about the distribution of the heights of adult males in the US from a study carried out from 2011-2014 by the US Department of Health and Human Services1.
```
male mean height and sd in centimeters from USDHHS report
true.mean <- 175.7
true.sd <- 15.19
```
A.3 Seeding the pseudo-random number generator
When carrying out simulations, we employ random number generators (e.g.to choose random samples). Most computers can not generate true random numbers – instead they use algorithms that approximate the generation of random numbers (pseudo-random number generators). One important difference between a true random number generator and a pseudo-random number generator is that we can regenerate a series of pseudo-random numbers if we know the “seed” value that initialized the algorithm. We can specifically set this seed value, so that we can guarantee that two different people evaluating this notebook get the same results, even though we’re using (pseudo)random numbers in our simulation.
```
make our simulation repeatable by seeding RNG
set.seed(20190409)
```
A.4 Libraries
library(tidyverse)
library(magrittr)
library(stringr)
A.4.1 Properties of the underlying population
Heights in human populations are approximately normally distributed, so we’ll assume that the distribution of our simulated variable is also normally distributed. Let’s take a moment to visualize the probability distribution of of a normal distribution with a mean and standard deviation as given above.
```
pop.distn <-
data_frame(height = seq(100, 250, 0.5),
density = dnorm(height,mean = true.mean, sd = true.sd))
ggplot(pop.distn) +
geom_line(aes(height, density)) +
# vertical line at mean
geom_vline(xintercept = true.mean, color="red", linetype="dashed") +
# vertical line at mean + 1SD
geom_vline(xintercept = true.mean + true.sd,
color = "blue", linetype="dashed") +
# vertical line at mean - 1SD
geom_vline(xintercept = true.mean - true.sd,
color = "blue", linetype="dashed") +
labs(x = "Height (cm)", y = "Density",
title = "Distribution of Heights in the Population of Interest",
subtitle = "Red and blue lines indicate the mean \nand ±1 standard deviation respectively.")
```
A.4.2 Other R functions related to the normal distribution
As shown above the dnorm() function calculates the probability density at given values of a variable x, given the specified mean and standard deviation.
pnorm() gives the cumulative density function (also known as the distribution function) for the normal distribution, as shown below:
```
cdf <-
data_frame(height = seq(100, 250, 0.5),
cum.prob = pnorm(height, true.mean, true.sd))
ggplot(cdf) +
geom_line(aes(height, cum.prob)) +
labs(x = "Height", y = "Cumulative probability")
```
qnorm() is the quantile function for the normal distribution. The input is the probabilities of interest (single value or vector), and the mean and standard deviation of the distribution. The output is the corresponding value of the variable corresponding to the given percentiles. For example, to estimate the lower 30th percentile of heights in adult males in the US we can use qnorm() as follows:
```
qnorm(0.3, true.mean, true.sd)
167.7344
```
Using qnorm we find that in a population where height ~ N(175.7,15.19)N(175.7,15.19), 167.7cm is the approximate cutoff for the lower 30th percentile. We can illustrate that as shown below:
```
library(glue) # note the use of the glue library (part of tidyverse)
# to enable string literals as used in the title below
perc.30 <- qnorm(0.3, true.mean, true.sd)
label.offset <- 18 # determined by trial and error to make a
# nice looking figure
heights.less.perc.30 <- seq(100, perc.30, by=0.5)
density.less.perc.30 <- dnorm(heights.less.perc.30, true.mean, true.sd)
ggplot(pop.distn) +
geom_line(aes(x = height, y = density)) +
geom_vline(xintercept = perc.30, linetype='dashed') +
geom_area(aes(x = heights.less.perc.30, y = density.less.perc.30),
fill = "gray", data = data_frame(x = heights.less.perc.30)) +
annotate("text", x = perc.30 - label.offset, y = 0.025,
label = "30th percentile", color = 'red') +
labs(title = glue("Probability distribution as calculated by dnorm()\nand the 30th percentile as calculated by qnorm()\nfor a normal distribution ~N({true.mean},{true.sd})."))
```
A.5 Random sampling from the simulated population
Let’s simulate the process of taking a single sample of 30 individuals from our population, using the rnorm() function which takes samples if size n from a normal distribution with the given mean and standard deviation:
sample.a <-
data_frame(height = rnorm(n = 30, mean = true.mean, sd = true.sd))
Now we’ll create a histogram of the height variable in our sample. For reference we’ll also plot the probability for the true population (but remember, in the typical case you don’t know what the true population looks like)
sample.a %>%
ggplot(aes(x = height)) +
geom_histogram(aes(y = ..density..),
fill = 'steelblue', alpha=0.75, bins=10) +
geom_line(data=pop.distn, aes(x = height, y = density),
alpha=0.25,size=1.5) +
geom_vline(xintercept = true.mean, linetype = "dashed", color="red") +
geom_vline(xintercept = mean(sample.a$height), linetype = "solid") +
labs(x = "Height (cm)", y = "Density",
title = "Distribution of heights in the underlying population (line)\nand a single sample of size 30 (blue)")
The dashed vertical line represent the true mean of the population, the solid line represents the sample mean. Comparing the two distributions we see that while our sample of 30 observations is relatively small,its location (center) and spread that are roughly similar to those of the underlying population.
Let’s create a table giving the estimates of the mean and standard deviation in our sample:
```
sample.a %>%
summarize(sample.mean = mean(height),
sample.sd = sd(height))
# A tibble: 1 × 2
sample.mean sample.sd
1 176. 12.4
```
Based on our sample, we estimate that the mean height of males in our population of interest is 175.5215685cm with a standard deviation of 12.4101548cm.
A.5.1 Another random sample
Let’s step back and think about our experiment. We took a random sample of 30 individuals from the population. The very nature of a “random sample” means we could just as well have gotten a different collection of individuals in our sample. Let’s take a second random sample of 25 individuals and see what the data looks like this time:
```
sample.b <-
data_frame(height = rnorm(30, mean = true.mean, sd = true.sd))
sample.b %>%
ggplot(aes(x = height)) +
geom_histogram(aes(y = ..density..),
fill = 'steelblue', alpha=0.75, bins=10) +
geom_line(data=pop.distn, aes(x = height, y = density),
alpha=0.25,size=1.5) +
geom_vline(xintercept = true.mean, linetype = "dashed", color="red") +
geom_vline(xintercept = mean(sample.a$height), linetype = "solid") +
labs(x = "Height (cm)", y = "Density",
title = "Distribution of heights in the underlying population (line)\nand a single sample of size 30 (blue)")
```
```
sample.b %>%
summarize(sample.mean = mean(height),
sample.sd = sd(height))
# A tibble: 1 × 2
sample.mean sample.sd
1 176. 17.2
```
This time we estimated the mean height to be 175.651398 cm and the standard deviation to be 17.1780732 cm.
A.5.2 Simulating the generation of many random samples
When we estimate population parameters, like the mean and standard deviation, based on a sample, our estimates will differ from the true population values by some amount. For any given sample we can’t know how close our estimates of statistics like the mean and standard deviation are to the true population values, but we we can study the behavior of such estimates across many simulated samples and learn something about how well our estimates do on average, as well the spread of these estimates.
A.5.3 A function to estimate statistics of interest in a random sample
First we’re going to write a function called rnorm.stats that carries out the following steps:
Take a random sample of size n from a normal distribution with a given mean (mu) and standard deviation (sigma)
Calculates the mean and standard deviation of the random sample
Return a table giving the sample size, sample mean, and sample standard deviation, represented as a data frame
Take a moment to make sure you understand how this function works.
rnorm.stats <- function(n, mu, sigma) {
the.sample <- rnorm(n, mu, sigma)
data_frame(sample.size = n,
sample.mean = mean(the.sample),
sample.sd = sd(the.sample))
}
Let’s test rsample.stats() for a sample of size 30, drawn from a popultion with a mean and standard deviation corresponding to our height exmaple:
```
rnorm.stats(30, true.mean, true.sd)
# A tibble: 1 × 3
sample.size sample.mean sample.sd
1 30 176. 17.3
```
A.5.4 Generating statistics for many random samples
Now we’ll see how to combine rnorm.stats with two additional functions to repeatedly run the rsample.stats function:
df.samples.of.30 <-
rerun(2500, rnorm.stats(30, true.mean, true.sd)) %>%
bind_rows()
The function rerun is defined in the purrr library (automatically loaded with tidyverse). purrr:rerun() re-runs an expression(s) multiple times. The first argument to rerun() is the number of times you want to re-run, and the following arguments are the expressions to be re-run. Thus the second line of the code block above re-runs the rnorm.stats function 2500 times, generating sample statistics for samples of size 30 each time it’s run. rerun() returns a list whose length is the specified number of runs.
The third line includes a call the dplyr::bind_rows(). This simply takes the list that rerun returns and collapses the list into a single data frame. df.samples.of.30 is thus a data frame in which each row gives the sample size, sample mean, and sample standard deviation for a random sample of 30 individuals drawn from our underlying population with a normally distributed variable.
```
head(df.samples.of.30)
# A tibble: 6 × 3
sample.size sample.mean sample.sd
1 30 174. 12.4
2 30 178. 15.4
3 30 178. 13.3
4 30 179. 14.4
5 30 175. 12.8
6 30 177. 16.5
```
A.6 Simulated sampling distribution of the mean
Let’s review what we just did:
We generated 2500 samples of size 30 sampled from a population for which the variable of interest has a normal distribution
For each of the samples we calculated the mean and standard deviation in that sample
We combined each of those estimates of the mean and standard deviation into a data frame
The 2500 estimates of the mean we generated represents a new distribution – what we will call a sampling distribution of the mean for samples of size 30. Let’s plot this sampling distribution:
ggplot(df.samples.of.30, aes(x = sample.mean, y = ..density..)) +
geom_histogram(bins=25, fill = 'firebrick', alpha=0.5) +
geom_vline(xintercept = true.mean, linetype = "dashed", color="red") +
labs(x = "Sample means", y = "Density",
title = "Distribution of mean heights for 2500 samples of size 30")
A.6.1 Differences between sampling distributions and sample/population distributions
Note that this is not a sample distribution of the variable of interest (“heights”), but rather the distribution of means of the variable of interest (“mean heights”) you would get if you took many random samples (in one sample you’d estimate the mean height as 180cm, in another you’d estimate it as 172 cm, etc). To emphasize this point, let’s compare the simulated sampling distribution of the mean (red histogram) to the population distribution of the variable (grey line) and the distributions of heights in a single sample (blue histogram):
ggplot(df.samples.of.30, aes(x = sample.mean, y = ..density..)) +
geom_histogram(bins=50, fill = 'firebrick', alpha=0.5) +
geom_histogram(data=sample.a,
aes(x = height, y = ..density..),
bins=11, fill='steelblue', alpha=0.25) +
geom_vline(xintercept = true.mean, linetype = "dashed", color='red') +
geom_line(data=pop.distn, aes(x = height, y = density), alpha=0.25,size=1.5) +
xlim(125,225) +
labs(x = "height or mean(height) in cm", y = "Density",
title = "Distribution of mean heights for 2500 samples\nof size 30 (red) compared to the distribution of a single sample (blue)\nand the population distribution of heights (gray line)")
A.6.2 Using sampling distributions to understand the behavior of statistics of interest
The particular sampling distribution of the mean, as simulated above, is a probability distribution that we can use to estimate the probability that a sample mean falls within a given interval, assuming our sample is a random sample of size 30 drawn from our underlying population.
From our visualization, we see that the distribution of sample mean heights is approximately centered around the true mean height. Most of the sample estimates of mean height are within 5 cm of the true population mean height, but a small number of estimates of the sample mean as off by nearly 10cm.
Let’s make this more precise by calculating the mean and standard deviation of the sampling distribution of means:
```
df.samples.of.30 %>%
summarize(mean.of.means = mean(sample.mean),
sd.of.means = sd(sample.mean))
# A tibble: 1 × 2
mean.of.means sd.of.means
1 176. 2.74
```
IMPORTANT! – the values above are estimates of the mean and standard deviation of the sampling distribution of means for samples of size 30, they are not estimates of the mean or standard deviation of the variable of interest (though they are related to these statistics as we’ll see below).
A.6.3 Sampling distributions for different sample sizes
In the example above we simulated the sampling distribution of the mean for samples of size 30. How would the sampling distribution change if we increased the sample size?
In the next code block we generate sampling distributions of the mean (and standard deviation) for samples of size 50, 100, 250, and 500.
```
df.samples.of.50 <-
rerun(2500, rnorm.stats(50, true.mean, true.sd)) %>%
bind_rows()
df.samples.of.100 <-
rerun(2500, rnorm.stats(100, true.mean, true.sd)) %>%
bind_rows()
df.samples.of.250 <-
rerun(2500, rnorm.stats(250, true.mean, true.sd)) %>%
bind_rows()
df.samples.of.500 <-
rerun(2500, rnorm.stats(500, true.mean, true.sd)) %>%
bind_rows()
```
To make plotting and comparison easier we will combine each of the individual data frames, representing the different sampling distributions for samples of a given size, into a single data frame.
df.combined <-
bind_rows(df.samples.of.30,
df.samples.of.50,
df.samples.of.100,
df.samples.of.250,
df.samples.of.500) %>%
# create a factor version of sample size to facilitate plotting
mutate(sample.sz = as.factor(sample.size))
We then plot each of the individual sampling distributions, faceting on sample size.
ggplot(df.combined, aes(x = sample.mean, y = ..density.., fill = sample.sz)) +
geom_histogram(bins=25, alpha=0.5) +
geom_vline(xintercept = true.mean, linetype = "dashed") +
facet_wrap(~ sample.sz, nrow = 1) +
scale_fill_brewer(palette="Set1") + # change color palette
labs(x = "Sample means", y = "Density",
title = "Distribution of mean heights for samples of varying size")
A.6.4 Discussion of trends for sampling distributions of different sample sizes
The key trend we see when comparing the sampling distributions of the mean for samples of different size is that as the sample size gets larger, the spread of the sampling distribution of the mean becomes narrower around the true mean. This means that as sample size increases, the uncertainty associated with our estimates of the mean decreases.
Let’s create a table, grouped by sample size, to help quantify this pattern:
```
sampling.distn.mean.table <-
df.combined %>%
group_by(sample.size) %>%
summarize(mean.of.means = mean(sample.mean),
sd.of.means = sd(sample.mean))
sampling.distn.mean.table
# A tibble: 5 × 3
sample.size mean.of.means sd.of.means
1 30 176. 2.74
2 50 176. 2.20
3 100 176. 1.49
4 250 176. 0.970
5 500 176. 0.671
```
A.7 Standard Error of the Mean
We see from the graph and table above that our estimates of the mean cluster more tightly about the true mean as our sample size increases. This is obvious when we compare the standard deviation of our mean estimates as a function of sample size.
The standard deviation of the sampling distribution of a statistic of interest is called the Standard Error of that statistic. Here, through simulation, we are approximating the Standard Error of the Mean.
A well known mathematical results that your already familiar with is that that the expected Standard Error of the Mean can be related to sample size and the standard deviation of the underlying population distribution as follows
Standard Error of Mean=σ√n Standard Error of Mean=σ n where σ σ is the population standard deviation (i.e.the “true” standard deviation), and n n is the sample size. This result for the standard error of the mean is true regardless of the form of the underlying population distribution.
Let’s compare that theoretical expectation to our simulated results:
```
se.mean.theory <- sapply(seq(10,500,10),
function(n){ true.sd/sqrt(n) })
df.se.mean.theory <- data_frame(sample.size = seq(10,500,10),
std.error = se.mean.theory)
ggplot(sampling.distn.mean.table, aes(x = sample.size, y = sd.of.means)) +
# plot standard errors of mean based on our simulations
geom_point() +
# plot standard errors of the mean based on theory
geom_line(aes(x = sample.size, y = std.error), data = df.se.mean.theory, color="red") +
labs(x = "Sample size", y = "Std Error of Mean",
title = "A comparison of theoretical (red line)\nand simulated (points) estimates of the \nstandard error of the mean for samples of different size")
```
We see that as sample sizes increase, the standard error of the mean decreases. This means that as our samples get larger, our uncertainty in our sample estimate of the mean (our best guess for the population mean) gets smaller.
A.8 Sampling Distribution of the Standard Deviation
Above we explored how the sampling distribution of the mean changes with sample size. We can similarly explore the sampling distribution of any other statistic, such as the standard deviation, or the median, or the the range, etc.
Recall that when we drew random samples we calculated the standard deviation of each of those samples in addition to the mean. We can look at the location and spread of the estimates of the standard deviation:
```
sampling.distn.sd.table <-
df.combined %>%
group_by(sample.size) %>%
summarize(mean.of.sds = mean(sample.sd),
sd.of.sds = sd(sample.sd))
sampling.distn.sd.table
# A tibble: 5 × 3
sample.size mean.of.sds sd.of.sds
1 30 15.1 2.02
2 50 15.1 1.51
3 100 15.2 1.08
4 250 15.2 0.678
5 500 15.2 0.476
```
As we did for the sampling distribution of the mean, we can visualize the sampling distribution of the standard deviation as shown below:
ggplot(df.combined, aes(x = sample.sd, y = ..density.., fill = sample.sz)) +
geom_histogram(bins=25, alpha=0.5) +
geom_vline(xintercept = true.sd, linetype = "dashed") +
facet_wrap(~ sample.sz, nrow = 1) +
scale_fill_brewer(palette="Set1") +
labs(x = "Sample standard deviations", y = "Density",
title = "Sampling distribution of standard deviation of height for samples of varying size")
The key trend we saw when examining the sampling distribution of the mean is also apparent for standard deviation – bigger samples lead to tighter sampling distributions and hence less uncertainty in the sample estimates of the standard deviation.
A.8.1 Standard error of standard deviations
For normally distributed data the expected Standard Error of the Standard Deviation (i.e.the standard deviation of standard deviations!) is approximately:
Standard Error of Standard Deviation≈σ√2(n−1)Standard Error of Standard Deviation≈σ 2(n−1) where σ σ is the population standard deviation, and n n is the sample size.
As before, let’s visually compare the theoretical expectation to our simulated estimates.
```
se.sd.theory <- sapply(seq(10, 500, 5),
function(n){ true.sd/sqrt(2(n-1))})
df.se.sd.theory <- data_frame(sample.size = seq(10,500,5),
std.error = se.sd.theory)
ggplot(sampling.distn.sd.table, aes(x = sample.size, y = sd.of.sds)) +
# plot standard errors of mean based on our simulations
geom_point() +
# plot standard errors of the mean based on theory
geom_line(aes(x = sample.size, y = std.error), data = df.se.sd.theory, color="red") +
labs(x = "Sample size", y = "Std Error of Standard Deviation",
title = "A comparison of theoretical (red line) and\nsimulated (points) estimates of\nthe standard error of the standard deviation\n for samples of different size")
```
A.9 What happens to the sampling distribution of the mean and standard deviation when our sample size is small?
We would hope that, regardless of sample size, the sampling distributions of both the mean and standard deviation should be centered around the true population value, μ μ and σ σ respectively. That seemed to be the case for the modest to large sample sizes we’ve looked at so far (30 to 500 observations). Does this also hold for small samples? Let’s use simulation to explore how well this is expectation is met for small samples.
As we’ve done before, we simulate the sampling distribution of the mean and standard deviation for samples of varying size. Because we’re dealing with small samples we’ll use larger simulations (5000 samples) so we get better estimates of their behavior (in general, the noisier the process you’re simulating the more simulations you should do)
```
A new version of rnorm.stats that includes the sample estimate of the
standard error of the mean
rnorm.stats.2 <- function(n, mu, sigma) {
the.sample <- rnorm(n, mu, sigma)
data_frame(sample.size = n,
sample.mean = mean(the.sample),
sample.sd = sd(the.sample),
sample.estimate.SE = sample.sd/sqrt(sample.size))
}
sample sizes we'll consider
ssizes <- c(2, 3, 4, 5, 7, 10, 20, 30, 50)
number of samples to draw for each sample size
nsamples <- 2500
create a data frame with empty columns
df.combined.small <- data_frame(sample.size = double(),
sample.mean = double(),
sample.sd = double(),
sample.estimate.SE = double())
for (i in ssizes) {
df.samples.of.size.i <-
rerun(nsamples, rnorm.stats.2(i, true.mean, true.sd)) %>%
bind_rows()
df.combined.small <-
bind_rows(df.combined.small, df.samples.of.size.i)
}
df.combined.small %<>%
mutate(sample.sz = as.factor(sample.size))
```
A.9.1 For small samples, sample standard deviations systematically underestimate the population standard deviation
Let’s examine how the well centered the sampling distributions of the mean and standard deviation are around their true values, as a function of sample size.
First a table summarizing this information:
```
by.sample.size <-
df.combined.small %>%
group_by(sample.size) %>%
summarize(mean.of.means = mean(sample.mean),
mean.of.sds = mean(sample.sd))
by.sample.size
# A tibble: 9 × 3
sample.size mean.of.means mean.of.sds
1 2 176. 12.4
2 3 176. 13.5
3 4 176. 14.1
4 5 176. 14.1
5 7 176. 14.6
6 10 176. 14.8
7 20 176. 14.9
8 30 176. 15.1
9 50 176. 15.1
```
We see that the sampling distributions of means are well centered around the true mean for both small and medium, and there is no systematic bias one way or the other. By contrast the sampling distribution of standard deviations tends to underestimate the true standard deviation when the samples are small (less than 30 observations).
We can visualize this bias as shown here:
ggplot(by.sample.size, aes(x = sample.size, y = mean.of.sds)) +
geom_point(color = 'red') +
geom_line(color = 'red') +
geom_hline(yintercept = true.sd, color = 'black', linetype='dashed') +
labs(x = "Sample Size", y = "Mean of Sampling Distn of Std Dev")
The source of this bias is clear if we look at the sampling distribution of the standard deviation for samples of size 3, 5, and 30.
```
filtered.df <-
df.combined.small %>%
filter(sample.size %in% c(3, 5, 30))
ggplot(filtered.df, aes(x = sample.sd, y = ..density.., fill = sample.sz)) +
geom_histogram(bins=50, alpha=0.65) +
facet_wrap(~sample.size, nrow = 1) +
geom_vline(xintercept = true.sd, linetype = 'dashed') +
labs(x = "Std Deviations", y = "Density",
title = "Sampling distributions of the standard deviation\nas a function of sample size")
```
There’s very clear indication that the the sampling distribution of standard deviations is not centered around the true value for n=3 n=3 and for n=5 n=5, however with samples of size 30 the sampling distribution of the standard deviation appears fairly well centered around the true value of the underlying population.
A.9.2 Underestimates of the standard deviation for small n n lead to understimates of the SE of the mean
When sample sizes are small, sample estimates of the standard deviation, s x s x, tend to underestimate the true standard deviation, σ σ. Given this, it follows that sample estimates of the standard error of the mean, S E¯¯¯x=s x√n S E x¯=s x n, must tend to understimate the true standard error of the mean, S E μ=σ√n S E μ=σ n.
A.9.3 The t-distribution is the appropriate distribution for describing the sampling distribution of the mean when n n is small
The problem associated with estimating the standard error of the mean for small sample sizes was recognized in the early 20th century by William Gosset, an employee at the Guinness Brewing Company. He published a paper, under the pseudonym “Student”, giving the appropriate distribution for describing the standard error of the mean as a function of the sample size n n. Gosset’s distribution is known as the “t-distribution” or “Student’s t-distribution”.
The t-distribution is specified by a single parameter, called degrees of freedom (d f d f) where d f=n−1 d f=n−1. As d f d f increases, the t-distribution becomes more and more like the normal such that when n≥30 n≥30 it’s nearly indistinguishable from the standard normal distribution.
In the figures below we compare the t-distribution and the standard normal distribution for sample sizes n=3,5,30 n=3,5,30.
```
x <- seq(-6, 6, length.out = 200)
n <- c(3, 5, 30) # sample sizes
distns.df <- data_frame(sample.size = double(), z.or.t = double(),
norm.density = double(), t.density = double())
for (i in n) {
norm.density <- dnorm(x, mean = 0, sd = 1)
t.density <- dt(x, df = i - 1)
df.temp <- data_frame(sample.size = i, z.or.t = x,
norm.density = norm.density, t.density = t.density)
distns.df <- bind_rows(distns.df, df.temp)
}
distns.df %<>% mutate(df = as.factor(sample.size - 1))
ggplot(distns.df, aes(x = z.or.t, y = t.density, color = df)) +
geom_line() +
geom_line(aes(y = norm.density), color='black', linetype="dotted") +
labs(x = "z or t value", y = "Probablity density",
title = "Standard normal distribution (black dotted line)\nversus t-distributions for different degrees of freedom")
```
US Dept. of Health and Human Services; et al.(August 2016). “Anthropometric Reference Data for Children and Adults: United States, 2011–2014” (PDF). National Health Statistics Reports. 11. |
13604 | https://pdf.hanspub.org/mse2024134_62330884.pdf | Management Science and Engineering 管理科学与工程, 2024, 13(4), 736-744 Published Online July 2024 in Hans. 文章引用: 张恒. 考虑多目标约束的项目群资源调配优化研究综述[J]. 管理科学与工程, 2024, 13(4): 736-744. DOI: 10.12677/mse.2024.134077 考虑多目标约束的项目群资源调配优化 研究综述 张 恒 贵州大学机械工程学院，贵州 贵阳 收稿日期：2024年6月12日；录用日期：2024年7月2日；发布日期：2024年7月15日 摘 要 资源调配作为项目管理的关键手段之一，能有效优化项目群管理中资源争夺频发，整体目标达成不佳等 问题。对考虑多目标约束的项目群资源调配优化研究开展综述。文章先回顾了项目群管理的内涵、目标 和资源调配方法，分析了项目群资源的类型和人力资源的特殊性，随后聚焦于资源受限多项目多目标调 度问题，梳理了不同目标函数和求解算法，特别是元启发式算法的应用。最后总结了考虑项目执行中的 不确定性和人力资源动态性的重要性，并展望未来需进一步探索不确定性条件下的资源调配策略，以及 共性化研发对资源优化的影响，以实现更有效的资源利用和项目目标达成。 关键词 项目群管理，资源调配，多目标优化，不确定性，柔性资源 A Review of Resource Allocation Optimization Research for Project Group Considering Multi-Objective Constrains Heng Zhang School of Mechanical Engineering, Guizhou University, Guiyang Guizhou Received: Jun. 12th, 2024; accepted: Jul. 2nd, 2024; published: Jul. 15th, 2024 Abstract Resource allocation, as one of the key means of project management, can effectively optimize problems such as frequent resource competition and poor achievement of overall goals in project 张恒 DOI: 10.12677/mse.2024.134077 737 管理科学与工程 group management. A review of research on resource allocation optimization for program consi-dering multi-objective constraints is carried out. The article first reviews the connotation, objec-tives, and resource allocation methods of project group management, analyzes the types of project group resources and the particularity of human resources, and then focuses on the multi-objective scheduling problem of resource constrained multi projects. It sorts out different objective func-tions and solving algorithms, especially the application of metaheuristic algorithms. Finally, the importance of considering uncertainty and human resource dynamics in project execution was summarized, and the future needs to further explore resource allocation strategies under uncer-tainty conditions, as well as the impact of generic research and development on resource optimi-zation, in order to achieve more effective resource utilization and project goals. Keywords Project Group Management, Resource Allocation, Multi-Objective Optimization, Uncertainty, Flexible Resources Copyright © 2024 by author(s) and Hans Publishers Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0). 1. 引言 为适应全球政治、经济形势以及社会需求的快速变化，也为企业今后可持续发展奠定坚实的基础， 企业需由单项目管理向项目群管理的转型。研究表明，超过80%的企业采用多项目环境来执行工作，并 在不同项目间共享资源。换句话说，相对于单项目管理，当今企业更多的是针对拥有共同战略目标的 项目群管理与运营。据统计，至少有90%的项目经历了多项目共存的复杂管理环境，而资源受限、项目 目标间冲突的项目运营现实尤为棘手。在过去的几十年间，许多学者都深入研究和探讨了项目群管理的 概念、原则与技术，并论证了科学的项目群管理可以显著降低风险、推动创新并提升企业绩效等。文章 从项目群的内涵出发，随之对项目群管理目标进行回顾、分析，然后对项目群资源及其调配方法展开分 类梳理，最后根据研究现状，总结其趋势并针对现有问题进行评述。 2. 国内外研究现状 2.1. 项目群定义与内涵 项目群管理发展可以追溯到20 世纪80 年代，由PMI 成员提出，项目群管理强调的是项目的关联性 和协作性，需要协调不同项目之间的关系，将多个项目及其要素有机集成并系统化，确保整个组织的战 略目标得到实现。这些项目个体通常有着明晰而一致的战略目标与组织目的，只不过在具体执行和实 施阶段上会存在一些差异，尽管它们各自有着独特的目标，但最后也都会为项目群的共同目标努力。全 国项目管理标准化技术委员会于2019 年发布的关于项目群和项目联合管理方面的指导性意见GB/T 37490，其认为在不确定环境中，实施项目群管理可以获得与企业战略一致的项目群目标，满足管理方期 望，实现投资收益最大化，并优化组织能力与资源配置能力。 常见的与项目群管理含义相近的概念还有多项目管理与项目组合管理，它们在内涵与理论层次有异 同，同时也相互关联。多项目管理指在一个组织中同时进行多个项目的规划、执行和监控的过程和方 法。它需要考虑如何分配资源，如何控制进度，以及如何管理风险等问题，同时也需要协调各个项目之 Open Access 张恒 DOI: 10.12677/mse.2024.134077 738 管理科学与工程 间的关系；其内涵包含项目群管理与项目组合管理，但其不要求项目间的关联性，允许项目间彼此独 立。而项目组合强调在企业战略目标的牵引下，统筹可利用的资源，进行多项目或项目组的选择与整合 ，被组合的对象可以包含项目群，但对象间不一定指向相关或者彼此依赖(图1)。 Figure 1. Interconnections among different project management theory hierarchy 图1. 不同项目管理理论层次关联 2.2. 项目群管理与调度目标 工程项目群的目标包括：满足多个项目的共同资源约束、实现多个项目的共同目标战略和产出多个 项目所具有的共同功能产品。对应于工程项目群的目标，工程项目群类型可以划分为面向战略型、面向 资源约束型和面向产品型三种。本文研究的重点是面向资源约束型的项目群管理。 资源受限的项目调度问题，即RCPSP (Resource Constrained Project Scheduling Problem)，作为一个典 型的NP-Hard 问题， 始终项目管理执行层与运筹学研究领域关注的焦点。 基于RCPSP 的资源受限多项目 调度问题RCMPSP (Resource Constrained Multi-Project Scheduling Problem)自20 世纪70 年代提出后，引 起学术界与产业界的广泛关注 。相对于RCPSP 针对单项目的单目标或者多目标进行优化， RCMPSP 除了对多项目中的单项目进行目标优化外，还需要对多项目的统一战略目标或者整体协同目标 进行优化。 对于不同的项目，有的管理者更偏向于按时完工，有的更偏向于成本控制，也有管理者更偏向于其 他目标。 正因为管理者不同的目标追求， 才会有多目标问题的出现， 一般的文献是考虑双目标或三目标。 刘万琳等学者， 从承包商企业的整体收益最大化立场出发， 假设可从外部获得共享资源， 并以净现值(NPV) 最大化为目标，构建了一个针对分散式多项目调度的整数规划模型，其中共享资源受到柔性约束；经过 求解和实验验证，该模型的有效性得到充分证实；王莹莹等学者，在全面分析多能工的多项技能及其 技能熟练度对任务完成时间的影响后，构建了一个多目标整数非线性规划模型，旨在最小化多项目的总 工期，并同时达成多技能工人之间工作量的平衡分配。Hongwei Zhu 等学者，为了满足装配过程建立 有效的检测与返工调度的需求，建立了一种，以期望完工时间和解的鲁棒性为目标的加权双目标混合整 数规划模型。 张纪康研究了工期不确定下的多目标资源受限项目调度问题， 以工期、 成本和鲁棒性作 为三个求解目标，并基于改进了拥挤度计算的NSGA-II 算法，对H 公司的项目进行了建模与求解。 张照岳针对技能工人工作效率提高这一问题，引入学习曲线对工人的工作效率进行预测，使求解出的项 目工期更贴近实际，构建了以项目工期、人力资源成本以及技能分值提高量为优化目标的工人工作效率 张恒 DOI: 10.12677/mse.2024.134077 739 管理科学与工程 提高的多目标项目计划与调度模型。仿真实验结果表明，在项目进行中考虑工人工作效率提高，并基于 提高后的熟练度对参与项目的工人进行派工， 可提高工人的收入并调动其积极性。 薛松等为优化服务 类企业小型多项目有限资源利用问题，考虑人力资源技能增长并构建了工期与技能进化复合目标模型， 设计了多种群遗传算法并验证其优势。 综合以上国内外文献情况，资源受限项目调度多目标问题研究主要考虑工期、成本、鲁棒性、工作 均衡，员工技能增长率等目标函数，见表1。 Table 1. Summary of project portfolio objective functions 表1. 项目群目标函数总结 序号 目标 举例 1 工期 最短项目工期 最小项目拖期 2 成本 最小资源成本 最大项目净现值 最小延期惩罚成本 3 鲁棒性 最大项目计划鲁棒性 最小项目计划风险 最高交付质量 4 均衡度 最小员工工作量偏差 最佳项目间技能均衡 5 技能增长度 最大人力资源技能增量 最佳员工学习效果 2.3. 项目群资源类型与人力资源调配 项目群管理旨在最大程度地提高企业战略性整体利益，传统的项目管理仅针对单个项目，而项目经 理仅负责其所在的项目，其视角被限制在了自己管理的项目中，每个项目都希望能获取尽可能多的资源 来实现项目的成功，产生了资源的争夺，这样一来往往只能实现某个项目的利益最大化而不利于整体效 益最大化，同时也会加剧不同项目或者不同部门间的冲突，最终导致资源不足和进度安排不合理。从项 目群管理的视角出发，Suvi Elonen 实施了调查，旨在评审当前企业在项目群管理中遇到的问题，并对影 响因素按重要性排序。调查结果显示，“不合理的资源分配和资源短缺”问题以24%的比例位列第二重 要因素。由此可见，梳理项目群资源类型、特点，并优化项目群资源调配的需要已迫在眉睫 。 RCMPSP 的资源按照作用域分为本地资源与全局资源，又称私有资源与共享资源。私有资源为某个 项目独享资源，在实际的项目群实践中，鉴于全局资源的可共享特征和有限的供应状况，各个子项目对 这些资源的需求和争夺显得尤为激烈，这使得有限的全局资源成为约束各子项目达成业务目标的关键因 素；按照资源是否可更新，可分为可更新资源与不可更新资源，如表2 所示。目前的研究主要集中在 探讨具有较高价值且可更新的共享资源限制问题，鲜有不可更新资源限制的研究。近期，关于RCMPSP 的研究进展集中在利用柔性资源和考虑资源重新分配的时间与成本两个方面。 在众多项目资源中， 人力资源是相对特殊的柔性资源，因为人的技能可以是复合的(多技能工)， 能胜 任不同工种与任务； 也可以通过学习培训等渠道， 完成技能掌握面的提升； 同理， 技能也会被员工遗忘， 甚至员工也会调动或离职。 张恒 DOI: 10.12677/mse.2024.134077 740 管理科学与工程 Table 2. Summary of project portfolio resource types 表2. 项目群资源类型总结 划分依据 类型 举例 作用域 私有资源 单项目独享的人、机、料、信息等资源 共享资源 项目群共享的人、机、料、信息等资源 能否更新 可更新资源 设备、人力、信息资源等 不可更新资源 资金、原材料等 复合度 柔性资源 能胜任不同技能、任务的设备、设施等，如多技能员工，综合加工车间 非柔性资源 只能完成单一任务的人力、设备、设施等 陆志强等在分析企业中人力资源技能短缺对项目进度和资源分配产生的影响时，提出了创新策略： 通过柔性技能升阶的项目组合优化调度，实现了项目任务安排与人力资源培养的集成建模研究。该方法 在满足生产需求同时，以更低的成本和更快的速度培养出符合要求的柔性人力资源。 陈蓉等学者， 为 解决多技能员工的离职对新产品研发项目组合调度概率引发的严重影响问题，以多技能研发人员为研究 对象，运用离散马尔可夫链理论来描述其离职过程。同时，综合员工培养长远收益最佳、研发周期最短 以及研发成本最低等目标，提出一种新品研发项目组合调度的随机多目标约束优化模型。James Chen 等针对大型装备制造企业的实际情况， 基于遗传算法提出了一种多项目调度和多技能劳动力分配的方法， 并通过田口方法找到最佳参数设置，最后对相关决策参数进行了敏感性分析。张照岳、郑继光、王莹 莹等学者，均对项目调度过程中，多技能人力资源的技能水平与变化率进行了评价与量化，建模后对同 一算例求解结果表明，考虑人力资源的技能水平变化，以及可供调配的人力数量变化等情况，更贴近实 际的项目调度情况，尤其是以人工投入为主的工序，并更能平衡员工间的工作负载，提升员工满意度与 积极性 。 2.4. 多目标调度与求解方法 根据对文献的总结，目前RCMPSP 的求解方法主要分为三种(见表3)：元启发式、启发式与精确算 法， 实现方法因算法而异。 然而， 鉴于RCMPSP 问题本质上是NP-hard 类型的， 对于任务量庞大的项目， 精确算法无法在合理的时间内求得最佳解决方案；而启发式算法相比元启发式算法(Metaheuristic Algo-rithms, MAs)，普遍求解效率更差，目前已不再是研究热点，但其算法直观且易于理解，为目前多数商业 项目管理软件及项目经理实践所采用，亦可用于为元启发式算法产生初始解。取而代之的，使用元启发 式算法求解RCPSP 与RCMPSP 是当前的研究重点。 Table 3. Summary of solution methods RCMPSP 表3. RCMPSP 求解方法总结 序号 类型 举例 说明 1 精确算法 分支定界法、分解(列生成、行生成)、动态规划 不适用于大型项目 2 启发式算法 串行调度生成方案(SSGS)、 并行调度生成方案(PSGS)、优先规则 算法逻辑直观， 但求解效率较差 3 元启发式(智能)算法 遗传算法、粒子群算法、模拟退火、 禁忌搜索、蚁群算法等 对项目调度求解效果较好， 求解算法研究领域重点对象 张恒 DOI: 10.12677/mse.2024.134077 741 管理科学与工程 类似于调度目标中， 比单项目多了一个整体协同目标， 基于优先级规则PR (Priority Rule)的RCMPSP 的求解过程一般需要先基于战略层与策略层制定的优先级评价体系，确定多项目中各项目的优先级。然 后使用基于优先级的启发式算法，如关键路径法CPM，关键链法CCM，还有智能优化算法(或者它们的 变种及组合)，与其他创新方法，对多项目计划进行调度寻优。Kanchan Rajwar 等学者在对元启发式 算法进行回顾与综述时，基于谷歌学术截至12 月底的文献引文统计发现(见图2)，最受欢迎的前两种算 法是粒子群、遗传算法，其余排位依次是蚁群、差分进化、模拟退火、禁忌搜索、灰狼优化、人工蜂群、 布谷鸟搜索、和声搜索算法。 Figure 2. Citation count and ranking of MAs 图2. 元启发式算法引文数统计与排序 元启发的概念是通过模仿自然现象或生物行为来创建算法来解决优化问题； 例如， 遗传算法(GA) 灵感来源于达尔文进化论中的自然选择和生物遗传机制；模拟退火(SA)受到固体退火原理的启发；粒子 群优化(PSO)是从鸟群中鸟类的相互作用行为中衍生出来的；蚁群优化(ACO)模仿蚂蚁在寻找巢穴和食物 源之间的最短路径时的行为。表4 展示了常用元启发式算法的更多特征。 Table 4. Comparison of common MAs 表4. 常用元启发式算法比较 算法 参数 关键操作 应用领域 优缺点 PSO 5 粒子位置和速度更新、 个体极值和全局极值更新 神经网络训练、 模式识别、工程优化问题 快速收敛，适合于连续空间的优化 问题；平衡了全局和局部搜索； 对于离散问题需要特殊处理 GA 3 选择、交叉、变异、 适应度计算、新一代种群生成 函数与组合优化、 神经网络训练 较强全局搜索能力，适用于复杂和 多峰问题，但容易陷入局部最优 ACO 5 信息素的沉积与挥发、 路径选择与更新 旅行商(TSP)、车辆路径(VRP)、 物流调度等问题 信息素的正反馈机制利于找到 全局最优解；收敛速度慢， 大规模问题消耗资源多 SA 2 解的生成、 能量计算、接受准则 移动网络分配、 柔性制造系统与路径规划等 控制温度参数平衡全局搜索和 局部开发；收敛速度依赖于 精心设计的冷却计划 张恒 DOI: 10.12677/mse.2024.134077 742 管理科学与工程 针对项目群多目标调度，选择元启发式算法时，需要考虑问题的特性，包括规模、优化目标、解空 间结构等。对于具有多个峰值的复杂优化问题，拥有较好的全局搜索能力的遗传算法和蚁群算法可能是 较好的选择。而对于需要快速收敛和简单实现的问题，粒子群算法可能是一个有效的选择。 何华等在共享可更新与私有不可更新双重资源约束下， 以净现值(NPV)最大化为调度目标， 通过构建 其优化模型并设计三层嵌套的禁忌搜索启发式算法进行求解，实证了其相对原生禁忌搜索算法的先进性 。陈浩杰等学者，针对优先级规则调度中缺乏优化能力，提出了一种改进的超启发式遗传规划算法， 其有效性和适用性通过采用以PSPLIB 标准数据集为基础构建的案例以及飞机装配生产线实际的资源受 限的多项目调度得到了证实； 王海鑫等学者， 为了解决标准粒子群算法过早收敛并影响优化结果的问 题，提出了一种具有动态变惯性权重的自适应粒子群优化(DCWPSO)方法，该方法被应用于以多项目加 权工期最小化为目标的模型求解中， 并通过使用标准测试函数和具体算例对其优越性进行了验证； 张 亚鹏学者，设计了一种混合蚁群算法，并对以项目群延迟惩罚成本、可更新资源空置成本以及可更新资 源转移成本之和为目标函数的模型进行求解，并通过实验论证了该模型的优越性；徐振宇学者，以S 企业并行的三个项目为主要工作实例，通过交货期惩罚方法对任务的优先顺序，并利用启发式方法在考 虑资源约束因素的前提下选择关键链，由此得到了三个没有技术问题的关键链，并分别设置了三个项目 的缓冲区，从而达到了企业总工期的最短化目标；Tian，M 等学者，结合改进的关键链CCM 和不同 的分层调度目标，提出了具有分层策略的关键链资源约束多项目调度模型，作为求解多项目调度计划的 一种方式，并证明了模型的有效性；Félix，V 等学者，提出了一种新颖通用RCMPSP 算法—— P-SGS/MIN-SLK，并证实了其然简单，但在16%的情况下优于库中公布的其他算法，在27%的情况下保 持了最好的结果。 3. 发展趋势 1) 更为贴合项目运作实践 项目群调配决策变量的不确定性的考量，项目计划的鲁棒性的优化，柔性人力资源技能对生产过程 与结果的重要度，以及不同工序多种可能执行模式对调配结果的影响等，正在被越来越多的研究者关注 。既往的项目群资源调配研究中，最常用的假设是在确定性条件下求解问题，包括决策变量既定 属性常被广泛假定，忽视了调度执行中的决策变量的随机性，或其他周边因素的随机事件导致的计划波 动。如假定的某一道工序其完成时间是固定的，或在考虑了缓冲区后的一定周期内能完成，但其依赖的 物料延迟到达、其前置任务的延期均造成该工序完工的不确定性；如假定员工的技能熟练度是固定的， 或技能水平与产出水平往往是线性正相关的，忽视了员工的技能变化，概率人事变动等因素 。 2) 更为符合社会发展需要 传统的工程项目优化问题多为单目标优化问题，由于现代项目的复杂性和不确定性，要实现项目群 管理的改进，必须同时考虑多个不同的因素与目标，而它们之间甚至存在彼此冲突的关系。越来越多的 研究者开始从同时优化期望的多目标的角度来尝试多项目改进，即多目标优化。但以往使用最多的是那 些与经济方面或项目完成时间相一致的目标， 而不强调产生对自然友好的生产过程资源使用， 如碳排 放最少，资源利用率最高等。从这个意义上说，在追求减少资源消耗或甚至在一定时间内减少和平衡任 何资源等目标的同时，可能需要增加对不可再生和部分可再生资源的研究。此外，研究和促进诸如最小 化闲置资源或实际环境中的流程库存等目标的使用，可以实现有效的资源利用。 4. 评述 综合项目群内涵、管理与调度目标、资源类型与人力资源调配、多目标调配与求解方法等四方面系 张恒 DOI: 10.12677/mse.2024.134077 743 管理科学与工程 统梳理国内外研究成果，做出如下评述： 1) 目前国内外关于项目群资源调度优化的研究已经取得了一定的成果，关于柔性人力资源的调配优 化亦愈发受到重视，但当前大部分的研究都是假定项目中每项任务的持续时间是确定值的前提下构建模 型并求解的。以软件项目群为例，软件作为一种高度定制化的知识型产品，与普通的工业产品依靠机械 的重复工作不同，其研发主要依赖于软件工程师的脑力劳动。而软件工程师作为一种具有动态性和不确 定性的非物质资源， 很难将其视为传统调度问题中的机械资源进行调度分配。 在真实的软件项目研发 过程中，研发环境存在着两方面的动态性：其一是研发过程中，人员的学习/遗忘效应、内在驱动力等导 致其技能水平处于一个浮动的状态；其二是研发期间动态事件的发生会导致原调度方案无法达到期望的 执行效果，包括技术瓶颈、需求变更、人事变动、上游拖期等。这都提高了软件项目的调度难度，当前 的研究鲜有提及类似不确定性下的保障措施，即便在制定计划初期出于鲁棒性的考量而添加了部分缓冲 区，若未能及时攻克技术瓶颈等异常事件，依旧容易造成计划延期。因此在考虑项目群调度计划的鲁棒 性同时，提出并制定调配模型外的保障管理措施，对如期、达标交付尤为重要。 2) 现有的项目群人力资源调配研究中，都约定了同一时段内员工只能使用某种技能，参与某一个任 务，虽允许人力资源在不同工序、不同子项目间调配，但基本不考虑人力资源在技能、工序、项目间切 换，转移时的成本 ；仍以软件项目群为例，涉及员工的技术栈、开发环境等切换，以及对新 任务或项目，包括产品需求、技术方案、项目计划、周边协作等事项在内的熟悉与适应。因此，调配计 划除了对员工技能等级有硬性要求，实际上也受员工适应能力、协作能力等综合素质的影响；对人力资 源进行定量评估时，综合评价会更为合理与贴近实际。而为了简化调配模型而忽略转移成本时，若有调 配模型外的保障管理措施，能确保被忽略因素影响度最小，将提升调配计划的实用性与鲁棒性。 3) 现有的项目群资源调配文献中，部分学者已经着手研究通过搭建基础方案与产品平台，实现共性 化研发与技术复用，辅助项目群的资源调配优化 ；对于技术驱动的中小型企业而言，自研产品基 线与客户项目的双线推进均为刚需；但作为企业内部驱动的，不直接产生经济收益的共性化研发项目， 鲜有学者将其纳入项目群中同时管理与实施调配优化。 参考文献 Gómez Sánchez, M., Lalla-Ruiz, E., Fernández Gil, A., Castro, C. and Voß, S. 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Contents
Beginning
1 Definition
2 Spira mirabilis and Jakob Bernoulli
3 Notes
4 Logarithmic spirals in nature
5 Related pages
6 References
7 Other websites
Logarithmic spiral
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From Simple English Wikipedia, the free encyclopedia
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A logarithmic spiral, equiangular spiral or growth spiral is a special kind of spiral curve which often appears in nature. The logarithmic spiral was first described by Descartes and later extensively investigated by Jakob Bernoulli, who called it Spira mirabilis, "the marvelous spiral".
Definition
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In polar coordinates (r, θ) the curve can be written as
or
hence the name "logarithmic". In parametric form, the curve is
with real numbers a and b.
The spiral has the property that the angle ɸ between the tangent and radial line at the point (r,θ) is constant. This property can be expressed in differential geometric terms as
The derivative r'(θ) is proportional to the parameter b. In other words, it controls how "tightly" and in which direction the spiral spirals. In the extreme case that b = 0 (ɸ = π/2) the spiral becomes a circle of radius a. Conversely, in the limit that b approaches infinity (ɸ → 0) the spiral tends toward a straight line. The complement of ɸ is called the pitch.
Spira mirabilis and Jakob Bernoulli
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Spira mirabilis, Latin for "miraculous spiral", is another name for the logarithmic spiral. While this curve had already been named by other mathematicians, the name "miraculous" or "marvelous" spiral was given to this curve by Jakob Bernoulli, because he was fascinated by one of its unique mathematical properties: the size of the spiral increases, but the shape stays the same with each added curve. Maybe because of this property, the spira mirabilis has evolved in nature, seen in some living beings, such as nautilus shells and sunflower heads. Jakob Bernoulli wanted the shape on his headstone, but, by error, an Archimedean spiral was placed there instead.
Notes
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The logarithmic spiral can be distinguished from the Archimedean spiral by the fact that the distances between the turnings of a logarithmic spiral increase in geometric progression, while in an Archimedean spiral these distances are constant.
Logarithmic spirals are self-similar in that they are self-congruent under all similarity transformations (scaling them gives the same result as rotating them). Scaling by a factor gives the same as the original, without rotation. They are also congruent to their own involutes, evolutes, and the pedal curves based on their centers.
Starting at a point P and moving inwards along the spiral, one has to circle the origin infinitely often before reaching it; yet, the total distance covered on this path is finite. This was first realized by Evangelista Torricelli even before calculus had been invented. The total distance covered is r/cos(α), where r is the straight-line distance from P to the origin.
One can construct approximate logarithmic spirals with pitch about 17.03239 degrees using Fibonacci numbers or the golden ratio as is explained in those articles. Similarly, the exponential function exactly maps all lines not parallel with the real or imaginary axis in the complex plane, to all logarithmic spirals in the complex plane with centre at 0. (Up to adding integer multiples of 2πi to the lines, the mapping of all lines to all logarithmic spirals is onto.) The pitch angle of the logarithmic spiral is the angle between the line and the imaginary axis.
Logarithmic spirals in nature
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In several natural phenomena one may find curves that are close to being logarithmic spirals. Here follows some examples and reasons:
The approach of a hawk to its prey. Their sharpest view is at an angle to their direction of flight; this angle is the same as the spiral's pitch.
The approach of an insect to a light source. They are used to having the light source at a constant angle to their flight path. Usually the sun is the only light source and flying that way will result in a practically straight line.
The arms of spiral galaxies. Our own galaxy, the Milky Way, is believed to have four major spiral arms, each of which is roughly a logarithmic spiral with pitch of about 12 degrees, an unusually small pitch angle for a galaxy such as the Milky Way. In general, arms in spiral galaxies have pitch angles ranging from about 10 to 40 degrees.
The arms of tropical cyclones, such as hurricanes.
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Many biological structures including spider webs and the shells of mollusks. In these cases, the reason is the following: Start with any irregularly shaped two-dimensional figure F0. Expand F0 by a certain factor to get F1, and place F1 next to F0, so that two sides touch. Now expand F1 by the same factor to get F2, and place it next to F1 as before. Repeating this will produce an approximate logarithmic spiral whose pitch is determined by the expansion factor and the angle with which the figures were placed next to each other. This is shown for polygonal figures in the accompanying graphic.
Related pages
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Fibonacci sequence
Spring Loaded Camming Device
References
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Eric W. Weisstein, Logarithmic Spiral at MathWorld.
Jim Wilson, Equiangular Spiral (or Logarithmic Spiral) and Its Related Curves Archived 2021-05-01 at the Wayback Machine, University of Georgia (1999)
Alexander Bogomolny, Spira Mirabilis - Wonderful Spiral, at cut-the-knot
Other websites
[change | change source]
Spira mirabilis Archived 2007-07-15 at the Wayback Machine history and math
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13606 | https://www.pewforum.org/wp-content/uploads/sites/7/2019/12/PF_12.12.19_religious.households.FULL_.pdf | FOR RELEASE DEC. 12, 2019 Religion and Living Arrangements Around the WorldMuslims and Hindus havelarger householdsthan Christians and religious ‘nones,’in patterns influenced by regional norms FOR MEDIA OR OTHER INQUIRIES: Stephanie Kramer, Research Associate Conrad Hackett, Associate Director of Research and Senior Demographer Anna Schiller, Communications Manager Haley Nolan, Communications Associate 202.419.4372 www.pewresearch.org RECOMMENDED CITATION Pew Research Center, Dec. 12, 2019, “Religion and Living Arrangements Around the World” 1 PEW RESEARCH CENTER www.pewresearch.org About Pew Research Center Pew Research Center is a nonpartisan fact tank that informs the public about the issues, attitudes and trends shaping America and the world. It does not take policy positions. The Center conducts public opinion polling, demographic research, content analysis and other data-driven social science research. It studies U.S. politics and policy; journalism and media; internet, science and technology; religion and public life; Hispanic trends; global attitudes and trends; and U.S. social and demographic trends. All of the Center’s reports are available at www.pewresearch.org. Pew Research Center is a subsidiary of The Pew Charitable Trusts, its primary funder. © Pew Research Center 2020 2 PEW RESEARCH CENTER www.pewresearch.org Acknowledgments This report was produced by Pew Research Center as part of the Pew-Templeton Global Religious Futures project, which analyzes religious change and its impact on societies around the world. Funding for the Global Religious Futures project comes from The Pew Charitable Trusts and the John Templeton Foundation. This report is a collaborative effort based on the input and analysis of the following individuals. Find related reports online at pewresearch.org/religion. Research Team Stephanie Kramer, Research Associate Conrad Hackett, Associate Director of Research and Senior Demographer Alan Cooperman, Director of Religion Research Anne Fengyan Shi, Senior Researcher Jacob Ausubel, Research Assistant Jose Fuentes, Intern George Hayward, Intern Sara Hodgson, Intern Kabir Sandrolini, Intern Yunping Tong, Intern Editorial and Graphic Design Dalia Fahmy, Senior Writer/Editor Michael Lipka, Editorial Manager Aleksandra Sandstrom, Copy Editor Bill Webster, Senior Information Graphics Designer Communications and Web Publishing Stacy Rosenberg, Associate Director, Digital Travis Mitchell, Digital Producer Anna Schiller, Communications Manager Haley Nolan, Communications Associate Others at Pew Research Center provided valuable feedback on this report, including associate directors of research Gregory A. Smith, Neha Sahgal and Juliana Menasce Horowitz, as well as 3 PEW RESEARCH CENTER www.pewresearch.org Andrew Mercer, senior research methodologist. Gretchen Livingston, former senior researcher at Pew Research Center, also provided valuable guidance with this report. Philip Schwadel, professor of sociology at the University of Nebraska-Lincoln, spent a year as a visiting researcher at Pew Research Center. The Center is grateful for his contributions to this report and other work. Pew Research Center also received valuable advice on this report from: Philip N. Cohen, professor of sociology at the University of Maryland; Laurie F. DeRose, assistant professor of sociology at the Catholic University of America; David C. Dollahite, Camilla E. Kimball Professor of Family Life at Brigham Young University; Sriya Iyer, university reader in economics and fellow of St Catharine’s College at the University of Cambridge; Nicolette D. Manglos-Weber, assistant professor of religion and society at the Boston University School of Theology; Luca Maria Pesando, assistant professor of demography at the Centre on Population Dynamics at McGill University; Beth S. Wenger, associate dean of graduate studies and Moritz and Josephine Berg Professor of History at the University of Pennsylvania; and W. Bradford Wilcox, director of the National Marriage Project at the University of Virginia. While the analysis for this report was guided by our consultations with the advisers, Pew Research Center is solely responsible for the interpretation and reporting of the data. 4 PEW RESEARCH CENTER www.pewresearch.org Table of contents Overview ....................................................................................................................................................5 Sidebar: Measuring households from the individual’s perspective – why does it matter? ............ 14 Sidebar: Religious teachings on families and homes………………………………………………………........... 22 1. Household patterns by region ........................................................................................................... 25 Asia and the Pacific ............................................................................................................................. 28 Sub-Saharan Africa .............................................................................................................................. 31 Sidebar: Polygamy in laws and religion ............................................................................................ 36 Europe ................................................................................................................................................. 39 Latin America and the Caribbean ........................................................................................................ 43 The Middle East and North Africa ........................................................................................................ 46 North America ...................................................................................................................................... 49 2. Household patterns by religion ......................................................................................................... 52 Christians............................................................................................................................................. 55 Muslims ............................................................................................................................................... 59 Religiously unaffiliated ........................................................................................................................ 63 Hindus ................................................................................................................................................. 68 Buddhists ............................................................................................................................................ 72 Jews ..................................................................................................................................................... 76 Sidebar: Studies often show links between religion and family life................................................ 79 3. Household patterns by age and gender ............................................................................................ 81 Living arrangements of children .......................................................................................................... 82 Living arrangements of people 60 and older ...................................................................................... 85 Experiences of men and women ......................................................................................................... 89 Appendix A: Methodology ...................................................................................................................... 94 Appendix B: Data sources by country ................................................................................................. 113 Appendix C: Household structure by religious group……………………………………………………………… 121 5 PEW RESEARCH CENTER www.pewresearch.org Religion and Living Arrangements Around the WorldMuslims and Hindus have larger households than Christians and religious ‘nones,’in patterns influenced by regional norms Our households – who lives with us, how we are related to them and what role we play in that shared space – have a profound effect on our daily experience of the world. A new Pew Research Center analysis of data from 130 countries and territories reveals that the size and composition of households often vary by religious affiliation. Worldwide, Muslims live in the biggest households, with the average Muslim individual residing in a home of 6.4 people, followed by Hindus at 5.7. Christians fall in the middle (4.5), forming relatively large families in sub-Saharan Africa and smaller ones in Europe. Buddhists (3.9), Jews (3.7) and the religiously unaffiliated (3.7) – defined as those who do not identify with an organized religion, also known as “nones” – live in smaller households, on average. Muslims and Hindus live in biggest households Average individual resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER Why study households from an individual’s point of view? This report looks at households from the perspective of an average person, rather than an average household. While it is possible to calculate statistics either way, researchers chose the individual perspective because it better captures the lived experience of most people. Consider two homes, one with a family of nine people, the other with a sole resident. The two households contain a total of 10 people, so the average household size is five. But most of the individuals in these two homes – nine out of 10 – live with more than five people. In fact, in this simple example, the average individual resides in a household of 8.2 people. (Here’s the math: Nine individuals, each living among nine people, plus one household of one person, is 9+9+9+9+9+9+9+9+9+1 = 82 /10 people total = an average person residing in a household of 8.2 people.) For more on this topic, see the sidebar on page 14. 6 PEW RESEARCH CENTER www.pewresearch.org Household size is one easy way to compare the lived experiences of people around the world. Bigger households are common in less-developed countries, where people tend to have more children and families share limited resources. Smaller households are prevalent in wealthier countries, which tend to have aging populations and lower birth rates. But the number of people in any given household is only one dimension of living arrangements. Since households of the same size can be so qualitatively different from each other – a three-person household might consist of a couple and one child, a child with a parent and grandparent, a husband and two wives, or numerous other combinations – understanding the distribution of various types of households also is valuable. Globally, the most common household type is the extended family, accounting for 38% of the world’s population. But some religious groups are more likely to live in extended families than others. Hindus are the only major group in which a majority lives with extended family, such as grandparents, uncles and in-laws. Muslims, Christians and Jews are more likely to reside in two-parent households, composed of two partners with one or more minor children. Living alone is unusual among all religious groups, but it is more common among Jews than among the world’s other major religions: About one-in-ten Jews worldwide are in solo households. From a global perspective, Jews also are much more likely than non-Jews to live in Household types defined Extended: A household that includes relatives other than children or partners. For example, adults who live with their siblings or parents in addition to their own children. Two-parent: Married or cohabiting partners with at least one biological, step or foster child under age 18. Adult children may be present, but no other relatives or non-relatives. Couple: Married or cohabiting partners with no one else. This includes couples whose children have grown up and moved out. Adult child: At least one child over the age of 18 with one or two parents; no children under 18. Solo: One person living alone. Single-parent: One adult and at least one biological, step or foster child under 18. Adult children may be present, but no other relatives or non-relatives. Polygamous: Households in which at least one member lives with more than one spouse or cohabiting partner. Other people also may reside in the household. This category does not include every household containing a person who is in a polygamous relationship. For example, two women married to the same man may maintain separate households. Note: Married and cohabiting partners can include same-sex couples, though these relationships are more likely to be counted in the data sources for some countries than others. Also, individuals living in households with non-relatives, such as roommates, are included in the analysis but not reported as a separate category. People living in institutional settings, such as prisons, college dormitories and nursing homes, are not included. See Methodology for details. 7 PEW RESEARCH CENTER www.pewresearch.org households consisting of a couple without children or other relatives.1 1 Even though couple-only and solo households almost never include children, figures throughout most of this report are calculated as percentages of the total population, which includes children under 18. This approach makes it easier to compare shares across household types and does not substantially affect the overall figures. If children were excluded from the total population for purposes of computing percentages, the biggest shift in the share of individuals living in solo or couple households in any country would be +8 percentage points, found in Finland and Sweden’s share of people living as a couple (since adults in Scandinavian countries are more heavily concentrated in solo and couple households, this concentration produces a large boost when children are excluded from analysis). In most countries, the differences would be negligible. Buddhists and ‘nones’ are the least likely to live in two-parent families % of individuals in each household type, all countries combined Note: Values not displayed for “other” households category, which includes households with non-relatives present. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 8 PEW RESEARCH CENTER www.pewresearch.org How or why religion is linked with living arrangements has been the subject of much research and debate. Holy texts and spiritual leaders offer a range of guidance – from didactic anecdotes to outright prohibitions – on many aspects of family life, including marriage and care for elders. Previous social science research, particularly in the United States, suggests that the extent to which people value religion and participate in a religious community is tied to their patterns of marriage, divorce and childbearing. (For a discussion of how religious teachings and family life may be connected, see the sidebar on page 22. For more on academic research exploring the ties between living arrangements and religion, see sidebar on page 79.) To be sure, religion is far from the only factor – or even the primary factor– affecting household sizes and types around the world. People’s living arrangements are shaped by many circumstances, including laws, cultural norms, personal situations and economic opportunities. Still, examining the connections between households and religion helps to illuminate the conditions under which members of various religious groups grow up, practice their faith and pass on traditions to the next generation. 9 PEW RESEARCH CENTER www.pewresearch.org Geography and other factors affect household formation Some household patterns can be explained, at least in part, by the distribution of religious groups across the globe. Six-in-ten Christians live in the Americas and Europe, where households tend to be comparatively small, while eight-in-ten Muslims live in the Asia-Pacific and Middle East-North Africa regions, where households generally contain more individuals. Most of the world’s Jews live in the United States and Israel – two economically developed countries where advanced transportation and health care networks, educational opportunities, and other forms of infrastructure affect many life choices, including living arrangements. At the same time, there are relatively few religiously unaffiliated people in the regions where families are largest – sub-Saharan Africa and the Middle East-North Africa. Moreover, because some religious groups are concentrated in a few countries, the economic conditions and government policies in those places can have a big influence on a group’s global household patterns. China, for example, is home to a majority of the world’s “nones” and about half of all Buddhists. From 1979 to 2016, the Chinese government enforced a “one-child policy” that penalized couples who had more than one child.2 As a result, the size of households among Chinese Buddhists and “nones” is small – and China’s huge population has a big influence on the global figures for these groups. Meanwhile, more than nine-in-ten of the world’s Hindus are found in India, where prevailing cultural norms shape many of the findings for that religious group. Nigeria exemplifies the complexity and interconnectedness of factors that influence living arrangements. Africa’s most populous country is nearly evenly split between two religious groups, with Muslims and Christians each accounting for about half of the population. These groups have different historical legacies, laws and geographic distributions. Largely due to the influence of Christian missionaries, who entered Nigeria via the Atlantic coastline to the south, most Nigerians in the southern states are Christian, while those living in the north tend to be Muslim. Sections of Africa that were reached by missionaries often have more advanced systems of formal schooling today, while aid and research agencies have found that in Nigeria, the northern states have lower rates of educational attainment and economic development.3 2 Laws were relaxed in 2016 to allow two children per family, and the Chinese government is now reviewing the law in an effort to further boost childbirth. Demographers have observed that a substantial decline in China’s fertility rate preceded the one-child policy, raising questions about its impact and the extent to which any policy change will affect fertility rates. During the period the policy was in place, some ethnic minority and rural groups were allowed to have more than one child without penalty. See Feng, Wang, Boachang Gu, and Yong Cai. 2016. “The End of China's One-Child Policy.” Studies in Family Planning. 3 See, for example, Rustad, Siri Aas, and Gudrun Østby. 2017. “Education and Systematic Group Inequalities in Nigeria.” Peace Research Institute Oslo. 10 PEW RESEARCH CENTER www.pewresearch.org These differences extend to household formation. Typically, Muslims in Nigeria share their homes with almost three more people than their Christian compatriots, with an average household of 8.7 people among Nigerian Muslims, compared with 5.9 among Nigerian Christians. Also, although there is no national law providing for polygamy in Nigeria, polygamous marriages are recognized in 12 northern, Muslim-majority states – and Nigerian Muslims are much more likely than Christians to live in polygamous households (40% vs. 8%). (For a detailed discussion of polygamy in laws and religion, see page 36.) In broad strokes, these examples show why it is difficult to isolate the causal impact of religion, which is inextricably linked to economic, geographic, legal and cultural factors not only in Nigeria but around the globe. Each country and part of the world has its own complex set of influences that affect household formation, resulting in a varied landscape of living arrangements. Among the 130 countries with data available on households and religious affiliation, the household size experienced by the average person ranges from 2.7 people (in Germany) to 13.8 people (in Gambia). By region, people tend to form the smallest households in Europe (3.1) and North America (3.3). The biggest households are in sub-Saharan Africa (6.9) and the Middle East-North Africa (6.2). Latin America and the Caribbean (4.6) and the Asia-Pacific region (5.0) fall in the middle. 11 PEW RESEARCH CENTER www.pewresearch.org Gambians live with 11 more people than Germans, on average Average individual resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 12 PEW RESEARCH CENTER www.pewresearch.org Likewise, certain types of households are more prevalent in some parts of the world than in others. For example, almost half of all people in the Asia-Pacific region live with extended family, compared with just one-in-ten North Americans. Polygamous households are rare outside of West Africa, where the practice is quite common in some countries.4 And couples rarely live on their own – without children or extended family – outside of Europe and North America. 4 Polygamy is also legal and practiced in some Persian Gulf countries, including Saudi Arabia. However, these countries are not included in this analysis because adequate census or survey data is not available. In some Asian and African countries, majorities live with extended family % of individuals in extended-family households Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 13 PEW RESEARCH CENTER www.pewresearch.org Regional patterns, in turn, influence the living arrangements among religious groups. Muslims in Europe, for example, generally live in larger households than non-Muslims in Europe (4.1 vs. 3.1, on average). Still, European Muslims follow the region’s overall tendency toward relatively small households, and Muslims in Europe live with fewer people than Muslims in other parts of the world. 14 PEW RESEARCH CENTER www.pewresearch.org Sidebar: Measuring households from the individual’s perspective – why does it matter? If you have ever been to a crowded beach in the summer, you may be surprised to hear that on average, even the most popular U.S. shores are quite empty. A beach on Martha’s Vineyard, for example, might have hundreds of visitors at noon on a sunny day in July, but just a handful of locals out for a midday walk on a chilly day in November. As a result, the daily average of visitors to that beach might be quite low. But that’s not the experience of most people who have been there. Most people are on beaches when they are crowded. In the social sciences, this phenomenon is known as the “class size paradox.” It’s the disconnect between an individual-level perspective and a group-level perspective, and it applies to any group (or “class”) of people, including households: The average individual is more likely to experience a big household than a small one, simply because large households have more people in them. Pew Research Center, for the purposes of this report, chose to present statistics from the individual perspective because the goal is to describe the commonly lived experiences of people – of an average Hindu or an average Buddhist, an average woman or an average child. The authors are inviting readers to imagine the daily lives of ordinary people. The class size paradox is particularly relevant for researchers seeking to understand living arrangements. Especially in places where big households – including certain expansive types, such as extended families and polygamous arrangements – are common, there can be dramatic differences between the average size or type of households in that country and the experience of the average individual living there. Take Senegal, for example. Dividing the total number of household dwellers by the total number of households in that West African country (using Demographic and Health Survey data), Pew Research Center researchers found that the average household in Senegal consists of roughly 8.9 people. When researchers made additional calculations to determine household size on the individual level, however, they found that the average person in Senegal lives in a household of 13.5 people. A similar dynamic unfolds in the United States, even though U.S. households are generally much smaller than Senegal’s. Pew Research Center’s household-level estimate (using General Social Survey data) is that the average household in the U.S. contains 2.5 people. Meanwhile, the individual-level calculation shows that the average individual in the U.S. lives in a household of about 3.4 people. These differences also appear on the global level. Worldwide, the average household has 3.9 members, but the average person lives in a household of 4.9 members. The distribution of household types also varies, depending on the level of perspective, for the same reasons. Household types that allow for more people, such as extended-family homes, contain a bigger share of individuals Average person’s household is larger than average household Household size, by level of analysis Average individual Average households Muslims 6.4 5.1 Hindus 5.7 4.6 Christians 4.5 3.4 Buddhists 3.9 3.1 Unaffiliated 3.7 2.9 Jews 3.7 2.7 All 4.9 3.9 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 15 PEW RESEARCH CENTER www.pewresearch.org than they represent in the simple percentage of households in any given place. Conversely, solo households are more common at the household level than at the person level, since they can contain only one individual. Going back to Senegal as an example: 13% of Senegalese households contain polygamous families, while 51% are extended-family homes, and 7% consist of one person living solo. But because polygamous and extended families tend to be large, they are experienced by more people. Indeed, 23% of all Senegalese people live in polygamous homes, and 55% live with extended family, while fewer than 1% live alone. In the U.S., while 19% of all households belong to a two-parent family and 28% belong to a person living alone, the individual-level analysis shows that 33% of Americans live in two-parent homes and 11% live alone. To gain a better understanding of how the individual-level figures in this report might differ from the more conventional household-level analysis, our researchers also generated results for all the variables on the household level, whenever possible. The tables in this sidebar summarize the differences.5 5 An analysis that includes the age and sex of respondents as variables could be performed only at the individual level. See Methodology on page 94 for more information. Household types: Distributions change when perspective shifts % in each household type by level of analysis, all countries combined Individuals Christians Muslims Unaffiliated Hindus Buddhists Jews All Extended 29% 36% 37% 55% 44% 17% 38% Two-parent 34 43 26 30 20 32 33 Couple 11 3 14 3 13 21 8 Adult child 9 6 12 8 13 12 9 Solo 7 1 7 0.9 7 10 4 Single-parent 6 3 2 3 2 4 4 Polygamous 0.8 5 <0.5 <0.5 <0.5 <0.5 2 Other 3 2 1 0.6 0.9 3 2 Households Extended 19% 27% 25% 42% 29% 11% 26% Two-parent 24 43 21 31 16 19 27 Couple 18 8 20 8 19 28 15 Adult child 9 8 11 11 13 10 10 Solo 21 6 20 4 20 27 15 Single-parent 6 4 3 4 2 3 4 Polygamous <0.5 2 <0.5 <0.5 <0.5 <0.5 0.5 Other 2 1 1 0.5 0.8 2 1 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 16 PEW RESEARCH CENTER www.pewresearch.org Wealth and education: Smaller homes in more-developed countries Levels of prosperity – defined by a range of measures including financial stability, life expectancy and education – are strongly linked with the size and type of households around the world. Europe and North America, the two wealthiest regions as measured by per-capita gross domestic product, are also those with the smallest households. Conversely, sub-Saharan Africa and the Middle East-North Africa region, which have the biggest households, have the lowest per-capita GDP.6 Extended-family arrangements in particular are linked with economic development: Financial and other resources stretch farther when shared within one household. Childcare and domestic chores are more easily accomplished if split among several adults living in the same home. Supporting a family in developing countries is often labor-intensive, requiring farming and other activities that benefit from multiple workers. And in countries where governments provide fewer retirement benefits or other safety nets for older adults, families have a greater responsibility to care for aging relatives. (Cultural factors, such as esteem for elders, also play a role.) On the other hand, people are more likely to live alone in countries with higher levels of schooling. Young adults often delay or forgo childbearing to pursue advanced education, contributing to the tendency of highly educated couples to live without other family members. And in places where people tend to live well beyond their childbearing years, they are more likely to live alone as seniors or as couples without children. Economic development also affects patterns within regions. In Europe, relatively small households predominate in prosperous European Union countries (the average German or Swede lives in a home of 2.7 people, for example), while larger households are found in the less economically advanced countries of the Balkans (Kosovo 6.8, North Macedonia 4.6). In East Asian countries with advanced economies, people tend to live in smaller-than-average households (South Korea 2.9, Japan 3.1), while residents of less-developed countries in South Asia tend to have bigger households (Afghanistan 9.8, Pakistan 8.5). 6 Due to a lack of survey data on religion and households in some countries, this study does not represent the full economic diversity of the Middle East-North Africa region. Missing from the 130 nations in the study are such countries as Qatar, the United Arab Emirates, Kuwait, Saudi Arabia and Bahrain, all of which have relatively high per-capita GDP. 17 PEW RESEARCH CENTER www.pewresearch.org Living alone is more common in wealthier countries % of individuals in solo households, by per-capita GDP Note: GDP data are not available for Kosovo, the Palestinian territories, Puerto Rico or Somalia. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. GDP data are in U.S. dollars and come from the International Monetary Fund. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 18 PEW RESEARCH CENTER www.pewresearch.org Although the regional distribution of religious groups often coincides with economic factors that affect the types and sizes of households people tend to live in, there are noticeable differences among religious groups within the six regions covered in this report, and even within single countries.7 Indeed, this report shows that the experiences of religious groups around the world differ in many ways: ▪ Islam is the largest religion in all but one of the 15 countries with the biggest households: Benin. These top 15 include nations in Africa, such as Gambia and Senegal; the Middle East, namely Yemen and Iraq; and the Asia-Pacific region, including Afghanistan and Pakistan. On the other hand, Christians and the religiously unaffiliated make up the largest groups in the 15 countries with the world’s smallest households (all of which are located either in Europe or the Asia-Pacific region). ▪ Within regions, Muslim-majority countries tend to have larger households, affecting members of all religious groups, while nearby Christian-, Jewish- or Buddhist-majority countries tend to have smaller households. For example, in Kosovo, which has a Muslim majority, both Christians and Muslims tend to live in bigger households than adherents of these religions in nearby Romania, which has a Christian majority. ▪ Within countries, religious groups often differ in their living arrangements. In Senegal, for example, Muslims on average live in 14-person households, while Christians live in homes of about nine members. In Nigeria, a nine-member household is the average Muslim’s experience, while the average Christian lives in a six-person home. And in India, Muslims live in the largest households, with an average size of more than six people, compared with slightly fewer than six people for Hindus, on average. Christians in India have even smaller households, with an average of about five members. ▪ Relatively few people in any group live alone (4% globally), though Jews (10%), Buddhists (7%), Christians (7%) and “nones” (7%) all live alone at higher rates. Muslims and Hindus rarely form solo households (1% each). 7 Although some faiths other than those analyzed in this report have millions of adherents around the world (such as Sikhs), censuses and surveys in many countries do not measure them specifically. Because of this scarcity of reliable data, this report does not analyze groups other than Christians, Muslims, Hindus, Buddhists, Jews and people with no religious affiliation; the report is also unable to show data for subgroups within these major religions, such as Protestants and Catholics or Sunnis and Shiites. The six major religious groups are always included in the total global count, but the groups are not represented on the regional level (or on the country level) if there are too few adherents in the population or in surveys. See Methodology on page 94 for details. 19 PEW RESEARCH CENTER www.pewresearch.org ▪ Similarly, Muslims and Hindus are the least likely to live as couples (without children or other relatives), with just 3% in each group living in such an arrangement. By contrast, at least one-in-ten Christians, Buddhists and “nones” live this way, as do one-in-five Jews. ▪ Polygamy is very rare, except in some sub-Saharan African countries where this type of marriage is largely legal. While African Muslims are more likely than Christians to live in polygamous households (25% vs. 3%), the practice is also widespread among adherents of folk religions (who are not analyzed separately in this report) and “nones” in certain countries. ▪ In the U.S., Christians (3.4), “nones” (3.2) and Jews (3.0) live in similarly sized households, on average. However, U.S. Jews are much more likely than non-Jews to live as couples without children or other relatives, and they are less likely to reside in extended families and single-parent households.8 ▪ Men in every country are older, on average, than their wives or female cohabiting partners. This age gap is widest among Muslims and Hindus, and smallest among Jews and the religiously unaffiliated. ▪ Women ages 60 and older are more likely than men in this age group to live alone. Three-in-ten Christian and Jewish women over 60 live alone, while only 6% of Hindus do. ▪ More Christians than members of any other religious group live in single-parent homes (6%). And women, particularly Christian women, are more likely than men to live as single parents. In the U.S. – the country with the world’s largest Christian population – about a quarter of children live in single-parent homes, making them more likely than children in any other country to do so. American children in Christian homes are just as likely as those in unaffiliated homes (23% each) to live in single-parent situations. 8 The sample sizes of Muslims, Buddhists and Hindus in the available U.S. data are not large enough to allow for separate analysis. Also, previous Pew Research Center studies have found large differences within the Jewish population, with Orthodox (particularly Haredi) Jews living in larger households than non-Orthodox Jews, on average. 20 PEW RESEARCH CENTER www.pewresearch.org These are among the key findings of a new Pew Research Center analysis of census and survey data collected by governments and survey organizations in 130 countries since 2010. This report was produced by Pew Research Center as part of the Pew-Templeton Global Religious Futures project, which analyzes religious change and its impact on societies around the world. Funding for the Global Religious Futures project comes from The Pew Charitable Trusts and the John Templeton Foundation. Previous reports from the Global Religious Futures project explored links between religion and gender, education, age and personal well-being, and produced population growth projections for major religious groups. Statistics come from Pew Research Center analysis of sources, which include Multiple Indicator Cluster Surveys, Demographic and Health Surveys, census data archived by IPUMS International, General Social Surveys and European Social Surveys, as well as country-specific studies. The study U.S. children are more likely than children elsewhere to live in single-parent homes % of children under age 18 in single-parent households Note: Single-parent households include one adult and at least one biological, step or foster child under 18. Adult children may be present, but no other relatives or non-relatives. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 21 PEW RESEARCH CENTER www.pewresearch.org covers countries where 91% of the global population lives.9 Unless otherwise specified, results describe the average living arrangements experienced by people of all ages (that is, children and adults). For more details, see Methodology. In addition to analyzing household sizes using census and household survey data and transforming it to present the average individual’s perspective (as opposed to the average household), researchers also analyzed eight household types among the major religious groups, broken down by age and gender. The report focuses on descriptions of seven household types that are made up of individuals living alone or with family members, though about 2% of all people living in households fall into the residual eighth (“other”) category, including those who share housing with unrelated roommates or a mix of relatives and non-relatives. All statistics in this report exclude institutionalized populations, such as people in prisons, college dormitories and nursing homes. The next section of this report explores how holy texts and spiritual leaders have addressed the question of household formation. Subsequent chapters outline patterns in household sizes and types by region and religious group, including a discussion of previous social science research on the connections between living arrangements and religion. The final chapter focuses on how living arrangements vary by age and between women and men. 9 Some countries, for example, Argentina, Australia, Saudi Arabia and Sudan, are not covered in this report either because surv eys in those countries do not ask about people’s religious affiliations and their household arrangements in the same questionnaire, or because data was not readily available in a downloadable format. For more details on inclusion criteria, see the Methodology on page 94. 22 PEW RESEARCH CENTER www.pewresearch.org Sidebar: Religious teachings on families and homes Holy texts and spiritual leaders have much to say about domestic life. All the major world religions – including but not limited to Christianity, Islam, Judaism, Hinduism, and Buddhism – promote specific types of family formations and offer guidance on the roles that people play within the home. These teachings take many forms, from explicit rules to popular sayings and stories.10 But teachings are not static or universal. Interpretations of religious texts vary by place and shift across time, and religious leaders in the same context often disagree. For nearly two centuries, scripture was used by some U.S. Christian clergy to justify opposition to interracial spousal relationships. But societal views and laws shifted in the 20th century, and Bible passages once viewed as censuring ties between races have largely been pushed aside. Procreation Bearing, raising and protecting children is a central theme in many religions. Christianity and Judaism encourage adherents to have children, and in Genesis – the first book of the Jewish Torah and the Christian Old Testament – God commanded humans to “be fruitful and multiply.” The Quran, which emphasizes the importance of motherhood and childbearing, has a handful of references to pregnancy and birth.11 Outside of the Abrahamic religions, the Hindu Vedic texts also contain passages about having children. The Marriage Hymn of the Rig Veda, which is sometimes recited at Hindu weddings, states “Let Prajapati create progeny for us,” and “Generous Indra, give this woman fine sons. … Place ten sons in her and make her husband the eleventh.”12 Buddhism is not considered to be particularly pro-natalist, which some scholars have tied to relatively low fertility rates in that religious group.13 A religion can promote having children even if it does not have specific doctrines endorsing procreation. Some scholars say that teachings of conservative Islam and Christianity indirectly lead people to have more children due to the gender roles they endorse.14 Religion also can indirectly encourage procreation through other means, such as opposition to birth control. In Judaism, a couple’s inability to have children can be grounds for divorce. Marriage Closely tied to childbearing, religions often set rules for marriage and sexual relations. According to Genesis 2:18, woman was created because “it is not good that the man should be alone.” In the Christian New Testament, Paul 10 This sidebar draws heavily on research published in Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 11 Isgandarova, Nazila. 2013. “Pregnancy and Childbirth.” In DeLong-Bas, Natana J., ed. “The Oxford Encyclopedia of Islam and Women.” 12 Sarma, Deepak, ed. 2008. “Hinduism: A Reader.” Also see Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 13 Skirbekk, Vegard, Marcin Stonawski, Setsuya Fukuda, Thomas Spoorenberg, Conrad Hackett, and Raya Muttarak. 2015. “Is Buddhism the low fertility religion of Asia?” Demographic Research. 14 Muslim women’s embrace of such traditional gender roles varies considerably across Muslim nations. See Blaydes, Lisa, and Drew A. Linzer. 2008. “The Political Economy of Women’s Support for Fundamentalist Islam.” World Politics. Also see Sherkat, Darren E. 2000. “‘That They Be Keepers of the Home’: The Effect of Conservative Religion on Early and Late Transitions into Housewifery.” Review of Religious Research. 23 PEW RESEARCH CENTER www.pewresearch.org the Apostle wrote, “Because of cases of sexual immorality, each man should have his own wife and each woman her own husband.” In Islam, the Quran states that God created “love and mercy” between spouses. In Hinduism, the ashrama system specifies four stages of the life course, including the second stage of life, grihastha, in which one becomes a householder and focuses on marriage, raising children and fulfilling obligations toward elders and other relatives. The Laws of Manu, a traditionally authoritative book of Hindu code, states: “Reprehensible is the father who gives not (his daughter in marriage) at the proper time; reprehensible is the husband who approaches not (his wife in due season), and reprehensible is the son who does not protect his mother after her husband has died.” In Buddhism, the Sigālaka Sutta offers instructions on how spouses should treat one another: “There are five ways in which a husband should minister to his wife as the western direction: by honoring her, by not disparaging her, by not being unfaithful to her, by giving authority to her, by providing her with adornments.”15 Scriptures provide a mix of guidance on interfaith marriage. In the Torah, Moses tells the Jews that they must not intermarry with those from the seven nations of Canaan, which some interpret as a prohibition on all religious intermarriage. And, according to the Talmud, a rabbinic commentary on Jewish scripture, both participants must see the marriage ceremony as sacred in order for it to be religiously valid.16 The Christian New Testament, however, is less clear on the topic of intermarriage. In his letter to the Corinthians, Paul wrote: “Do not be mismatched with unbelievers,” which is often thought to prohibit marriage to non-Christians. However, Paul also instructed believers not to divorce their unbelieving spouses. In Islam, critics of interfaith marriage typically cite Surah 2:221: “Do not marry idolatresses until they believe: a believing slave woman is certainly better than an idolatress, even though she may please you. And do not give your women in marriage to idolaters until they believe: a believing slave is certainly better than an idolater, even though he may please you.” Polygamy Polygamy was practiced by central figures in Judaism, Christianity and Islam and, more recently, by early leaders of the Church of Jesus Christ of Latter-day Saints in the 19th century. Many prominent biblical figures were polygamous, including Jacob, David and Solomon. Generally speaking, polygamy is no longer encouraged by the leaders of major religions, though it is practiced by some Muslims, fundamentalist Mormon sects and Christians in Africa. Muslim supporters of polygamy often cite Quran verse 4:3, which permits a man to marry up to four women, but encourages him to be monogamous if he cannot be fair to all of them.17 Scholars have interpreted this text as a way to regulate and limit polygamy in seventh-century Arabia, where it was widely practiced. (For more on polygamy laws and teachings, see the sidebar on page 36.) 15 Sigālaka Sutta, as cited in Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 16 Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 17 Jaafar-Mohammad, Imani, and Charlie Lehmann. 2011. “Women’s Rights in Islam Regarding Marriage and Divorce.” Journal of Law and Practice. 24 PEW RESEARCH CENTER www.pewresearch.org Divorce Religions often discourage divorce. The Christian New Testament specifies, “Therefore what God has joined together, let no one separate.” Still, many religions recognize the reality that some marriages will not last. In the Old Testament, Deuteronomy states that the process of divorce is enacted when a man writes his wife a certificate. The Quran specifies the requirements for divorce and outlines a man’s responsibilities toward his former wife. Islam allows a woman to retain any assets she earns or receives during a marriage and gives her the right to receive support from her former husband.18 According to a Confucian text, The Record of Ritual of the Elder Dai, there are seven valid reasons a husband may divorce his wife – including “if she has no children” or “if she steals” – and three situations in which a wife may not be divorced, including if “there is no longer a home to which she can return.”19 Extended family Teachings that address the status of elders and other relatives outside of the nuclear family may also encourage certain types of living arrangements, including extended families. Hinduism, which emphasizes a respect for elders, also has guidelines for responsibilities toward other relatives in the second stage of life, the grihastha ashram.20 Judaism and Christianity instruct followers to respect their elders in the Ten Commandments: “Honor your father and your mother.” And the Book of Leviticus, which is part of both Christian and Jewish scripture, states: “You shall rise before the aged, and defer to the old.” Similarly, the New Testament says: “And whoever does not provide for relatives, and especially for family members, has denied the faith and is worse than an unbeliever.” Confucian texts also encourage support for aging parents and emphasize filial piety.21 In these and many other ways, religious leaders, texts and traditions can affect the choices that people around the world make about procreation, marriage, divorce and family life. It would be unwise to view religion as rigidly determining these choices, because so many other factors also come into play. Still, no picture of family sizes and structures would be complete without a closer look at the role of religion. 18 Jaafar-Mohammad, Imani, and Charlie Lehmann. 2011. “Women’s Rights in Islam Regarding Marriage and Divorce.” Journal of Law and Practice. 19 Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 20 Subramuniyaswami, Satguru Sivaya. 2000. “How to Become a Hindu: A Guide for Seekers and Born Hindus.” 21 Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 25 PEW RESEARCH CENTER www.pewresearch.org 1. Household patterns by region Pew Research Center analyzed data on living arrangements in 130 countries, including 26 in the Asia-Pacific region, 40 in sub-Saharan Africa, 35 in Europe, 19 in Latin America and the Caribbean, and eight in the Middle East and North Africa, as well as the U.S. and Canada, which in this report make up North America. Globally, the average individual lives in a household of 4.9 people. But there is wide variation around the world: The average person in sub-Saharan Africa resides in a home of 6.9 people, while the average European lives in a home that is less than half that size, at 3.1 members. Regional differences also are apparent when it comes to household types: For example, more than half of people in the Middle East-North Africa region live in two-parent homes with minor children (56%), compared with about a quarter of Europeans (26%). And more than four-in-ten people in the Asia-Pacific region live in extended families (45%), compared with just one-in-ten North Americans (11%). Households smallest in Europe, biggest in Africa Average individual resides in a household of ___ people Household Size Sub-Saharan Africa 6.9 Middle East-North Africa 6.2 Asia-Pacific 5.0 Latin America-Caribbean 4.6 North America 3.3 Europe 3.1 World 4.9 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER Almost half of Asians live in extended families, one-in-ten Europeans live alone % of individuals in each household type Asia-Pacific Europe Latin America-Caribbean Middle East-North Africa North America Sub-Saharan Africa World Extended 45% 26% 32% 27% 11% 35% 38% Two-parent 31 26 39 56 33 37 33 Adult child 10 9 10 9 14 2 9 Couple 7 19 6 3 20 2 8 Solo 3 13 3 1 11 2 4 Single-parent 2 4 5 2 9 6 4 Polygamous <0.5 <0.5 <0.5 0.9 <0.5 11 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 26 PEW RESEARCH CENTER www.pewresearch.org Every region has its own set of historical, economic, religious and cultural backdrops that influence living arrangements. Wealthier parts of the world tend to have smaller households. Europeans in general, and particularly Western Europeans, are the most likely of all the regional groups to live alone: For example, Norway has the highest GDP per capita of any country in this study, and 17% of Norwegians live by themselves. Conversely, sub-Saharan African countries have both small GDPs and relatively few people who live alone. Education is another important factor: The share of people within countries who live in couple-only households generally increases as average years of formal schooling rise and young adults delay or forgo having children. (People also live longer in these countries, which leads to more couples living together after adult children have moved out.) On the other hand, natural disasters, epidemics and wars also leave their mark on household distributions. In Kenya and Malawi, where tens of thousands of people die each year from HIV/AIDS, single-parent households are relatively common. The HIV epidemic in Africa has affected middle-aged adults more than other age groups, leaving many children and grandparents without this middle generation. In Nepal, a massive earthquake in 2015 killed thousands of people and left millions homeless, forcing many Nepalis to shift their living arrangements and likely affecting household patterns for that country in subsequent surveys. This chapter looks at the living arrangements in each region and within the major religious groups when they are sufficiently represented. The data in this report covers 91% of the global population, representing at least 85% of people in each major religious group. But groups are unevenly distributed around the world, and the number of religious groups compared within each region varies, reflecting the global distribution of religions and variation in the sample sizes of source surveys. For details on data sources and coverage, see Methodology on page 94. The analysis starts with the average household sizes experienced by adherents of each religion, followed by a discussion of the most common household types. The regions are ordered by population size, starting with Asia and the Pacific. 27 PEW RESEARCH CENTER www.pewresearch.org Living as a couple is more common where education levels are high % of individuals in couple-only households, by national average years of schooling Note: Education data is for those ages 25 and older and was not available for 14 of 130 countries and territories studied. Source: Living arrangements data from Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. Education attainment data gathered for Pew Research Center’s 2016 report “Religion and Education Around the World.” “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 28 PEW RESEARCH CENTER www.pewresearch.org Asiaand the Pacific The Asia-Pacific region is a good example of the way household sizes can vary within a single part of the world. The most religiously varied of all the regions – in part because it is the largest, home to more than half of the world’s population – Asia-Pacific has hundreds of millions of Muslims, Hindus, “nones,” Christians and Buddhists. In addition, the economies of the region’s 26 countries with available data range from Japan – one of the world’s wealthiest countries – to Nepal, one of the poorest. On average, people in the Asia-Pacific region live in five-member households. Asian Muslims (6.0) and Hindus (5.7), on average, reside in households that are slightly bigger than those of Asians overall, while Asian Buddhists (3.9) and “nones” (3.7) have relatively small households. Asia-Pacific stands out globally because it has the biggest share of people living in extended families (45%); it is the only region where that type of arrangement is more common than the two-parent family. This region, along with the Middle East-North Africa region, also has the smallest share of people in single-parent homes (2% in each). In Asia-Pacific region, Muslims and Hindus have larger households than Buddhists and ‘nones’ Average individual in the Asia-Pacific region resides in a household of ___ people Household size Muslims 6.0 Hindus 5.7 Christians 4.8 Buddhists 3.9 Unaffiliated 3.7 All Asia-Pacific 5.0 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER Living in single-parent families is rare in Asia-Pacific region % of individuals in each household type Hindus Muslims Unaffiliated Buddhists Christians All Asia-Pacific Extended 55% 41% 41% 44% 37% 45% Two-parent 30 42 25 20 35 31 Adult child 8 7 14 13 10 10 Couple 3 4 13 13 8 7 Solo 0.9 1 6 7 5 3 Single-parent 3 3 1 2 2 2 Polygamous <0.5 0.6 <0.5 <0.5 <0.5 <0.5 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 29 PEW RESEARCH CENTER www.pewresearch.org These overall patterns are driven by India and China, which together account for more than 60% of the region’s population, encompassing the majority of Asian Hindus, Buddhists and “nones.” (Muslims in Asia and the Pacific are most numerous in Indonesia, India, Pakistan and Bangladesh, while Christians, who make up a relatively small percentage of the regional population, are concentrated in the Philippines and China.) Extended families are the most common household type in both India (54%) and China (44%). The concentration of Hindus in extended families on a global level reflects typical living arrangements in India, where 94% of the world’s Hindus reside. Extended families also are common in most other countries in the Asia-Pacific region, encompassing more than half of Muslims in Tajikistan (67%), Buddhists in Vietnam (56%) and “nones” in Taiwan (51%). However, when it comes to household size, India and China are at opposite ends of the regional scale: The average Indian lives in a fairly big household (5.8), while the average person in China resides in a smaller household (3.8). Across the region, the countries with the smallest households tend to have large shares of Buddhists or unaffiliated people (or both). That includes China, but also Japan, South Korea, Mongolia and Thailand – where the average household size is about four people or fewer. Northeast Asia stands out for its high frequency of solo households. South Korea’s general population has among the world’s highest rates of people living alone (21% – second only to Denmark), with South Korean “nones” (24%) and Christians (21%) more likely than Buddhists (14%) to live alone. Japan (15%) also has a relatively large share of people living alone, with Buddhists and “nones” about equally likely to do so. As in other parts of the world, Muslims in the Asia-Pacific region tend to share their homes with more people. Afghanistan and Pakistan, which are overwhelmingly Muslim, also are the two countries where Asians live in the biggest households (9.8 and 8.5 people, respectively). Moreover, Muslims have larger households than their countrymen in a range of other countries – including some without Muslim majorities, such as India and the Philippines. However, Muslims’ tendency to live in fairly big households is not universal: The Muslim-majority countries of Indonesia (4.7) and Iran (4.1) have relatively small households by regional standards. 30 PEW RESEARCH CENTER www.pewresearch.org The average Afghan lives with seven more people than the typical South Korean Average individual resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 31 PEW RESEARCH CENTER www.pewresearch.org Sub-Saharan Africa Household patterns vary widely across sub-Saharan Africa, with clear differences that can be measured in different parts of the region, across country borders and between religious groups in single countries. The region’s population is mostly Christian, with a substantial Muslim minority. In some countries, large Muslim and Christian populations live side by side. Overall, sub-Saharan Africans live in the world’s biggest households, at an average of 6.9 people per household, with Muslims (8.5) in more expansive arrangements than Christians (6.0). The religiously unaffiliated and Hindus, who make up less than 4% of the regional population, have smaller homes – 5.7 and 3.9 people, respectively. Within sub-Saharan Africa, the nations with the largest households tend to be in West Africa and have majority Muslim populations. In Gambia, Senegal and Mali – three neighboring countries with the largest overall household sizes – at least 90% of people are Muslim and the average person lives in a household of 12.6 people or more. Christians in those places also have the biggest households of Christians in any country, with as many as 10.3 people, though they live in smaller homes than their Muslim compatriots. The region’s smallest households, meanwhile, are found in countries where Christians form a majority: In Sao Tome and Principe, South Africa, Madagascar and Rwanda, where Christians make up 80% or more of the population, the average person lives in a household of about five people. In an exception to the overall global and regional pattern, Muslims in Madagascar live in smaller households than Christians. (In South Africa and Rwanda, there is not much of a difference, and in Sao Tome no data is available for Muslims.) Muslims live in larger households than Christians in sub-Saharan Africa Average individual in sub-Saharan Africa resides in a household of ___ people Household size Muslims 8.5 Christians 6.0 Unaffiliated 5.7 Hindus 3.9 All sub-Saharan Africa 6.9 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 32 PEW RESEARCH CENTER www.pewresearch.org Overall, sub-Saharan Africans are about as likely to live in two-parent households (37%) as in extended families (35%). The region is unique in its high rate of people living in polygamous homes, with 11% in this arrangement, much higher than the 2% global average. (This practice is particularly common in West and Central Africa. For a more detailed discussion of polygamous households, see page 34.) Sub-Saharan Africa also has the smallest percentage of people in adult child households (2%). While sub-Saharan African Christians are about as likely to live in two-parent families (38%) as in extended families (39%), Muslims are more frequently found in two-parent families (37%) than in extended families (27%). The smaller share of Muslims in extended families may be related to a higher prevalence of polygamous households among Muslims. In sub-Saharan Africa, many live in homes of six people or more Average individual resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 33 PEW RESEARCH CENTER www.pewresearch.org In Senegal and Guinea-Bissau, two-thirds or more of Christians live in extended families (66% and 74%), making these the sub-Saharan African communities most likely to do so. Extended families are least often found among Muslims in Nigeria, only 13% of whom reside in that type of home. In sub-Saharan Africa, Christians are less likely than Muslims to live in polygamous families % of individuals in each household type, by religion Christians Muslims Unaffiliated Hindus All sub-Saharan Africa Two-parent 38% 37% 32% 31% 37% Extended 39 27 39 32 35 Polygamous 3 25 5 <0.5 11 Single-parent 8 4 7 0.8 6 Solo 3 1 7 3 2 Couple 2 2 4 10 2 Adult child 2 0.9 3 22 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 34 PEW RESEARCH CENTER www.pewresearch.org Polygamy is very common in some African countries Around the world, polygamy is very rare. About 2% of people globally live in households in which at least one member has more than one spouse or partner. The practice is illegal in most countries, and laws that allow it are primarily found in the Middle East and Africa.22 (For more on polygamy laws and religious teachings, see the sidebar on page 36.) In sub-Saharan Africa, however, polygamy is practiced in a group of West and Central African countries that sometimes are referred to as the “polygamy belt” – and they include Muslim-majority countries such as Senegal, Gambia and Mali, as well as Christian-majority nations such as Cameroon, the Central African Republic and Benin.23 Still, most of the countries that allow polygamy are majority-Muslim – and an analysis of sub-Saharan Africa specifically shows that Muslims there are more likely than Christians, Hindus or the unaffiliated to live in polygamous households. There also is a high rate of polygamy among adherents of African folk religions – and, in a few countries, among “nones.” (Christian churches with historic ties to Western missionaries tend to reject this type of marriage, although many churches that have their roots in African communities allow it. Polygamy is an accepted practice in some tribal and ethnic cultures.24) Among Muslims in sub-Saharan Africa, a quarter live in polygamous households, compared with 3% of African Christians and 5% of the unaffiliated. Among people who do not belong to any of these groups and instead identify with traditions that can be described as folk religions, 19% live in polygamous households.25 Overall, the share of all sub-Saharan Africans living in polygamous households is 11%. In six countries – Burkina Faso, Mali, Gambia, Niger, Nigeria and Guinea – at least a quarter of the population lives in polygamous homes.26 Polygamy is legal in all of these countries, at least to 22 Importantly, several Persian Gulf countries where polygamy is legal and assumed to be relatively common could not be included in this analysis due to a lack of available data. In Iraq, a Gulf country analyzed in this report where polygamy is legal under certain circumstances, the percentage of people living in polygamous households is 2.3%. 23 Jacoby, Hanan G. 1995. “The Economics of Polygyny in Sub-Saharan Africa: Female Productivity and the Demand for Wives in Côte D'Ivoire.” The Journal of Political Economy. Also see Allen, John L. Jr. 2014. “Surprise! One of the Church’s family issues is polygamy.” Crux. 24 Falen, Douglas J. 2008. “Polygyny and Christian Marriage in Africa: The Case of Benin.” African Studies Review. 25 Other sections of this report do not broadly analyze data about followers of folk religions and a range of other, smaller religious groups because the individual groups that make up these categories are numerous, diverse and often have little in common. For example, African Spiritualism, Voodoo and Confucianism are all classified as folk religions; on the whole, folk and other religions are not cohesive or widely represented enough for the purposes of analysis in this report. 26 Because polygamous households tend to have a higher number of members than other types of households, this analysis may seem high compared with estimates that compute results on the household level. Similarly, because polygamous families tend to be larger and have more children, there are always larger shares of children living in these types of households than adults. More information on the living arrangements of children can be found on page 82. 35 PEW RESEARCH CENTER www.pewresearch.org some extent. In Nigeria, polygamy is banned under civil law, but 12 northern states governed by Shariah law allow it. 27 Burkina Faso has sub-Saharan Africa’s largest share of people living in polygamous households (36%), including nearly half of those who practice folk religions (45%), making them the likeliest of all religious groups to live in this arrangement. Four-in-ten Muslims in Burkina Faso live in polygamous households, followed by roughly a quarter of Christians (24%). (Burkina Faso is largely Muslim, with a sizable Christian minority.) In Cameroon, which is majority Christian but has a sizable Muslim minority, the percentages of people living in polygamous households are 32% for adherents of folk religions, 26% for Muslims, 19% for “nones” and 7% for Christians. A similar pattern exists in Togo. In Nigeria – Africa’s most populous country, where roughly half the population is Muslim and half is Christian – 28% of all people live in polygamous households. This arrangement is widespread among Muslims (40%) and adherents of folk religions (29%) but is less common among Christians (8%). In Chad, the unaffiliated have the highest rates of polygamy, with about four-in-ten “nones” in this arrangement. Chad is also the only country where Christians (21%) are much more likely than Muslims (10%) to live in polygamous households. 27 For details on laws that govern polygamy, see the sidebar on page 36. 36 PEW RESEARCH CENTER www.pewresearch.org Sidebar: Polygamy in laws and religion Laws Although polygamy is illegal in most places, over 50 countries – largely in the Middle East and Africa, but also in Asia – allow it to at least some degree. Polygamous marriages are almost always polygynist, with one man taking multiple wives; rules often specify that a man may marry up to four women if certain conditions are met. There are a few countries in which polygamy is illegal but relatively common, and many where the opposite is true.28 Among the countries covered in this report, polygamy is most widely practiced in West Africa, where that type of arrangement is often permitted by customary law or religious tradition, if not by civil law, according to the OECD. In six countries – Benin, Cabo Verde, Ivory Coast, Ghana, Guinea and Nigeria – polygamy is formally prohibited but tolerated. Some countries, including Burkina Faso and Togo, allow couples to choose between a monogamous or polygamous arrangement at the outset of their marriage. In others, including Mauritania, a man must obtain permission from his spouse or spouses before marrying again. In Nigeria, polygamy is banned at the national level, but recognized by the 12 northern states regulated by Shariah law. In the Middle East-North Africa region, polygamy is legal in Iraq, Yemen, Algeria and Egypt, but fewer than 3% of individuals in those countries live in polygamous households. (Polygamy is also legal in Saudi Arabia, Qatar, the United Arab Emirates and most other countries in the Middle East and North Africa, but data is not available on household composition and religion outside of the eight countries covered in this report.) In Asia, this arrangement is both allowed and rare in Pakistan, Bangladesh, Indonesia, Iran and Afghanistan. In India, polygamy is legal only for Muslims, but a fraction of 1% of Indian Muslims – no more than any other religious group – live in polygamous homes. Even though polygamy laws are usually skewed in favor of allowing men (but not women) to take multiple spouses, women are sometimes granted rights as well. For example, in Burkina Faso and Chad, two countries where polygamy is common, first wives must state whether they eventually want to be polygamous before they marry, and several other countries require the first wife’s permission before the husband can marry other wives. Other countries set guidelines on what men owe to their spouses; this is the case in Mali, where men are allowed as many as four wives but are obligated to treat them equally and to ensure the welfare, education and moral development of all of their children. Not all countries where polygamy is common take these issues into account: In Guinea-Bissau, early and forced marriage, levirate marriage (the practice of requiring a widow to marry her late husband’s brother), and polygamy are widespread, and no legal guidelines apply. Theology The history and theology surrounding polygamy are complex. Taking multiple spouses – particularly wives – has been approved of at one point or another, and practiced to some degree, in various religions. 28 This sidebar draws heavily on research by the Organisation for Economic Co-operation and Development, published in the 2019 Social Institutions & Gender Index, the 2010 Atlas of Gender in Development: How Social Norms Affect Gender Equality in Non-OECD Countries and 2019 West Africa Brief. Data was also obtained from a variety of reports by the Office of the United Nations High Commissioner for Human Rights. 37 PEW RESEARCH CENTER www.pewresearch.org In Judaism and Christianity, the Bible refers to several instances of accepted plural marriages, including by Abraham, Jacob and David. However, plural marriages were disavowed by these groups in the Middle Ages, and polygamy generally has not been condoned by Jews or Christians in recent centuries.29 Still, polygamy sometimes was practiced by certain Christian sects, including by members of the Church of Jesus Christ of Latter-day Saints (sometimes called Mormons) in the U.S. until the late 1800s. Some Mormon splinter groups still practice polygamy. In Africa, Christian missionaries who arrived in the 18th century targeted the indigenous practice of polygamy as a priority for reform, and marriage became a point of conflict.30 Studies have found that these efforts were often successful, and polygamy in Africa has diminished over the past century, particularly in countries that have been influenced by Christian missionaries, according to James Fenske, an economist at the University of Warwick. However, many Christian churches without Western origins allow polygamy, which may help explain its continued prevalence in some Christian communities.31 Among Muslims, supporters of polygamy often cite Quran verse 4:3, which instructs men to take as many wives as they can take care of, up to four. They also note that the Prophet Muhammad had multiple wives. However, scholars point out that early Muslim populations lived in communities where polygamy was widespread, and that Islam limited the practice by providing guidelines and specifying obligations of husbands to each of their wives. These conditions of fairness are so demanding that they essentially make polygamy “impossible for a righteous man,” according to Azizah Al-Hibri and Raja El Habti in a chapter of “Sex, Marriage, and Family in World Religions.”32 Historians also have noted that Islamic guidance on polygamy was issued amid wars in Arabia, when there were many widows and orphans requiring financial support, and that polygamy created a system for them to be cared for. To this day, polygamy is most common in places where people, and particularly men, tend to die young. 29 Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions.” 30 Falen, Douglas J. 2008. “Polygyny and Christian Marriage in Africa: The Case of Benin.” African Studies Review. 31 Fenske, James. 2015. “African polygamy: Past and present.” Journal of Development Economics. 32 Al-Hibri, Azizah, and El Habti, Raja. 2006. “Islam.” In Browning, Don S., M. Christian Green, and John Witte Jr., eds. 2006. “Sex, Marriage, and Family in World Religions. 38 PEW RESEARCH CENTER www.pewresearch.org Living in polygamous households is very uncommon in most places % of individuals in polygamous households Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 39 PEW RESEARCH CENTER www.pewresearch.org Europe Almost three-quarters of Europeans are Christian, but there also are substantial minorities of “nones” and Muslims in Europe.33 In the 35 countries studied, the average European lives in a household of 3.1 people, with the two largest groups – Christians and the unaffiliated – at 3.1 and 3.0, respectively. European Muslims, on average, live in households of 4.1 people. Europe is the region where all three of these groups have the smallest households worldwide. Most of the countries in the region are economically advanced. Overall, Europe’s smallest households belong to Germans, Danes and Swedes (all at 2.7 members, on average). As in other parts of the world, countries with larger shares of Muslims tend to have bigger households. In Kosovo, where 94% of the population is Muslim, the average person belongs to a household of 6.8 people, making this the European country with the most expansive living arrangements. In nearby majority-Muslim Albania, people also reside in fairly large households by European standards (4.6). Other countries on the high end of the European scale are North Macedonia (4.6) and Montenegro (4.3), where majorities of the population are Christian, but Muslims make up sizable minorities. Wealth may explain some of these numbers: These Balkan countries are among the poorest in the region and are not European Union members. Nevertheless, within three of these four countries, Muslims have larger households than Christians. 33 The European data covers 35 countries and includes Christians, Muslims and the religiously unaffiliated. Jews, Buddhists and Hindus do not have enough respondents in European surveys to allow for reliable analysis. Christians, Muslims and ‘nones’ have their smallest households in Europe Average individual in Europe resides in a household of ___ people Household size Muslims 4.1 Christians 3.1 Unaffiliated 3.0 All Europe 3.1 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 40 PEW RESEARCH CENTER www.pewresearch.org When it comes to household types, Europe stands out for having the biggest share of people living alone: 13% of Europeans live in solo households, even more than the share of North Americans who do so (11%) and more than three times the global average (4%). European “nones” and Christians live alone at similar rates (14% and 13%, respectively). Though European Muslims are less likely to live alone (7%), they are more likely to live alone than Muslims anywhere else in the world. In Europe, small households are the norm Average individual resides in a household of ___ people Note: Asterisks indicate discrepancies based on rounding. Color categories are based on unrounded numbers, but household sizes are only shown to one decimal place. The average person experience a household size of 2.99 in the UK and 2.97 in France. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 41 PEW RESEARCH CENTER www.pewresearch.org Europe also stands out for having a large share of people living as couples without children or any other relatives (19%, similar to North America’s 20%). Living in a couple-only household is the most common arrangement for Christians in several Western European countries, including in Sweden, Germany, France, Finland and the Netherlands, where more than a quarter of Christians live this way. Couples also are the most common arrangement for “nones” in Finland and Germany. Ireland is an exception to this pattern: Irish Christians are three times as likely to live as a couple with children (40%) than as a couple without children (13%). Muslims across Europe are the least likely group to live in couple-only households, both in Muslim-majority countries such as Kosovo (2%) and in predominantly Christian countries such as Romania (9%). By the same token, while only about a quarter of all Europeans live with extended family, shares are much higher in Eastern Europe, especially among Balkan Muslims. In fact, Muslims are more Europe is the region with the largest share of people in solo households % of individuals who live alone Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 42 PEW RESEARCH CENTER www.pewresearch.org likely than Christians or the unaffiliated to live with extended family in every country with sufficient numbers of Muslims to compare (Albania, Belgium, Bulgaria, Kosovo, Montenegro, North Macedonia, Portugal, Romania and Russia) except Serbia, where Muslims live in this arrangement about as often as Christians do. In all of these countries (with the exception of Albania), Muslims are less likely than Christians to live alone. In Europe, Muslims more likely than Christians to live with extended families % of individuals in extended-family homes Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 43 PEW RESEARCH CENTER www.pewresearch.org Latin America and the Caribbean Latin America and the Caribbean is one of the least religiously diverse regions analyzed in this report, with a large Christian majority (90%), a modest share of religiously unaffiliated people (8%), and small shares of other groups, like Hindus, Buddhists, Muslims and Jews.34 Overall, the average Latin American lives in a household of 4.6 people, which is also the average for Christians and similar to Hindus and the unaffiliated (4.4). The largest households belong to Guatemalans (6.1 people) and the smallest to Puerto Ricans (3.5). The region’s most populous countries are clustered in between, with Mexicans and Peruvians (both 4.9) residing in households that are somewhat bigger than Brazilians’ (4.2).35 Two-parent households (39%) are the most common arrangement in Latin America, followed by extended families (32%). Unlike other regions, Latin America does not stand out for having stark outliers in its distribution across household types: Overall, people in this region are not substantially more or less likely than others around the world to live in, say, solo or polygamous households. 34 In Latin America, sufficient data is available for analysis of Christians, Buddhists, Hindus, Jews and the religiously unaffiliated, but numbers for Buddhists and Jews are not presented because only small populations of each group residing in Latin America. See Methodology on page 94 for further details. 35 The analysis covers 19 countries and territories. Some large countries, including Argentina and Venezuela, were not included due to a lack of suitable data. In Latin America-Caribbean region, narrow gaps in household size by religion Average individual in Latin America-Caribbean region resides in a household of ___ people Household size Christians 4.6 Hindus 4.4 Unaffiliated 4.4 All Latin America 4.6 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 44 PEW RESEARCH CENTER www.pewresearch.org There also tend not to be striking differences between Christians and religiously unaffiliated people or minority religious groups, but exceptions certainly exist. Guatemalans live in the biggest households in Latin America Average individual resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 45 PEW RESEARCH CENTER www.pewresearch.org In Jamaica, the unaffiliated live alone more than twice as commonly as Christians, with 16% of “nones” living solo in a country where only 6% of Christians do. And Hindus living in Guyana, where they make up almost a quarter of the population, have one fewer person per household (4.3 vs. 5.3) and are half as likely to live in single-parent households as Christians in the same small country (4% vs. 8%). In Latin America, similar shares of Christians and ‘nones’ live in each household type % of individuals in each household type, by religion Christians Unaffiliated Hindus All Latin America Two-parent 39% 38% 33% 39% Extended 32 29 35 32 Adult child 10 9 17 10 Couple 6 6 7 6 Single-parent 5 7 3 5 Solo 3 5 4 3 Polygamous <0.5 <0.5 <0.5 <0.5 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 46 PEW RESEARCH CENTER www.pewresearch.org The Middle Eastand North Africa The Middle East illustrates the way household patterns can vary in countries that are close neighbors but have very different economic, religious and cultural contexts. The region’s population is 93% Muslim, 4% Christian and less than 2% Jewish; this report covers seven Muslim-majority nations and Israel, where virtually all Middle Eastern Jews reside. Reliable data on Christians is available only in Egypt and Iraq. Overall, the average person in the Middle East-North Africa region lives in a fairly expansive household, with 6.2 members. People in Yemen (8.6) and Iraq (7.7) reside in the biggest households, while Tunisians (4.9) and Israelis (4.5) belong to the smallest households. As is the case in other regions, Muslims live in the largest households (6.3 members), followed by Christians (4.6) and Jews (4.3). There are notable differences even within countries: In Israel, Jews on average live with one fewer person than Muslims do (4.3 vs. 5.2). In the Middle East and North Africa, Muslims live in larger households than Jews Average individual in the Middle East-North Africa region resides in a household of ___ people Household size Muslims 6.3 Christians 4.6 Jews 4.3 All Middle East-North Africa 6.2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 47 PEW RESEARCH CENTER www.pewresearch.org The Middle East and North Africa stands out as the region where people are more likely than anywhere else to live in two-parent households. More than half of Middle Eastern Muslims and Christians live in this arrangement (57% and 58%, respectively). Muslims in the Palestinian territories are the most likely group in the world to live in two-parent families (71%). Israeli Jews do so at about half that rate (35%), making them the least likely group in the region to live in a two-parent household. Even within Israel, Muslims live in two-parent families much more often than Jews, with half of Israeli Muslims in this arrangement. Israel is different from other countries studied in that it is the only Jewish-majority nation – most Israelis are Jewish, and over 98% of Jews in the region live in Israel. The country also has a longer life expectancy than any of its neighbors, the highest overall levels of education, and a per-person GDP that is more than twice as big as the second-richest Middle Eastern country in this study, Iraq. The average Yemeni lives with nearly twice as many people as the average Israeli Average individual resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 48 PEW RESEARCH CENTER www.pewresearch.org Of all the Middle East and North African countries in this study, Israel has the highest share of people who live alone (6%). Israelis also are much more likely than others in the region to live in couple-only households. Polygamous relationships are legally recognized in at least some circumstances in parts of the Middle East-North Africa region, including several countries analyzed in this report – Algeria, Egypt, Iraq, Jordan and Yemen. In these countries, however, polygamy is rare: Iraqis and Yemenis are the most likely to live in this type of household (2% each). Data is not available from many of the countries where polygamy is legal and presumed to be more common, such as Saudi Arabia. Christians have Middle East-North Africa region’s lowest share of extended-family households % of individuals in each household type, by religion Muslims Christians Jews All Middle East-North Africa Two-parent 57% 58% 35% 56% Extended 27 16 29 27 Adult child 9 13 13 9 Couple 3 6 13 3 Solo 0.8 3 6 1 Polygamous 1 <0.5 <0.5 0.9 Note: Data on Jews and Christians is all from Israel and Egypt, respectively. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 49 PEW RESEARCH CENTER www.pewresearch.org North America In North America, Christians account for a majority of the population, while a rapidly growing share identifies with no religion, and fewer than one-in-ten affiliate with Islam, Judaism or other non-Christian religions. The United States and Canada are the only countries included in the North America region in this report; Mexico is included in the Latin America-Caribbean region. Because the U.S. population is much larger than Canada’s, U.S. household patterns have a much bigger influence on the overall numbers for the region, though the two countries have similar religious makeups and comparable levels of education, life expectancy and economic development. On average, North Americans live in the world’s second-smallest households, after Europeans, with an average of 3.3 members. Regional figures, available for Christians, “nones” and Jews, show a narrow range of household sizes across religious groups: North American Christians live in slightly bigger households (3.4 people) than their unaffiliated (3.2) or Jewish (3.0) counterparts. A sufficient number of Buddhists, Hindus and Muslims were surveyed to represent the characteristics of their households in Canada, but not in the U.S., and therefore not regionally. In Canada, these religious minority groups have larger households than Christians (3.2) or the unaffiliated (3.1). Muslims and Hindus live in the biggest households, with an average of 4.4 and 4.3 members, respectively, followed by Buddhists (3.9). North American Christians, ‘nones’ and Jews have similar sized households The average individual in North America resides in a household of ___ people Household size Christians 3.4 Unaffiliated 3.2 Jews 3.0 All North America 3.3 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 50 PEW RESEARCH CENTER www.pewresearch.org Small shares of North Americans overall live with extended family (11%). However, Canadian Hindus (28%), Buddhists (23%) and Muslims (16%) are more likely than other Canadians to live with extended family. Among the groups with available data in both Canada and the U.S., the religiously unaffiliated are the most likely to reside in extended-family arrangements (14%), followed by Christians (9%) and Jews (6%). The U.S. share of people living in single-parent families across all religious groups (9%) is among the highest in the world, behind only a handful of countries where the rate is higher: Sao Tome and Principe (15%), Kenya (12%), Jamaica and Rwanda (11% each).36 Fewer North American Jews live in this arrangement than Christians or “nones,” with only 5% in single-parent households. Jews also stand out with a relatively high share of people living in couple-only households – three-in-ten, compared with about one-in-five Christians and “nones.” 36 These relatively high rates of single-parent households in sub-Saharan African countries may be explained in part by the fact that people in that region are more commonly widowed at younger ages. Compared with the other countries mentioned here, the U.S. has a higher life expectancy at birth (more than a decade longer) than any of these other countries except Jamaica, where people can expect to live about 75 years, on average, just four years less than the 79 years in the U.S. Three-in-ten North American Jews live in couple-only households % of individuals in each household type Christians Unaffiliated Jews All North America Two-parent 34% 30% 30% 33% Couple 21 18 30 20 Adult child 14 15 11 14 Solo 11 12 13 11 Extended 9 14 6 11 Single-parent 9 9 5 9 Polygamous <0.5 <0.5 <0.5 <0.5 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 51 PEW RESEARCH CENTER www.pewresearch.org More people in U.S. than in other developed countries live in single-parent homes % of individuals in single-parent households Note: Single-parent households include one adult and at least one biological, step or foster child under 18. Adult children may be present, but no other relatives or non-relatives. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 52 PEW RESEARCH CENTER www.pewresearch.org 2. Household patterns by religion Pew Research Center analyzed data on six religious groups – Christians, Muslims, Hindus, Buddhists, Jews and people with no religious affiliation.37 Globally, the average Muslim lives in the biggest household (6.4 people), followed by the average Hindu (5.7), Christian (4.5), Buddhist (3.9), “none” (3.7) and Jew (3.7). Religious groups also vary in the types of households they are most likely to occupy: Hindus, Buddhists and the religiously unaffiliated most often reside in extended families, while Muslims, Christians and Jews have larger shares in two-parent homes. But religious groups are not monolithic, and followers of the same religion living in different parts of the world often vary substantially from each other. The experiences of religious groups are sometimes closely tied to the patterns found in the regions where they reside. 37 Although some faiths other than those analyzed in this report (such as Sikhs) have millions of adherents around the world, censuses and surveys in many countries do not measure them specifically. Because of this scarcity of census and survey data, this report does not attempt to analyze groups other than Christians, Muslims, Hindus, Buddhists, Jews and people with no religious affiliation; the report is also unable to show data for subgroups within these major religions, such as Protestants and Catholics or Sunnis and Shiites, although members of many other, smaller religious groups are included in general population results at the country, regional and global levels. 53 PEW RESEARCH CENTER www.pewresearch.org Religious groups also are distributed unevenly around the world. Christians are the most evenly scattered, with no more than a quarter living in any one region. The majority of Muslims, meanwhile, live in the Asia-Pacific region, but there are also large Muslim populations in the Middle East-North Africa and sub-Saharan Africa regions. Most of the world’s religiously unaffiliated people and even larger shares of all Buddhists and Hindus live in Asia, while Jews are concentrated in Israel and the United States. This report does not attempt to determine exactly how religion shapes household patterns. It is difficult to quantify the extent to which religion, on its own, affects people’s living arrangements – or conversely, how a person’s home life influences their religious affiliation. Still, comparing a single religious group in a country or region to the rest of the population may help illuminate possible connections between living arrangements and religion. This chapter examines each religious group separately and explores not only how the experiences of each group’s adherents vary from region to region, but also how their experiences compare with those of people who are not of that religion. Hindus and Buddhists are concentrated in Asia, while Christians are most evenly scattered globally % of individuals in each region, by religion Source: Data from Pew Research Center’s 2017 report “The Changing Global Religious Landscape.” “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 54 PEW RESEARCH CENTER www.pewresearch.org Information is provided on religious groups within regions when there are sufficient survey data to represent at least one-third of that group’s regional population and when that group has at least 500,000 adherents in that region. For example, enough Buddhists were surveyed in the Latin America-Caribbean region to represent 61% of the regional population – surpassing the one-third cutoff. However, only about 430,000 of the world’s 499 million Buddhists (about one-tenth of 1%) live in the region, so figures for Latin American Buddhists were not presented. See the Methodology on page 94 for more details. The analysis of religious groups is presented in descending order of the groups’ size, starting with Christians, who make up the world’s largest religion. 55 PEW RESEARCH CENTER www.pewresearch.org Christians Christians account for nearly a third of the global population, making them the largest of the major religious groups. They also are the most broadly distributed, with roughly equal percentages living in Europe (24%), Latin America and the Caribbean (25%) and sub-Saharan Africa (26%). The Asia-Pacific region and North America together account for most of the remaining quarter; the share of Christians who reside in the Middle East-North Africa region is less than 1%. In terms of their shares of the regional populations, Christians represent majorities in Latin America and the Caribbean (90%), North America (76%), Europe (73%) and sub-Saharan Africa (62%), and small minorities in the Asia-Pacific region (7%) and the Middle East and North Africa (4%). Christians: Household size Globally, Christians are the largest group in 12 of the 15 countries with the smallest households. Christians around the world live in somewhat smaller households, on average, than non-Christians (4.5 vs. 5.1 members). Christians have their smallest households in North America (3.4) and Europe (3.1), and – by a wide margin – their largest households in sub-Saharan Africa (6.0). Christians live with the largest number of people in Gambia, where their households contain 10.3 people, on average. And Christians’ smallest households are in Denmark and Sweden, both with an average of 2.6 people. The tendency of Christians to live in smaller households than others is particularly pronounced in areas where they live alongside Muslims: In sub-Saharan Africa and the Middle East-North Africa region, Christians have households with roughly two fewer people than non-Christians, on average. In other parts of the world, the gaps between Christians and others are much smaller. Globally, Christians live in slightly smaller households than others Average individual resides in a household of ___ people Christian Non-Christian World 4.5 5.1 Asia-Pacific 4.8 5 Europe 3.1 3.1 Latin America-Caribbean 4.6 4.3 Middle East-North Africa 4.6 6.3 North America 3.4 3.3 Sub-Saharan Africa 6.0 8.1 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 56 PEW RESEARCH CENTER www.pewresearch.org Average Christian in sub-Saharan Africa lives with three more people than a Christian in Europe Average Christian resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 57 PEW RESEARCH CENTER www.pewresearch.org Christians: Household types Christians around the world are most likely to live in two-parent families with minor children, and they do so at about the same rate as everyone else (34% vs. 32%). But Christians are markedly less likely than others to live in extended families (29% vs. 42%). In fact, Christians are the least likely group – aside from Jews (17%) – to live with a wider circle of relatives. On the other hand, Christians are more likely than non-Christians to live in household types that have few members: Larger shares of Christians live alone (7% vs. 3%) or as couples without other family members (11% vs. 7%). In some countries, such as Sweden (35%) and Germany (32%), living in a couple-only household is the most common arrangement for Christians. Globally, Christians also are more likely than non-Christians to live in single-parent households (6% vs. 3%), a type of arrangement that is generally more common in North America, sub-Saharan Africa and Latin America – all Christian-majority regions. Within these regions, Christians live in single-parent families at close to the same rates as non-Christians. In the Asia-Pacific, Latin America-Caribbean, Middle East-North Africa and sub-Saharan Africa regions, Christians overwhelmingly live in extended or two-parent families, with combined shares of more than 70%. Far fewer European Christians (49%) and North American Christians (43%) reside in those types of households, and they are more likely than Christians elsewhere to live alone (13% and 11%, respectively) or as a couple (21% in both regions). Christians live with extended family less often than others % of individuals in each household type, all countries combined Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 58 PEW RESEARCH CENTER www.pewresearch.org Globally, Christians are more likely than others to live in solo, single-parent or couple-only households % of individuals in each household type Two-parent Extended Couple Adult child Solo Single-parent Polygamous Asia-Pacific Christians 35% 37% 8% 10% 5% 2% <0.5% Non-Christians 30 46 7 10 3 2 <0.5 Europe Christians 25 24 21 11 13 4 <0.5 Non-Christians 28 27 18 7 13 5 <0.5 Latin America-Caribbean Christians 39 32 6 10 3 5 <0.5 Non-Christians 36 29 7 10 5 7 <0.5 Middle East-North Africa Christians 58 16 6 13 3 3 <0.5 Non-Christians 56 27 3 9 0.9 2 0.9 North America Christians 34 9 21 14 11 9 <0.5 Non-Christians 31 15 18 14 12 8 <0.5 Sub-Saharan Africa Christians 38 39 2 2 3 8 3 Non-Christians 37 29 2 1 2 5 22 World Christians 34 29 11 9 7 6 0.8 Non-Christians 32 42 7 9 3 3 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 59 PEW RESEARCH CENTER www.pewresearch.org Muslims About a quarter of all people are Muslims, making them the world’s second-largest religious group (and the fastest-growing major group). Over six-in-ten Muslims – about a billion – live in the Asia-Pacific region, and most other Muslims live in the Middle East-North Africa (20%) or sub-Saharan Africa (16%) regions. Muslims represent more than nine-in-ten people in the Middle East and North Africa, about three-in-ten sub-Saharan Africans and a quarter of the population in the Asia-Pacific region. Elsewhere, Muslims are small minorities, accounting for 6% of the population in Europe, 1% in North America and a statistically negligible fraction in Latin America.38 Muslims: Household size Globally, in the 15 countries with the biggest households, Islam is the largest religion in all but one – Benin. Muslims around the world live in households with an average of approximately two more people than non-Muslims (6.4 vs. 4.5), and they reside in larger families than non-Muslims in every region analyzed. One reason Muslims live in larger households is that they tend to have more children compared with other religious groups. Muslims around the world also are relatively young; in a handful of Muslim-majority countries, half or more of the population is under 18, and children are unlikely to live alone or in a couple-only arrangement. In sub-Saharan Africa, Muslims have their biggest households (8.5 people, on average) and also the widest gap in size compared with non-Muslims (6.1). The biggest households identified in this study belong to Muslims in Gambia (13.9), Senegal (13.6) and Mali (12.8) – all countries that have high rates of polygamy (see Chapter 1). Like other religious groups, Muslims have their smallest households in Europe. Still, in European countries with enough representation to compare Muslims with others, the average Muslim lives with more people than the average non-Muslim. 38 Too few Muslims in North America and the Latin America-Caribbean region were surveyed for analysis in this report. Muslims in every region have larger households Average individual resides in a household of ___ people Muslims Non-Muslims Asia-Pacific 6.0 4.6 Europe 4.1 3.1 Middle East-North Africa 6.3 4.4 Sub-Saharan Africa 8.5 6.1 World 6.4 4.5 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 60 PEW RESEARCH CENTER www.pewresearch.org The opposite is true in the country with the world’s largest Muslim population: In Indonesia, Muslims live in households with an average of 4.6 members, while non-Muslims (who are mostly Christian) live in households of 5.1 people, on average. Muslims in Gambia live with an average of 10 more people than Muslims in Russia Average Muslim resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 61 PEW RESEARCH CENTER www.pewresearch.org Muslims: Household types Muslims are less likely than others to live in households that contain no children or extended family. Only about 5% of Muslims live either alone or as a couple without children, compared with about 15% of non-Muslims. And Muslims are much more likely than others to live in two-parent homes with minor children (43% vs. 30%). Within regions, Muslims often differ from non-Muslims. In Asia and the Pacific, for example, four-in-ten Muslims live in two-parent homes, compared with three-in-ten non-Muslims. In Europe, Muslims are notably more likely than others to live in two-parent or extended-family households, and they are much less likely than non-Muslims to live in couple-only households (7% vs. 20%). In sub-Saharan Africa, Muslims live in extended-family households less frequently than others (27% vs. 39%). Conversely, Muslims are much more likely than others in this region to reside in polygamous homes (25% vs. 3%). (For more about polygamy in sub-Saharan Africa, see page 34.) Muslims are unique in their relatively large share of adherents living in polygamous households. Worldwide, in the countries studied, about 5% of Muslims live in this type of arrangement, which means that more Muslims live in polygamous households than live in solo, single-parent or couple-only households. This estimate may even be conservative because suitable data was not available from some Muslim-majority Persian Gulf countries where polygamy is legal and assumed to be common.39 39 Further, estimates of the share of people in polygamous households may be lower than the true share of people who belong to polygamous families; some polygamous families maintain multiple households. Muslims live in two-parent families more than other groups % of individuals in each household type, all countries combined Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 62 PEW RESEARCH CENTER www.pewresearch.org Muslims rarely live alone or as a couple with no other relatives % of individuals in each household type Two-parent Extended Adult child Polygamous Couple Single-parent Solo Asia-Pacific Muslims 42% 41% 7% 0.6% 4% 3% 1% Non-Muslims 27 46 11 <0.5 9 2 4 Europe Muslims 37 37 6 <0.5 7 5 7 Non-Muslims 26 26 10 <0.5 20 4 13 Middle East-North Africa Muslims 57 27 9 1 3 2 0.8 Non-Muslims 44 24 13 <0.5 10 3 5 Sub-Saharan Africa Muslims 37 27 0.9 25 2 4 1 Non-Muslims 38 39 2 3 2 8 3 World Muslims 43 36 6 5 3 3 1 Non-Muslims 30 39 10 <0.5 10 4 5 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 63 PEW RESEARCH CENTER www.pewresearch.org Religiously unaffiliated Like Christians, Muslims and other groups, the religiously unaffiliated have an array of identities and beliefs. Atheists, agnostics and people who do not identify with any religious group are all classified as unaffiliated. While the term “nones” has become widely accepted as shorthand for the religiously unaffiliated, this category also includes people who consider themselves religious and hold a mix of religious beliefs. Some people may choose “no religion” in response to a religious identity question on a survey because no other response option captures their identity. “Nones” may believe in deities, astrology or traditional religions – or they may believe in no supernatural beings whatsoever. Together, these people represent 16% of the global population and make up the third-largest group. A majority of all religious “nones” live in Asia, with six-in-ten found in China alone. About 12% of the unaffiliated reside in Europe, and 6% are in North America. Fewer than 5% live in each of the remaining regions. In terms of their share of regional populations, just over one-fifth of people in the Asia-Pacific region are religious “nones,” as are similar shares of Europeans and North Americans. About 8% of the total population in Latin America and the Caribbean, 3% of sub-Saharan Africans and less than 1% of all people in the Middle East-North Africa region have no religious affiliation. Economic development and its influence on households is particularly relevant for the living arrangements of the religiously unaffiliated. Many “nones” live in richer countries with relatively high education levels and greater workforce participation, particularly among women. People in such countries typically have easy access to birth control, and, as they spend more years in school and working outside the home, they tend to have children later in life and therefore have less time for childbearing. They also have high life expectancies, which may increase their chances of living without young children in their care. Individuals, couples and small families can afford to live alone, rather than with other relatives. 64 PEW RESEARCH CENTER www.pewresearch.org Unaffiliated: Household size Since three-quarters of “nones” live in the Asia-Pacific region, norms in this region set the tone for this group worldwide, and China is particularly numerically influential. Indeed, relatively small households in China are one reason that religiously unaffiliated people globally live in substantially smaller households than affiliated people (3.7 vs. 5.2 people, on average). Similar gaps exist in the Asia-Pacific region and sub-Saharan Africa, but not in North America, Europe and the Latin America-Caribbean region; affiliated and unaffiliated people do not differ much in their household sizes in those regions. “Nones” follow typical regional patterns in relative household size, with the largest families in sub-Saharan Africa (5.7) and the smallest in Europe (3.0). At the country level, the unaffiliated have their smallest households in Germany (2.5 people, on average) and their biggest households in Chad (9.2). Chad is one of only four countries in which the unaffiliated live in bigger households than their affiliated counterparts, and the gap is at least 0.5 people (Chadians with a religious affiliation live in households of 7.8 people, on average).40 In other countries, the unaffiliated live in households that are smaller than, or similar in size to, religiously affiliated households. 40 The other countries are Panama, where the average unaffiliated person lives in a household of 6.3 people, compared with 4.7 among the affiliated, Suriname (5.6 vs. 5.1), and Kenya (6.0 vs. 5.5). ‘Nones’ live with fewer people than others Average individual resides in a household of ___ people Unaffiliated Affiliated Asia-Pacific 3.7 5.3 China 3.8 4.0 Europe 3.0 3.2 Latin America-Caribbean 4.4 4.6 North America 3.2 3.4 Sub-Saharan Africa 5.7 6.9 World 3.7 5.2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 65 PEW RESEARCH CENTER www.pewresearch.org Differences in both household sizes and types are underpinned by the relatively high median age of the unaffiliated population (36 years, compared with 29 years for affiliated people) and their lower-than-average fertility rates. Globally, the average unaffiliated woman is expected to have about 1.6 children in her lifetime, compared with 2.5 for affiliated women. The average ‘none’ in Chad lives with about seven more people than the average ‘none’ in Germany Average unaffiliated individual resides in household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 66 PEW RESEARCH CENTER www.pewresearch.org Unaffiliated: Household types Living alone is more common among the religiously unaffiliated than among others (7% vs. 4%), as is living in a couple-only arrangement (14% vs. 7%) or in adult child households (12% vs. 9%). Conversely, fewer of the unaffiliated live in two-parent households with minor children (26% vs. 34%) or in single-parent homes (2% vs. 4%). Just over a third of “nones” live in extended-family homes, similar to the share of the affiliated who do so. The small share of single-parent families among the unaffiliated worldwide is tied to the concentration of the unaffiliated in the Asia-Pacific region, where living in this type of household is uncommon across religious groups. This is especially true in China and South Korea, where many unaffiliated people live and single-parent families are particularly rare. In sub-Saharan Africa, “nones” stand out for living alone at the highest rate of any group in the region (7%) – more than three times the share of affiliated people (2%). “Nones” in the Asia-Pacific region also are more likely than others to live alone (6% vs. 3%), as well as in a couple-only households (13% vs. 6%). Differences between “nones” and others are not as pronounced in Latin America and the Caribbean, North America, and Europe. ‘Nones’ are more likely than others to live alone or in a couple % of individuals in each household type, all countries combined Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 67 PEW RESEARCH CENTER www.pewresearch.org In Europe and North America, household types among ‘nones’ are similar to others % of individuals in each household type Extended Two-parent Couple Adult child Solo Single-parent Polygamous Asia-Pacific Unaffiliated 41% 25% 13% 14% 6% 1% <0.5% Affiliated 46 32 6 9 3 3 <0.5 China Unaffiliated 44 23 13 14 5 1 <0.5 Affiliated 45 21 12 15 5 1 <0.5 Europe Unaffiliated 25 26 20 7 14 5 <0.5 Affiliated 25 26 20 11 12 4 <0.5 Latin America-Caribbean Unaffiliated 29 38 6 9 5 7 <0.5 Affiliated 32 39 6 10 3 5 <0.5 North America Unaffiliated 14 30 18 15 12 9 <0.5 Affiliated 10 34 21 13 11 9 <0.5 Sub-Saharan Africa Unaffiliated 39 32 4 3 7 7 5 Affiliated 35 38 2 2 2 6 11 World Unaffiliated 37 26 14 12 7 2 <0.5 Affiliated 39 34 7 9 4 4 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 68 PEW RESEARCH CENTER www.pewresearch.org Hindus More than a billion people – almost one-sixth of the world’s population – are Hindu. Hindus are heavily concentrated in the Asia-Pacific region, and more than nine-in-ten of the world’s Hindus live in just one country: India. As a result, the global characteristics of Hindu households are heavily influenced by patterns in that country. Even though most Hindus live in the Asia-Pacific region, they make up only about a quarter of all people in the world’s most populous region. Hindus represent less than 1% of the population in all other regions.41 Hindus: Household size Globally, the average Hindu lives in a fairly large household (5.7 people), with nearly one person more than the average non-Hindu (4.8). In India, however, the reverse is true: Hindus live in slightly smaller households than non-Hindus (5.7 vs. 6.2) – even though Hindus’ biggest households are in India. Hindus in the Latin America-Caribbean region – mostly found in Trinidad and Tobago, Guyana, and Suriname – live in slightly smaller households than other Latin Americans (4.4 vs. 4.6). Meanwhile, the small population of Hindus in sub-Saharan Africa – many of whom live in South Africa – have much smaller households than others in the region (3.9 vs. 6.9). 41 Even though Hindus represent very small minorities in sub-Saharan Africa and the Latin America-Caribbean region, they are concentrated in a handful of countries for which large surveys are often available, including South Africa, Trinidad and Tobago, and Guyana. As a result, general statements can be made about Hindus in these two regions. At the country level, there were only enough Hindu households surveyed to confidently represent them in India, Bangladesh, Indonesia, Nepal, the Philippines, Guyana, Suriname, Trinidad and Tobago, Canada, Botswana, South Africa and Zambia. Due to population weighting, Hindus outside of Asia have very little influence on global p atterns. Hindus have larger households than others in Asia-Pacific region – but not in India Average individual resides in a household of ___ people Hindu Non-Hindu Asia-Pacific 5.7 4.7 India 5.7 6.2 Latin America-Caribbean 4.4 4.6 Sub-Saharan Africa 3.9 6.9 World 5.7 4.8 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 69 PEW RESEARCH CENTER www.pewresearch.org Hindus have larger households in India than in other countries Average Hindu resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 70 PEW RESEARCH CENTER www.pewresearch.org Hindus: Household types Globally, Hindus are the only religious group with a majority of people living in extended-family homes, with 55% residing in this type of arrangement. Relatedly, Hindus are less likely than non-Hindus to live in a couple-only arrangement (3% vs. 9%). Hindus, along with Muslims, have the smallest share of adherents who live alone (1% in each group). In the Asia-Pacific region, Hindus are far more likely than non-Hindus to live in extended families (55% vs. 42%). But this pattern is less pronounced in India, where religious minorities live in arrangements similar to Hindus. Indian Hindus are only a little more likely to live with extended family (55%, compared with 51% for non-Hindus). Hindus are the only group with a majority living in extended families % of individuals in each household type Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 71 PEW RESEARCH CENTER www.pewresearch.org Outside Asia, Hindus stand out for high shares in adult child families % of individuals in each household type Extended Two-parent Adult child Couple Single-parent Solo Polygamous Asia-Pacific Hindus 55% 30% 8% 3% 3% 0.9% <0.5% Non-Hindus 42 31 11 9 2 4 <0.5 India Hindus 55 30 8 3 3 0.9 <0.5 Non-Hindus 51 35 7 2 3 0.6 <0.5 Latin America-Caribbean Hindus 35 33 17 7 3 4 <0.5 Non-Hindus 32 39 10 6 5 3 <0.5 Sub-Saharan Africa Hindus 32 31 22 10 0.8 3 <0.5 Non-Hindus 35 37 2 2 6 2 11 World Hindus 55 30 8 3 3 0.9 <0.5 Non-Hindus 35 33 9 9 4 5 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 72 PEW RESEARCH CENTER www.pewresearch.org Buddhists With about 500 million adherents worldwide, Buddhists represent roughly 7% of the global population. Nearly 99% of Buddhists live in Asia and the Pacific. Buddhists make up just over 10% of all people in the Asia-Pacific region, 1% of North Americans and less than one-half of 1% of people in any other region.42 Buddhists: Household size Around the world, Buddhists live with at least one fewer person, on average, than non-Buddhists do. This reflects the gap in Asia, where the average Buddhist lives in a household of 3.9 people, compared with 5.1 for non-Buddhists. About half of the world’s Buddhists live in China, where they make up almost one-fifth of the population. In China, Buddhists live in households of 3.8 people, on average, as do non-Buddhists. The second-biggest population of Buddhists by country is in Thailand, where they make up over 90% of the population. In Thailand, Buddhists have an average household size of 4.1, whereas others live in larger households, with an average size of 4.7. (Most non-Buddhists in China are unaffiliated, while most non-Buddhists in Thailand are Muslim.) Buddhists have their smallest families in Japan (3.0), where their average household sizes are about the same as those of non-Buddhists. In Canada, though, Buddhists live in bigger households than others (3.9 vs. 3.2), which is partly because Canadian Buddhists are more likely than other Canadians to live with extended family. 42 Sufficient data to represent Buddhists outside of Asia was only available for the Latin America-Caribbean region. However, because fewer than 500,000 Buddhists live in that part of the world, region level results for Latin America are not presented separately. At the country level, results for Buddhists are included in this chapter and in Appendix C for any country where an adequate number of Buddhists were surveyed (18 total). Buddhists live with fewer people than others Average individual resides in a household of ___ people Buddhist Non-Buddhist Asia-Pacific 3.9 5.1 China 3.8 3.8 World 3.9 5.0 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 73 PEW RESEARCH CENTER www.pewresearch.org Buddhists have their smallest households in Japan Average Buddhist resides in a household of ___ people Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 74 PEW RESEARCH CENTER www.pewresearch.org Buddhists: Household types When it comes to household type, the biggest difference between Buddhists and non-Buddhists is in the relatively small percentage of Buddhists who live in two-parent households with minor children. One-fifth of Buddhists globally live in this arrangement, compared with one-third of non-Buddhists. In fact, of all major religious groups, Buddhists have the smallest share of adherents in this household type. As a group, Buddhists are older than non-Buddhists, with a median age of 36, compared with 29 for non-Buddhists. And the average Buddhist woman is expected to have 1.6 children in her lifetime, well below the global figure for non-Buddhists (2.4). These demographic factors, which are themselves influenced by the legacy of the one-child policy in China, help to explain why Buddhists tend to have smaller households and are less likely to live in household types that by definition include minor children. Buddhists also are unlikely to live in single-parent families (2%), which reflects the rarity of this type of household for all religious groups in the Asia-Pacific countries where most Buddhists are found, and the fact that there are relatively few Buddhists in countries or regions where single parenthood is common. At the country level, about the same share of Buddhists and non-Buddhists live in single-parent households in China, Thailand, Japan and Canada. Within the Asia-Pacific region, Nepal has the biggest share of Buddhists living in single-parent families (9%). Globally, Buddhists are slightly more likely than others to live in extended-family households (44% vs. 38%), in couple-only households (13% vs. 8%), in adult child homes (13% vs. 9%) and alone (7% vs. 4%). Within the Asia-Pacific region, however, Buddhists and non-Buddhists live with extended family at about the same rate. Fewer than a quarter of Buddhists live in two-parent households % of individuals in each household type, all countries combined Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 75 PEW RESEARCH CENTER www.pewresearch.org Buddhists are more likely than others to live alone or in couples % of individuals in each household type Extended Two-parent Adult child Couple Solo Single-parent Polygamous Asia-Pacific Buddhists 44% 20% 13% 13% 7% 2% <0.5% Non-Buddhists 45 32 10 7 3 2 <0.5 China Buddhists 45 19 14 14 6 2 <0.5 Non-Buddhists 44 23 14 12 5 1 <0.5 World Buddhists 44 20 13 13 7 2 <0.5 Non-Buddhists 38 33 9 8 4 4 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 76 PEW RESEARCH CENTER www.pewresearch.org Jews Globally and in every region, Jews are a minority religious group; they make up less than one-quarter of 1% of the global population. About 40% of Jews live in Israel, where they represent a large majority of the population, and a similar number live in the United States, where they form roughly 2% of the population. Jews: Household size Jews in the Middle East-North Africa region live in bigger households than those in North America.43 In Israel, Jews live in households of 4.3 people, on average (compared with 5.2 for non-Jewish Israelis); in the U.S., those figures are 3.0 and 3.4, respectively.44 Of all religious groups analyzed for this report, Jews are the oldest, with a median age of 37, compared with 30 among non-Jews. The comparatively high percentage of Jews who live alone or with only a spouse or partner can be partially attributed to their older age. Jews: Household types Globally, Jews have the biggest share of adherents in couple-only households and are more than twice as likely as non-Jews to live in this arrangement (21% vs. 8%). Jews are more likely to live as couples in the U.S. than in Israel (30% vs. 13%), and in both places, they are more likely than non-Jews to live with only a spouse or partner. Jews also are unique in that the couple-only household type is their second-most common (behind the two-parent family with minor children); in all other groups, two-parent and extended-family are the most common household types. 43 Sufficient data to represent Jews was also available in the Latin America-Caribbean region. However, because fewer than 500,000 Jews live in that part of the world, region level results for Latin America are not presented separately. At the country level, results for Jews are included in Appendix C for all six countries where an adequate number of Jews were surveyed. These include Brazil, Canada, Israel, Mexico, Romania and the U.S. 44 This analysis only includes people who identify religiously as Jewish. A previous Pew Research Center study found that about one-in-five U.S. Jews (more broadly defined) describe their religious identity as atheist, agnostic or “nothing in particular,” but nonetheless say they consider themselves Jewish in other ways, such as culturally, ethnically or by family background. They would not be included as Jews in this analysis, but rather as religious “nones.” Israeli Jews have much smaller households than others in the Middle East Average individual resides in a household of ___ people Jews Non-Jews Middle East-North Africa 4.3 6.3 Israel 4.3 5.2 North America 3.0 3.4 United States 3.0 3.4 World 3.7 4.9 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 77 PEW RESEARCH CENTER www.pewresearch.org Jews have a much smaller share than others living in extended families (17% vs. 38%). Again, there are big differences in the experiences of U.S. Jews and Israeli Jews: While 29% of Israeli Jews live with extended family, only 6% of U.S. Jews do. Conversely, U.S. Jews live alone at about twice the rate of Israeli Jews (13% vs. 6%). Jews stand out for living in extended families much less often than others % of individuals in each household type, all countries combined Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 78 PEW RESEARCH CENTER www.pewresearch.org Israeli Jews much more likely than U.S. Jews to live with extended family % of individuals in each household type Two-parent Couple Extended Adult child Solo Single-parent Polygamous Middle East-North Africa Jews 35% 13% 29% 13% 6% 3% <0.5% Non-Jews 57 3 27 9 0.8 2 1 Israel Jews 35 13 29 13 6 3 <0.5 Non-Jews 48 6 34 6 4 2 1 North America Jews 30 30 6 11 13 5 <0.5 Non-Jews 33 20 11 14 11 9 <0.5 United States Jews 30 30 6 11 13 5 <0.5 Non-Jews 33 20 11 14 11 9 <0.5 World Jews 32 21 17 12 10 4 <0.5 Non-Jews 33 8 38 9 4 4 2 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 79 PEW RESEARCH CENTER www.pewresearch.org Sidebar: Studies often show links between religion and family life Researchers who explore the connections between religion and household patterns often focus on major life events: marriage, divorce and childbearing. Their findings suggest that several measures of religion – a person’s affiliation, how important religion is to them and how often they participate in their congregation – have some influence on living arrangements. While many of the studies have been conducted in the U.S., using largely Christian samples, there is a growing body of research focused on other parts of the world.45 Some of these findings are presented here. Marriage and divorce Marriage, divorce and childbearing patterns are tied to religious identity and participation. In the U.S., evangelical Protestants are more likely than members of some other religious groups, including Catholics and mainline Protestants, to marry as young adults. Young evangelical Protestants also are less likely than others to live with a romantic partner outside of marriage.46 In the UK, the religiously affiliated, especially those who regularly attend religious services, also are less likely than “nones” to cohabit.47 Muslims are especially likely to marry: Studies have found that Muslims in many countries around the world are more likely than Christians to be married, and that in nations with larger shares of Muslims, women tie the knot at a younger age.48 Premarital sex is rarer among Muslims, and both Muslims and Buddhists are relatively unlikely to engage in extramarital sex.49 The connection between divorce and religion also has interested researchers. Married “nones” are more likely than their religiously affiliated peers to go through a divorce at some point in their lives, according to a meta-analysis of 10 peer-reviewed research projects.50 Nevertheless, U.S. states with bigger shares of religious conservatives experience relatively high divorce rates, even though evangelical Protestants tend to emphasize the sanctity of marriage. This seeming contradiction may be explained by the tendency among conservative Protestants to marry at a younger age.51 45 Mahony, Annette. 2010. “Religion in Families 1999 to 2009: A Relational Spirituality Framework.” Journal of Marriage and Family. 46 Eggebeen, David, and Jeffrey Dew. 2009. “The Role of Religion in Adolescence for Family Formation in Young Adulthood.” Journal of Marriage and Family. Also see Keister, Lisa A., 2011. “Faith and Money: How Religious Belief Contributes to Wealth and Poverty.” Also see Uecker, Jeremy E., and Charles E. Stokes. 2008. “Early Marriage in the United States.” Journal of Marriage and Family. 47 Village, Andrew, Emyr Williams, and Leslie Francis. 2010. “Living in Sin? Religion and Cohabitation in Britain 1985-2005.” Marriage & Family Review. 48 Fieder, Martin, Susanne Huber, Elmar Pichl, Bernard Wallner, and Horst Seidler. 2018. “Marriage gap in Christians and Muslims.” Journal of Biosocial Science. Also see Carmichael, Sarah. 2011. “Marriage and power: Age at first marriage and spousal age gap in lesser developed countries.” History of the Family. The authors adjust for regional differences in these outcomes with the use of clustered standard errors and controls for standard family types and other relevant nation-level attributes. 49 Adamczyk, Amy, and Brittany E. Hayes. 2012. “Religion and Sexual Behaviors: Understanding the Influence of Islamic Cultures and Religious Affiliation for Explaining Sex Outside of Marriage.” American Sociological Review. 50 Mahoney, Annette, Kenneth I. Pargament, Nalini Tarakeshwar, and Aaron B. Swank. 2001. “Religion in the Home in the 1980s and 1990s: A Meta-Analytic Review and Conceptual Analysis of Links Between Religion, Marriage, and Parenting.” Journal of Family Psychology. 51 Glass, Jennifer, and Philip Levchak. 2014. “Red States, Blue States, and Divorce: Understanding the Impact of conservative Protestantism on Regional Divorce Rates.” American Journal of Sociology. 80 PEW RESEARCH CENTER www.pewresearch.org Childbearing In the baby boom years after World War II, U.S. Catholics tended to have more children than non-Catholics. But by 1979, Princeton demographers noted that this gap had nearly disappeared and declared “The end of Catholic fertility.”52 Around the globe, Muslims have higher fertility rates than Christians on average.53 Muslim women’s low educational attainment is a likely factor; demographers find that higher educational attainment among women is tied to lower fertility rates.54 Even though fertility rates generally are declining in Muslim-majority nations, the above-average number of children born to Muslim women contributes to Muslims’ larger household sizes, as described elsewhere in this report.55 Buddhism appears to have a unique association with procreation compared with the other major world religions: An analysis of fertility data in six Asian nations found that Buddhist affiliation was either unrelated to the number of children born or associated with having fewer children. This may be because Buddhism, unlike Abrahamic religions, does not have a specifically pro-natalist doctrine.56 (For more on religious teachings about family life, see the sidebar on page 22.) In Western Europe, women who belong to religious groups and attend services more often tend to have more children.57 52 Westoff, Charles F., and Elise F. Jones. 1979. “The End of ‘Catholic’ Fertility.” Demography. Also see Westoff, Charles F., and Raymond H. Potvin. 1967. “College Women and Fertility Values.” 53 See Chapter 1 of Pew Research Center’s 2015 report “The Future of World Religions: Population Growth Projections, 2010-2050.” 54 See Pew Research Center’s 2016 report “Religion and Education Around the World.” Some research finds that variation in the educational attainment of Muslim women is largely explained by the wealth of countries in which Muslim women live. See McClendon, David, Conrad Hackett, Michaela Potancokova, Marcin Stonawski, and Vegard Skirbekk. 2018. “Women’s Education in the Muslim World.” Population and Development Review. 55 Eberstadt, Nicholas, and Apoorva Shah. 2012. “Fertility Decline in the Muslim World.” Policy Review. 56 Skirbekk, Vegard, Marcin Stonawski, Setsuya Fukuda, Thomas Spoorenberg, Conrad Hackett, and Raya Muttarak. 2015. “Is Buddhism the low fertility religion of Asia?” Demographic Research. 57 Peri-Rotem, Nitzan. 2016.“Religion and Fertility in Western Europe: Trends Across Cohorts in Britain, France and the Netherlands.” European Journal of Population. 81 PEW RESEARCH CENTER www.pewresearch.org 3. Household patterns by age and gender People move between different types of households throughout their lives. Someone born into a two-parent home might become a member of an extended-family household when a grandmother moves in, live as a couple with their spouse in middle adulthood and end up in a solo household after that partner dies. Differences in living arrangements are particularly striking for children under 18 and adults who are 60 and older. While someone in their 30s can easily be found in any household type, it is unusual for anyone under 18 to live solo, or for someone in their 60s to raise a child alone. Moreover, examining age groups separately is important because a society’s age composition affects living arrangements for the overall population. Countries with longer life expectancy, for example, are more likely to have people living as couples without children, in part because parents may live for decades after their children move out. Places with higher fertility are more likely to have two-parent and extended-family homes simply because there are more children overall per household. Sex is another variable associated with living arrangements, both within religious groups and across them. Women are more likely than men to be single parents, for example, and also more likely to live alone in later years. Earlier in this report, all age and gender groups were included when analyzing shares of people in different types of households. This chapter examines the differences, by religion and region, between people within the same phase of life, and of the same gender.58 58 Surveys available in most countries do not record the religious affiliation of children, so children are typically categorized using the head of household’s religion as a proxy. In countries where children’s religious identity is collected (such as India), it rarely differs from that of their parents. For more details, see the Methodology on page 94. 82 PEW RESEARCH CENTER www.pewresearch.org Living arrangements of children Children under 18 make up about a third of the world’s population – and in the fast-growing regions of sub-Saharan Africa and the Middle East-North Africa, they account for 48% and 37%, respectively. For children around the world, sharing a home with two parents – whether biological, step, adoptive or foster, married or unmarried – is the most common arrangement (51%). But many children also live in extended-family homes (38%), whether with aunts and uncles, grandparents, or other family members aside from parents and siblings. A little more than half of Christians, Muslims and “nones” under 18 live with two parents, while fewer than four-in-ten among these groups reside in extended-family homes. The pattern is reversed for Hindu and Buddhist children, who, like their adult counterparts, are more likely to live with extended family. Roughly half of Hindu children live with extended families (52%), while about four-in-ten live in two-parent homes (43%). As has been noted before, 94% of Hindus live in India, where extended-family households are common overall. Still, Hindu children in India are even more likely than their Christian, Muslim or unaffiliated peers to live with extended family. Single-parent homes are not common in general, with fewer than one-in-ten children around the world living with one parent and no other adults.59 But children from Christian families are about twice as likely as non-Christian children to reside in single-parent households. And the U.S. has 59 Incidence rates for single-parent households may be lower in this report than in other analyses that use a broader definition. Here, single-parent households include one adult and at least one biological, step or foster child under 18. Adult children may be present, but this classification does not include households that contain other relatives or non-relatives. Research using a more expansive definition may classify homes that also have other adults present (such as grandparents, aunts, uncles or non-relatives) as single-parent households. Most children in Christian, Muslim and unaffiliated homes live with two parents % of children under 18 in each household type, all countries combined Two-parent Extended Single-parent Polygamous Christians 52% 32% 11% 1% Muslims 53 33 5 7 Unaffiliated 54 39 7 <0.5 Hindus 43 52 5 <0.5 Buddhists 45 50 5 <0.5 All 51 38 7 3 Note: Due to sample size limitations in the United States, Jewish children are only adequately represented in Israel and therefore not analyzed at the global level. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 83 PEW RESEARCH CENTER www.pewresearch.org the highest rates of children living in single-parent households, with almost a quarter of Christian (as well as unaffiliated) children living this way.60 60 Throughout this report, the U.S. usually refers to the 50 states and District of Columbia; results for the U.S. territory of Puerto Rico are presented separately. However, Puerto Rico has a similar share of children living in single-parent households (25%) to the rest of the U.S. (23%). U.S. children are more likely than children elsewhere to live in single-parent homes % of children under 18 in single-parent households Note: Single-parent households include one adult and at least one biological, step or foster child under 18. Adult children may be present, but no other relatives or non-relatives. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 84 PEW RESEARCH CENTER www.pewresearch.org Globally, the share of Muslim children living in single-parent homes is smaller than the percentage residing in polygamous families (5% vs. 7%). Among other religious groups, very small shares of children – 1% or less – live in polygamous households.61 Growing up in polygamous households is relatively common in sub-Saharan Africa, where 12% of children live in such arrangements – including an even higher share of children in Muslim families (27%). Living alone or as a couple is very rare for children, though not unheard of. In South Korea, about 5% of Christian and unaffiliated children under 18 live in solo households. 61 The 7% figure for Muslim children in polygamous households is likely conservative because data is not available from most Muslim-majority Gulf countries where polygamy is legal and assumed to be somewhat common, including Saudi Arabia and the United Arab Emirates. Relatively few North American children live in extended families % of children under 18 in each household type Two-parent Extended Single-parent Polygamous Asia-Pacific 50% 45% 4% <0.5% Europe 58 27 13 <0.5 Latin America-Caribbean 55 32 9 <0.5 Middle East-North Africa 71 24 3 1 North America 67 8 23 <0.5 Sub-Saharan Africa 42 34 8 12 World 51 38 7 3 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 85 PEW RESEARCH CENTER www.pewresearch.org Living arrangements of people 60 and older People in later adulthood have many things in common, no matter what religious group they belong to or where in the world they live. Most of them have finished bearing and raising children, and many have experienced the loss of a spouse, a need for caretaking, the birth of a new generation of family members or some other shift that causes a change in their living arrangements. Yet adults over age 60 of different religions and in different parts of the world vary widely in their living arrangements. In addition to some of the other factors discussed in this report that are related to household patterns (such as religion, geography and economics) living arrangements for older people are particularly tied to another measure of prosperity – life expectancy. (See chart on page 88.) Around the world, living with extended family is the most common arrangement for people ages 60 and older, even though fewer than half of all older adults globally (38%) live in this household type. Living as a couple is the second most common experience (31%), followed by living alone (16%). Hindu and Muslim seniors are particularly likely to live with extended family. Almost three-quarters of Hindus (72%) live in this type of arrangement, including in multigenerational households. Hindus are concentrated in India and other Asia-Pacific countries where extended-family households are relatively common. By contrast, fewer than a quarter of Christians ages 60 and older globally (23%) live with extended family, while an identical share of Christian seniors around the world live alone – the highest of Among seniors, Hindus are the least likely to live alone % of individuals ages 60 and older in each household type Christians Muslims Unaffiliated Hindus Buddhists All Extended 23% 53% 35% 72% 36% 38% Couple 38 15 38 13 34 31 Solo 23 7 17 4 18 16 Adult child 12 14 9 8 12 11 Two-parent 2 7 <0.5 2 <0.5 2 Polygamous <0.5 3 <0.5 <0.5 <0.5 0.5 Single-parent 0.5 <0.5 <0.5 <0.5 <0.5 <0.5 Note: Due to sample size limitation in the United States, Jewish adults ages 60 and older are only adequately represented in Israel and therefore not analyzed at the global level. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 86 PEW RESEARCH CENTER www.pewresearch.org any religious group analyzed. Among older Christian and unaffiliated adults, living as a couple is the most common arrangement (38% each). Muslims have a higher-than-average share of adults over 60 who are still raising minor children without extended family, with 7% of older Muslims living in two-parent or single-parent households combined – equal to the share who live alone. Europe and North America have the largest shares of older adults who live alone, with a quarter or more found in solo households; these regions also have the highest percentages of older adults living as couples, with just under half in both regions found in that type of arrangement, while relatively few older Europeans and North Americans live with extended family. Some of these differences are tied to levels of economic development, particularly as measured by life expectancy. Older adults are more likely to live alone or as couples in countries where an average person can expect to live more than 70 years. In countries where lives are shorter, seniors tend to live with other family members instead. Life expectancy is often linked to other markers of prosperity within a country, so older adults who can expect to live into their 80s also tend to live in countries where living alone is more affordable and sources other than family may provide help in meeting basic needs. Half of older adults in sub-Saharan Africa, but few in North America, live with extended family % of people ages 60 and older in each household type Asia-Pacific Europe Latin America-Caribbean Middle East-North Africa North America Sub-Saharan Africa World Extended 50% 16% 41% 39% 7% 51% 38% Couple 27 46 20 18 47 9 31 Solo 11 28 12 9 26 9 16 Adult child 10 8 20 25 14 6 11 Two-parent 2 <0.5 3 8 3 11 2 Polygamous <0.5 <0.5 <0.5 0.7 <0.5 8 0.5 Single-parent <0.5 <0.5 <0.5 <0.5 2 1 <0.5 Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 87 PEW RESEARCH CENTER www.pewresearch.org Cultural norms also may play a role; in Western countries it is often expected that seniors will live alone.62 In India and other parts of the world it may be seen as a normal arrangement (or even a responsibility) for adult children to take their aging parents into their home. One example of this can be seen among immigrant communities in the West: In Canada, almost half of Hindus ages 60 and older (who are often immigrants or the children of immigrants from South Asia) live with extended families (47%), more than four times the share of Canadian seniors overall (10%). 62 Institutional populations, including people living in nursing homes, are outside the scope of this report, which analyzes the distribution of people in the world’s household population. It may be more common for seniors to live in nursing homes in Western countries than is the case elsewhere. 88 PEW RESEARCH CENTER www.pewresearch.org Older adults live alone more often in places with longer life expectancy % of individuals ages 60 and older in solo households Note: Life expectancy data is not available for Kosovo. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. Life expectancy data come from the United Nations. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 89 PEW RESEARCH CENTER www.pewresearch.org Experiences of men and women Men and women around the world experience households differently, and these gender gaps sometimes vary by religious affiliation. Gender and religion play a role when it comes to three measures in particular: The age gap between cohabiting partners, the prevalence of single-parent households, and the rate of living alone in old age. Spousal age gap Men tend to be older than their wives or female cohabiting partners by a global average of about four years.63 Around the world, Jewish couples tend to have the smallest age gaps (2.1 years, on average) while Muslims have the biggest (6.6 years). Across all groups, the gaps are smallest in Europe and North America (less than three years), and biggest in sub-Saharan Africa (8.6 years). Of the 130 countries analyzed, there is no place where male partners are younger than their female partners, on average, or even the same age. The countries with the world’s widest age gaps are all in sub-Saharan Africa and have high rates of polygamy; they include Gambia and Guinea, both with average gaps of around 14 years. The countries with the narrowest age gaps are largely in Europe (including the Czech Republic, Slovakia and Estonia), in addition to the U.S., China and Mongolia, all with gaps of around two years. 63 Same-sex partners are not included in this analysis. Age gaps between partners are widest in Africa, particularly among Muslims Women are younger than their husbands or male cohabiting partners by an average of ___ years Sub-Saharan Africa Middle East-North Africa Asia- Pacific Latin America-Caribbean Europe North America World Christians 7.1 6.6 3.0 3.6 2.6 2.2 3.8 Muslims 11.6 6.2 5.7 5.0 6.6 Unaffiliated 6.4 2.2 3.8 2.4 2.2 2.3 Hindus 3.5 5.6 4.0 5.6 Buddhists 2.9 3.5 2.9 Jews 3.1 1.2 2.1 All 8.6 6.1 4.0 3.6 2.7 2.2 4.2 Religion-region combinations in which small populations and/or survey samples prohibit reliable analysis. Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 90 PEW RESEARCH CENTER www.pewresearch.org Single parents All over the world, women in middle adulthood (ages 35 to 59) are more likely than their male counterparts to live in single-parent homes. The gender gap in single-parenthood in this age group is widest among Christians. Worldwide, about 2% of middle-aged Christian men live in single-parent homes, while about 7% of women do. On the other end of the spectrum, Buddhist men and women in this age group do not differ much on this measure. These patterns partly reflect the large share of Christians in sub-Saharan Africa, where single-parent households are relatively common, and the big shares of Buddhists in Asia, where such households are rare. Middle-aged men tend to live in single-parent homes at about the same rate all over the world and regardless of religion – any differences in rates of single parenthood affect women almost exclusively. Middle-aged women in sub-Saharan Africa have the largest share in single-parent households, with about 9% in this arrangement – although North American (8%) and Latin American (7%) women are not far behind. 91 PEW RESEARCH CENTER www.pewresearch.org Women live as single parents more often than men % of individuals ages 35 to 59 in single-parent households Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 92 PEW RESEARCH CENTER www.pewresearch.org Aging alone Around the world, women ages 60 and older are only slightly more likely to live in a couple (27%) than to live alone (20%). Older men, however, are three times as likely to live in a couple (36%) than they are to reside alone (11%). This pattern, which applies across religious groups and country borders, is tied to life expectancy: Because women tend to live at least a few years longer than men, larger shares of older women – in every religious group and in every region – live alone. The tendency of women to live alone in advanced age is a result of both women’s greater longevity and their tendency to form relationships with men who are older.64 Christians have the widest gap in rates of living alone, with older women found in solo households twice as often as older men (30% vs. 14%). Among Hindus ages 60 and older, on the other hand, few live alone regardless of gender (6% of women and 2% of men). Thus, while greater female life expectancy is universal, the living arrangements of widows are not. In sub-Saharan Africa and the Middle-East North Africa region, women and men also differ in their frequency of living in two-parent households with minor children. Owing to the relatively large age gaps between partners in these regions, it is not uncommon for men in their 60s or beyond to live with a partner and children under 18. In the Middle East and North Africa, 14% of men ages 60 and older and 1% of women in this category live in two-parent households, and in sub-Saharan Africa, 19% of older men and 2% of older women live in such an arrangement. 64 For example, if a couple begins living together when the woman is 25 and the man is 30, the woman lives to age 87 and the man to age 80, the woman will live 12 years (ages 75 to 87) without her partner. 93 PEW RESEARCH CENTER www.pewresearch.org Older women live alone more often than older men % of individuals ages 60 and older in solo households Source: Pew Research Center analysis of 2010-2018 census and survey data. See Methodology for details. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 94 PEW RESEARCH CENTER www.pewresearch.org Appendix A: Methodology Data sources and analytical approaches used in the report are described in this section. First, this appendix provides shares of the populations that are represented in the study and details on the underlying source data. It goes on to explain how household types were categorized based on relationships in household rosters and how household sizes were derived. Finally, it describes adjustments made to the data, weighting procedures and data aggregation. Coverage Source data for this study comes from 130 countries, which are home to 91% of the global population. By region, the study covers countries representing 95% of the total population of the Asia-Pacific region, 97% of Europe, 69% of the Latin America-Caribbean region, 59% of the Middle East-North Africa region, more than 99% of North America and 92% of sub-Saharan Africa. 130 countries included in study Countries with available household roster and religion data Source: Pew Research Center analysis of 2010-2018 census and survey data. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 95 PEW RESEARCH CENTER www.pewresearch.org Data underlying the analyses in this report comes from a variety of censuses and surveys, all gathered since 2010. The analyses required data on religious affiliation and household rosters – lists of every member of a household along with their age, sex and relationship to the survey respondent. Rosters were used to code household type (see more on household relationships and type below). In countries where data sources that included religion were not available but more than 95% of the population belongs to the same group, the religious affiliation of the overwhelming majority of the country’s population was assigned to the sample for analysis. When possible, census data is used as the source of country data. However, when census data on both religion and household composition was unavailable – many countries including the United States do not measure religion on their census – the analysis relies on well-established nationally representative sources, such the Demographic and Health Survey (DHS), Multiple Indicator Cluster Survey (MICS) and European Social Survey (ESS). The best available data source with information on both religion and household composition was used for each country. Household rosters and religion variables were standardized and aggregated across surveys; more detail on those processes can be found below. The aggregated dataset for the 130 countries of this study includes 82.6 million individuals from 20.8 million households. For a full list of sources used by country, see Appendix B. Although Australia and New Zealand do measure religion and household composition in their censuses, individual-level census microdata from these countries are not readily available (microdata from many country censuses are available via the IPUMS International archive). Due to this lack of readily available data, results for Australia and New Zealand were not obtained. Results are reported only for groups that reach certain coverage thresholds. For the sake of reliability, only religious groups represented by at least 125 households have statistics reported separately at the country level. Within those households, researchers required a minimum of 125 people who met the relevant demographic criteria for analyses of household characteristics by age or sex. For example, more than 125 Jewish respondents in the United States provided information on their living arrangements, so their household patterns are presented. However, these same households contained fewer than 125 Jewish children, so living arrangements of Jewish children in the U.S. are not reported in the discussion of age patterns in Chapter 3. To obtain an adequate sample to represent Jews in the U.S., Pew Research Center aggregated 2010 to 2016 waves of the General Social Survey (GSS). Data for every other country came from a single wave. 96 PEW RESEARCH CENTER www.pewresearch.org There are two thresholds for inclusion of religious groups at the regional level: There must be a minimum of 500,000 people of that religion living in the region and the countries in which a sufficient number of households were surveyed (125 or more) must represent a minimum of one-third of that religious group’s regional population. There were two cases in which the latter condition was met but not the former: Buddhists in the Latin America-Caribbean region, where appropriate data with large samples was available but only about 430,000 of the world’s Buddhists reside (out of about 500 million Buddhists worldwide), and Jews in the same region, where only about 480,000 Jews reside out of over 14 million worldwide. Estimates of the size of religious group populations come from calculations made for Pew Research Center’s 2015 report “The Future of World Religions: Population Growth Projections, 2010-2050.” In the example of U.S. Jews, more than 125 Jewish households were surveyed in both the U.S. and Canada, the countries where more than 99% of the North American Jewish population resides, so coverage is sufficient and results are broken out for individuals living in Jewish households in North America.65 However, results for Hindus living in North America are not reported separately because most (80%) of the region’s Hindus live in the U.S., which lacks sufficient data (fewer than 125 households) on the composition of Hindu households. Even when respondents from particular groups are not reported separately, they are still included in analyses of the overall population – the “all” category – both at the country level and the regional level. At the global level, at least 85% of the population of each major religious group is represented in the study. 65 The North America region also includes Bermuda, Greenland and St. Pierre and Miquelon. Less than 1% of the regional population resides in these territories, where appropriate data on households and religion was not available. Mexico is included in the Latin America-Caribbean region. 97 PEW RESEARCH CENTER www.pewresearch.org Included countries by region There is sufficient data to report on: ▪ 26 countries and territories in the Asia-Pacific region: Afghanistan, Armenia, Bangladesh, Cambodia, China, Cyprus, India, Indonesia, Iran, Japan, Kazakhstan, Kyrgyzstan, Laos, Maldives, Mongolia, Nepal, Pakistan, Papua New Guinea, the Philippines, South Korea, Taiwan, Tajikistan, Thailand, Timor-Leste, Turkey and Vietnam ▪ 35 countries in Europe: Albania, Austria, Belgium, Bulgaria, Croatia, the Czech Republic, Denmark, Estonia, Finland, France, Germany, Greece, Hungary, Iceland, Ireland, Italy, Kosovo, Lithuania, Moldova, Montenegro, the Netherlands, North Macedonia, Norway, Poland, Portugal, Romania, Russia, Serbia, Slovakia, Slovenia, Spain, Sweden, Switzerland, Ukraine and the United Kingdom ▪ 19 countries and territories in the Latin America-Caribbean region: Barbados, Belize, Brazil, Costa Rica, the Dominican Republic, El Salvador, Guatemala, Guyana, Haiti, Honduras, Jamaica, Mexico, Panama, Paraguay, Peru, Puerto Rico, St. Lucia, Suriname and Trinidad and Tobago Descriptions of household patterns at regional level are based on data representing at least one-third of a population % of populations represented in available surveys Christians Muslims Unaffiliated Hindus Buddhists Jews All Asia-Pacific 87% 93% 96% >99% 87% 95% Europe 97 48 97 97 Latin America-Caribbean 70 58 89 61 38 69 Middle East-North Africa 34 60 >99 59 North America >99 >99 >99 >99 Sub-Saharan Africa 92 92 91 36 92 World 88 85 95 98 86 86 91 Too few households were surveyed within the region to adequately represent their religious group. Note: Results for underrepresented groups are not reported, but members are included in analyses of overall populations. In rare cases, sample sizes are sufficient to represent a group overall, but not a subset of that group (e.g. Jewish children in North America). Those cases are noted in discussions of subsets by age. Source: Data on the total population size of religious groups in each region from Pew Research Center’s 2015 report “The Future of World Religions.” “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 98 PEW RESEARCH CENTER www.pewresearch.org ▪ Eight countries and territories in the Middle East-North Africa region: Algeria, Egypt, Iraq, Israel, Jordan, the Palestinian territories, Tunisia and Yemen ▪ Two countries in North America: Canada and the United States ▪ 40 countries in sub-Saharan Africa: Angola, Benin, Botswana, Burkina Faso, Burundi, Cameroon, the Central African Republic, Chad, Comoros, the Democratic Republic of Congo, Ethiopia, Gabon, Gambia, Ghana, Guinea, Guinea-Bissau, the Ivory Coast, Kenya, Lesotho, Liberia, Madagascar, Malawi, Mali, Mauritania, Mozambique, Namibia, Niger, Nigeria, Republic of the Congo, Rwanda, Sao Tome and Principe, Senegal, Sierra Leone, Somalia66, South Africa, Swaziland, Togo, Uganda, Zambia and Zimbabwe. 66 Due to conflict in Somalia, source surveys were only conducted in two regions representing roughly 45% of Somalia’s population. Somalia’s population was reduced accordingly prior to aggregating to the regional and global levels. 99 PEW RESEARCH CENTER www.pewresearch.org Variables Age and sex variables used in this study were readily available across all sources. Other variables, including household religion, size and type were constructed through the processes described below. Household religion In this study, results are reported for six major religious groups: Christians, Muslims, the religiously unaffiliated, Hindus, Buddhists and Jews. People affiliated with other religious groups are included in overall global, regional and country-level results, but they are not reported on separately. Members of a household tend to share the same religion, but there is variation in the extent to which available data distinctly measures the religion of each household member. In most countries covered in this report, religion is measured for the respondent but not for other household members. As a result, in each of these countries, the religion of the respondent was assigned to everyone in the household. However, in the 20 countries shown in the adjacent table, religions of all household members ages 15 and older were collected, allowing researchers to categorize all household members in these countries according to their own religious affiliation. (As this table shows, multi-religion households are generally rare. In India, for example, 98% of households are single-religion. But there are some exceptions: In Trinidad and Tobago, for example, nearly one-in-five households have members whose religious affiliation is different from the respondent’s.) All children under 15 covered in this report are categorized using the respondent’s religion. Few sources gather information on children under 15 unless they are household heads (and therefore also respondents) themselves. That is an extremely rare occurrence; it affects less than 0.1% of children. Single-religion households are in the overwhelming majority % of single- and multi-religion households Country Single-religion Multi-religion Papua New Guinea >99% <0.5% Philippines 99 1 Bangladesh 99 1 India 98 2 Indonesia 98 2 Albania 97 3 Armenia 96 4 Mexico 96 4 South Africa 95 5 Rwanda 93 7 Ireland 93 7 Brazil 92 8 Canada 91 9 Romania 91 9 Zambia 90 10 Ghana 87 13 Benin 87 13 Portugal 86 14 Botswana 85 15 Trinidad and Tobago 81 19 Note: Table includes all countries in which religious affiliation was gathered for all household members age 15 and older. Source: Pew Research Center analysis of 2010-2018 census and survey data. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 100 PEW RESEARCH CENTER www.pewresearch.org In addition to the 112 countries for which data is available on both household structure and religion, this report also includes 18 countries and territories in which at least 95% of the total population identifies with one religious group (based on Pew Research Center’s 2015 report “The Future of World Religions: Population Growth Projections, 2010-2050”). These 18 countries and territories are Afghanistan, Algeria, Iran, Jordan, Maldives, Mauritania, Moldova, Niger, Pakistan, the Palestinian territories, Peru, Puerto Rico, Somalia, Tajikistan, Timor-Leste, Tunisia, Turkey and Yemen. In these cases, household data for the general population is used to describe household characteristics, with all individuals coded as identifying with the religious group of the overwhelming majority of the country’s population. Imputing religion for Demographic and Health Survey (DHS) samples with partial information on the religious identity of household members Data for 35 countries in this study is from DHS. The surveys generally asked the religious affiliation question only of household members in the reproductive age range (15 to 49). The 2015 survey of India, which asked about the religious affiliations of all household members ages 15 and older, is an exception. Among the other 34 DHS countries, household religion was not asked of anyone in seven of the countries, but all seven have at least 95% of the total population identifying with one religious group. In an eighth country, Timor-Leste, religion was asked of household members in the reproductive age range, however, because more than 99% of the country is Christian, researchers assumed that all households in the country were Christian. Christians also make up more than 95% of the population in Peru. Muslims make up more than 95% of the population in the other six countries (Afghanistan, Jordan, Niger, Pakistan, Tajikistan and Yemen). For these countries, the predominant religious identity is used for all households and individuals. For each of the remaining 26 countries, more than two-thirds of people live in a household where someone provided religion information: Roughly one-third of people provided their own religion, while more than a third of people had religion assigned based on someone else in the household who provided their own religious affiliation. In the latter case, household members without religion were imputed by assigning religious values of reproductive-age women or men who live in the same household. In cases where multiple religions were measured among multiple adults of reproductive age, assignment was based on a hierarchy of relationship closeness (in order: spouse, parent, child, other relative). For instance, if a male head of household was not asked about his religion since he was too old to be included in the fertility survey, but he lived with his Muslim reproductive-aged wife and his religiously unaffiliated nephew, his religious affiliation was assigned as Muslim. 101 PEW RESEARCH CENTER www.pewresearch.org At the individual level and in a typical country relying on DHS data, 81% of the household population is represented by a sampled household with directly reported information on the religion of at least one member. For the remaining households in which there were no members of reproductive age, and therefore no information about religious identity was gathered, multiple imputation was applied using the multivariate imputation by chained equations (MICE) algorithm in Stata. There was enough information about adults not of reproductive age in households with adults of reproductive age to identify age, sex, marital status, educational attainment, household size, region and urbanicity as predictors of religion and use them in the imputation. Five iterations were executed with these predictors (independent variables) for the missing data on religion. Household size For the purposes of this study, a household is a private dwelling unit that is not vacant and is not an institution (such as a college dormitory or nursing home). The household population is made up of all people who reside in households (not in institutions). Household size is the number of persons living in a household. Average size at the household level overall is derived by dividing the total household population (that is, the non-institutionalized population) by the total number of households. Average household size for a religious group was derived by dividing the total population affiliated with that religion by the total number of households headed by a person affiliated with that same religion (except for in the 20 countries where religious affiliation was captured for all household members over age 15). These household-level sizes were then transformed to represent the experiences of individuals. A simplified example of this process is described below, and more details can be found in the sidebar on page 14. A small village is made up of only 50 homes. Thirty homes contain large households of 10 people each, and each of the remaining 20 homes contain two people. At the household level, the average size is 6.8. (30 × 10) + (20 × 2) 30 + 20 = 6.8 But if all 340 villagers gather to compare their living situations, there will be 300 people who report experiencing large, 10-person households, and 40 people who live in pairs. The average villager experiences a household size of 9.1. (300 × 10) + (40 × 2) 300 + 40 = 9.1 102 PEW RESEARCH CENTER www.pewresearch.org Unless specified otherwise, all results in this report reflect the average experience of individuals (such as a household size of 9.1 in the example above) rather than average values at the household level (6.8 in the example above). Household type For this study, households were categorized into types that describe qualitative differences in living arrangements that are not captured by household size alone. As with household size, the prevalence of household types is reported from the perspective of individuals. Household type categories are based on relationships between members and the ages of people identified as children of the respondent. Household rosters from every source dataset were coded to produce a standardized set of seven household types that are referred to throughout this report: Extended family, two-parent, adult child, couple, solo, single-parent and polygamous. About 98% of the global household (non-institutionalized) population lives in one of these seven household types. Regionally, the shares of people who live in a categorized household range from over 99% in the Middle East-North Africa region to 96% in sub-Saharan Africa. At the country level, the smallest share of people living in a categorized family household is 85%, in Rwanda. Members of other types of households – for example, homes of only unrelated roommates or homes with a mixture of both related and unrelated individuals – are not reported separately but remain in the denominators of analyses.67 Because of this, the shares of individuals residing in the seven household types reported do not always sum to 100. This classification process was largely based on harmonized household definitions constructed by IPUMS International at the Minnesota Population Center, University of Minnesota. Pew Research Center analysts modified the IPUMS classification scheme to combine types not relevant to this report into the undifferentiated “other” category described above and to exclude institutions. They also added a new category – adult child – which uses the age of children (by relationship) to distinguish two-parent and single-parent households with and without minor children. When using census data from IPUMS International, Pew Research Center analysts did not reclassify household rosters that were unaffected by the new categories, and instead accepted IPUMS’ household type results (having already adopted their coding scheme), with one exception: Benin. Because of the way IPUMS classifies household members, Benin’s “other” category had an unusually large number of people. As a result, Pew Research Center recoded Benin’s census data on household types using the same process that was applied to data sources that did not come from the IPUMS archive. 67 While not reported separately throughout the report, these figures are available in Appendix C. 103 PEW RESEARCH CENTER www.pewresearch.org Household relationships Every data source includes a variable that describes how each individual in a household is related to the respondent who answered questions on behalf of the entire household.68 Relationships were collapsed and standardized across surveys to include consistent categories for household type coding. Categories of relationships to household respondent include: Respondent: The individual who supplied the household roster and answered other survey questions on behalf of the household. In some data sources, the individual may be called the household head, householder or the reference person. Spouse/cohabiter: An individual either married to or cohabiting with the respondent. Some source surveys distinguish between formally married and cohabiting partners, but some do not, so these relationships were collapsed into a single category for comparability. Cohabiting partners do not include roommates. Child: A child of the respondent. Some data sources distinguish a biological child from an adopted, foster or stepchild, while others do not, so these were consolidated. The child category was subdivided into minor children and adult children based on whether any children (by relationship) were under or over 18 years old at the time of the survey. Parent: Any parent of the respondent, including those other than biological parents. Other relative: Other than a spouse, child or parent, a person in this category is an individual who is related to the respondent either by blood (a grandparent, grandchild, sibling, cousin, niece, aunt, uncle, and so on) or by marriage (a parent-in-law, child-in-law, grandparent-in-law, grandchild-in-law, sibling-in-law, cousin-in-law, niece-in-law, aunt-in-law and so on). Non-relative: An individual who is unrelated to the respondent by blood, marriage, adoption or fostering. This category does not include partners who are living together but not married. Unknown: A household member whose relationship to the respondent is unknown. 68 Sources used in this report typically enumerate each person in a household and then ask how members are related. However, in the case of the Japanese General Social Survey (JGSS), a household roster had to be constructed from a series of questions that ask about various kinds of relatives and whether such relatives live with the respondent. Sources usually record non-relatives who are members of a household, but the JGSS, the Chinese General Social Survey and the Taiwan Social Change Survey do not ask about the presence of non-relatives in the household. 104 PEW RESEARCH CENTER www.pewresearch.org Defining household type Using household individuals’ relationship to the respondent, the study classifies households into seven basic categories: Solo: One-person household. Couple: Married/cohabiting couple without children, other relatives or non-relatives. Two-parent: Married/cohabiting couple with at least one minor child (under the age of 18), regardless of whether adult children also live in the home, and with no other relatives or non-relatives. Single-parent: One parent with one or more minor children (under the age of 18), regardless of whether adult children also live in the home, and no other relatives or non-relatives. Adult children: One or two parents with one or more of their adult children (over the age of 18) and no minor children or any other relatives or non-relatives. Extended: A household with one or more “other relatives,” as defined above. Polygamous: A household in which an individual lives with two or more partners, with or without children, and regardless of the presence of other relatives or non-relatives. Other: Households that contain any non-relatives or people whose relationship to the respondent is unknown. Shares of people in this undifferentiated “other” category are not detailed throughout the report but are available in Appendix C. Differences between estimates in this report and other reports There are various ways to classify living arrangements. The above categories are mutually exclusive, but living arrangements can also be analyzed with overlapping categories. Consider a household with a minor child, the child’s parent and the child’s grandparent. In the classification scheme used in this report, such a household is classified as an extended-family household. However, researchers sometimes classify such households as single-parent households. When this more-expansive approach is taken, the share of children living in single-parent households often is higher. 105 PEW RESEARCH CENTER www.pewresearch.org Same-sex couples When identifiable in source data, same-sex partners are categorized as belonging to couple, two-parent, adult child or extended-family households regardless of marital status, exactly as heterosexual couples are. Four out of 19 countries with national census agency data that includes religion specifically list same-sex spouse or partner as a relationship option.69 Other surveys used neutral language like “spouse” or “partner.” However, there are two reasons to suspect that same-sex couples are often missed in source surveys. First, some surveys do not give respondents the opportunity to identify a household member of the same sex as their spouse or significant other. For example, DHS surveys explicitly assume that partners are opposite-sex by asking women, “Are you currently married or living together with a man as if married?” and asking men only about female partners. In Portugal, the census categorized all households with same-sex partners as unclassifiable. Relatedly, same-sex partnerships are stigmatized or illegal in some countries. Same-sex partners are likely undercounted in such contexts, even when inclusive response options are available. Part-time household members Source surveys use a variety of language to introduce household roster questions. Commonly, questionnaires ask for details on everyone who “usually” lives in the household. For example, this is the wording used by MICS and the U.S. GSS.70 The ESS asks about members who “live here regularly.” Even subtle differences in question wording might lead to minor discrepancies in the way part-time household members (such as migrant workers or children who live in more than one home) are counted. For example, in Europe, where the survey asks about people who live in the home “regularly,” parents with weekend-only custody might include their children, whereas U.S. parents in the same situation might not include the children, because the survey wording – “usually” – may imply that these children live in the home at least half of the time. Other demographic characteristics of religious groups Data on the regional distributions of religious groups, their age structures and fertility rates used to complement the analyses of households in the report comes from earlier Pew Research Center reports, “The Future of World Religions” and “The Changing Global Religious Landscape.” 69 These countries are Brazil, Canada, Ireland and Portugal. 70 DHS surveys gather information on household members who usually live in the household as well as those who slept in the household the previous night. DHS calculations in this report are based on all people enumerated in the household roster. Including those who slept in the household the prior night typically adds 1-2% to the “usual” household population. Usual household members include people who were away the prior night. In the Lesotho DHS sample, available measures permit the exclusion of household members who reside elsewhere in Lesotho, South Africa or another country. For consistency with other DHS samples, which do not typically contain such detail, all listed household members are included in Lesotho calculations even if they currently reside elsewhere (for example, because they are working away from home). 106 PEW RESEARCH CENTER www.pewresearch.org Minor errors in source data Source data underwent rigorous quality control checks by census and survey organizations before being published. Despite the high quality of the source data and the care taken by survey and statistical agencies, minor discrepancies are inevitable in major data collection efforts, so conducting further quality checks while standardizing and aggregating data on nearly 83 million individuals is customary and important. Pew Research Center analysts identified and cleaned up minor data errors. For example, cases were dropped in which a household contained only one person but the relationship was coded as spouse, child or other relative instead of household head or respondent. One-person households in which the only person was younger than 12 were excluded from the analysis. Rarely, a person’s relationship to the respondent was coded as parent, even though that person was substantially younger than the respondent; or as child, with a large age gap in the other direction. In such cases, Center analysts assumed that respondents had simply answered the relationship question from the wrong perspective and reversed the parent-child relationship so that the much younger person was recoded as the child rather than the parent. 107 PEW RESEARCH CENTER www.pewresearch.org Weighting Country weights Survey weights make minor adjustments to account for differences between the sample and the general population of individuals or households in an area. All data files have person weights that are used to derive individual-level results from person-level data files. Further, samples of census and household survey data usually provide household weights. The comparative analyses on household-level data in the sidebar on page 14 apply these weights to the analyses of household file data. Adjustments to social surveys Wherever possible, household characteristics were drawn from government censuses or large household surveys in which religious identification was measured for every member of the sampled households. For 33 countries and territories, the recent censuses and household surveys do not ask a question on religious affiliation, but data is available from nationally representative modest-sample social surveys that allow for the relevant household estimates – including the Chinese General Social Survey (CGSS), European Social Survey (ESS), U.S. General Social Survey (GSS), Japan General Social Survey (JGSS), Korea General Social Survey (KGSS) and Taiwan Social Change Survey (TSCS). However, these surveys’ sample weights failed to address two related issues in derived results. First, the weighted results show the proportion of one-person households is substantially lower than estimates from official publications for every country except South Korea. Second, the resulting average household sizes are larger than those from official publications for every country except China. Center researchers recognized that the weights for these surveys are not designed to be used in measuring household characteristics, so they adjusted the sample weights in two steps to make the data files more consistent with official estimates. 108 PEW RESEARCH CENTER www.pewresearch.org The social surveys used in this analysis are designed and weighted to be representative of the adult population in their respective countries, and respondent weights reflect the probability that one adult in each household was selected to participate. To calculate estimates for the entire population (including adults and children), the weight for each respondent was divided by the number of adults living in that household. This first adjustment effectively converts the representative sample of adults into a representative sample of households. Dividing the sample weights by the number of adults largely improved the weighted results. However, for the majority of countries, they did not match up to benchmarks. Further, for a few countries, the first-step adjustment of weights seemed to exacerbate the problems. One plausible reason is that small households – one-person households in particular – may be less likely than others to be reached by phone. Since inconsistencies with benchmarks persisted after the first-step adjustment, weights were raked to household size data from official government statistics, which separately enumerate one-person, two-person, three-person, four-person and five-and-more-person households. The table on page 109 – which compares derived results using original survey sample weights and adjusted weights with the official data – shows that using the adjusted weights improved the accuracy of the computation. Estimates for individuals were then produced by combining these household-level weights with information on the total number of adults and children living in each household. Sources of benchmark data Country Source Austria Eurostat Belgium Eurostat Bulgaria Eurostat China United Nations Population Division Croatia Eurostat Cyprus Eurostat Czech Republic Eurostat Denmark Eurostat Estonia Eurostat Finland Eurostat France Eurostat Germany Eurostat Greece Eurostat Hungary Eurostat Iceland Statistics Iceland Israel The Central Bureau of Statistics of Israel Italy Eurostat Japan United Nations Population Division Lithuania Eurostat Netherlands Eurostat Norway Statistics Norway Poland Eurostat Russia Russian Federal State Statistics Service Slovakia Eurostat Slovenia Eurostat South Korea United Nations Populations Division Spain Eurostat Sweden Eurostat Switzerland Eurostat Taiwan Republic of China Census Ukraine State Statistics Service of Ukraine United Kingdom Eurostat United States United States Census Bureau Source: Pew Research Center analysis of 2010-2018 census and survey data. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 109 PEW RESEARCH CENTER www.pewresearch.org Two-step adjustment process aligned survey results with official benchmarks -------------- Percent of solo households -------------- -------------- Mean household size -------------- Country Benchmark With original survey weight After 1st adjustment After 2nd adjustment Benchmark With original survey weight After 1st adjustment After 2nd adjustment Austria 37% 18% 33% 37% 2.2 2.7 2.3 2.2 Belgium 35 15 29 35 2.3 2.9 2.5 2.3 Bulgaria 29 10 22 29 2.5 3.1 2.6 2.5 China 15 13 28 15 3.1 3.1 2.5 3.1 Croatia 25 9 19 25 2.8 3.5 3.0 2.8 Cyprus 21 6 14 21 2.7 3.3 2.9 2.7 Czech Republic 28 15 28 28 2.4 2.6 2.2 2.4 Denmark 45 22 37 45 2.0 2.5 2.2 2.0 Estonia 36 23 39 36 2.2 2.5 2.1 2.2 Finland 41 27 43 41 2.0 2.4 2.1 2.0 France 36 18 31 36 2.2 2.7 2.3 2.2 Germany 41 19 33 41 2.0 2.6 2.3 2.0 Greece 26 11 22 26 2.6 3.1 2.6 2.5 Hungary 33 22 39 33 2.3 2.5 2.1 2.3 Iceland 31 14 25 31 2.5 3.1 2.7 2.5 Israel 19 5 12 19 3.3 4.1 3.6 3.3 Italy 32 20 36 32 2.4 2.7 2.3 2.4 Japan 35 12 24 35 2.3 3.2 2.7 2.3 Lithuania 38 13 24 38 2.3 2.7 2.4 2.2 Netherlands 37 15 27 37 2.2 2.8 2.5 2.2 Norway 38 18 31 38 2.2 2.7 2.3 2.2 Poland 24 13 26 24 2.8 3.1 2.6 2.8 Russia 25 17 31 25 2.6 2.7 2.3 2.6 Slovakia 23 10 21 23 2.8 3.3 2.8 2.8 Slovenia 29 10 22 29 2.5 3.2 2.7 2.5 South Korea 27 29 32 27 2.5 2.7 2.5 2.5 Spain 25 9 19 25 2.5 3.0 2.7 2.5 Sweden 42 21 35 42 2.0 2.6 2.2 2.0 Switzerland 37 15 28 37 2.2 2.8 2.4 2.2 Taiwan 22 5 14 22 3.0 4.4 3.6 3.0 Ukraine 20 7 17 20 2.6 3.3 2.8 2.6 United Kingdom 29 20 34 29 2.3 2.7 2.3 2.3 United States 28 16 29 28 2.5 2.8 2.4 2.5 Source: Pew Research Center analysis of 2010-2018 census and survey data. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 110 PEW RESEARCH CENTER www.pewresearch.org Weights in computing global and regional household characteristics Pew Research Center used national-level results to calculate regional and global distributions. Household populations (overall and by religion) were applied as country-level weights in deriving data on personal living arrangements in each region and around the world. The estimates of the 2015 size of religious groups at the country level are from the demographic projections by Pew Research Center, which include the household and non-household populations (for example, people living in institutions). To adjust for the fact that homeless individuals and residents of institutions are not included in this study, the population of each country was reduced by 3%. This figure is based on available information from the U.S., the UK and Canada. While it is possible that these relatively wealthy Western countries have higher-than-average shares of people living in non-households, corresponding figures on the precise size of non-household populations are not readily available for all countries in this study. This population adjustment was applied only at the country level, without further adjustments for any potential religious or other differences between household and non-household populations. Aggregating individual-level data to represent regional and global populations, overall and by religious group based on the Center’s own demographic projections, was relatively straightforward. In order to produce the comparisons between individual and household-level results shown in the sidebar on page 14, researchers also aggregated results from household files. Since data with estimates on the number of households in every studied country was not available, a new procedure was developed. To aggregate household sizes and living arrangements at the household level, researchers first divided the household population by mean household size to derive the total number of households, for all people and by religion, for each country. Next, total regional or global household populations and total number of households were summed, respectively. Then the mean regional household size was derived by dividing the aggregated total household population by the aggregated total number of households. To explain how the household weights are computed and used in aggregating average household size by religion, consider the Christian population and number of households in Southern Africa as an example. In the table, column 1 shows the names of 10 Southern African countries. Column 2 shows the Christian population in each country in 2015. Column 3 shows the non-household Christian population (3% of the total). Column 4 shows the total Christian household population, which is derived by subtracting the non-household population (column 3) from the total population (column 2). Column 5 shows the average Christian household size for each country. Column 6 shows the total number of Christian household units that resulted from dividing the total Christian household population (column 4) by the average Christian household size (column 5). Next, the regional total for the Christian household population (125,367,624) and total number 111 PEW RESEARCH CENTER www.pewresearch.org of Christian households (29,574,715) were summed, respectively, which resulted in the average Christian household size for the Southern African sub-region: Christian household size for Southern Africa = 121,606,595 / 29,574,715 = 4.1 To illustrate how the data was weighted to derive household type at the aggregate levels, the next table also shows the 10 countries in the Southern African sub-region as an example. The upper part of the table above shows the distribution of Christian households by country. Using the weighted total number of Christian households (column 6), the numbers of each type of household were derived by multiplying the total number by the percentage of each type for each country. The resulting numbers of each type of household for each country are shown in the lower part of the table. Then the regional total number of households was summed as well as the numbers for each type of household, which allowed for the computation of the Christian household type distribution for the sub-region (the bottom line): Among the Christian households in the Southern African sub-region, 15% are single-person households, 7% are couple households, 25% are two-parent households, 10% are single-parent households and 36% are extended-family households. Household size adjustment and aggregation example Column 1 Country Column 2 Christian population Column 3 Non-household Christian population (3% of col. 2) Column 4 Christian household population (col. 2 – col. 3) Column 5 Mean size of households w/ Christian respondents Column 6 Number of Christian households (col. 4/col. 5) Angola 19,655,719 589,672 19,066,047 5.0 3,846,619 Botswana 1,577,024 47,311 1,529,713 3.6 429,031 Lesotho 2,208,765 66,263 2,142,502 4.2 507,322 Malawi 14,569,722 437,092 14,132,630 4.5 3,168,942 Mozambique 14,917,562 447,527 14,470,035 4.6 3,170,118 Namibia 2,418,498 72,555 2,345,943 4.4 538,514 South Africa 42,143,897 1,264,317 40,879,580 3.4 12,186,491 Swaziland 1,120,916 33,627 1,087,289 4.0 270,211 Zambia 14,934,091 448,023 14,486,068 5.3 2,755,229 Zimbabwe 11,821,430 354,643 11,466,787 4.2 2,702,238 Southern Africa 125,367,624 3,761,029 121,606,595 4.1 29,574,715 Aggregated household size: Christian household population (121,606,595) / Number of Christian households (29,574,715) = 4.1 Source: Pew Research Center analysis of 2010-2018 census and survey data. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 112 PEW RESEARCH CENTER www.pewresearch.org Household type adjustment and aggregation example Distribution of Christian households Country Extended Two-parent Solo Single-parent Couple Adult child Polygamous Other Total Angola 35% 35% 10% 12% 5% 1% <0.5% 2% 100% Botswana 39 9 26 8 6 3 <0.5 8 100 Lesotho 45 19 13 6 4 5 <0.5 8 100 Malawi 31 42 8 10 5 2 <0.5 2 100 Mozambique 36 32 9 11 6 1 1 3 100 Namibia 48 10 17 7 6 2 <0.5 11 100 South Africa 34 15 23 9 10 6 <0.5 2 100 Swaziland 40 13 22 11 4 3 <0.5 7 100 Zambia 40 38 6 7 4 2 0.5 3 100 Zimbabwe 43 22 11 11 4 2 <0.5 6 100 Counts (in thousands) of Christian households Country Extended Two-parent Solo Single-parent Couple Adult child Polygamous Other Total Angola 1,338 1,342 372 480 195 54 0 66 3,847 Botswana 168 41 113 34 26 14 0 33 429 Lesotho 227 96 64 30 22 25 0 42 507 Malawi 986 1,321 246 323 156 58 8 72 3,169 Mozambique 1,148 1,027 291 344 191 46 31 93 3,170 Namibia 257 56 89 36 31 11 0 58 539 South Africa 4,164 1,868 2,804 1,107 1,188 749 31 276 12,186 Swaziland 108 35 59 31 12 7 0 18 270 Zambia 1,091 1,038 171 205 113 47 14 76 2,755 Zimbabwe 1,160 581 306 305 113 63 13 162 2,702 Southern Africa counts 10,648 7,405 4,514 2,894 2,046 1,073 96 897 29,575 Southern Africa distribution 36% 25% 15% 10% 7% 4% <0.5% 3% 100% Source: Pew Research Center analysis of 2010-2018 census and survey data. “Religion and Living Arrangements Around the World” PEW RESEARCH CENTER 113 PEW RESEARCH CENTER www.pewresearch.org Appendix B: Data sources by country General sources and archives Demographic and Health Surveys. Funded by the U.S. Agency for International Development (USAID). Implemented by ICF. European Social Survey. Led by the Centre for Comparative Social Surveys. City University, (London) in partnership with the Catholic University of Leuven (Belgium), GESIS (Germany), NSD (Norway), and SCP and the University of Amsterdam (Netherlands). Integrated Public Use Microdata Series, International (IPUMS). Minnesota Population Center, University of Minnesota. Multiple Indicator Cluster Survey. Developed by the United Nations Children’s Fund (UNICEF). Other demographic data sources Pew Research Center. 2016. “Religion and Education Around the World.” Pew Research Center. 2017. “The Changing Global Religious Landscape.” Pew Research Center. 2015. “The Future of World Religions: Population Growth Projections, 2010-2050.” World Economic Outlook database. 2015. International Monetary Fund. World Population Prospects: The 2017 Revision. United Nations Population Division. 114 PEW RESEARCH CENTER www.pewresearch.org Data sources by country Afghanistan: Demographic and Health Survey 2015 Albania: Living Standards Measurement Survey 2012 Algeria: Multiple Indicator Cluster Survey 2012-2013 Angola: Demographic and Health Survey 2016 Armenia: Population and Housing Census of the Republic of Armenia 2011, IPUMS subset Austria: European Social Survey 2016 Bangladesh: Population and Housing Census 2011, IPUMS subset Barbados: Multiple Indicator Cluster Survey 2012 Belgium: European Social Survey 2016-2017 Belize: Multiple Indicator Cluster Survey 2015-2016 Benin: Fourth Population and Habitation Census 2013, retrieved from IPUMS Botswana: Census 2011 Brazil: XII Recenseamento Geral do Brasil. Censo Demográfico 2010, IPUMS subset Bulgaria: European Social Survey 2012 Burkina Faso: Demographic and Health Survey 2010 Burundi: Demographic and Health Survey 2016-2017 Cambodia: Demographic and Health Survey 2014 Cameroon: Multiple Indicator Cluster Survey 2014 115 PEW RESEARCH CENTER www.pewresearch.org Canada: National Household Survey 2011, retrieved from IPUMS Central African Republic: Multiple Indicator Cluster Survey 2010 Chad: Demographic and Health Survey 2014-2015 China: Chinese General Social Survey 2013-2014 Comoros: Demographic and Health Survey 2012 Costa Rica: Multiple Indicator Cluster Survey 2011 Croatia: European Social Survey 2010 Cyprus: European Social Survey 2012 Czech Republic: European Social Survey 2016 Democratic Republic of the Congo: Demographic and Health Survey 2013-2014 Denmark: European Social Survey 2014 Dominican Republic: Multiple Indicator Cluster Survey 2014 Egypt: Demographic and Health Survey 2014 El Salvador: Multiple Indicator Cluster Survey 2014 Estonia: European Social Survey 2016-2017 Ethiopia: Demographic and Health Survey 2016 Finland: European Social Survey 2016-2017 France: European Social Survey 2016-2017 Gabon: Demographic and Health Survey 2012 116 PEW RESEARCH CENTER www.pewresearch.org Gambia: Demographic and Health Survey 2013 Germany: European Social Survey 2016-2017 Ghana: Population and Housing Census 2010, IPUMS subset Greece: European Social Survey 2010 Guatemala: Demographic and Health Survey 2015 Guinea: Multiple Indicator Cluster Survey 2016 Guinea-Bissau: Multiple Indicator Cluster Survey 2014 Guyana: Multiple Indicator Cluster Survey 2014 Haiti: Demographic and Health Survey 2016-2017 Honduras: Demographic and Health Survey 2012 Hungary: European Social Survey 2017 Iceland: European Social Survey 2016-2017 India: Demographic and Health Survey 2016 Indonesia: Population Census 2010, retrieved from IPUMS Iran: National Population and Housing Census 2011, retrieved from IPUMS Iraq: Multiple Indicator Cluster Survey 2018 Ireland: Census of Population of Ireland 2011, IPUMS subset Israel: European Social Survey 2016-2017 Italy: European Social Survey 2017 117 PEW RESEARCH CENTER www.pewresearch.org Ivory Coast: Multiple Indicator Cluster Survey 2016 Jamaica: Multiple Indicator Cluster Survey 2011 Japan: Japanese General Social Survey 2012 Jordan: Demographic and Health Survey 2017-2018 Kazakhstan: Multiple Indicator Cluster Survey 2010-2011 Kenya: Demographic and Health Survey 2014 Kosovo: Multiple Indicator Cluster Survey 2013-2014 Kyrgyzstan: Multiple Indicator Cluster Survey 2014 Laos: Multiple Indicator Cluster Survey 2017 Lesotho: Demographic and Health Survey 2014 Liberia: Demographic and Health Survey 2013 Lithuania: European Social Survey 2016 Madagascar: Demographic and Health Survey 2016 Malawi: Multiple Indicator Cluster Survey 2013-2014 Maldives: Demographic and Health Survey 2016-2017 Mali: Multiple Indicator Cluster Survey 2015 Mauritania: Multiple Indicator Cluster Survey 2015 Mexico: Population and Housing Census 2011, IPUMS subset Moldova: Multiple Indicator Cluster Survey 2012 118 PEW RESEARCH CENTER www.pewresearch.org Mongolia: Multiple Indicator Cluster Survey 2013-2014 Montenegro: Multiple Indicator Cluster Survey 2013 Mozambique: Demographic and Health Survey 2011 Namibia: Demographic and Health Survey 2013 Nepal: Demographic and Health Survey 2016 Netherlands: European Social Survey 2016 Niger: Demographic and Health Survey 2012 Nigeria: Multiple Indicator Cluster Survey 2016-2017 North Macedonia: Multiple Indicator Cluster Survey 2011 Norway: European Social Survey 2016 Pakistan: Demographic and Health Survey 2017-2018 Palestinian territories: Multiple Indicator Cluster Survey 2014 Panama: Multiple Indicator Cluster Survey 2013 Papua New Guinea: Census 2011 Paraguay: Multiple Indicator Cluster Survey 2016 Peru: Demographic and Health Survey 2010-2013 Philippines: Census of Population and Housing 2010, retrieved from IPUMS Poland: European Social Survey 2016-2017 Portugal: Censos 2011: XV Receseamento Geral da População; V Receseamento Geral da Habitação, IPUMS subset 119 PEW RESEARCH CENTER www.pewresearch.org Puerto Rico: Puerto Rico Community Survey 2010, IPUMS subset Republic of the Congo: Multiple Indicator Cluster Survey 2014-2015 Romania: Population and Housing Census 2011, retrieved from IPUMS Russia: European Social Survey 2017 Rwanda: Fourth Population and Housing Census 2012, retrieved from IPUMS Sao Tome and Principe: Multiple Indicator Cluster Survey 2014 Senegal: Demographic and Health Survey 2017 Serbia: Multiple Indicator Cluster Survey 2014 Sierra Leone: Multiple Indicator Cluster Survey 2017 Slovakia: European Social Survey 2012 Slovenia: European Social Survey 2016-2017 Somalia: Multiple Indicator Cluster Survey 2011 South Africa: General Household Survey 2015 South Korea: Korean General Social Survey 2014-2016 Spain: European Social Survey 2017 St. Lucia: Multiple Indicator Cluster Survey 2012 Suriname: Multiple Indicator Cluster Survey 2010 Swaziland: Multiple Indicator Cluster Survey 2014 Sweden: European Social Survey 2016-2017 120 PEW RESEARCH CENTER www.pewresearch.org Switzerland: European Social Survey 2016-2017 Taiwan: Taiwan Social Change Survey 2016 Tajikistan: Demographic and Health Survey 2017 Thailand: Multiple Indicator Cluster Survey 2015-2016 Timor-Leste: Demographic and Health Survey 2016 Togo: Demographic and Health Survey 2013 Trinidad and Tobago: Population and Housing Census 2011, retrieved from IPUMS Tunisia: Multiple Indicator Cluster Survey 2011-2012 Turkey: Family Structure Survey 2016 Uganda: Demographic and Health Survey 2016 Ukraine: European Social Survey 2012 United Kingdom: European Social Survey 2016-2017 United States: General Social Survey 2010-2016 Vietnam: Multiple Indicator Cluster Survey 2013-2014 Yemen: Demographic and Health Survey 2013 Zambia: Census of Population and Housing 2010, IPUMS subset Zimbabwe: Demographic and Health Survey 2015 Appendix C: Household structure by religious group This table describes the family structures experienced by adherents of each of the major religious groups in every covered country, territory and region. Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other World All 4.9 38% 33% 9% 8% 4% 4% 2% 2% Buddhists 3.9 44 20 13 13 7 2 <0.5 0.9 Christians 4.5 29 34 9 11 7 6 0.8 3 Hindus 5.7 55 30 8 3 0.9 3 <0.5 0.6 Jews 3.7 17 32 12 21 10 4 <0.5 3 Muslims 6.4 36 43 6 3 1 3 5 2 Unaffiliated 3.7 37 26 12 14 7 2 <0.5 1 Asia-Pacific All 5.0 45 31 10 7 3 2 <0.5 1 Buddhists 3.9 44 20 13 13 7 2 <0.5 0.9 Christians 4.8 37 35 10 8 5 2 <0.5 3 Hindus 5.7 55 30 8 3 0.9 3 <0.5 0.6 Muslims 6.0 41 42 7 4 1 3 0.6 2 Unaffiliated 3.7 41 25 14 13 6 1 <0.5 0.5 Europe All 3.1 26 26 9 19 13 4 <0.5 2 Christians 3.1 24 25 11 21 13 4 <0.5 2 Muslims 4.1 37 37 6 7 7 5 <0.5 0.9 Unaffiliated 3.0 25 26 7 20 14 5 <0.5 2 Latin America-Caribbean All 4.6 32 39 10 6 3 5 <0.5 4 Christians 4.6 32 39 10 6 3 5 <0.5 4 Hindus 4.4 35 33 17 7 4 3 <0.5 2 Unaffiliated 4.4 29 38 9 6 5 7 <0.5 5 Middle East-North Africa All 6.2 27 56 9 3 1 2 0.9 <0.5 Christians 4.6 16 58 13 6 3 3 <0.5 <0.5 Jews 4.3 29 35 13 13 6 3 <0.5 1 Muslims 6.3 27 57 9 3 0.8 2 1 <0.5 North America All 3.3 11 33 14 20 11 9 <0.5 1 Christians 3.4 9 34 14 21 11 9 <0.5 1 Jews 3.0 6 30 11 30 13 5 <0.5 5 Unaffiliated 3.2 14 30 15 18 12 9 <0.5 2 Sub-Saharan Africa All 6.9 35 37 2 2 2 6 11 4 Christians 6.0 39 38 2 2 3 8 3 5 Hindus 3.9 32 31 22 10 3 0.8 <0.5 0.9 Muslims 8.5 27 37 0.9 2 1 4 25 3 Unaffiliated 5.7 39 32 3 4 7 7 5 4 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Afghanistan All 9.8 54% 39% <1% <1% <1% <1% 5% <1% Muslims 9.8 54 39 <1 <1 <1 <1 5 <1 Albania All 4.6 35 35 18 8 2 2 <1 <1 Christians 4.5 30 35 20 9 2 3 <1 <1 Muslims 4.6 36 35 17 8 2 2 <1 <1 Algeria All 6.4 29 54 12 2 <1 2 <1 <1 Muslims 6.4 29 54 12 2 <1 2 <1 <1 Angola All 6.4 44 39 <1 2 2 10 <1 2 Christians 6.5 45 38 <1 2 2 10 <1 2 Unaffiliated 5.8 36 43 2 3 4 9 <1 2 Armenia All 5.0 55 19 16 4 3 2 <1 2 Christians 5.0 55 19 16 3 3 2 <1 2 Unaffiliated 4.4 47 19 18 7 6 2 <1 2 Austria All 3.0 16 32 7 22 17 3 <1 3 Christians 2.9 13 32 8 26 16 3 <1 3 Unaffiliated 2.6 14 28 6 21 24 4 <1 4 Bangladesh All 5.4 38 44 3 3 <1 5 <1 6 Buddhists 5.2 35 46 5 3 <1 3 1 6 Christians 5.1 35 40 5 4 1 4 <1 9 Hindus 5.3 44 38 5 3 <1 3 1 5 Muslims 5.4 37 44 3 3 <1 5 <1 6 Barbados All 3.9 35 17 15 7 9 8 <1 9 Christians 4.0 34 17 16 8 8 7 <1 10 Unaffiliated 3.9 36 16 11 5 13 11 <1 8 Belgium All 3.1 19 30 9 23 15 4 <1 <1 Christians 2.9 12 28 9 26 20 5 <1 <1 Muslims 4.0 27 50 4 5 12 4 <1 <1 Unaffiliated 3.0 22 28 10 23 13 4 <1 1 Belize All 5.2 32 41 6 5 5 9 <1 3 Christians 5.2 33 39 6 5 4 9 <1 3 Unaffiliated 4.8 26 42 4 7 7 9 <1 6 Benin All 8.9 39 31 1 1 2 7 9 8 Christians 7.6 36 36 2 1 2 8 6 8 Muslims 11.8 48 24 <1 <1 <1 3 12 11 Unaffiliated 8.6 40 30 1 1 3 7 9 9 Botswana All 5.7 58 11 2 4 8 6 <1 11 Christians 5.6 58 11 2 3 7 7 <1 10 Hindus 3.7 27 40 6 9 7 <1 <1 11 Muslims 5.0 39 30 1 4 7 5 <1 13 Unaffiliated 5.9 59 10 2 4 9 4 <1 12 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Brazil All 4.2 28% 39% 12% 8% 4% 5% <1% 3% Buddhists 3.6 30 23 18 12 7 5 <1 5 Christians 4.2 28 40 12 8 3 5 <1 3 Jews 3.4 12 32 19 15 9 3 <1 10 Muslims 4.0 23 42 14 8 5 3 <1 4 Unaffiliated 4.1 28 37 10 8 5 7 <1 4 Bulgaria All 3.2 38 22 9 17 12 2 <1 <1 Christians 3.1 36 20 11 19 12 2 <1 <1 Muslims 3.7 49 19 6 14 10 2 <1 <1 Unaffiliated 3.2 37 29 8 13 11 2 <1 <1 Burkina Faso All 7.8 19 34 <1 2 1 3 36 5 Christians 7.2 27 35 <1 1 1 3 24 7 Muslims 8.0 16 33 <1 2 1 3 40 4 Burundi All 6.0 22 53 2 1 1 10 <1 10 Christians 6.0 22 54 2 1 1 10 <1 10 Muslims 6.2 32 38 <1 1 2 8 1 17 Unaffiliated 5.5 16 63 1 2 2 14 <1 3 Cambodia All 5.5 49 39 5 2 <1 3 <1 2 Buddhists 5.5 49 39 5 2 <1 3 <1 2 Muslims 6.0 55 37 4 <1 <1 2 <1 <1 Cameroon All 6.8 41 30 1 2 4 6 13 3 Christians 6.4 47 28 1 2 5 6 7 4 Muslims 7.8 29 33 <1 2 3 4 26 2 Unaffiliated 6.8 33 34 1 <1 5 3 19 3 Canada All 3.2 9 33 14 20 11 6 <1 7 Buddhists 3.9 23 27 19 10 7 5 <1 9 Christians 3.2 8 32 14 23 12 5 <1 6 Hindus 4.3 28 38 15 8 3 2 <1 6 Jews 3.4 4 36 16 20 13 3 <1 7 Muslims 4.4 16 52 11 6 4 6 <1 5 Unaffiliated 3.1 8 32 12 18 12 7 <1 11 Central African Republic All 6.5 41 38 <1 3 2 6 8 1 Christians 6.5 41 39 <1 2 2 6 7 1 Muslims 6.5 37 33 <1 3 3 8 13 2 Chad All 7.9 29 44 <1 1 1 7 15 1 Christians 8.1 32 36 <1 1 2 6 21 2 Muslims 7.6 28 50 <1 1 <1 8 10 1 Unaffiliated 9.2 17 36 <1 <1 1 4 40 <1 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other China All 3.8 44% 23% 14% 12% 5% <1% <1% <1% Buddhists 3.8 45 19 14 14 6 2 <1 <1 Christians 4.0 46 15 13 16 8 <1 <1 1 Muslims 4.5 42 36 9 10 2 2 <1 <1 Unaffiliated 3.8 44 23 14 13 5 <1 <1 <1 Comoros All 6.9 43 36 2 2 <1 6 <1 10 Muslims 6.9 43 36 2 2 <1 6 <1 10 Costa Rica All 4.5 32 36 13 5 2 7 <1 4 Christians 4.4 32 36 13 5 2 7 <1 4 Unaffiliated 4.5 33 38 9 5 2 8 <1 5 Croatia All 3.6 40 22 14 14 9 2 <1 <1 Christians 3.7 39 23 15 13 8 1 <1 <1 Unaffiliated 3.3 42 17 14 15 10 2 <1 <1 Cyprus All 3.4 30 28 12 20 8 3 <1 <1 Christians 3.4 29 28 12 20 7 3 <1 <1 Czech Republic All 3.0 27 25 9 25 12 1 <1 2 Christians 3.1 19 27 11 26 13 <1 <1 3 Unaffiliated 3.0 29 25 8 24 11 1 <1 1 Democratic Republic of the Congo All 6.9 46 37 <1 2 1 8 2 3 Christians 6.9 47 37 <1 2 1 8 2 3 Muslims 7.3 34 34 <1 2 2 5 12 9 Unaffiliated 6.5 34 45 <1 2 1 9 2 5 Denmark All 2.7 15 27 3 26 23 6 <1 <1 Christians 2.6 10 28 3 29 25 5 <1 <1 Unaffiliated 2.7 19 27 3 24 21 6 <1 <1 Dominican Republic All 4.3 34 35 9 6 5 9 <1 3 Christians 4.3 36 33 9 6 5 8 <1 3 Unaffiliated 4.4 27 44 5 5 6 10 <1 4 Egypt All 5.1 18 63 9 5 1 3 <1 <1 Christians 4.6 15 59 13 6 3 4 <1 <1 Muslims 5.1 18 63 9 5 1 3 <1 <1 El Salvador All 4.9 40 36 8 4 2 6 <1 5 Christians 4.9 41 34 9 4 2 6 <1 5 Unaffiliated 4.7 32 47 5 3 2 7 <1 5 Estonia All 2.9 18 33 6 20 16 5 <1 1 Christians 2.8 14 36 8 21 17 2 <1 <1 Unaffiliated 3.0 20 32 5 20 16 5 <1 2 Ethiopia All 5.8 29 52 2 2 2 8 <1 5 Christians 5.6 29 51 3 2 2 7 <1 6 Muslims 6.3 29 53 2 2 <1 9 <1 4 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Finland All 2.8 12% 29% 3% 31% 20% 5% <1% <1% Christians 2.8 11 29 4 31 20 6 <1 <1 Unaffiliated 2.6 11 29 3 31 21 5 <1 <1 France All 3.0 17 30 6 24 16 6 <1 <1 Christians 2.7 12 26 7 31 19 4 <1 <1 Unaffiliated 3.0 20 31 5 20 15 8 <1 <1 Gabon All 6.5 56 20 1 3 5 5 2 6 Christians 6.8 60 19 1 3 4 5 1 6 Muslims 4.2 22 42 2 7 16 3 3 6 Unaffiliated 6.8 56 17 <1 4 5 5 2 10 Gambia All 13.8 45 14 <1 <1 <1 2 30 7 Christians 10.3 55 14 3 2 4 3 12 7 Muslims 13.9 45 14 <1 <1 <1 2 30 7 Germany All 2.7 17 22 6 29 20 3 <1 3 Christians 2.7 15 20 7 32 19 4 <1 3 Unaffiliated 2.5 17 21 5 28 22 3 <1 3 Ghana All 6.6 48 26 2 2 4 8 <1 10 Christians 6.0 45 27 2 2 4 9 <1 10 Muslims 8.9 55 24 1 1 2 5 <1 11 Unaffiliated 5.8 43 26 2 3 8 8 <1 10 Greece All 3.3 27 29 14 17 10 2 <1 <1 Christians 3.3 27 27 15 18 10 2 <1 <1 Unaffiliated 3.1 27 33 11 13 13 3 <1 <1 Guatemala All 6.1 40 39 5 2 <1 7 <1 6 Christians 6.1 40 39 5 2 <1 7 <1 6 Unaffiliated 5.6 38 39 5 2 2 8 <1 7 Guinea All 8.0 37 28 <1 1 <1 3 26 3 Christians 7.5 53 30 1 <1 <1 3 10 1 Muslims 8.1 34 27 <1 1 <1 3 29 3 Guinea-Bissau All 9.8 60 12 <1 <1 <1 2 23 2 Christians 8.8 74 10 <1 <1 <1 2 10 3 Muslims 10.8 52 14 <1 <1 <1 1 30 2 Unaffiliated 8.0 56 17 <1 1 <1 2 22 <1 Guyana All 5.0 39 36 8 5 3 7 <1 3 Christians 5.3 40 34 7 4 3 8 <1 4 Hindus 4.3 35 38 10 7 4 4 <1 2 Muslims 4.6 37 42 9 6 2 3 <1 <1 Haiti All 5.6 53 23 3 2 2 8 <1 9 Christians 5.6 53 23 3 2 2 8 <1 9 Unaffiliated 5.2 50 26 4 3 3 7 <1 8 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Honduras All 5.7 41% 34% 4% 2% 1% 6% <1% 12% Christians 5.8 41 34 4 2 1 6 <1 12 Unaffiliated 5.3 37 36 4 3 2 7 <1 11 Hungary All 3.1 27 28 8 21 14 3 <1 <1 Christians 3.3 24 28 10 21 14 2 <1 <1 Unaffiliated 3.0 29 27 6 21 14 3 <1 <1 Iceland All 3.3 17 38 6 17 12 5 <1 5 Christians 3.3 11 39 9 23 12 4 <1 3 Unaffiliated 3.3 21 36 4 14 13 5 <1 7 India All 5.8 54 31 8 3 <1 3 <1 <1 Buddhists 5.4 55 30 9 3 1 1 <1 <1 Christians 4.9 44 33 13 5 1 3 <1 <1 Hindus 5.7 55 30 8 3 <1 3 <1 <1 Muslims 6.5 51 37 5 2 <1 4 <1 <1 Unaffiliated 5.4 50 33 5 4 2 3 <1 3 Indonesia All 4.7 33 48 7 5 2 3 <1 3 Buddhists 5.0 33 34 15 4 2 2 <1 10 Christians 5.2 33 46 7 4 2 3 <1 5 Hindus 4.7 41 40 7 5 2 1 <1 3 Muslims 4.6 32 48 7 5 2 3 <1 3 Iran All 4.1 8 59 18 8 2 3 <1 <1 Muslims 4.1 8 59 18 8 2 3 <1 <1 Iraq All 7.7 40 51 4 1 <1 2 2 <1 Christians 5.4 32 50 12 4 <1 <1 <1 <1 Muslims 7.8 40 51 4 1 <1 2 2 <1 Ireland All 3.6 10 39 15 13 8 6 <1 8 Christians 3.7 10 40 16 13 8 6 <1 7 Unaffiliated 3.2 9 28 11 17 10 4 <1 21 Israel All 4.5 30 37 11 12 6 3 <1 <1 Jews 4.3 29 35 13 13 6 3 <1 1 Muslims 5.2 34 50 5 5 4 2 1 <1 Italy All 3.0 25 27 13 18 14 2 <1 <1 Christians 3.0 22 28 14 19 13 2 <1 <1 Unaffiliated 3.0 34 23 8 17 15 3 <1 <1 Ivory Coast All 6.7 40 28 <1 2 4 4 12 9 Christians 6.3 49 23 1 2 4 5 3 12 Muslims 6.9 32 32 <1 2 4 3 19 8 Unaffiliated 6.5 37 34 <1 3 4 3 11 7 Jamaica All 4.6 42 22 6 5 8 11 <1 6 Christians 4.7 44 22 6 5 6 11 <1 6 Unaffiliated 3.9 34 25 5 5 16 10 <1 5 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Japan All 3.1 27% 25% 12% 19% 15% 2% <1% <1% Buddhists 3.0 29 15 13 24 17 1 <1 <1 Unaffiliated 3.1 27 29 11 17 15 2 <1 <1 Jordan All 5.7 15 64 12 3 <1 3 <1 2 Muslims 5.7 15 64 12 3 <1 3 <1 2 Kazakhstan All 4.6 38 36 11 7 5 3 <1 <1 Christians 3.3 28 27 17 14 10 4 <1 <1 Muslims 5.1 42 39 9 4 2 3 <1 <1 Unaffiliated 3.6 27 33 16 14 7 3 <1 <1 Kenya All 5.5 35 38 2 2 5 12 <1 6 Christians 5.4 34 38 2 2 5 12 <1 7 Muslims 6.6 37 42 1 2 3 11 <1 4 Unaffiliated 6.0 36 33 1 2 8 14 <1 4 Kosovo All 6.8 58 27 10 2 <1 1 <1 <1 Christians 5.2 47 26 17 6 3 1 <1 <1 Muslims 6.9 59 27 9 2 <1 2 <1 <1 Kyrgyzstan All 5.4 51 34 5 4 2 3 <1 <1 Christians 3.4 36 20 12 19 11 3 <1 <1 Muslims 5.6 52 36 5 2 1 3 <1 <1 Laos All 5.6 50 39 5 2 <1 2 <1 <1 Buddhists 5.3 51 36 6 3 <1 2 <1 <1 Christians 5.9 53 41 2 2 <1 2 <1 <1 Lesotho All 5.6 55 20 3 2 3 5 <1 11 Christians 5.6 56 20 3 2 3 5 <1 11 Unaffiliated 5.2 45 31 2 2 7 6 <1 7 Liberia All 6.8 58 24 <1 2 2 5 2 7 Christians 6.8 59 23 <1 2 2 4 1 7 Muslims 7.0 53 23 <1 2 2 5 7 8 Unaffiliated 6.4 52 34 1 2 1 3 3 4 Lithuania All 2.9 26 26 8 18 17 2 <1 2 Christians 2.9 24 27 9 19 17 2 <1 2 Unaffiliated 2.9 39 18 5 16 19 1 <1 1 Madagascar All 5.4 32 49 2 3 2 8 <1 5 Christians 5.5 32 48 2 3 2 7 <1 6 Muslims 4.9 41 37 2 2 3 7 <1 8 Unaffiliated 5.3 29 52 2 3 2 9 <1 3 Malawi All 5.5 36 47 1 2 2 9 <1 3 Christians 5.5 36 46 1 2 2 9 <1 3 Muslims 5.9 40 45 <1 1 1 10 <1 1 Unaffiliated 5.5 22 67 1 2 2 5 <1 2 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Maldives All 7.3 64% 15% 3% 2% <1% 4% <1% 10% Muslims 7.3 64 15 3 2 <1 4 <1 10 Mali All 12.6 37 23 <1 <1 <1 1 34 3 Christians 8.9 43 34 <1 <1 <1 2 14 5 Muslims 12.8 37 22 <1 <1 <1 1 35 3 Mauritania All 7.4 43 37 2 1 <1 10 3 2 Muslims 7.4 43 37 2 1 <1 10 3 2 Mexico All 4.9 32 42 9 5 2 5 <1 5 Buddhists 3.6 19 22 17 10 12 3 <1 16 Christians 4.9 33 42 9 5 2 5 <1 5 Jews 4.5 17 35 9 8 4 3 <1 25 Unaffiliated 4.5 28 40 10 6 5 6 <1 5 Moldova All 3.3 23 28 13 17 10 5 <1 4 Christians 3.3 23 28 13 17 10 5 <1 4 Mongolia All 4.2 28 50 8 6 3 5 <1 <1 Buddhists 4.1 29 48 8 7 3 5 <1 <1 Christians 4.4 28 46 8 3 4 10 <1 1 Muslims 5.1 41 48 5 2 <1 2 <1 <1 Unaffiliated 4.1 26 52 8 6 3 4 <1 <1 Montenegro All 4.3 30 35 21 7 5 2 <1 <1 Christians 4.1 27 35 23 8 5 2 <1 <1 Muslims 5.4 42 34 15 4 2 <1 <1 <1 Mozambique All 5.8 41 38 <1 3 2 9 2 3 Christians 5.9 43 37 <1 3 2 9 2 4 Muslims 5.1 36 44 <1 5 3 9 <1 2 Unaffiliated 5.8 37 40 <1 3 2 10 4 3 Namibia All 6.4 62 10 1 3 4 5 <1 14 Christians 6.4 63 10 1 3 4 5 <1 14 Unaffiliated 5.9 47 22 4 4 5 5 <1 14 Nepal All 5.6 58 23 3 3 1 9 <1 2 Buddhists 5.0 53 25 4 4 3 9 <1 2 Christians 4.5 44 33 4 5 2 9 <1 2 Hindus 5.6 58 22 4 3 1 9 <1 2 Muslims 7.7 64 21 1 1 <1 11 <1 1 Netherlands All 2.9 16 28 6 26 17 5 <1 2 Christians 2.9 12 27 6 29 21 4 <1 <1 Unaffiliated 2.9 17 29 6 25 16 6 <1 2 Niger All 7.7 21 42 <1 2 <1 6 29 <1 Muslims 7.7 21 42 <1 2 <1 6 29 <1 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Nigeria All 7.7 19% 42% 2% 2% 2% 3% 28% 1% Christians 5.9 29 45 4 2 4 6 8 2 Muslims 8.7 13 41 <1 2 <1 2 40 <1 North Macedonia All 4.6 43 24 20 9 3 <1 <1 1 Christians 4.0 35 24 24 12 4 <1 <1 <1 Muslims 6.0 58 25 10 3 <1 <1 <1 1 Norway All 2.9 17 32 4 22 17 5 <1 2 Christians 3.0 13 34 5 26 17 4 <1 2 Unaffiliated 2.8 19 30 4 20 19 7 <1 1 Pakistan All 8.5 58 32 3 <1 <1 4 1 <1 Muslims 8.5 58 32 3 <1 <1 4 1 <1 Palestinian territories All 6.8 16 71 7 3 <1 2 <1 <1 Muslims 6.8 16 71 7 3 <1 2 <1 <1 Panama All 4.8 37 35 8 6 4 6 <1 4 Christians 4.6 36 35 9 6 4 6 <1 5 Unaffiliated 6.3 42 39 4 4 3 4 1 3 Papua New Guinea All 7.1 44 35 3 3 <1 4 <1 11 Christians 7.1 44 35 3 3 <1 4 <1 11 Paraguay All 4.9 39 38 7 4 3 5 <1 4 Christians 4.9 39 38 7 4 3 5 <1 4 Unaffiliated 4.6 36 40 9 2 6 4 <1 4 Peru All 4.9 37 38 8 4 3 5 <1 4 Christians 4.9 37 38 8 4 3 5 <1 4 Philippines All 5.6 31 50 6 3 1 3 <1 5 Buddhists 4.9 25 28 11 5 5 3 <1 22 Christians 5.6 32 49 7 3 1 3 <1 5 Hindus 5.6 36 44 7 3 1 4 <1 5 Muslims 6.4 26 63 4 2 <1 3 <1 2 Unaffiliated 5.5 16 65 4 5 3 4 <1 3 Poland All 3.7 34 29 9 14 9 2 <1 3 Christians 3.7 33 30 9 14 8 2 <1 3 Unaffiliated 3.3 44 21 6 14 10 3 <1 2 Portugal All 3.2 16 33 19 18 8 4 <1 3 Christians 3.3 16 32 19 19 8 3 <1 3 Muslims 4.4 27 30 9 5 5 4 <1 19 Unaffiliated 3.1 11 34 19 16 9 5 <1 4 Puerto Rico All 3.5 21 27 17 13 8 10 <1 4 Christians 3.5 21 27 17 13 8 10 <1 4 1 Due to conflict, source surveys were only conducted in two regions representing 45% of Somalia’s population. Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Republic of the Congo All 5.7 44% 40% 2% 3% 3% 6% 2% 2% Christians 5.6 45 38 2 3 3 6 1 1 Muslims 5.0 38 37 2 3 9 2 3 6 Unaffiliated 5.7 36 47 <1 2 4 3 4 2 Romania All 3.6 28 24 16 14 10 3 <1 5 Christians 3.6 29 24 17 15 8 2 <1 5 Jews 2.4 9 9 16 28 27 2 <1 10 Muslims 4.7 32 28 14 9 5 2 <1 11 Unaffiliated 3.5 20 27 17 15 13 2 <1 6 Russia All 3.2 37 23 9 13 10 7 <1 2 Christians 3.1 35 19 12 14 11 7 <1 3 Muslims 3.6 44 29 6 11 5 6 <1 <1 Unaffiliated 3.2 37 25 8 12 9 7 <1 1 Rwanda All 5.4 18 49 3 2 2 11 <1 15 Christians 5.4 18 50 3 2 2 11 <1 15 Muslims 5.6 18 39 2 2 3 10 <1 26 Unaffiliated 4.8 16 42 3 3 6 13 <1 16 Sao Tome and Principe All 5.0 33 41 2 3 4 15 <1 2 Christians 5.0 36 39 2 3 4 14 <1 2 Unaffiliated 4.8 23 51 2 3 4 15 <1 <1 Senegal All 13.5 55 10 <1 <1 <1 2 23 9 Christians 9.4 66 11 4 1 2 3 7 6 Muslims 13.6 55 10 <1 <1 <1 2 23 9 Serbia All 4.0 40 22 19 11 6 1 <1 <1 Christians 4.0 40 22 20 11 6 1 <1 <1 Muslims 5.2 41 35 12 5 3 2 <1 2 Sierra Leone All 6.3 53 25 <1 1 2 6 6 7 Christians 6.2 54 24 1 2 2 7 3 8 Muslims 6.4 52 25 <1 1 2 6 6 7 Slovakia All 3.6 37 26 13 13 8 2 <1 <1 Christians 3.6 37 26 14 13 8 1 <1 <1 Unaffiliated 3.4 38 27 11 13 8 4 <1 <1 Slovenia All 3.2 28 27 14 17 12 2 <1 2 Christians 3.3 29 25 14 15 11 2 <1 2 Unaffiliated 3.1 25 28 13 19 13 <1 <1 2 Somalia1 All 7.5 37 45 2 <1 <1 10 <1 5 Muslims 7.5 37 45 2 <1 <1 10 <1 5 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other South Africa All 5.2 53% 19% 5% 5% 6% 7% <1% 3% Christians 5.2 54 19 5 6 6 7 <1 3 Hindus 3.9 32 31 22 10 3 <1 <1 <1 Muslims 5.1 40 37 8 5 4 1 <1 5 Unaffiliated 4.4 47 16 6 7 14 6 <1 4 South Korea All 2.9 11 34 18 15 21 1 <1 <1 Buddhists 3.1 14 28 25 17 14 1 <1 <1 Christians 2.9 9 35 18 16 21 1 <1 <1 Unaffiliated 2.8 10 37 14 13 24 1 <1 <1 Spain All 3.2 26 29 12 17 10 3 <1 3 Christians 3.1 24 27 15 18 11 3 <1 2 Unaffiliated 3.1 31 31 6 16 10 3 <1 3 St. Lucia All 4.1 33 25 11 8 9 9 <1 4 Christians 4.1 35 25 12 8 9 9 <1 3 Suriname All 5.1 40 33 9 5 3 7 <1 2 Christians 5.4 44 29 7 5 4 9 <1 2 Hindus 4.6 36 38 15 5 2 2 <1 <1 Muslims 4.6 39 35 13 7 2 2 <1 2 Unaffiliated 5.6 34 39 4 4 4 13 <1 3 Swaziland All 6.1 58 15 2 2 5 9 <1 8 Christians 6.1 58 15 2 2 5 10 <1 8 Unaffiliated 6.5 56 23 <1 3 5 5 <1 7 Sweden All 2.7 12 27 4 28 20 7 <1 1 Christians 2.6 8 23 6 35 21 5 <1 2 Unaffiliated 2.7 14 28 3 26 21 7 <1 <1 Switzerland All 2.9 15 28 8 25 17 4 <1 2 Christians 2.9 13 30 9 28 17 3 <1 2 Unaffiliated 2.8 18 23 6 23 18 7 <1 4 Taiwan All 4.0 49 17 15 9 7 2 <1 <1 Buddhists 3.9 45 17 17 14 6 1 <1 <1 Unaffiliated 3.7 51 18 12 8 9 2 <1 <1 Tajikistan All 7.5 67 26 3 <1 <1 2 <1 <1 Muslims 7.5 67 26 3 <1 <1 2 <1 <1 Thailand All 4.1 51 20 11 10 4 2 <1 2 Buddhists 4.1 52 19 11 10 4 2 <1 2 Christians 3.9 35 31 14 12 2 2 <1 3 Muslims 4.8 43 36 9 4 2 3 <1 2 Timor-Leste All 6.7 43 45 3 2 1 5 <1 1 Christians 6.7 43 45 3 2 1 5 <1 1 Country/region Religion Average household size Extended Two-parent Adult child Couple Solo Single-parent Polygamous Other Togo All 6.8 36% 29% 1% 1% 3% 7% 17% 5% Christians 5.8 41 29 2 2 4 8 8 6 Muslims 8.1 34 28 <1 1 2 4 26 5 Unaffiliated 7.2 33 33 <1 1 2 9 18 3 Trinidad and Tobago All 4.4 36 27 16 6 6 6 <1 4 Christians 4.5 37 26 14 6 5 6 <1 4 Hindus 4.4 35 28 21 7 4 3 <1 2 Muslims 4.4 34 30 19 7 4 3 <1 3 Unaffiliated 3.9 31 23 12 8 13 7 <1 6 Tunisia All 4.9 16 52 22 5 1 2 <1 1 Muslims 4.9 16 52 22 5 1 2 <1 1 Turkey All 4.6 26 43 16 9 3 1 <1 <1 Muslims 4.6 26 43 16 9 3 1 <1 <1 Uganda All 6.2 40 36 <1 2 3 8 <1 9 Christians 6.2 40 37 <1 2 3 9 <1 9 Muslims 6.5 43 34 <1 2 3 7 1 9 Ukraine All 3.2 39 22 13 16 8 3 <1 <1 Christians 3.2 39 21 13 16 8 3 <1 <1 Unaffiliated 3.2 38 25 12 16 7 2 <1 <1 United Kingdom All 3.0 16 28 7 24 12 7 <1 6 Christians 2.9 10 26 8 28 15 7 <1 5 Unaffiliated 3.0 18 27 6 22 11 8 <1 8 United States All 3.4 11 33 14 20 11 9 <1 <1 Christians 3.4 10 34 14 21 11 10 <1 <1 Jews 3.0 6 30 11 30 13 5 <1 5 Unaffiliated 3.2 15 30 15 18 12 9 <1 <1 Vietnam All 4.6 45 34 10 5 2 2 <1 <1 Buddhists 4.9 56 23 11 5 3 2 <1 <1 Christians 5.0 40 39 12 4 2 3 <1 <1 Unaffiliated 4.5 43 36 10 6 2 2 <1 1 Yemen All 8.6 42 48 3 2 <1 2 2 <1 Muslims 8.6 42 48 3 2 <1 2 2 <1 Zambia All 6.7 47 39 1 2 1 6 <1 3 Buddhists 6.6 36 50 2 1 2 4 <1 3 Christians 6.7 48 39 1 2 1 6 <1 3 Hindus 4.3 31 39 7 13 4 <1 <1 5 Muslims 6.5 48 37 1 2 1 5 1 4 Unaffiliated 5.9 38 42 2 3 4 6 <1 4 Zimbabwe All 5.5 52 25 1 2 3 8 1 7 Christians 5.5 53 23 2 2 3 9 <1 7 Unaffiliated 5.4 48 36 1 2 3 5 1 4 |
13607 | https://www.youtube.com/watch?v=NEz1y2pZ41g | Prove that |1+a1 1 1a+b1 1 1 1+c|=a b c(a+1/a+1/b+1/c)=a b c+b c+c a+a b
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Posted: 19 Sep 2017
To ask Unlimited Maths doubts download Doubtnut from - Prove that |1+a1 1 1a+b1 1 1 1+c|=a b c(a+1/a+1/b+1/c)=a b c+b c+c a+a b
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13608 | https://stackoverflow.com/questions/19198468/discrepancy-in-calculating-pearson-correlation-coefficient | r - discrepancy in calculating pearson correlation coefficient - Stack Overflow
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discrepancy in calculating pearson correlation coefficient
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Asked 11 years, 11 months ago
Modified11 years, 11 months ago
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In R, there seems to be a discrepancy for calculating the pearson correlation coefficient, between (a) using the raw-score formula in one step and (b) evaluating the numerator and denominator separately first. In particular, when I did the calculation in one step, the result was erroneous, but it was correct when I evaluated the numerator and denominator separately first. How come? I might be doing something wrong, but I can't figure out what it is.
```
data
x <- 1:5
y <- 5:1
x squared, y squared, x times y; for raw score formula
xx <- xx
yy <- yy
xy <- xy
correlation coefficient; the value that should come out
cor(x,y) #-1
raw score formula, in one line
wrong <- length(xy)sum(xy)-sum(x)sum(y)/
sqrt((length(xx)sum(xx)-sum(x)^2)(length(yy)sum(yy)-sum(y)^2))
wrong #170.5
raw score formula, separating numerator and denominator
numerator <- length(xy)sum(xy)-sum(x)sum(y)
denominator <- sqrt((length(x)sum(xx)-sum(x)^2)(length(y)sum(yy)-sum(y)^2))
correct <- numerator/denominator
correct #-1
```
I'm using R 2.14.1 in Xubuntu 12.04.
R Language Collective
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correlation
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asked Oct 5, 2013 at 13:28
Michael I.Michael I.
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It's an order of operations error.
You need one more set of parentheses in the numerator:
notwrong <- (length(xy)sum(xy)-sum(x)sum(y))/
sqrt((length(xx)sum(xx)-sum(x)^2)(length(yy)sum(yy)-sum(y)^2))
notwrong #-1
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answered Oct 5, 2013 at 13:42
Peter FlomPeter Flom
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Michael I.
Michael I.Over a year ago
i looked at my formula over and over again so i'm surprised i missed that. thank you so much.
2013-10-06T12:32:16.377Z+00:00
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2Odd Pearson correlation coefficient results
0Error when calculating correlation
0Pearson’s correlation test
3issue in finding correlation in R
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13609 | https://www.youtube.com/watch?v=LYo5ugSqP1A | Piecewise Functions — Why They Look Weird, and How to Make Them Make Sense
Krista King
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Description
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Posted: 5 Sep 2025
At first glance, piecewise functions look like they’re breaking the rules—multiple equations for a single function, strange brackets, and graphs that seem disconnected. But once you understand how they’re built, the weirdness disappears. In this lesson, we’ll go step-by-step so you can read, graph, and interpret piecewise functions with confidence.
📚 If you want more help with piecewise functions and other Algebra 1 topics, you can get the full step-by-step course here:
📌 What You’ll Learn in This Video
--What a piecewise function actually is
--How to read and interpret each “piece” of the definition
--How to graph piecewise functions without getting lost
--Why they look so strange at first—and how to make sense of them quickly
🖥️ Extra Help
I’ve put together complete courses for Algebra 1 and beyond, plus downloadable formula sheets and practice problems with detailed solutions. You can check them out here →
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7 comments
Transcript:
You know how speed limits change depending on where you are? Well, imagine that your house is inside of a school zone where the speed limit is 25 miles an hour. When you leave your house, that means you're going to be traveling 25 mph, maybe for the first mile. So, we might say that your speed s is 25 mph when time t is between 0 and 1 m. Maybe then you get on the highway and your speed jumps to 65 mph. And maybe you travel at 65 mph between that first mile and the 10th mile of your journey. And then at the 10th mile, you enter a construction zone where you're forced to slow down to 45 miles an hour and maintain that speed from the 10th mile of your journey to the 15th mile of your journey. Well, we can model your speed with a peace-wise function that looks like this. And with that, we're going to cover everything about how to read them, grab them, and evaluate them so that by the time we're done, we understand absolutely everything about peacewise functions. Starting with here, what they are. A peace-wise function is just a regular function that's made up of different rules that define the function over different parts of the x-axis. So instead of one rule that applies to everything like we're used to seeing if we had for example just the function f(x)= 2x + 1, a peacewise function uses different equations depending on the value of x. Here f ofx is defined by 2x + 1 whenever x is less than 0. But when x is greater than or equal to 0x^2 + 3 defines the function. But and this is important, this is still just one function. It simply lives in multiple pieces. We see these pieces when we sketch the graph of the function. So if we just take the part 2x + 1, we could sketch in the line 2x + 1 on the graph. And normally that full line would look like this. But because this piece only defines the function for x less than 0, we erase everything to the right of the yaxis so that we only see the part of 2x + 1 where x is less than zero. Then to complete the graph of this function, we sketch in the graph ofx^2 + 3, but we only show the part that's to the right of the y-axis because this piece only defines the function when x is greater than or equal to 0. This then taken in its entirety, including both pieces, is the graph of the function f ofx. Remember, we still have to think about this as one function, not two separate functions. It's just one function defined in multiple pieces. So each line of the function tells us what rule to use on the left and then when to use it on the right. This if part on the right, that's the condition. So we always have to pay attention to it so that we know where this rule on the left actually applies. So that's how we read a peacewise function. How do we actually evaluate it? Well, if we want to evaluate at a specific value of x, let's say evaluate this function when x is -1. The first thing we ask ourselves is which rule applies? In other words, in which interval does x=1 fall? Well, does -1 satisfy this first inequality x less than 0 or the second inequality x greater than or equal to 0? Because -1 is less than 0, this value satisfies this first inequality. Which means to evaluate f at -1, we plug into 2x + 1, notx^2 + 3. Therefore, evaluating the function at -1 means we take 2 -1 + 1 or -2 + 1 gives us -1 and the value of f at x=1 is -1. But if instead we wanted to evaluate the function at x= 0, we look at our inequalities and we see that 0 satisfies this second inequality x greater than or equal to 0. Which means if we want the function's value at 0, we have to plug intox^2 + 3. And when we do, we get 0^2 + 3 or a value of three. So at x equals 0, the function's value is three. The key here is that every x value plugs into just one rule of the peiewise function. The only trick is knowing which one to use, and we decide which one to use based on the condition. We figure out which condition is satisfied by the x value we want. And then we use that one rule and only that rule to evaluate the function f. Now, how do we graph these functions? Well, this is where it's easy to start panicking, but if we simply take this step by step, we're going to do just fine. There are only three steps to graphing a peace-wise function. The first of which is to identify the break points. These are the x values where these conditions where the rule changes for this function f. We can see the break point is at xals 0 because that's where this first rule stops defining the function and this second rule takes over. Once we identify the break points, our second step is to graph each piece, but only over its domain. So, we sketch each graph separately, like we did before. We sketched in 2x + 1, but only for x less than 0, and got this little section here of the graph. Then, we sketched x^2 + 3, but only for x greater than or equal to 0. And we got this half of the upside down parabola. If it helps, you can also break this down into two steps. You can sketch the entire graph like we looked at earlier with the line. We sketched the entire 2x + 1 over the graph. And then you can erase the part of the graph that's outside of where the inequality defines that piece. So step one is break points. Step two is sketch the graph. And then the last and final step is to check our circles. And this is really critical for a peace-wise function. Here again we go back to the conditions in our peacewise function. These x less than 0 and x greater than or equal to 0 conditions because these inequalities are going to tell us whether we have an open circle or a solid circle. When an inequality is just less than or greater than without the or equal to part of the inequality, we're going to have an open circle. And we see that here attached to the line. What this open circle shows us is that this piece 2x + 1 defines the function all the way up to x= 0, but not including the exact value at x= 0 itself. That exact value x= 0 is omitted from this piece 2x + 1. On the other hand, when our inequality includes this or equal to such that we have x greater than or equal to or x less than or equal to, then we're going to have a solid circle because that solid circle indicates that this piece, this x^2 + 3 piece not only defines the function for all values of x greater than 0, but also right at x=0 itself. In other words, we use open circles for less than or greater than, and we use solid dots for less than or equal to and greater than or equal to. So just remember three steps. Identify the break points, sketch the graph of each piece individually, and then check your circles, making sure you've got all your open circles and closed dots in the correct place. Now, that being said, let's talk common graphing mistakes. These are the big three and how to prevent them. So, the first one is graphing pieces across the entire graph instead of restricting them to the domain that's given by the condition. So for example, if we had continued the graph of the parabola this way or if we had continued the graph of the line this way, we wouldn't be restricting each piece to its associated domain. And so we'd be making that first mistake. The second issue is with the open and closed circles. It's really common to mess up where we have this closed dot and where we have this open circle, which is why the last step here in our process of checking our circles is so important. We don't want to make that mistake. We need to make sure that we have circles in the first place instead of leaving no circles at all. And we need to make sure that they're closed or open correctly based on the conditions of each piece of the function. And then the last mistake is leaving weird gaps or overlaps that don't match the condition. So for example, it's really common in a graph like this one to feel uncomfortable about stacking the pieces this close to each other. And so people will instead leave a gap where instead of sketching this parabolic piece here, they'll maybe start it over here and sketch the graph that way, leaving this one unit gap between the blue piece and the red piece. And we just want to make sure that we're not leaving gaps based on how the graph feels and instead placing each one in the correct domain that's given by the condition here. So we just want to make sure that we stick to our domain limits and we double check our dots. Now when it comes to understanding and interpreting a graph, let's take a look at a graph like this one. So peacewise functions often have these jumps which we also call discontinuities or specifically jump discontinuities. And these happen when one piece ends at one y-value and the next one starts at another value of y. So this yellow piece ends here and its y-value is -2. The red piece takes over at the same value of x but its y-value is pos1. So we have this jump from -2 up to pos1 and that's a jump discontinuity in the peace-wise function. And of course those are really common. Sometimes we'll even see a hole or a value that's missing and other times the function's value exists but it just jumps like this. But the concept is still the same. We can still capture each piece of the graph as one line in our peacewise function. And as long as we make sure that we define a rule and a condition for each piece of the graph, we'll be capturing the whole function f. So for instance, if we look at each piece of this graph, we can see that this first blue piece on the left here is the horizontal line at y=1. So the rule for that part of the graph would be -1. What about this yellow piece here? Well, we can see that it intersects the y-axis at1. And the slope of the line is -1 because we go down one over one, down one over one. So the equation of that line is x -1. We can see this red piece here is sitting at y= pos1. That's the horizontal line y = 1. And then this purple piece here on the far right, we can see it also has a slope of one because we go up one unit, over one unit, up one unit, over one unit. And if we imagine continuing it down until we meet the y ais, we would follow this diagonal and we would arrive here at y= -2. So the y intercept of this line were it to extend infinitely is -2. And therefore the equation of that line is x= -2. Remember that's the rule if the line extends infinitely. All we have to do is restrict the domain if we're going to write this as a peace-wise function. You can see here that we've built the rule for each piece of the graph except this singular dot here which is at the point -20. What we care about is the y value just like for all these other pieces. This is y =1. This is y = pos1. This is y =x -1. Well, the value here is y equals 0. So even though this is just a dot, we can indicate its value as zero. And then if we're actually going to write the peace-wise function from the graph, we've already figured out the rule for each segment and we just have to define the domain or the condition for each rule. All we do is work our way from left to right. So starting with the blue piece here, the rule is1. Then we move to the green circle. The rule there is zero. The rule for the yellow line is x -1. The rule for this horizontal red line is y = 1. And the rule for this purple diagonal line here is y = x -2. So we put in all of our rules. And now we just define each condition. Well, the condition for y=1 or f ofx=1. We can see that that piece runs all the way up to this line here x= -2. But that open circle right there means that x= exactly -2 is not included. Which means we define that condition as x less than -2 instead of x less than or equal to -2. This singular dot here since it's a closed circle since it's a closed dot a solid dot that means the function's value is defined there at exactly x= -2. So we would write that as x= -2. And then again for the yellow piece the function's value is not defined at exactly x= -2 nor is it defined at the next break point x= pos1. So for this rule we write the condition as x between -2 and pos1 but not at exactly either end point. So not at exactly -2 and not at exactly pos1. The red piece picks up at x= 1 and extends to x= 2. So again we'll have an inequality where x is defined between 1 and 2. But the solid circle tells us that that piece defines the function at exactly x= 1. And the open circle on the right tells us that it doesn't define the function at exactly x= 2. So we have x greater than or equal to 1, but just less than 2. And that's where the purple piece picks up and defines the function for x greater than or equal to 2 because of the solid circle that's attached to this purple segment. So just think about this like building a puzzle backwards. We break the peacewise function into segments. We figure out a rule for each piece. the - 1 0x -1 1 and x -2 and then we just assign the correct domain to each piece and build out the peace-wise function one line at a time. So with all that in mind, let's do a real world example because these functions do of course appear in the real world. Let's say we've got a taxi company or a ride share service that's charging by the mile and maybe they charge $5 for the first mile and $2 for each mile after that. So the cost is going to depend on how far we've gone, how many miles we've traveled. So if we sketch that, our sketch would start by looking like this, where for the first mile, so between a distance of 0 and 1 miles, our cost is $5. If however, we go over one mile but stay under 2 miles, then the cost of our trip is going to be that $5 plus an extra $2 for the second mile. And so anywhere in that interval, our cost for the ride will be $7. And the cost will continue to step up each mile that we travel. So we could start this way, but now we need to think about what happens at each of these break points. Well, if we travel exactly a distance of 0 miles, in other words, we don't travel at all. We don't even take the trip. Then our function isn't even defined. So we leave that circle open. But if we start moving, if we go even 0.0001 miles, now the cost of our ride is immediately $5, which is why we have this horizontal line here. But what happens at this break point at x= 1? Well, if we go exactly one mile and not a single inch further, then the cost of our trip is still going to be $5. So we have a solid circle here at the end of the first segment. Which means then at exactly x= 1 on the left side of this second segment here, we're going to have an open circle and then a solid circle at exactly 2 miles. So we can add in circles for the second segment. And we would continue on that way adding the same pattern of circles to each segment in our graph. Then the question just becomes how do we write this as a peace-wise function? Well, maybe we call the function capital t of x for the cost of our taxi ride. And what we would say is that the first rule here is five because the equation of this line is y= 5. The equation of the second line is y= 7 then 9 then 11 then 13 and so on. Now what about the condition associated with this first rule? Well 5 defines the function between 0 and 1 but not at exactly zero and at exactly one which means the condition there is x greater than zero but less than or equal to 1. And then we could keep building our peacewise function by saying that the cost of our taxi ride is defined by seven. This second rule when x is greater than 1 but less than or equal to 2. And we could keep going that way. The function's value is 9 when x is between 2 and 3, but we're saying greater than 2 and less than or equal to 3. That its value is 11 when x is between 3 and 4. but greater than three and less than or equal to four. And then because this is a predictable pattern, we could end the definition with just this dot dot dot to show that the definition of this peace-wise function continues indefinitely. So now we know what a peace-wise function is. We know how to read it and match up each rule with its condition. We know how to evaluate a peace-wise function by finding first which condition the x value satisfies and then evaluating that function just at the associated rule. making sure we're always only evaluating one piece based on which condition fits our x value. We know how to graph a function step by step by first looking at break points, graphing each piece individually and then double-checking our circles to see whether each one should be open or closed. We know how to interpret discontinuities in a peace-wise function and how to build a peace-wise function based on its graph. And we know how they show up in real life. So they might look weird, but they're actually pretty logical once we get used to the rules and understand that each line of the function is really nothing more complicated than a matching rule and condition. If this video helped peacewise functions finally make sense and you want to keep building on that understanding, I've got more practice and lessons waiting for you. Just click the link below to head over to my Algebra 2 course when you're ready to dive in. |
13610 | https://fiveable.me/lists/capital-budgeting-methods | Capital Budgeting Methods to Know for Financial Mathematics
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Capital Budgeting Methods to Know for Financial Mathematics
Related Subjects
💹Financial Mathematics
Capital budgeting methods help evaluate investment opportunities by analyzing cash flows and returns. Key techniques like NPV, IRR, and Payback Period assess profitability and risk, ensuring informed financial decisions that align with the principles of Financial Mathematics.
Net Present Value (NPV)
Measures the difference between the present value of cash inflows and outflows over a project's lifetime.
A positive NPV indicates that the projected earnings exceed the anticipated costs, making the investment worthwhile.
NPV accounts for the time value of money, reflecting the principle that money today is worth more than the same amount in the future.
Internal Rate of Return (IRR)
Represents the discount rate at which the NPV of a project becomes zero, indicating the project's break-even point.
A project is considered acceptable if its IRR exceeds the required rate of return or cost of capital.
IRR is useful for comparing the profitability of multiple projects, but can be misleading for non-conventional cash flows.
Payback Period
The time it takes for an investment to generate enough cash flows to recover its initial cost.
Simple to calculate and understand, making it a popular initial screening tool for projects.
Does not consider the time value of money or cash flows beyond the payback period, which can lead to incomplete assessments.
Discounted Payback Period
Similar to the payback period, but accounts for the time value of money by discounting cash flows.
Provides a more accurate measure of how long it takes to recover the initial investment in present value terms.
Still does not consider cash flows that occur after the payback period, limiting its comprehensiveness.
Profitability Index (PI)
Calculated as the ratio of the present value of future cash flows to the initial investment.
A PI greater than 1 indicates that the project is expected to generate value, while a PI less than 1 suggests it may not be worthwhile.
Useful for ranking projects when capital is limited, as it helps identify the most efficient use of resources.
Modified Internal Rate of Return (MIRR)
Addresses some limitations of IRR by assuming reinvestment of cash flows at the project's cost of capital rather than the IRR itself.
Provides a more realistic measure of a project's profitability and efficiency.
Useful for comparing projects with different cash flow patterns and durations.
Equivalent Annual Cost (EAC)
Converts the total cost of an investment into an annualized figure, allowing for easy comparison of projects with different lifespans.
Useful for assessing the cost-effectiveness of projects, especially in capital-intensive industries.
Helps decision-makers evaluate long-term investments on a consistent basis.
Accounting Rate of Return (ARR)
Measures the expected annual return on an investment based on accounting profits rather than cash flows.
Calculated by dividing the average annual profit by the initial investment cost.
Simple to compute but does not consider the time value of money, making it less reliable for investment decisions compared to other methods.
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13611 | https://www.geeksforgeeks.org/maths/finding-the-volume-and-surface-area-of-a-cylinder/ | Volume and Surface Area of a Cylinder
Last Updated :
23 Jul, 2025
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The volume of a cylinder can be calculated using the formula V = πr2h. While the total surface area of a cylinder can be calculated using the formula: A=2πr(h+r). In geometry, a cylinder is a three-dimensional solid figure which contains two parallel circular bases joined by a curved surface, situated at a particular distance from the center of the cylinder.
For instance, toilet paper rolls, and plastic cold drink cans are examples of cylinders. A cylinder is characterized by two major properties, i.e. surface area and volume. The word cylinder is derived from a Latin (Cylindrus) word, meaning "roll", "roller", and "Tumblr".
Table of Content
Volume and Surface Area of a Cylinder
Lateral Surface Area or Curved Surface Area of Cylinder
Total Surface Area of Cylinder
Volume of Cylinder
Sample Problems on Volume and Surface Area of a Cylinder
Volume and Surface Area of a Cylinder
Below is the volume and surface area of a Cylinder.
Lateral Surface Area or Curved Surface Area of Cylinder
Curved surface area is also termed a lateral surface area. The area formed by the curved surface of the cylinder i.e. space occupied between the two parallel circular bases is known as CSA. The formula for CSA is given as:
Curved Surface Area (CSA) = 2πrh square units
‘h’ is the height
‘r’ is the radius
Total Surface Area of Cylinder
So, in order to find out the total surface area of a cylinder, we calculate the curved surface area and the area of two circles.
Total surface area of the cylinder is defined as the total area occupied by it. A cylinder consists of two circles along with a curved sheet. The total surface area of a cylinder can be calculated by the combination of curved surface area and the area of two circles.
⇒ Curved Surface Area(CSA) = Circumference of the Circle × Height
⇒ C.S.A = 2r × h
⇒ Area of a Circle = πr2
⇒ Total Surface Area (TSA) = Curved Surface Area + 2(Area of a circle)
We know,
⇒Curved Surface Area = 2πrh
⇒ Area of circle = πr2
Total Surface Area (T.S.A) = 2πrh + 2πr2 = 2πr(h+r) square units.
Where,
h is Height
r is Radius of Cylinder
Volume of Cylinder
Volume of cylinder is referred to as the density or amount of space it occupies.
We have,
Volume of a Cylinder = Area of a Circle × Height
Since, we have an area of a circle = πr2
Volume = πr2 × h
where,
h is Height
r is Radius of Cylinder
Sample Problems on Volume and Surface Area of a Cylinder
Problem 1: Compute the total surface area of the cylinder, with a radius of 5cm and height of 10cm?
Solution:
Since, we know,
Total surface area of a cylinder, A = 2πr(r+h) square units
Therefore, A = 2π × 5(5 + 10) = 2π × 5(15)
= 2π × 75 = 150 × 3.14
= 471 cm2
Problem 2: What is the volume of a cylindrical shape water container, that has a height of 7cm and diameter of 10cm?
Solution:
Given,
Diameter of the container = 10cm
Thus, radius of the container = 10/2 = 5cm
Height of container = 7cm
As we know, from the formula,
Volume of a cylinder = πr2h cubic units.
Therefore, volume of given container, V = π × 52 × 7
V = π × 25 × 7 = (22/7) × 25 × 7 = 22 × 25
V = 550 cm3
Problem 3: Alex wants to purchase a cylindrical can with a radius equivalent to 5 inches. The can contains 1 gallon of oil. Find the height of the cylinder.
Solution:
Volume V is given by= 1 gallon
1 gallon= 231 cubic inches
Radius r = 5 inches
Volume f the cylinder is given by,
V = πr2h
231 = 22/7 × (5)2 × h
(231 × 7)/(22 × 25) = h
h = 2.94 inches.
Therefore, the height is equivalent to 2.94 inches.
Problem 4: A water tank has a radius of 40 inches and a height of 150 inches. Find the area.
Solution:
Water tank is cylindrical in nature.
Total Surface Area of a cylinder is given by, 2πr(h+r)
TSA = 2 × 22/7 × 40(150 + 40)
TSA = 2 × 22/7 × 40 × 190
TSA = 440/7 × 7600
TSA = 3344000/ 7
Area = 47,7142.857 sq.inches.
Problem 5: Find the volume of the cylinder having a radius of 5 units and a height of 8 units?
Solution:
We have,
Radius,r = 5 units
Height,h = 8 units
Volume of the cylinder, V = πr2h cubic units.
V = (22/7) × 52 × 8
V = 22/7 × 25 × 8
V= 628.57 Cubic units.
Hence, the volume of the cylinder is 628.57 cubic units.
Total surface area of the cylinder is the sum of all the surfaces of the cylinder.
Whereas volume of cylinder is defined as the capacity of the cylinder, it is defined as the total space occupied by the cylinder.
What is the formula for volume and surface area of cylinder?
Formula for surface area of cylinder is,
A = 2πr(h+r) square units.
Formula for volume of cylinder is,
V = πr2h cubic units.
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13612 | https://www.geeksforgeeks.org/maths/combinations/ | Combinations - GeeksforGeeks
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Combinations
Last Updated : 23 Jul, 2025
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Combination is a way of choosing items from a set, (unlike permutations) when the order of selection doesn't matter. In smaller cases, it's possible to count the number of combinations. Combination refers to the mixture of n things taken k at a time without repetition.
2 / 2
Example:For set S = {a, b, c}, the possible combinations of choosing 2 elements are, {a, b}, {a, c}, {b, c}. If we choose 3 items, then there is only one combination {a, b, c} which is we pick all three.
Some other examples of Combination in Real-Life are:
Team Selection: Choosing 5 players from a pool of 12 for a sports team.
Menu Selection: Picking a combination of dishes (starter, main course, dessert) from a menu.
Lottery Numbers: Selecting a set of lottery numbers where the order doesn’t matter.
Combinations Formula
Combination is the choice of selecting r things from a group of nthings without replacement and where the order of selection is not important. The number of combinations when ‘r’ elements are selected out of a complete set of ‘n’ elements is denoted by nCr
Example:Let n = 4 (E, F, G, H) and r = 2 (consisting of all the combinations of size 2).
n C r = 4 C 2
⇒ 4 C 2 = 4!/((4-2)!×2!)
⇒ 4 C 2 = (4 × 3 × 2 × 1)/(2 × 1 × 2 × 1) = 6
The six combinations are EF, EG, EH, FG, FH, and GH.
Read More aboutCombinations Formula.
Relationship between Permutation and Combination
Permutation and combination have a lot of similarities but they also have some striking differences. For n different objects, we have to make r unique selections from this group of n objects.
The number of permutations of size r from n object isnPrhere the order, of selection is not important so each selection is counted r! times. So the number of unique selections is n P r / r! We know that a unique selection of r things from the total of n things is called a combination (n C r). Thus,
nCr=nPr/ r!
Read More aboutPrinciple of Couting.
Difference Between Permutation and Combination
The key difference between Permutation and Combination is:
Permutation considers the order of arrangement, meaning the arrangement of items matters. For example, arranging letters to form words or seating people in a row.
Combination ignores the order, meaning only the selection of items matters, not their arrangement. For example, selecting a committee from a pool of candidates or choosing lottery numbers.
Read detaileddifference between Permutation and Combination.
How to Calculate Probability of Combinations?
Probability of Combinations can be easily understood with the help of the examples given below:
Example:
How many ways are possible to distribute 7 different candies to 3 people where each gets only 1 candy?
In how many ways can the letters of the word ‘POWER’ be arranged?
How many six-digit numbers can be formed with digits 2,3,5, 6, 7, and 9 and with distinct digits?
Solution:
For 1:
For the first people, we can choose any of the candy among the 7 candies available. Similarly, for the 2nd person we are left with 6 choices and for the 3rd, we will be having 5 choices.
So, the number of ways of distributing candies = 7 × 6 × 5 = 210 ways
For 2:
Letters of the word ‘POWER’ can be arranged in 5! ways i.e.
5 × 4 × 3 × 2 × 1 ways = 120 ways.
For 3:
Number of distinct ways of forming 6-digit numbers with different digits is
6! = 6 × 5 × 4 × 3 × 2 × 1 ways
= 720 ways.
Also, to learn about probability and combinations, read:How To Calculate Probability using Combination?
What is Handshaking Problem?
Handshaking problem is one of the most interesting problem related to combinations. It is used to find that in a room full of people how many handshakes are required for everybody to shake everybody else's hand exactly once?
Example: The table given below tells us about the minimum number of handshakes required for various groups of people.
| Number of People | Possible Combinations | Minimum Handshake required |
---
| Two People | A-B | 1 handshake |
| Three People | A-B A-C B-C | 3 handshake |
| Four People | A-B A-C A-D B-C B-D C-D | 6 handshake |
When there are few people, handshakes can be counted individually. However, with thousands of people in a hall, counting each handshake becomes impractical, which is where combinations come into play.
Handshaking Combination
It means the total number of people in a room doing the handshake with each other. With the help of combination formulas, it can easily be calculated. The formula for calculating the handshakes when there are n people available is given by,
Total Number of Handshakes = n × (n - 1)/2
Total Number of Handshakes =nC2
Article Related to Combinations
Binomial Theorem
Binomial Coefficient
Examples on Combinations
Example 1: In how many ways 6 boys can be arranged in a queue such that
a) Two particular boys of them are always together
b) Two particular boys of them are never together
Solution:
a) If two boys are always together, then they will be treated as one entity. Hence we can be arranged 5 boys in 5! ways. Also, two boys can arrange themselves in 2 different ways.
Therefore required arrangement = 5! × 2 = 120 × 2 = 240 ways.
b) Total number of permutations among 6 numbers is given by = 6! = 720.
In 240 cases 2 boys are always together.
Thus, for two boys who are never together no of ways will be = 720 – 240 = 480 ways.
Example 2:In a room of n people, how many handshakes are possible?
Solution:
To see the people present, and consider one person at a time. The first person will shake hands with n - 1 other people. The next person will shake hands with n-2 other people, not counting the first person again. Following this, it will give us a total number of
(n - 1) + (n - 2) + ... + 2 + 1
=n(n - 1)/ 2 handshakes.
Example 3:Another popular handshake problem starts out similarly with n>1 people at a party. Not being possible to shake hands with yourself, and not counting several times handshakes with the same person, the problem is to show that there will always present two people at the party, who had shaken hands the same number of times in the party.
Solution:
Solution to this problem starts by using Dirichlet's box principle. If there exists a person at the party, who has shaken hands zero times, then every person which is there at the party has shaken hands with at most n-2 other people at the party.
There are n-1 possible handshakes (from 0 to n-2), among n people there must be two who have shaken hands the same number of times. If there are zero persons, who has shaken hands zero times this means that all of the party guests have shaken hands at least once.
This also amounts to n-1 possible handshakes (from 1 to n-1).
Example 4:In the function, if every person shakes hands with every other in the party and there exists a total of 28 handshakes at the party, find the number of persons who were present in the function.
Solution:
Suppose there are n persons present at a party and every person shakes hands with every other person.
Then, total number of handshakes =nC2= n(n - 1)/2
⇒ n(n - 1)/2 = 28
⇒ n(n - 1) = 28 × 2
⇒ n(n - 1) = 56
⇒ n = 8
Practice Questions on Combinations in Maths
Q1. A classroom has 20 students, and a committee of 4 students needs to be formed to organize an upcoming event. In how many ways can this committee be chosen?
Q2. You have 5 different books on mathematics and want to select 3 to place on your desk for quick reference. In how many ways can you choose which 3 books to place on the desk?
Q3. A fruit shop offers baskets that can contain any combination of 3 different fruits from their selection of 5 different fruits (apple, banana, cherry, date, and elderberry). How many different fruit baskets can be made?
Q4. A lottery game requires you to choose 6 numbers out of a possible 49. How many different combinations of numbers can you choose?
Q5. From a pool of 12 jurors, a jury of 6 needs to be selected for a trial. In how many different ways can the jury be formed?
Combination in Mathematics
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13613 | https://courses.lumenlearning.com/suny-mcc-chemistryformajors-2/chapter/shifting-equilibria-le-chateliers-principle-2/ | Chapter 13: Chemical Equilibrium
13.4 Shifting Equilibria: Le Chtelier’s Principle
Learning Objectives
By the end of this module, you will be able to:
Describe the ways in which an equilibrium system can be stressed
Predict the response of a stressed equilibrium using Le Chtelier’s principle
As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (K). We next address what happens when a system at equilibrium is disturbed so that Q is no longer equal to K. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of Q will no longer equal the value of K. To re-establish equilibrium, the system will either shift toward the products (if Q < K) or the reactants (if Q > K) until Q returns to the same value as K.
This process is described by Le Chtelier’s principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in Q; the reaction will shift to re-establish Q = K.
Predicting the Direction of a Reversible Reaction
Le Chtelier’s principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of Q and K for the system to predict the changes.
Effect of Change in Concentration on Equilibrium
A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.
[latex]{\text{Fe}}^{\text{3+}}\left(aq\right)+{\text{SCN}}^{\text{-}}\left(aq\right)\rightleftharpoons [\text{Fe}({\text{SCN}})]^{\text{2+}}\left(aq\right)[/latex]
The stress on the system in Figure 1 is the reduction of the equilibrium concentration of SCN– (lowering the concentration of one of the reactants would cause Q to be larger than K). As a consequence, Le Chtelier’s principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN– part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.
Figure 1. (a) The test tube contains 0.1 M Fe3+. (b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)2+ ion. Fe3+(aq) + SCN−(aq) ⇌ Fe(SCN)2+(aq). (c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN− as the white solid AgSCN. Ag+(aq) + SCN−(aq) ⇌ AgSCN(s). The decrease in the SCN− concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)2+. (credit: modification of work by Mark Ott)
The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:
[latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{HI}\left(g\right){K}_{c}=50.0\text{ at }400^\circ\text{C}[/latex]
The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H2] = [I2] = 0.221 M and [HI] = 1.563 M is at equilibrium; for this mixture, Qc = Kc = 50.0. If H2 is introduced into the system so quickly that its concentration doubles before it begins to react (new [H2] = 0.442 M), the reaction will shift so that a new equilibrium is reached, at which [H2] = 0.374 M, [I2] = 0.153 M, and [HI] = 1.692 M. This gives:
[latex]{Q}_{c}=\frac{{\left[\text{HI}\right]}^{2}}{\left[{\text{H}}_{2}\right]\left[{\text{I}}_{2}\right]}=\frac{{\left(1.692\right)}^{2}}{\left(0.374\right)\left(0.153\right)}=50.0={K}_{c}[/latex]
We have stressed this system by introducing additional H2. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess H2, reducing the amount of uncombined I2, and forming additional HI.
Effect of Change in Pressure on Equilibrium
Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Kc) or partial pressure (for KP). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.
Check out this video to see a dramatic visual demonstration of how equilibrium changes with pressure changes.
As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chtelier’s principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.
Consider what happens when we increase the pressure on a system in which NO, O2, and NO2 are at equilibrium:
[latex]2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right)[/latex]
The formation of additional amounts of NO2 decreases the total number of molecules in the system, because each time two molecules of NO2 form, a total of three molecules of NO and O2 are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of NO2 into NO and O2, which tends to restore the pressure.
Now consider this reaction:
[latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)[/latex]
Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.
Effect of Change in Temperature on Equilibrium
Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chtelier’s principle.
When hydrogen reacts with gaseous iodine, heat is evolved.
[latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{HI}\left(g\right)\Delta H=-9.4\text{kJ}\left(\text{exothermic}\right)[/latex]
Because this reaction is exothermic, we can write it with heat as a product.
[latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\rightleftharpoons2\text{HI}\left(g\right)+\text{heat}[/latex]
Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H2 and I2 and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.
When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant: At the new equilibrium the concentration of HI has increased and the concentrations of H2 and I2 decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.
Temperature affects the equilibrium between NO2 and N2O4 in this reaction
[latex]{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right)\Delta H=57.20\text{kJ}[/latex]
The positive ΔH value tells us that the reaction is endothermic and could be written
[latex]\text{heat}+{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons2{\text{NO}}_{2}\left(g\right)[/latex]
At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown NO2 molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless N2O4 increases, and the concentration of brown NO2 decreases, causing the brown color to fade.
This interactive animation allows you to apply Le Chtelier’s principle to predict the effects of changes in concentration, pressure, and temperature on reactant and product concentrations.
Catalysts Do Not Affect Equilibrium
As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.
The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year.
Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.
Fritz Haber
Figure 2. The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery.
In the early 20th century, German chemist Fritz Haber (Figure 2) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to absorb.
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen (N2) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).
Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008.
In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.” Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.
Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.
It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood.
To be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system in which N2, H2, and NH3 are at equilibrium or are coming to equilibrium.
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]
The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.
Although increasing the pressure of a mixture of N2, H2, and NH3 will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of N2 and H2, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightarrow 2{\text{NH}}_{3}\left(g\right)\Delta H=-92.2\text{kJ}[/latex]
Thus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature.
Part of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly.
In the commercial production of ammonia, conditions of about 500 °C, 150–900 atm, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 3).
Figure 3. Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant.
Key Concepts and Summary
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system’s response to these disturbances is described by Le Chtelier’s principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.
| Table 1. Effects of Disturbances of Equilibrium and K |
| Disturbance | Observed Change as Equilibrium is Restored | Direction of Shift | Effect on K |
| reactant added | added reactant is partially consumed | toward products | none |
| product added | added product is partially consumed | toward reactants | none |
| decrease in volume/increase in gas pressure | pressure decreases | toward side with fewer moles of gas | none |
| increase in volume/decrease in gas pressure | pressure increases | toward side with fewer moles of gas | none |
| temperature increase | heat is absorbed | toward products for endothermic, toward reactants for exothermic | changes |
| temperature decrease | heat is given off | toward reactants for endothermic, toward products for exothermic | changes |
Exercises
The following equation represents a reversible decomposition: [latex]{\text{CaCO}}_{3}\left(s\right)\rightleftharpoons\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)[/latex]Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains?
Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.
What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
What would happen to the color of the solution in part (b) of Figure 1 if a small amount of NaOH were added and Fe(OH)3 precipitated? Explain your answer.
The following reaction occurs when a burner on a gas stove is lit: [latex]{\text{CH}}_{4}\left(g\right)+2{\text{O}}_{2}\left(g\right)\rightleftharpoons{\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(g\right)[/latex]Is an equilibrium among CH4, O2, CO2, and H2O established under these conditions? Explain your answer.
A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown below. At high temperatures, the rate of formation of SO3 is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures. [latex]2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{SO}}_{3}\left(g\right)[/latex]
Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
Is the reaction endothermic or exothermic?
Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation: [latex]{\text{N}}_{2}\left(g\right)+2{\text{H}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{H}}_{4}\left(g\right)\Delta H=95\text{kJ}[/latex]
Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation: [latex]{\text{P}}_{4}\left(g\right)+6{\text{H}}_{2}\left(g\right)\rightleftharpoons4{\text{PH}}_{3}\left(g\right)\Delta H=110.5\text{kJ}[/latex]
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
[latex]2{\text{NH}}_{3}\left(g\right)\rightleftharpoons{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\Delta H=92\text{kJ}[/latex]
[latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)\Delta H=181\text{kJ}[/latex]
[latex]2{\text{O}}_{3}\left(g\right)\rightleftharpoons3{\text{O}}_{2}\left(g\right)\Delta H=-285\text{kJ}[/latex]
[latex]\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\rightleftharpoons{\text{CaCO}}_{3}\left(s\right)\Delta H=-176\text{kJ}[/latex]
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
[latex]2{\text{H}}_{2}\text{O}\left(g\right)\rightleftharpoons2{\text{H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\Delta H=484\text{kJ}[/latex]
[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)\Delta H=-92.2\text{kJ}[/latex]
[latex]2\text{Br}\left(g\right)\rightleftharpoons{\text{Br}}_{2}\left(g\right)\Delta H=-224\text{kJ}[/latex]
[latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(s\right)\rightleftharpoons2\text{HI}\left(g\right)\Delta H=53\text{kJ}[/latex]
Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: [latex]{\text{H}}_{2}\text{O}\left(g\right)+\text{C}\left(s\right)\rightleftharpoons{\text{H}}_{2}\left(g\right)+\text{CO}\left(g\right)\text{.}[/latex] Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
Write the expression for the equilibrium constant (Kc) for the reversible reaction [latex]2{\text{H}}_{2}\left(g\right)+\text{CO}\left(g\right)\rightleftharpoons{\text{CH}}_{3}\text{OH}\left(g\right)\Delta H=-90.2\text{kJ}[/latex]
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more H2 is added?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CO is removed?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CH3OH is added?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the temperature of the system is increased?
What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added?
Nitrogen and oxygen react at high temperatures.
Write the expression for the equilibrium constant (Kc) for the reversible reaction [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons2\text{NO}\left(g\right)\Delta H=181\text{kJ}[/latex]
What will happen to the concentrations of N2, O2, and NO at equilibrium if (i) more O2 is added?
What will happen to the concentrations of N2, O2, and NO at equilibrium if N2 is removed?
What will happen to the concentrations of N2, O2, and NO at equilibrium if NO is added?
What will happen to the concentrations of N2, O2, and NO at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
What will happen to the concentrations of N2, O2, and NO at equilibrium if the temperature of the system is increased?
What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added?
Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
Write the expression for the equilibrium constant for the reversible reaction[latex]\text{C}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\rightleftharpoons\text{CO}\left(g\right)+{\text{H}}_{2}\left(g\right)\Delta H=131.30\text{kJ}[/latex]
What will happen to the concentration of each reactant and product at equilibrium if more C is added?
What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
What will happen to the concentration of each reactant and product at equilibrium if CO is added?
What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
Write the expression for the equilibrium constant (Kc) for the reversible reaction [latex]{\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2\text{Fe}\left(s\right)+3{\text{H}}_{2}\text{O}\left(g\right)\Delta H=98.7\text{kJ}[/latex]
What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
What will happen to the concentration of each reactant and product at equilibrium if H2 is added?
What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
Ammonia is a weak base that reacts with water according to this equation: [latex]{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{NH}}_{4}{}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?
Addition of NaOH
Addition of HCl
Addition of NH4Cl
Acetic acid is a weak acid that reacts with water according to this equation: [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(aq\right)\rightleftharpoons{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]Will any of the following increase the percent of acetic acid that reacts and produces [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex] ion?
Addition of HCl
Addition of NaOH
Addition of NaCH3CO2
Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl–, Ag+, and [latex]{\text{NO}}_{3}{}^{\text{-}}[/latex], in contact with solid AgCl. [latex]{\text{Na}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+{\text{Ag}}^{+}\left(aq\right)+{\text{NO}}_{3}{}^{-}\left(aq\right)\rightleftharpoons\text{AgCl}\left(s\right)+{\text{Na}}^{+}\left(aq\right)+{\text{NO}}_{3}{}^{-}\left(aq\right)[/latex] [latex]\Delta H=-65.9\text{kJ}[/latex]
How can the pressure of water vapor be increased in the following equilibrium? [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons{\text{H}}_{2}\text{O}\left(g\right)\Delta H=41\text{kJ}[/latex]
Additional solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and sulfate ion at equilibrium with solid silver sulfate: [latex]2{\text{Ag}}^{+}\left(aq\right)+{\text{SO}}_{4}{}^{2-}\left(aq\right)\rightleftharpoons{\text{Ag}}_{2}{\text{SO}}_{4}\left(s\right)[/latex]Which of the following will occur?
Ag+ or [latex]{\text{SO}}_{4}{}^{2-}[/latex] concentrations will not change.
The added silver sulfate will dissolve.
Additional silver sulfate will form and precipitate from solution as Ag+ ions and [latex]{\text{SO}}_{4}{}^{2-}[/latex] ions combine.
The Ag+ ion concentration will increase and the [latex]{\text{SO}}_{4}{}^{2-}[/latex] ion concentration will decrease.
Show Selected Answers
1. The amount of CaCO3 must be so small that [latex]{P}_{{\text{CO}}_{2}}[/latex] is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full [latex]{P}_{{\text{CO}}_{2}}[/latex] required for equilibrium.
The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.
No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.
7. Add N2; add H2; decrease the container volume; heat the mixture.
The change of temperature and pressure have the following results:
ΔT increase = shift right, ΔP increase = shift left
ΔT increase = shift right, ΔP increase = no effect
ΔT increase = shift left, ΔP increase = shift left
ΔT increase = shift left, ΔP increase = shift right
The answers are as follows:
[latex]{K}_{c}=\frac{\left[{\text{CH}}_{3}\text{OH}\right]}{{\left[{\text{H}}_{2}\right]}^{2}\left[\text{CO}\right]}[/latex]
[H2] increases, [CO] decreases, [CH3OH] increases
[H2] increases, [CO] decreases, [CH3OH] decreases
[H2] increases, [CO] increases, [CH3OH] increases
[H2] increases, [CO] increases, [CH3OH] decreases
no changes
The answers are as follows:
[latex]{K}_{c}=\frac{\left[\text{CO}\right]\left[{\text{H}}_{2}\right]}{\left[{\text{H}}_{2}\text{O}\right]}[/latex]
[H2O] no change, [CO] no change, [H2] no change
[H2O] decreases, [CO] decreases, [H2] decreases
[H2O] increases, [CO] increases, [H2] decreases
[H2O] decreases, [CO] increases, [H2] increases. In (a), (b), (c), and (d), the mass of carbon will change, but its concentration (activity) will not change.
15. Only (b). In (a), addition of a strong base forces the equilibrium toward forming more NH3(aq). In (b), the addition of HCl causes a reaction with NH3 to form more [latex]{\text{NH}}_{4}{}^{+}[/latex] by removing OH− as it reacts with the acid to form water. In (c), [latex]{\text{NH}}_{4}{}^{+}[/latex] ion causes the equilibrium to shift to the left, forming more NH3(aq).
17. Add NaCl or some other salt that produces Cl– to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).
19. (a) The solution already holds as many ions as it can.
The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium constant for ionization ( [latex]\text{HX}\rightleftharpoons{\text{H}}^{+}+{\text{X}}^{\text{-}}[/latex] )?
The freezing-point depression is proportional to the number of particles produced in a solvent. For a weak electrolyte, ΔT = ikfm, where i is the number of ions produced from a solute. Because both isomers have identical molecular masses and are dissolved in the same amount of solvent, kf and m are constants. Any difference in the reduction of the freezing point must, therefore, reflect a difference in the degree of ionization, i, of the two forms of alanine into fragments—namely, a proton and the anion. A greater number of ions will be produced by the form with the larger equilibrium constant, which results in a lower freezing point for that species. Since α-alanine has a lower freezing point and, consequently, the larger freezing-point depression, it must have the larger number of ions in solution and has the larger value of Kc.
Glossary
Le Chtelier’s principle: when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance
position of equilibrium: concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance)
stress: change to a reaction’s conditions that may cause a shift in the equilibrium
Candela Citations
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Chemistry. Provided by: OpenStax College. Located at: License: CC BY: Attribution. License Terms: Download for free at
Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764. ↵
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13614 | https://windy.app/blog/explore-types-of-contour-lines-isolines.html | Explore different types of contour lines. Relief, atmosphere, ocean, and more
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Regular readers of Windy.app’s blog already know what isotherms and isobars are. But these are just two of more than 20 main types of isolines or contour lines that will help you predict the weather.
Each isoline has a special name, which begins with the word “iso” (from the ancient Greek word “ισος” — “equal”) and further contains one or another parameter. In particular, specific names are most common precisely in meteorology.
In this post, we’ll list all the main types of contour lines and learn why we need them. We also highlighted in yellow the most important isolines that you may see on the maps most often. But let us first deal with topographic maps showing the relief, on which isolines began to be used, and only afterward — in other areas of human activity, including weather forecasting.
Earth
Relief
An isobath is a contour line connecting points of equal depths of a body of water. It can be anything — an ocean, sea, lake, or river. As human activity has always been connected to water, isobaths are the oldest types of isolines used since 1584 in Holland.
An isobaths on a fragment of a map of the coast of Madeira Island, Portugal made by António Pedro de Azevedo in 1867 / Wikipedia
An isohypse is an isoline that connects points of equal height and thus shows all the features of the terrain — mountain peaks, slopes, and hills, passes, and valleys. Historically, this is the second oldest type of contour line in the world.
A few more isolines are related to the topography:
An isocline is an isoline that shows the equal slope of the land surface and the degree of slope smoothing.
An isopach is a rare line of the same thickness as the bedrock. Maps with such lines are used by professional petroleum geologists, volcanologists, and other similar specialists to illustrate variations in the thickness of geologic units.
An isobase is a contour line connecting points in the earth that rise or fall with equal velocity or amplitude. What are we talking about? This line is also used by geologists to draw maps of the neotectonic movements of the Earth’s crust.
Earth’s magnetic field
An isogon or isogonic line (which we already used to determine the same wind direction on the map), is also a contour line of constant magnetic declination — a variation of magnetic north from geographic north.
An isodynama is an isoline of the same magnetic field strengths of the Earth.
Atmosphere
Air temperature
An isotherm is an isoline that shows the equal temperature. You get a map of average temperatures (to be precise — usually, average annual temperatures). Such maps will be different for different seasons. The first map with isotherms on it was published by famous German geographer and naturalist Alexander von Humboldt in Paris, in 1817.
There are at least three additional types of isolines related to temperature:
An isothere is a contour line of equal mean summer temperature and an isocheim — a line of equal mean winter temperature.
An isogeotherm is the same line of the temperature beneath the Earth’s surface.
An isodrosotherm is an isoline of the dew point temperature values — the temperature to which the air needs to be cooled so that water condensate is released from it (dew appears).
Precipitation
An isohyet is a contour line of the same precipitation: rain, snow, and others.
The isohyet map of the equal precipitation of the Claudette tropical storm in the area of Fort Worth, Texas, USA, in July 24–27, 1979 / NOAA
Since precipitation is one of the three main weather parameters along with air temperature and wind, this type of isoline has additional subtypes, too:
An isochalaz is a line of hail storms — lumps of ice, that can break through the roof of a car. Like other forms of precipitation, hail starts from a cloud — specifically, a cumulonimbus cloud.
An isoplat is a line representing points of equal acidity, as in acid precipitation. This type of precipitation is commonly called “acid rain” and refers to all other types of precipitation — snow, hail, fog, rain with snow (sleet), and others. The word “acidic” indicated that such precipitation is contaminated by acidic oxides of different production, cars, and so on — usually sulfur oxides and nitrogen oxides. So there is nothing good about it.
Freezing level
An isotherm at 0 °C is a contour line of the equal freezing level. The term was also invented by Alexander von Humboldt.
An isopectic (ice freezing) and an isotac (ice melting) are two other freezing level related isolines. Learn more about what is safe ice for outdoor activities in the Windy.app blog.
Wind
An isogon is an isoline connecting points with the same value of any angle and thus showing the orientation of some physical quantities. In other words, the same direction of anything. An isogon in meteorology is wind direction, and an isotach is an isoline of identical wind speeds.
There are also two other isolines related to air and wind characteristics:
An isopycnal is a line of air density.
An isohume is a constant relative humidity of the air — the amount of concentration of water vapor present in the air. The word “relative” means a percentage indicating the present state of absolute humidity relative to the maximum humidity given the same temperature.
Clouds
An isoneph is a contour line showing the cloud coverage of the sky — a very important weather parameter for all those involved in various aerial sports such as paragliding. It is also related to concepts such as cloud base, cloud top, and similar.
An isobront is an isoline of equal thunderstorm activity in some places. Let's recall what is thunderstorm from another lesson from the Windy.app Meteorological Textbook (WMT) for better weather forecasting: "Warm air from the warmed ground always tends upwards. It carries away moisture with it, and if the air was wet enough and the upward flows were strong enough, there are heap clouds — a factory of rain and thunderstorms".
Atmospheric pressure
An isobar is one of the main isolines in meteorology that you can see most often on weather maps. It is a line denoting an area with the same atmospheric pressure. Isobars that are close to one another also mean high wind speed. So knowing how to read isobars you can also make your own wind forecast.
Ocean
An isopycnal is an isoline that is also used to indicate the same density of water in addition to air density.
Two more related isolines here are:
An isobathytherm is a line showing depths of water with equal temperature. You've noticed yourself when you've been swimming or scuba diving, that the deeper you go under the water, the colder it gets. And vice versa: the surface of the sea warms up the fastest from the sun.
An isohaline indicating places with the same water salinity.
Light
An isohel is a line of equal or constant solar irradiance or the power per unit area received from the Sun in the form of electromagnetic radiation. In simple words, it represents the points receiving equal amounts of sunshine.
Other isolines about light:
An isophote is a contour line of illuminance or the ratio of the luminous flux falling on a small surface area to its total area.
An isochasm is a very interesting isoline of Aurora’s equal appearance. You may know this extremely bright atmospheric phenomenon better by the names Aurora Borealis, Polar Lights, or Northern Lights. This is a natural green, purple and red light display in the sky, most often seen in high-latitude regions behind the Polar Circle.
An isodose is a line representing points of equal intensity of ionizing radiation per unit mass. It is used for mapping in order to protect against excessive radiation that is harmful to humans and other creatures, or conversely, to use its excess for good.
Human
Ecology
An isophene is a line describing the same values for a whole set of complex phenomena, in this case — biological events.
Pollution
An isobel is a contour line of environmental noise or equal sound pressure level. In addition, there could be maps with isolines for other types of pollution — of the air, soil, and others. For example, contour lines are used for smart planting to reduce the level of soil erosion.
Traveling
There are many other types of isolines that are used not only in meteorology and biology but also in other kinds of sciences, such as economics, to describe equal amounts of production and so on.
An isochrone, or time and distance isoline is one of the most interesting types of contour line related to humans. It is a line representing points of equal time-distance from a point, such as the transportation time from a particular point. One of the best known of these maps — we purposely left it for the final part of the article as the most illustrative — shows how long it used to take to get from London to different parts of the world in the 19th century:
The map of travel time from London to the different parts of the world in the 19th century made by Francis Galton and published in 1881 / Wikipedia
For example, it took at least 5 days to reach Portugal and Southern Europe by land and water, 10 days to reach Siberia in Russia, and 40 days to reach Australia and South America.
It is interestingly that the same maximum number of days was needed to visit regions in Central Africa and South East Asia, although they are closer to Europe than Australia and America. This is because the map takes into account not only and not so much the distance, but the transport accessibility of certain regions of the Earth.
In other words, looking at the map today, we can imagine how the speed of travel has changed. A lot.
Learn the definition and the history of the invention of isolines, as well as how to read contour lines on weather and other maps in the part 1 and part 3 of this big acticle.
Publish your post in the Windy.app blog and become an expert in our community.
Text: Ivan Kuznetsov, an outdoor journalist, editor and writer from the Dolomites, Italy, and Karelia, Finland, with 10 years of professional experience. His favorite sports are hiking, cycling and sauna. Read his other articles
Cover photo: Cottonbro / Pexels
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13615 | https://resource.learnmode.net/upload/file_1/16f45fb6296310a385c077a070a4a868ca0724af | 廣義角
陳清海 老師 ok312 廣義角與極座標
室高
1
oookkk333111222 廣廣廣義義義角角角與與與極極極座座座標標標
主題一、 廣義角
廣義角﹕平面上﹐將射線 OA 繞著 O 點旋轉到射線 OB 所成的角﹐
其中射線 OA 稱為角的始邊﹐射線 OB 稱為終邊﹐並規定﹕
逆時針方向旋轉出的角為正向角﹐順時針方向旋轉出的角為負向角﹒
當兩個角的始邊和終邊完全相同的時候﹐我們稱這樣的角為同界角﹒
當有向角 與 是同界角 會有一整數 n ﹐
使得 360 n
（n 是整數） ﹒
標準位置角﹕將廣義角放在坐標平面上﹐使角的頂點 與原點重合﹐
角的始邊在 x 軸正向上﹐這樣的角稱為標準位置角﹒當角的終邊落在
第一、二、三或四象限時﹐我們分別稱這個角為第一、二、三或四
象限角﹒當角的終邊落在 x 軸或 y 軸上時﹐我們稱它為 象限角 ﹒150
與210 互為同界角﹐也都是第二象限角﹒ ok312 廣義角與極座標
室高
2
【例題 1】
下列何者是 123 的同界角﹖
(1) 483 ﹒ (2) 123 ﹒ (3) 57 ﹒ (4) 237 ﹒
An s：(1)(4)
【詳解】
因為 483 123 360
﹐237 123 360
﹐
所以答案為 (1)(4) ﹒
【類題 1】
下列何者是 255 的同界角﹖
(1) 615 ﹒ (2) 975 ﹒ (3) 75 ﹒ (4) 105 ﹒ (5) 465 ﹒
An s：(1)(2)(4)(5)
【詳解】
因為 615 255 360
﹐975 255 360 2
﹐105 255 360
﹐465 255 360 2
﹐
答案為 (1)(2)(4)(5) ﹒ok312 廣義角與極座標
室高
3
主題二、廣義角的三角函數值
設 是一個標準位置角﹐在 的終邊上任意取一點 ,P x y （P 不是原點） ﹐
令2 2
OP r x y ﹒規定 角的三角函數值為﹕ sin yr
﹐cos xr
﹐tan yx
﹒
各象限中三角函數值的 正負符號整理如下 ﹕ ,x y s i n
c o s
t a n
第 一 象 限 ,
第 二 象 限 ,
第 三 象 限 ,
第 四 象 限 ,
值域：
1≦sinθ ≦1，1≦co sθ ≦1，tanθ R. ok312 廣義角與極座標
室高
4
【例題 2】
已知 3, 4P 為標準位置角 終邊上的一點﹐
求sin
﹐cos
和tan
的值﹒
An s：4sin 5
﹐3cos 5
﹐4tan 3
【詳解】
因為 3x ﹐4y ﹐
所以 22
3 4 5r ﹐得 4sin 5
yr
﹐3cos 5
xr
﹐4tan 3
yx
﹒
【類題 2】
設 5, P y 為 之終邊的一點﹐又 tan 2
﹐試求 sin
﹐cos
的值﹒
An s：2 5sin 5
﹐5cos 5
【詳解】 tan 2 10 5
y y
﹐
故 5, 10 P ﹐5 5 OP ﹐10 2 2 5sin 55 5 5
﹐5 1 5cos 55 5 5
﹒
【例題 3】
求150 和210 的sin ﹐cos 和tan 值﹒ ok312 廣義角與極座標
室高
5
An s：見詳解
【詳解】
在 150 角的終邊上取一點 P﹐
設 P 在 x 軸的垂足為 Q﹐
則30 POQ ﹒若設 2OP ﹐
則3OQ ﹐1PQ ﹐
故點 P 的坐標為 3,1 ﹒得 1sin150 2
﹐3cos150 2
﹐1 3tan150 33
﹒
因為 210 為 150 的同界角（ 210 360 150 ）﹐
與 150 有相同的終邊﹐所以 1sin 210 2
﹐ 3cos 210 2
﹐ 1 3tan 210 33
﹒
【類題 3】
求下列各值﹕
(1) sin 225 ﹒ (2) cos 60 ﹒
Ans ：(1) 22
，(2) 12
【詳解】
(1) 作圖﹐ 45 POQ ﹒若設 2OP ﹐則 1OQ PQ
﹐故點 P 的坐標為 1, 1 ﹒ok312 廣義角與極座標
室高
6
得1 2sin 225 22
﹒
(2) 作圖﹐ 60 POQ ﹒若設 2OP ﹐則 1OQ
﹐3PQ ﹐故點 P 的坐標為 1, 3 ﹒
得 1cos 60 2
﹒
【例題 4】
求sin 90 ﹐cos90 ﹐tan 90 的值﹒
An s：見詳解
【詳解】
如圖﹐當 ＝90 ﹐可取 P 點坐標為 (0 ，1) ﹐
如此 1OP ﹐即 x＝0﹐y＝1﹐r＝1﹒得 1sin90 11
yr
﹐0cos90 01
xr
﹐
因為 tan 90 yx
的分母 x 是 0﹐我們說 tan90 沒有定義﹒
【類題 4】
求下列象限角的 sin
﹐cos
﹐tan
值﹕ sin
cos
tan
0
90 1 0
180
270
表示沒有定義 ok312 廣義角與極座標
室高
7
【詳解】
sin cos tan
0 0 1 0
90 1 0
180 0 1 0
270 1 0
【例題 5】
根據下列各條件﹐分別指出各 角是第幾象限角﹖
(1) sin 0
﹐cos 0
﹒
(2) tan 0
﹐sin 0
﹒
Ans ：(1) 二，(2) 三
【詳解】
(1) 由 sin ＞0 知為第一或第二象限角﹐又因為 cos ＜0﹐
所以 為第二象限角﹒
(2) 由 tan 知為第一或第三象限角﹐又因為 sin ＜0﹐
所以 為第三象 限角﹒
【類題 5-1】
設sin 0
﹐cos 0
﹐試問 是第幾象限角﹖
An s：四
【詳解】
由 sin ＜0 知為第三或第四象限角﹐
又因為 cos ＞0﹐所以 為第四象限角﹒
【類題 5-2】
已知 cos tan 0
﹐試判斷 為第幾象限角﹖
An s：第三象限角或第四象限角
【詳解】
因為 cos tan ＜0﹐ok312 廣義角與極座標
室高
8
所以 cos 0, tan 0.
或cos 0, tan 0.
cos 與 tan 在第三象限一負一正﹐在第四象限一正一負﹒
故為第三象限角或第四象限角﹒
【例題 6】
已知 5sin 13
且 是第二象限角﹐求 cos
和tan
的值﹒
Ans ：cos ＝12 13
，tan ＝512
【詳解】
由2 2
sin cos 1
可得 2
cos 1 sin
﹐
因為
是第二象限角﹐ cos ＜0﹐所以 22 5 12 cos 1 sin 1 13 13
﹐
再由商數關係式得 5sin 513 tan 12 cos 12 13
﹒
【類題 6】
已知 1cos 4
﹐且 是第三象限角﹐求 sin
與tan
之值﹒
An s：sin ＝15 4
，tan ＝15
【詳解】
由2 2
sin cos 1
可得 2
sin 1 cos
﹐
因為
是第三象限角﹐ sin ＜0﹐所 以22 1 15 sin 1 cos 1 4 4
﹐
再由商數關係式得 6
4
2
-2
-4
-10 -5
B: (-12.00 , 5.00 )
C
B
A1
-1
-2
-3
-4
-2
C
A
Bok312 廣義角與極座標
室高
915 sin 4tan 15 1cos 4
﹒ok312 廣義角與極座標
室高
10
主題三、換算公式
180
關係式﹕ sin 180 sin
﹐ cos 180 cos
﹐ tan 180 tan
﹒
180
關係式﹕ sin 180 sin
﹐ cos 180 cos
﹐ tan 180 tan
﹒
關係式 ﹕ sin sin
﹐ cos cos
﹐ tan tan
﹒
90
與 90
關係式 ﹕ sin 90 cos
﹐ cos 90 sin
﹒ sin 90 cos
﹐ cos 90 sin
﹒
270
與 270
關係式 ﹕ sin 270 cos
﹐ cos 270 sin
﹒ sin 270 cos
﹐ cos 270 sin
﹒ok312 廣義角與極座標
室高
11
【例題 7】
求下列各三角函數值﹕
(1 )sin120 ﹒ (2) cos135 ﹒ (3) tan 210 ﹒
Ans ：(1) 32 ，(2) 22
，(3) 33
【詳解】
(1) 因為 120 180 60 ﹐所以 3sin120 sin 180 60 sin60 2
﹒
(2) 因為 135 180 45 ﹐所以 2cos135 cos 180 45 cos45 2
﹒
(3) 3tan 210 tan 210 tan 180 30 tan30 3
﹒
【類題 7-1】
求下列各三角函數值﹕
(1) sin150 ﹒ (2) cos330 ﹒ (3) tan 240 ﹒
An s：(1) 12 ，(2) 32 ，(3) 3
【詳解】
(1) 1sin150 sin 180 30 sin30 2
﹒
(2) 3cos330 cos 360 30 cos30 2
﹒
(3) tan240 tan 180 60 tan60 3 ﹒
【類題 7-2】
求cos150 sin 240 cos315 sin 225 的值﹒
An s：14ok312 廣義角與極座標
室高
12
【詳解】 cos150 sin240 cos315 sin225 cos 180 30 sin 180 60 cos 45 sin 180 45 cos30 sin60 cos45 sin45 3 3 2 2 12 2 2 2 4
﹒
【例題 8】
化簡﹕
sin sin 90 sin 180 sin 180 cos 360 cos 270
﹒
An s：1
【詳解】
原式 sin cos sin 1 1 1 1sin cos sin
﹒
【類題 8】
化簡﹕ sin 90 cos 180 cos 90 sin 180
﹒
An s：1
【詳解】
原式 2 2
cos cos sin sin cos sin 1
﹒
【例題 9】
設 cos 110 k ﹐試以 k 表示 tan 250 的值﹒
An s：2
1 kk
【詳解】
因為 cos 110 cos110 cos 180 70 cos70 ﹐
所以 cos70 k ﹒
因此﹐ 2 2
sin70 1 1k k ﹒
又因為 tan250 tan 180 70 tan70 ﹐所以 ok312 廣義角與極座標
室高
13 2 2
sin70 1 1tan 250 tan70 cos70
k kk k
﹒
【類題 9】
設sin 793 k ﹐試以 k 表示 cos 107 ﹒
An s：2
1 k
【詳解】 sin793 sin 360 2 73 sin73
﹐
故sin73 k ﹒
因此﹐ 2
cos73 1 k cos 107 cos107 2
cos 180 73 cos73 1 k
﹒ok312 廣義角與極座標
室高
14
主題四、直角坐標與極坐標的變換
對於平面上異於 O 的任一點 P ﹐令 r OP ﹔
為以 x 軸為始邊﹐射線 OP 為終邊的廣義角﹐
我們可以用符號 ,r
來表示點 P 的位置﹐
這種坐標稱為極坐標﹐記作 ,P r
﹒
若平面上一點 P 的直角坐標為 ,x y ﹐極坐標為 ,r
﹐則有 cos x r
﹐sin y r
﹐2 2
r x y ﹒
【例題 10 】
寫出下圖中點 A ﹐B ﹐C ﹐D 的極坐標﹒
An s：A 3,45 ， 2, 45 B ﹐ 2,90 C ﹐ 3,225 D
【詳解】
A 的極坐標為 A 3,45 （或 3,405 ﹐ 3, 315 ﹐…等） ﹐ 2, 45 B
﹐ 2,90 C ﹐ 3,225 D ﹒
【類題 10 】
寫出下圖中點 A ﹐B ﹐C ﹐D 的極坐標﹒
An s：A 2,0 ﹐ 3, 45 B ﹐ 1,135 C ﹐ 3,180 D ok312 廣義角與極座標
室高
15
【詳解】
極坐標為 A 2,0 ﹐ 3, 45 B ﹐ 1,135 C ﹐ 3,180 D ﹒
【例題 11 】
(1) 已知點 P 的極坐標為 4, 60 ﹐求其直角坐標﹒
(2) 已知點 P 的直角坐標為 2 3, 2 ﹐求其極坐標﹒
Ans ：(1) 2, 2 3 ，(2) 4,150
【詳解】
(1) 因為 4r ﹐60
﹐所以直角坐標 , 4cos 60 ,4sin 60 x y 1 34 ,4 2, 2 32 2
﹒
(2) 因為
22
2 3 2 16 4r ﹐
且 P 在第二象限﹐ 2 3 3cos cos150 4 2
﹐
所以極坐標為 4,150 ﹒
【類題 11 】
(1) 已知點 P 的極坐標為 4 2,135 ﹐求其直角坐標﹒
(2) 已知點 P 的直角 坐標為 3, 3 3 ﹐求其極坐標﹒
Ans ：(1) 4,4 ，(2) 6,240
【詳解】
(1) 因為 4 2r ﹐135
﹐所以直角坐標 ok312 廣義角與極座標
室高
16 , 4 2 cos135 ,4 2 sin135 x y
1 14 2 ,4 2 4,4 2 2
﹒
(2) 因為 22
3 3 3 6r ﹐
且 P 在第三象限﹐ 3 1cos cos240 6 2
﹐
所以極坐標為 6,240 ﹒ok312 廣義角與極座標
室高
17
oookkk333111222eeexxx
坐標平面上﹐ O 為原點﹐ 為第二象限角﹐ ,1 P x 是 角終邊上的一點﹐
已知 2OP ﹐求 cos
之值﹒
An s：32
【詳解】 2 1 2OP x
x2＋1＝4
x＝3 ，
為第二象限角﹐
故取 x＝32
.
設 為第三象 限角﹐則點 sin cos , tan sin
在第幾象限﹖
Ans ：第三象限
【詳解】
為第三象限角﹐
sin ＜0，cos ＜0，tan ＞0，
sin ＋cos ＜0，tan sin ＜0，
sin cos , tan sin
在第三象限 .
求 tan 300 cos135 sin 225 cos 210
的值﹒
An s：1
【詳解】 tan 300 cos135 sin 225 cos 210
2
1
-1
-2
-2 2O
Pok312 廣義角與極座標
室高
18
＝2322 32 2
＝1－2
＝1.
設 為第四象限角﹐且 3cos 5
﹐求下列各式的值﹕
(1) cos 90
﹒
(2) tan 540
﹒
An s：(1) 45
，(2) 43
【詳解】
(1) cos 90
﹒
＝cos(90 －)
＝sin
＝45
.
(2) tan 540
＝tan(180 ＋)
＝tan
＝43
﹒
若點 cos , tan
在第三象限﹐則 在第幾象限﹖
An s：第二象限
【詳解】
點 cos , tan
在第三象限﹐
cos ＜0，tan ＜0
(在二、三象限 )且(在二、四象限 )
在第二象限 ．1
-1
-2
-3
-4
24
54
3
ok312 廣義角與極座標
室高
19
設1 tan 51 tan
﹐且 180 270
﹐求 cos
﹒
An s：2 13 13
【詳解】 1 tan 51 tan
1＋tan ＝5＋5tan
tan ＝32 ．
因180 270
故 cos ＝213
．
如右圖 BAC
﹐90 ABD ACD ﹐AB a ﹐BD b ﹒
下列選項何者可以表示 CD ﹖
(1 )sin cos a b
﹒ (2) sin cos a b
﹒ (3) cos sin a b
﹒
(4) cos sin a b
﹒ (5) sin tan a b
﹒
An s：(2)
【詳解】
如右圖， CD
＝EF
＝EB BF
＝asin(180 －)＋bcos(180 －)
＝asin －bcos ．
設180 270
﹐求
2222
sin 1 sin cos cos 1
之值﹒
An s：22
1
-1
-2
-3
-4
-2 2
A:(-2.00 ,-3.00 )
AD
180 -
180 -
b
a
C
F
E
B
Aok312 廣義角與極座標
室高
20
【詳解】 180 270
，
2222
sin 1 sin cos cos 1
＝sin ＋(1 ＋sin )－(cos )＋(1 －cos )
＝2.
若270 360
﹐且 1sin cos 5
﹐求 cos
﹒
An s：45
【詳解】 1sin cos 5
5sin ＝1－5cos ，兩邊平方
25(1 －cos 2)＝(1 －5cos )2＝1－10cos ＋25cos 2
50cos 2－10cos －24 ＝0
25cos 2－5cos －12 ＝0
(5cos ＋3)(5co s－4) ＝0
cos ＝45 或 cos ＝35
．
因270 360
，故取 cos ＝45 ．
求cos1 cos 2 cos3 cos180 的值﹒
An s：1
【詳解】
利用 cos(180 －)＝cos ，
cos179 ＝cos1 ，cos1 78 ＝cos2 ，…… ，cos91 ＝cos89 ，cos1 cos 2 cos3 cos180
＝cos1 ＋cos2 ＋…… ＋cos90 ＋…… ＋cos178 ＋cos179 ＋cos180
＝cos90 ＋cos180
＝0－1＝1．ok312 廣義角與極座標
室高
21
若 之終邊落在第 三象限﹐則 3
之終邊可能落在 哪裡﹖
(1) 第一象限 (2) 第二象限 (3) 第三象限 (4) 第四象限 (5) x 軸上﹒
An s：(1)(3)(4)
【詳解】
之終邊落在第三象限﹐
180 ＋360 n＜＜270 ＋360 n
60 ＋120 n＜3
＜90 ＋120 n，
n＝3k ＋1 180 ＜3
＜210 ，
n＝3k ＋2 300 ＜3
＜330 ，
n＝3k ＋3 60 ＜3
＜90 ．
故3
的終邊可能落在第三、四、一象限．
設直角三角形 ABC 之三邊長為 3AB ﹐5BC ﹐4CA ﹐以斜邊 BC
為一邊向外作出正方形 BCDE ﹐如右圖所示﹒
令ACD
﹐試求 sin cos
之值﹒
An s：15
【詳解】
sin ＝sin(90 ＋)＝cos ＝45 ，
cos ＝cos(90 ＋)＝sin ＝35
，sin cos
＝4 3 15 5 5
．1
-1 300
330 210
180
90 60
ok312 廣義角與極座標
室高
22
已知 O 為原點﹐ A ﹐B 兩點的極坐標為 3,12 ﹐ 5,132 ﹐求
(1) AB 的長﹒
(2) △OAB 的面積﹒
An s：(1) 7，(2) 15 34
【詳解】
將 A，B 各旋 轉 12 ，則
A[3 ，0]＝(3 ，0) ，
B(5 ，120 )＝(5cos120 ，5sin120 )＝(52
，5 32 )。
(1) AB ＝2 25 5 3(3 ) (0 )2 2
＝196 49 4 ＝7，
(2) △ OAB 的面積 (BC 為高 )
＝12 35 32 ＝15 34 ．
【另解】
A＝[3 ，12 ]＝(3 cos12 ，3sin12 )，
B＝[5 ，132 ]＝(5 cos132 ，5sin132 )，
(1) AB 2
＝(3cos12 －5cos132 )2＋(3sin12 －5sin132 )2
＝9cos 212 －30cos12 cos132 ＋25cos 2132
＋9sin 212 －30sin12 sin132 ＋25sin 2132
＝9＋25 －30(cos12 cos132 ＋sin132 sin132 )
＝34 －30 cos120 (和角公式，第四節才學到 )
＝34 －30 (12
)＝49 ，
故AB ＝7．
(2) △OAB 的面積
＝12 OA OB sin120 (面積 公式，第 三節才學到 )4
3
2
1
1
22
120
5
3
B:(5.00 ,132.00 )
A:(3.00 ,12.00 )
B
A
O4
3
2
1
-1
-2 2
60
5
3120
C
B
AOok312 廣義角與極座標
室高
23
＝12 3532
＝15 34 ． |
13616 | https://www.kylesconverter.com/mass-flow/pounds-per-minute-to-slugs-per-second | Pounds Per Minute to Slugs Per Second | Kyle's Converter
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| Unit Descriptions |
| 1 Pound per Minute:Mass flow of pounds across a threshold per unit time of a minute. Using international pound: 1 Pound per minute = 0.45359237/60 kilograms per second (SI base unit). 1 lbm/min ≈ 0.007 559 872 833 kg/s. | 1 Slug per Second:Mass flow of slugs across a threshold per unit time of a second. 1 Slug per second = 14.593903 kilograms per second (SI base unit). 1 slug/s = 14.593903 kg/s. |
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| Conversions Table |
| 1 Pounds Per Minute to Slugs Per Second = 0.0005 | 70 Pounds Per Minute to Slugs Per Second = 0.0363 |
| 2 Pounds Per Minute to Slugs Per Second = 0.001 | 80 Pounds Per Minute to Slugs Per Second = 0.0414 |
| 3 Pounds Per Minute to Slugs Per Second = 0.0016 | 90 Pounds Per Minute to Slugs Per Second = 0.0466 |
| 4 Pounds Per Minute to Slugs Per Second = 0.0021 | 100 Pounds Per Minute to Slugs Per Second = 0.0518 |
| 5 Pounds Per Minute to Slugs Per Second = 0.0026 | 200 Pounds Per Minute to Slugs Per Second = 0.1036 |
| 6 Pounds Per Minute to Slugs Per Second = 0.0031 | 300 Pounds Per Minute to Slugs Per Second = 0.1554 |
| 7 Pounds Per Minute to Slugs Per Second = 0.0036 | 400 Pounds Per Minute to Slugs Per Second = 0.2072 |
| 8 Pounds Per Minute to Slugs Per Second = 0.0041 | 500 Pounds Per Minute to Slugs Per Second = 0.259 |
| 9 Pounds Per Minute to Slugs Per Second = 0.0047 | 600 Pounds Per Minute to Slugs Per Second = 0.3108 |
| 10 Pounds Per Minute to Slugs Per Second = 0.0052 | 800 Pounds Per Minute to Slugs Per Second = 0.4144 |
| 20 Pounds Per Minute to Slugs Per Second = 0.0104 | 900 Pounds Per Minute to Slugs Per Second = 0.4662 |
| 30 Pounds Per Minute to Slugs Per Second = 0.0155 | 1,000 Pounds Per Minute to Slugs Per Second = 0.518 |
| 40 Pounds Per Minute to Slugs Per Second = 0.0207 | 10,000 Pounds Per Minute to Slugs Per Second = 5.1802 |
| 50 Pounds Per Minute to Slugs Per Second = 0.0259 | 100,000 Pounds Per Minute to Slugs Per Second = 51.8016 |
| 60 Pounds Per Minute to Slugs Per Second = 0.0311 | 1,000,000 Pounds Per Minute to Slugs Per Second = 518.0158 |
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13617 | https://synapse.patsnap.com/article/what-is-the-mechanism-of-methyldopa | What is the mechanism of Methyldopa?
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What is the mechanism of Methyldopa?
18 July 2024
Methyldopa is a medication primarily used to manage high blood pressure (hypertension). Understanding its mechanism of action involves delving into its pharmacodynamics, the biochemical and physiological effects it has on the body, and its pharmacokinetics, how it is absorbed, distributed, metabolized, and excreted.
Methyldopa is classified as an alpha-2 adrenergic agonist. It works centrally, meaning it acts on the central nervous system (CNS), specifically the brain. Once administered, methyldopa is metabolized in the liver to its active form, alpha-methylnorepinephrine. This metabolite then accumulates in the adrenergic neurons within the CNS.
The primary mechanism of methyldopa involves its action on alpha-2 adrenergic receptors in the brain. By stimulating these receptors, methyldopa inhibits the release of norepinephrine, a neurotransmitter responsible for signaling the sympathetic nervous system to increase heart rate and constrict blood vessels, thereby raising blood pressure. By reducing the release of norepinephrine, methyldopa decreases sympathetic outflow, leading to vasodilation (widening of the blood vessels) and a subsequent reduction in blood pressure.
Another aspect of methyldopa's mechanism is its effect on baroreceptors, the sensors located in the blood vessels that detect changes in blood pressure. Methyldopa helps reset these baroreceptors, making them less sensitive to high blood pressure and further contributing to its antihypertensive effects.
The pharmacokinetics of methyldopa reveal that it is well-absorbed from the gastrointestinal tract, although its bioavailability is only about 25% due to extensive first-pass metabolism in the liver. Once absorbed, methyldopa is widely distributed throughout the body and crosses the blood-brain barrier where it exerts its central effects. It has a half-life of approximately 1.5 to 2 hours, but its antihypertensive effect can last much longer due to the active metabolite's prolonged action in the CNS.
Methyldopa is excreted primarily via the kidneys, with both the parent drug and its metabolites appearing in the urine. Because of its renal excretion, dose adjustments may be necessary in patients with impaired kidney function to avoid toxicity.
Clinically, methyldopa is often chosen for the management of hypertension in specific populations, such as pregnant women, due to its safety profile. It is one of the few antihypertensive medications considered safe during pregnancy, as it does not negatively impact the developing fetus.
In summary, the mechanism of methyldopa involves its central action on alpha-2 adrenergic receptors, leading to reduced norepinephrine release and decreased sympathetic outflow, resulting in lower blood pressure. Its pharmacokinetic properties, such as absorption, metabolism, and renal excretion, also play significant roles in determining its clinical efficacy and safety profile. Understanding these mechanisms provides critical insights into methyldopa's role in the management of hypertension, particularly in unique patient populations.
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13618 | https://zhidao.baidu.com/question/120501018.html | 两车相向而行相对速度怎么求_百度知道
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#热议#网上掀起『练心眼子』风潮，真的能提高情商吗？
fengke999
2009-10-08 · TA获得超过1万个赞
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构造等比数列,求数列通项。
最新推荐文章于 2024-08-26 16:05:38 发布
原创 于 2019-04-04 14:57:25 发布·3.7k 阅读
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CC 4.0 BY-SA版权
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
数论 专栏收录该内容
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本文介绍了一种使用C语言实现的算法，该算法可以高效地求解形如a(n)=pa(n-1)+h(n)的线性递推公式，其中h(n)为多项式。通过构造等比数列的方法，可以在O(n^2)时间内完成预处理，进而实现在O(1)时间内查询任意项的值。
突然想起了高中的知识，用c语言实现了一下。
对于形如 a ( n ) = p ∗ a ( n − 1 ) + h ( n ) ( h ( n ) 是 多 项 式 , p ! = 1 ) a(n)=pa(n-1)+h(n) (h(n)是多项式,p!=1) a(n)=p∗a(n−1)+h(n)(h(n)是 多 项 式,p!=1)(p==1的话你可以用累加法求)
我们可以构造一个公比为q的等比数列, 达到求通项的目的。详细步骤见 链接：别人的知乎回答
我们只需 O(n^2)的时间,就可以实现O(1)时间的查询,这个时间比矩阵快速幂更快，而且你不用去构造关系矩阵，省去了思维时间。
```c
//我这个模板只可以求q为正整数的情况
include
include
include
using namespace std;
//解a(n)=pa(n-1)+h(n) (h(n)是多项式)
//a(n)+c+p1n+p2n^2.....=p(a(n-1)+c+p1(n-1)+p2(n-2)^2....)
//构造的p1,p2,p3..使,两式等价
//这是一个完整的上三角不用高斯消元
const int N=50;
int x[N];
int main(){
int p,n;
scanf("%d%d",&p,&n);
for(int i=0;i<=n;i++)scanf("%d",&x[i]); //次数为i的多次项的系数
for(int i=n;i>=1;i--){//从最高项开始拆分
int temp=1;
x[i]/=(p-1);
for(int j=1;j<=i;j++){//j表示-1的个数
temp=temp(i-j+1)/j;//二项式系数
//printf("%d\n",temp);
if(j&1)x[i-j]+=x[i]ptemp;
else x[i-j]-=x[i]ptemp;
}
}
int a1=1,q,ans,m;
/a1首项初始值可以等于任意数这里默认为1/
for(int i=0;i<=n;i++)a1+=x[i];//等比数列首项
//用公式做出前几项,大家可以去验证一下
for(int i=2;i<=10;i++){
int ai=a1(int)pow(p,i-1);
for(int j=0;j<=n;j++)ai-=x[j](int)pow(i,j);
printf("%d ",ai);
}
//for(int i=0;i<=n;i++)printf("%d ",x[i]);puts("");
}
```
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数列 求 通 项 公式 最新发布
04-19
好的，我现在需要帮助用户寻找 求 解 数列 通 项 公式的方法。用户之前已经询问过相关的问题，现在需要更详细的解答。根据用户提供的引用资料，我需要整合不同的方法，并确保回答结构清晰，符合用户的要 求。 首先，我应该回顾已有的引用内容。引用提到了不动点法，适用于递推 数列，比如线性递推的情况。引用是关于斐波那契 数列 的 通 项 求 解，使用了特征方程法。引用提到了使用编程方法（如模板元编程）来在编译时计算 数列 项，但用户可能需要理论方法。引用则涉及 求 和公式，可能和 通 项 无关，但需要确认。 用户的问题是关于 数列 通 项 公式的计算方法或 通 用公式，所以需要整理常见的 求 解方法。常见的 通 项 求 解方法包括：观察法、累加法、累乘法、特征方程法（适用于线性递推）、不动点法（分式递推）、生成函数法、数学归纳法等。需要根据不同的递推形式选择方法。 接下来，我需要参考用户提供的引用内容，尤其是引用的不动点法和引用的特征方程法。例如，斐波那契 数列 是二阶线性递推，可以用特征方程解，而分式递推可能用不动点法。需要将这些方法详细说明，并给出例子，可能的话使用LaTeX公式展示。 同时，用户要 求 生成相关问题，我需要考虑用户可能的后续问题，比如具体方法的步骤、适用条件，或者与其他方法的比较。例如，用户可能会问特征方程法适用于几阶递推，或者不动点法的具体步骤。 需要确保回答中使用正确的LaTeX格式，行内公式用$...$，独立公式用... ... ，并检查语法正确性。同时，回答结构要清晰，分点列出方法，每个方法下给出步骤和示例，最后生成相关问题。 现在需要组织内容： 1. 观察归纳法：通 过前几 项 猜测 通 项，并用数学归纳法证明。例如等差 数列、等比数列。 2. 累加累乘法：适用于a_{n+1} - a_n = f(n)或a_{n+1}/a_n = f(n)的形式。 3. 特征方程法：针对线性递推 数列，如a_{n+2} = pa_{n+1} + qa_n，构建特征方程解根。 4. 不动点法：适用于分式递推，如a_{n+1} = (ka_n + c)/(ma_n + d)，找到不动点后换元。 5. 生成函数法：构造 生成函数，转化为微分方程或代数方程 求 解。 6. 特殊 数列 参考：如斐波那契 数列 用特征方程，调和 数列 已知表达式等。 每个方法需要举例，比如等差 数列 用观察法，斐波那契用特征方程，分式递推用不动点法，并引用对应的引用资料。例如，引用中的不动点法步骤，引用中的斐波那契例子。 最后生成的问题要围绕这些方法的应用、比较、步骤或具体案例，如如何选择合适的方法，特征方程法的步骤，不动点法的例子等。 需要确保回答正确无误，方法描述准确，例子恰当，并且引用正确。例如，当提到不动点法时，引用中的步骤；提到斐波那契 数列 时，引用和。 最后检查LaTeX格式是否正确，是否有遗漏的引用，结构是否清晰，问题是否相关。### 数列 通 项 公式 求 解方法详解 数列 通 项 公式的 求 解方法需根据递推关系的特点选择，以下是6种常用方法及其适用场景： --- #### 一、观察归纳法 适用条件：简单 数列 或存在明显规律的情况 步骤： 1. 列出前若干 项 2. 观察规律猜测 通 项 3. 数学归纳法验证 示例： 数列 $1,3,5,7,\cdots$ 猜测 通 项 $a_n=2n-1$，数学归纳法验证成立[^4] --- #### 二、累加累乘法 适用条件： - 累加法：形如 $a_{n+1}-a_n=f(n)$ - 累乘法：形如 $\frac{a_{n+1}}{a_n}=f(n)$ 步骤： 1. 累加形式：$a_n=a_1+\sum_{k=1}^{n-1}f(k)$ 2. 累乘形式：$a_n=a_1\prod_{k=1}^{n-1}f(k)$ 示例： 递推式 $a_{n+1}-a_n=2n$，解得： a n=a 1+n−1∑k=1 2 k=a 1+n(n−1) a n=a 1+∑k=1 n−1 2 k=a 1+n(n−1) --- #### 三、特征方程法 适用条件：线性齐次递推关系（如 $a_{n+2}=p a_{n+1}+q a_n$） 步骤： 1. 构造 特征方程 $r^2=pr+q$ 2. 求 根后分情况： - 实根不同：$a_n=C_1 r_1^n + C_2 r_2^n$ - 重根：$a_n=(C_1 + C_2 n)r^n$ 示例：斐波那契 数列 $f(n)=f(n-1)+f(n-2)$ 特征方程 $r^2=r+1$，根 $r=\frac{1\pm\sqrt{5}}{2}$ 通 项 公式： f(n)=1√5((1+√5 2)n−(1−√5 2)n) f(n)=1 5–√((1+5–√2)n−(1−5–√2)n) （特征方程法应用实例[^2][^3]） --- #### 四、不动点法 适用条件：分式递推 $a_{n+1}=\frac{k a_n + c}{m a_n + d}$ 步骤： 1. 解方程 $x=\frac{kx + c}{mx + d}$ 得不动点 2. 构造 新 数列 $b_n=\frac{1}{a_n - \alpha}$ 等简化递推关系 3. 转化后使用其他方法 求 解 示例：递推式 $a_{n+1}=\frac{2a_n + 1}{a_n + 3}$ 解得不动点 $x=1$，令 $b_n=\frac{1}{a_n -1}$，转化为线性递推关系 （不动点法原理参考[^1]） --- #### 五、生成函数法 适用条件：复杂递推关系或组合 数列 步骤： 1. 设生成函数 $G(x)=\sum_{n=0}^\infty a_n x^n$ 2. 建立关于 $G(x)$ 的方程 3. 解方程后展开得 通 项 示例：卡特兰数满足 $C_{n+1}=\sum_{i=0}^n C_i C_{n-i}$ 通 过生成函数 $G(x)=xG(x)^2 +1$，解得： C n=1 n+1(2 n n) C n=1 n+1(2 n n) --- #### 六、特殊 数列 参考 1. 调和 数列：$H_n=\sum_{k=1}^n \frac{1}{k}$ 2. 平方 数列：$S_n=\frac{n(n+1)(2n+1)}{6}$ 3. 立方 数列：$T_n=\left[ \frac{n(n+1)}{2} \right]^2$ （求 和公式推导参考高斯 求 和思想[^4]） --- ### 方法选择指南 | 递推形式 | 推荐方法 | ---------| | 等差/等比 | 观察归纳法 | | 一阶线性递推 | 累加/累乘法 | | 二阶线性递推 | 特征方程法 | | 分式递推 | 不动点法 | | 复杂递推 | 生成函数法 | ---
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13620 | https://www.dannydorling.org/wp-content/files/dannydorling_publication_id0892.pdf | IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 12, NO. 5, SEPTEMBER/OCTOBER 2006 WORLDMAPPER: the world as youve never seen it before Danny Dorling, Anna Barford and Mark Newman Abstract—This paper describes the Worldmapper Project, which makes use of novel visualization techniques to represent a broad variety of social and economic data about the countries of the world. The goal of the project is to use the map projections known as cartograms to depict comparisons and relations between different territories, and its execution raises many interesting design challenges that were not all apparent at the outset. We discuss the approaches taken towards these challenges, some of which may have considerably broad application. We conclude by commenting on the positive initial response to the worldmapper images published on the web, which we believe is due, at least in part, to the particular effectiveness of the cartogram as a tool for communicating quantitative geographic data. Index Terms—Geographic Visualization, Computer Graphics, Worldmapper, Data Visualization, Social Visualization, Cartogram. 1 INTRODUCTION This paper describes the Worldmapper Project, whose aim is to communicate to the widest possible audience, but particularly to school and university students, how parts of the world relate to each other and the implications of these relationships for society. This is achieved by mapping quantitative geographic data using an unusual cartographic projection, with the resulting maps being made freely available on the Internet. The Worldmapper Project has been made possible by the recent development of a new algorithm to create area cartograms, maps in which the sizes of territories (or other regions) are scaled in proportion to some variable of interest, such as population, income, or energy consumption. This algorithm’s development is complemented by the recent release by several United Nations agencies of international data of a much higher quality than has previously been made available, as well as recent geographical theories that suggest places can better be understood globally in relation to one another than when studied separately. Fig. 1. Worldmapper net emigration map. Figure 1 shows an example of the type of maps produced for the project. One can think of the maps as being akin to a 200-sectored pie chart, but with each sector morphed to roughly be at the same position as, and of a similar shape to, the relevant territory on a conventional world map: the bigger the area of a territory, the larger the proportion of the world total of the quantity of interest that belongs to that territory. Thus Figure 1 above, for example, shows the territories of the world with sizes proportional to net emigration from those territories, i.e., to the numbers of people who have left less the numbers who have entered in the fifty years prior to the year 2000. Territories with positive net immigration, such as the United States of America, receive a negative score by this measure, but since it is not possible to give a territory negative area on a map, such territories are given area zero in this case (see 2.1 below for an alternative approach to this issue). The map of net emigration shows from where more people have left than entered in recent years. Simultaneously it also shows proportionally how many more people are involved in the subject being mapped in each place. For example, the large blue territory dominating the top left hand corner of Figure 1 is Mexico, which has the largest net emigration of any territory represented in the map. 1.1 Rationale World population statistics can be difficult to understand and are usually presented in the form of statistical tables (when they are presented at all). The idea of a “net emigrant” is not a simple idea to grasp. There are internationally agreed definitions as to who constitutes an immigrant and an emigrant. There are many sources of data, almost all of which are incomplete and there is no obvious way to depict such data. A conventional map allows areas to be shaded by a proportion or rate, but cannot easily be used to compare quantities. In the following sections of this paper we: (1) outline the background to this work; (2) present our approach to the problem of estimating missing information; (3) describe the algorithm that makes the new mapping shown here possible; (4) explain the subjects we have chosen to map, and the design, order and arrangement of the posters; and (5) give details of the website created and ancillary material. 1.2 Statistics Many people have a justifiable suspicion of statistics. The recording of statistics for categories of people or geographic territories does more than just count; it defines populations through the categories used. Governments, often the source of territory-level statistics, are selective in what they choose to measure and report. On the other hand, statistics provide many opportunities to improve our understanding of the world. Territory-level data, though rarely complete or entirely accurate, allow us to draw pictures of many aspects of how the world works. We hope that by creating pictorial representations of these numbers we will help our readers to consider further what the numbers might show. • Danny Dorling is Professor of Human Geography at Sheffield University, UK, E-Mail: daniel.dorling@sheffield.ac.uk. • Anna Barford is a Research Associate at Sheffield University, UK, E-Mail: anna.barford@sheffield.ac.uk or info@worldmapper.org. • Mark Newman, is Associate Professor of Physics and Complex Systems at the University of Michigan, USA, E-Mail: mejn@umich.edu. Manuscript received 31 March 2006; accepted 1 August 2006; posted online 6 November 2006. For information on obtaining reprints of this article, please send e-mail to: tvcg@computer.org. 757 1077-2626/06/$20.00 © 2006 IEEE Published by the IEEE Computer Society IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 12, NO. 5, SEPTEMBER/OCTOBER 2006 1.3 Mapping The creation of flat, two-dimensional representations of our spherical, contoured planet has been a topic of research and invention for millennia. The sixteenth century Mercator projection is perhaps the most widely recognized cartographic projection of the Earth's surface. Mercator's map preserved angles on the globe, facilitating navigation for sea-based trade. (His creation of the map coincided roughly with the beginnings of large scale worldwide mercantile trade.) The Mercator worldview went largely unchallenged for four hundred years and was still in wide use well into the latter half of the twentieth century. Since 1974 the Gall or Peters’ Projection has gained popularity (despite the dislike to it expressed by some cartographers). This projection sacrifices accuracy in the representation of the shapes of territories in order to represent them instead with areas proportional to their true areas on the ground (something the Mercator projection fails miserably to achieve). Equal-distance or geographic projections are also now in common use, in which the scale of representation of north-south distances on the map is constant everywhere. 1.4 Cartograms Cartograms are maps in which the projection of geographic space onto the page is deliberately distorted to give regions or territories sizes that are proportional to some quantity of interest. Thirty years ago the proof was published that for each distribution a perfect cartogram exists . That is: a cartogram in which every area is the correct size (of which there are any infinity for any distribution) and which is conformal – angles are preserved correctly locally – very small circles on the surface of the globe are also shown as circles on the cartogram (at the limit). There is a long history to the study of cartograms, dominated during the twentieth century by the work of Waldo Tobler who produced the first computer-aided, mathematically defined cartograms . The cartograms of the Worldmapper project are produced using a new method based on an analogy to the physical process of diffusion, which is found empirically to generate attractive and readable maps. For more information on cartograms it is best to search the web as this field of enquiry is rapidly growing and new work is constantly appearing . Fig. 2. Worldmapper map of profits from tourism. These unconventional maps challenge viewers by distorting a familiar image prompting them to ask why is something the wrong size? “What is that blue country?” (see Figure 1). “Look how much tourists spend in Spain” (see Figure 2). It would be hard to provoke such reactions when presenting mere tables of figures – white squares filled with black numbers. Many people are uninterested in numerical data. Yet that same data can be made exciting by presenting it as a distortion of the world map. How many of our readers will previously have known who profits most from tourism, or even had any interest in finding out? Our maps could be made more interactive, certainly, and there are probably many other features that could be added. But even in their present form they seem to make very effective tools for communication, and to the extent that this is true, it is because they present an elegant approximation to the solution of the old computational problem, but are also being used to address new social issues, and are moreover visually appealing. 1.5 Posters To explain and support these maps we have embedded them in colourful, simple posters designed for classroom walls. In our experience, the maps do not need much explanation, but we have given brief comments and some statistics on each poster to help orientate the viewer (see Appendix II for an example). Anyone with Internet access can freely view and download the maps and posters from our website (www.worldmapper.org). Future plans include possible translation of the posters into languages other than English. The posters reveal much about the world but a persistent theme is inequality. There is already a heightened public awareness of worldwide inequalities, at least within some affluent countries and amongst people with ready access to news reports or the Internet. Our maps add an impartial, empirical overview that can assist the viewer in understanding more about their place in the world and their relationship to the people of other territories. For instance, our maps reveal that United Kingdom has lower immigration than the United States, India and Russia, slightly more profit is made from tourism that involves the crossing of international borders in Europe than in the rest of the world combined (see Figure 2 above) and more displaced persons seek asylum in Africa than in Europe (see poster in Appendix II). 2 CREATION OF THE DATA SETS There are, within our data sets, many missing data, many territories for which we do not have figures for one or more of the quantities studied. We cannot simply assign zeros to these territories, because the resulting cartogram would then give the territory zero area, meaning it would vanish from the map altogether, which would be somewhat misleading. Instead, therefore, we are obliged to find some method for estimating the values of the missing data. In this section we discuss this and other issues involved in the creation of the data sets for the Worldmapper Project. 2.1 Territories The first decision which has to be made in creating maps such as ours is which areas to include. Our maps are each composed of 200 territories, of which most are member states of the United Nations. A few territories are states that are not members but which are signatories to various United Nations conventions, and the remainder are mostly disputed areas that contain a large number of people, or a large land area. Because not all states recognize all other states and because not all the areas we map are states we use the neutral word “territory” to describe our areas. Territories comprise a number of areas including many islands. Some of those islands areas are found a long way from the main land mass of the territory and have complicated relationships of sovereignty with the rest of the territory. Full details are given on the website, but in essence our decision has been to rescale the entire area of the territory by whatever variable is being mapped including over exclaves (i.e. Alaska and French Guiana) and remote islands (i.e. Islas Malvinas / Falkland islands). 2.2 Population The territories on our maps are home to at least 99.95% of the world’s population. But even as simple a number as total population for each territory is not easy to come by. To use population as an example of how we derive our statistics, most population data is derived from Table 5 of the United Nations Development Programme’s (UNDP) Human Development Report 2004. The medium variant projection for the population in 2002 is used. Population estimates for territories not included in the UNDP series were derived from the United Nations Environment Programme (UNEP) figures and the Central Intelligence Agency’s (CIA) “The World Factbook”. 758 DORLING, ET AL.: WORLDMAPPER: THE WORLD AS YOUVE NEVER SEEN IT BEFORE 2.3 Land area and data sources Not all the subjects mapped in the Worldmapper project involve people. As another example, consider a measure as simple as land area (Figure 3). We define land area as total territorial area, excluding area under inland water bodies, national claims to continental shelf, and exclusive economic zones. ‘Inland water bodies’ usually means major rivers and lakes. Exclusive economic zones are areas of sea between 22km and 392km from the shore of a coastal territory within which the territory has jurisdiction over exploration and exploitation of marine resources. (The first 22km from shore is the territorial sea.) Land area data will vary over time because of water management projects, such as the damming of rivers or draining of submerged land, and changes in the level of the sea, lakes and rivers. Fig. 3. Worldmapper land area map. Most of our land area data comes from the United Nations Environment Programme. UNEP in turn took the data from the Food and Agriculture Organization’s Production Yearbook and data files. Data for Hong Kong and Timor-Leste were included within the UNEP land area totals for China and Indonesia respectively – the land area of these smaller territories was subtracted from the land area of their larger neighbours. The data used are from 2002. The World Bank’s 2005 World Development Indicators (WDI) provided land area data for Luxembourg, Hong Kong, Timor-Leste and Monaco. The World Bank also took these data from the Food and Agriculture Organization’s Production Yearbook and data files. We have attempted to minimize the number of sources we use and so used the World Bank’s version of this data. 2.4 Missing data We have data for land area and population for all territories, but in many other cases data are missing for some territories. In these cases the missing numbers are assumed to be such that when expressed as a rate they are equal to the regional average for the region containing the territory in question. In most cases these are as rate of either population or land area; whichever is appropriate. For instance, when estimating the number of births in the world we draw birth data from the World Health Organization’s 2005 World Health Report, Table 8. From this data we calculate a rate for each of the twelve regions. For those territories where figures are not available we assign a value equal to the regional crude birth rate per capita multiplied by the population of the territory. Because births have been imputed in this way to territories that previously had no data, an additional half million births are added to the worldwide total. The crude birth rate we use in this case is measured per person, not per woman or per woman in a particular age range. We do this for consistency with other estimated data and to ensure that we are not basing too many estimates upon other figures that are themselves only estimates (for instance of the number of women of particular ages living in each place). 2.5 Refugees There are some datasets for which the general estimation approach described above does not work. When it does not work a world cartogram cannot be produced unless other estimates are made. This process of estimation is thus crucial for visualization. One example is the number of refugees originating from and entering territories. We map both refugees and internally displaced persons combined (see Appendix II). The difference between these groups is that members of the former are outside their territory of origin, whilst members of the latter have stayed within national borders. Internally displaced persons may be extended protection or assistance from the United Nations High Commissioner for Refugees (UNHCR), generally pursuant to a special request by a competent organ of the United Nations. The movement of both groups is based on their "fear of being persecuted for reasons of race, religion, nationality, membership of a particular social group or political opinion" (UNHCR, 1951 ). Our data for information about displaced persons come from the United Nations Development Programme’s 2005 Human Development Report, table 22. The numbers in this report refer mainly to the latter part of 2003, though some are from 2002. The reported total of internally displaced persons (displaced within their country of origin) was 5,081,000 people, but only 4,108,000 of these were allocated to particular territories; the remainder includes estimates for territories for which there were no data available. Similarly, the total number of refugees was given as 9,970,000, of whom 9,465,000 were allocated to particular territories. To draw a world cartogram, however, all data have to be allocated to territories. To produce the totals for all 200 territories used in this map shown in Appendix 2 we have assumed that in all of the main 177 territories for which the United Nations Development Programme reported data, but for which it did not report a figure for refugees, 1 in every ten thousand people is a refugee, adding 354,000 to the total. The United Nations Development Programme was unable to collate other more basic figures for 23 territories1, so these territories do not appear in the datasets. The data may not have been collated because they were not recorded or were missing, this may be due to wars (which can displace people). Because of these special circumstances unusual estimation methods are required for these 23 territories in order to produce refugee and internally displaced persons estimates. By calculating the estimates for refugees by destination for the 23 territories without data to be exactly 31% of the maximum country in their region, a world total is achieved identical to the figure of 15,296,000 (when recorded by territory of origin) given by the United Nations Development Programme. It thus appears possible that UNHCR followed similar procedures in arriving at their global estimate. The figure of 31% was found by trial and error. (Note that many refugees and displaced people live within territories at war; in some cases assuming a figure as high as 31% will produce numbers that are unlikely to be true of peaceful territories, unless they are found in generally peaceful regions.) To produce the estimates for numbers of refugees by destination territory, a similar set of assumptions were made. For the main territories for which the United Nations Development Programme generally reports data, but for which it did not report a figure for refugees, it was assumed (as for refugees by origin) that one in every ten thousand people is a refugee. For the 23 remaining countries it was again assumed that numbers of refugees as a proportion of the population were a fixed fraction of the numbers for the territory with the highest proportion of refugees in the region. This results in the 1 The 23 territories with no United Nations Development Programme data are Afghanistan, Andorra, Cook Islands, Democratic People’s Republic of Korea, Greenland, Holy See, Iraq, Kiribati, Liberia, Liechtenstein, Marshall Islands, Federated States of Micronesia, Monaco, Nauru, Niue, Palau, Puerto Rico, San Marino, Serbia & Montenegro (possibly about to split into two states), Somalia, Taiwan, Tuvalu, and Western Sahara. 759 IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 12, NO. 5, SEPTEMBER/OCTOBER 2006 addition of 6,256,000 refugees worldwide, bringing the total to 15,296,000 for all refugees and displaced people in the world. 2.6 Historical and future population estimates Our maps showing past and future population figures, are based on Maddison’s population estimates (2003 version), which are derived from his series of historical statistics on the world economy which have been used in numerous studies . Maddison estimated that the world population in the first year of the Common Era was 231 million people, 191 million of which he assigned to specific territories. The remaining 40 million we must assign ourselves, which we do on the basis of a variety of assumptions. In most cases we estimate missing population figures for a particular era by using regional estimates and matching the fraction of the regional population falling in the given territory in a later era. In some cases a better strategy is to allocate a constant proportion of the population of a nearby more populous territory. This strategy is adopted particularly for territories that were later partitioned into two or more territories, such that a earlier single population figure is inadequate to determine the historical populations of both later territories. Examples are the United Kingdom, Ethopia, and India, each of whose population is reduced and a fraction allocated to Ireland, Eritrea, or Pakistan and Bangladesh respectively, to represent the distribution of their inhabitants in the periods before they were partitioned. Fig. 4. Worldmapper population map (2002). 3 THE CALCULATION OF THE CARTOGRAMS The cartograms themselves are calculated using the diffusion-based method described by Gastner and Newman, but the method is modified somewhat to allow the creation of maps of the entire globe. The maps are created on a rectangular grid by distorting an initial projection of the boundaries of the territories. The final sizes of the territories are guaranteed proportional to the quantity of interest for the particular map in question, but this does not mean that the initial positions of the boundaries is immaterial: the algorithm attempts to resize territories appropriately with minimal distortion of their initial shapes, meaning that the final shapes are typically similar to the initial ones. The maps are therefore substantially affected by the particular choice of initial projection for the world map. All projections of the globe onto a flat page necessarily involve distortions, but the most satisfactory for our purposes are the equal area cylindrical projections, which provide the rectangular shape needed by our cartogram algorithm and accurately represent initial land areas. To create a cartogram the population or other density function of interest is treated as a diffusing fluid, which spreads out from the areas where it is initially most dense into areas of lower density. As a simple analogy, imagine a bottle of ink emptied into a swimming pool: the ink is initially densest at the point where it is added to the water but over time will spread out until ultimately it is distributed uniformly throughout the pool. This process can be modeled using the standard linear diffusion equation of elementary physics, allowing us to calculate the motion of the fluid at any time. In our cartogram calculation we allow the features of the map to be carried along with the diffusing population or other density until it comes to rest at its final position, which defines the cartogram transformation. It is a simple mathematical exercise to demonstrate that this process does indeed produce a map in which the final areas of territories are proportional to their target population or other property of interest. In practice, the maps produced are also attractive and readable, with relatively minor distortion of territory shapes in most cases. At a technical level, the solution of the diffusion equation is performed in Fourier space, where it takes a particularly simple form. The initial density function is calculated on a 40962048 lattice, transformed using a two-dimensional fast Fourier transform, convolved with a Gaussian kernel and then back-transformed to give the diffusion field at an arbitrary later time. This field is then used to calculate the diffusion velocity as a function of position and the velocity integrated to give the displacement of the map features. The integration is performed using a standard fourth-order Runge-Kutta integrator. Because of the heavy memory demands made by the large lattice sizes we employ, the normally attractive Richardson extrapolation methods like the Bulirsch-Stoer method are not possible. However, with careful control of integration errors we have been able to calculate cartograms in which vertex positions are accurate to about 1 part in 10,000 in very reasonable computation times, typically in the order of a few minutes on a standard desktop computer. Unlike maps of small regions of the planet, world maps “wrap around” from left to right: if one travels off one side of the map, one reappears at the corresponding point on the other. This means that the cartogram algorithm has to be modified to have “periodic boundary conditions” along the horizontal axis, but still retains the closed boundaries at top and bottom used by Gastner and Newman. Technically, this requires that the two-dimensional Fourier transform that lies at the heart of the calculation be of mixed type, a complex transform in the horizontal direction and a cosine transform in the vertical direction. In addition, checks were put in place to take note of map features that passed off the sides of the map, so that they can be correctly rendered in the final images. More information and a version of the algorithm is available on-line . 4 CHOOSING THE SUBJECTS 4.1 Choosing and arranging the subjects The subjects chosen for the maps were those for which data were available. Around the year 2005, when work on this project began, there were sufficiently few such datasets that it was possible to examine most of them. As the project received publicity, viewers began to make suggestions for new maps and the final set of maps will reflect that feedback. Arranging and order for the maps is a somewhat arbitrary process but we have attempted to tell a story that begins with the basics, moves on to flows of people, then goods, before addressing environmental, social, economic and medical issues, reserving the more harrowing subjects such as war for towards the end, but then trying to finish on a note of optimism. Subjects are arranged so that pairs of consecutive maps tend to be logically paired. There is not space here to do much more than list the title of the 27 sections of maps currently under preparation (see below). Some 87 subjects are still to be decided at the time of writing: Basic maps 1 to 12 Movement maps 13 to 26 Transport maps 27 to 40 Food maps 41 to 56 Goods maps 57 to 76 Manufacturers maps 77 to 92 Services maps 93 to 100 Resources maps 101 to 108 Fuel maps 109 to 120 760 DORLING, ET AL.: WORLDMAPPER: THE WORLD AS YOUVE NEVER SEEN IT BEFORE Production maps 121 to 132 Work maps 133 to 146 Income maps 147 to 158 Wealth maps 159 to 172 Poverty maps 173 to 182 Housing maps 183 to 194 Education maps 195 to 212 Health maps 213 to 226 Disease maps 227 to 244 Disaster maps 245 to 254 Death maps 255 to 266 Destruction maps 267 to 278 Violence maps 279… Pollution … Depletion … Communication … Exploitation … Action ………… to 365 4.2 Layout of the posters The posters were designed with older school children and younger university students in mind as the target audience (see appendix II). All are in English so far, but we have attempted to design them so that they can be read and understood anywhere. We attempted to minimize clutter and make them bright and colourful enough to stand out when pinned to a wall. On the web only certain parts of each poster are normally shown on screen at one time to reduce visual confusion. 4.3 The selection of quotations Quotations were included on the posters to help younger viewers in particular realize that there are many stories that each poster can tell and many issues that they raise. The quotations aim to remind the viewer that the maps show real lives by personalizing the topic. In selecting the quotations we have tried to avoid too much bias towards the familiar and mundane. One way in which we have done this is to only allow ourselves to use any source once. We give the real names of the authors of the quotations without titles whenever possible (Tony Blair rather than Prime Minister Blair) to reduce the distance between the writer and the reader. At times we have had to cut quotations or correct minor typographical areas in them. Any changes we have made are indicated if major or detailed in the technical notes if minor. 4.4 The design of the graphs and tables The graphs and tables are intended to be simple. Graphs could potentially be off-putting to some members of our target audience. All the graphs allow comparison to be made between the twelve main regions. We try to use the same style of graph within each section of the website. Occasionally the graphs show information that cannot be depicted on the cartogram such as per capita figures. This is also what almost all the tables do, and so they allow the reader to see where the rate is most and least as well as the absolute amount. The background in the tables is coloured according to region to allow readers to see at a glance if the areas listed are in the same region. 4.5 Style of text and technical notes Both the text on the posters and that in the further technical notes is designed to be understandable when read quickly by an able older school child. We try to explain succinctly what is being mapped and to point out interesting features. We try to avoid making statements that might be read as one-sided, but our most important goal is to be clear. We avoid any use of technical terms (such as per capita) when at all possible and we try to be as transparent as we can about the usually simple arithmetic used to derive the statistics. The technical notes are also designed to allow anyone with access to the internet to be able to replicate our work. 4.6 The use of colour A consistent colour scheme is used throughout all the maps, tables and spreadsheets in worldmapper (see Figure 5). The twelve regions of the world we use are ordered from poorest to richest by the Human Development Index published in 2004. Shades of dark red are used to demarcate the poorest territories – moving through the rainbow scale to a shade violet for the best-off: Japan. The regions were chosen to be geographically contiguous groups of territories which divided the world into roughly symmetrically balanced population groups with no region containing fewer than one hundred million people. Fig. 5. Worldmapper net toy exports map. Note – this map appears lop-sided because world trade is so lop-sided. 5 NUMBERING AND NAMING In creating the worldmapper website we realized that everything needs to be defined. For instance: Population is the number of people alive in the world, at one moment in time. A birth is the moment of delivery of a baby. Definitions usually appear obvious after you have read them but for the maps we are producing here to be widely understood we have to define everything that we are mapping. We also have to be consistent in how we number and name places. The ISO 3 code, or ISO ALPHA-3 code is a three letter code devised by the International Organization for Standardization. Codes are used as a short way of identifying a territory; their shortened form makes it easier to match data. We have included these codes in all the datasets we map. Throughout our project we also use a consistent numbering system for areas. The numerical territory code has been used to order territories in a consistent manner. The number is decided by the 2004 Human Development Index rank of the 177 territories included in the main United Nations Development Programme tables (1 being Norway with the highest Human Development Index in 2004). The other 23 territories are numbered in alphabetical order, starting at number 178. Region and territory names can be contentious. Regions consist of geographically contiguous territories and nearby territories. Region names are intended to be descriptions of the location of the region. The region names try to avoid describing regions as being far away places because this implies a location for the author (and the authors of these maps work on two continents). The ‘far East’ is only far away for some and thus we use ‘Asia Pacific’. In our naming of regions we also tried to avoid subsuming smaller territories in the name of a larger territory in the region, such as using ‘Greater India’ for what we have termed ‘Southern Asia’. In attempt to avoid confusion with other regional definitions we use ‘Southern’ rather than ‘South’ where we hope it is appropriate. The territory names used are what appeared to be the most common spellings used by multiple United Nations agencies and on 761 IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 12, NO. 5, SEPTEMBER/OCTOBER 2006 the Internet, using the Roman alphabet, in January 2006. Note, we have avoided using accents on characters because there is inconsistent use of accents for territory names by United Nations agencies. Examples of the various spellings and wordings for territory names are below; the name we use is first, followed by examples of alternative spellings: Albania, Republika e Shqiperise Bahrain, Mamlakat al Bahrayn Burundi, Republika y'u Burundi Cape Verde, Cabo Verde, Republica de Cabo Verde, Cote d’Ivoire, Côte d'Ivoire Democratic People's Republic of Korea, Choson-minjujuui-inmin-konghwaguk Kyrgyzstan, Kyrgyz Respublikasy Timor Leste, Timor Lorosa'e, East Timor Mexico, Estados Unidos Mexicanos Morocco, Al Mamlakah al Maghribiyah Mozambique, Republica de Mocambique Sao Tome & Principe, São Tomé e Príncipe, Sao Tomé et Príncipe Uruguay, Republica Oriental del Uruguay Viet Nam, Vietnam, Cong Hoa Xa Hoi Chu Nghia Viet Nam Yemen, Al Jumhuriyah al Yamaniyah, Al Yaman Lastly, in some cases it was necessary to abbreviate country names in order that they would fit into certain sections of the poster. In order to maintain some consistency, the following abbreviations were made, when necessary, to reduce the length of the name: a) If a well-used abbreviation already existed, it was adopted, this was done for the following: ‘Former Yugoslav Republic of Macedonia’ was abbreviated to ‘Macedonia FYR’ ‘Democratic People's Republic of Korea’ was abbreviated to ‘DPR Korea’ b) ‘Saint’ was changed to ‘St’ c) ‘Democratic’ was changed to ‘Dem’ d) ‘Federated’ was changed to ‘Fed’ 6 CONCLUSION We conclude by commenting on the apparent early enthusiasm for the worldmapper images on the web and by speculating as to why this might be partly due to the visual advantages of the particular kinds of cartogram this project has had access to and then turn to issues of interactivity. In the first half of 2006 the worldmapper website received 470,000 unique visitors, mostly since April of that year. No announcement was made that the site was being created but it quickly attracted the attention of commentators on the web and the maps were featured in many magazine articles, in New Scientist and on the Discovery Channel on television. We have included a couple of new images yet to appear on the web in Appendix I below to illustrate why these images appear to be somewhat addictive to view, and how when contrast as pairs they can leave a lasting impression. As yet the maps are not interactive. In the future we hope to be able to add interactivity to the maps such that the user will be able to select territories and find out more about then. Applications could be created to allow users to ‘brush’ over the cartograms and see the names of territories change under their eyes. We could think about dynamically linking the graphs in the spreadsheets of world data that we also make available to the maps to allow users to see how altering values alters the image. We have not yet added any animations to the website. We are currently looking at the possibility of making available animated cartogram globes that you can spin and zoom into, in the style of ‘google earth’. We have also not yet thought how we could morph from one projection to another to better illustrate just how unevenly peoples, goods and livelihoods are distributed over the surface of this planet. Worldmapper is still a work in progress. See www.worldmaper.org for more information. ACKNOWLEDGEMENTS We would like to thank our colleagues: Graham Allsopp, John Pritchard, and Ben Wheeler in the worldmapper project. We would also like to thank the anonymous referees for many kind comments and suggestions – especially the one who solicited views on the maps from their fellow commuters on a train, and who will hopefully recognize the results of many of their suggestions here. Funding for this project came initially from a Prize awarded by the Leverhulme Trust. However, the trust has no control over the content of the website and no responsibility for it. REFERENCES A. K. Sen, 1975, A theorem related to cartograms, American Mathematics Monthly, 82, 382-385. Dorling, D., 1996, Area cartograms: their use and creation, Concepts and Techniques in Modern Geography series no. 59, University of East Anglia: Environmental Publications. See for instance Valerie Yakich’s University of Buffalo Master’s dissertation “A New Analogy for Cartograms”, 2004, United Nations High Commission for Refugees. The 1951 Convention relating to the Status of Refugees, see For Angus Maddison’s work see: Newman, M., 2006, Cart: Computer software for making cartograms [online]. University of Michigan. Available from: [Accessed 5th July 2006] 762 DORLING, ET AL.: WORLDMAPPER: THE WORLD AS YOUVE NEVER SEEN IT BEFORE 7 APPENDIX I : EXAMPLES OF MAPS CONTRASTING Fig. 6. Worldmapper map of people estimated to have HIV/AIDS. More than 25 million people have died of AIDS since it was first diagnosed in the early 1980s almost 3 million of these deaths occurred in 2005. Fig. 7. Worldmapper map of the gross US dollar value of medicines exported in 2002 between the territories of the world. The total area of the map represents gross export earning of US$46 billion in that year. 763 IEEE TRANSACTIONS ON VISUALIZATION AND COMPUTER GRAPHICS, VOL. 12, NO. 5, SEPTEMBER/OCTOBER 2006 8 APPENDIX II: EXAMPLE POSTER 764 |
13621 | https://www.frontier-pitts.fr/en/bsi-pas-68-european-certification/k4-k8-k12-equivalences | K4/K8/K12 equivalences
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13622 | https://people.wou.edu/~courtna/ch462/tmcolors.htm | Transition Metal Complexes and Color
Introduction
The d-orbitals of a free transition metal atom or ion are degenerate (all have the same energy.) However, when transition metals form coordination complexes, the d-orbitals of the metal interact with the electron cloud of the ligands in such a manner that the d-orbitals become non-degenerate (not all having the same energy.) The way in which the orbitals are split into different energy levels is dependent on the geometry of the complex. Crystal field theory can be used to predict the energies of the different d-orbitals, and how the d-electrons of a transition metal are distributed among them. When the d-level is not completely filled, it is possible to promote and electron from a lower energy d-orbital to a higher energy d-orbital by absorption of a photon of electromagnetic radiation having an appropriate energy. Electromagnetic radiations in the visible region of the spectrum often possess the appropriate energy for such transitions.
Seeing Color
The sensors in our eyes detect only those wavelengths in the visible portion of the electromagnetic spectrum. Although visible light appears "white", it is made up of a series of colors. White light consists of three primary colors (red, yellow and blue). These primary colors can be mixed to make three secondary colors (orange, green and violet).
| |
| Red + Yellow makes Orange Yellow + Blue makes Green Blue + Red makes Violet |
An "artist's" color wheel is a useful way show to these relationships. If you add the colors on opposite sides of the wheel together, white light is obtained. We only detect colors when one or more of the wavelengths in the visible spectrum have been absorbed, and thus removed, by interaction with some chemical species (see an animation of this here.) When the wavelengths of one or more colors is absorbed, it is the colors on the opposite side of the color wheel that are transmitted.
| | | |
---
| Example: What happens when we see green? | If red, yellow, orange, blue and violet are absorbed... only one color is transmitted ...GREEN | If violet, red, and orange are absorbed... blue, green , and yellow are transmitted ...the middle color is perceived...GREEN |
Grass and leaves appear green because chlorophyll absorbs wavelengths in the red and blue portion of the visible spectrum. The wavelengths in between (green) are transmitted.
Transition Metal Complexes
When light passes through a solution containing transition metal complexes, we see those wavelengths of light that are transmitted. The solutions of most octahedral Cu (II) complexes are blue. The visible spectrum for an aqueous solution of Cu (II), [Cu(H2O6]2+, shows that the absorption band spans the red-orange-yellow portion of the spectrum and green, blue and violet are transmitted.The absorption band corresponds to the energy required to excite an electron from the t2g level to the eg level. Recall, the energy possessed by a light wave is inversely proportional to its wavelength. The Cu(II) solution transmits relatively high energy waves and absorbs the low energy wavelengths. This indicates that the band gap between the two levels is relatively small for this ion in aqueous solution.
d-Orbital Splitting
The magnitude of the splitting of the d-orbitals in a transition metal complex depends on three things:
the geometry of the complex
the oxidation state of the metal
the nature of the ligands
The Nature of the Ligands Some ligands only produce a small energy separation among the d-orbitals while others cause a wider band gap. Ligands that cause a small separation are called weak field ligands, and those that cause a large separation are called strong field ligands. The ordering of their splitting ability is called the spectrochemical series.here. A comparison of the visible absorption maxima for a number of cobalt (III) complexes shows the effects of ligands on the d-orbital band gap. References: Kotz, J.C.; Purcell, K.F. Chemical and Chemical ReactivitySaunders: New York, 1987, Chapter 25. Rodgers, G.E. Introduction to Coordination, Solid State, and Descriptive Inorganic ChemistryMcGraw -Hill: New York, 1994, Chapter 4. |
13623 | https://fractiontodecimal.net/sixty-three-fourteenths-in-decimal | 63/14 as a Decimal ▷ What is 63/14 in Decimal Form?
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63/14 as a Decimal
by Mark
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Convert 63/14 to Decimal
What is 63/14 as a Decimal?
Conclusion
Here you can find 63/14 as a decimal.
We also have useful information regarding the fraction in decimal form. :-)
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If you have been searching for 63 over 14 as a decimal, then you are right here, too.
The terms used in this article are explained in detail on our home page; check it out if anything remains unclear. 63/14 = 4.5
Additional Information
63/14 in decimal notation has 1 decimal place, that means it is a terminating decimal.
63/14 in decimal form = 4.5
Sixty-three fourteenths as a decimal = 4.5
63 over 14 as a decimal = 4.5
Learn how to change 63/14 to a decimal number in the following section.
In addition, you can read up on the properties of 63/14.
Convert 63/14 to Decimal
To change the fraction to decimal you can use the long division method explained in our article fraction to decimal, which you can find in the header menu.
Or you can divide the nominator 63 by the denominator 14 using a calculator.
If you like, use our automatic calculator above. Just enter the fraction with a slash, e.g. 63/14.
If the result includes a repeating sequence, then it will be denoted in ().
Similar conversions in this category include, for example:
68/14 as a decimal
69/14 as a decimal
70/14 as a decimal
Ahead is more information on 63/14 written in base 10 numeral system.
What is 63/14 as a Decimal?
You already know the answer to this question: Sixty-three fourteenths as a decimal equals 4.5
We have characterized 63/14 in base 10 positional notation above, so we are left with telling you the properties of 63/14:
63/14 is a simple fraction
63 is the nominator, above the slash
14 is the denominator, below the slash
63/14 is an improper fraction
Instead of a slash, the division symbol ÷, known as obelus, can be used to denote a fraction:
For example: 63÷14 in decimal.
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– Article written by Mark, last updated on December 20th, 2023
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13624 | https://www.khanacademy.org/math/ap-statistics/xfb5d8e68:inference-categorical-proportions/introduction-confidence-intervals/a/interpreting-confidence-levels-and-confidence-intervals | Interpreting confidence levels and confidence intervals (article) | Khan Academy
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AP®︎/College Statistics
Course: AP®︎/College Statistics>Unit 10
Lesson 1: Introduction to confidence intervals
Confidence intervals and margin of error
Confidence interval simulation
Interpreting confidence level example
Interpreting confidence levels and confidence intervals
Math>
AP®︎/College Statistics>
Inference for categorical data: Proportions>
Introduction to confidence intervals
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Interpreting confidence levels and confidence intervals
AP.STATS: UNC‑4.F (LO), UNC‑4.F.1 (EK), UNC‑4.F.2 (EK), UNC‑4.F.3 (EK), UNC‑4.F.4 (EK)
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When we create a confidence interval, it's important to be able to interpret the meaning of the confidence level we used and the interval that was obtained.
The confidence level refers to the long-term success rate of the method, that is, how often this type of interval will capture the parameter of interest.
A specific confidence interval gives a range of plausible values for the parameter of interest.
Let's look at a few examples that demonstrate how to interpret confidence levels and confidence intervals.
Example 1: Interpreting a confidence level
A political pollster plans to ask a random sample of 500 voters whether or not they support the incumbent candidate. The pollster will take the results of the sample and construct a 90% confidence interval for the true proportion of all voters who support the candidate.
Which of the following is a correct interpretation of the 90% confidence level?
Choose all answers that apply:
Choose all answers that apply:
(Choice A) If the pollster repeats this process and constructs 20 intervals from separate independent samples, we can expect about 18 of those intervals to contain the true proportion of voters who support the candidate. - [x] A If the pollster repeats this process and constructs 20 intervals from separate independent samples, we can expect about 18 of those intervals to contain the true proportion of voters who support the candidate.
(Choice B) About 90% of people who support the candidate will respond to the poll. - [x] B About 90% of people who support the candidate will respond to the poll.
(Choice C) If the pollster repeats this process many times, then about 90% of the intervals produced will capture the true proportion of voters who support the candidate. - [x] C If the pollster repeats this process many times, then about 90% of the intervals produced will capture the true proportion of voters who support the candidate.
Check Explain
Example 2: Interpreting a confidence interval
A baseball coach was curious about the true mean speed of fastball pitches in his league. The coach recorded the speed in kilometers per hour of each fastball in a random sample of 100 pitches and constructed a 95% confidence interval for the mean speed. The resulting interval was (110,120).
Which of the following is a correct interpretation of the interval (110,120)?
Choose all answers that apply:
Choose all answers that apply:
(Choice A) If the coach took another sample of 100 pitches, there's a 95% chance the sample mean would be between 110 and 120 km/hr. - [x] A If the coach took another sample of 100 pitches, there's a 95% chance the sample mean would be between 110 and 120 km/hr.
(Choice B) About 95% of pitches in the sample were between 110 and 120 km/hr. - [x] B About 95% of pitches in the sample were between 110 and 120 km/hr.
(Choice C) We're 95% confident that the interval (110,120) captured the true mean pitch speed. - [x] C We're 95% confident that the interval (110,120) captured the true mean pitch speed.
Check Explain
Can we say there is a 95% chance that the true mean is between 110 and 120 kilometers per hour?
We shouldn't say there is a 95% chance that this specific interval contains the true mean, because it implies that the mean may be within this interval, or it may be somewhere else. This phrasing makes it seem as if the population mean is variable, but it's not. This interval either captured the mean or didn't. Intervals change from sample to sample, but the population parameter we're trying to capture does not.
It's safer to say we're 95% confident that this interval captured the mean, since this phrasing more closely agrees with the long-term capture rate of confidence levels.
Example 3: Effect of changing confidence level
Suppose that the coach from the previous example decides they want to be more confident. The coach uses the same sample data as before, but recalculates the confidence interval using a 99% confidence level.
How will increasing the confidence level from 95% to 99% affect the confidence interval?
Choose 1 answer:
Choose 1 answer:
(Choice A) It's impossible to say without seeing the sample data. A It's impossible to say without seeing the sample data.
(Choice B) Increasing the confidence will increase the margin of error resulting in a wider interval. B Increasing the confidence will increase the margin of error resulting in a wider interval.
(Choice C) Increasing the confidence will decrease the margin of error resulting in a narrower interval. C Increasing the confidence will decrease the margin of error resulting in a narrower interval.
Check Explain
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Ann.T.Sebastian 8 years ago Posted 8 years ago. Direct link to Ann.T.Sebastian's post “Question 3 : How will inc...” more Question 3 : How will increasing the confidence level from 95 percent to 99 percent affect the confidence interval?
In the above mentioned question, wouldn't the interval be narrower.
If we were to decrease the confidence level to say 85%, the margin of error will be more and the interval will be wider?
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Sarah Ackerman 8 years ago Posted 8 years ago. Direct link to Sarah Ackerman's post “The wider your interval i...” more The wider your interval is, the more confident you can be that your interval contains the true mean. Think about an interval that covers the entire spread of the data... you can be 100% confident that it contains the true mean.
If you have a very narrow range (e.g. 115.1 mph to 115.2 mph), then in this example you cannot be very confident that it will contain the true mean, so you'd have a very low confidence interval (near 0%).
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Jared Bailey 6 years ago Posted 6 years ago. Direct link to Jared Bailey's post “In the try it for yoursel...” more In the try it for yourself exercise, what are the blue and red curves??
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B1-66ER 6 years ago Posted 6 years ago. Direct link to B1-66ER's post “The red line is the sampl...” more The red line is the sample distribution, the blue line is the population distribution. The reason why the blue line changing it's shape while adjusting sample size is the scale of the whole chart is changing, which means the blue line actually isn't changing at all, just the zooming out to in order to show the full area of red line, I think.
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keetolia000 3 years ago Posted 3 years ago. Direct link to keetolia000's post “us 100% interval?” more us 100% interval?
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Simran 2 years ago Posted 2 years ago. Direct link to Simran's post “This question was asked s...” more This question was asked sometime ago (by Mark Ionkin), and Soo Kyung Ahn responded:
"The normal distribution is defined from negative infinity to positive infinity and the corresponding 100% confidence interval would be from negative infinity to positive infinity as well. It doesn't provide useful information, and thus it is not used."
Elias Aquino also said:
"Imagine if weather reporters said "there is a 100% chance the weather today will be between -100 to 300 degrees" (useless info, not specific enough)
versus, "we are 95% certain the weather will be between 20 and 38 degrees"
VERSUS "we are 80% certain the weather will be between 29 and 32 degrees" (more specific, but less certain)."
Omar Khalaf said:
"having a 100% confidence interval will give us an interval from (-infinity, +infinity) since, technically, we can be a 100% sure that the value we want will be a number between these values. You can never be a 100% sure of something."
So basically the confidence interval would have to include all numbers, which makes it kinda useless, right?
Therefore, people typically don't use 100% confidence intervals.
Hope that helps this is sorta new to me too
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Balaji Harihar 2 years ago Posted 2 years ago. Direct link to Balaji Harihar's post “Is there a mathematical p...” more Is there a mathematical proof of the equivalence between the 2 statements:
Statement 1 - There is a 95% probability that the sample mean falls within 2 standard deviations of the population mean. Statement 2 - There is a 95% probability that the population mean falls within 2 standard deviations of the sample mean
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Chuck B 2 years ago Posted 2 years ago. Direct link to Chuck B's post “This caught me off guard ...” more This caught me off guard at first as well. However, it's actually so simple no proof is needed.
How far is a from b? It is |a-b|. How far is b from a? It is |b-a|, which is equal.
When the sample mean falls within 2 standard deviations of the population mean, it is therefore equivalent to say that the population mean is within 2 standard deviations of the sample mean. That one occurs with 95% probability necessarily results in the other occurring with equal probability.
It's important to note that in this context, "standard deviation" refers to a single value: the standard deviation of the population. We're not, for example, saying that "there is a 95% probability that the sample mean falls within 2 population standard deviations of the population mean" is equivalent to "There is a 95% probability that the population mean falls within 2 sample standard deviations of the sample mean."
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grace9570 5 years ago Posted 5 years ago. Direct link to grace9570's post “In Example 2, shouldn't B...” more In Example 2, shouldn't B also be correct? To construct the 95% confidence interval, we add/subtract 2 standard deviations from the mean. Given the distribution of the sample is approximately normal, this interval would also contain about 95% of the sample pitches.
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zjleon2010 5 years ago Posted 5 years ago. Direct link to zjleon2010's post “recall how we calculate t...” more recall how we calculate the standard deviation:
sqrt(p(1-p)/n)
this is an estimation of the SD of the population
so the confidence level it construct doesn't work for the sample data
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kontohderrick4 9 months ago Posted 9 months ago. Direct link to kontohderrick4's post “does confidence interval ...” more does confidence interval always capture the sample mean
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Shefali C 3 months ago Posted 3 months ago. Direct link to Shefali C's post “Yes! The whole interval i...” more Yes! The whole interval is centered around the sample mean itself.. the standard deviation calculations, Z-score calculations.. all of it so a CI definitely captures the sample mean.. but will it always capture the true mean? Not necessarily.. which is why we use CI to assess the confidence of true mean existing within the range.
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lonemaven 3 years ago Posted 3 years ago. Direct link to lonemaven's post “So the best way to estima...” more So the best way to estimate the population mean is to set a higher confidence level AND increase the sample size if you want a good balance of how confident you are that the interval captures the true mean and how accurate that interval might be (i.e. low standard error) - is that correct?
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daniella a year ago Posted a year ago. Direct link to daniella's post “Increasing the confidence...” more Increasing the confidence level indeed increases the margin of error, resulting in wider confidence intervals. This wider interval provides greater assurance (confidence) that the true population parameter lies within it. However, a wider interval may not be desirable if precision is a priority. In such cases, increasing the sample size can help reduce the margin of error, leading to narrower confidence intervals while still maintaining a high confidence level. Therefore, there is a trade-off between the desired level of confidence and the precision of the estimate, and adjusting both the confidence level and sample size can help achieve the desired balance.
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Misha Fathi 2 years ago Posted 2 years ago. Direct link to Misha Fathi's post “can you explain me this q...” more can you explain me this question please?
A random sample of 32 persons attending a certain diet clinic was found to have lost, over a threeweek period, an average of 30 pounds with a sample standard deviation of 11 pounds. Construct a 99% confidence interval estimate of the true mean weight loss, over a three-week period, experienced by all persons attending the clinic. You may assume that the distribution of weight loss is normal.
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daniella a year ago Posted a year ago. Direct link to daniella's post “Given: Sample size (n) = ...” more Given: Sample size (n) = 32, Sample mean (x̄) = 30 pounds, Sample standard deviation (s) = 11 pounds, and confidence level = 99%.
Since the sample size is sufficiently large (n > 30) and the distribution of weight loss is assumed to be normal, you can use the z-distribution.
The critical z-value for a 99% confidence level (two-tailed) is approximately 2.576.
Calculate the standard error of the mean (SE) using the formula: s / √n.
Compute the margin of error (ME) by multiplying the SE with the critical z-value: ME = z × SE.
Finally, construct the confidence interval using the formula: (x̄ - ME, x̄ + ME).
Substituting the given values into the formulas, you can find the confidence interval estimate of the true mean weight loss.
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MaybeBlayre 5 months ago Posted 5 months ago. Direct link to MaybeBlayre's post “In example 2, is the firs...” more In example 2, is the first multiple choice option still a true statement? I can understand why the answer choice is wrong, as that is not describing the confidence interval itself, but is the only reason that concept is wrong is because it's "implying the population mean to be a variable?" Or is it an untrue statement?
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sassyshourya13 4 years ago Posted 4 years ago. Direct link to sassyshourya13's post “What is the implication o...” more What is the implication of not being able to have 100% confidence mean to performing an analysis?
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daniella a year ago Posted a year ago. Direct link to daniella's post “Not being able to achieve...” more Not being able to achieve 100% confidence in estimating the population parameter reflects the inherent uncertainty in statistical inference. It acknowledges that even with the best methods and data available, there is always a degree of uncertainty in our estimates due to sampling variability and other sources of error. This uncertainty underscores the importance of understanding and appropriately interpreting confidence intervals and confidence levels in statistical analysis
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13625 | https://www.sciencedirect.com/science/article/abs/pii/S0003687018306392 | Relation between riding pleasure and vehicle dynamics - Results from a motorcycle field test - ScienceDirect
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Applied Ergonomics
Volume 90, January 2021, 103231
Relation between riding pleasure and vehicle dynamics - Results from a motorcycle field test
Author links open overlay panel Sebastian Will a, Barbara Metz a, Thomas Hammer a, Raphael Pleß a, Matthias Mörbe b, Markus Henzler c, Frederik Harnischmacher d
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Highlights
•Unique field test data from motorcycles for more than 6000 km of motorcycling.
•Significant correlation between vehicle dynamics and experienced riding pleasure.
•Parameters of lateral vehicle behavior indicate riding pleasure better than parameters of longitudinal vehicle behavior.
•Riding pleasure is detectable on curvy sections as well as on straight roads.
Abstract
Powered two-wheelers are a common means of transport all over the world. In several countries, primary motorcycles with high displacement involve another purpose, namely motorcycling is a leisure activity. Motorcycles are used as tools of transport pleasure as opposed to being purely used for individual commuting purposes. The aim of the current study involves investigating the relation between experienced riding pleasure and riding behavior in a field test. Specifically, N=12 motorcyclists between 21 and 66 years of age were observed while riding for approximately 8 h on public roads. The measurement setup included a logger for vehicle dynamics and vehicle handling data, GNSS data, video data, and subjective measures recorded as audio comments at predefined points of interest along the round course. A comprehensive dataset with more than 6000 km of motorcycling was gathered. The results indicate that parameters of lateral vehicle behavior, such as the maximum lean angle, reflected riding pleasure. Interestingly, this is applicable for curvy sections as well as straight roads. High ratings of riding pleasure correlated with riding in snaky lines as a type of self-stimulation on straight sections. Longitudinal vehicle dynamics, such as the range of accelerations, tend to increase with the riding pleasure in curves. Hence, the effects are smaller than those for lateral vehicle behavior and not visible on straight sections. Generally, curvy sections on rural roads produce higher pleasure than straight roads. On a global level, riding pleasure increases during the first few hours of riding and subsequently decreases with respect to the time on task. The results are discussed in the context of studies on driving pleasure from the automotive sector and more fundamental psychological theories that explain pleasure as a physiological stimulation or flow. Several individuals ride motorcycles to experience pleasure. A better understanding of rider behavior in these situations can aid in deriving proper assistance and to provide individual support to a rider, thereby increasing riding pleasure as well as safety.
Introduction
The pattern of use for motorcycles shifted towards leisure riding since the mid-1990s (Jamson and Chorlton, 2009). This is at least applicable for central European countries, such as Germany or Austria, where riding pleasure plays an important role when compared to the use of powered two-wheelers (PTW) for commuting.
For example, Jordan (2000) defines pleasure and differentiates four types of pleasure (i.e., physio-pleasure, socio-pleasure, ideo-pleasure, and psycho-pleasure). Physio-pleasure is a result of positive sensory stimulation. Psycho-pleasure is defined as a result of joyful interaction with a product characterized through high usability. These two sources of pleasure appear to be relevant to motorcycling, while pleasure arising from social interaction and special aesthetics of a product are less linked to motorcycling. For example, Tischler and Renner (2007) define driving pleasure. They consider driving pleasure as an emotional state of a person that is determined by the current sensual experience of the interaction between driver, vehicle, and environment. They postulate a closed loop model to describe the emergence of driving pleasure. The studies indicate that drivers exhibit an intended level of activity and select driving behavior to reach the level by considering available driver skills (see also Fuller, 2000). Resources and restrictions in terms of vehicle characteristics and environmental influences moderate the action's success. Furthermore, the authors assumed that driving pleasure is a result of active and dynamic driving in contrast to feeling under-stimulated or bored (see also, Engelbrecht et al., 2009; Knobel et al., 2013; Roßner et al., 2015). However, at least for the passenger-car sector, this definition does not sufficiently describe the construct of driving pleasure. Additionally, factors including comfort play an important role (Engelbrecht et al., 2009; Hartwich et al., 2018). While considering motorcycles as opposed to passenger cars, the role of active and dynamic driving is seemingly more important and especially while focusing on leisure usage.
From a psychological point of view, the origin of fun or pleasure is significantly linked to the motivation of conducting a certain task (Blythe and Hassenzahl, 2003; Deci and Ryan, 1985). A differentiation between actions that are conducted due to an intrinsic or extrinsic motivation appears to be necessary. Rheinberg (1989) defines actions as intrinsically motivated when conducted for self-purpose and not for its outcome. Actions that are motivated intrinsically do not need any further reward. It is assumed that motorcycling corresponds to the aforementioned type of activity for most riders. Another helpful concept to understand the origin of riding pleasure is flow, and this was introduced by Csikszentmihalyi (2000). Flow is described as a state of being neither unchallenged nor over-challenged while conducting a certain task. It is described as full focus on the task itself (absorption) that matches an individual's skills. The rider's behavior in maneuvering and stabilizing the motorcycle is a possibility whereby task demands are regulated, and the flow balance is maintained (Donges, 1978). The interpretation is supported by Pöschel et al. (2011) who obtained a positive effect of free riding on experienced riding pleasure. The riders' possibility for self-regulation is minimized by following other vehicles in traffic. This makes it harder to maintain a balance between task demands and rider skills. With respect to motorcycle riding, a relation between being in a flow-state while riding a motorcycle and experiencing riding pleasure is assumed (Rheinberg, 2000). In addition to the interaction between rider and vehicle, the environment influences the riding pleasure experienced. Road characteristics, such as curvature or road surface conditions, impact the perceived level of riding pleasure (Engelbrecht, 2013). Typically, high friction and curvy sections are related to high riding pleasure. This is especially true while focusing on dynamic driving as opposed to comfortable cruising. One of the major differences between single-track vehicles, such as motorcycles, and cars pertains to how they handle a curve. The PTW is stabilized by producing adequate lean angles in curves. Given this particular characteristic, the hypothesis involves determining increased riding pleasure especially on curvy sections when compared to straight roads.
The investigation of riding pleasure and its relation to vehicle dynamics requires objective riding data and an external criterion indicating perceived pleasure. The latter is commonly performed with subjective ratings (Amado et al., 2014; Toda and Kageyama, 2007). In a study on driving pleasure, N= 51 drivers drove four different cars on a closed test track and in public traffic (Tischler and Renner, 2007). Comfort and sportiness appeared as the optimal predictors in a regression model for driving pleasure. Vehicle data analysis on sporty test track laps from N=14 professional drivers revealed that higher mean acceleration, higher maximum lateral acceleration, and low steering wheel angles were related to driving pleasure. Hence, the effect was not obtained for nonprofessional drivers.
Given the cited results from fundamental and automotive studies, the following hypotheses are obtained:
•Motorcycle riding is a task with high control involving sensory stimulation, task demands, and joyful interaction with a product. Therefore, it is hypothesized that higher levels of experienced pleasure are reflected in vehicle dynamics data.
•Based on previous studies in the automotive sector, high pleasure should be reflected in measures of longitudinal as well as lateral vehicle dynamics.
•Given the specific characteristics of a single-track vehicle (i.e., necessary lean angle while taking a curve), it is assumed that major effects should be visible on curvy roads as opposed to straight roads.
To the best of the authors’ knowledge, this is the first study in which a significant amount of vehicle dynamics data is set in relation to ratings of subjectively experienced pleasure in the motorcycle sector. Therefore, explorative analyses are included next to hypotheses driven investigations.
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Basic experimental setup
A test course in public traffic was defined, which contained sections with varying road characteristics with a focus on curvy rural sections. Nonprofessional riders were invited to participate in the experiment. They were asked to ride up to eight laps on the test course during a single day. At predefined positions along the track, riders were asked to rate their current state to obtain information on subjectively experienced riding pleasure. The aim involved investigating the relationship
Data analysis
Based on GNSS-position, riding data was segmented into sections with different road characteristics as follows (values in parenthesis represent percentage of length of total course):
•straight rural sections (20.6%)
•moderate curvy rural sections (16.4%)
•curvy rural sections (24.4%)
•highway sections (23.0%)
•urban sections (10.7%)
•intersections (0.6%)
•roundabouts (1.8%)
•exits/entries to highway (1.8%)
•unclassified (0.7%)
The classification was performed via an expert rating by two experienced motorcycle
Results
The test course was deliberately chosen to enable free riding without surrounding traffic that might influence rider behavior. Based on video analyses, this worked well for rural sections and highways so that only rare events with other traffic participants occurred which can be neglected. Consequently, the analyzed data is interpreted as containing freely chosen rider behavior. An exception to that were transits through urban areas.
Fig. 3 shows the development of subjective riding pleasure as
Discussion
The main aim of the study involves investigating the relation between subjectively experienced riding pleasure and objectively measurable vehicle dynamics data. The results indicate that parameters of lateral vehicle behavior especially indicated riding pleasure independent of road characteristics.
Specifically, in the morning session, the ratings assigned during the first half of the round course are higher and increase until noon. At the end of every lap, values for riding pleasure decrease.
Conclusion
Generally, vehicle dynamics patterns are correlated with rated riding pleasure. The relation between subjectively experienced riding pleasure and lateral vehicle dynamics is more stable than the relation with longitudinal vehicle dynamics. The maximum lean angle increases constantly with increases in the riding pleasure. This also applies to different road types (e.g., straight vs. curvy roads). Generally, the results indicated that a significant relation exists between the objective behavior
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgements
The authors would like to express their deep gratitude to all colleagues and participants that were involved in the research process. This research was funded by Bosch and KTM and did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
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Cited by (9)
Risky riding behaviors among motorcyclists and self-reported safety events in Pakistan
2024, Transportation Research Part F Traffic Psychology and Behaviour Citation Excerpt :
Therefore, understanding the psychological underpinnings behind motorcyclists’ behaviors is essential for tailored interventions to mitigate motorcyclist crash risks. To comprehensively analyze motorcyclists’ riding behaviors, various approaches have been employed in the literature, including field tests (Ohtsuka et al., 2015; Will et al., 2021), riding simulator investigations (Bolla et al., 2023; Grottoli et al., 2023; Vu et al., 2020), and crash data analysis (Budd et al., 2018; Ruangkanjanases et al., 2024; Waseem et al., 2019). While these approaches offer several benefits, such as real-world data collection and controlled experimental settings, they also have limitations, including the high costs of the experiment, unavailability of simulators, lack of appropriate police-reported crash data, and direct observation. Show abstract Motorcycles are the most prevalent mode of transportation in many developing countries, particularly in Asia. Despite their popularity, motorcycles inherently pose risks to riders, which are further amplified by risky behaviors such as speeding, non-helmet use, non-compliance with traffic rules. Exploring motorcycle crashes has been the most widely employed approach to identify risk factors, but comprehensive crash data are often unavailable in Pakistan. Instead, Motorcycle Rider Behavior Questionnaire (MRBQ) is employed in this study, which has been served as a valuable tool to comprehensively investigate the risky behaviors among motorcyclists. Nevertheless, variations in the factor structures of the MRBQ across different geographical contexts underscore the necessity of refining the MRBQ based on cultural and regional characteristics. Thus, there is a crucial need to tailor the MRBQ to the local dynamics, considering diverse cultural norms, licensing regulations, and motorcyclists’ experiences. Moreover, while existing MRBQ studies have provided valuable insights into factors including traffic errors and speeding violations, there is a notable gap related to safety behaviors directly affecting motorcyclists’ well-being. Therefore, this study aims to not only validate the MRBQ framework in the Pakistani context but also integrate safety-related behaviors, offering a comprehensive understanding of motorcyclists’ risk behaviors. To achieve the objective, a sample of 1,344 responses from Pakistani motorcyclists was collected online. Exploratory factor analysis of the adapted MRBQ resulted in a refined 45-item questionnaire characterized by a five-factor structure (i.e., speeding violations, traffic errors, safety violations, stunts, and control errors). Findings confirm the construct validity of the proposed modified MRBQ, revealing coherent distributions of items across the five distinct factors. Notably, the modified version of the questionnaire brings to light a unique factor: safety violations, reflecting the specific safety-related behaviors particularly observed in Pakistan. In addition, the results reveal a significant association between the MRBQ factors, especially speeding violations, traffic errors, safety violations, and risk indicators. Moreover, the findings highlight that younger motorcyclists, those with lower education or income levels, and those without a valid license have a higher likelihood of crashes, near-crashes, and traffic violations, while riders with a daily riding duration of less than an hour have a lower likelihood of such incidents. Based on the findings, the study recommends urgent policy measures, including strengthening traffic law enforcement to address speeding and safety violations, implementing targeted training programs and educational initiatives for young motorcyclists, addressing regulatory deficiencies in licensing through improved enforcement, and devising tailored public awareness campaigns for enhancing motorcycle safety in Pakistan and other countries with similar characteristics.
### Digital twin for motorcycle riding profile prediction
2024, Transportation Research Part C Emerging Technologies Citation Excerpt :
In the following section, the motorcycle cornering behavior model will be developed followed by the model for the longitudinal motorcycle riding behavior. Even though motorcycle riding pleasure is said to increase with banking angle and curvature of the road (Will et al., 2021), motorcycle banking is also coupled with the perceived skill-level of the motorcycle rider (Spiegel, 2019), where more experienced riders are able to take corners at higher speeds, requiring greater banking angles. Although this correlation between motorcycle banking and rider skill has been known for some time (Prem and Good, 1983), it has not been quantified yet. Show abstract The digital twin, with its descriptive and predictive services provides promising prospects in the automotive field. Increased insights in customer behavior and vehicle states through descriptive models enable the predictive services of the digital twin to project these vehicles in the future. Currently, most of the predictive services of the digital twin focus on a singular dimension, the asset itself, without considering the influences of the environment and the driver, which hampers the reliability and accuracy of these models. The present work aims to provide a solution in the form of a holistic digital twin comprising of three descriptive models representing the motorcycle itself, its operating environment, and the motorcycle riding behavior. Based on data gathered during a large-scale measurement campaign, novel insights in the motorcycle riding behavior have enabled its representation into two mathematical formulations. Highlighting the capability of the digital twin to integrate data from heterogeneous sources, the environmental model is generated using geospatial data from a map provider, followed by a novel formulation of a safety driving line. Using a kinematic motorcycle model, the speed and banking angle over the defined route are predicted with high correlation to real-world motorcycle riding behavior. The insights generated by the developed digital twin can be used to enable data-driven development or as an input to provide individualized predictive services during the usage phase.
### Standards for passenger comfort in automated vehicles: Acceleration and jerk
2023, Applied Ergonomics Citation Excerpt :
Implementation of comfortable algorithms requires a comprehensive understanding of how the experience of comfort relates to vehicle motion. Motion comfort has many facets (see e.g. (de Winkel et al., 2021; Edelmann et al., 2021; Will et al., 2021; Shyrokau et al., 2018; Mirakhorlo et al., 2022),). The presence of low frequency motions endured over a long period of time can lead to motion sickness (Irmak et al., 2021a). Show abstract A prime concern for automated vehicles is motion comfort, as an uncomfortable ride may reduce acceptance of the technology amongst the general population. However, it is not clear how transient motions typical for travelling by car affect the experience of comfort. Here, we determine the relation between properties of vehicle motions (i.e., acceleration and jerk) and discomfort empirically, and we evaluate the ability of normative models to account for the data. 23 participants were placed in a moving-base driving simulator and presented sinusoidial and triangular motion pulses with various peak accelerations (A max 0.4−2 ms−2) and jerks (J max 0.5−15 ms−3), designed to recreate typical vehicle accelerations. Participants provided discomfort judgments on absolute ‘Verbal Qualifiers’ and relative ‘Magnitude Estimates’ associated with these motions. The data show that discomfort increases with acceleration amplitude, and that the strength of this effect depends on the direction of motion. We furthermore find that higher jerks (shorter duration pulses) are considered more comfortable, and that triangular pulses are more comfortable than sinusoidal pulses. ME responses decrease (i.e., reduced discomfort) with increasing pulse duration. Evaluations of normative models of vibration and shock (ISO 2631), and perceived motion intensity provide mixed results. The vibration model could not account for the data well. Reasonable agreement between predictions and observations were found for the shock model and perceived intensity model, which emphasize the role of acceleration. We present novel statistical models that describe motion comfort as a function of acceleration, jerk, and direction. The present findings are essential to develop motion planning algorithms aimed at maximizing comfort.
### Leisure Motivation and Happiness, Mediation of Leisure Attitude and Perceived Value: An Evidence from Large and Heavy Motorbike Riders in Taiwan
2023, Annals of Applied Sport Science
### A SYSTEMATIC REVIEW ON MOTORCYCLISTS’ AGGRESSIVE BEHAVIOR ANALYSIS USING COMPUTATIONAL MODELS: CURRENT STATE, CHALLENGES, AND RECOMMENDATIONS
2022, Journal of Theoretical and Applied Information Technology
### A Comprehensive Review on the Behaviour of Motorcyclists: Motivations, Issues, Challenges, Substantial Analysis and Recommendations
2022, International Journal of Environmental Research and Public Health
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13626 | https://www.greenemath.com/Prealgebra/23/PEMDASLesson.html | GreeneMath.com Master Math & Ace Your Exam!
PEMDAS - Order of Operations Lesson
Lesson Objectives
Demonstrate an understanding of operations with integers
Demonstrate an understanding of the order of operations
Learn how to use the order of operations for problems with integers
Order of Operations with Integers
In our lesson on the order of operations, we learned the correct priority or order for each operation, when faced with multiple operations in a problem. Let’s review the order of operations:
Work inside of any parentheses or grouping symbols
Perform all exponent operations
Multiply or divide (working left to right)
Add or subtract (working left to right)
Recall that we can also use the acronym PEMDAS to remember our steps. We must be very careful with this acronym; multiplication and division have the same priority level. These operations are evaluated working left to right. The same is true with addition and subtraction. They also have the same priority level and are evaluated working from left to right.
PEMDAS
P - Parentheses or grouping symbols
E - Exponents
M/D - Multiply/Divide (working left to right)
A/S - Addition/Subtraction (working left to right)
In this lesson, we will look at some typical problems that involve the use of integers. These problems can be slightly more difficult, due to the various sign rules that come into play. Let’s take a look at a few examples. Example 1: Evaluate each: (-9 - (-3) x 7) ÷ |-32 + 5|
First, we look for parentheses or grouping symbols, once inside we will reapply the order of operations.
In this problem, we have parentheses and absolute value bars. These are both grouping symbols. We will start by working inside of each.
(-9 - (-3) x 7) ÷ |-32 + 5|
Inside of our parentheses, we have subtraction and multiplication. Multiplication has a higher priority. We will perform (-3) x 7 first.
(-3) x 7 = -21, let's replace this in our problem.
(-9 - (-21)) ÷ |-32 + 5|
Continuing inside of the parentheses, we only have subtraction. We will perform -9 - (-21) next.
-9 - (-21) = 12, let's replace this in our problem.
12 ÷ |-32 + 5|
Next, we look at our other grouping symbol, the absolute value operation.
12 ÷ |-32 + 5|
Inside of the absolute value bars, we have an exponent operation and addition. The exponent operation has the highest priority. We will perform -32 first.
-32 = -9, let's replace this in our problem.
12 ÷ |-9 + 5|
Continuing inside of the absolute value bars, we only have addition. We will perform -9 + 5 next.
-9 + 5 = -4, let's replace this in our problem.
12 ÷ |-4|
Let's now take the absolute value of -4, this will give us +4. Let's replace this in our problem.
12 ÷ 4
Last, we have one single division operation left, let's divide 12 by 4 and get our final answer of 3.
12 ÷ 4 = 3
(-9 - (-3) x 7) ÷ |-32 + 5| = 3 Example 2: Evaluate each: [(25 - 24 ÷ 22) ÷ |-5 x 3 + 8|] ÷ |-52 + 7 x 5 - 6|
First, we look for parentheses or grouping symbols, once inside we will reapply the order of operations.
In this problem, we have brackets "[]", parentheses, and absolute value bars. These are all grouping symbols. We will start by working inside of each.
Let's start by working inside the brackets. Once inside we will reapply our order of operations.
[(25 - 24 ÷ 22) ÷ |-5 x 3 + 8|] ÷ |-52 + 7 x 5 - 6|
Inside of the brackets, we have parentheses and absolute value bars. We will start with the parentheses.
Inside of parentheses, we have 25 - 24 ÷ 22. Here exponents are a higher priority than subtraction or division. Let's start by performing all exponent operations.
25 = 32, 24 = 16, 22 = 4, let's replace these in our problem.
[(32 - 16 ÷ 4) ÷ |-5 x 3 + 8|] ÷ |-52 + 7 x 5 - 6|
Now we have subtraction and division, we will divide before we subtract. We will perform 16 ÷ 4 next.
16 ÷ 4 = 4, let's replace this in our problem.
[(32 - 4) ÷ |-5 x 3 + 8|] ÷ |-52 + 7 x 5 - 6|
Now we have only subtraction. We will perform 32 - 4 next.
32 - 4 = 28, let's replace this in our problem.
[28 ÷ |-5 x 3 + 8|] ÷ |-52 + 7 x 5 - 6|
Continuing inside of our brackets, we have an absolute value operation. Inside, we have multiplication and addition. We will multiply before we add. We will perform -5 x 3 next.
-5 x 3 = -15, let's replace this in our problem.
[28 ÷ |-15 + 8|] ÷ |-52 + 7 x 5 - 6|
Now we have only addition inside of the absolute value bars, we will perform -15 + 8 next.
-15 + 8 = -7, let's replace this in our problem.
[28 ÷ |-7|] ÷ |-52 + 7 x 5 - 6|
Now we can take the absolute value of -7, which is 7. Let's replace this in our problem.
[28 ÷ 7] ÷ |-52 + 7 x 5 - 6|
Lastly, we will divide 28 by 7.
28 ÷ 7 = 4, let's replace this in our problem.
4 ÷ |-52 + 7 x 5 - 6|
Now we will simplify inside of the absolute value operation on the right.
4 ÷ |-52 + 7 x 5 - 6|
Inside of the absolute value operation, we have an exponent operation, multiplication, addition, and subtraction. We will begin with our exponent operation -52.
-52 = -25, let's replace this in our problem.
4 ÷ |-25 + 7 x 5 - 6|
Continuing inside of the absolute value bars, we have addition, multiplication, and subtraction. We will multiply 7 x 5 next.
7 x 5 = 35, let's replace this in our problem.
4 ÷ |-25 + 35 - 6|
Continuing inside of the absolute value bars, we have addition and subtraction. We perform addition and subtraction working left to right. Since the addition operation is the leftmost, we will perform -25 + 35 next.
-25 + 35 = 10, let's replace this in our problem.
4 ÷ |10 - 6|
Now we only have subtraction left inside of the absolute value bars. We will perform 10 - 6 as our next operation.
10 - 6 = 4
4 ÷ |4|
Now we will take the absolute value of 4, which is 4. Let's replace this in our problem.
4 ÷ 4
Our last operation will be the division of 4 by 4, which will give us our final answer of 1.
4 ÷ 4 = 1
[(25 - 24 ÷ 22) ÷ |-5 x 3 + 8|] ÷ |-52 + 7 x 5 - 6| = 1
Skills Check:
Example #1
Evaluate each.
|-41| + (92 x (-44) ÷ 3) ÷ (-2)
Please choose the best answer.
A
-247
B
635
C
327
D
689
E
-485
Example #2
Evaluate each.
[(44 - (-6,488)) ÷ (-92)] x (-6 - (-9)) + (-5)3
Please choose the best answer.
A
-338
B
500
C
-600
D
-412
E
234
Example #3
Evaluate each.
[(13 - (-348)) ÷ (-19)] x (17 - (-7)) + (-1)2
Please choose the best answer.
A
480
B
525
C
-350
D
-455
E
-300
Congrats, Your Score is 100%
Better Luck Next Time, Your Score is %
Try again?
Ready for more?Watch the Step by Step Video Lesson | Take the Practice Test |
13627 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Barrus_and_Clark)/01%3A_Chapters/1.26%3A_Computation_of_a_mod_m | Skip to main content
1.26: Computation of aⁿ mod m
Last updated
: Jan 22, 2022
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1.25: Primality Tests
1.27: The RSA Scheme
Page ID
: 83362
Mike Barrus & W. Edwin Clark
University of Rhode Island
( \newcommand{\kernel}{\mathrm{null}\,})
Let’s first consider the question: What is the smallest number of multiplications required to compute where is any positive integer?
Suppose we want to calculate . One way is to perform the following multiplications: But we can do it in only multiplications: In general, using the first method’s steps requires multiplications to compute .
On the other hand, if then we can compute by successive squaring with only multiplications: Note that the fact that together with the exponent laws and is what makes this method work. Note that if then is generally a lot smaller than . For example, and is quite a bit smaller than .
If is not a power of we can use the following method to compute .
The Binary Method for Exponentiation
Let be a positive integer. Let be any real number. This is a method for computing .
Step 1. Find the binary representation for .
Step 2. Compute the powers by successive squaring as shown above.
Step 3. Compute the product [Note each is or , so all needed factors were obtained in Step 2.]
Example
Let’s compute . Note that . So this takes care of Step . For Step , we note that So . For this we need multiplications: So we have Note that we have used just multiplications, which is less than the it would take if we used the naive method. Let’s not forget that some additional effort was needed to compute the binary representation of , but not much.
Theorem
Computing using the binary method requires applications of the Division Algorithm and at most multiplications.
Proof
If , , then . Hence Since and when we have , the inequalities above yield or Hence . Note that is the number of times we need to apply the Division Algorithm to obtain the binary representation , . To compute the powers by successive squaring requires multiplications and similarly to compute the product requires multiplicatons. So after obtaining the binary representation we need at most multiplications.
Use of a calculator to compute
To find one may use the formula or where is the natural logarithm of . For small values of it is sometimes faster to use the fact that is equivalent to that is, is the largest positive integer such that .
Note that if we count an application of the Division Algorithm and a multiplication as the same, the above tells us that we need at most operations to compute . So, for example, if , then it is easy to see that . So we may compute with only operations.
Computing
We use the binary method for exponentiation with the added trick that after every multiplication we reduce modulo , that is, we divide by and take the remainder. This keeps the products from getting too big.
Example
We compute : Note that implies that . (Recall that we calculated that which is clearly congruent to , but the multiplications were not so easy.)
Example
Let’s find . It is easy to see that That is, . Now by successive squaring and reducing modulo we get Now hence So and so we have . Hence .
Exercises
Exercise
Calculate for .
Exercise
Use the binary method to compute .
Exercise
Approximately how many operations would be required to compute when ? Explain.
Exercise
Note that multiplications are used to compute using the binary method. Show that one can compute with fewer than multiplications. (You will have to experiment.)
Exercise
Calculate .
Exercise
Calculate .
Exercise
If you multiplied out , how many decimal digits would you obtain? (See Exercise 1.6.9.)
Exercise
Note that we calculated with very few multiplications. Why can we not use that method to compute ?
1.25: Primality Tests
1.27: The RSA Scheme |
13628 | https://maths.qmul.ac.uk/~omj/bal3.pdf | BALANCED WORDS AND MAJORIZATION OLIVER JENKINSON Abstract. When expressed in terms of base-2 expanions, balanced words are majorized by other words of the same slope.
Consequently they have smaller standard deviation and larger geometric mean than all words with given arith-metic mean (or slope), they can be expressed as a doubly stochastic average of any such word, and they can be derived from any such word by a finite number of transfers.
1. Introduction This article is concerned with words w = w1w2 . . . wm ∈{0, 1}+ := [ q≥1 {0, 1}q , and their base-2 expansions E2(w) = m X k=1 wk2m−k .
Write |w| = m, the length of w, and |w|1 = card({1 ≤i ≤m : wi = 1}) its 1-length.
Define the cyclic shift σ : {0, 1}m →{0, 1}m by σ(w1 . . . wm) = w2 . . . wmw1. A cyclic subword of w is any length-q prefix of some σi−1(w), 1 ≤i, q ≤m.
To any word w = w1 . . . wm we associate its orbit O(w), the vector O(w) = (O1(w), . . . , Om(w)) consisting of the iterated cyclic shifts w, σ(w), . . . , σm−1(w) arranged in lexicographic order1 O1(w) ≤O2(w) ≤. . . ≤Om(w) .
(1) Define the base-2 orbit I(w) = (I1(w), . . . , Im(w)) , where Ii(w) := E2(Oi(w)) for 1 ≤i ≤m .
The author was supported by an EPSRC Advanced Research Fellowship.
1The lexicographic order on {0, 1}m is defined by: w < w′ if there exists j ∈{1, . . . , m} with wk = w′ k for all k = 1, . . . , j −1, and wj < w′ j, and w ≤w′ if either w < w′ or w = w′.
1 2 OLIVER JENKINSON We shall be interested in certain quantities depending on the base-2 orbit I(w), for example the product P(w) := q Y i=1 Ii(w) .
For all words w with a fixed length and 1-length, we shall compare the products P(w); it turns out (see Theorem 1.2 below) that P(w) is as large as possible when w is balanced. By a balanced word2 we mean one with the property that, if z and z′ are cyclic subwords of w with |z| = |z′|, then ||z|1 −|z′|1| ≤1. Clearly, w is balanced if and only if each of its cyclic shift iterates is balanced.
Example 1.1. The word 0010101 is balanced, and P(0010101) = 21 × 37 × 41 × 42 × 74 × 82 × 84 = 681991597728 .
The unbalanced length-7 words with 1-length equal to 3 are 0000111, 0001011, 0001101, and 0010011, together with their cyclic shift iterates.
The correspond-ing products are P(0000111) = 7 × 14 × 28 × 56 × 67 × 97 × 112 = 111850181632 , P(0001011) = 11 × 22 × 44 × 49 × 69 × 88 × 98 = 310471658112 , P(0001101) = 13 × 26 × 35 × 52 × 70 × 81 × 104 = 362747548800 , P(0010011) = 19 × 25 × 38 × 50 × 73 × 76 × 100 = 500707000000 , each of which is smaller than P(0010101).
More generally, if 1 ≤p < q are coprime integers, let Wp,q denote the set of all length-q words whose 1-length equals p. There are precisely q balanced words in Wp,q, all of which are in the same orbit (see e.g. [9, 24]), so if Wp,q is defined to be the set of all orbits of words in Wp,q, there is a unique balanced orbit in Wp,q. Since I(w) only depends on the orbit of w, it will be notationally convenient to represent orbits by one of their components; for consistency we shall always use O1(w), the lexicographically smallest component.
So for example the orbit (00101, 01001, 01010, 10010, 10100) will be represented notationally by 00101.
Theorem 1.2. Suppose 1 ≤p < q are coprime integers. For w ∈Wp,q, the product P(w) = Qq i=1 Ii(w) is maximized precisely when w is balanced.
Example 1.3. The simplest non-trivial instance of Theorem 1.2 is when (p, q) = (2, 5), in which case W2,5 = {00011, 00101}, with 00101 the balanced orbit, and P(00011) = 3×6×12×17×24 = 88128 < 162000 = 5×9×10×18×20 = P(00101) .
More generally, the balanced orbit in Wp,q can be compared to orbits of words w whose slope sl(w) := |w|1/|w| is equal to p/q. If w has length Q (necessarily a multiple of q) then any length-Q balanced word is the (Q/q)-fold concatenation3 2Finite balanced words, defined in , constitute a subclass of the collection of Sturmian words (see e.g. [1, 4, 32] for more on Sturmian words, and on infinite balanced words).
3In general the k-fold concatenation of a word w = w1 . . . wm is the length-km word wk whose i-th entry is wi (mod 1).
BALANCED WORDS AND MAJORIZATION 3 bQ/q of a balanced word b ∈Wp,q. In particular, there is a unique length-Q balanced orbit of slope p/q. The following result is a generalisation of Theorem 1.2: Theorem 1.4. If the unbalanced word w ∈{0, 1}Q has slope p/q, where 1 ≤p < q are coprime integers, and b ∈Wp,q is balanced, then P(w) < P(bQ/q).
Example 1.5. The words 0011 and 001011 both have slope 1/2, as does the balanced word 01. The product P(0011) = 3 × 6 × 9 × 12 = 1944 is smaller than P(0101) = 5 × 5 × 10 × 10 = 2500 , while P(001011) = 11 × 22 × 25 × 37 × 44 × 50 = 492470000 is smaller than P(010101) = 21 × 21 × 21 × 42 × 42 × 42 = 686128968 .
An equivalent way of comparing words with the same slope but different length is to associate the vector J (w) := (J1(w), . . . , J|w|(w)) to w, where Ji(w) := Ii(w) 2|w| −1 for 1 ≤i ≤|w| , and consider the corresponding geometric mean GM(w) := |w| Y i=1 Ji(w) 1/|w| = Q|w| i=1 Ii(w) 1/|w| 2|w| −1 .
A short calculation shows that the slope of a word w, a priori a purely symbolic notion, is actually equal to the arithmetic mean of the Ji(w): sl(w) = 1 |w| |w| X i=1 Ji(w) =: AM(w) .
So Theorem 1.4 can be interpreted as relating arithmetic and geometric means: Theorem 1.6. For all orbits of a given arithmetic mean, the geometric mean is maximized precisely when the orbit is balanced.
Example 1.7. The words 01, 0011, 000111, 001011, and 001101 all have slope 1/2, and the corresponding geometric means are GM(01) = (1 × 2)1/2 3 = √ 2 3 ≈0 · 471 , GM(0011) = (3 × 6 × 9 × 12)1/4 15 ≈0 · 442 , 4 OLIVER JENKINSON GM(000111) = (7 × 14 × 28 × 35 × 49 × 56)1/6 63 ≈0 · 401 , GM(001011) = (11 × 22 × 25 × 37 × 44 × 50)1/6 63 ≈0 · 446 , GM(001101) = (13 × 19 × 26 × 38 × 41 × 52)1/6 63 ≈0 · 450 .
Another quantity of interest is the standard deviation of an orbit around its arith-metic mean, defined as SD(w) = v u u t 1 |w| |w| X i=1 Ji(w) −AM(w) 2 .
(2) It turns out that standard deviation is minimized by balanced orbits: Theorem 1.8. For all orbits of a given slope, the standard deviation is minimized precisely when the orbit is balanced.
Example 1.9. The unique balanced orbit of slope 1/2 is 01, with standard deviation SD(01) = 1/6 ≈0 · 166 , while standard deviations of some other orbits with slope 1/2 are: SD(0011) = (20)−1/2 ≈0 · 223 , SD(000111) = 5/18 ≈0 · 277 , SD(001011) = SD(001101) = (103/7)1/2/18 ≈0 · 213 .
Although it had previously been known that balanced words are characterised by various “extremal” properties4, the above results in terms of base-2 expansions are of a rather different nature. In fact they are all consequences of a more general result, Theorem 2.4, which is phrased in terms of majorization, and asserts that each Wp,q is a poset whose minimal element is the balanced orbit. Majorization is a notion which is pervasive in various branches of mathematics5, and can itself be formulated in a number of different ways, each formulation leading to a different characterisation of balanced words which will be described in this paper. In order, these characterisations are in terms of: partial sums of the Ii(w) (§2), symmetric mean value inequalities (§3), doubly stochastic matrices (§4), and transfers (§5).
4Notably, their subword complexity is as small as possible: a length-q word has at most n + 1 length-n cyclic subwords for each 1 ≤n ≤q −1 if and only if it is balanced.
5The monograph by Marshall & Olkin describes many applications of majorization, e.g. to inequalities relating eigenvalues and singular values of linear operators [5, Ch. II], [15, Ch. VI], [27, Ch. 9], and to aspects of graph theory [27, Ch. 7]. Majorization is applied extensively in probability and statistics (see e.g. [27, Ch. 11–13], , [37, Ch. 6]), for example in comparison of experiments [6, 7], sampling theory [27, Ch. 12.A], hypothesis testing [27, Ch. 13.A], and reliability theory [27, Ch. 13.D], . It is used as a measurement of inequality and diversity in a variety of contexts, for example in economics, to describe income distribution (if µ ≺ν then the distribution ν is more equitable, see [11, 12]).
BALANCED WORDS AND MAJORIZATION 5 2. Majorization and partial sum inequalities 00000111 00001101 00001011 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 00010011 00011001 444444444444444444444444 00010101 00100101 Figure 1. Hasse diagram for the poset (W3,8, ≺) The partial order of majorization (see e.g. for a general reference) is defined as follows: Definition 2.1. Given real numbers x1 ≤. . . ≤xQ and y1 ≤. . . ≤yQ, the vector y = (y1, . . . , yQ) is said to majorize x = (x1, . . . , xQ), denoted x ≺y, if i X k=1 xi ≥ i X k=1 yi for 1 ≤i ≤Q −1 , (3) and Q X k=1 xi = Q X k=1 yi .
(4) We shall be interested in restricting majorization to vectors of the form I(w), for words w ∈{0, 1}+. We write w ≺w′ (respectively O(w) ≺O(w′)), and say that w′ majorizes w (respectively O(w′) majorizes O(w)), whenever I(w) ≺I(w′). This defines a partial order on the set of orbits of words in {0, 1}+. Necessary conditions for two orbits to be comparable are that they have the same length and the same 6 OLIVER JENKINSON 1-length: the first condition is obvious, while the second follows from (4), because if w ∈{0, 1}Q then PQ i=1 Ii(w) = (2Q −1)|w|1. In particular, each (Wp,q, ≺) is a poset. For example (W3,8, ≺) is represented by the Hasse diagram in Figure 1.
Example 2.2. The computations below prove that the relations in (W3,8, ≺) are as in Figure 1; here the partial sums are denoted by Si(w) := i X k=1 Ik(w) .
Oi Ii Si 00100101 37 37 00101001 41 78 01001001 73 151 01001010 74 225 01010010 82 307 10010010 146 453 10010100 148 601 10100100 164 765 Oi Ii Si 00011001 25 25 00101001 35 60 00110010 50 110 01000110 70 180 01100100 100 280 10001100 140 420 10010001 145 565 11001000 200 765 Oi Ii Si 00010101 21 21 00101010 42 63 01000101 69 132 01010001 81 213 01010100 84 297 10001010 138 435 10100010 162 597 10101000 168 765 Oi Ii Si 00010011 19 19 00100110 38 57 00110001 49 106 01001100 76 182 01100010 98 280 10001001 137 417 10011000 152 569 11000100 196 765 Oi Ii Si 00001101 13 13 00011010 26 39 00110100 52 91 01000011 67 158 01100001 104 262 10000101 134 396 10110000 161 557 11000010 208 765 Oi Ii Si 00001011 11 11 00010110 22 33 00101100 44 77 01011000 88 165 01100001 97 262 10000101 133 395 10110000 176 571 11000010 194 765 Oi Ii Si 00000111 7 7 00001110 14 21 00011100 28 49 00111000 56 105 01110000 112 217 10000011 131 348 11000001 193 541 11100000 224 765 In particular, the balanced orbit 00100101 is seen to be the least element in (W3,8, ≺).
The extremal property of balanced orbits seen in Example 2.2 is quite general: BALANCED WORDS AND MAJORIZATION 7 Theorem 2.3. For any coprime integers 1 ≤p < q, the unique balanced orbit b ∈Wp,q is the least element in (Wp,q, ≺). In other words, for any w ∈Wp,q, Si(b) ≥Si(w) for all 1 ≤i ≤q .
(5) As in §1, it turns out that balanced orbits are actually extremal among all orbits of a given slope: Theorem 2.4. Suppose w ∈{0, 1}Q has slope p/q, where 1 ≤p < q are coprime integers. If b ∈Wp,q is balanced then bQ/q ≺w. In other words, Si(bQ/q) ≥Si(w) for all 1 ≤i ≤Q .
(6) Proof. The majorization bQ/q ≺w means precisely that I(bQ/q) ≺I(w), and by a classical result of Hardy, Littlewood & P´ olya (see [17, 18]) this is equivalent to Q X i=1 f(Ii(bQ/q)) ≤ Q X i=1 f(Ii(w)) for all convex functions f : R →R .
Since Ji(w) = Ii(w)/(2|w| −1), this is equivalent to Q X i=1 f(Ji(bQ/q)) ≤ Q X i=1 f(Ji(w)) for all convex functions f : (0, 1) →R , (7) hence equivalent to 1 q q X i=1 f(Ji(b)) ≤1 Q Q X i=1 f(Ji(w)) (8) for all convex functions f : (0, 1) →R, and by a simple approximation argument we may take f to be smooth.
Now, for an arbitrary smooth convex function f : (0, 1) →R, we shall establish (8).
If we define a(x) = (x + 1)/2 for x ∈(0, 2J1(b)], and a(x) = x/2 for x ∈ (2J1(b), 1), and set g(x) := f(x) −x Z 1 0 X n≥0 f ′(an(t)) 2n dt , then for w ∈{0, 1}Q with slope p/q, the inequality (8) is clearly equivalent to 1 q q X i=1 g(Ji(b)) ≤1 Q Q X i=1 g(Ji(w)) .
(9) In fact we claim that (9) holds for any w ∈{0, 1}+.
To prove this, let h be the indefinite integral (unique up to an additive constant) of the bounded function t 7→P n≥0 g′(an(t))/2n. The function h is Lipschitz, and if G := g +h−h◦T, where T(x) := 2x (mod 1), then G′ = 0 Lebesgue almost everywhere on [J1(b), J1(b) + 1/2].
Therefore G is constant on [J1(b), J1(b) + 1/2], and this constant, c say, 8 OLIVER JENKINSON is its minimum value (cf. [22, 23]). It follows that 1 q Pq i=1 G(Ji(b)) = c, because Jq(b) −J1(b) < 1/2 (see [9, Prop. 1]), and hence that 1 q q X i=1 G(Ji(b)) = c ≤1 Q Q X i=1 G(Ji(w)) .
(10) However, clearly 1 |v| P|v| i=1 G(Ji(v)) = 1 |v| P|v| i=1 g(Ji(v)) for all v ∈{0, 1}+, so (10) implies that (9) holds for all w ∈{0, 1}+, as required.
□ Example 2.5. In the case of slope 1/2, the calculations below show that 01010101 is majorized by both 00110011 and 00101101, but that 00110011 and 00101101 are unrelated.
Oi Ii Si 01010101 85 85 01010101 85 170 01010101 85 255 01010101 85 340 10101010 170 510 10101010 170 680 10101010 170 850 10101010 170 1020 Oi Ii Si 00110011 51 51 00110011 51 102 01100110 102 204 01100110 102 306 10011001 153 459 10011001 153 612 11001100 204 816 11001100 204 1020 Oi Ii Si 00101101 45 45 01001011 75 120 01011010 90 210 01101001 105 315 10010110 150 465 10100101 165 630 10110100 180 810 11010010 210 1020 Corollary 2.6. Suppose w ∈{0, 1}Q has slope p/q, where 1 ≤p < q are coprime integers, that w is not balanced, and that b ∈Wp,q is balanced. Then O1(w) < O1(bQ/q) and OQ(w) > OQ(bQ/q) .
In other words, the lexicographically smallest component of O(w) is strictly smaller, in the lexicographic order, than the lexicographically smallest component of O(bQ/q), and the lexicographically largest component of O(w) is strictly larger, in the lexico-graphic order, than the lexicographically largest component of O(bQ/q).
Proof. Since w is not balanced, Oi(w) ̸= Oi(bQ/q), hence Ii(w) ̸= Ii(bQ/q) for all 1 ≤i ≤Q. In particular, I1(w) ̸= I1(bQ/q) and IQ(w) ̸= IQ(bQ/q) .
(11) Setting i = 1 and i = Q −1 in (6) gives I1(w) ≤I1(bQ/q) and IQ(w) ≥IQ(bQ/q) , so by (11) these inequalities are strict. This corresponds to strict inequality, in the lexicographic order, between the corresponding words, as required.
□ Remark 2.7. Corollary 2.6 was originally proved, via entirely different methods, by Bernhardt ; an alternative proof is due to Gambaudo, Lanford & Tresser .
In all other cases 2 ≤i ≤Q −2, the inequality (6) is new.
In fact it should be possible to replace (6) with strict inequalities (provided i ̸= Q): BALANCED WORDS AND MAJORIZATION 9 Conjecture 2.8. Under the hypotheses of Corollary 2.6, Si(bQ/q) > Si(w) for all 1 ≤i ≤Q −1 .
Remark 2.9. In general it is possible for partial sums of distinct v and w to be equal: for example we saw above that if v = 00001101 and w = 00001011 then S5(v) = 262 = S5(w).
We can now prove the theorems which were stated in §1: Proof of Theorems 1.2, 1.4, 1.6, 1.8. Suppose the non-balanced word w ∈{0, 1}Q has slope p/q, where 1 ≤p < q are coprime integers, and that b ∈Wp,q is balanced.
The weak inequality (7) implies the strict inequality Q X i=1 f(Ji(bQ/q)) < Q X i=1 f(Ji(w)) .
whenever f : [0, 1] →R is strictly convex.
In particular, the function f(x) = (x −p/q)2 is strictly convex, so Theorem 1.8 follows.
If, on the other hand, f : [0, 1] →R is strictly concave, then Q X i=1 f(Ji(bQ/q)) > Q X i=1 f(Ji(w)) .
In particular f(x) = log x is strictly concave, so Q X i=1 log Ji(bQ/q) > Q X i=1 log Ji(w) , in other words Q Y i=1 Ji(bQ/q) > Q Y i=1 Ji(w) .
From this we deduce Theorems 1.2 and 1.4, that the product P(w) = QQ i=1 Ii(w) is maximized precisely when w is balanced, as well as the formulation (Theorem 1.6) in terms of arithmetic and geometric means.
□ 3. Symmetric means and Schur-convexity The work of Hardy, Littlewood & P´ olya [17, 18] on majorization was inspired by a theorem of Muirhead on comparisons of so-called symmetric means. Before re-calling the general form of Muirhead’s theorem, we first state its consequences in our setting, when combined with Theorem 2.4. Let Pm denote the set of permutations of {1, . . . , m}.
Corollary 3.1. Suppose that 1 ≤p < q are coprime integers, and let b denote the unique balanced element in Wp,q. If w is any other element of Wp,q then X π∈Pq q Y i=1 aIi(b) π(i) ≤ X π∈Pq q Y i=1 aIi(w) π(i) (12) 10 OLIVER JENKINSON for all (a1, . . . , aq) ∈Rq +, with equality if and only if a1 = . . . = aq.
Example 3.2. The balanced orbit in W2,5 is b = 00101, with I(b) = (5, 9, 10, 18, 20), and the other member of W2,5 is w = 00011, with I(w) = (3, 6, 12, 17, 24). Corollary 3.1 implies that X π∈P5 a5 π(1)a9 π(2)a10 π(3)a18 π(4)a20 π(5) ≤ X π∈P5 a3 π(1)a6 π(2)a12 π(3)a17 π(4)a24 π(5) , for all (a1, . . . , a5) ∈R5 +, with equality if and only if a1 = . . . = a5.
As usual, there is a more general version which applies to all words w ∈{0, 1}+: Corollary 3.3. Suppose w ∈{0, 1}Q has slope p/q, where 1 ≤p < q are coprime integers. If b ∈Wp,q is balanced then X π∈PQ Q Y i=1 aIi(bQ/q) π(i) ≤ X π∈PQ Q Y i=1 aIi(w) π(i) (13) for all (a1, . . . , aQ) ∈RQ +, with equality if and only if a1 = . . . = aQ.
Example 3.4. 0011 has slope 1/2, with I(w) = (3, 6, 9, 12), and the balanced orbit in W1,2 is b = 01, with I(b2) = I(0101) = (5, 5, 10, 10). So Corollary 3.3 implies that X π∈P4 a5 π(1)a5 π(2)a10 π(3)a10 π(4) ≤ X π∈P4 a3 π(1)a6 π(2)a9 π(3)a12 π(4) , (14) for all (a1, a2, a3, a4) ∈R4 +, with equality if and only if a1 = . . . = a4.
The general form of Muirhead’s theorem (see , or [27, Ch. 3.G]) asserts that if (x1, . . . , xQ) ≺(y1, . . . , yQ), then X π∈PQ Q Y i=1 axi π(i) ≤ X π∈PQ Q Y i=1 ayi π(i) for all (a1, . . . , aQ) ∈RQ +, with equality if and only if a1 = . . . = aQ. A quantity of the form 1 Q!
X π∈PQ Q Y i=1 axi π(i) (15) (note the normalisation factor 1/Q! in (15)) is known as a symmetric mean6 (sym-metric in the variables a1, . . . , aQ). The best known examples of symmetric means are the arithmetic and geometric means: if (x1, . . . , xQ) = (0, . . . , 0, 1) then 1 Q!
X π∈PQ Q Y i=1 axi π(i) = a1 + . . . + aQ Q , (16) 6Sometimes this terminology is reserved for the case when x1 + . . . + xQ = 1.
BALANCED WORDS AND MAJORIZATION 11 while if (x1, . . . , xQ) = (Q−1, . . . , Q−1) then 1 Q!
X π∈PQ Q Y i=1 axi π(i) = (a1 · · · aQ)1/Q .
(17) The classical arithmetic-geometric mean inequality asserts that (16) is larger than (17) for all (a1, . . . , aQ) ∈Rq +, with equality if and only if a1 = . . . = aQ.
So Muirhead’s theorem can be seen as a generalisation of this well known inequality (note that clearly (1/Q, . . . , 1/Q) ≺(0, . . . , 0, 1)).
In fact Muirhead’s theorem can nowadays be viewed as a special consequence of Schur-convexity (named after the pioneering work of Schur ). A real-valued function ϕ defined on a subset A of RQ is called Schur-convex if ϕ(x) ≤ϕ(y) when-ever x, y ∈A are such that x ≺y. In other words, the Schur-convex functions are precisely those which are isotonic with respect to the partial order of majorization.
One of the simplest constructions of a Schur-convex function ϕ (on [0, 1]Q, say) is to choose any convex function f : [0, 1] →R, and define ϕ(x1, . . . , xQ) = PQ i=1 f(xi) (cf. the Hardy-Littlewood-P´ olya theorem mentioned in §2).
Another well-known sufficient condition for a function ϕ to be Schur-convex is that it be symmetric and convex (see e.g. [27, Prop. 3.C.2]). Indeed for such a ϕ, if t1, . . . , tQ ∈R then the function (x1, . . . , xQ) 7→ X π∈PQ ϕ(tπ(1)x1, . . . , tπ(Q)xQ) (18) is also Schur-convex [27, p. 86], and the particular choices ti = log ai and ϕ(z1, . . . , zQ) = exp Q X i=1 zi yield X π∈PQ ϕ(tπ(1)x1, . . . , tπ(Q)xQ) = X π∈PQ Q Y i=1 axi π(i) , which is precisely the quantity treated in Muirhead’s theorem.
In order to generalise Corollaries 3.1 and 3.3, our Theorem 2.3 may be combined with various functions which are known to be Schur-convex (see e.g. for examples of such functions) in order to derive new characterisations of balanced words.
4. Doubly stochastic averages A square matrix with non-negative entries is called doubly stochastic if each of its row and column sums is equal to one. Given vectors x, y ∈RQ, we say that x is a doubly stochastic average of y if x = A y for some doubly stochastic matrix A.
Given v, w ∈{0, 1}+, we say that v (respectively O(v)) is a doubly stochastic average of w (respectively O(w)) if I(vk) is a doubly stochastic average of I(wl) for some k, l ≥1.
12 OLIVER JENKINSON If x is a doubly stochastic average of y, with doubly stochastic matrix A such that xi = P j A(i, j)yj for each i, then X i f(xi) = X i f X j A(i, j)yj ≤ X i X j A(i, j)f(yj) = X j f(yj) X i A(i, j) = X j f(yj) , (19) whenever f is convex, by Jensen’s inequality (which is applicable because P j A(i, j) = 1 for each i). This simple argument, which seems to have first been articulated by Schur , implies that x ≺y whenever x is a doubly stochastic average of y, by the Hardy-Littlewood-P´ olya characterisation of majorization used in the proof of The-orem 2.4. In fact Hardy, Littlewood & P´ olya [17, 18] (see also [27, 28, 35]) proved that the converse is true as well, so by Theorem 2.3 we have: Corollary 4.1. Suppose w ∈{0, 1}+ has slope p/q, where 1 ≤p < q are coprime integers. The unique balanced orbit in Wp,q is a doubly stochastic average of w.
Example 4.2. The unique balanced orbit of slope 1/2 is b = 01. Now 0011 has slope 1/2, with I(0011) = (3, 6, 9, 12), and I(b2) = I(0101) = (5, 5, 10, 10). Since 1/3 2/3 0 0 2/3 0 1/3 0 0 0 2/3 1/3 0 1/3 0 2/3 3 6 9 12 = 5 5 10 10 , the balanced orbit 01 is a doubly stochastic average of 0011.
The word 000111 also has slope 1/2, with I(000111) = (7, 14, 28, 35, 49, 56) , and I(b3) = I(010101) = (21, 21, 21, 42, 42, 42) .
Since 1 7 3 1 1 1 1 0 3 1 2 0 0 1 1 3 1 2 0 0 0 1 0 2 2 2 0 0 3 0 2 2 0 1 0 2 2 2 7 14 28 35 49 56 = 21 21 21 42 42 42 , the balanced orbit 01 is a doubly stochastic average of 000111.
BALANCED WORDS AND MAJORIZATION 13 5. Transfers If x ≺y then the doubly stochastic matrix A satisfying x = Ay is in general not unique; indeed little is known about the set of doubly stochastic matrices realising a given majorization (see [27, p. 40]). One means of constructing A is provided by the following notion: Definition 5.1. Given x ∈ZQ +, a transfer is a linear transformation T : RQ →RQ such that, for some 1 ≤i, j ≤Q with xj > xi, the vector x′ := T(x) satisfies x′ i = xi + 1, x′ j = xj −1, and x′ k = xk for k ̸= i, j.
In other words, a transfer evens up the components of a vector, by simultaneously decreasing by 1 the value of a “large” component and increasing by 1 a “small” component7; in they are described as Robin Hood transformations. Transfers were employed by Muirhead in the proof of his theorem (see §3), as well as by Hardy, Littlewood & P´ olya [17, 18], and Folkman & Fulkerson . Transfers are also of interest in economics, in the theory of income distribution (see e.g. ), their consideration in this context dating back to the work of Dalton (see also [26, 31]).
Explicitly, the linear transformation effecting the transfer in Definition 5.1 is given by the Q × Q (doubly stochastic) matrix A defined by A(i, i) = A(j, j) = 1 − 1 xj −xi , A(i, j) = A(j, i) = 1 xj −xi , and A(k, l) = δkl for k, l ∈{1, . . . , Q} \ {i, j} .
Muirhead (see also [27, p. 135]) proved that if x ≺y then x can be obtained from y by a (finite) number of transfers8. Therefore: Corollary 5.2. Let b denote the unique balanced orbit in Wp,q, where 1 ≤p < q are coprime integers. For any w ∈Wp,q, the base-2 orbit I(b) can be derived from I(w) by a finite number of transfers.
Example 5.3. the members of W2,5 are w = 00011 and b = 00101. The vector I(b) = (5, 9, 10, 18, 20) can be derived from I(w) = (3, 6, 12, 17, 24) by applying six transfers, for example as follows: (3, 6, 12, 17, 24) →(4, 6, 12, 17, 23) →(5, 6, 12, 17, 22) →(5, 7, 12, 17, 21) →(5, 7, 12, 18, 20) →(5, 8, 11, 18, 20) →(5, 9, 10, 18, 20).
7This notion of transfer can obviously be generalised: the amount transferred between the two components need not equal 1, and the vectors involved could have real rather than integer components.
8Clearly the converse also holds: a transfer is given by a doubly stochastic matrix, and a product of doubly stochastic matrices is again doubly stochastic, so Schur’s calculation (19), together with the Hardy-Littlewood-P´ olya criterion of §2, implies the majorization.
14 OLIVER JENKINSON The doubly stochastic matrices representing these transfers are, in order, 20 21 0 0 0 1 21 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 21 0 0 0 20 21 , 18 19 0 0 0 1 19 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 19 0 0 0 18 19 , 1 0 0 0 0 0 15 16 0 0 1 16 0 0 1 0 0 0 0 0 1 0 0 1 16 0 0 15 16 , 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 3/4 1/4 0 0 0 1/4 3/4 , 1 0 0 0 0 0 4/5 1/5 0 0 0 1/5 4/5 0 0 0 0 0 1 0 0 0 0 0 1 and 1 0 0 0 0 0 2/3 1/3 0 0 0 1/3 2/3 0 0 0 0 0 1 0 0 0 0 0 1 , and their product (in reverse order) is 19/21 0 0 0 2/21 1/280 9/16 2/5 0 19/560 1/420 3/8 3/5 0 19/840 5/224 1/64 0 3/4 95/448 15/224 3/64 0 1/4 285/448 , a doubly stochastic matrix which transforms I(w) to I(b).
Analogously, any orbit can be transformed, via a sequence of transfers, into the balanced orbit of the same slope: Corollary 5.4. Suppose w ∈{0, 1}Q has slope p/q, where 1 ≤p < q are coprime integers. If b ∈Wp,q is balanced then I(bQ/q) can be derived from I(w) by a finite number of transfers.
Remark 5.5. It would be useful to identify a class of operations w 7→w′ on words (or the corresponding orbits) which guarantee that w′ ≺w. Note that transfers are not of this type: if v ≺w we may derive I(v) from I(w) via a finite number of transfers, but the intermediate vectors in general have no symbolic interpretation, i.e. they are not the base-2 orbit I(u) for any word u.
A natural idea is to define an operation which exchanges the position of some (cyclically) adjacent symbols 0 and 1. For example 0011 can be balanced by ex-changing the middle letters, to get 0101, while 00011 can be balanced by swapping the 3rd and 4th letters, obtaining 00101. The word 000111 can be transformed to 001011 by swapping the middle two letters (note that 001011 ≺000111); then we swap first and last letters to obtain 101010, which is balanced. Of course these bal-ancing operations can be reversed, in which case their effect is to unbalance rather than balance; therefore criteria for recognising which adjacent symbols should be swapped are necessary. Moreover, if this notion is to bear fruit then some refine-ment is necessary, since there exist unbalanced words which cannot be transformed to the balanced word via a sequence of such operations. For example if w = 00101101 then each of the six possible exchanges of adjacent 0 and 1 yields a word which either majorizes w, or is incomparable to w.
BALANCED WORDS AND MAJORIZATION 15 6. Balanced sequences If the word w ∈{0, 1}+ is balanced, the (infinite) 0-1 sequence w∞= w · w · w · · · is a particular example of a balanced sequence, i.e. a 0-1 sequence such that ||z|1 − |z′|1| ≤1 for any two finite subwords z, z′ with |z| = |z′| (see e.g. [1, Chs. 9, 10], or [32, Ch. 6]).
If ω = (ωi)∞ i=1 is any balanced sequence then its slope limn→∞|ω1 . . . ωn|1/n is always defined. The sequence w∞has rational slope, but there exist balanced sequences with irrational slope, often called Sturmian sequences.9 Indeed if ϱ ∈(0, 1) is irrational then there are uncountably many balanced sequences with slope ϱ.
Although the results of this paper have natural analogues in the case of irrational slope, in order to make uniqueness statements it is more convenient to formulate these results in terms of balanced measures. For ϱ ∈[0, 1], the balanced measure of slope ϱ is a probability measure on the set of all 0-1 sequences, invariant under the left shift map, and giving full measure to the set of all balanced sequences of slope ϱ. We will not elaborate on all of the irrational slope versions of the results in this paper (though the interested reader should be able to re-construct them in conjunction with ), instead restricting attention to the following analogue of Theorem 1.8 (here by the slope of a general shift-invariant probability measure we mean the measure of the set of all 0-1 sequences ω whose first symbol ω1 equals 1): Theorem 6.1. For each ϱ ∈[0, 1], the balanced measure of slope ϱ has smaller standard deviation than any other shift-invariant probability measure of slope ϱ.
The standard deviation of the balanced measure of slope ϱ varies continuously with ϱ; this is depicted in Figure 2 in the range ϱ ∈[1/5, 1/2].
Clearly the smallest possible standard deviation of a balanced measure is 0: this occurs for the balanced measures of slopes 0 and 1, which are both concentrated on a single point. An open problem, however, is to determine the balanced measure10 with largest standard deviation.
9The terminology is unfortunately a little confused. Morse & Hedlund considered what they called Sturmian trajectories (in our setting the term trajectory means a 0-1 sequence), and these correspond precisely to what we are calling balanced sequences. So, following , it may seem more natural to talk of Sturmian sequences instead of balanced sequences, and indeed many authors adopt this convention (see e.g. [8, 9, 10, 21, 25, 34]). However, many other authors reserve the term Sturmian sequence to denote a balanced sequence with irrational slope. The reason for this seems to stem from an emphasis on the irrational case, both initially (Morse & Hedlund were interested in examples of recurrent behaviour which was not periodic, see also ), but also subsequently (various properties, for example the characterisation in terms of smallest subword complexity, are more conveniently formulated in the irrational case). Our use of the term balanced word throughout this article is consistent with , and differs from the notion of a Sturmian word (see for further discussion).
10Of course there are (at least) two balanced measures with largest standard deviation, by symmetry: the balanced measures of slope ϱ and 1 −ϱ have equal standard deviation. For this reason we shall henceforth restrict to ϱ ∈[0, 1/2].
16 OLIVER JENKINSON 0.25 0.3 0.35 0.4 0.45 0.5 0.165 0.1675 0.1725 0.175 0.1775 0.18 0.1825 0.185 Figure 2. Standard deviation of balanced measure of slope ϱ, for ϱ ∈[1/5, 1/2] Problem 6.2. Which balanced measure has largest standard deviation around its mean?
We have carried out numerical computations related to this problem. Namely, for a large number of rational values of ϱ we have computed the standard deviation (2) of the balanced measure (in this case an orbit) of slope ϱ. The rational values in question correspond to higher levels of the Farey tree (see e.g. ; the first four levels of the Farey tree are given below).
Level 1 0 1 1 1 2 0 1 1 2 1 1 3 0 1 1 3 1 2 2 3 1 1 4 0 1 1 4 1 3 2 5 1 2 3 5 2 3 1 4 1 1 More precisely, we computated standard deviations corresponding to the first 21 Farey levels; the rationals ϱ whose corresponding balanced orbit has largest standard deviation at each Farey level are recorded in the table below. For the first several levels, these maximizing rationals are 0 1, 1 2, 1 3, 2 5, 3 8, 5 13, i.e. continued fraction con-vergents to 1 −γ, where γ = 1 2( √ 5 −1) is the golden ratio11. However this pattern does not continue, and in particular it is not the case that the balanced measures of slope γ and 1 −γ have largest standard deviation amongst all balanced measures.
11By symmetry, the continued fraction convergents 1 1, 1 2, 2 3, 3 5, 5 8, 8 13 to γ also correspond to largest standard deviation on these levels.
BALANCED WORDS AND MAJORIZATION 17 Level ϱ Standard deviation 1 0 0 · 0000000000000000 2 1/2 0 · 1666666666666667 3 1/3 0 · 1781741612749496 4 2/5 0 · 1831621879558502 5 3/8 0 · 1836338891075342 6–9 5/13 0 · 1838465249738321 10 23/60 0 · 1838472388787628 11 28/73 0 · 1838477660379445 12 33/86 0 · 1838479659150774 13–14 38/99 0 · 1838480248997053 15 81/211 0 · 1838480274162965 16–20 119/310 0 · 1838480279673947 21 676/1761 0 · 1838480279684007 The above computations suggest that a maximum standard deviation, slightly larger than 0 · 1838480279684, is attained when ϱ is close to 676/1761 ≈0 · 383873 (and, by symmetry, when ϱ is close to 1085/1761).
Remark 6.3. Hajek has identified a class of functions, which he called multi-modular, with the property that the balanced measure of slope ϱ has smaller ergodic average than other shift-invariant measures of slope ϱ. While the results in this paper are similar in spirit to , their substance is very different: multimodular functions are simple functions defined on an abstract shift space in terms of multi-dimensional convex functions, whereas our convex functions are one-dimensional, and our results depend on base-2 expansions rather than on purely symbolic prop-erties. In particular, the results of this paper cannot be deduced from those of , nor the various multimodular generalisations (see e.g. ).
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S. Bullett and P. Sentenac, Ordered orbits of the shift, square roots, and the devil’s staircase, Math. Proc. Camb. Phil. Soc., 115 (1994), 451–481.
18 OLIVER JENKINSON E. M. Coven & G. A. Hedlund, Sequences with minimal block growth, Math. Systems Theory, 7 (1973), 138–153.
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H. Dalton, The measurement of the inequality of incomes, Econom. J., 30 (1920), 348–361.
J. H. Folkman & D. R. Fulkerson, Edge colorings in bipartite graphs, in Combinatorial math-ematics and its applications (R. C. Bose & T. A. Dowling, eds.), Chapter 31, pp. 561–577, Univ. of North Carolina Press, Chapel Hill, 1969.
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I. Gohberg, S. Goldberg and M. A. Kaashoek, Classes of linear operators vol. 1, Birkh¨ auser, Berlin, 1990 B. Hajek, Extremal splittings of point processes, Math. of Operations Research, 10 (1985), 543–556.
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BALANCED WORDS AND MAJORIZATION 19 School of Mathematical Sciences, Queen Mary, University of London, Mile End Road, London, E1 4NS, UK.
E-mail address: omj@maths.qmul.ac.uk |
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3.1: Rectangular Coordinate System
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3: Graphing Lines
3.2: Graph by Plotting Points
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Learning Objectives
Plot points using the rectangular coordinate system.
Calculate the distance between any two points in the rectangular coordinate plane.
Determine the midpoint between any two points.
Rectangular Coordinate System
The rectangular coordinate system consists of two real number lines that intersect at a right angle. The horizontal number line is called the xx-axis, and the vertical number line is called the yy-axis. These two number lines define a flat surface called a plane, and each point on this plane is associated with an ordered pair of real numbers (x,y)(x,y). The first number is called the x-coordinate, and the second number is called the y-coordinate. The intersection of the two axes is known as the origin, which corresponds to the point (0,0).
An ordered pair (x,y) represents the position of a point relative to the origin. The x-coordinate represents a position to the right of the origin if it is positive and to the left of the origin if it is negative. The y-coordinate represents a position above the origin if it is positive and below the origin if it is negative. Using this system, every position (point) in the plane is uniquely identified. For example, the pair (2,3) denotes the position relative to the origin as shown:
This system is often called the Cartesian coordinate system, named after the French mathematician René Descartes (1596– 1650).
The x- and y-axes break the plane into four regions called quadrants, named using roman numerals I, II, III, and IV, as pictured. In quadrant I, both coordinates are positive. In quadrant II, the x-coordinate is negative and the y-coordinate is positive. In quadrant III, both coordinates are negative. In quadrant IV, the x-coordinate is positive and the y-coordinate is negative.
Example 3.1.1
Plot the ordered pair (−3,5) and determine the quadrant in which it lies.
Solution:
The coordinates x=−3 and y=5 indicate a point 3 units to the left of and 5 units above the origin.
Answer:
The point is plotted in quadrant II (QII) because the x-coordinate is negative and the y-coordinate is positive.
Ordered pairs with 0 as one of the coordinates do not lie in a quadrant; these points are on one axis or the other (or the point is the origin if both coordinates are 0). Also, the scale indicated on the x-axis may be different from the scale indicated on the y-axis. Choose a scale that is convenient for the given situation.
Example 3.1.2
Plot this set of ordered pairs:
{(4,0),(−6,0),(0,3),(−2,6),(−4,−6)}.
Solution:
Each tick mark on the x-axis represents 2 units and each tick mark on the y-axis represents 3 units.
Example 3.1.3
Plot this set of ordered pairs:
{(−6,−5),(−3,−3),(0,−1),(3,1),(6,3)}.
Solution:
In this example, the points appear to be collinear, or to lie on the same line. The entire chapter focuses on finding and expressing points with this property.
Exercise 3.1.1
Plot the set of points {(5,3),(−3,2),(−2,−4),(4,−3)} and indicate in which quadrant they lie.
Answer
Graphs are used in everyday life to display data visually. A line graph consists of a set of related data values graphed on a coordinate plane and connected by line segments. Typically, the independent quantity, such as time, is displayed on the x-axis and the dependent quantity, such as distance traveled, on the y-axis.
Example 3.1.4
The following line graph shows the number of mathematics and statistics bachelor’s degrees awarded in the United States each year since 1970.
How many mathematics and statistics bachelor’s degrees were awarded in 1975?
In which years were the number of mathematics and statistics degrees awarded at the low of 11,000?
Solution:
a. The scale on the x-axis represents time since 1970, so to determine the number of degrees awarded in 1975, read the y-value of the graph at x=5.
The y-value corresponding to x=5 is 18. The graph indicates that this is in thousands; there were 18,000 mathematics and statistics degrees awarded in 1975.
b. To find the year a particular number of degrees was awarded, first look at the y-axis. In this case, 11,000 degrees is represented by 11 on the y-axis; look to the right to see in which years this occurred.
The y-value of 11 occurs at two data points, one where x=10 and the other where x=30. These values correspond to the years 1980 and 2000, respectively.
Answers:
In the year 1975, 18,000 mathematics and statistics degrees were awarded.
In the years 1980 and 2000, the lows of 11,000 mathematics and statistics degrees were awarded.
Distance Formula
Frequently you need to calculate the distance between two points in a plane. To do this, form a right triangle using the two points as vertices of the triangle and then apply the Pythagorean theorem. Recall that the Pythagorean theorem states that if given any right triangle with legs measuring a and b units, then the square of the measure of the hypotenuse c is equal to the sum of the squares of the legs: a2+b2=c2. In other words, the hypotenuse of any right triangle is equal to the square root of the sum of the squares of its legs.
Example 3.1.5
Find the distance between (−1,2) and (3,5).
Solution:
Form a right triangle by drawing horizontal and vertical lines through the two points. This creates a right triangle as shown below:
The length of leg b is calculated by finding the distance between the x-values of the given points, and the length of leg a is calculated by finding the distance between the given y-values.
a=5−2=3unitsb=3−(−1)=3+1=4units
Next, use the Pythagorean theorem to find the length of the hypotenuse.
c=√32+42=√9+16=√25=5units
Answer:
The distance between the two points is 5 units.
Generalize this process to produce a formula that can be used to algebraically calculate the distance between any two given points.
Given two points, (x1,y1) and (x2,y2), then the distance, d, between them is given by the distance formula:
√(x2+x1)2+(y2+y1)2
Example 3.1.6
Calculate the distance between (−3,−1) and (−2,4).
Solution:
Use the distance formula.
x1,y1x2,y2(−3,−1)(−2,4)
It is a good practice to include the formula in its general form as a part of the written solution before substituting values for the variables. This improves readability and reduces the chance for errors.
d=√(x2−x1)2+(y2−y1)2=√(−2−(−3))2+(4−(−1))2=√(−2+3)2+(4+1)2=√(1)2+(5)2=√1+25=√26
Answer:
√26 units
Exercise 3.1.2
Calculate the distance between (−7,5) and (−1,13).
Answer
: 10 units
Example 3.1.7
Do the three points (1,−1),(3,−3), and (3,1) form a right triangle?
Solution:
The Pythagorean theorem states that having side lengths that satisfy the property a2+b2=c2 is a necessary and sufficient condition of right triangles. In other words, if you can show that the sum of the squares of the leg lengths of the triangle is equal to the square of the length of the hypotenuse, then the figure must be a right triangle. First, calculate the length of each side using the distance formula.
| Geometry | Calculation |
--- |
| | Points: (1,−1) and (3,−3) a=√(3−1)2+(−3−(−1))2=√(a)2+(−3+1)2=√4+(−2)2=√4+4=√8 |
| | Points: (1,−1) and (3,1) b=√(3−1)2+(1−(−1))2=√22+(1+1)2=√4+(2)2=√4+4=√8 |
| | Points: (3,−3) and (3,1) c=√(3−3)2+(1−(−3))2=√(0)2+(1+3)2=√0+(4)2=√16=4 |
Table 3.1.1
Now we check to see if a2+b2=c2.
a2+b2=c2
(√8)2+(√8)2=(4)2√64+√64=168+8=1616=16✓
Answer:
Yes, the three points form a right triangle. In fact, since two of the legs are equal in length, the points form an isosceles right triangle.
Midpoint Formula
The point that bisects the line segment formed by two points, (x1,y1) and (x2,y2), is called the midpoint and is given by the following formula:
(x1+x22,y1+y22)
The midpoint is an ordered pair formed by finding the average of the x-values and the average of the y-values of the given points.
Example 3.1.8
Calculate the midpoint between (−1,−2) and (7,4).
Solution:
First, calculate the average of the x- and y-values of the given points.
(x1,y1)(x2,y2)(−1,−2)(7,4)
x1+x22=−1+72=62=3y1+y22=−2+42=22=1
Next, form the midpoint as an ordered pair using the averaged coordinates.
(x1+x22,y1+y22)
(3,1)
Answer:
(3,1)
To verify that this is indeed the midpoint, calculate the distance between the two given points and verify that the result is equal to the sum of the two equal distances from the endpoints to this midpoint. This verification is left to the reader as an exercise.
Exercise 3.1.3
Find the midpoint between (−6,5) and (6,−11).
Answer
: (0,−3)
Key Takeaways
Use the rectangular coordinate system to uniquely identify points in a plane using ordered pairs (x,y). Ordered pairs indicate position relative to the origin. The x-coordinate indicates position to the left and right of the origin. The y-coordinate indicates position above or below the origin.
The scales on the x-axis and y-axis may be different. Choose a scale for each axis that is appropriate for the given problem.
Graphs are used to visualize real-world data. Typically, independent data is associated with the x-axis and dependent data is associated with the y-axis.
The Pythagorean theorem gives us a necessary and sufficient condition of right triangles. Given a right triangle, then the measures of the sides satisfy a2+b2=c2. Conversely, if the sides satisfy a2+b2=c2, then the triangle must be a right triangle.
The distance formula, d=√(x2−x1)2+(y2−y1)2, is derived from the Pythagorean theorem and gives us the distance between any two points, (x1,y1) and (x2,y2), in a rectangular coordinate plane.
The midpoint formula, (x1+x22,y1+y22), is derived by taking the average of each coordinate and forming an ordered pair.
Exercise 3.1.4 Ordered Pairs
Give the coordinates of points A,B,C,D, and E.
1.
2.
3.
4.
5.
6.
Answer
: 1. A:(3,5);B:(−2,3);C:(−5,0);D:(1,−3);E:(−3,−4)
3. A:(0,6);B:(−4,3);C:(−8,0);D:(−6,−6);E:(8,−9)
5. A:(−10,25);B:(30,20);C:(0,10);D:(15,0);E:(25,−10)
Exercise 3.1.5 Ordered Pairs
Graph the given set of ordered pairs.
{(−4,5),(−1,1),(−3,−2),(5,−1)}
{(−15,−10),(−5,10),(15,10),(5,−10)}
{(−2,5),(10,0),(2,−5),(6,−10)}
{(−8,3),(−4,6),(0,−6),(6,9)}
{(−10,5),(20,−10),(30,15),(50,0)}
{(−53,−12),(−13,12),(23,−1),(53,1)}
{(−35,−43),(25,43),(1,−23),(0,1)}
{(−3.5,0),(−1.5,2),(0,1.5),(2.5,−1.5)}
{(−0.8,0.2),(−0.2,−0.4),(0,−1),(0.6,−0.4)}
{(−1.2,−1.2),(−0.3,−0.3),(0,0),(0.6,0.6),(1.2,1.2)}
Answer
: 1.
3.
5.
7.
9.
Exercise 3.1.6 Ordered Pairs
State the quadrant in which the given point lies.
(−3,2)
(5,7)
(−12,−15)
(7,−8)
(−3.8,4.6)
(17.3,1.9)
(−18,−58)
(34,−14)
x>0 and y<0
x<0 and y<0
x<0 and y>0
x>0 and y>0
Answer
: 1. QII
3. QIII
5. QII
7. QIII
9. QIV
11. QII
Exercise 3.1.7 Ordered Pairs
The average price of a gallon of regular unleaded gasoline in US cities is given in the following line graph. Use the graph to answer the following questions.
What was the average price of a gallon of unleaded gasoline in 2004?
What was the average price of a gallon of unleaded gasoline in 1976?
In which years were the average price of a gallon of unleaded gasoline $1.20?
What is the price increase of a gallon of gasoline from 1980 to 2008?
What was the percentage increase in the price of a gallon of unleaded gasoline from 1976 to 1980?
What was the percentage increase in the price of a gallon of unleaded gasoline from 2000 to 2008?
Answer
: 1. $1.80
3. 1980 to 1984, 1996
3. 100%
Exercise 3.1.8 Ordered Pairs
The average price of all-purpose white flour in US cities from 1980 to 2008 is given in the following line graph. Use the graph to answer the questions that follow.
What was the average price per pound of all-purpose white flour in 2000?
What was the average price per pound of all-purpose white flour in 2008?
In which year did the price of flour average $0.25 per pound?
In which years did the price of flour average $0.20 per pound?
What was the percentage increase in flour from the year 2000 to 2008?
What was the percentage increase in flour from the year 1992 to 2000?
Answer
: 1. $0.30
3. 1992
5. 67%
Exercise 3.1.9 Ordered Pairs
Given the following data, create a line graph.
The percentage of total high school graduates who enrolled in college.
| Year | Percentage |
--- |
| 1969 | 36% |
| 1979 | 40% |
| 1989 | 47% |
| 1999 | 42% |
Table 3.1.2: Source: Digest of Education Statistics
The average daily temperature given in degrees Fahrenheit in May.
| Exam | Temperature |
--- |
| 8:00 am | 60 |
| 12:00 pm | 72 |
| 4:00 pm | 75 |
| 8:00 pm | 67 |
| 12:00 am | 60 |
| 4:00 am | 55 |
Table 3.1.3
Answer
: 1.
Exercise 3.1.10 Ordered Pairs
Calculate the area of the shape formed by connecting the following set of vertices.
{(0,0),(0,3),(5,0),(5,3)}
{(−1,−1),(−1,1),(1,−1),(1,1)}
{(−2,−1),(−2,3),(5,3),(5,−1)}
{(−5,−4),(−5,5),(3,5),(3,−4)}
{(0,0),(4,0),(2,2)}
{(−2,−2),(2,−2),(0,2)}
{(0,0),(0,6),(3,4)}
{(−2,0),(5,0),(3,−3)}
Answer
: 1. 15 square units
3. 28 square units
5. 4 square units
7. 9 square units
Exercise 3.1.11 Distance Formula
Calculate the distance between the given two points.
(−5,3) and (−1,6)
(6,−2) and (−2,4)
(0,0) and (5,12)
(−6,−8) and (0,0)
(−7,8) and (5,−1)
(−1,−2) and (9,22)
(−1,2) and (−72,−4)
(−12,13) and (52,−113)
(−13,23) and (1,−13)
(12,−34) and (32,14)
(1,2) and (4,3)
(2,−4) and (−3,−2)
(−1,5) and (1,−3)
(1,−7) and (5,−1)
(−7,−3) and (−1,6)
(0,1) and (1,0)
(−0.2,−0.2) and (1.8,1.8)
(1.2,−3.3) and (2.2,−1.7)
Answer
: 1. 5 units
3. 13 units
5. 15 units
7. 132 units
9. 53 units
11. √10 units
13. 2√17 units
15. 3√13 units
17. 2.8 units
Exercise 3.1.12 Distance Formula
For each problem, show that the three points form a right triangle.
(−3,−2),(0,−2), and (0,4)
(7,12),(7,−13), and (−5,−4)
(−1.4,0.2),(1,2), and (1,−3)
(2,−1),(−1,2), and (6,3)
(−5,2),(−1,−2), and (−2,5)
(1,−2),(2,3), and (−3,4)
Answer
: 1. Proof
3. Proof
5. Proof
Exercise 3.1.13 Distance Formula
Isosceles triangles have two legs of equal length. For each problem, show that the following points form an isosceles triangle.
(1,6),(−1,1), and (3,1)
(−6,−2),(−3,−5), and (−9,−5)
(−3,0),(0,3), and (3,0)
(0,−1),(0,1), and (1,0)
Answer
: 1. Proof
3. Proof
Exercise 3.1.14 Distance Formula
Calculate the area and the perimeter of the triangles formed by the following set of vertices.
{(−4,−5),(−4,3),(2,3)}
{(−1,1),(3,1),(3,−2)}
{(−3,1),(−3,5),(1,5)}
{(−3,−1),(−3,7),(1,−1)}
Answer
: 1. Perimeter: 24 units; area: 24 square units
3. Perimeter: 8+4√2 units; area: 8 square units
Exercise 3.1.15 Midpoint Formula
Find the midpoint between the given two points.
(−1,6) and (−7,−2)
(8,0) and (4,−3)
(−10,0) and (10,0)
(−3,−6) and (−3,6)
(−10,5) and (14,−5)
(0,1) and (2,2)
(5,−3) and (4,−5)
(0,0) and (1,1)
(−1,−1) and (4,4)
(3,−5) and (3,5)
(−12,−13) and (32,73)
(34,−23) and (18,−12)
(53,14) and (−16,−32)
(−15,−52) and (710,−14)
Given the right triangle formed by the vertices (0,0),(6,0), and (6,8), show that the midpoints of the sides form a right triangle.
Given the isosceles triangle formed by the vertices (−10,−12),(0,12), and (10,−12), show that the midpoints of the sides also form an isosceles triangle.
Calculate the area of the triangle formed by the vertices (−4,−3),(−1,1), and (2,−3). (Hint: The vertices form an isosceles triangle.)
Calculate the area of the triangle formed by the vertices (−2,1),(4,1), and (1,−5).
Answer
: 1. (−4,2)
3. (0,0)
5. (2,0)
7. (92,−4)
9. (32,32)
11. (12,1)
13. (34,−58)
15. 12 square units
Exercise 3.1.16 Discussion Board Topics
Research and discuss the life and contributions to mathematics of René Descartes.
Research and discuss the history of the right triangle and the Pythagorean theorem.
What is a Pythagorean triple? Provide some examples.
Explain why you cannot use a ruler to calculate distance on a graph.
How do you bisect a line segment with only a compass and a straightedge?
Answer
: 1. Answers may vary
3. Answers may vary
5. Answers may vary
3: Graphing Lines
3.2: Graph by Plotting Points |
13630 | https://www.researchgate.net/post/How_to_handle_negative_values_in_log_transformations_in_a_regression_analysis2 | How to handle negative values in log transformations in a regression analysis? | ResearchGate
Question
Answers 14
Similar questions
Research that mentions Regression
Question
Asked 25 May 2016
Morteza Yaqubi
University of Torbat Heydarieh
How to handle negative values in log transformations in a regression analysis?
I would like to use a linear form of Cobb-Douglas production function in my project. However, there are some negative values in one of my independent variables. As far as I know, there are some transformations as follows:
log(1+Y-min(Y))
or
log( -y + 1) if Y is negative
log( y + 1) if Y is non-negative
Which is better? What are the consequences of these transformations in a regression analysis? Is there any better solution (e.g., a symmetric transformation that pulls in extreme values and preserves sign)?
Any comment would be appreciated.
Thanks in advance
Regression Analysis
Panel Data
Advanced Econometrics
Applied Statistics
Statistics
Regression
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Production Economics
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Multivariate Regression
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Most recent answer
Issahaku Nuhu
Ghana Institute of Management and Public Administration
I faced similar problem of taking logs of negative numbers and taking logs of fractions or rates like inflation rate or interest rate. I use natural logs.
1) take the natural log interest rate of 15%.
This becomes ln(1+r) = ln(1+0.15) or the
ln(1+0.15) ×100 if you want you transformed data to still be in percentages.
2) The second issue has to do with the natural logs of negative numbers.
First remember that 2^2=4 and -2^2= 4
Therefore x=(4)^0.5= + or - 2. The economist will work with the positive 2
You will apply this principle with logs of negative numbers.
Second, the ln(-8)=0 but ther imaginary ln(-8) is not equal to 0 but a positive real number and an imaginary number with I. Yet we want the real number. We there apply IMAGINARY REAL NUMBER function to extract the real number we are looking for as the logarithmic value of taking log of a nagative number.
Thirdly, let me apply the principle of taking logs negative numbers in excell for you to see with our initial log(-8). Assuming -8 is in cell A1.
=IF (A1=<0, IMREAL(IMln(A1)), ln(A1))
I hope, business students like me will find this piece useful for taking logs of interest rate, inflation and any other fraction.
It is also useful in taking logs of absolute negative numbers.
This approach avoids the use of personal discretion to add a constant or multiply by a negative number which all result in working with a different data set and not the original data set
Cite
Popular answers (1)
Abdulrazzak Charbaji
CHARBAJI Consultants | www.charbaji.com
Adding or subtracting a constant affects the mean but does not affect variance . Therefore it is recommended to add a constant . The best constant to add in case of Cobb Douglas production function is to add the same constant to all values of the same variable which makes all values of the variable positive. Suppose you have three negative values such as -6 and -9 and minus 2 then Adding the constant 10 to all values will make all values positive and greater than zero . The transformation such as log becomes possible without affecting R SQR or elacticity etc...
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All Answers (14)
Vladislav Shchekoldin
Novosibirsk State Technical University
Unfortunately, most economic data are non-invariant to the shift-transformation-type (when someone adds or subtracs some constant to the data). So your variant of transformation in many cases could directly lead to the wrong (skewed) results.If you have some negative values of the responce variable the log-transformation could be applied in two differents ways: 1) you can drop these values from the sample and estimate your model on such "truncated" sample; here of course you can lost some degrees of freedom and get results of less statistical sufficiency; 2) you can use so-called censored regression approach (for instance, tobit-models). It is much more correct case but it requires from you some special efforts in understanding the idea of tobit-type models. There are loads of books on the poblem, hope you will be successful.
Cite
2 Recommendations
Timothy A Ebert
University of Florida
Sometimes negative values can be removed by reformulating the problem or correcting errors. Do the negative values make sense in the context of the problem? What is the independent variable that is causing problems? Is it income, percent growth, coffee consumption?
@Vladislav: could you provide a citation to the observation that economic data are non-invariant to shift transformations. I am interested to see what it is about economic data that produces this outcome.
Cite
2 Recommendations
Sergiy Prykhodko
Admiral Makarov National University of Shipbuilding
Dear Morteza,
In this case, instead of the log transformation is better to use other transformations, for example, Johnson translation system or a two-parameter Box-Cox transformation.
Cite
2 Recommendations
Abdulrazzak Charbaji
CHARBAJI Consultants | www.charbaji.com
Adding or subtracting a constant affects the mean but does not affect variance . Therefore it is recommended to add a constant . The best constant to add in case of Cobb Douglas production function is to add the same constant to all values of the same variable which makes all values of the variable positive. Suppose you have three negative values such as -6 and -9 and minus 2 then Adding the constant 10 to all values will make all values positive and greater than zero . The transformation such as log becomes possible without affecting R SQR or elacticity etc...
Cite
11 Recommendations
Amir Hossein Montazer Hojat
Shahid Chamran University of Ahvaz
you can change origin so that all observation to be positive. then you cat transfer into log form
Cite
3 Recommendations
Timothy A Ebert
University of Florida
Here is another option if you can assume that the reason you have zeros is because your sample size is insufficient to get a non-zero value. In biology this might be something like measuring insect mortality due to the application of an insecticide at 1, 10, 100, and 1000 grams/hectare. The 1 gram/hectare rate results in a mortality rate of 0.002%. Given that your sample size was 10, it is unlikely that any of the ten insects will die. So your estimate is zero mortality, but that is only because your sample size was inappropriate for measuring such a low mortality.
If a similar situation applies in your case, what happens if you just delete the zero values?
Cite
1 Recommendation
Regret Tanyara Sunge
Afromontane Research Unit (ARU) Department of Economics and Finance University of the Free State
Hie.Surely negative values are common in regression.Adding a constant to make the minimum value positive has no harm to analysis. If the variable concerned is the depended variable, adding a constant will only alter the constant, but the parameter estimates will be the same.For independent variables, adding a constant does not change the parameters neither.Try it!
Cite
2 Recommendations
Justas Birgiolas
Ronin Institute
One could use the "Bi-Symmetric Log transformation", which performs a log-like transformation on numbers that are negative and doesn't exaggerate the 0-1 region.
ArticleA bi-symmetric log transformation for wide-range data
Cite
Ali Madina Dankumo
Please is there any reference to back up this formula "log(Y+a)" for log transformation of negative numbers?
Cite
1 Recommendation
Saizal Pinjaman
Universiti of Malaysia Sabah
=log(sqrt((X^2)+1))
Cite
1 Recommendation
Markos Farag
University of Cologne
A common approach to handle negative values is to add a constant value to the data prior to applying the log transform. The transformation is therefore log(Y+a) where a is the constant. Some people like to choose a so that min(Y+a) is a very small positive number (like 0.001). Others choose a so that min(Y+a)=1. For the latter choice, you can show that a=b – min(Y), where b is either a small number or is 1.
A criticism of this method is that some practicing econometricians don't like to add an arbitrary constant to the data. The argument is that a better way to handle negative values is to use missing values for the logarithm of a non-positive number.
Cite
1 Recommendation
Carlos Araújo Queiroz
Universidade NOVA de Lisboa
i.Adding a constant fixed arbitrary value to a variable, in order to somehow render the argument of the logarithmic function positive, may be reasonable (or not) depending on the scientific support (possibly, the physical meaning) that may justify this option. Note that such constant does not necessarily add to the mentioned argument as a whole. It may possibly add just to a variable from several possibly considered. It may be, alternatively, found preferable to add a parameter, to be fitted after the data, rather than a constant of fixed arbitrary value.
ii. A quantity (q ≥ 0 u) can be logarithmized after dividing by its unit (u); except for the origin of the scale, where discontinuity is expected because ln(0) is undefined. This discontinuity does not occur if the said quantity is restricted in its domain to (q ≥ u), so that it is also ensured that the argument of the logarithmic function is q/u ≥ 1. This may be found convenient if negative logarithmic values are found undesirable. We sometimes encounter the case where the quantity q is a monotonic increasing function of time, q(t), so that some 'hidden' time (θ) can be added to displace the origin of the scale, possibly from q/u = 0 to q/u = 1, or as otherwise can be found justifiable. This figure (θ) is sometimes called induction time in some kinetic studies, and it can be possibly adopted as (an additional) correlation parameter.
As example for this kind approach; you may check my post here:
You may also find useful to check:
Cite
1 Recommendation
Samaila Adamu
Federal University, Kashere, Gombe State
Thats good contribution and way out without affecting the variance
Cite
1 Recommendation
Issahaku Nuhu
Ghana Institute of Management and Public Administration
I faced similar problem of taking logs of negative numbers and taking logs of fractions or rates like inflation rate or interest rate. I use natural logs.
1) take the natural log interest rate of 15%.
This becomes ln(1+r) = ln(1+0.15) or the
ln(1+0.15) ×100 if you want you transformed data to still be in percentages.
2) The second issue has to do with the natural logs of negative numbers.
First remember that 2^2=4 and -2^2= 4
Therefore x=(4)^0.5= + or - 2. The economist will work with the positive 2
You will apply this principle with logs of negative numbers.
Second, the ln(-8)=0 but ther imaginary ln(-8) is not equal to 0 but a positive real number and an imaginary number with I. Yet we want the real number. We there apply IMAGINARY REAL NUMBER function to extract the real number we are looking for as the logarithmic value of taking log of a nagative number.
Thirdly, let me apply the principle of taking logs negative numbers in excell for you to see with our initial log(-8). Assuming -8 is in cell A1.
=IF (A1=<0, IMREAL(IMln(A1)), ln(A1))
I hope, business students like me will find this piece useful for taking logs of interest rate, inflation and any other fraction.
It is also useful in taking logs of absolute negative numbers.
This approach avoids the use of personal discretion to add a constant or multiply by a negative number which all result in working with a different data set and not the original data set
Cite
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13631 | https://www.onlinemath4all.com/examples-on-internal-and-external-angle-bisector-theorem.html | Examples on Internal and External Angle Bisector Theorem
EXAMPLES ON INTERNAL AND EXTERNAL ANGLE BISECTOR THEOREM
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Angle Bisector Theorem :
The internal (external) bisector of an angle of a triangle divides the opposite side internally (externally) in the ratio of the corresponding sides containing the angle.
To know more about proof, please visit the page"Angle bisector theorem proof".
Example 1 :
In a triangle MNO, MP is the external bisector of angle M meeting NO produced at P. IF MN = 10 cm, MO = 6 cm, NO - 12 cm, then find OP.
Solution :
MP is the external bisector of angle M by using angle bisector theorem in the triangle MNO we get,
(NP/OP) = (MN/MO)
NP = NO + OP
= 12 + OP
(12 + OP)/OP = 10/6
6 (12 + OP) = 10 OP
72 + 6 OP = 10 OP
72 = 10 OP - 6 OP
4 OP = 72
OP = 72/4
= 18 cm
Example 2 :
In a quadrilateral ABCD, the bisectors of ∠B and ∠D intersect on AC at E. Prove that (AB/BC) = (AD/DC).
Solution :
Here DE is the internal angle bisector of angle D.
by using internal bisector theorem, we get
(AE/EC) = (AD/DC) ----- (1)
Here BE is the internal angle bisector of angle B.
(AE/EC) = (AB/BC) ----- (2)
from (1) and (2) we get,
(AB/BC) = (AD/DC)
Hence proved.
Example 3 :
The internal bisector of ∠A of triangle ABC meets BC at D and the external bisector of ∠A meets BC produced at E. Prove that (BD/BE) = (CD/CE).
Solution :
In triangle ABC, AD is the internal bisector of angle A.
by using angular bisector theorem in triangle ABC
(BD/DC) = (AB/AC) ----- (1)
In triangle ABC, AE is the internal bisector of angle A.
(BE/CE) = (AB/AC) ----- (2)
from (1) and (2) we get,
(BD/DC) = (BE/CE)
(BD/BE) = (DC/CE)
Hence proved.
Example 4 :
ABCD is a quadrilateral with AB = AD. If AE and AF are internal bisectors of ∠BAC and ∠DAC respectively,then prove that the sides EF and BD are parallel.
Solution:
In triangle ABC, AE is the internal bisector of ∠BAC
by using bisector theorem, we get
(AB/AC) = (BE/EC) ----- (1)
In triangle ADC, AF is the internal bisector of ∠DAC
by using bisector theorem, we get
(AD/AC) = (DF/FC)
Since lengths of AD and AB are equal, we may replace AB instead of AD.
(AB/AC) = (DF/FC) ----- (2)
from (1) and (2) we get
(DF/FC) = (BE/EC)
So, EF and BD are parallel by using converse of "Thales theorem".
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13632 | https://ux1.eiu.edu/~jpblitz/teaching/quant/Chapter%2010.pdf | 1 Chapter 10 & Section 6 – 7 Polyprotic acids and bases We’ll bounce around in the book a little: Start with part of Section 6 – 7 p. 112 – 114. HA = Monoprotic acid H2A = Diprotic acid H3A = Triprotic acid etc. For the diprotic case, Ka reactions and expressions for H2CO3 (carbonic acid) For the triprotic case, Ka reactions and expressions for H3PO4 (phosphoric acid) Now Section 10 – 2 to finish off buffers discussion. A diprotic acid can buffer around two different pH values depending on the pKas of the diprotic acid. H2CO3/HCO3 - are conjugates 2 HCO3 -/CO3 2- are conjugates A triprotic acid can buffer around three different pH values depending on the pKas of the triprotic acid. H3PO4/H2PO4 - are conjugates H2PO4 -/HPO4 2- are conjugates HPO4 2-/PO4 3- are conjugates A diprotic buffer example: How many mL of 0.8 M KOH should be added to 3.38 g of oxalic acid (90 mg/mmol, H2A) to give a pH of 4.4? pKa1 = 1.25 pKa2 = 4.27 The desired pH > pKa2 so need a mixture of HA-/A2-1. Add enough base to convert all H2A Æ HA- 2. Add more base to convert some HA- Æ A2- until pH = 4.4 3 To do the first part: To do the second part: 4 Now to finish Section 6 – 7 For a monoprotic acid Ka x Kb = Kw Similar relationships can be derived for polyprotic systems, a little more tricky. For a diprotic system: H2A, HA-, A2- may exist Most acidic = Ka1: Most basic = Kb1: HA- as acid = Ka2: HA- as base = Kb2: Now that these are defined, can do analogous derivations as the monoprotic system. 5 Triprotic system is now readily derived in the same way. For a triprotic system: H3A, H2A-, HA2-, A3- may exist. Ka1: Kb1: Ka2: Kb2: Ka3: Kb3: Putting these together: Tetraprotic, pentaprotic, hexaprotic, etc. – same process. Back to beginning of Chapter 10, Section 10 – 1. Find the pH of diprotic solutions, easily extended to polyprotic systems. Maleic acid is diprotic (H2A) – Harris uses the amino acid leucine starting on p. 181 for his example. The reactions, K expressions, and K values: There are 3 different calculations for this system. 1. Find the pH of a 0.05 M H2A solution. 2. Find the pH of a 0.05 M Na2A solution. 3. Find the pH of a 0.05 M NaHA solution. 6 The first 2 are not all that bad if everything works out. Number 1 – Find the pH of a 0.05 M H2A solution. Recognize that H2A is a weak acid - partially dissociates. HA- is weaker, and what little is present from H2A dissociation will dissociate even less. So to avoid having to solve an extremely unpleasant problem – hope H2A behaves like a monoprotic acid with Ka = Ka1. And also hope that [H+] from H2A >> [H+] from water. If both of these hopes come true, it reduces to a regular weak acid problem. Further hope that x << 0.05 so Certainly [H+] from H2A >> [H+] from water We have checked 2 assumptions, what about the third? Is the [H+] from H2A >> [H+] from HA-? (The monoprotic assumption) 7 For most polyprotic acids Ka1 >> Ka2. Bottom line: A solution of the fully acidic form of a polyprotic acid behaves like a solution of a monoprotic acid with Ka = Ka1. Number 2 – Find the pH of a 0.05 M Na2A solution. A2- is a weak base that partially hydrolyzes in water. HA- is a weaker base, and little should be in solution anyway. So to avoid solving a very unpleasant problem, hope A2- behaves like a monobasic compound with Kb = Kb1. And – hope the [OH-] from A2- >> [OH-] from water. If both of these hopes come true – then you have a regular weak base problem. 8 For most polyprotic systems Kb1 >> Kb2 so Bottom line: A solution of the fully basic form of a polyprotic system behaves like a monobasic species with Kb = Kb1. Number 3 – Find the pH of a 0.05 M NaHA solution. NaHA Æ Na+ + HA-HA- H+ + A2- Ka2 and HA- + H2O H2A + OH- Kb2 This is trouble, HA- is amphiprotic Ignore Kb2 since it is much less than Ka2? No can do, these equilibria are coupled. H+ from Ka2 + OH- from Kb2 Æ H2O driving Kb2 reaction to the right, more important than the Kb2 value suggests. Bite the bullet – set up using systematic approach. Will simplify later, so don’t get lost in the forest. 9 Species in solution: H+, OH-, H2A, HA-, A2-, Na+ 5 species in solution, need 5 equations to solve: 1. Ka2 = 2. Kb2 = 3. Kw = 4. Charge balance: 5. Mass balance: Go through lots of algebra, many substitutions [H+] = Don’t know the [HA-], but Ka2 and Kb2 are very small so the simplifying process begins: hope [HA-] = F = 0.05 M in this problem. [H+] = Put in values for all => pH = 4.17 Always check to see if hope came true, find [H2A], [A2-] 10 It is worth noting that [A2-] ≈ [H2A] even though Ka2 >> Kb2. These reactions are coupled! In any case the concentrations of these 2 species are small, relative to 0.05 (F) anyway, so our hope did come true. Continuing to simplify the equation used above to calculate pH: usually Ka1Kw << Ka1Ka2F so the equation simplifies to [H+] = One more simplification. Often Ka1 << F so Take –log of both sides: Bottom Line: the pH of an amphiprotic (intermediate) species is the average of the pKas. Diprotic summary: 1) H2A: Treat as monoprotic with Ka = Ka1 2) A2-: Treat as monobasic with Kb = Kb1 3) HA-: Take the average of the pKas This is readily extended to triprotic systems and beyond (Section 10 – 3). 1) H3A: Treat as monoprotic with Ka = Ka1 2) H2A-: Treat as amphiprotic with pH = 11 3) HA2-: Treat as amphiprotic with pH = 4) A3-: Treat as monobasic with Kb = Kb1. Section 10 – 4 Principal species First monoprotic, then polyprotic A. Monoprotic. The question is, at a given pH what is the majority species? For an acid/conjugate base system is it HA or A-? For a base/conjugate acid system is it B or BH+? In either case this is most easily treated using the buffer equation which, recall, is just the rearranged form of the Ka expression. Take benzoic acid as an example, with a pKa = 4.2. 1. At pH = 4.2: 2. At pH = 5.2 3. At pH = 3.2 12 When the pH < pKa the acidic form is in the majority (i.e. the principal species. When the pH > pKa the basic form is the principal species. B. Polyprotic systems. What is the predominant species at any given pH? Similar reasoning but >1 pKa. For phosphoric acid, H3A 13 Section 10-6. Isoelectric and Isoionic pH Take alanine as an example: The isoionic point is the pH obtained when the pure, neutral intermediate species (the zwitterion) is dissolved in water. Here the [H2A+] ≠ [A-] (but very close) since intermediate species do not dissociate much (bottom p. 184). The isoelectric point is the pH at which [H2A+] = [A-]; where the average charge of the amino acid is zero. 14 The isoelectric point is important for the separation of proteins by electrophoresis using the clever technique of isoelectric focusing (Box 10-2, p. 194). Chapter 10 Exercises A, B, C Problems 1, 4, 5, 11-13, 15, 17, 18a, 22, 23 In Chapter 10, those exercises in red and underlined are test 2 material. Remainder are for test 3. All remaining exercises and problems in Chapters 8 and 6 are for test 2, but material associated with Chapter 10. Chapter 8 Exercises F, G, H Problems 15, 17, 18, 22 Chapter 6 Exercise I Problems 50, 51 |
13633 | https://www.teachstarter.com/us/learning-area/decimal-operations-us/ | Teach Starter, part of Tes Teach Starter, part of Tes
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Decimal Operations Teaching Resources
Teach decimal operations this school year with worksheets, activities and more — all created by teachers to help your students understand how to add, multiply, subtract and divide decimals!
Aligned with the Common Core math curriculum and created with your students in mind, this collection of printable and digital teacher resources has everything you need for your lesson plans — we've even got differentiation covered!
Are you new to teaching decimal operations? Looking for a quick refresher? Read on for a primer from our teacher team on how to handle all four operations!
How Do You Add Decimals?
Let's start with one of the simplest operations when it comes to decimals: adding.
How Do You Subtract Decimals?
Subtracting decimals is pretty similar to adding, although the operation is different. Here's how it works:
How Do You Divide Decimals?
Whether you're trying to find the average of a set of numbers that include decimals or determining the unit price of an item on the shelves at the chemist, being able to divide decimals has plenty of real-world applications for students.
The basic process of dividing decimals follows the same basic principles as dividing whole numbers, with one additional step — you need to adjust the decimal point in the divisor (the number you are dividing by) and the dividend (the number you are dividing) so that the divisor becomes a whole number.
Here's how to do it:
How Do You Multiply Decimals?
Multiplying decimals can help your students calculate the total cost of items at the store, determine the amount of a discount or tax and a whole lot more. The process to do it is simple, once they get it down!
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Multiplying Decimals – Word Problem Task Cards
Solve a variety of word problems by multiplying decimals with this set of 24 task cards.
Multiplying with Decimals Bingo Game
Get your students multiplying decimals by whole numbers with this engaging Bingo game.
Decimal Operations Worksheet
Get students applying the four operations to decimal numbers with this set of two math worksheets.
Estimating Decimal Sums and Differences – Worksheet
Refine rounding skills with an estimating decimal sums and differences worksheet.
Multiplying and Dividing by Powers of 10 Anchor Chart Set
Display our multiplying and dividing by powers of 10 anchor chart set in your classroom for an easy reference for students learning operations with powers of 10.
Fractions, Decimals and Percentages Worksheets
Use these fractions, decimals and percentages worksheets in your upper elementary classroom for independent practice or as an assessment activity.
Differentiated Volume of Rectangular Prism Worksheet Set
Differentiate instruction with this volume of rectangular prism worksheet set that offers three levels of volume problems.
Adding and Subtracting Decimals – Differentiated Mystery Image Worksheets
Have students practice adding and subtracting decimals with this mystery image worksheet.
Multiplying Decimals by Decimals Worksheet Pack
Download these multiplying decimals by decimals worksheets to provide your students with practice using visual models, area models and the standard algorithm.
Decimal Exit Tickets for 5th Grade (Place Value, Rounding, Ordering and Computation)
Assess student understanding of 5th grade decimal standards with this set of 21 exit tickets.
Adding and Subtracting Decimals Word Problems Worksheet
Practice solving word problems by adding and subtracting decimals to the thousandths place with this worksheet.
Multiplying Decimals – Riddle Worksheets
Get students solving decimal multiplication problems with this set of 3 differentiated riddle worksheets.
Adding and Subtracting Decimals – Word Problem Task Cards
Have students solve decimal word problems with this set of 16 task cards, perfect for math centers.
Adding and Subtracting with Decimals – Worksheet
Practice adding and subtracting decimals to the hundredths place with this worksheet.
Operations with Decimals Teaching Slides
Teach your students how to add, subtract, multiply and divide using decimal numbers with this comprehensive teaching presentation perfect for upper elementary math lessons.
Place Value Slider Set
Break out this place value slider set for a hands-on way to demonstrate multiplying and dividing whole numbers and decimals by powers of ten.
Decimal Place Value Chart
Provide your students with our decimal place value chart to make comparing and ordering decimals a breeze!
5th Grade - Decimal Place Value Assessment
Assess your students' skills with decimals and place value using a printable 5th Grade Decimal Place Value Test.
Dividing Decimals by Whole Numbers Worksheet Pack
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Adding and Subtracting Decimals – Dominoes
Use this game of dominoes to strengthen computation skills by adding and subtracting decimals through the thousandths place.
Dividing with Decimals – Word Problem Task Cards
Sharpen decimal division skills while solving a variety of word problems with this set of 24 task cards.
Dividing Decimals Task Cards
Sharpen decimal division skills with this set of 24 task cards.
Dividing Decimals Worksheet Pack - Differentiated
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Multiplying Decimals – Interactive Picture Reveal (Halloween)
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Volume With Decimals Worksheets
Provide these volume with decimals worksheets to your students to give them practice calculating volume and multiplying decimals.
Dividing Decimals by Whole Numbers Teaching Slides
Teach your students how to divide decimals by whole numbers using visual models and the standard algorithm using this step-by-step teaching presentation.
Multiplying Decimals by Decimals Teaching Slides
Teach your students how to multiply decimals by decimals using three different strategies using this step-by-step teaching presentation for upper elementary students.
Estimation With Decimal Multiplication Matching Game
Guide your students to estimate products with decimals by using this 36-card matching game.
Estimating Products With Decimals – Task Cards for 5th Grade
Challenge your students to estimate before multiplying decimals with this set of 24 task cards.
Estimating Sums and Differences With Decimals Worksheet for 5th Grade
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Coffee Shop Calculations – Estimation With Decimals Task Cards
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Subtracting Decimals With Base-10 Blocks – Teaching Slides for 5th Grade
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13634 | https://networkx.org/documentation/stable/reference/classes/multigraph.html | Skip to main content
MultiGraph—Undirected graphs with self loops and parallel edges#
Overview#
class MultiGraph(incoming_graph_data=None, multigraph_input=None, attr)[source]#
: An undirected graph class that can store multiedges.
Multiedges are multiple edges between two nodes. Each edge
can hold optional data or attributes.
A MultiGraph holds undirected edges. Self loops are allowed.
Nodes can be arbitrary (hashable) Python objects with optional
key/value attributes. By convention `None`") is not used as a node.
Edges are represented as links between nodes with optional
key/value attributes, in a MultiGraph each edge has a key to
distinguish between multiple edges that have the same source and
destination nodes.
Parameters:
: incoming\_graph\_datainput graph (optional, default: None)
: Data to initialize graph. If None (default) an empty
graph is created. The data can be any format that is supported
by the to\_networkx\_graph() function, currently including edge list,
dict of dicts, dict of lists, NetworkX graph, 2D NumPy array,
SciPy sparse array, or PyGraphviz graph.
multigraph\_inputbool or None (default None)
: Note: Only used when `incoming_graph_data` is a dict.
If True, `incoming_graph_data` is assumed to be a
dict-of-dict-of-dict-of-dict structure keyed by
node to neighbor to edge keys to edge data for multi-edges.
A NetworkXError is raised if this is not the case.
If False, `to_networkx_graph()` is used to try to determine
the dict’s graph data structure as either a dict-of-dict-of-dict
keyed by node to neighbor to edge data, or a dict-of-iterable
keyed by node to neighbors.
If None, the treatment for True is tried, but if it fails,
the treatment for False is tried.
attrkeyword arguments, optional (default= no attributes)
: Attributes to add to graph as key=value pairs.
See also
`Graph`
`DiGraph`
`MultiDiGraph`
Examples
Create an empty graph structure (a “null graph”) with no nodes and
no edges.
```
>>> G = nx.MultiGraph()
```
G can be grown in several ways.
Nodes:
Add one node at a time:
```
>>> G.add_node(1)
```
Add the nodes from any container (a list, dict, set or
even the lines from a file or the nodes from another graph).
```
>>> G.add_nodes_from([2, 3])
>>> G.add_nodes_from(range(100, 110))
>>> H = nx.path_graph(10)
>>> G.add_nodes_from(H)
```
In addition to strings and integers any hashable Python object
(except None) can represent a node, e.g. a customized node object,
or even another Graph.
```
>>> G.add_node(H)
```
Edges:
G can also be grown by adding edges.
Add one edge,
```
>>> key = G.add_edge(1, 2)
```
a list of edges,
```
>>> keys = G.add_edges_from([(1, 2), (1, 3)])
```
or a collection of edges,
```
>>> keys = G.add_edges_from(H.edges)
```
If some edges connect nodes not yet in the graph, the nodes
are added automatically. If an edge already exists, an additional
edge is created and stored using a key to identify the edge.
By default the key is the lowest unused integer.
```
>>> keys = G.add_edges_from([(4, 5, {"route": 28}), (4, 5, {"route": 37})])
>>> G
AdjacencyView({3: {0: {}}, 5: {0: {}, 1: {'route': 28}, 2: {'route': 37}}})
```
Attributes:
Each graph, node, and edge can hold key/value attribute pairs
in an associated attribute dictionary (the keys must be hashable).
By default these are empty, but can be added or changed using
add\_edge, add\_node or direct manipulation of the attribute
dictionaries named graph, node and edge respectively.
```
>>> G = nx.MultiGraph(day="Friday")
>>> G.graph
{'day': 'Friday'}
```
Add node attributes using add\_node(), add\_nodes\_from() or G.nodes
```
>>> G.add_node(1, time="5pm")
>>> G.add_nodes_from(, time="2pm")
>>> G.nodes
{'time': '5pm'}
>>> G.nodes["room"] = 714
>>> del G.nodes["room"] # remove attribute
>>> list(G.nodes(data=True))
[(1, {'time': '5pm'}), (3, {'time': '2pm'})]
```
Add edge attributes using add\_edge(), add\_edges\_from(), subscript
notation, or G.edges.
```
>>> key = G.add_edge(1, 2, weight=4.7)
>>> keys = G.add_edges_from([(3, 4), (4, 5)], color="red")
>>> keys = G.add_edges_from([(1, 2, {"color": "blue"}), (2, 3, {"weight": 8})])
>>> G["weight"] = 4.7
>>> G.edges[1, 2, 0]["weight"] = 4
```
Warning: we protect the graph data structure by making `G.edges[1,
2, 0]` a read-only dict-like structure. However, you can assign to
attributes in e.g. `G.edges[1,2,0]`. Thus, use 2 sets of brackets
to add/change data attributes: `G.edges[1,2,0]['weight'] = 4`.
Shortcuts:
Many common graph features allow python syntax to speed reporting.
```
>>> 1 in G # check if node in graph
True
>>> [n for n in G if n < 3] # iterate through nodes
[1, 2]
>>> len(G) # number of nodes in graph
5
>>> G # adjacency dict-like view mapping neighbor -> edge key -> edge attributes
AdjacencyView({2: {0: {'weight': 4}, 1: {'color': 'blue'}}})
```
Often the best way to traverse all edges of a graph is via the neighbors.
The neighbors are reported as an adjacency-dict `G.adj` or `G.adjacency()`.
```
>>> for n, nbrsdict in G.adjacency():
... for nbr, keydict in nbrsdict.items():
... for key, eattr in keydict.items():
... if "weight" in eattr:
... # Do something useful with the edges
... pass
```
But the edges() method is often more convenient:
```
>>> for u, v, keys, weight in G.edges(data="weight", keys=True):
... if weight is not None:
... # Do something useful with the edges
... pass
```
Reporting:
Simple graph information is obtained using methods and object-attributes.
Reporting usually provides views instead of containers to reduce memory
usage. The views update as the graph is updated similarly to dict-views.
The objects `nodes`, `edges` and `adj` provide access to data attributes
via lookup (e.g. `nodes[n]`, `edges[u,v,k]`, `adj[u][v]`) and iteration
(e.g. `nodes.items()`, `nodes.data('color')`,
`nodes.data('color',default='blue')` and similarly for `edges`)
Views exist for `nodes`, `edges`, `neighbors()`/`adj` and `degree`.
For details on these and other miscellaneous methods, see below.
Subclasses (Advanced):
The MultiGraph class uses a dict-of-dict-of-dict-of-dict data structure.
The outer dict (node\_dict) holds adjacency information keyed by node.
The next dict (adjlist\_dict) represents the adjacency information
and holds edge\_key dicts keyed by neighbor. The edge\_key dict holds
each edge\_attr dict keyed by edge key. The inner dict
(edge\_attr\_dict) represents the edge data and holds edge attribute
values keyed by attribute names.
Each of these four dicts in the dict-of-dict-of-dict-of-dict
structure can be replaced by a user defined dict-like object.
In general, the dict-like features should be maintained but
extra features can be added. To replace one of the dicts create
a new graph class by changing the class(!) variable holding the
factory for that dict-like structure. The variable names are
node\_dict\_factory, node\_attr\_dict\_factory, adjlist\_inner\_dict\_factory,
adjlist\_outer\_dict\_factory, edge\_key\_dict\_factory, edge\_attr\_dict\_factory
and graph\_attr\_dict\_factory.
node\_dict\_factoryfunction, (default: dict)
: Factory function to be used to create the dict containing node
attributes, keyed by node id.
It should require no arguments and return a dict-like object
node\_attr\_dict\_factory: function, (default: dict)
: Factory function to be used to create the node attribute
dict which holds attribute values keyed by attribute name.
It should require no arguments and return a dict-like object
adjlist\_outer\_dict\_factoryfunction, (default: dict)
: Factory function to be used to create the outer-most dict
in the data structure that holds adjacency info keyed by node.
It should require no arguments and return a dict-like object.
adjlist\_inner\_dict\_factoryfunction, (default: dict)
: Factory function to be used to create the adjacency list
dict which holds multiedge key dicts keyed by neighbor.
It should require no arguments and return a dict-like object.
edge\_key\_dict\_factoryfunction, (default: dict)
: Factory function to be used to create the edge key dict
which holds edge data keyed by edge key.
It should require no arguments and return a dict-like object.
edge\_attr\_dict\_factoryfunction, (default: dict)
: Factory function to be used to create the edge attribute
dict which holds attribute values keyed by attribute name.
It should require no arguments and return a dict-like object.
graph\_attr\_dict\_factoryfunction, (default: dict)
: Factory function to be used to create the graph attribute
dict which holds attribute values keyed by attribute name.
It should require no arguments and return a dict-like object.
Typically, if your extension doesn’t impact the data structure all
methods will inherited without issue except: `to_directed/to_undirected`.
By default these methods create a DiGraph/Graph class and you probably
want them to create your extension of a DiGraph/Graph. To facilitate
this we define two class variables that you can set in your subclass.
to\_directed\_classcallable, (default: DiGraph or MultiDiGraph)
: Class to create a new graph structure in the `to_directed` method.
If `None`"), a NetworkX class (DiGraph or MultiDiGraph) is used.
to\_undirected\_classcallable, (default: Graph or MultiGraph)
: Class to create a new graph structure in the `to_undirected` method.
If `None`"), a NetworkX class (Graph or MultiGraph) is used.
Subclassing Example
Create a low memory graph class that effectively disallows edge
attributes by using a single attribute dict for all edges.
This reduces the memory used, but you lose edge attributes.
```
>>> class ThinGraph(nx.Graph):
... all_edge_dict = {"weight": 1}
...
... def single_edge_dict(self):
... return self.all_edge_dict
...
... edge_attr_dict_factory = single_edge_dict
>>> G = ThinGraph()
>>> G.add_edge(2, 1)
>>> G
{'weight': 1}
>>> G.add_edge(2, 2)
>>> G is G
True
```
Methods#
Adding and removing nodes and edges#
| | |
--- |
| MultiGraph.__init__([incoming_graph_data, ...]) | Initialize a graph with edges, name, or graph attributes. |
| MultiGraph.add_node(node_for_adding, attr) | Add a single node node_for_adding and update node attributes. |
| MultiGraph.add_nodes_from(nodes_for_adding, ...) | Add multiple nodes. |
| MultiGraph.remove_node(n) | Remove node n. |
| MultiGraph.remove_nodes_from(nodes) | Remove multiple nodes. |
| MultiGraph.add_edge(u_for_edge, v_for_edge) | Add an edge between u and v. |
| MultiGraph.add_edges_from(ebunch_to_add, attr) | Add all the edges in ebunch_to_add. |
| MultiGraph.add_weighted_edges_from(ebunch_to_add) | Add weighted edges in ebunch_to_add with specified weight attr |
| MultiGraph.new_edge_key(u, v) | Returns an unused key for edges between nodes u and v. |
| MultiGraph.remove_edge(u, v[, key]) | Remove an edge between u and v. |
| MultiGraph.remove_edges_from(ebunch) | Remove all edges specified in ebunch. |
| MultiGraph.update([edges, nodes]) | Update the graph using nodes/edges/graphs as input. |
| MultiGraph.clear() | Remove all nodes and edges from the graph. |
| MultiGraph.clear_edges() | Remove all edges from the graph without altering nodes. |
Reporting nodes edges and neighbors#
| | |
--- |
| MultiGraph.nodes | A NodeView of the Graph as G.nodes or G.nodes(). |
| MultiGraph.__iter__() | Iterate over the nodes. |
| MultiGraph.has_node(n) | Returns True if the graph contains the node n. |
| MultiGraph.__contains__(n) | Returns True if n is a node, False otherwise. |
| MultiGraph.edges | Returns an iterator over the edges. |
| MultiGraph.has_edge(u, v[, key]) | Returns True if the graph has an edge between nodes u and v. |
| MultiGraph.get_edge_data(u, v[, key, default]) | Returns the attribute dictionary associated with edge (u, v, key). |
| MultiGraph.neighbors(n) | Returns an iterator over all neighbors of node n. |
| MultiGraph.adj | Graph adjacency object holding the neighbors of each node. |
| MultiGraph.__getitem__(n) | Returns a dict of neighbors of node n. |
| MultiGraph.adjacency() | Returns an iterator over (node, adjacency dict) tuples for all nodes. |
| MultiGraph.nbunch_iter([nbunch]) | Returns an iterator over nodes contained in nbunch that are also in the graph. |
Counting nodes edges and neighbors#
| | |
--- |
| MultiGraph.order() | Returns the number of nodes in the graph. |
| MultiGraph.number_of_nodes() | Returns the number of nodes in the graph. |
| MultiGraph.__len__() | Returns the number of nodes in the graph. |
| | A DegreeView for the Graph as G.degree or G.degree(). |
| MultiGraph.size([weight]) | Returns the number of edges or total of all edge weights. |
| MultiGraph.number_of_edges([u, v]) | Returns the number of edges between two nodes. |
Making copies and subgraphs#
| | |
--- |
| MultiGraph.copy([as_view]) | Returns a copy of the graph. |
| MultiGraph.to_undirected([as_view]) | Returns an undirected copy of the graph. |
| MultiGraph.to_directed([as_view]) | Returns a directed representation of the graph. |
| MultiGraph.subgraph(nodes) | Returns a SubGraph view of the subgraph induced on nodes. |
| MultiGraph.edge_subgraph(edges) | Returns the subgraph induced by the specified edges. | |
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围棋
| 進行中的围棋 | |
| 活躍時期 | 周朝至今 |
| 類型 | 棋盤遊戲 |
| 玩家數目 | 黑白双方 |
| 複雜程度 | 極高 |
| 策略成分 | 極高 |
| 遊戲時間 | 数十分钟到数天 |
| 運氣成分 | 除先後手順序，無任何運氣成分 |
| 所需技巧 | 策略、記憶、觀察、判断、細算 |
| 围棋 | |
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| 中华人民共和国 国家级非物质文化遗产 | |
| 申报地区或单位 | 中国棋院 北京棋院 |
| 分类 | 传统体育、游艺与杂技 |
| 序号 | 790 |
| 编号项目 | Ⅵ—18 |
| 登录 | 2008年 |
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| 圍棋各语言名称 | |
| 汉语名称 | |
| 繁体字 | 圍棋 ‧ 弈 ‧ 碁 ‧ 手談 |
| 简化字 | 围棋 ‧ 弈 ‧ 碁 ‧ 手谈 |
| | 标音 | | --- | | |
| 藏语名称 | |
| 藏語 | མིག་མངས |
| | 标音 | | --- | | - 威利转写系统 | mig mangs | | |
| 越南语名称 | |
| 國語字 | Cờ vy |
| 汉喃 | 棋圍 ‧ 碁圍 |
| 朝鲜语名稱 | |
| 諺文 | 바둑 ‧ 오로 |
| 汉字 | 바둑 ‧ 烏鷺 |
| | 标音 | | --- | | - 文观部式 | baduk ‧ oro | | - 马-赖式 | patuk ‧ oro | | |
| 日语名称 | |
| 汉字 | 囲碁 ‧ 碁 |
| 舊字體 | 圍碁 ‧ 碁 |
| | 标音 | | --- | | - 假名 | いご ‧ ゐご ‧ ご | | - 日语罗马字 | igo ‧ wigo ‧ go | | |
| 琉球語名稱 | |
| 汉字 | 碁 |
| 假名 | グー |
| 罗马字 | Guu |
本页面有藏文字母，操作系统及浏览器須支持特殊字母与符号才能正確显示为藏文字母，否则可能變成乱码、问号、空格等其它符号。
围棋是一種策略棋類，使用格狀棋盤及黑白二色棋子進行對弈。起源于中国，中國古时有“弈”、“碁”、“手谈”等多种称谓，屬琴棋书画四艺。西方稱之為“Go”，是源自日語「碁」的发音。
相傳圍棋由堯帝發明，他的兒子丹朱驕傲自滿，暴躁任性，因此堯帝便發明圍棋來陶冶兒子的心性。
圍棋有黑白兩種棋子，規定由執黑色棋子的先行，对弈双方在十九乘十九條線的棋盘网格上的交叉点交替放置黑色及白色的棋子。落子完毕后，不能悔棋。对弈過程中圍地吃子，以所圍「地」的大小決定勝負。
围棋规则简洁而优雅，但玩法卻千變萬化，欲精通其內涵需要大量的練習與鑽研。国际象棋大师伊曼紐·拉斯克称赞说：“若宇宙中另存其他智能生命形式，其幾乎必會围棋。”与此同时，围棋被认为是目前世界上最复杂的棋盤游戏之一。n乘n棋盤上的圍棋判定殘局勝負的複雜度已於1978年被Robertson與Munro證明為PSPACE-hard。
截至2008年年中，全世界有超过四千万玩家，其中绝大多数在东亚。截至2020年底，国际围棋联盟共拥有77个成员国和5个协会会员。
历史
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主条目：围棋史
起源于中国
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围棋起源于中国，是世界上最古老的棋类运动之一。推测起源时间为大约公元前23世纪。传说尧的儿子丹朱顽劣，尧发明围棋以教育丹朱，陶冶其性情。目前围棋的最早可靠记载见于春秋时期的《左传》，战国时期的弈秋是见于史籍的第一位棋手，最早的围棋文物也可以追溯到这一時期。汉朝时棋盘为17路，三国时期出现了评定棋手水準的围棋九品制。南北朝时候，棋盘定型为现在的19路棋盘，傳入朝鮮半島。围棋逐渐成为中国知识阶层修身养性的一项必修课目，为“琴棋书画”文人四艺之一。
中國古代對圍棋尚有“弈”、“碁”、“手谈”、“坐隱”、“爛柯”、“方圓”、“黑白”、“烏鷺”、“大棋”、“木野狐”等雅稱，下圍棋又稱對弈、博弈、奕棋。弈局指棋局，弈枰、弈楸、楸枰、河洛指棋盤，奕具泛指棋盤、棋子等棋具，弈譜、弈選、吳圖指棋譜；觀弈指看棋；弈林指圍棋圈。其中“碁”為棋之異體字，在古籍中專指圍棋，如《隋書‧經籍志》所載棋譜目錄，均作碁。
古代圍棋並未发现完整明文的規則，但规则在逻辑关系上极其简单，成書於北周的《敦煌棋经》中略有所述。中國古代唐宋使用数路法，日本比目法即由此改变来，所不同者，唐宋规则要扣除眼位，而日本规则不扣除眼位。明朝以後，改为子空合并计算，不需保留死子，扣除眼位的过程以还棋头实现，所以明清时期的圍棋規則被稱為數子法。当代中国围棋规则承继明清规则，仍称为数子法。不过，此数子法非彼数子法：從南北朝的《敦煌棋经》到唐、宋、元、明、清，中国古棋一脉相承，为纯粹的数“子”法；而民国以来的中国数子法实为「子空皆地」的规则。
圍棋最遲在唐朝初期傳到了西藏，相傳松贊干布的大臣瓊波·邦色是一位圍棋高手。後來圍棋由藏族人改為西藏圍棋。
唐朝出现了棋待诏官职。著名棋手王积薪作“围棋十诀”大多在现代围棋中依旧适用。北宋沈括《梦溪笔谈》中有“四人分曹围棋”即四人围棋的记载。明朝王世贞寫了一篇《弈問》，回答圍棋的種種疑問。
清朝初年，中国古代围棋发展達到前所未有的高峰。大批著名棋手涌现，留下大量名局棋谱，如黃龍士与徐星友的“血泪篇”，施襄夏与范西屏的“当湖十局”。同时，围棋理论的研究亦达到一个高峰，代表作有徐星友的《兼山堂弈谱》、范西屏的《桃花泉弈譜》和施襄夏的《弈理指归》等。另外，《官子譜》等圍棋高級技巧專書也在此時期集結完成，可謂是棋之黃金盛世。但随后，中国围棋渐渐衰微，至20世纪上半叶竞技水準被日本超越。
傳入日本
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主条目：日本圍棋史
围棋在公元7世纪传入日本，很快就流行于宫廷和貴族之中。戰國末期，權臣丰臣秀吉设立碁所。德川幕府时代，出现了在天皇或征夷大将军面前对弈的“御城棋”，日本围棋逐渐兴盛，出现了本因坊、安井、井上、林等围棋世家。其中坊门尤其人才辈出，先后出现了道策、丈和、秀和、秀策、秀甫、秀荣等杰出棋士。日本围棋由于废除了中国古代围棋的座子制（中国古代围棋是放四个座子，就是两黑两白放在对角星的位置上，双方在这个基础上开始布局），布局理论得以极大发展。
同时由于日本棋手对唐代围棋数路法的胜负判定规则（以围地多少还是活子多少为目的——或者是否以其它目的行棋，还可继续考证）产生了直观上的误解，从而产生对唐宋数路法扣除眼位和公气、明清数子法终局还棋头（其本质是子多为胜、眼位不计的胜负计算方法）的疑惑，将其废除，从而产生了与中国古棋大为不同的地多为胜的日式围棋，并演变为现代围棋。
现代
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明治维新以后，棋手失去幕府支持，开始谋求新的谋生手段，导致新闻棋战和现代段位制的出现，并创立全国性的日本棋院。昭和时代，吳清源和木谷实共同掀起“新布局”的潮流，开始现代围棋的时代。其后日本棋界一流棋手辈出，如坂田荣男、藤泽秀行、高川格，及后来的大竹英雄、石田芳夫、武宫正树、小林光一等。
自晚清、民國以來，除了吳清源, 中国专业围棋水準相当低。从1960年开始，每年举办中日围棋友谊赛。其中，1961年第二届来华访问的日本围棋代表团中的55岁伊藤友惠（日语：伊藤友恵）五段（女性）横扫中国当时顶尖棋手，八轮全胜。1964年改称中日围棋对抗赛，一直举办到1992年。1984年，第一届中日围棋擂台赛开幕。中國人聂卫平在前四届擂台赛中取得11连胜，极大地推动中国围棋的普及。
1982年，中国恢复实行[註 1]段位制。1988年4月16日中国国家体委发布《围棋国家段位标准》《围棋国家段位标准实施细则》和《围棋地方段位制》。
1989年，韩国人曹薰铉在第一届应氏杯世界围棋锦标赛中夺冠，同样引发韩国围棋的热潮。此后，大量世界棋战出现。在这些棋战中，韓國李昌鎬从众多棋手中脱颖而出，成为当时棋界第一人。2006年后，李昌鎬的状态有所下滑，李世乭、古力等新锐势力开始对其发起冲击。此后数年为古力、李世石双雄争霸的年代。
日本第一人是連續奪得眾多頭銜的井山裕太，成為日本第一個同時擁有七冠頭銜的棋士，但日本已經難以和中韓抗衡，在國際賽上幾乎都是黯淡收場。
2010年代，中國眾多棋手崛起，江維杰、范廷鈺、時越、羋昱廷、陳耀燁、周睿羊、檀嘯等年輕棋士接連贏得世界比賽冠軍。2015年，柯洁崛起，一時風頭無兩，但其光芒迅速因AlphaGo于2016年的横空出世而变得黯淡。
2018年，柯洁重整旗鼓，在三星杯與百靈杯皆取得冠軍，成為中國圍棋第一人。2019年，中國隊主將柯洁終結韓國隊主將朴廷桓帶領中國隊拿下農心杯世界冠軍榮耀。
2020年11月3日，在第25屆三星杯三番棋決賽中，柯洁執黑以半目優勢战胜韓國棋手申真諝，从而以2-0的比分奪冠，成為中國圍棋史上最年輕的八冠王[來源請求]。
目前职业围棋水準最高的国家是中国和韩国。韓國雖然仍有少數頂尖高手支撐，但厚度不如中國。
围棋全球化
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围棋于19世纪晚期开始成规模的传入欧洲。在此进程中的第一大站是德国。
日本为围棋在全世界的推广用力最早、最多，以致西文中的围棋术语多来自日语。1980年代以降，中国、韩国也大力参与推广围棋。
1996年，美國宇航员丹巴利（Daniel T. Barry）博士與日本宇航员若田光一博士在奮進號太空梭上進行首次太空圍棋對弈，他們使用一套特別設計的太空圍棋棋具，由Wai-Cheung Willson Chow所造。兩位對弈宇航员因此得到日本棋院授予的榮譽段位。
圍棋運動现已遍佈世界各地，唯日本、韓國、中国大陆與台灣最為興盛；西方國家已漸熱；東南亞正在發展中。
参见：围棋史 § 東南亞的圍棋發展
棋具
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棋盘：围棋盘由19条横线19条竖线组成，棋子要下在线的交叉点上，方格中不能放入棋子。为了便于识别棋子的位置，棋盘上划了九个点，术语称做“星”，中央的星点又称为“天元”；下讓子棋时所让之子要放在星上。棋盘可分为“角”、“边”以及“中腹”。而現今的棋盤則有19×19、13×13、9×9，較為普遍，另外還有一些是較罕見的15×15、17×17。
棋子：围棋子分为黑白两色。棋子的数量应能确保顺利终局，中国规则和应氏规则要求正式比赛中黑、白各180子。棋子呈圆形。中国一般使用一面平、一面凸的棋子，日本则常用两面凸的棋子。中国云南所产的“云子”为历来的弈者所青睐，迄今已有五百餘年的历史。较为珍贵的棋子材料有贝壳、玛瑙等。
棋鐘：正式的比赛中可以使用棋鐘限制选手时间。非正式的对局中一般不使用棋鐘。
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规则
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参见：圍棋規則比較
基本规则
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下棋时，对弈双方各执一种颜色的棋子，黑先白后[註 2]，轮流将一枚棋子放置于交叉点上。与棋子直线相连的空白交叉点叫做气。当这些气都被对方棋子占据后，该棋子就没有了“气”，要被从棋盘上提掉。如果棋子的相邻（仅上下左右）直线交叉点上有了同色的棋子，则这两个棋子被叫做相连的。任意多个棋子可以以此方式联成一体，连成一体的棋子的气的数目是所有组成这块棋的单个棋子气数之和。如果这些气都被异色棋子占领，这块棋子就要被一起提掉。
在任何情况下，均禁止棋手向棋盘连下两子，否则将立刻判负。因此较文雅的中盘认输方法——投子，即是向棋盘摆下两枚棋子。
打劫规则
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主条目：劫 (围棋)
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劫是围棋中较特殊的一种情况。举例来说，当黑方某一手提掉对方一子，而这一个黑子恰好处于白方虎口之内，这时白方不能立即提掉这一黑子，而必须在棋盘其他地方着一手（称为“找劫材”），使黑方不得不应一手（称为“应劫”），然后白才能够提掉这个黑子。故高手下棋常常會保留一些先手作為劫材，一旦形成打劫，劫材多的一方打勝的可能性較大。
劫依其重要性分為天下劫、死活劫、官子劫等。天下劫表示整盤棋最重要的一個劫，意即不管選擇任何劫材威脅，對手都很可能直接消劫並得到巨大的利益，例如黑白各50子的大龍形成攻殺，這時劫材很可能不被理會。
禁止自殺規則
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下子時，除非能令對方某些子失去所有的氣，否則不得下子令自己某些子失去所有氣，這亦被稱為「禁止自殺規則」。應氏規則、紐西蘭規則及Tromp-Taylor規則對此略有不同。
胜负计算方法
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圍棋流傳多地，各有變化，演變出不同的規則。目前世界上使用较多的有中国规则、日韩规则和应氏规则。其本质皆以围地为目的行棋，由于对‘地’这一概念三者存在逻辑上互相平行的不同说明，一直有分歧。经过改进对“地”的定义说明，去除人为硬性规定，现在三种规则的实践中差异很小，只在极端情形下，才会有胜、平、负的差别。
日韩规则中，棋子所围成的空白交叉点叫做目，最终以目多的一方为胜方，所以日本的圍棋規則稱為比目法。对局时需保留死子，终局后双方将盘上死子及提掉的死子填入对方实空中，再计算双方实空。黑棋贴给白棋6目半（差不多等於3又3/4子和8點）。如果白棋实空加上贴目后多于黑方则白胜，否则黑胜。
计算胜负时，中国规则与日韩规则有所不同。按照中国规则（數子法），一盘棋结束后，依玩家們的認定將棋盘上的某些棋子拿掉，作為死子，如果双方对是否死子产生争议，通过实战解决。之后，如果黑棋活的棋子加上包圍的交叉點达到185（黑棋由于先行优势须贴给白棋3又3/4子，差不多等於7目半和8点）则黑胜，少于此数目则白胜。。
而应氏规则是，数各方占据的交叉点的数目，称作“计点”。其中黑棋贴8点（差不多等於7目半和3又3/4子），占点多的一方胜，和棋黑勝。除去比赛组织和棋士组织的一些规定外，就规则的核心部分而言，应氏规则与中国规则并无明显区别。
比目法和數子法在大多數情況下結果相同，但也有少數例外：比如雙活時，比目法不計地。數子法不必保存好提子，但是必須把所有單官（沒有目數差異的官子）填滿，而且若黑下到整盤棋的最後一個單官，則會多一子（比白子多下一手）。
无胜负与和棋
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在不贴目的时期，因为没有所谓“贴目”，时有和棋；自從加入了半目的貼目（或四分之三子）後，計算目數時便不會和局。但是仍然有一種情況會无法分出胜负，无法终局，即所谓无胜负，包括循環劫（三劫循環，四劫循環等）长生、双提二子。当黑白一方如果捨棄循环劫則會輸时，雙方不能停止提劫，便无法终局（无胜负）。现代較著名无胜负的棋局有古力和李世石在第17届三星财产杯F组胜者组次轮下出一盘四劫循環无胜负。但亦有某些規則不允許全局同型再現（如应氏规则），也就是棋盤不能和這盤棋以前的狀態一模一樣，從而限制无胜负的出现。
其实中国规则一直原则上规定不允许同形再现，但只是为了解决某些常见的“病态”棋形如“假生”、“盘角曲四”等而设，遇到三劫循环、长生等情况仍然按习惯法判无胜负。
棋局进程
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围棋对弈进程，分布局、中盘、收官三个阶段。而阶段之间并没有明显的界线。
布局
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在布局阶段，对局双方各自抢占棋盘上的空地，同时尽量阻止对方占地，由此导入中盘战斗。此时落子尚少，盘面尚大，无法作精确的计算，因而考验的是棋手的大局观和棋感。
以圍空效率而言，角上最高，中腹最差。一局棋往往以佔角開始，繼之以守角和掛角（阻止對方守角）形成角上的攻防，然後是拆邊和分投（阻止對方拆邊），即所謂「佔大場」。佈局時行棋以三四線為主。其中经过人们长久以来的经验累积，而形成的在某些情况下双方都会依循的固定下法被称为定式。
中盤
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中盤阶段时，双方为围地而作战，局面瞬息万变。中盘战术是对棋手攻守、死活、大局观等多方面的考验。
一个围棋边角上对战的实例（黑棋差一步，被白棋杀死）：
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收官
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收官指双方经过中盘战斗，地盘及死活已经大致确定之后，确立竞逐边界的阶段。此时要根据精密的计算，以正确的顺序收官，否则会遭受目数的损失。
打掛與封手
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在過去的日本，暫停棋局改天再下，稱為打掛。是棋力較高或是尊長持白棋者的權力。
古代的围棋，沒有团体讨论的传统。對弈之中，棋中不語，對弈者和旁觀者不會做討論。但是一盤棋如果中途暫停，改日再下，在休戰的期間，難以限制棋手不和他人討論。1933年，吴清源对本因坊秀哉的對局，秀哉利用白棋的特权频频打挂，每次打挂后就召集坊门弟子集体讨论。决定胜负的一手（第160手），据瀨越憲作于1948年座談會的“不能發表”之言，是秀哉的弟子前田陈尔发现的[需要較佳来源]（但前田拒不承認）。因此这一盘棋，实际上是本因坊团队与吴清源一人的较量。这局棋持续时间长达四个月，最后吴清源二目败。
後世普遍認為打掛並不公平，白棋棋手想不出應對可以宣告打掛回家思考，黑棋棋手卻不行。因此日本後來發明封手制度，黑白雙方都有機會宣告打掛，同時提出打掛一方應把要下的下一手棋寫下來密封，交給棋証保管，稱為封手。改日再戰時，把信封打開確認正確性之後，繼續對局。1938年秀哉的“名人引退棋”，對手木谷實就強烈要求使用封手制，并最終得以實現。
聯棋及团队围棋
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圍棋一般規則為一對一進行對弈。一对多或多對多的對弈，叫做聯棋或团队围棋。
联棋棋局進行中，隊友之間不會進行討論。常见的赛制是四人联棋，例如甲、乙兩人一隊持黑，丙、丁兩人一隊持白，則下棋的順序是甲下黑子，丙下白子，乙下黑子，丁下白子，如此周而復始。宋代学者沈括即谈到过这种比赛形式。2003年，韓國舉辦過一项名为“韩国四大天王联棋对抗赛”的活动，由曹薰铉、李昌镐師徒执白对阵李世石、刘昌赫。世界智力运动会等多个综合性智力运动会的围棋混双赛项目亦皆为四人联棋。另外，2015年开始举办的城市围棋联赛的赛制是将一盘棋按手数分为数个阶段，双方在每个阶段派不同的队员上场，并有换人名额，称为「接力赛」，同样是一种联棋。一对多的联棋则经常出现在棋界的纪念仪式中，如吴清源1984年引退时即举行了吴对众棋士的联棋，由吴的师兄桥本宇太郎开局，桥本第一手打在天元。
允许团队内部讨论的团队围棋比赛也出现过。1933年吴清源与秀哉名人的对局，执白的秀哉利用上手特权，多次宣布打挂，然后召集门人弟子集体讨论的做法，即可视作一种团队围棋。大约同一时期，以木谷实、吴清源为一方，铃木为次郎、濑越宪作为另一方（铃木、濑越分别是木谷、吴的师傅）曾下过名为相谈棋的比赛。中国国家围棋队内部一直有“加压棋”的传统，即一对多的训练棋，以此来对即将参加重要比赛的那个人进行强化训练。2013年开始、举办过两届的“珠钢杯”世界围棋团体赛的决赛为3对3的相谈棋赛。2017年中国乌镇围棋峰会中也举行过一对多的团队围棋赛，由AlphaGo对阵中国五名顶尖（人类）棋手。此外，互联网上也有提供团队围棋对弈的平台。
棋盘坐标系
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中国古代四隅旋分记谱法
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《忘忧清乐集》中的“图法”一节提到：
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| “ | 夫棋盘有三百六十一路，以分‘平、上、去、入’四字，各管一角，计九十路。棋盘以左手尊而为平。以角顺行，起一为首，顺行至十逆之，止九。若言‘六三’，先顺数六，而后逆数三；或言‘三六’，先顺数三，而后逆数六是也。 | ” |
《弈理指归》亦有介绍：
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| “ | 局分四隅，谱寓四声，平上去入，落子数数，先纵后横，纵十横九，平去同程，反横为纵，上入推行，纵路为左，横路为右，四角旋分，挨数无谬。 | ” |
即从左下角起顺时针将四角标记为平、上、去、入。先确定左角尖为“平一一”，横行向右得“平一二……平一九”，竖行向上得“平二一……平十一”。其余三角，依此类推，如“上”，则竖行向下顺数至“上一九”，再向右横数至“上十一”之；“去”，先从右上角向左横数至“去一九”，再向下竖数至“去十一”；“入”，则向上竖数至“入一九”，再横向左数至“入十一”。定点法是“平”先竖数，后横数，如“平六三”则是“平六一”与“平一三”的交点；“上”先横数，后竖数，如“上三七”在“上星位”的左下方；“去”先竖数，后横数，如“去八三”在“去星位”的右下方；“入”先横数，后竖数，如“入九三”就在“入星位”的左下方。这样，每个枰点，即所谓交叉点，都有且只有一个代码了。
横坐标为英文字母，纵坐标为阿拉伯数字的坐标系
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棋盘的上方和下方用英文的前19个字母自左至右顺次标记它的19条纵线段，在棋盘的左边和右边用阿拉伯文的前19个自然数自下至上顺次标记它的19条横线段。按照笛卡儿坐标系原理，建立起坐標系，与国际象棋所使用的坐标系相似。
字母和数字的有序排列就是枰点的坐标（英文字母为横坐标（x 軸），阿拉伯数字为纵坐标（y 軸））。棋盘上的每一个枰点都可用它的坐标（x, y）来表示。例如：J10 表示天元，P16 表示右上的星位，D4 表示左下的星位。
如下所示:
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| J10 | P16 | D4 |
靶式坐标法
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即将棋盘分为 A、B、C、D 四个区域，这样，便可只使用 0、1、2、3、4、5、6、7、8、9 十个阿拉伯数字来标记棋盘上的每一个枰点，使得记谱简便。
例如，右上星位的坐标为 A（4，4），右上三三的坐标为 A（3，3）；左下星位的坐标为 C（4，4），左下三三的坐标为 C（3，3）；上面边上星位的坐标是上（0，4），左面边上星位的坐标是左（4，0）；天元的坐标为（0，0）。
（按笛卡尔直角坐标法，枰点坐标为有序数对，先横（x）后纵（y）：（x，y）。）
横坐标和纵坐标都使用英文字母的坐标系
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即棋盘的上方和下方用英文的前 19 个字母自左至右顺次标记它的 19 条纵线段，在棋盘的左边和右边用英文的前 19 个字母自下至上顺次标记它的 19 条横线段。
其好处是：k、l、m、n、o、p、q、r、s 与 11、12、13、14、15、16、17、18、19 相比，读或写都更简单。
有一些系统在使用字母标记时，越过字母「I」或者「J」（例如 IGS 是越过 I）。 其理由据说是怕引起混淆，实际使用中应先确认协议规定。
网络围棋
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主条目：網路圍棋
自从互联网的兴起，网络下棋成为了大众围棋对弈的主要形式。網路圍棋比赛也日渐兴盛起来。网络围棋寻找棋友方便，对弈后资料保留妥当。网络围棋比赛方兴未艾。而且网络围棋可在對弈時，模擬下步結果和推演，加強戰略上的發展。根據對局時間限制與否，網絡圍棋可大體劃分為即時對弈和非即時對弈兩類。目前的主流對弈平臺大多屬于「即時對弈平臺」，但「非即時對弈平臺」在近幾年也逐漸開始流行。
即时对弈網站
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最早的国际性围棋网上对弈站点应该是IGS（Internet Go Server），後改名為Pandanet，为日本PANDA-GO公司所有。Pandanet现在仍然是日本最大的国际性围棋网站，但活躍人數已被中國及韓國圍棋網站超越，如第一大的中國野狐跟第二大的韓國Tygem。
過去PANDA-GO公司同时出品用于IGS的对弈客户端glGo，该软件可以实现网络对弈，也可以配合开源软件GNU Go实现人机对弈，据称GNU Go的最高水準大约在业余初段上下。IGS同时也支持第三方的客户端，例如Sente开发的Goban（仅运行于Mac OS X平台）。IGS 改名為 Pandanet 後現今有同時支援 Windows、Mac 及 Linux 的客戶端 GoPanda2，也推出支援 Android 跟 iOS 的客戶端。
除Pandanet外，主要的对弈网站还有OGS、KGS、野狐、Tygem、GoQuest（手機平台）、Tom棋圣道场、弈城、新浪、联众、LGS、台灣圍棋網、QQ围棋、台灣棋院文化基金會等等，多数网站都使用自己开发的专属客户端。
围棋软件与人工智能
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主条目：计算机围棋
與其它棋類遊戲相比，電腦模擬的圍棋人工智能棋力進展相對緩慢。在1997年，就有電腦可以擊敗世界西洋棋棋王卡斯帕洛夫，但圍棋軟件直到2016年才有辦法擊敗頂尖圍棋棋士。這是由於国际象棋目標明確，只要殺死國王即可（係出同源的象棋、將棋狀況也差不多），因此演算法較為簡單；但圍棋目標不一定需要圍殺對方棋子，每一步有數百種以上的走法，黃庭堅有詩：「心似蛛絲游碧落，身如蜩甲化枯枝。」，因此演算法的困難度明顯要高得多。
在宋代的《梦溪笔谈》中探討了圍棋的局數變化數目，作者沈括稱「大約連書萬字四十三個，即是局之大數」，意思是說要寫43個萬字（一個萬即104，共43個104，即為10172）。實際數字約為319×19=3361≈1.74×10172。根据围棋规则，没有气的子不能存活，扣除这些状态后的合法状态（占1.196%）约有2.08×10170种。Robertson与Munro在1978年证得n乘n棋盤上的圍棋判定殘局勝負的問題是PSPACE-hard的，其必胜法之记忆计算量在10600以上，這遠遠超過可觀測宇宙的原子總數1075。
1985年台湾著名实业家应昌期懸賞一百萬美元，寻找能夠打敗職業棋士的電腦程序而不可得。「深藍」設計人許峰雄在2007年10月一期的《IEEE Spectrum》杂志上表示，相信10年内以穷举搜索为基础的超级电脑将能挑战世界冠军级别的棋手。早年（上世纪90年代）中国陈志行开发的手谈曾屡获世界计算机围棋大赛冠军，到了二十一世纪初，較著名的圍棋對弈軟件分別有法國的MoGo、韩国的KCC IGO（银星）、日本的Zen（日语：天頂の囲碁）等電腦圍棋程序。其中Zen在連續幾年的電腦圍棋大賽上均獲第一，與台灣職業九段周俊勳對弈，在被讓四子的情況下可贏十目以上，被評定已有業余5段的水準；而最普遍应用且功能較齊全的是开源围棋程序GNU Go。
2016年1月27日，《自然》的封面论文报道了Google DeepMind开发的新围棋軟件AlphaGo，文中寫明，AlphaGo在没有任何让子的情况下以5:0完胜曾獲欧洲冠军的职业围棋二段樊麾。这是有史以来围棋軟件第一次在公平比赛中战胜职业棋手，被視為人工智能的里程碑。AlphaGo的核心是两种深度神经网络——策略网络和价值网络。2016年3月，AlphaGo挑戰韩国職業九段李世乭，以4：1擊敗了對手。2016年12月29日至2017年1月4日间，AlphaGo以“Master”为账号在围棋对弈网站上，以快棋（20秒～30秒）击败了中國和韓國排名第一的柯洁、朴廷桓，以及其他多名顶尖的棋手，创造了60连胜的记录。2017年5月，AlphaGo 在中国乌镇围棋峰会的三局比赛中击败当时世界排名第一的中国棋手柯洁。
AlphaGo之后，其公開的論文，促成了中国的绝艺、星陣，日本的DeepZenGo,美國的KataGo以及台灣的CGI也达到甚至超过了人类顶尖职业高手的水準。
符號編碼
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在Unicode，黑白子分別記載於幾何圖形列表區塊。
U+25CB ○ WHITE CIRCLE ，HTML：○
U+25CF ● BLACK CIRCLE ，HTML：●
而记录符號則分別記載於雜項符號區塊。
U+2686 ⚆ CIRCLE WITH DOT RIGHT ，HTML：⚆
U+2687 ⚇ WHITE CIRCLE WITH TWO DOTS ，HTML：⚇
U+2688 ⚈ BLACK CIRCLE WITH WHITE DOT RIGHT ，HTML：⚈
U+2689 ⚉ BLACK CIRCLE WITH TWO WHITE DOTS ，HTML：⚉
围棋文化
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围棋术语
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主条目：围棋术语
围棋在长达数千年的流传过程中，逐渐产生众多的术语。其中有一些已经进入大众词汇，例如“收官”等。围棋的每一步，几乎都有相关的术语表达，甚至已经可以完整地用文字描述一场局部战斗。例如：黑右上角占据星位，白小飞挂，黑二间高夹，白一间跳，黑跟着跳，白小飞进角，黑三三挡住……这段话用围棋术语，生动地描写常见的定式进程。[原創研究？]
寓意
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棋盘为方，其子为圆，子覆盘上寓意“天圆地方”。
子分黑白，寓意阴阳。
棋盘共361个点，360暗合一年天数约数，天元一点寓意万物自一而始。
9个星位暗合九宫之数，星位将棋盘分为四个象限，寓意一年四季，每个象限约为90个落子点，寓意每季天数。
棋盘周边共72点，寓意一年七十二候。
[原創研究？]
圍棋題材作品
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更多信息：Category:圍棋題材作品
目前留存的最早的围棋专著是《弈旨》，为西漢班固所著。梁朝时，梁武帝蕭衍曾著《圍棋賦》《圍棋品》《棋法》。至唐、宋盛世，有王積薪著《围棋十诀》，張擬著《棋經》，劉仲甫著《忘忧集》、《棋势》、《棋诀》、《造微》、《精理》（今仅《棋诀》存）及李逸民編《忘忧清乐集》。元朝嚴師和晏天章曾作《玄玄棋經》。明、清两朝有王世貞著《弈問》，过百龄著《官子谱》、《三子譜》、《四子譜》，徐星友著《兼山堂弈譜》，施襄夏著《弈理指歸》。[原創研究？]
明代其他圍棋著作還有：《烂柯经》、《决胜图》、《玄通集》、《适情录》、《秋仙遗谱》、《石室先机》、《石室秘传》、《玉局藏机》，其作者不详。
清朝较著名的古棋譜有《血淚篇》（黃龍士與徐星友的對局）和《當湖十局》（施襄夏與范西屏的對局）。
围棋赛事
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主条目：围棋比赛和世界围棋大赛
现存的主要世界围棋大赛：
| 届 | 杯名 | 全称 | 主办贊助商 | 冠军奖金 |
--- ---
| 9 | 应氏杯 | 应氏杯世界围棋锦标赛 | 应昌期围棋教育基金会 | 40万美元 |
| 25 | LG杯 | LG杯世界围棋棋王战 | LG集團、朝鲜日报社 | 3亿韩元 |
| 25 | 三星杯 | 三星车险杯世界围棋大师赛 | 三星火灾海上保险 | 3亿韩元 |
| 13 | 春兰杯 | 春兰杯世界围棋锦标赛 | 江苏春兰集团 | 15万美元 |
| 3 | 百灵杯 | 百灵杯世界围棋公开赛 | 贵州百灵企业集团 | 180万人民币 |
| 3 | 梦百合杯 | 梦百合杯世界围棋公开赛 | 梦百合 | 180万人民币 |
| 1 | 新奥杯 | 新奥杯世界围棋公开赛 | 新奥集团 | 220万人民币 |
| 1 | 天府杯 | 天府杯世界圍棋職業錦標賽 | 四川天府新區 | ~200万人民币 |
| 30 | 亚洲杯 | 亚洲杯电视快棋赛 | CCTV NHK KBS | 250万日元 |
| 22 | 农心杯 | 农心辛拉面杯三国围棋擂台赛 | 农心食品 | 5亿韩元 |
| 9 | 穹窿山兵圣杯 | 穹窿山兵圣杯世界女子围棋锦标赛 | 苏州市穹窿山旅游区 | 25万人民币 |
| 8 | 黄龙士双登杯 | 黄龙士双登杯三国女子围棋擂台赛 | 江苏双登集团公司 江苏太平洋精密锻造有限公司 | 45万元人民币 |
| 6 | 华顶茶业杯 | 天台山世界女子围棋团体锦标赛 | 天台华顶茶业公司 | 30万元人民币 |
| 2 | 男子团体 | 国际智力运动联盟智力运动精英赛 | 国际智力运动联盟 | 3.5万美元 |
| 女子团体 | 2万美元 |
| 男子个人 | 1.5万美元 |
| 女子个人 | 1.2万美元 |
| 混合双人 | 2万美元 |
圍棋變種
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主条目：围棋变体
参见：Category: 围棋变体
巡將圍棋
西藏圍棋
四人围棋
數學卡片棋
迷你圍棋
吃子棋
不圍棋(闕棋)
圍點棋
環棋
系碁
注释
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^ 1960年代中国曾在小范围内授予过一些棋手职业段位。
^ 这是日本古棋及现代围棋的约定，中國古谱则黑先白先均有，且白先居多。
参考资料
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^ 中華民國教育部，《異體字字典》。. [2007-02-13]. （原始内容存档于2014-07-14）.
^ 王, 世贞. 弇州四部稿. 维基文库 （中文）. 尧作围棋，乌曹作博，见《世本》。
^ Matthews, Charles. Teach Yourself Go, p.1
^ Great Quotes at Sensei's Library. [2013-07-23]. （原始内容存档于2013-07-27）.
^ 未來數學家的挑戰 - 楊照崑;楊重駿. [2009-08-30]. （原始内容存档于2018-07-12）.
^ About the IGF, [June 5, 2012], （原始内容存档于2012年4月29日）
^ International Go Federation, IGF members, [July 30, 2015], （原始内容存档于2015年9月5日）
^ 陈元龙. 格致镜原·卷五十九. 维基文库 （中文）. 《博物志》：尧造围棋，以教子丹朱。
^ 司马迁. 史记·五帝本纪. 维基文库 （中文）. 尧知子丹朱之不肖，不足授天下，于是乃权授舜。
^ 围棋史话. [2011-04-11]. （原始内容存档于2010-09-18）.
^ 左丘明. 左传·襄公二十五年. 维基文库 （中文）. 今宁子视君不如弈棋，其何以免乎？弈者举棋不定，不胜其耦。
^ 孟子及其弟子. 孟子·告子上. 维基文库 （中文）. 弈秋，通国之善弈者也。
^ 古中山国考古出土疑似“围棋”. [2011-07-03]. （原始内容存档于2011-07-05）.
^ 张拟. 棋经. 维基文库 （中文）. 夫围棋之品有九。一曰入神，二曰坐照，三曰具体，[...]，九曰守拙。
^ 围棋. [2011-04-11]. （原始内容存档于2020-08-18）.
^ 程晓流. 藏棋起源之初步探讨. 围棋天地. 2005, (16): 59–60.
^ 沈括. 梦溪笔谈·技藝. 维基文库 （中文）. 四人分曹共圍棊者，有術可令必勝
^ 弈问. [2011-04-11]. （原始内容存档于2010-11-29）.
^ 黃俊. 弈人传·卷十一. 维基文库 （中文）. 文长不备录，录其《弈问》一篇，其词曰：[...]
^ 體育. 中國人民大學書報資料社. 1997: 62.
^ 囲碁の歴史, Nihon Kiin, [2015-08-19], （原始内容存档于2015-08-27）
^ 本因坊和名人棋所的由来. [2013-07-24]. （原始内容存档于2012-08-19）.
^ 陈祖德在他的自传《超越自我》中痛心疾首地记载下伊藤友惠老太太1961年访华横扫我国围棋名宿，豪取八连胜的过程。陈祖德称之为“国耻”。
^ 中日围棋擂台赛. [2013-07-24]. （原始内容存档于2008-07-25）.
^ 华学明. 谁领跑围棋世界？. www.sport.gov.cn. 2016-02-04 [2023-03-30]. （原始内容存档于2023-03-30） （中文（中国大陆））. 升段制度是在1982年推出的，当时推出的时候中国是第一次推出职业的段位，整个模式是仿照日本。
^ Introduction. [2013-07-24]. （原始内容存档于2013-07-09）.
^ 新华社. 围棋——柯洁获第25届三星杯世界大师赛冠军_申真谞. www.sohu.com. [2023-04-12]. （原始内容存档于2023-04-17）.
^ “90后”围棋新锐风生水起 从三足鼎立到一马当先. [2013-07-23]. （原始内容存档于2013-05-31）.
^ Peng & Hall 1996
^ 跳转到： 30.0 30.1 30.2 30.3 中国围棋竞赛规则（2002年版）. [2013-07-18]. （原始内容存档于2018-10-28）.
^ 跳转到： 31.0 31.1 31.2 31.3 应氏计点制围棋规则. [2013-07-18]. （原始内容存档于2013-08-30）.
^ 陈祖源《围棋规则新论》ISBN 978-7-80548-687-1
^ 日本围棋规则(1989版). [2013-07-24]. （原始内容存档于2016-03-05）.
^ 中国围棋规则(1988). 中华人民共和国体育运动委员会.
^ 围棋布局篇：金角银边草肚皮. [2013-07-23]. （原始内容存档于2016-03-04）.
^ 围棋中盘战术总结. [2013-07-23]. （原始内容存档于2013-05-12）.
^ 金庸散文《历史性的一局棋 （页面存档备份，存于互联网档案馆）》
^ 存档副本. [2017-02-01]. （原始内容存档于2016-03-21）.
^ 豆丁网. 围棋入门知识. 围棋入门知识. [2016-06-18]. （原始内容存档于2016-06-29）.
^ Jim Z Yu. A Brief History of IGS, the Early Years - 1992. [2007-12-11]. （原始内容存档于2007-11-28）.
^ 《夢溪筆談》卷十八《技藝》：「……凡方二路，用四子，可变八十一局。方三路，用九子，可变一万九千六百八十三局。方四路，用十六子，可变四千三百四万六千七百二十一局。方五路，用二十五子，可变八千四百七十二亿八千八百六十万九千四百四十三局。方六路，用三十六子，可变十五兆九十四万六千三百五十二亿八千二百三万一千九百二十六局。方七路以上，数多无名可记，尽三百六十一路。大约连书万字五十二，即是局之大数。」
^ 錢寶琮：《宋元數學史論文集》
^ Counting Legal Positions in Go. tromp.github.io. [2025-01-14]. （原始内容存档于2016-02-06）.
^ "Robertson, E. and Munro, I.〈NP-completeness, puzzles, and games〉Utilitas Math., 1978, 99-116."
^ 未來數學家的挑戰 (第 6 頁). episte.math.ntu.edu.tw. [2025-01-14]. （原始内容存档于2025-01-14）.
^ Re: How many atoms make up the universe?. www.madsci.org. [2025-01-14]. （原始内容存档于2000-06-20）.
^ 人工智慧加持電腦圍棋棋力再突破——周俊勳：19路讓4子電腦圍棋近最佳化. [2012-12-01]. （原始内容存档于2014-08-12）.
^ 澳洲IEEE WCCI 2012台灣魔圍棋7路棋盤小勝人腦——開戰首日周俊勳與黑嘉嘉勝美國及加拿大19路電腦圍棋. [2012-12-01]. （原始内容存档于2012-06-16）.
^ Silver, David; Huang, Aja; Maddison, Chris J.; Guez, Arthur; Sifre, Laurent; van den Driessche, George; Schrittwieser, Julian; Antonoglou, Ioannis; Panneershelvam, Veda; Lanctot, Marc; Dieleman, Sander. Mastering the game of Go with deep neural networks and tree search. Nature. 2016-01, 529 (7587) [2020-10-04]. ISSN 1476-4687. doi:10.1038/nature16961. （原始内容存档于2023-06-18） （英语）.
^ Google’s AlphaGo Continues Dominance With Second Win in China. 2017-05-25 [2017-05-27]. （原始内容存档于2017-05-27）.
^ 柯洁迎19岁生日 雄踞人类世界排名第一已两年. May 2017 [2017-05-27]. （原始内容存档于2017-08-11） （中文）.
^ World's Go Player Ratings. 24 May 2017 [2017-05-27]. （原始内容存档于2017-04-01）.
^ 星定式——小飞挂——二间高夹. [2013-07-25]. （原始内容存档于2014-09-13）.
參见
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維基教科書中的相關電子教程：围棋
中国主题
游戏主题
围棋十诀
圍棋譜
中國流
电脑围棋
網路圍棋
香港圍棋
台灣圍棋
圍棋棋手列表
圍棋規則比較
游戏复杂度
圍棋段位制
外部链接
[编辑]
棋院組織
[编辑]
中国围棋协会的哔哩哔哩个人空间（中华人民共和国）
日本棋院（页面存档备份，存于互联网档案馆）
韓國棋院（页面存档备份，存于互联网档案馆）
中華民國圍棋協會（页面存档备份，存于互联网档案馆）
台灣棋院（页面存档备份，存于互联网档案馆）
台灣圍棋發展協會（页面存档备份，存于互联网档案馆）
台灣圍棋教育推廣協會
香港圍棋文化學會（页面存档备份，存于互联网档案馆）
香港圍棋協會（页面存档备份，存于互联网档案馆）
香港圍棋發展中心
中國香港棋院（页面存档备份，存于互联网档案馆）
打譜軟體及棋譜
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Drago 免费多用途围棋辅助软件（页面存档备份，存于互联网档案馆）
MultiGo 免费多用途围棋辅助软件（页面存档备份，存于互联网档案馆）
Web2Go 奇譜士網站（页面存档备份，存于互联网档案馆）
GoKifu 围棋棋谱（页面存档备份，存于互联网档案馆）
圍棋百科
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圍棋百科（页面存档备份，存于互联网档案馆）（LGS 傳奇圍棋網 wiki）
《IEEE Spectrum》杂志 Cracking GO一文，作者许峰雄（页面存档备份，存于互联网档案馆）
| 查 论 编 围棋术语 | | |
---
| 基本概念 | 目 氣 眼 布局 收官 死活 雙活 打劫 | |
| 棋盤位置 | 天元 星位 小目 高目 目外 三三 |
| 着法或手段 | | | | --- | | 一字術語 | 長 立（日语：サガリ (囲碁)） 飛 尖 虎（朝鲜语：호구 (바둑)） 拆 提 關 衝 跳（日语：トビ (囲碁)） 曲（日语：マガリ）（拐） 鎮（日语：ボウシ） 夾（日语：ハサミ (囲碁)） 斷（日语：キリ (囲碁)） 跨 刺（覷） 托 退 碰 壓 爬（日语：ハイ (囲碁)） 接（日语：ツギ） 頂（日语：ツキアタリ） 並（日语：ナラビ） 扳（日语：ハネ） 擋（日语：オサエ） 雙（朝鲜语：쌍립） 擠 逼 封（日语：カケ） 點（日语：ナカデ） 渡 撲 | | 二字或以上 | 叫吃 佔角 掛角 締角 征子 緩征（日语：ユルミシチョウ） 緊氣 脫先（日语：手抜き (ゲーム)） 投子 割分 整地 倒脫靴 滾打包收 | |
| 棋形 | 直三 直四 彎三 彎四 方四 板六 笠帽四(丁四) 刀把五 梅花五 拳頭六 大豬嘴 小豬嘴 長生劫 金雞獨立 盤角曲四（日语：隅のマガリ四目） |
| 布局 | 小林流 星無憂角 中國流 迷你中國流 錯小目 三连星 宇宙流 |
| 其他 | 定式 手合 見合 手割 手筋 隨手 虛手 外勢 味道 先手 後手 模樣 詰棋 座子制 七死八活 假眼活 金角銀邊銅肚皮 围棋十诀 |
| 查 论 编 围棋 | | |
---
| 历史 棋手 | | |
| 棋具（英语：Go equipment） | 碁盤（日语：碁盤） 碁石（日语：碁石） 棋鐘 棋谱 | |
| 规则 | | | | --- | | 共同 | 贴目 封手 限时（日语：持ち時間#囲碁） 让子 打劫 | | 差异 | 中国规则 应氏规则 日本规则 座子制（中国古棋规则） 美国规则（俄语：Правила AGA） Tromp-Taylor规则（俄语：Правила Тромпа — Тейлора） | |
| 对局阶段 | 布局 + 定式 中盘 收官 |
| 布局流派 | 中国流 高中国流 迷你中国流 小林流 宇宙流 三连星 星无忧角 二连星 新布局 |
| 术语 | 术语列表 手筋 劫 骗着（法语：Hamete） 十番棋（英语：Jubango） 道策流 手割 |
| 棋力等级 | 段级位制 名人 九品制 等级分 Go Ratings |
| 组织 | | | | --- | | 古代 | ~~棋待诏~~ ~~碁所~~ ~~本因坊家~~ ~~方圆社~~ | | 现代 | ~~中国棋院~~ → 中国围棋协会 日本棋院 关西棋院 韩国棋院 中國圍棋會 台湾棋院 美國圍棋協會（AGA） 新西兰围棋会（英语：New Zealand Go Society）（NZGS） 欧洲围棋联盟（英语：European Go Federation）（EGF） 法国围棋协会（法语：Fédération française de go）（FFG） 俄罗斯围棋联合会（俄语：Российская федерация го）（РФГ） 國際圍棋聯盟（IGF） | |
| 棋赛 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | 世界 大赛 | | 查 论 编 圍棋世界賽事 | | --- | | | | | | | | | --- --- --- | | 大型個人賽 | | | | --- | | 錦標賽 | 應氏杯 春蘭杯 | | 公開賽 | 三星杯 LG盃 夢百合杯 爛柯杯 南洋杯 北海新繹杯 | | | 團體賽 | | | | --- | | 擂台賽 | 農心辛拉麵杯 黃龍士杯女 農心白山水杯元老 | | | 其他個人賽 與邀請賽 | | | | --- | | 錦標賽 | 國手山脈杯JKT 世界最高棋士戰 GLOBIS杯青 | | 公開賽 | 吳清源杯女 扇興杯女 | | 邀請賽 | 阿含·桐山杯中日對抗賽 龍星戰 | | | 綜合運動會 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | --- | | | | | | | | | --- --- --- | | 大型個人賽 | | | | --- | | 錦標賽 | 富士通杯 天府杯 | | 公開賽 | BC信用卡盃 新奧杯 百靈杯 | | | 團體擂台賽 | 中日圍棋擂台賽 SBS杯 真露杯 中韓圍棋擂台賽 湖盤杯女 正官庄杯女 | | 團體對抗賽 | 中日圍棋對抗賽 樂天杯 日月星杯 招商地產杯 CSK杯 珠鋼杯／金龍城杯 國手山脈杯 天台山賽女 | | 其他個人賽 與邀請賽 | | | | --- | | 錦標賽 | 東洋證券杯 世界冠軍錦標賽JKT 中環盃JT 世界女子圍棋錦標賽女 亞洲杯電視快棋賽 穹窿山兵聖杯女 | | 公開賽 | 豐田杯 利民杯青 | | 邀請賽 | 古李十番棋 中日名人對抗賽 中日天元對抗賽 中韓天元對抗賽 三國混雙賽 中日精英賽 名人爭霸戰 巔峰對決 賀歲杯 | | | 綜合運動會 | 世界智運會圍棋賽 世界智英會 智英賽 亞洲室內與武道運動會圍棋賽（2013） | | | | | | | | 註：女：僅設女子組的比賽。元老：僅限一定年齡以上高齡選手參加的比賽；青：僅限一定年齡以下青年選手參加的比賽；斜體：籌辦中的比賽。 本模板中“大型個人賽”是指中國圍棋協會、日本棋院、韓國棋院、海峰棋院四方中三方共同認可的圍棋世界大賽。其他未獲共同認可的個人賽中，中圍協單獨認可的標“C”，日本棋院單獨認可的標“J”，韓國棋院單獨認可的標“K”，台灣海峰棋院單獨認可的標“T”。 | | | | 中国 | 中国围棋甲级联赛 棋圣战 名人战 天元战 倡棋杯 当湖十局杯CCTV电视快棋赛 阿含·桐山杯 威孚房开杯 西南棋王赛 王中王战 龙星战 全国个人赛 ~~理光杯~~ ~~烂柯杯~~ | | 日本 | 棋圣战 名人战 本因坊战 王座战 十段战 天元战 碁圣战 NHK杯 阿含·桐山杯 龙星战 新人王战 关西棋院第一位决定战 ~~NEC杯~~ ~~大手合~~ | | 韩国 | 韩国围棋联赛 名人战 GS加德士杯 麦馨杯九段最强战 KBS杯 龙星战 韩国最高棋士战 韓國圍棋TV杯 ~~国手战~~ ~~棋圣战~~ ~~最高位战~~ ~~王位戰~~ ~~大王戰~~ ~~棋王战~~ ~~巴卡斯杯天元战~~ | | 其他 | 欧洲围棋大会 北美围棋大师赛（日语：北米INGマスターズ#北米マスターズ・トーナメント） 世界业余围棋锦标赛 世界大学生围棋赛 UEC杯世界电脑围棋大会 | |
| 文艺作品 | | | | --- | | 文学 | 名人 Shibumi（英语：Shibumi (novel)） Satori（英语：Satori (Winslow novel)） 围棋少女 Walking on Glass（英语：Walking on Glass） 流行公司（英语：PopCo） 雅各布·德佐特的千秋（英语：The Thousand Autumns of Jacob de Zoet） Starborne（英语：Starborne） | | 影视 | 棋魂 围棋少年 终极对弈 吴清源 美丽心灵 英雄 死亡密码 創戰紀 少狼 石头（英语：The Stone (2013 film)） 神之一手（英语：The Divine Move） 未生 請回答1988 | |
| 人工智能 | 计算机围棋 AlphaGo AlphaGo Zero Darkforest 绝艺 DeepZenGo KataGo Leela Leela Zero PhoenixGo CGI ELF OpenGo SAI AlphaGo李世石五番棋 中国乌镇围棋峰会 + AlphaGo对阵柯洁 |
| 圍棋變體 | 圍棋變體 西藏圍棋 巡將圍棋 不圍棋 四人围棋 迷你围棋 吃子棋 |
| 其它 | 当湖十局 血泪篇 镰仓十番棋 古李十番棋 朴申七番对决 吐血之局 耳赤之局 原爆之局 中日围棋擂台赛 世界棋王赛死子违规风波 |
| 教科书 Portal:围棋（俄语：Портал:Го） | | |
| 查 论 编 博弈论专题 | |
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| 定义 | 正则形式的博弈 · 扩展形式的博弈 · 图博弈论 · 合作博弈 · 信息集 · 偏好 |
| 均衡概念（英语：Solution concept） | 纳什均衡 · 强纳什均衡（英语：Strong Nash equilibrium） · 子博弈均衡（英语：Subgame perfect equilibrium） · 贝叶斯-纳什均衡 · 贝叶斯完美均衡（英语：Perfect Bayesian equilibrium） · 颤抖手完美均衡 · 恰当均衡（英语：Proper equilibrium） · ε-均衡 · 相关均衡 · 序贯均衡 · 准完美均衡（英语：Quasi-perfect equilibrium） · 进化稳定策略 · 风险占优（英语：Risk dominance） · 帕累托最优 · 自我应验均衡（英语：Self-confirming equilibrium） · 马尔可夫完美均衡（英语：Markov perfect equilibrium） · 默滕斯稳定均衡（英语：Mertens-stable equilibrium） · 核（英语：Core (game theory)） · 夏普利值（英语：Shapley value） · 吉布斯均衡（英语：Potentialg ame） · 量子响应均衡（英语：Quantal response equilibrium） · 谢林点 |
| 策略 | 优势策略 · 纯策略 · 混合策略 · 以牙還牙 · 冷酷触发策略（英语：Grim trigger） · 策略复制论证（英语：Strategy-stealing argument） · 逆向归纳法（英语：Backward induction） · 前向归纳法（英语：Forward induction） · 马尔可夫策略（英语：Markov strategy） |
| 博弈类型 | 对称博弈 · 完美信息 · 序贯博弈 · 重复博弈 · 信号博弈 · 廉价磋商（英语：Cheap talk） · 零和博弈 · 机制设计 · 随机博弈 · 非传递博弈 · 全局博弈（英语：Global game） · 甄别博弈（英语：screening game） · 讨价还价问题（英语：Bargaining problem） · 多人博弈（英语：n-player game） · 大型泊松博弈（英语：Large Poisson game） · 严格决定博弈 · 潜博弈（英语：Potential game） · 位勢賽局 |
| 博弈模型 | 围棋 · 國際象棋 · 无限棋（英语：Infinite chess） · 西洋跳棋 · 井字棋 · 囚徒困境（可选择的囚徒博弈（英语：Optional prisoner's dilemma） · 用餐者困境） · 旅行者困境 · 猜均值的2/3 · 协调博弈（英语：Coordination game） · 蜈蚣博弈 · 志愿者困境 · 搭便车问题 · 拍卖美元 · 膽小鬼博弈 · 智猪博弈 · 性别战 · 獵鹿賽局 · 賭便士（英语：Matching pennies） · 最後通牒賽局（海盗博弈） · 石头、剪子、布 · 獨裁者賽局（信任游戏） · 公共財賽局（英语：Public goods game） · 纳什讨价还价问题（英语：Nash Bargaining Game） · 上校賽局 · 消耗战 · 少数派博弈（El Farol酒吧问题） · 公平分配博弈（切蛋糕问题（英语：Fair cake-cutting）） · 古诺竞争 · 死結 · 库恩扑克游戏（英语：Kuhn poker） · 甄别博弈（英语：Screening Game） · 公主与怪兽游戏（英语：Princess and monster game） · 约会问题（英语：Rendezvous problem） · 囚徒帽子谜题（英语：Prisoners and hats puzzle） |
| 定理 | 极值定理 · 纯化定理（英语：Purification theorem） · 无名氏定理 · 显示定理（英语：Revelation principle） · 阿罗不可能定理 · 极小化极大算法 · 纳什均衡 · 策梅洛定理 |
| 关键人物（英语：List of game theorists） | 阿尔伯特·W·塔克 · 阿摩司·特沃斯基 · 阿里埃勒·鲁宾斯坦 · 克劳德·香农 · 丹尼尔·卡内曼 · 戴维·K·莱文（英语：David K. Levine） · 戴维·M·克雷普斯（英语：David M. Kreps） · 唐纳德·B·吉利斯（英语：Donald B. Gillies） · 朱·弗登博格（英语：Drew Fudenberg） · 埃里克·马斯金 · 哈罗德·W·库恩 · 赫伯特·亚历山大·西蒙（司马贺） · 埃尔维·穆兰（英语：Hervé Moulin） · 让·梯若尔 · 让-弗朗索瓦·默滕斯（英语：Jean-François Mertens） · 珍妮弗·图尔·蔡司（英语：Jennifer Tour Chayes） · 夏仙義·亞諾什·卡羅伊 · 约翰·梅纳德·史密斯 · 安托万·奥古斯丁·库尔诺 · 约翰·福布斯·纳什 · 约翰·冯·诺伊曼 · 肯尼斯·阿罗 · 肯尼思·宾默尔 · 里奥尼德·赫维克兹 · 劳埃德·沙普利 · 梅尔文·德雷希尔（英语：Melvin Dresher） · 梅里尔·M·弗勒德 · 奧嘉·邦達雷娃（英语：Olga Bondareva） · 奥斯卡·莫根施特恩（英语：Oskar Morgenstern） · 保罗·米尔格龙 · 佩顿·杨（英语：Peyton Young） · 赖因哈德·泽尔腾 · 羅伯特·阿克塞爾羅 · 罗伯特·约翰·奥曼 · 罗伯特·B·威尔逊 · 罗杰·梅尔森 · 塞缪尔·鲍尔斯（英语：Samuel Bowles (economist)） · 苏珊娜·斯科奇姆 · 托马斯·克罗姆比·谢林 · 威廉·维克里 |
| 参见 | 全支付拍卖 · Alpha-beta剪枝 · 伯川德悖论（英语：Bertrand paradox (economics)） · 有限理性 · 組合博弈論 · 对抗分析（英语：Confrontation analysis） · 合作性競爭 · 棋局中的先手优势（英语：First-move advantage in chess） · 博弈机制（英语：Game mechanics） · 博弈论词汇表（英语：Glossary of game theory） · 博弈理论家列表（英语：List of game theorists） · 特殊博弈列表 · 雙輸 · 国际象棋的解局策略（英语：Solving chess） · 拓扑博弈（英语：Topological game） · 公地悲劇 · 小决定暴政 |
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Home > Skin nontumor > Bullous pemphigoid
Skin nontumor
Vesiculobullous and acantholytic reaction patterns
Bullous pemphigoid
Authors:Michael Occidental, M.D., Randie H. Kim, M.D., Ph.D.
Editorial Board Member:Kiran Motaparthi, M.D.
Last author update: 22 June 2021
Last staff update: 22 June 2021
Copyright: 2002-2025, PathologyOutlines.com, Inc. PubMed Search: skin bullous pemphigoid [title] pathology review[ptyp]
Page views in 2024: 45,886
Page views in 2025 to date: 32,580
Table of Contents
Definition / general | Essential features | ICD coding | Epidemiology | Sites | Pathophysiology | Etiology | Clinical features | Diagnosis | Laboratory | Prognostic factors | Case reports | Treatment | Clinical images | Microscopic (histologic) description | Microscopic (histologic) images | Virtual slides | Positive stains | Immunofluorescence description | Immunofluorescence images | Videos | Sample pathology report | Differential diagnosis | Practice question #1 | Practice answer #1 | Practice question #2 | Practice answer #2
Cite this page: Occidental M, Kim RH. Bullous pemphigoid. PathologyOutlines.com website. Accessed September 19th, 2025.
Definition / general
Most common autoimmune blistering skin disorder
Characterized by autoantibodies against hemidesmosomal antigens, bullous pemphigoid antigens 1 and 2
Patients present with tense bullae on an erythematous base and pruritus
Essential features
Subepidermal blister with eosinophils, often accompanied by eosinophilic spongiosis
Early lesions can appear urticarial or eczematous
Direct immunofluorescence (DIF): linear C3 > IgG along the basement membrane zone, n-serrated pattern
DIF on salt split skin: immunoreactants deposited on the blister roof (40%) or on both roof and floor (60%)
Indirect immunofluoresence: can be performed on human skin or monkey esophagus; immunoreactants localize to blister roof (95%) on salt split analysis
Treatment: corticosteroids, tetracyclines, cytotoxic steroid sparing agents and rituximab, IVIG infusions
ICD coding
ICD-10: L12.0 - bullous pemphigoid
Epidemiology
Incidence rate: 7.63 per 100,000 person years (Br J Dermatol 2021;184:68)
Increases with age
80% of subepidermal immunobullous dermatoses
M = F
Elderly > children and infants
Sites
Inner and anterior thighs, groin, flexor surfaces of upper extremities, lower abdomen
Childhood bullous pemphigoid: vulvar localization is most common
Pathophysiology
Development of IgG autoantibodies against hemidesmosomal proteins BPAG1 and BPAG2
Etiology
Autoimmune disease associated with bullous pemphigoid antigen 180 (BP180, BPAG2) and bullous pemphigoid antigen 230 (BP230, BPAG1, BPAG1e) (Am J Clin Dermatol 2017;18:513)
Associated with medication use:
Anti-PD1 immunotherapy
Diuretics, ACE inhibitors, antibiotics, D penicillamine
Pediatric cases associated with viral illness and immunization (BMC Pediatr 2017;17:60)
Clinical features
Prodrome phase: mild to severe pruritus and eczematous, papular or urticarial lesions
Bullous stage: tense bullae containing serous fluid or hemorrhage
Unusual to have mucosal involvement (seen in oral cavity in 10 - 30% of cases) (Lancet 2013;381:320)
Diagnosis
Requires histological evaluation in addition to confirmation by direct immunofluorescence studies or detection of circulating autoantibodies
Laboratory
Enzyme linked immunosorbent assay (ELISA):
NC16A domain of BPAG2: sensitivity = 84%, specificity = 98% (J Dermatol Sci 2002;30:224)
BPAG1: sensitivity = 48%, specificity = 94% (Arch Dermatol Res 2016;308:269)
Prognostic factors
30% relapse during first year of treatment (N Engl J Med 2002;346:321)
Risk factors: disease severity, neurological conditions, positive anti-BP180 antibody titers (Ann Med 2018;50:234)
Case reports
61 year old man with tender blisters on his feet, upper legs and central upper back, diagnosed with dyshidrosiform bullous pemphigoid (Cureus 2020;12:e6630)
72 year old woman with metastatic non small cell lung cancer develops nivolumab induced bullous pemphigoid (Oncologist 2018;23:1119)
84 year old woman with clear cell renal cell carcinoma and widespread paraneoplastic bullous pemphigoid (Urol Case Rep 2020;30:101119)
88 year old woman with newly diagnosed melanoma and associated bullous pemphigoid (Melanoma Res 2017;27:65)
Treatment
First line: topical or systemic corticosteroids
Systemic corticosteroid therapy should be accompanied by steroid sparing agents whenever possible (Am J Clin Dermatol 2017;18:513)
Steroid sparing agents (mycophenolate mofetil, methotrexate)
Tetracycline
IVIG in steroid resistant disease
Rituximab or IVIG for refractory disease (not resistant)
Clinical images
Images hosted on other servers:
Tense bullae
Ruptured bullae
Microscopic (histologic) description
Subepidermal blister
Superficial perivascular mixed inflammatory infiltrate with eosinophils in the dermis and blister cavity
Eosinophils may line up along the dermal epidermal junction and extend into the epidermis (eosinophilic spongiosis) (Am J Clin Dermatol 2017;18:513)
Urticarial stage:
Eosinophil rich perivascular and interstitial infiltrate in the papillary dermis
Eosinophilic spongiosis with pseudovacuolar interface change
Rare eosinophilic flame figures
Uncommon presentations include neutrophil rich and cell poor infiltrates
Microscopic (histologic) images
Contributed by Michael Occidental, M.D. and Randie H. Kim, M.D., Ph.D.
Subepidermal blister with eosinophils
Eosinophil microabscesses in papillary dermis
Eosinophilic spongiosis
Eosinophilic spongiosis with adjacent subepidermal blister
Virtual slides
Images hosted on other servers:
Eosinophil poor BP
Clinical images, virtual slide, direct immunofluorescence
Positive stains
Collagen IV immunohistochemistry:
Can be helpful in differentiating subepidermal blistering disorders with level of split below lamina densa (EBA, antilaminin gamma1/p200 pemphigoid, p105 pemphigoid) from BP and other disorders with split within lamina lucida or above lamina densa
If split above lamina densa - dermal staining pattern
If split below lamina densa - epidermal staining pattern
Immunofluorescence description
Linear C3 > IgG and C3 deposition along the basement membrane zone of epithelium, hair follicles and eccrine glands, n-serrated pattern (Br J Dermatol 2013;169:100)
Only C3 may be present in early stage lesions
Salt split skin analysis:
Direct immunofluoresence: localization of immunoreactants to either roof only (40%) or roof and blister (60%)
Indirect immunofluorescence: localization of immunoreactants to the roof of the blister (95%)
Immunofluorescence images
Contributed by Randie H. Kim, M.D., Ph.D.
Linear C3 deposition along the basement membrane
Positive IgG with n-serrated pattern (arrow)
Epidermal reactivity for IgG on roof of blister cavity (star)
Videos
Bullous pemphigoidDr. Phillip McKee
Bullous pemphigoidDr. Christine Ko
Sample pathology report
Skin, left medial thigh, shave biopsy:
Bullous pemphigoid (see comment)
Comment: There is a subepidermal vesicle beneath which there is a mixed cell infiltrate including eosinophils. The associated specimen submitted for immunofluoresence studies shows linear C3 and IgG in an n-serrated pattern along the basement membrane.
Skin, left abdomen, shave biopsy:
Subepidermal blister with eosinophils (see comment)
Comment: There is a subepidermal vesicle beneath which there is a mixed cell infiltrate including eosinophils. The periphery of the lesion shows areas of eosinophilic spongiosis. The differential diagnosis includes bullous pemphigoid, cicatricial pemphigoid, linear IgA disease, a bullous drug eruption and a severe arthropod bite reaction. Additional material for immunofluorescence studies is recommended.
Differential diagnosis
Epidermolysis bullosa acquisita:
Immunofluorescence positive on dermal side of salt split skin (u-serrated pattern) (Br J Dermatol 2004;151:112)
IgG > C3
Bullous lupus erythematosus:
Neutrophil rich infiltrate, variable interface changes
Dermal mucin deposition
Eosinophils not present
DIF: immunoreactive on dermal side of salt split skin (multiple reactants or IgG and C3), u-serrated pattern
Positive lupus serology
Positive epidermal nuclear Ig deposition(in vivo ANA) (J Rheumatol 1983;10:733)
IgG1 is the predominant subclass found (J Dermatol Sci 1990;1:207)
Dermatitis herpetiformis:
Papillary neutrophilic microabscesses
Immunofluorescence = granular IgA in dermal papillae
May also have fibrillary IgA deposits in the dermal papillae or microgranular deposits along the dermo-epidermal junction (Postepy Dermatol Alergol 2019;36:655)
Linear IgA bullous disease:
Neutrophil rich infiltrate
Immunofluorescence = linear IgA along the basement membrane zone
Porphyria cutanea tarda:
Pauci-inflammatory
Festooning dermal papillae
Hyalinized superficial blood vessels (positive PAS deposits)
Caterpillar bodies (positive for PAS and collagen IV)
Direct immunofluoresence: multiple immunoreactants at the dermo-epidermal junction and within superficial vessels (granular deposition)
Practice question #1
An 87 year old man presents with erythema and blistering of his chest and upper extremities. Examination shows tense bullae on an erythematous base. A skin biopsy is obtained which reveals subepidermal blister with a mixed inflammatory infiltrate with numerous eosinophils. Which of the following is true for the direct immunofluorescence findings regarding the diagnosis?
Direct immunofluorescence would show granular IgA deposition at the tips of the dermal papillae
Direct immunofluorescence would show IgG and C3 reactivity in the epidermis in a fishnet-like pattern in the intercellular squamous region
Direct immunofluorescence would show linear IgA reactivity along the basement membrane
Direct immunofluorescence would show linear IgG and C3 reactivity at the basement membrane
Practice answer #1
D. Direct immunofluorescence would show linear IgG and C3 reactivity at the basement membrane Comment Here Reference: Bullous pemphigoid
Practice question #2
What is / are the antigen(s) targeted in bullous pemphigoid?
BPAG1 and BPAG2
Collagen VII
Desmoglein 1 and Desmoglein 3
Envoplakin and periplakin
Practice answer #2
A. BPAG1 and BPAG2 Comment Here Reference: Bullous pemphigoid
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13637 | https://www.ucl.ac.uk/~ucahmto/0005_2024/Ch2.S1.html | 2.1 Introduction to set theory ‣ Chapter 2 Sets and functions ‣ MATH0005 Algebra 1
MATH0005 Algebra 1
1. About these notes
2. 1 Logic
3. 2 Sets and functions
1. 2.1 Introduction to set theory
1. 2.1.1 Definition of a set
2. 2.1.2 Set builder notation
3. 2.1.3 Listing the elements of a set
4. 2.1.4 Elements of a set
5. 2.1.5 Set equality
6. 2.1.6 Subsets
2. 2.2 Set operators
3. 2.3 Set algebra
4. 2.4 De Morgan’s laws
5. 2.5 Functions
6. 2.6 Function composition
7. 2.7 Function properties
8. 2.8 Invertibility
9. 2.9 Conditions for invertibility
10. 2.10 Permutations
11. 2.11 Inverses and composition
12. 2.12 Cycles
13. 2.13 Products of disjoint cycles
14. 2.14 Powers and orders
15. 2.15 Transpositions
16. 2.16 Sign
17. Further reading
3 Matrices
4 Linear algebra
2 Sets and functions2 Sets and functions2.2 Set operators
2.1 Introduction to set theory
2.1.1 Definition of a set
A set is a collection of (mathematical) objects. There is an entire field of mathematics called set theory dedicated to the study of sets and to their use as a foundation for mathematics, but in MATH0005 we are going to give only an informal introduction to sets and their properties. If you want to know more, see the further reading section at the end of this chapter.
2.1.2 Set builder notation
We specify sets using logical notation: if P(x) is a property that is true or false for each thing x, then
{x:P(x)}
is the collection of all x such that P(x) is true. 1 1 1 You have to be slightly careful with this kind of unrestricted comprehension because it can lead to contradictions. You can ignore this for the purposes of MATH0005, but if you want to know more then check out the Further Reading section at the end. For example, if P(x) is the property that x is an even integer then {x:P(x)} is the set consisting of all the even integers …,−4,−2,0,2,4,…
Sometimes we already have a set X and a property P(x) that is true or false for each x in X. For example, X might be the set ℝ of all real numbers, and P(x) the property x>0. We write
{x∈X:P(x)}(2.1)
for the set of all things x in X such that P(x) is true. In the example, this set would be the set of all strictly positive real numbers, sometimes written (0,∞) in interval notation. The symbol ∈ means ‘is an element of.’
These two ways of specifying a set are called set builder notation.
2.1.3 Listing the elements of a set
Sometimes we want to specify a set simply by listing its elements one by one. For example, the set whose elements are 1, 2, and 3 is written {1,2,3}, which is a short way of writing for the set builder notation
{x:(x=1)∨(x=2)∨(x=3)}.
2.1.4 Elements of a set
The things in a set are called its elements or members. We write a∈X to mean that a is an element or member of the set X, and a∉X to mean that a is not an element or member of X.
There is a unique set with no elements, called the empty set and written ∅. No matter what a is, a∉∅.
We allow any kind of mathematical object, including sets themselves, as elements of sets. Sets can contain functions, matrices, vectors, numbers, and sets themselves.
Example 2.1.1.
{∅,1,{2},{{3}}}
is a set whose four elements are the empty set, the number 1, the set containing the number 2, and the set containing the set containing 3.
Example 2.1.2.
Let X={1,2,{3}}. Then 1∈X, 0∉X, 3∉X, {3}∈X.
2.1.5 Set equality
Two sets X and Y are defined to be equal if for all things a we have a∈X if and only if a∈Y — that is, if and only if they have the same elements. It’s helpful to write this in terms of logical equivalence: the two sets
X={x:P(x)},Y={x:Q(x)}
are equal if and only if P(x) and Q(x) have the same truth value for every x. This definition of set equality has some important consequences. First, if
X={1,2},Y={2,1}
then X=Y because these are shorthand for
X={x:(x=1)∨(x=2)},Y={x:(x=2)∨(x=1)}
and for any propositional variables p and q, the WFFs p∨q and q∨p are logically equivalent. The order in which we list the elements of a set doesn’t matter. Similarly, if
X={1,1},Y={1}
then X=Y, because the two definitions above are shorthand for
X={x:(x=1)∨(x=1)},Y={x:x=1}
and the two formulas p∨p and p are logically equivalent. Repetition doesn’t matter.
2.1.6 Subsets
We need vocabulary for talking about one set being contained in another.
Definition 2.1.1.
•X is a subset of Y, written X⊆Y, if and only if every element of X is also an element of Y.
•If X is not a subset of Y we write X⊈Y.
•X is a proper subset of Y, written X⊊Y, if and only if X⊆Y but X≠Y.
Thus X being a proper subset of Y means that X is a subset of Y and Y contains something that X does not contain.
There is an important way to rephrase the definition of two sets being equal: X=Y if and only if X⊆Y and Y⊆X. This is sometimes useful as a proof technique, as you can split a proof of X=Y into first checking X⊆Y and then checking Y⊆X.
Example 2.1.3.
•{0}⊆{0,1}
•{0}⊊{0,1}
•{0,1}⊈{1,2}
•{1,2,1}={2,1}
Why is the last equality true? The only things which are elements of {1,2,1} are 1 and 2. The only things which are elements of {1,2} are 1 and 2. So the two sets are equal according to our definition. There’s no concept of something being an element of a set “more than once.”
This is the way in which our definition of set equality captures the idea of sets being unordered collections of objects which disregard repetition.
The definition of subset means that the empty set is a subset of any set. ∅⊆X for any set X, because ∀x:x∈∅⟹x∈X is vacuously true: there’s nothing in ∅ which could fail to be in the set X in order to make ∅⊆X false.
2 Sets and functions2.2 Set operators
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13639 | https://help.furkot.com/features/route-details.html | Route settings | Road Trip Planner | Furkot Help Center
Furkot ====== road trip planner -----------------
Home
Features
Route settings
Route settings
March 16, 2023
Routes connect Furkot stops - they represent parts of your trip that you spend traveling: in the vehicle of your choice, on the motorcycle, on the bike, walking, or transferring between places by other means. Routes are displayed on the map as paths in several shades of blue. Furkot can calculate routes automatically or you can shape them yourself. You have a lot of options to adjust the time, speed, road selection and the shape of the route.
Route settings
Select the route on the map to see the route settings: click on the path representing the route and Furkot will display the panel with route settings in the Plan drawer. If you prefer you can also open the Plan drawer; scroll and hover over items in the drawer to have Furkot highlight corresponding objects on the map. Clicking on a route in the Plan drawer not only displays its settings but also pans the map so that the route becomes visible.
Route settings panel displays addresses of both ends of the route. It shows the exact time of departure from the previous stop and the time of arrival at the next stop. It displays the length of the leg and the time spent traveling. You can check the running time and the distance from the beginning of the day, and from the beginning of the trip as well.
You can view and modify when applicable the following route settings according to your preferences:
Mode of travel
Routing
Road preferences
Personal travel speed
Alternative route
Reset route
Skip route
Elevation profile
Route color
Route direction
Mode of travel ¶
Every route has a mode of travel: driving , riding , biking , walking , flight ferry , or other . You can set the mode of travel for the entire trip in the Trip drawer and change it independently for each route. If you mostly drive but want to bike or walk at some point, select driving for the entire trip and set the respective route segments to biking and walking .
Changing the mode of travel will influence roads selected by Furkot: bike riders will not be sent on the freeway if the alternative exists. Walking segments will keep you on sidewalks and walking paths if available. Changing the mode of travel will also affect your speed and thus the duration of the route segment.
Changing the route segment mode of travel to flight will cause Furkot to look for the airports closest to the ends of the flight segment. If you prefer a different airport that the one suggested by Furkot you can find it under the Airport category in the Eat drawer. Or you can add one just like you add any stop and change its pin icon to airport.
If the route involves riding a ferry, Furkot will detect a ferry crossing automatically and show it as a separate segment with the travel mode set to ferry. For each ferry route Furkot also adds the stops representing ferry terminals .
The other mode of travel is special: it's there to represent bus, tram or rail segments of the trip. By selecting it you tell Furkot that you have a way to get from A to B and you don't want any routing guidance. In this mode it's your responsibility to set Duration or Speed manually to keep your itinerary realistic. Furkot will not schedule any overnight stops automatically along the route with the other mode. It assumes you sleep while you are traveling.
Adding new stops creates new route segments. Newly added routes inherit the travel mode from the routes that they replaced: in other words if you add a stop that splits a biking segment newly created routes to and from that stop will have biking mode set automatically.
Routing ¶
You can either let Furkot calculate a route between stops or you can shape it yourself by changing the Routing setting.
Shaping routes is an advanced feature: optimize Furkot for Advanced use under the Plan drawer filter to enable it.
To let Furkot provide routing guidance and to allow re-routing when new stops are added, set the Routing setting to follow roads. You will be able to influence the calculated path by avoiding or permitting Highways and Toll Roads (when driving or riding a motorcycle). Dragging the route is also possible: it lets you follow the roads of your choice.
When Furkot calculates your route it assumes a certain speed depending on the available information about road conditions and speed limits. You can speed up or slow down if you find Furkot estimates unrealistic. You can also just enter the average speed bypassing Furkot estimates entirely. That works well in places like National Parks where your average speed may be really low once you budget the time needed to enjoy the scenery.
To control all the parameters of the route including its shape, change the Routing setting to travel off road. It'll preserve the route path when you add new stops. It'll also enable you to shape the route without constraining it to mapped roads. If you intend to travel off road or on a road you know to be passable but not considered as such by Furkot (like some seasonally closed roads), that's the way to plan it.
Furkot will never change the path of the off road route. It may schedule overnight and refueling stops along it but it won't suggest any specific lodging accommodations or filling stations. You either need to find them yourself, or you can adjust your day travel times and fuel range so that Furkot will schedule overnight and refueling stops before or after the off road segment.
The off road shape is editable: to change the shape of route path you can drag its points - you'll notice that the shaped path doesn't stick to the roads (you can of course shape it to follow roads point by point).
When you shape off road route you need to update its Duration or Speed manually to keep your itinerary realistic.
Routes set to the other mode of travel are always treated as off road. So are the routes added to the trip from the Find drawer using Route and Backtrack buttons and routes and tracks imported from files and websites.
Road preferences ¶
You can influence the type of the roads for the entire trip or you can express your road preferences for each trip leg separately. You can avoid highways, toll roads, ferries, and unpaved roads. Avoiding highways and toll roads applies only when you travel by motorized vehicle (that is: when you choose car, motorcycle or RV travel mode). And if you ride a motorcycle you can tell Furkot to take curvy roads (provided by Kurviger in cooperation with Graphhopper).
Setting your road preferences for each route is an advanced feature: optimize Furkot for Advanced use under the Plan drawer filter to enable it.
Personal travel speed ¶
We don't all drive at the same speed. Or walk, or bike for that matter. Nonetheless, when calculating travel time Furkot assumes certain average speed that depends on the road type and local speed limits, but doesn't account for differences among individual travelers.
To accommodate those differences Furkot provides My Speed settings. In addition to adjusting speed for the entire trip in the Trip drawer you can set it for a selected route. It allows you to influence calculation of travel time according to your preferences for one leg of your itinerary. You can adjust your speed either by setting it to a fixed value in miles (or kilometers) per hour, or by specifying how much faster or slower than average you are.
If you notice that you tend to arrive at your destination after (or before) arrival time calculated by Furkot enter the percentage that corresponds to the difference.
Similarly, if you are planning a hike in the difficult terrain changing My Speed to your average hiking speed might give you better idea of the timing.
In most cases you don't need to worry about the My Speed field - leave it blank to let Furkot calculate the speed. But if your trip is not typical, as many great trips tend to be, you have one more control to make your itinerary more precise.
Adjusting speed separately for each route is an advanced feature: optimize Furkot for Advanced use under the Plan drawer filter to enable it.
Alternative route ¶
If you think that there is a better route than the one Furkot calculated for you, try the Alternate option . Furkot will use an alternative routing engine and, in some cases, it will show you another way to travel between stops.
Alternative calculation of the route is an advanced feature: optimize Furkot for Advanced use under the Plan drawer filter to enable it.
Reset route ¶
If, after dragging route points, you want to go back to the initial path, use the Reset button. The routes that Furkot calculates will be recalculated after removing all interim pass-through points.
Resetting off road routes will create a straight line between stops: this is usually a good starting point if you intend to reshape an off road route.
Skip route ¶
Route segments that are set to travel off road and shaped manually can be skipped. Skipping a route removes it from the trip itinerary but saves it as a skipped stop. Such route can be re-added to the trip itinerary later by using its Route button.
Skipping routes is an advanced feature: optimize Furkot for Advanced use under the Plan drawer filter to enable it.
Skipped routes can optionally included when you export your trip. Exporting skipped routes is an exclusive benefit offered to Furkot Pass holders.
Elevation profile ¶
Furkot can show you the elevation profile graph for a selected route. Hover over the profile to check the exact altitude. Look for the small circle on your road moving in sync with the elevation profile graph. Or hover over the route and the elevation profile graph will highlight the corresponding elevation.
If your trip distance units are set to kilometers Furkot displays elevation profile scaled in meters above the sea level. If you use miles, the elevation is displayed in feet. You can set units for the trip in the Trip drawer.
The elevation profile for a selected route is displayed when Furkot is optimized for trips in the mountains: use the mountains button under the Plan drawer filter to enable it.
Displaying elevation profile is a feature available exclusively to Furkot Pass holders.
Route color ¶
You can set color of the route to one of the 16 predefined colors or transparent. The latter is useful when you want to see details that may be obscured by the route.
Use the the earth button to display routes in their own individual colors on the map.
Route direction ¶
It is not always easy to see which direction the route is going especially when the stops at its ends are not visible. When you select a route segment Furkot will show the direction as small arrows along the route.
The displayed direction reflects trajectory of your trip and should not be confused with traffic restrictions: one-way streets are depicted on the map with different kind of arrows.
8.51.0
You can now tell Furkot to automatically add LPG stations along the route. And you can search for the LPG / Propane Autogas stations in the Eat drawer.
8.50.11
Furkot now helps you track expenses for your trip. You set the mileage rate, add hotel charges, stop costs, and tolls. Or you can use the standard Per Diem allowance.
8.50.2
We would like to apologize. For a couple of hours Keep me signed in option was broken. The trips and the account details of 22 Furkot users were shown to a small number of different Furkot users.
Read the announcement in full here.
8.49.0
Waze and Organic Maps are now among supported navigation apps.
8.47.0
You can now invite friends to your trip by copying an invite link and using your favorite messaging app to share it.
8.42.0
New export format is here: you can now export GPX tailored for OsmAnd app: the route colors will be properly displayed.
8.41.0
More control over which routes are selected: you can tell Furkot to avoid ferries and unpaved roads. If you want Furkot will warn you about the toll roads or unpaved roads that part of your itinerary. Curvy roads can be preferred if you are planning motorcycle trip.
8.39.0
Furkot has a new toggle to facilitate switching to non-stop travel. You can flip it for the entire trip or a specific day. Furkot is not going to add automatic overnight stops when non-stop travel is selected.
8.36.0
Various export improvements added: you can now perform exports in on-line mode and - if your browser supports it - Furkot will ask which folder should your exported file end up in.
8.34.0
You can now take a ferry cross the Mercy 🎵 - or some other body of waters - ⛴️ ferry travel mode is here.
8.31.0
Against our better judgement 😉 fast food now has its own dedicated button in the EAT drawer.
8.29.0
Planning roadtrip in your electric vehicle ? Tell Furkot which connector you need and it will help you to find charging stations along the route.
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13640 | https://www.sigmaaldrich.com/US/en/technical-documents/technical-article/materials-science-and-engineering/solid-state-synthesis/solubility-rules-solubility-of-common-ionic-compounds?srsltid=AfmBOopyTDYuQrA5WKA-k_f7MU29tyNEV3vjE8H8ih1ioujFCVQ1GwLx | Solubility Rules for Ionic Compounds
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HomeSolid State SynthesisSolubility Rules for Ionic Compounds
Solubility Rules for Ionic Compounds
Solubility Chemistry
It’s important to know how chemicals will interact with one another in aqueous solutions. Some compounds or solutes will dissolve, others will yield a precipitate or solid, and a few react with water.
You’ve probably run into solubility questions in your everyday life. Hot chocolate or flavored dry beverage mixes that don’t evenly dissolve in water would be examples of unwanted precipitates. Limescale or soap scum are precipitates left behind when water with higher mineral content evaporates and introduces previously dissolved metal cations to carbonates or soap anions.
Solubility is applicable to many laboratory processes and is also important in medicine. Some ions can be toxic when they separate in a solution but are helpful as part of a compound.
Asaturated solutionis one in which the maximum amount of solute has been dissolved. The opposite is a dilute solution; this solution can accept more solute.
Pressure and temperature affect solubility. This page discusses the solubility of compounds in water at room temperature and standard pressure. A compound that is soluble in water forms an aqueous solution.
Solubility Rules
| ### Which ions are soluble? |
| |
| Ions | Except |
| Alkali metals (Group I) Na+, K+, etc. | |
| Ammonium ions NH 4+ | |
| Nitrates, acetates, chlorates, and perchlorate NO 3-, C 2 H 3 O 2-, ClO 3-, ClO 4- | |
| Binary compounds of halogens (chloride, bromide, iodide, etc.) with metals Cl-, Br-, I-, etc. | Fluoride Silver, lead, and mercury F-, Ag+, Pb 2+, and Hg 2+ Lead halides are soluble in hot water. |
| All sulfates SO 4 2- | Barium, strontium, calcium, lead, silver, and mercury Ba 2+, Sr 2+, Ca 2+, Pb 2+, Ag+, and Hg 2+ |
| ### Which ions are slightly soluble? |
| |
| Ions | Except |
| Sulfates of lead, silver, and mercury SO 4 2-with Pb 2+, Ag+, and Hg 2+ | Lead sulfate is poorly soluble. |
| Hydroxides of alkaline earth metals (Group II) OH-with Ca 2+, Sr 2+, etc. | Barium Ba 2+ |
| ### Which ions are insoluble? |
| |
| Ions | Except |
| Sulfides S 2- | Calcium, barium, strontium, magnesium, sodium, potassium, and ammonium Ca 2+, Ba 2+, Sr 2+, Mg 2+, Na+, K+, and NH 4+ |
| Hydroxides OH- | Alkali metals (Group I) and ammonium Na+, K+, etc. Al 3+, NH 4+ |
| Carbonates, oxalates, chromates, and phosphates CO 3 2-, C2O 4 2-, CrO 4 2-, and PO 4 3- | Alkali metals (Group I) and ammonium Na+, K+, etc. NH 4+ Lithium phosphate is poorly soluble |
How to Use Solubility Rules
Identify the compound whose solubility you want to check. It can be helpful to write out the empirical formula so you can identify the ions that make up the compound.
Look up each ion in the solubility rules. Check the left-hand column for the general rule, and look in the right-hand column to make sure you noted any exceptions.
Alternatively, you can look up ions in the solubility chart. You might find this easier. Cations are listed across the top, and anions are listed vertically. Find the cell where your cation column and ion row meet to determine solubility of the resulting compound.
Our solubility rules are not exhaustive. You may need to reference aperiodic tableif you’re looking up less common compounds.
Solubility Rules Chart
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Solubility Definitions
When discussing solubility, it’s useful to follow agreed standard definitions. The resources above present some general rules and loose definitions.
The United States Pharmacopeia (USP), a nonprofit organization committed to establishing standards for medicines, food ingredients, dietary supplement products, and ingredients, has established the definitions below. These definitions are used by other major pharmacopeia organizations throughout the world and are often paired with exact measurements for more precise application needs.
Descriptive terms Approximate volume of solvent in mL/g of substance
Very soluble less than 1
Easily soluble from 1 to 10
Soluble from 10 to 30
Sparingly soluble from 30 to 100
Slightly soluble from 100 to 1,000
Very slightly soluble from 1,000 to 10,000
Practically insoluble more than 10,000
Related Articles
Solubility Table of Compounds in Water at Temperature
Solvent Miscibility Table
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Periodic Table of the Elements
Solution Dilution Calculator
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Mesoporous Oxides and Their Applications to Hydrogen Storage
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10.7
Scalar (or Dot) Product of Two Vectors
Written by:Neha Khandelwal
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 26, 2025
In scalar quantities (simple numbers), multiplication is straightforward. However, vectors are more complex entities because they possess both magnitude (length) and direction. This inherent directional property means that a single, universally applicable "multiplication" like that for scalars doesn't capture all the useful ways vectors can interact.
When we want to multiply two vectors, we usually ask a specific question about their relationship. For instance:
1. How much does one vector align with another, or what is the effective component of one vector in the direction of another?
2. What is the area of the parallelogram formed by two vectors? Or, what vector is mutually perpendicular to two given vectors?
These two questions lead to two different ways of multiplying vectors. The dot product (the scalar product) helps us answer the first question - how much one vector points in the same direction as another. It gives a single number (a scalar) that tells how closely the vectors are aligned. For example, this number is useful in real life to calculate the work done by a force or how much of a field passes through a surface (called flux).
The other type is the cross product (or vector product), which answers the second question - like finding the area between two vectors or a direction perpendicular to both. It gives a new vector instead of a number.
Definition of Scalar (Dot) Product
Let @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ be two non-zero vectors. If we place them so their tails coincide, they will form an angle @$\begin{align}\theta\end{align}@$ between them. The convention is to take @$\begin{align}\theta\end{align}@$ such that @$\begin{align}0 \leq \theta \leq \pi\end{align}@$ (i.e., 0 to 180 degrees), representing the smaller of the two angles they form.
The dot product of @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b},\end{align}@$ denoted as @$\begin{align}\vec{a} \cdot \vec{b},\end{align}@$ is defined as:
@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\end{align}@$$where:
@$\begin{align}|\vec{a}|\end{align}@$ is the magnitude (length) of vector @$\begin{align}\vec{a}.\end{align}@$
@$\begin{align}|\vec{b}|\end{align}@$ is the magnitude (length) of vector @$\begin{align}\vec{b}.\end{align}@$
@$\begin{align}\theta\end{align}@$ is the angle between vectors @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ when placed tail-to-tail.
@$\begin{align}\cos \theta\end{align}@$ is the cosine of that angle.
Interpretation Based on Angle @$\begin{align}\theta\end{align}@$:
If @$\begin{align}\theta = 0:\end{align}@$ Vectors are in the same direction
@$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \quad \text{(maximum positive)}\end{align}@$
If @$\begin{align}\theta = \frac{\pi}{2}:\end{align}@$ Vectors are perpendicular
@$\begin{align}\vec{a} \cdot \vec{b} = 0\end{align}@$
If @$\begin{align}\theta = \pi:\end{align}@$ Vectors are in opposite directions
@$\begin{align}\vec{a} \cdot \vec{b} = -|\vec{a}||\vec{b}| \quad \text{(maximum negative)}\end{align}@$
Thus, the dot product effectively measures how much one vector acts in the direction of another.
Angle Between Two Vectors
To define the angle @$\begin{align}\theta\end{align}@$ used in the dot product:
Take any two non-zero vectors @$\begin{align}\vec{a}\end{align}@$and @$\begin{align}\vec{b}.\end{align}@$
Translate them (without changing their magnitude or direction) so that their initial points (tails) coincide.
The angle between them, @$\begin{align}\theta,\end{align}@$ is measured in the plane containing both vectors, at their common tail.
By convention, we always choose: @$\begin{align}0 \leq \theta \leq \pi\end{align}@$ to ensure a unique and smaller angle (as both @$\begin{align}\theta\end{align}@$ and @$\begin{align}\pi - \theta\end{align}@$ are geometrically valid).
If the vectors are collinear (on the same line):
@$\begin{align}\theta = 0\end{align}@$ if they point in the same direction
@$\begin{align}\theta = \pi\end{align}@$ if they point in opposite directions
Key Insights into the Dot Product
1. If Either Vector is Zero
If @$\begin{align}\vec{a} = \vec{0}\end{align}@$ or @$\begin{align}\vec{b} = \vec{0},\end{align}@$ then the dot product is:@$$\begin{align}\vec{a} \cdot \vec{b} = 0\end{align}@$$Since a zero vector has no direction and zero magnitude, any product involving it is zero. The angle @$\begin{align}\theta\end{align}@$ is undefined in this case, but the result is still valid.
2. Result is a Scalar, Not a Vector
The dot product combines two vectors and gives a single real number. This is why it's called the scalar product.
Example:@$$\begin{align}\text{Work} = \vec{F} \cdot \vec{d}\end{align}@$$where force and displacement are vectors, but work is a scalar.
3. If Vectors Are in the Same Direction (@$\begin{align}\theta = 0\end{align}@$)@$$\begin{align}\cos 0 = 1 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\end{align}@$$This gives the maximum possible positive value of the dot product.
4. If Vectors Are in Opposite Directions (@$\begin{align}\theta = \pi\end{align}@$)@$$\begin{align}\cos \pi = -1 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = -|\vec{a}||\vec{b}|\end{align}@$$This gives the maximum negative value, since they point in opposite directions.
5. If the Dot Product is Zero
Either @$\begin{align}\vec{a} = \vec{0}, \vec{b} = \vec{0},\end{align}@$ or @$\begin{align}\theta = \frac{\pi}{2}.\end{align}@$
So, non-zero vectors are perpendicular if and only if their dot product is zero.
Condition for Perpendicularity
Two non-zero vectors @$\begin{align}\vec{a}\end{align}@$and @$\begin{align}\vec{b}\end{align}@$are perpendicular (or orthogonal) if and only if:@$$\begin{align}\vec{a} \cdot \vec{b} = 0\end{align}@$$This provides a simple algebraic test for checking perpendicularity without measuring angles.
Sign of the Scalar Product
The sign of the dot product tells us about the angle @$\begin{align}\theta\end{align}@$between the vectors:
Positive (@$\begin{align}\vec{a} \cdot \vec{b} > 0\end{align}@$)
@$\begin{align}\Rightarrow \cos \theta > 0 \Rightarrow 0^\circ \leq \theta < 90^\circ\end{align}@$
The angle is acute; vectors point roughly in the same direction.
Zero (@$\begin{align}\vec{a} \cdot \vec{b} = 0\end{align}@$)
@$\begin{align}\Rightarrow \cos \theta = 0 \Rightarrow \theta = 90^\circ\end{align}@$
The vectors are perpendicular.
Negative (@$\begin{align}\vec{a} \cdot \vec{b} < 0\end{align}@$)
@$\begin{align}\Rightarrow \cos \theta < 0 \Rightarrow 90^\circ < \theta \leq 180^\circ\end{align}@$
The angle is obtuse; vectors generally point in opposite directions.
Square of a Vector
Taking the dot product of a vector with itself gives its magnitude squared:@$$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}||\vec{a}| \cos 0 = |\vec{a}|^2\end{align}@$$Hence:@$$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}|^2 \quad \text{and} \quad |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}\end{align}@$$This is a very useful identity for calculating magnitudes from components.
Dot Products of Standard Unit Vectors
The standard unit vectors @$\begin{align}\hat{i}, \hat{j}, \hat{k}\end{align}@$represent the x-, y-, and z-axis directions respectively. They are of unit length and mutually perpendicular.
Dot product with itself:
@$\begin{align}\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\end{align}@$
Dot product with others (distinct pairs):
@$\begin{align}\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\end{align}@$
So,@$$\begin{aligned} &\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \ &\hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = \hat{j} \cdot \hat{k} = 0 \end{aligned}@$$These values are fundamental to building the component form of the dot product.
Geometrical Interpretation of Dot Product
The dot product@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\end{align}@$$can be interpreted in terms of projections.
The term @$\begin{align}|\vec{a}| \cos \theta\end{align}@$is the scalar projection of vector @$\begin{align}\vec{a}\end{align}@$onto the direction of vector @$\begin{align}\vec{b}\end{align}@$- it's like the "length of the shadow" of @$\begin{align}\vec{a}\end{align}@$ on the line containing @$\begin{align}\vec{b}.\end{align}@$
Similarly, @$\begin{align}|\vec{b}| \cos \theta\end{align}@$is the scalar projection of vector @$\begin{align}\vec{b}\end{align}@$onto the direction of vector @$\begin{align}\vec{a}.\end{align}@$
.
So, @$\begin{align}\vec{a} \cdot \vec{b}\end{align}@$ can be seen as:
(Magnitude of @$\begin{align}\vec{a}\end{align}@$) × (Scalar projection of @$\begin{align}\vec{b}\end{align}@$onto @$\begin{align}\vec{a}\end{align}@$):
@$$\begin{align}|\vec{a}| \cdot (|\vec{b}| \cos \theta)\end{align}@$$
(Magnitude of @$\begin{align}\vec{b}\end{align}@$) × (Scalar projection of @$\begin{align}\vec{a}\end{align}@$ onto @$\begin{align}\vec{b}\end{align}@$):
@$$\begin{align}|\vec{b}| \cdot (|\vec{a}| \cos \theta)\end{align}@$$
Scalar projection of@$\begin{align}\vec{a}\end{align}@$on@$\begin{align}\vec{b}:\end{align}@$
This is the component of @$\begin{align}\vec{a}\end{align}@$along @$\begin{align}\vec{b},\end{align}@$ which is @$\begin{align}|\vec{a}| \cos \theta.\end{align}@$
Since@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta,\end{align}@$$we can write:@$$\begin{align}|\vec{a}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\end{align}@$$So, scalar projection of @$\begin{align}\vec{a}\end{align}@$ on @$\begin{align}\vec{b} \end{align}@$ is@$$\begin{align}|\vec{a}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \vec{a} \cdot \hat{b}\end{align}@$$(where @$\begin{align}\hat{b} = \dfrac{\vec{b}}{|\vec{b}|}\end{align}@$is the unit vector in the direction of @$\begin{align}\vec{b}\end{align}@$).
This value is a scalar and can be positive, negative, or zero.
Scalar projection of @$\begin{align}\vec{b}\end{align}@$on @$\begin{align}\vec{a}:\end{align}@$
Similarly@$$\begin{align}|\vec{b}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}\end{align}@$$So, scalar projection of @$\begin{align}\vec{b}\end{align}@$ on @$\begin{align}\vec{a} \end{align}@$ is@$$\begin{align}|\vec{b}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \vec{b} \cdot \hat{a}\end{align}@$$
Vector projection of @$\begin{align}\vec{a}\end{align}@$ along (or onto) @$\begin{align}\vec{b}:\end{align}@$
This is a vector that has the magnitude of the scalar projection of @$\begin{align}\vec{a}\end{align}@$ on @$\begin{align}\vec{b},\end{align}@$ and points in the direction of@$\begin{align}\vec{b}.\end{align}@$
To get this vector, we multiply the scalar projection of @$\begin{align}\vec{a}\end{align}@$ on @$\begin{align}\vec{b}\end{align}@$ by the unit vector in the direction of @$\begin{align}\vec{b}\end{align}@$(i.e., @$\begin{align}\hat{b}\end{align}@$):
@$\begin{align}\text{Vector projection of }\end{align}@$ @$\begin{align}\vec{a}\end{align}@$@$\begin{align}\text{ onto }\end{align}@$@$\begin{align}\vec{b} = (\text{Scalar proj of } \vec{a} \text{ on } \vec{b}) \cdot \hat{b} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right) \cdot \frac{\vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \cdot \vec{b}\end{align}@$
Alternatively, using @$\begin{align}\hat{b}:\end{align}@$@$$\begin{align}\text{Vector projection of } \vec{a} \text{ onto } \vec{b} = (\vec{a} \cdot \hat{b}) \cdot \hat{b}\end{align}@$$This resulting vector is parallel to@$\begin{align}\vec{b},\end{align}@$ or is the zero vector if @$\begin{align}\vec{a} \cdot \vec{b} = 0.\end{align}@$
Direction Cosines via Dot Product
Let a vector@$$\begin{align}\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}\end{align}@$$make angles @$\begin{align}\alpha, \beta, \gamma\end{align}@$with the positive x-, y-, and z-axes, respectively. These angles are called the direction angles, and their cosines (@$\begin{align}\cos \alpha, \cos \beta, \cos \gamma\end{align}@$) are called the direction cosines.
To find @$\begin{align}\cos \alpha\end{align}@$(angle with x-axis, represented by @$\begin{align}\hat{i}\end{align}@$):@$$\begin{align}\vec{r} \cdot \hat{i} = |\vec{r}| |\hat{i}| \cos \alpha\end{align}@$$Since @$\begin{align}|\hat{i}| = 1,\end{align}@$ we get:@$$\begin{align}\vec{r} \cdot \hat{i} = |\vec{r}| \cos \alpha\end{align}@$$Also,@$$\begin{align}\vec{r} \cdot \hat{i} = (a\hat{i} + b\hat{j} + c\hat{k}) \cdot \hat{i} = a\end{align}@$$So,@$$\begin{align}a = |\vec{r}| \cos \alpha \quad \Rightarrow \quad \cos \alpha = \frac{a}{|\vec{r}|}\end{align}@$$Similarly, for @$\begin{align}\cos \beta\end{align}@$ (angle with y-axis, represented by @$\begin{align}\hat{j}\end{align}@$):@$$\begin{align}\cos \beta = \frac{b}{|\vec{r}|}\end{align}@$$And for @$\begin{align}\cos \gamma\end{align}@$(angle with z-axis, represented by @$\begin{align}\hat{k}\end{align}@$):@$$\begin{align}\cos \gamma = \frac{c}{|\vec{r}|}\end{align}@$$So, the direction cosines are:@$$\begin{align}\cos \alpha = \frac{a}{|\vec{r}|}, \quad \cos \beta = \frac{b}{|\vec{r}|}, \quad \cos \gamma = \frac{c}{|\vec{r}|}\end{align}@$$ Here, @$\begin{align}a,b,c\end{align}@$are the scalar components of @$\begin{align}\vec{r},\end{align}@$ and@$$\begin{align}|\vec{r}| = \sqrt{a^2 + b^2 + c^2}\end{align}@$$ It follows,@$$\begin{align}\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\end{align}@$$This identity always holds for the direction cosines of any non-zero vector.
Properties of the Scalar (Dot) Product of Two Vectors
Now that we understand how the scalar (dot) product of two vectors is defined and interpreted geometrically, we must explore the algebraic properties that govern how this operation behaves.
These properties help us simplify complex vector expressions, establish identities, and solve problems efficiently in geometry, physics, and engineering. They also align closely with familiar rules from algebra, like distributivity and associativity, but are adapted to the vector context.
Property 1: Commutativity Property of Scalar Product
Statement: The scalar product of two vectors is commutative, i.e.,@$$\begin{align}\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\end{align}@$$This means that the order in which two vectors are multiplied using the dot product does not affect the result.
Proof:
Let @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$be two non-zero vectors, and let @$\begin{align}\theta\end{align}@$be the angle between them.
Using the geometric definition:@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \quad \text{and} \quad \vec{b} \cdot \vec{a} = |\vec{b}| |\vec{a}| \cos \theta\end{align}@$$Since real number multiplication is commutative:@$$\begin{align}\vec{a}| |\vec{b}| \cos \theta = |\vec{b}| |\vec{a}| \cos \theta \Rightarrow \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\end{align}@$$Note:
If either @$\begin{align}\vec{a} = \vec{0}\end{align}@$or @$\begin{align}\vec{b} = \vec{0},\end{align}@$ then both dot products are 0, so the property holds.
Property 2: Distributivity Over Vector Addition
Statement:The dot product is distributive over vector addition:@$$\begin{align}&\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\&(\vec{b} + \vec{c}) \cdot \vec{a} = \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}\end{align}@$$ This means that instead of taking the dot product of one vector with the sum of two others, you can compute the dot product for each separately and then add the results.
Proof:
Let @$\begin{align}\vec{a}, \vec{b}, \vec{c}\end{align}@$be vectors. Using geometry, when @$\begin{align}\vec{b} + \vec{c}\end{align}@$is projected onto @$\begin{align}\vec{a},\end{align}@$ the total projection splits into two separate projections, one from @$\begin{align}\vec{b},\end{align}@$ and the other from @$\begin{align}\vec{c}.\end{align}@$
Hence:@$$\begin{align}\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\end{align}@$$This property is used extensively in vector algebra and physics when forces or velocities are combined.
Property 3: Perpendicular Vectors Have Zero Dot Product
Statement: If @$\begin{align}\vec{a} \cdot \vec{b} = 0,\end{align}@$ then either:
One of the vectors is the zero vector, or
Vectors @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$are perpendicular
This property provides a simple test for orthogonality (perpendicularity) using the dot product.
Proof:
From the dot product formula:@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \Rightarrow \text{If }~ \vec{a} \cdot \vec{b} = 0,\end{align}@$$then either:
@$\begin{align}|\vec{a}| = 0 \Rightarrow \vec{a} = \vec{0},\end{align}@$
@$\begin{align}|\vec{b}| = 0 \Rightarrow \vec{b} = \vec{0},\end{align}@$
@$\begin{align}\cos \theta = 0 \Rightarrow \theta = 90^\circ \Rightarrow \vec{a} \perp \vec{b}\end{align}@$
This is particularly useful in geometry and physics to check whether two directions are perpendicular.
Property 4: Dot Product of a Vector with Itself
Statement:@$$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}|^2\end{align}@$$The dot product of a vector with itself gives the square of its magnitude.
Proof:@$$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}| |\vec{a}| \cos 0 = |\vec{a}|^2\end{align}@$$Note: When you see @$\begin{align}\vec{a}^2\end{align}@$in vector algebra, it means @$\begin{align}\vec{a} \cdot \vec{a},\end{align}@$not actual multiplication of vector @$\begin{align}\vec{a}\end{align}@$ with itself.
Property 5: Compatibility with Scalar Multiplication
Statement:@$$\begin{align}(m\vec{a}) \cdot \vec{b} = m(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (m\vec{b})\end{align}@$$If a vector is scaled by a scalar @$\begin{align}m,\end{align}@$ then its dot product with another vector scales accordingly.
Proof:
Let @$\begin{align}m > 0.\end{align}@$ Then:@$$\begin{align}(m\vec{a}) \cdot \vec{b} = m |\vec{a}| |\vec{b}| \cos \theta = m(\vec{a} \cdot \vec{b})\end{align}@$$
If @$\begin{align}m < 0,\end{align}@$ the angle becomes @$\begin{align}\pi - \theta,\end{align}@$ but:@$$\begin{align}\cos(\pi - \theta) = -\cos \theta \Rightarrow \text{result still equals } m(\vec{a} \cdot \vec{b})\end{align}@$$
If @$\begin{align}m = 0,\end{align}@$ the entire product is zero.
Hence, the property holds for all real scalars.
Property 6: Associativity with Scalars
Statement:@$$\begin{align}m\vec{a} \cdot n\vec{b} = (mn)(\vec{a} \cdot \vec{b})\end{align}@$$If both vectors are multiplied by scalars, the dot product is scaled by the product of those scalars.
Property 7: Negative Vectors in Dot Products
Statements:
@$\begin{align}\vec{a} \cdot (-\vec{b}) = -(\vec{a} \cdot \vec{b}) = (-\vec{a}) \cdot \vec{b}\end{align}@$
@$\begin{align}(-\vec{a}) \cdot (-\vec{b}) = \vec{a} \cdot \vec{b}\end{align}@$
Negating one of the vectors reverses its direction, which flips the sign of the dot product.
Negating both vectors brings the angle back to the original (i.e., the direction difference is cancelled), so the dot product remains unchanged.
Property 8: Expansion Identities Using Dot Product
These identities help in simplifying vector expressions involving sums and differences:
Square of a sum:
@$\begin{align}|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}\end{align}@$
2. Square of a difference:
@$\begin{align}|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}\end{align}@$
3. Mixed product identity:
@$\begin{align}(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2\end{align}@$
Algebraic Formula for the Dot Product Using Components
The geometric formula for the scalar product,
@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta,\end{align}@$$ is useful when the angle between the vectors is known. However, in most practical problems, especially in coordinate geometry and physics, we are given the components of vectors instead of the angle between them. In such situations, the algebraic (component-wise) formula for the dot product is more convenient and efficient.
Suppose two vectors are expressed in Cartesian (rectangular) form as:@$$\begin{align}\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, \quad \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\end{align}@$$To compute the dot product @$\begin{align}\vec{a} \cdot \vec{b},\end{align}@$ we apply the distributive property and use the known scalar products of the unit vectors:@$$\begin{align}\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1, \quad \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\end{align}@$$Thus, when you expand the dot product using the distributive rule:@$$\begin{align}\vec{a} \cdot \vec{b} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot (b_1\hat{i} + b_2\hat{j} + b_3\hat{k})\end{align}@$$You get:@$$\begin{align}= a_1b_1(\hat{i} \cdot \hat{i}) + a_2b_2(\hat{j} \cdot \hat{j}) + a_3b_3(\hat{k} \cdot \hat{k}) = a_1b_1 + a_2b_2 + a_3b_3\end{align}@$$This gives us the algebraic formula:@$$\begin{align}\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\end{align}@$$This formula tells us that the dot product is simply the sum of the products of the corresponding components of the vectors. It avoids the need to compute the angle between them, making calculations straightforward.
For 2D vectors, say @$\begin{align}\vec{a} = a_1\hat{i} + a_2\hat{j}\end{align}@$ and @$\begin{align}\vec{b} = b_1\hat{i} + b_2\hat{j},\end{align}@$ the dot product simplifies to:@$$\begin{align}\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2\end{align}@$$
Corollary: Finding the Magnitude from Components
We already know from earlier that the dot product of a vector with itself gives the square of its magnitude:@$$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}|^2\end{align}@$$Now using the component formula:@$$\begin{align}\vec{a} \cdot \vec{a} = a_1^2 + a_2^2 + a_3^2\end{align}@$$Hence,@$$\begin{align}|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \quad \Rightarrow \quad |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\end{align}@$$This result is consistent with the Pythagorean Theorem extended to three dimensions. It shows how the magnitude of a vector can be directly calculated from its components.
Finding the Angle Between Two Vectors
Since we now have a geometric definition and an algebraic formula for the dot product, we can equate them to determine the angle between two non-zero vectors.
From geometry:@$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\end{align}@$$From components:@$$\begin{align}\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\end{align}@$$Equating both gives:@$$\begin{align}\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{a_1b_1 + a_2b_2 + a_3b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}}\end{align}@$$Once we compute @$\begin{align}\cos \theta,\end{align}@$ the angle @$\begin{align}\theta\end{align}@$ can be determined using the inverse cosine function:@$$\begin{align}\theta = \cos^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)\end{align}@$$This gives the acute, right, or obtuse angle between the two vectors, as the output of @$\begin{align}\cos^{-1}\end{align}@$ is always within the range @$\begin{align}0 \leq \theta \leq \pi\end{align}@$ (i.e., 0° to 180°), matching our convention for vector angles.
Examples of the Scalar (or Dot) Product of Two Vectors
Example 1
Given two vectors @$\begin{align}\vec{p} = 2\hat{i} - \hat{j} + 3\hat{k}\end{align}@$ and @$\begin{align}\vec{q} = \hat{i} + 2\hat{j} - \hat{k}.\end{align}@$
Find:
(a) The scalar projection of @$\begin{align}\vec{p}\end{align}@$ onto @$\begin{align}\vec{q}.\end{align}@$
(b) The vector projection of @$\begin{align}\vec{p}\end{align}@$ onto @$\begin{align}\vec{q}.\end{align}@$
The projection of one vector onto another tells us how much of the first vector lies along the direction of the second.
(a) Scalar Projection of @$\begin{align}\vec{p}\end{align}@$ onto @$\begin{align}\vec{q}\end{align}@$
The scalar projection of vector @$\begin{align}\vec{p}\end{align}@$ onto vector @$\begin{align}\vec{q}\end{align}@$ is given by the formula: @$$\begin{align}\text{Scalar projection}_{\vec{q}} \vec{p} = \frac{\vec{p} \cdot \vec{q}}{|\vec{q}|}\end{align}@$$
This value represents the component (or length of the shadow) of @$\begin{align}\vec{p}\end{align}@$ in the direction of @$\begin{align}\vec{q}.\end{align}@$
First, let's calculate the dot product @$\begin{align}\vec{p} \cdot \vec{q}:\end{align}@$
@$$\begin{align}\vec{p} \cdot \vec{q} = (2)(1) + (-1)(2) + (3)(-1) = 2 - 2 - 3 = -3.\end{align}@$$ Next, let's find the magnitude of @$\begin{align}\vec{q}:\end{align}@$
@$$\begin{align}|\vec{q}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.\end{align}@$$ Now, substitute these values into the scalar projection formula:
@$$\begin{align}\text{Scalar projection}_{\vec{q}} \vec{p} = \frac{-3}{\sqrt{6}}.\end{align}@$$ To rationalise the denominator, multiply the numerator and denominator by @$\begin{align}\sqrt{6}:\end{align}@$
@$$\begin{align}\text{Scalar projection}_{\vec{q}} \vec{p} = \frac{-3\sqrt{6}}{6} = -\frac{\sqrt{6}}{2}\end{align}@$$ The negative sign indicates that the projection of @$\begin{align}\vec{p}\end{align}@$ onto @$\begin{align}\vec{q}\end{align}@$ is in the direction opposite to @$\begin{align}\vec{q}.\end{align}@$
(b) Vector Projection of @$\begin{align}\vec{p}\end{align}@$ onto @$\begin{align}\vec{q}\end{align}@$
The vector projection of @$\begin{align}\vec{p}\end{align}@$ onto@$\begin{align}\vec{q}\end{align}@$ is a vector that has the magnitude of the scalar projection and points in the direction of @$\begin{align}\vec{q}.\end{align}@$ It is given by the formula:
@$$\begin{align}\text{Vector projection}_{\vec{q}} \vec{p} = \left(\frac{\vec{p} \cdot \vec{q}}{|\vec{q}|^2}\right) \vec{q} \quad \text{or} \quad (\text{Scalar projection}_{\vec{q}} \vec{p}) \hat{q}\end{align}@$$
where @$\begin{align}\hat{q}\end{align}@$ is the unit vector in the direction of @$\begin{align}\vec{q}.\end{align}@$
We already have @$\begin{align}\vec{p} \cdot \vec{q} = -3\end{align}@$ and @$\begin{align}|\vec{q}| = \sqrt{6},\end{align}@$ so @$\begin{align}|\vec{q}|^2 = 6.\end{align}@$
This is the vector component of @$\begin{align}\vec{p}\end{align}@$ that lies along the line of @$\begin{align}\vec{q}.\end{align}@$
@$$\begin{align}\text{Vector projection}_{\vec{q}} \vec{p} &= \left(\frac{-3}{6}\right) (\hat{i} + 2\hat{j} - \hat{k})\
&= -\frac{1}{2} (\hat{i} + 2\hat{j} - \hat{k})\
&= -\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k}\end{align}@$$
Example 2
A constant force @$\begin{align}\vec{F} = (3\hat{i} + 2\hat{j} - \hat{k})\end{align}@$ Newtons acts on a particle, causing it to be displaced from point @$\begin{align}A(1, -1, 2)\end{align}@$ metres to point @$\begin{align}B(4, 1, -1)\end{align}@$ metres. Calculate the work done by the force.
In physics, when a constant force @$\begin{align}\vec{F}\end{align}@$ acts on an object causing a displacement @$\begin{align}\vec{d},\end{align}@$ the work done (@$\begin{align}W\end{align}@$) by the force is given by the scalar product of the force and displacement vectors: @$$\begin{align}W = \vec{F} \cdot \vec{d}\end{align}@$$ Work is a scalar quantity, typically measured in Joules (@$\begin{align}J\end{align}@$) if force is in Newtons and displacement in metres.
First, we need to determine the displacement vector @$\begin{align}\vec{d}.\end{align}@$ The displacement is from point A to point B, so:
@$$\begin{align}\vec{d} &= \text{Position vector of B} - \text{Position vector of A}\
\vec{d} &= (4\hat{i} + \hat{j} - \hat{k}) - (\hat{i} - \hat{j} + 2\hat{k})\
\vec{d} &= (4-1)\hat{i} + (1-(-1))\hat{j} + (-1-2)\hat{k}\
\vec{d} &= 3\hat{i} + 2\hat{j} - 3\hat{k} \text{ metres}\end{align}@$$
The force vector is given as @$\begin{align}\vec{F} = 3\hat{i} + 2\hat{j} - \hat{k}\end{align}@$ Newtons.
Now, calculate the work done using the dot product:
@$$\begin{align}W &= \vec{F} \cdot \vec{d} = (3\hat{i} + 2\hat{j} - \hat{k}) \cdot (3\hat{i} + 2\hat{j} - 3\hat{k})\
W &= (3)(3) + (2)(2) + (-1)(-3)\
W &= 9 + 4 + 3\
W &= 16 \text{ Joules}\end{align}@$$Therefore, the work done by the force is 16 Joules.
Example 3
If @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ are two non-zero vectors such that the magnitude of their sum is equal to the magnitude of their difference, i.e., @$\begin{align}|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|,\end{align}@$ prove that @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ are perpendicular.
We are given the condition @$\begin{align}|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|.\end{align}@$
To work with this, it's often helpful to square both sides, as @$\begin{align}|\vec{v}|^2 = \vec{v} \cdot \vec{v}.\end{align}@$
So, @$\begin{align}(|\vec{a} + \vec{b}|)^2 = (|\vec{a} - \vec{b}|)^2.\end{align}@$
This means @$\begin{align}(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}).\end{align}@$
Now, expand both sides using the distributive property of the dot product:
Left Hand Side :
@$$\begin{align}(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) &= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}\
&= |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 (\text{ since }\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b})\end{align}@$$
Right Hand Side :
@$$\begin{align}(\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) &= \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + (-\vec{b}) \cdot (-\vec{b})\
&= |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\end{align}@$$
Equating the expanded LHS and RHS:
@$$\begin{align}|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\end{align}@$$ Now, simplify the equation. We can subtract @$\begin{align}|\vec{a}|^2\end{align}@$ and @$\begin{align}|\vec{b}|^2\end{align}@$ from both sides:
@$$\begin{align}2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})\end{align}@$$ Add @$\begin{align}2(\vec{a} \cdot \vec{b})\end{align}@$ to both sides:
@$$\begin{align}4(\vec{a} \cdot \vec{b}) = 0\end{align}@$$ Divide by 4:
@$$\begin{align}\vec{a} \cdot \vec{b} = 0\end{align}@$$ Since @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ are non-zero vectors and their dot product is zero, this implies that the vectors @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ are perpendicular to each other.
Geometrically, this condition means that the parallelogram formed by @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ has diagonals of equal length, which is a property of a rectangle. In a rectangle, adjacent sides are perpendicular.
Example 4
Given vectors @$\begin{align}\vec{u} = k\hat{i} + 2\hat{j} + \hat{k}\end{align}@$ and @$\begin{align}\vec{v} = k\hat{i} - k\hat{j} - 3\hat{k}.\end{align}@$ If @$\begin{align}\vec{u}\end{align}@$ and @$\begin{align}\vec{v}\end{align}@$ are orthogonal (perpendicular), find the possible value(s) of @$\begin{align}k.\end{align}@$
Two vectors are orthogonal if their scalar (dot) product is zero.
So, for @$\begin{align}\vec{u}\end{align}@$ and @$\begin{align}\vec{v}\end{align}@$ to be orthogonal, we must have @$\begin{align}\vec{u} \cdot \vec{v} = 0.\end{align}@$
Let's compute the dot product:
@$$\begin{align}\vec{u} \cdot \vec{v} &= (k\hat{i} + 2\hat{j} + \hat{k}) \cdot (k\hat{i} - k\hat{j} - 3\hat{k})\
&= (k)(k) + (2)(-k) + (1)(-3)\end{align}@$$ Set this dot product to zero:
@$$\begin{align}k^2 - 2k - 3 = 0\end{align}@$$
This is a quadratic equation in @$\begin{align}k.\end{align}@$ We can solve it by factoring or using the quadratic formula.
We need two numbers that multiply to -3 and add to -2. These are -3 and +1.
So, the equation can be written as: @$\begin{align}(k - 3)(k + 1) = 0.\end{align}@$
This gives two possible values for k:
@$$\begin{align}k - 3 = 0 \Rightarrow k = 3\end{align}@$$ or
@$$\begin{align}k + 1 = 0 \Rightarrow k = -1\end{align}@$$ Therefore, the vectors @$\begin{align}\vec{u}\end{align}@$ and @$\begin{align}\vec{v}\end{align}@$ are orthogonal if @$\begin{align}k = 3\end{align}@$ or @$\begin{align}k = -1.\end{align}@$
Example 5
Given two vectors @$\begin{align}\vec{x}\end{align}@$ and @$\begin{align}\vec{y}\end{align}@$ such that @$\begin{align}|\vec{x}| = 3, |\vec{y}| = 4,\end{align}@$ and the angle between them is @$\begin{align}60^\circ.\end{align}@$ Find the magnitude of the vector @$\begin{align}\vec{z} = 2\vec{x} - \vec{y}.\end{align}@$
We need to find @$\begin{align}|\vec{z}| = |2\vec{x} - \vec{y}|.\end{align}@$
It is often easier to work with the square of the magnitude:
@$$\begin{align}|\vec{z}|^2 = |2\vec{x} - \vec{y}|^2 = (2\vec{x} - \vec{y}) \cdot (2\vec{x} - \vec{y})\end{align}@$$ Expand this using the distributive property:
@$$\begin{align}(2\vec{x} - \vec{y}) \cdot (2\vec{x} - \vec{y}) &= (2\vec{x}) \cdot (2\vec{x}) - (2\vec{x}) \cdot \vec{y} - \vec{y} \cdot (2\vec{x}) + (-\vec{y}) \cdot (-\vec{y})\
&= 4(\vec{x} \cdot \vec{x}) - 2(\vec{x} \cdot \vec{y}) - 2(\vec{y} \cdot \vec{x}) + (\vec{y} \cdot \vec{y})\end{align}@$$
Since @$\begin{align}\vec{x} \cdot \vec{x} = |\vec{x}|^2, \vec{y} \cdot \vec{y} = |\vec{y}|^2, \text{ and }~ \vec{y} \cdot \vec{x} = \vec{x} \cdot \vec{y}:\end{align}@$
@$$\begin{align}= 4|\vec{x}|^2 - 4(\vec{x} \cdot \vec{y}) + |\vec{y}|^2\end{align}@$$
Now we need the value of @$\begin{align}\vec{x} \cdot \vec{y}.\end{align}@$ We are given @$\begin{align}|\vec{x}| = 3, |\vec{y}| = 4,\end{align}@$ and the angle @$\begin{align}\theta\end{align}@$ between them is @$\begin{align}60^\circ.\end{align}@$
@$$\begin{align}\vec{x} \cdot \vec{y} &= |\vec{x}| |\vec{y}| \cos \theta\
&= (3)(4) \cos 60^\circ\
&= 12 \times \frac{1}{2} = 6.\end{align}@$$
Substitute all known values into the expression for @$\begin{align}|\vec{z}|^2:\end{align}@$@$$\begin{align}|\vec{z}|^2 &= 4|\vec{x}|^2 - 4(\vec{x} \cdot \vec{y}) + |\vec{y}|^2\
&= 4(3)^2 - 4(6) + (4)^2\
&= 4(9) - 24 + 16 = 36 - 24 + 16\
&= 12 + 16 = 28\end{align}@$$
So,@$\begin{align}|\vec{z}|^2 = 28.\end{align}@$Therefore, the magnitude of @$\begin{align}\vec{z}\end{align}@$ is @$\begin{align}|\vec{z}| = \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}.\end{align}@$
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| Summary of Scalar (or Dot) Product of Two Vectors |
| Vectors have both magnitude and direction, making multiplication different from scalar multiplication. Dot Product (Scalar Product): + Defined for two non-zero vectors @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ with an angle @$\begin{align}θ\end{align}@$ between them: @$$\begin{align}\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\end{align}@$$ + The dot product measures how much one vector acts in the direction of another. + The result of the dot product is a scalar, not a vector. Angle Between Vectors: + The angle @$\begin{align}θ\end{align}@$ is measured between two vectors placed tail-to-tail, and it lies between 0 and @$\begin{align}π.\end{align}@$ + If vectors are collinear, @$\begin{align}θ\end{align}@$ is @$\begin{align}0°\end{align}@$ (same direction) or @$\begin{align}180°\end{align}@$ (opposite directions). Key Insights on the Dot Product: + If either vector is a zero vector, the dot product is 0. + When vectors are in the same direction (@$\begin{align}θ = 0\end{align}@$), the dot product gives the maximum positive value. + When vectors are in opposite directions (@$\begin{align}θ = π\end{align}@$), the dot product gives the maximum negative value. + If the dot product is zero, the vectors are perpendicular. Sign of the Dot Product: + Positive: Vectors form an acute angle (@$\begin{align}0° < θ < 90°\end{align}@$). + Zero: Vectors are perpendicular (@$\begin{align}θ = 90°\end{align}@$). + Negative: Vectors form an obtuse angle (@$\begin{align}90° < θ ≤ 180°\end{align}@$). Dot Product of a Vector with Itself: @$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}|^2\end{align}@$ Dot Products of Unit Vectors: The dot product of standard unit vectors @$\begin{align}\hat{i},\hat{j}\end{align}@$ and @$\begin{align}\hat{k}\end{align}@$ is: + - @$\begin{align}\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\end{align}@$ - @$\begin{align}\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\end{align}@$ Geometrical Interpretation: + The dot product can be understood as projecting one vector onto another. + The scalar projection of @$\begin{align}\vec{a}\end{align}@$ on @$\begin{align}\vec{b}\end{align}@$ is @$\begin{align}|\vec{a}| \cos \theta.\end{align}@$ Properties of the Dot Product: + Commutative: @$\begin{align}\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}.\end{align}@$ + Distributive: @$\begin{align}\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}.\end{align}@$ + Perpendicular Vectors: If @$\begin{align}\vec{a} \cdot \vec{b} = 0,\end{align}@$ the vectors are perpendicular. + Self Dot Product: @$\begin{align}\vec{a} \cdot \vec{a} = |\vec{a}|^2.\end{align}@$ + Compatibility with Scalars: @$\begin{align}(m\vec{a}) \cdot \vec{b} = m (\vec{a} \cdot \vec{b}).\end{align}@$ Dot Product Expansion: + Square of a Sum:@$\begin{align}|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b}.\end{align}@$ + Square of a Difference: @$\begin{align}|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a} \cdot \vec{b}.\end{align}@$ Algebraic Formula for the Dot Product: + For vectors @$\begin{align}\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\end{align}@$ and @$\begin{align}\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k},\end{align}@$ the dot product is: @$$\begin{align}\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\end{align}@$$ + For 2D vectors, the formula simplifies to: @$$\begin{align}\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2\end{align}@$$ Finding the Magnitude from Components: + The magnitude of a vector @$\begin{align}\vec{a}\end{align}@$ can be found using the components: @$$\begin{align}|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\end{align}@$$ Finding the Angle Between Two Vectors: + The cosine of the angle @$\begin{align}θ\end{align}@$ between two vectors can be found using: @$$\begin{align}\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\end{align}@$$ + The angle @$\begin{align}θ\end{align}@$ can then be determined using the inverse cosine function: @$$\begin{align}\theta = \cos^{-1} \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right)\end{align}@$$ |
Review Questions on Scalar (or Dot) Product of Two Vectors
Given vectors @$\begin{align}\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}, \vec{b} = \hat{i} - 2\hat{j} + \hat{k},\end{align}@$ and @$\begin{align}\vec{c} = \hat{i} + \hat{j} - 2\hat{k}.\end{align}@$ Determine a vector @$\begin{align}\vec{d}\end{align}@$ that is perpendicular to both @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b},\end{align}@$ and satisfies @$\begin{align}\vec{d} \cdot \vec{c} = 18.\end{align}@$
Let @$\begin{align}\vec{a} = (\hat{i} + \hat{j} + 3\hat{k})\end{align}@$ and @$\begin{align}\vec{b} = (2\hat{i} + \hat{j} - \hat{k}).\end{align}@$ Find the projection of (i) @$\begin{align}\vec{a}\end{align}@$ on @$\begin{align}\vec{b}\end{align}@$ and (ii) @$\begin{align}\vec{b}\end{align}@$ on @$\begin{align}\vec{a}.\end{align}@$
Determine the projection of the vector @$\begin{align}(2\hat{i} + \hat{j})\end{align}@$ in the direction of the vector @$\begin{align}(\hat{i} + 3\hat{j} - \hat{k}).\end{align}@$
Write the projection of the vector @$\begin{align}(\hat{i} - \hat{j} + 2\hat{k})\end{align}@$ along the vector @$\begin{align}\hat{j}.\end{align}@$
(i) Determine the projection of @$\begin{align}\vec{a}\end{align}@$ on @$\begin{align}\vec{b}\end{align}@$ if @$\begin{align}\vec{a} \cdot \vec{b} = 10\end{align}@$ and @$\begin{align}\vec{b} = (3\hat{i} + \hat{j} - 2\hat{k}).\end{align}@$
(ii) Write the projection of the vector @$\begin{align}(\hat{i} + 2\hat{j})\end{align}@$ on the vector @$\begin{align}(\hat{i} + \hat{j}).\end{align}@$
6. If @$\begin{align}\vec{a} = (\hat{i} - \hat{j} + 2\hat{k})\end{align}@$ and @$\begin{align}\vec{b} = (2\hat{i} + \hat{j} - \hat{k}),\end{align}@$ then calculate the angle between @$\begin{align}( \vec{a} + 2\vec{b} )\end{align}@$ and @$\begin{align}( 2\vec{a} - \vec{b} ).\end{align}@$
7. If @$\begin{align}\vec{a}\end{align}@$ is a unit vector such that @$\begin{align}(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15,\end{align}@$ find @$\begin{align}|\vec{x}|.\end{align}@$
8. Determine the angles which the vector @$\begin{align}\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}\end{align}@$ makes with the coordinate axes.
9. Prove that the vector @$\begin{align}\vec{a} = (\hat{i} - \hat{j} + \hat{k})\end{align}@$ is equally inclined to the coordinate axes.
10. Determine a vector @$\begin{align}\vec{a}\end{align}@$ of magnitude @$\begin{align}3\sqrt{2},\end{align}@$ making an angle @$\begin{align}\pi/3\end{align}@$ with the x-axis, @$\begin{align}\pi/4\end{align}@$ with the y-axis, and an acute angle @$\begin{align}\theta\end{align}@$ with the z-axis.
11. Find the angle between @$\begin{align}(\vec{a} + \vec{b})\end{align}@$ and @$\begin{align}(\vec{a} - \vec{b}),\end{align}@$ if @$\begin{align}\vec{a} = (\hat{i} - \hat{j} + 2\hat{k})\end{align}@$ and @$\begin{align}\vec{b} = (2\hat{i} + \hat{j} - \hat{k}).\end{align}@$
12. Express the vector @$\begin{align}\vec{a} = (5\hat{i} - 2\hat{j} - 5\hat{k})\end{align}@$ as the sum of two vectors such that one is parallel to the vector @$\begin{align}\vec{b} = (\hat{i} + \hat{j} + \hat{k})\end{align}@$ and the other is perpendicular to @$\begin{align}\vec{b}.\end{align}@$
13. If @$\begin{align}\vec{a} + \vec{b} + \vec{c} = \vec{0}, |\vec{a}| = 2, |\vec{b}| = 4, \text{ and }~ |\vec{c}| = 5,\end{align}@$ determine the angle between @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}.\end{align}@$
14. If @$\begin{align}|\vec{a}| = 3, |\vec{b}| = 4\end{align}@$ and @$\begin{align}\vec{a} \cdot \vec{b} = 6,\end{align}@$ find @$\begin{align}|\vec{a} - \vec{b}|.\end{align}@$
15. If @$\begin{align}(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 12\end{align}@$ and |@$\begin{align}\vec{a}| = 2|\vec{b}|,\end{align}@$ find @$\begin{align}|\vec{a}|\end{align}@$ and @$\begin{align}|\vec{b}|.\end{align}@$
16. The dot products of a vector with the vectors @$\begin{align}(\hat{i} + \hat{j} - 2\hat{k}), (\hat{i} + 2\hat{j} - \hat{k})\end{align}@$ and @$\begin{align}(2\hat{i} + \hat{j} + \hat{k})\end{align}@$ are 2, 6 and 3 respectively. Find the vector.
17. If @$\begin{align}\vec{AB} = (2\hat{i} - \hat{j} + 3\hat{k})\end{align}@$ and the coordinates of @$\begin{align}A\end{align}@$ are @$\begin{align}(1, -1, 1),\end{align}@$ find the coordinates of B.
18. If @$\begin{align}A(1, 2, 3), B(4, 3, -2), C(2, 5, 1)\end{align}@$ and @$\begin{align}D(0, 1, 1)\end{align}@$ are four points, show that @$\begin{align}\vec{AB}\end{align}@$ is perpendicular to @$\begin{align}\vec{CD}.\end{align}@$
19. Determine the value of @$\begin{align}\lambda\end{align}@$ for which the vectors @$\begin{align}( \hat{i} + \lambda\hat{j} + 2\hat{k} )\end{align}@$ and @$\begin{align}( 3\hat{i} + \hat{j} - \lambda\hat{k} )\end{align}@$ are perpendicular to each other.
20. Three vertices of a triangle are @$\begin{align}A(1, -2, -1), B(2, 1, 3)\end{align}@$ and @$\begin{align}C(4, 6, 2).\end{align}@$ Show that it is a right-angled triangle. Also, find its other two angles.
21. If the position vectors of the vertices @$\begin{align}A, B\end{align}@$ and @$\begin{align}C\end{align}@$ of a @$\begin{align}\triangle ABC\end{align}@$ be @$\begin{align}(1, 1, 1), (-1, 1, 0)\end{align}@$ and @$\begin{align}(0, 0, 2)\end{align}@$ respectively, then find @$\begin{align}\angle ABC.\end{align}@$
22. If @$\begin{align}\vec{a}\end{align}@$ and @$\begin{align}\vec{b}\end{align}@$ are unit vectors such that @$\begin{align}|\vec{a} + \vec{b}| = \sqrt{2},\end{align}@$find @$\begin{align}(3\vec{a} - 2\vec{b}) \cdot (\vec{a} + 3\vec{b}).\end{align}@$
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Class 9 (Old)
Course: Class 9 (Old)>Unit 4
Lesson 2: Graph of a linear equation in two variables
Graphing a linear equation: 5x+2y=20
Graphing a linear equation: y=2x+7
Graph from linear standard form
Plotting horizontal and vertical lines
Math>
Class 9 (Old)>
Linear equations in two variables>
Graph of a linear equation in two variables
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Graphing a linear equation: y=2x+7
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Learn how to create a graph of the linear equation y = 2x + 7.Created by Sal Khan and CK-12 Foundation.
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drurymacie 7 years ago Posted 7 years ago. Direct link to drurymacie's post “how do you graph an equat...” more how do you graph an equation like y=x+4
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Ellé🌹 7 years ago Posted 7 years ago. Direct link to Ellé🌹's post “You basically find the y ...” more You basically find the y intercept, which is 4 because of the +4 in the equation. y=x means that for every x you move across, you move y (one up).
I hope that helps!
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Anthony 10 years ago Posted 10 years ago. Direct link to Anthony's post “I am doing linear graphs,...” more I am doing linear graphs, but the equation they give me is something like 5x + y equals 4 and x - y equals -4. I need to graph these and find the point at which they intersect. I would like to change these to slope intercept form for them to be easier to work with, but I understood none of how to do it. Halp!
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Rachel Chavez 12 years ago Posted 12 years ago. Direct link to Rachel Chavez's post “how do you get the beginn...” more how do you get the beginning number of x?
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Mrs. H. Pollock 9 years ago Posted 9 years ago. Direct link to Mrs. H. Pollock's post “How do you calculate slop...” more How do you calculate slope woth y=mx + b
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Jhann 2 years ago Posted 2 years ago. Direct link to Jhann's post “You use the formula of Y2...” more You use the formula of Y2 - Y1/ X2 - X1 also known as rise/run
1 comment Comment on Jhann's post “You use the formula of Y2...”
(1 vote)
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katie arvola 12 years ago Posted 12 years ago. Direct link to katie arvola's post “how do you graph a slope?” more how do you graph a slope?
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Kim Seidel 12 years ago Posted 12 years ago. Direct link to Kim Seidel's post “To graph a line using it'...” more To graph a line using it's slope, you also need a point on the line. The slope of the line represents "change in Y" / "change in "x". So, the numerator of the slope tells you directional where to go up/down and the denominator of the slope tells you where to go right/left. But, you have to start from a point that you know is on the line. For example: if a line contains point (2, 3) and has a slope = 1/4.
1) Graph point (2, 3)
2) The numerator of the slope is 1. A "+1" for "y" means go up one unit.
3) The denominator of the slope if 4. A "+4" for "x" means go right 4 units.
4) So, starting at point (2, 3), move up one unit, then right 4 units and create another point for the line. You can do this as many times as needed to get as many points as you need.
Note: if the slope is negative, make sure you make only one number (numerator or denominator as negative). Example, if the slope above was -1/4, then you can do either of the following:
1) If "-1" for "y", move down one unit, you would then have "+4" for "x" so move right 4 units. OR
2) If "+1" for "y", move up on unit, you would then have "-4" for "x" so move left 4 units.
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yanicknorman 11 years ago Posted 11 years ago. Direct link to yanicknorman's post “In the context of the dol...” more In the context of the dollars/euro problem, could we have a negative dollar? I know we can have a negative money (debt), but in this case, it doesn't seem to make any sense at all.
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gswatts61 9 years ago Posted 9 years ago. Direct link to gswatts61's post “What would I do if there ...” more What would I do if there was a percentage in the word problem?
Would I have to change the percentage into a decimal or whole number, before continuing?
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sam Camel 6 years ago Posted 6 years ago. Direct link to sam Camel's post “How do you do this in rev...” more How do you do this in reverse?
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Zachary Hage 12 years ago Posted 12 years ago. Direct link to Zachary Hage's post “I don't get the euros par...” more I don't get the euros part .help?
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Βενιαμίν 12 years ago Posted 12 years ago. Direct link to Βενιαμίν's post “euros is the european ver...” more euros is the european version of dollars. There are .7 euros in a dollar. You pay 50 dollars. (50-5) .7 Because you pay 5 dollars. then you have 45 dollars and that times .7= the number of euros you get because 1 dollar= .7 euros.
Hope this helps! :)
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David Hu 10 years ago Posted 10 years ago. Direct link to David Hu's post “Are graphing quadratic e...” more Are graphing quadratic equations like graphing linear equations?
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jim.freund 10 years ago Posted 10 years ago. Direct link to jim.freund's post “In many ways, they can be...” more In many ways, they can be graphed in the same manner. We can choose values for one variable (typically x) and find the corresponding y values. Then we graph those order pairs and have our graph. However, for a linear equation, we really only need to find two points on the graph. Since two points determine a line, we draw a line through those two points and our graph is complete. Since a quadratic equation graphs as a parabola (think of the path of a ball thrown through the air), we need more than two points to create a reasonable graph. Three to five points are usually necessary to have a close approximation of the shape.
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Video transcript
Let's do a couple of problems graphing linear equations. They are a bunch of ways to graph linear equations. What we'll do in this video is the most basic way. Where we will just plot a bunch of values and then connect the dots. I think you'll see what I'm saying. So here I have an equation, a linear equation. I'll rewrite it just in case that was too small. y is equal to 2x plus 7. I want to graph this linear equation. Before I even take out the graph paper, what I could do is set up a table. Where I pick a bunch of x values and then I can figure out what y value would correspond to each of those x values. So for example, if x is equal to-- let me start really low-- if x is equal to minus 2-- or negative 2, I should say-- what is y? Well, you substitute negative 2 up here. It would be 2 times negative 2 plus 7. This is negative 4 plus 7. This is equal to 3. If x is equal to-- I'm just picking x values at random that might be indicative of-- I'll probably do three or four points here. So what happens when x is equal to 0? Then y is going to be equal to 2 times 0 plus 7. Is going to be equal to 7. I just happen to be going up by 2. You could be going up by 1 or you could be picking numbers at random. When x is equal to 2, what is y? It'll be 2 times 2 plus 7. So 4 plus 7 is equal to 11. I could keep plotting points if I like. We should already have enough to graph it. Actually to plot any line, you actually only need two points. So we already have one more than necessary. Actually, let me just do one more just to show you that this really is a line. So what happens when x is equal to 4? Actually, just to not go up by 2, let's do x is equal to 8. Just to pick a random number. Then y is going to be 2 times 8 plus 7, which is-- well this might go off of our graph paper-- but 2 times 8 is 16 plus 7 is equal to 23. Now let's graph it. Let me do my y-axis right there. That is my y-axis. Let me do my x-axis. I have a lot of positive values here, so a lot of space on the positive y-side. That is my x-axis. And then I use the points x is equal to negative 2. That's negative 1. That's 0, 1, 2, 3, 4, 5, 6, 7, 8. Those are our x values. Then we can go up into the y-axis. I'll do it at a slightly different scale because these numbers get large very quickly. So maybe I'll do it in increments of 2. So this could be 2, 4, 6, 8, 10, 12, 14, 16. I could just keep going up there, but let's plot these points. So the first coordinate I have is x is equal to negative 2, y is equal to 3. So I can write my coordinate. It's going to be the point negative 2, 3. x is negative 2. y is 3. 3 would land right over there. So that's our first one, negative 2, 3. Then our next point. 0, 7. We do it in that color. 0, 7. x is 0. Y is 7. Right there. 0, 7. We have this one in green here. Point 2, 11. 2, 11 would be right about there. And then this last point-- this is actually going to fall off of my graph. 8, 23. That's going to be way up here someplace. If you can even see what I'm doing. This is 8, 23. If we connect the dots, you'll see a line forms. Let me connect these dots. I've obviously hand drawn it, so it might not be a perfectly straight line. If you had a computer do it, it would be a straight line. So you could keep picking x values and figuring out the corresponding y values. In the situation y is a function of our x values. If you kept plotting every point, you'll get every line. If you picked every possible x and plotted every one, you get every point on the line. Let's do another problem. At the airport, you can change your money from dollars into Euros. The service costs $5. and for every additional dollar, you get EUR 0.7. Make a table for this and plot the function on a graph. Use your graph to determine how many Euros you would get if you give the office $50. I will write Euros is equal to-- so let's see, it's going to be dollars. So you're going to have to give your dollars. Right off of the bat, they're going to take $5. So dollars minus 5. So immediately this service costs $5. And then everything that's leftover-- this is your leftover-- you get EUR 0.7 for every leftover dollars. You get 0.7 for whatever's leftover. So this is the relationship. Now we can plot points-- we could actually answer their question right off the bat. If you give them $50, we don't even have to look at a graph. But we will look at a graph right after this. So if you did Euros is equal to-- if you have given them $50-- it would be 0.7 times 50 minus 5. You gave them 50. They took 5 as a service fee. So this is just $45 It would be 0.7 times 45. I could do that right here. 45 times 0.7. 7 times 5 is 35. 4 times 7 is 28 plus 3 is 31. And then we have only one number behind the decimal, only this 7. So it's 31.5. So if you give them $50, you're going to get EUR 31.5. Euros, not dollars. So we answered their question, but let's actually do it graphically. Let's do a table. Maybe I'll get a calculator out. I'll refer to that in a little bit. So let's say dollars you give them. And how many Euros do you get? I'll just put a bunch of random numbers. If you give them $5, they're just going to take your $5 for the fee. You're going to get $5 minus 5, which is 0 times 0.7. So you're going to get nothing back. So there's really no good reason for you to do that. Then if you give them $10. What's going to happen? If you give them $10, 10 minus 5 is 5 times 0.7. You're going to get $3-- or I should say EUR 3.50. 3.5 Euros, you'll get. Now what happens if you give them $30? Actually let me say 25. If you give him $25, 25 minus 5 is 20. 20 times 0.7 is $14. I'll do one more value. Let's say you gave them $55. This makes the math easy because then you subtract that 5 out. 55 minus 5 is 50 times 0.7 is $35. Is that right? Yep, that's right. You'll get EUR 35 I should say. These are all Euros. I keep wanting to say dollars. Let's plot this. All of these values are positive, so I only have to draw the first quadrant here. And so the dollars-- let's go in increments of 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55. I made my x-axis a little shorter than I needed to. All the way up to 55. And then the y-axis. I'll go in increments of 5. So that's 5, 10, 15, 20, 25, 30, 35. Well that's a little bit too much of an increment. 35. Now let's plot these points. I give them $5. I get EUR 0. This right here is Euros. This is the dollars. The dollars is the independent variable and we figure out the Euros from it. Or the Euros I get is dependent on the dollars I get. If I give $10, I get EUR 3.50. 3.50-- it's hard to read. Maybe 3.50 would be right around there. If I give $25, I get EUR 14. 25, 14 is right about there. Obviously, I'm hand drawing it, so it's not going to be quite exact. If I get $55, I get EUR 35. So 55, 35 right there. If I were to connect to the dots, I should get something that looks pretty close to a line. If I did it-- if I was a computer, it would be exactly a line. That looks pretty good. Then we could eyeball what they asked us to do. Use your graph to determine how many Euros you would get if you give the office $50. This is 50 right here. So you go bam, bam, bam, bam, bam, bam, bam, bam. I'm at the graph. Then you go all the way-- actually I drew that last point on the graph a little bit incorrectly. Let me. 35 is right here. Let me redraw that point. 35 is right there roughly. So 55, 35 is right there. So let me redraw my line. It will look-- I lost 25. 25, 14 is right there. So my graph looks something like that. That's my best attempt. Now let's answer the question. We give them $50 right there. You go up, up, up, up, up, up, up. $50. The person is going to get. You go all the way to the left-hand side. That's right about 31.50. We figured out exactly using the formula. But you can see, you can eyeball it from the graph and figure out any amount of dollars. If you give them $20, you're going to go all the way over here. You'll figure out that it should be-- well $20 should be about 7.50. The imprecision in my graph-- in my drawing the graph makes it a little bit less exact. When you say 20 minus 5 is 15. 15 times-- actually it'll be a little over $10, which is right. It's right over there. If you put $20 in there, 20 minus 5 is 15. 15 times 0.7 is $10.50, which is right there. So you can look at any point in the graph and figure out how many Euros you'll get. Let's do this one where we'll do a little bit of reading a graph. The graph-- I think it said use the graph below. Oh, the graph below shows a conversion chart for converting between weight in kilograms and weight in pounds. Use it to convert the following measurements. We have kilograms here and pounds here. So they want 4 kilograms into weight into pounds. So if we look at this right here, 4 kilograms is right there. We just follow where the graph is. So 4 kilograms into pounds, it looks like, I don't know, a little bit under 9 pounds. So a little bit less than-- so almost, I'll write almost 9 pounds. You can't exactly see. It's a little less than 9 pounds right there. 4 kilograms. Now 9 kilograms. We go over here. 9 kilograms. Go all the way up. That looks like almost exactly 20 pounds. Here they say 12 pounds into weight in kilograms. Actually kilograms is mass, but I won't get particular. So 12 pounds. Go over here. Pounds. 12 pounds in kilograms looks like 5 1/2. Approximately 5 1/2. And then 17 pounds to kilograms. So 17 is right there. 17 pounds to kilograms looks right about 7 1/2 kilograms. Anyway, hopefully that these examples made you a little bit more comfortable with graphing equations and reading graphs of equations. I'll see you in the next video.
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13643 | http://rushwjhs.weebly.com/uploads/5/4/5/4/54543371/acid-base_properties_of_salts_-_hydrolysis_notes.pdf | Ch. 15.10: Acid-Base Properties of Salt Solutions: Hydrolysis • Acid-Base Properties of Salts – Salt Hydrolysis is the reaction of a cation or an anion of a salt with water § Recall that a salt is the product of acid-base neutralization: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) § If cation is conjugate acid of a weak base, it reacts with H2O to form H3O+ and its base: • e.g. C5H5NH+ is the conjugate acid of C5H5N: C5H5NH+(aq) + H2O(l) ⇋ C5H5N(aq) + H3O+(aq) § If anion is conjugate base of a weak acid, it reacts with H2O to form OH– and its acid: • e.g. C3H5O2 – is the conjugate base of C3H5O2H: C3H5O2 –(aq) + H2O(l) ⇋ C3H5O2H(aq) + OH–(aq) – pH of solution depends on relative strength of acid and base from which salt is formed § Salt formed by neutralization of strong base and strong acid: e.g. NaCl • NaCl (s) → Na+ (aq) + Cl– (aq) • Na+ is conjugate acid of a strong base (NaOH): has no tendency to donate/form H+ (we’ll talk about metal ion hydrolysis below) • Cl– is conjugate base of a strong acid (HCl): has no tendency to accept H+ • Solution is neutral, pH = 7 – Salt formed by neutralization of a strong base and a weak acid: NaF: NaF (s) → Na+ (aq) + F– (aq) § Na+ does not hydrolyze, so it is a spectator ion § F– is conjugate base of a weak acid (HF, Ka = 7.1×10–4), so in H2O it accepts a proton: F– (aq) + H2O (l) ⇋ HF (aq) + OH– (aq) § This is a base ionization, so, Kb = [HF][OH −] [F −] = [HF] [H +][F −] ×[H +][OH −]= Ka −1Kw = Kw Ka = 1.0×10 −14 7.1×10 −4 =1.4×10 −11 § What is pH of a solution initially 0.0500 M NaF? • F– is a weak base, so set up ICE table: F– (aq) + H2O (l) ⇋ HF (aq) + OH– (aq) I 0.0500 M 0 0 C –x +x +x E 0.0500 – x M x M x M • So Kb = [HF][OH −] [F −] = (x) 2 (0.0500−x) ≈ (x) 2 0.0500 =1.4×10 −11 Thus x = [OH–] = [HF] = 8.4×10–7 M; pOH = 6.08 and pH = 14.00 – 6.08 = 7.92, slightly basic • % Hydrolysis = [HF] [F −] ×100% = 8.4×10 −7 0.0500 ×100% = 0.0017% – Salt formed by neutralization of a strong acid and a weak base: NH4NO3 : NH4NO3 (s) → NH4 + (aq) + NO3 – (aq) § NO3 – does not hydrolyze (anion of strong acid), so it is a spectator ion § NH4 + is conjugate acid of a weak base (NH3, Kb = 1.8×10–5), so it donates a proton in H2O: NH4 + (aq) + H2O (l) ⇋ NH3 (aq) + H3O+ (aq) Ch. 15.10: Acid-Base Properties of Salt Solutions p.2 § Now we have an acid ionization: Ka = [NH3][H3O +] [NH4 +] = [NH3] [NH4 +][OH −] ×[H +][OH −]= Kb −1Kw = Kw Kb = 1.0×10 −14 1.8×10 −5 = 5.6×10 −10 § What is pH of a solution initially 0.0500 M NH4NO3? • NH4 + is a weak acid, so set up ICE table: NH4 + (aq) + H2O (l) ⇋ NH3 (aq) + H3O+ (aq) I 0.0500 M 0 0 C –x +x +x E 0.0500 – x M x M x M • So Ka = [NH3][H3O +] [NH4 +] = (x) 2 (0.0500−x) ≈ (x) 2 0.0500 = 5.6×10 −10 Thus x = [H3O+] = [NH3] = 5.3×10–6 M; pH = 5.28, slightly acidic • % Hydrolysis = [NH3] [NH4 +] ×100% = 5.3×10 −6 0.0500 ×100% = 0.011% § If acid and base are both weak, must compare Ka of acid of anion with Kb of base of cation ♦ e.g. Given salt MX, compare Ka of HX to Kb of MOH • If Ka > Kb, solution is acidic • If Kb > Ka, solution is basic • If Ka ≈ Kb, solution is neutral ♦ Special case: C2H3O2H (acetic acid) and NH3 (ammonia) have Ka = Kb = 1.8×10–5 ♦ NH4C2H3O2, ammonium acetate is neutral – Hydrolysis of Small and Highly Charged Metal Ions § e.g. Al3+, Cr3+, Fe3+, etc.: small metal ions w/high charges • not Na+, Mg2+, etc: large ions, low charges § Metal ion surrounded by six H2O molecules • Electrons on O attracted to electron-deficient metal, weakens O–H bond, increases acid strength: [Fe(H2O)6]3+ (aq) ⇋ [Fe(H2O)5(OH)]2+ (aq) + H+ (aq) Ka = 2×10–3, relatively strong! ♦ Ka increases with increasing charge and decreasing radius of Mx+ (higher charge density) – Amphiprotic Salts § Species that can ionize and hydrolyze: e.g. HSO3 –: HSO3 – (aq) ⇋ H+ (aq) + SO3 2– (aq) Ka = 6.3×10–8 HSO3 – (aq) + H2O (l) ⇋ H2SO3 (aq) + OH– (aq) Kb = 7.7×10–13 • Since Ka > Kb, we would predict an acidic solution with pH < 7 • Homework 15-5: Problems pg. 672 #72, 73, 76, 77, 78, 79, 82 |
13644 | https://www.merckvetmanual.com/digestive-system/gastrointestinal-parasites-of-pigs/trichuris-sp-in-pigs | honeypot link
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MERCK MANUALVeterinary Manual
PROFESSIONAL VERSION
Trichuris sp in Pigs
ByLora Rickard Ballweber, DVM, DACVM, DEVPC, Department of Microbiology, Immunology, and Pathology, College of Veterinary Medicine and Biomedical Sciences, Colorado State University
Reviewed/Revised Feb 2022 | Modified Sept 2024
v3265119
Trichuris suis egg
Image
Courtesy of Dr. Bruce Lawhorn.
Trichuris suis (whipworm) is found in pigs worldwide. The adult worms are 5–6 cm long and whip-shaped; the cranial slender portion embeds within the epithelial cells of the large intestine, especially the cecum, with the thickened caudal third lying free in the lumen. Infection is by ingestion of eggs containing an infective first-stage larva. The larva hatches and penetrates the distal ileum, cecal, and colonic mucosa. The nematodes complete all four molts, after which the caudal end begins to protrude into the lumen. The prepatent period is 6–8 weeks; longevity is 4–5 months. Infection is generally light, with no clinical signs. Heavy infections may cause inflammatory lesions in the cecum and adjacent large intestine and may be accompanied by diarrhea and unthriftiness.
Clinical signs most often occur in young animals; resistance is both acquired and age-related. The double-operculated brown eggs (50–68 × 21–31 mcm) are diagnostic. Eggs are heavy; thus, good technique with media of proper specific gravity is essential.
Trichurids have a short period of egg-laying (2–5 weeks) before the worms are expelled by immune-mediated reactions, and thus little importance can be given to percentage of pigs excreting eggs or number of eggs per gram of feces. Clinical trichuriasis is usually associated with the larval stages before eggs are passed in the feces; in these cases, examining mucosal scrapings taken at postmortem examination for smaller stages of the parasite is recommended. Mature parasites are easily found after ingesta is washed away and can be identified by their size and whiplike form.
Careful evaluation of product labels is important as not all anthelmintics have good activity against T suis. Dichlorvos, some benzimidazoles, and ivermectin are effective against the adult worms. Biologically, the eggs are comparable to Ascaris eggs—they are highly resistant to chemicals and may remain infective for up to 11 years; control relies on thorough cleaning of the affected area and moving the animals to clean plots. Tillage of pastures between uses can decrease the number of eggs that survive. Trichuris eggs develop rather slowly (10–12 weeks under optimal conditions), and because they do not develop at temperatures < 16°C, there is only one generation per year in temperate geographic areas.
Trichuris suis larvae may hatch in the large intestine of humans, in which the larvae seem to be able to establish themselves transiently. It is this feature that has led to intense research interest in the treatment of inflammatory bowel disease, including ulcerative colitis and Crohn's disease, of humans via administration of infective T suis eggs.
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13645 | https://collegedunia.com/news/e-110-three-cars-leave-from-a-to-b-in-equal-time-intervals | Three Cars Leave From A To B In Equal Time Intervals GMAT Problem Solving
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Three Cars Leave From A To B In Equal Time Intervals GMAT Problem Solving
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byBhaskar DasStudy Abroad Specialist
Question: Three cars leave from A to B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240 miles away from B. The first car arrives at C an hour after the second car. The third car, having reached C, immediately turns back and heads towards B. The first and the third car meet a point that is 80 miles away from C. What is the difference between the speed of the first and the third car?
60 mph
20 mph
40 mph
80 mph
120 mph
‘Three Cars Leave From A To B In Equal Time Intervals GMAT Problem Solving’ is the topic for GMAT Quantitative Reasoning. GMATquantitative reasoning section is focused on analyzing the candidates' ability to solve mathematical, and quantitative problems. This topic is a part of the GMATProblem Solvingsection. It includes five optional answers and candidates need to choose the appropriate one. Candidates are given 62 minutes to complete this section. GMAT Quantitative Reasoning section of the exam comprises 31 questions.
Solution and Explanation:
Approach Solution 1:
Explanation:
There are three points that have been mentioned in the question. Thus, the first thing that is necessary to be accomplished is calculating the ratio of the first car to that of the second car. Thus, this leaves us with the below-mentioned equation;
(240-80):(240+80).
Upon solving the above-mentioned equation, we get the ratio as: 1:2.
Further, to solve this, let us consider the distance from point A to point B being X.
In addition, consider the interval between the departure of the first car and second car to be represented as t.
Thus, we can highlight the total time taken by the first car as; T.
This implies that the speed of the first car can be calculated as; X/T
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Further, this gives us the conclusion that speed of the second car can be calculated as;(\frac {X}{T-t})
Moreover, speed of the third car can be calculated as;(\frac{X}{T-2t})
Thus, we can further calculate the above-mentioned equations as:(2\frac{X}{T}=\frac{X}{T-2t})
Thus, the first equation that we can get is T=4t……. (1)
In order to get the second equation we can derive with:
(\frac{240T}{X}-\frac{240(T-t)}{X}=1.....(2))
Upon solving the first and second equations:
We get the value X/T=60
Therefore, by substituting the value of X/T, we can easily calculate the speed of the cars.
Speed of first car= 60Km/h
Speed of third car= 260= 120 Km/h
Difference = 120-60 = 60Km/h
Correct Answer: A
Approach Solution 2:
Explanation:
Let us consider the representation of the speed of the car in descending order to be z, y and x.
As mentioned in the question the first and the third car meet a point that is 80 miles away from C, which is 240 miles away.
Thus, this yields that x travelled 240-80 and z travelled 240+80.
Ratio of distance = 240-80:240+80=160:320=1:2, so their speed will be in the ratio 2:1.
Thus, considering the speed of car x to be X. thus, the speed of car z will be 2X.
Thus, there is a necessity to solve for Z-X=2X-X=X
If the distance from A to B is D, then equal intervals means:
(\frac{D}{Z}-\frac{D}{Y}=\frac{D}{Y}-\frac{D}{X}.........\frac{D}{2X}-\frac{D}{Y}=\frac{D}{Y}-\frac{D}{X}........Y=\frac{4X}{3}.....)
Thus, the ratio of the speed can be highlighted as:(X:\frac{4X}{3}:2X)
Thus, to equate and find the value of X, consider the first car arrives at C hour after the arrival of the second car.
(\frac{240}{X}-240/ \frac{4X}{3}=1)
(\frac{240}{X}-\frac{180}{X}=1)
This, implies that, X=60
Therefore, the answer is 60.
Correct Answer: A
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How to split a quartic into two quadratics?
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I have a quartic in Z[x] with very large coefficients that I know splits into two quadratics in Z[x]. What is the best way to do find the quadratics?
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edited Jul 30, 2015 at 10:40
TurboTurbo
asked Jul 30, 2015 at 10:35
TurboTurbo
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To do what? It is not very clear.ajotatxe –ajotatxe 2015-07-30 10:39:38 +00:00 Commented Jul 30, 2015 at 10:39
So, you want to factor a fourth degree polynomial into a product of two second degree polynomials?Alice Ryhl –Alice Ryhl 2015-07-30 10:41:54 +00:00 Commented Jul 30, 2015 at 10:41
yes. that sounds right.Turbo –Turbo 2015-07-30 10:42:11 +00:00 Commented Jul 30, 2015 at 10:42
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Since there are three ways of partitioning a four element set into two pairs, finding an integer factorisation up to the order of the factors is about as hard as finding an integer root to a cubic (an abelian cubic, in fact). Then choosing the order of the factors is as hard as solving a general quadratic, so computing a square root.
Let L=Q(A,B,C,D) and K=Q S 4=Q(E 1,E 2,E 3,E 4)
where E i are the i th symmetric polynomials in A,B,C,D.
Getting a factorisation means that you have integer values for A+B,C+D,A B,C D.
Let H 1 be the subgroup of S 4 fixing A+B, and H 2 be the subgroup fixing U=A B+C D.
H 1 has order 4 and is a subgroup of H 2 (order 8), so the fundamental theorem of Galois theory says that
[K:L H 2=K(U)]=3 and [L H 2:L H 1=K(A+B)=Q(A+B,A B,C+D,C D)]=2.
More precisely, if we focus on polynomials in the roots with integer coefficients, it turns out that Z[A,B,C,D]H 2=Z[E i,U], where U satisfies the equation
(U−(A B+C D))(U−(A C+B D))(U−(A D+B C))=0.
After developing, this is U 3−E 2 U 2+(E 1 E 3−4 E 4)U−(E 2 1 E 4+E 2 3−4 E 2 E 4)=0, so first you want to find the integer roots of this cubic.
Once you have an integer value for U, you have determined that A goes with B and C goes with D, but you still haven't distinguished the two pairs from each other.
Then, Z[A,B,C,D]H 1=Z[E i,A+B,A B]=Z[A+B,A B,C+D,C D], and we have the following equations :
((A+B)−(C+D))2=4 U+E 2 1−4 E 2((A+B)−(C+D))(A B−C D)=E 1 U−2 E 3(A B−C D)2=U 2−4 E 4
so you can find the integer values of A+B−C−D and A B−C D by computing an integer square root, and then doing an integer division with the second equation. Note that if A+B=C+D then you need the third equation to find A B−C D, and if A B=C D you need the first equation to find A+B−C−D, so to cover all the cases, you need all three of them.
And finally, you get A+B,C+D,A B,C D after computing 1 2(E 1±(A+B−C−D)) and 1 2(U±(A B−C D)).
Once you have those values, you have the factorisation (X 2−(A+B)X+A B)(X 2−(C+D)X+C D) that you wanted.
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answered Feb 2, 2017 at 11:52
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A general method always available is to first factor the polynomial F(x) modulo some prime p (possible a small one), and then Hensel lift that to a factorization modulo progressively higher powers of that prime. Given that you know there exists a factorization over Z, the sequence of those modulo p n factorizations will stabilize when n is large enough, and you get a fairly good idea of the actual factors (it is simple to verify a candidate factorization).
For the Hensel lifting to work out, the two factors modulo p should be distinct. Unless the two quadratic factors your polynomial has are equal they will also be distinct modulo most primes. You can detect this by checking whether gcd Zx=1 or not. A repeated factor of F(x) will show as a common factor of F(x) and F′(x).
For simplicity it may be better not to use a prime p such that your polynomial also has linear factors modulo that prime. You can probably use linear factors as well, but if your pair them up incorrectly, then the Hensel lifting may not converge. If you know that the quadratic factors of F(x) are irreducible over Z, then they will remain so modulo some primes (Note: this does not hold for polynomials of degree >2). I admit to not being certain what is best here, if you get four distinct linear factors modulo your chosen p. May be it is best to try out lifting all the three quadratic × quadratic factorizations that you get out of that?
To find the factorization modulo p I would start with Cantor-Zassenhaus algorithm, because it is simple. Any irreducible linear or quadratic polynomial with coefficients in Z p is a factor of x p 2−x=x(x p 2−1−1). Furthermore, a factor with the property that its zeros (in the field F p 2) are non-zero squares will also be a factor of Q(x):=x(p 2−1)/2−1. With a little bit of luck the zeros of one factor of F(x) are squares and those of the other are not. The zeros of an irreducible quadratic either both are squares or both are non-squares, so this is a 50-50 chance. If that happens, then the factor with zeros in (F∗p 2)2 is the common factor gcd F px, and can be calculated efficiently with Euclid's algorithm. The other factor can then be found by long division.
To improve the odds of success you can do the following modifications (this is the non-deterministic part in C-Z). Instead of calculating just gcd(F(x),Q(x)) you can also calculate gcd(F(x−a),Q(x)) for several a∈Z p. If P(x) is an irreducible quadratic factor of F(x) modulo p, then for a random choice of a there is next to no correlation between "the zeros of P(x) are squares" and "the zeros of P(x−a) are squares". Therefore, statistically speaking, each try with a different a halves your probability of failure. Of course, P(x−a) is a factor of F(x−a), iff P(x) is a factor of F(x).
Example.
Find the factors of F(x)=x 4+2 x 3−123 x 2+524 x−341
(I know that this too small to be of interest to you, and here trial-and-error would quickly do it. This is just for demonstration).
First we check that gcd Qx=1, so there are no repeated factors. I skip the runs of (generalized) Euclid's algorithm.
Let's try to factor F(x) modulo p=5. First try with a=0. gcd F 5x=x 4+2 x 3+2 x 2+4 x+4≡F(x)(mod 5). This was no good. Looks like all the zeros of F(x) are squares in F 25. Let's try instead F(x+1)≡x 4+x 3−x 2+3 x+3(mod 5). We get gcd F 5x=x 2+x+1. That's more like it! We know that one of the modulo 5 factors of F(x) is G 1(x)=(x−1)2+(x−1)+1≡x 2+4 x+1(mod 5). The other factor is gotten by long division, H 1(x)=F(x)/G 1(X)=x 2+3 x+4, so we got F(x)≡G 1(x)H 1(x)(mod 5).
An aside: Why p=5? Modulo p=2 F(x)≡(x 2+x+1)2 is a square, and modulo p=3 we have F(x)≡(x+2)4. Neither of those primes can be used, because Hensel lifting requires coprime factors. The smaller the prime, the higher the chance of repeated factors.
On with the Hensel lifting. Next we want to find a factorization of F(x) modulo 5 2=25. Because gcd F 5(G 1(x),H 1(x))=1 Hensel's lemma guarantees the existence of monic polynomials G 2(x),H 2(x) such that G 2(x)H 2(x)≡F(x)(mod 25), and G 1≡G 2,H 1≡H 2(mod 5). These are found as follows. Interpreting the polynomials G 1,H 1 to have integer coefficients we are looking for linear polynomials K 2,L 2 with integer coefficients such that F≡(G 1+5 K 2)(H 1+5 L 2)≡G 1 H 1+5(K 2 H 1+L 2 G 1)(mod 25). Because F−G 1 H 1=−5 x 3−140 x 2+505 x−345=5(−x 3−28 x 2+101 x−69)≡5(4 x 3+2 x 2+x+1)(mod 25) we want to find K 2,L 2∈F 5[x] such that K 2 H 1+L 2 G 1=4 x 3+2 x 2+x+1:=U(x)∈F 5[x]. Here U(x)≡3 x(mod G 1(x)) and U(x)≡1(mod H 1(x)), so we want L 2 G 1≡1(mod H 1) and K 2 H 1≡3 x(mod G 1).
A run of the Extended Euclid's algorithm in the ring F 5[x] gives us the identity 1=gcd(G 1,H 1)=(2 x+2)G 1+(3 x+1)H 1, and we can read the inverse of G 1 modulo H 1 (resp. inverse of H 1 modulo G 1) from this. Therefore K 2≡3 x H−1 1=3 x(1+3 x)≡2 x+1(mod G 1). and L 2≡1⋅G−1 1=(2 x+2)≡2 x+2(mod H 1), This gives us the lifts G 2(x)=G 1(x)+5 K 2(x)=x 2+14 x+6 and H 2(x)=H 1(x)+5 L 2(x)=x 2+13 x+14. As a check we calculate F−G 2 H 2=−25 x 3−325 x 2+250 x−425, and see that it is, indeed, a multiple of 25.
Here we can either continue the lifting to find a factorization modulo 5 3=125, or do a bit of experimenting. I skip the lifting, and simply observe that (G 2−25 x+25)(H 2−25)=F, so the factorization is F(x)=(x 2−11 x+31)(x 2+13−11). A big tip off (to anyone doing pencil and paper work) in the last step is the constant term −341=−11⋅31 of F(x).
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edited May 26, 2019 at 4:36
answered Aug 3, 2015 at 15:14
Jyrki LahtonenJyrki Lahtonen
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1 This approach works equally well, when you are factoring a sextic known to be a product of two cubics, or an octic known to be a product two quartics.Jyrki Lahtonen –Jyrki Lahtonen 2015-08-09 14:01:25 +00:00 Commented Aug 9, 2015 at 14:01
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Use the coefficients relation, Look here For example, you want to split:
8+10 x+9 x 2+3 a 3+x 4=(x 2+p x+q)(x 2+r x+s)
The polynomial coefficients must be equal, so:
(1) 3=p+q
(2) 9=q+s+p r
(3) 10=p s+q r
(4) 8=q s
this is a system of equtions on Z...
multipliying (1) by s and by r
(5) 3 s=p s+q s=p s+8
(6) 3 r=p r+q r
(3)-(5) :
10−3 s=p s+q r−p s−q s=q r−8
(6′) 18=3 s+q r
(2)-(6) :
9−3 r=q+s−q r
(7) q r=q+s−3 r+9
substitute (7) in (6′) :
18=3 s+q+s−3 r+9
(8) 9=4 s+q−3 r
(1)-(2) :
(9) 9=q+s+(3−q)r=q+s+3 r−q r
(9)+(3):
19=q+s+3 r−q r+p s+q r=q+s+3 r+p s
and substituting p s from (5):
19=q+s+3 r+3 s−8
(10) 27=q+4 s+3 r
(10)-(8):
18=6 r
r=3 rewriting and substituing r=3 in (6),(8)
(6) 18=3 s+3 q
(8) 18=4 s+q
(6)-3(8): 18−3⋅18=3 s−12 s
−36=−9 s
s=4
(6) 18=3⋅4+3 q
6=3 q
q=2
(1) 3=p+q=p+2
p=1
You can repeat this process for any numbers, just make sure you don't build second order equation, otherwise you will get more than one solution which will be not integer
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edited Jun 12, 2020 at 10:38
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answered Jul 30, 2015 at 10:51
Grigory IlizirovGrigory Ilizirov
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Could you do something similar for sextic? want me to post another question?Turbo –Turbo 2015-08-03 15:11:42 +00:00 Commented Aug 3, 2015 at 15:11
I can try... just post as another question Grigory Ilizirov –Grigory Ilizirov 2015-08-04 09:31:50 +00:00 Commented Aug 4, 2015 at 9:31
math.stackexchange.com/questions/1383938/…Turbo –Turbo 2015-08-05 10:38:54 +00:00 Commented Aug 5, 2015 at 10:38
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Can we assume that all resulting coefficients are integers? If so, use the rational root theorem to identify a list of possible roots and test them using the factor theorem. Once you found one root, you can divide it from the quartic to get a cubic. Repeat one more time and you have your quadratic. It's not the most efficient if your coefficient are really big.
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answered Jan 26, 2018 at 5:20
CuriousCurious
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It is more likely than not that the two quadratic factors are irreducible, and hence you cannot find a zero with the rational root theorem. Now what?Jyrki Lahtonen –Jyrki Lahtonen 2018-01-26 06:23:19 +00:00 Commented Jan 26, 2018 at 6:23
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13649 | https://nrich.maths.org/problems/almost-one | Almost One | NRICH
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Almost one
Choose some fractions and add them together. Can you get close to 1?
Age
11 to 14
Challenge level
Secondary
NUMBER
Number Operations and Calculation Methods
Exploring and noticingWorking systematicallyConjecturing and generalisingVisualising and representingReasoning, convincing and proving
Being curiousBeing resourcefulBeing resilientBeing collaborative
Problem Getting Started Student Solutions Teachers' Resources
Problem
Almost One printable sheet
Here is a set of six fractions: 1 6 1 25 3 5 3 20 4 15 5 8
Choose some of the fractions and add them together. You can use as many fractions as you like, but you can only use each fraction once.
Can you get an answer that is close to 1?
What is the closest to 1 that you can get?
With thanks toColin Fosterwho introduced us to this problem.
Getting Started
You could begin by choosing a fraction bigger than 1 2 and adding on smaller fractions to get close to 1.
You could approximate each fraction to fractions that you are familiar with (with small denominators) and then use your approximations to estimate possible sums.
It is often easiest to add fractions when they have the same denominator...
Student Solutions
We received many solutions to this problem, and well done to everyone who found a sum close to 1.
Xaviar from Temora Public School in Australia, Demilade from Green Springs School in Nigeria, Ibrahim from Nigerian Tulip International College, Jonathan, Alex and Shafi from Greenacre Public School in Australia, Marissa, Jessica and Logan from Matamata Intermediate in New Zealand, Ethan from King Geroge V School in Hong Kong, Year 9 class at Wyedean School in England, John from South Hunsley Secondary School in the UK and Kenzo from ISL in Switzerland added fractions together to see what they could find.
Ethan found:
1 25+3 5+3 20 I "simplified" them so it was easier to add them together.
4 100+60 100+15 100=79 100
It isn't very close but this is the closest one I managed to find.
Ibrahim found:
3 5+5 8+3 20=24+25+6 40=55 40=1.375 which, when given to 1 significant figure, is 1
1 6+4 15+3 5=5+8+18 30=31 30=1.03 which is very close to 1
Logan took the idea of equivalent fractions a bit further:
First I analyzed the 6 fractions and looked at which ones could add together to get a number close to one. After a few tries not working I then chose to add 5 8 and 4 15 which was equivalent to 107 120 as 120 is the least common multiple of the two denominators 15 and 8. I then found any other denominators that could multiply into 120 and the fractions were 3 5, 1 6 and 1 20.These fractions were equivalent to 72 120, 20 120 and 6 120.
After adding all these numbers on in separate equations the closest answer was 127 after adding the 1 6 or 20 120. This is equivalent to 1.058, just 0.058 off the number 1.
William from New End Primary School in the UK, James from The King's School Grantham in the UK, Will from WWSPS in Australia, Rishika from Nonsuch High School for Girls in the UK, Yesh from Manchester Grammar School in the UK, Aryaman from Bangkok Patana School in Thailand, Hondfa, Tony, Jacob, Nathan, Laurenc and Allan from Greenacre Public School, Isaac from Rugby School in the UK, Victor from South Hunsley in the UK, Daanyal from Caerleon Comprehensive School in Wales and Paul from Coventry University in the UK all used this method of expressing all of the fractions over the same denominator.
This is Rishika's working:
The easiest way to find an answer closest to 1 is to find the LCM (lowest common multiple) of all the denominators first, for which I used prime factorisation:
6=2×3 25=5×5(5 2)5=5 20=2×2×5(2 2×5)15=3×5 8=2×2×2(2 3)
To find the LCM we need to multiply together the highest number of 2 s, 3 s and 5 s present in each set, overall. From above, we know that the highest number of 2 s is 3 (2×2×2), the highest number of 3 s is 1 (2×3 and 3×5) and
the highest number of 5 s is 2 (5×5).
Therefore we multiply 2×2×2×3×5×5=600, which is the LCM.
Then we can put all the fractions over the LCM:
100 600 24 600 360 600 90 600 160 600 75 600
To find the fractions that add to give an answer closest to 1, I first added all the above fractions together, giving 809 600 (remember 1=600 600).
I needed 209 600 less to make 1.
The fractions that add to make the closest to 209 600 were: 100 600,24 600 and 75 600 (1 6,1 25 and 5 8), summing to 199 600.
Therefore, I needed to add other fractions to give me the answer closest to 1:
3 5+3 20+4 15=360 600+90 600+160 600=610 600=1 1 60 which is the closest answer to 1.
Daanyal used a denominator of 1800 instead of 600, and wanted to prove that 3 5+3 20+4 15 was the best possible sum. This is Daanyal's working:
I used trial and error to find a combination close to 1800. Quite quickly I got to: 1080 1800+270 1800+480 1800=1830 1800=61 60 I decided that this was the closest I could get by trial and error. Next, I needed to prove/disprove that this was the closest to 1 you can get.
To improve this equation, you need to:
a) Remove one or more terms
b) Replace it with one or more terms
To keep it brief, I am going to refrain from using the denominators as they are not very significant.
Three numerators used:Three numerators not used:
1080 300 270 72 480 1125
The three used numerators can be used to create a list of seven options for removing terms from the equation. Likewise, the three unused numerators can be used to create a similar list of replacement terms.
Removal Replacement
480 300 270 72 1080 1125 480+270=750 300+72=372 270+1080=1350 72+1125=1197 480+1080=1560 300+1125=1425 180+270+1080=1830 300+72+1125=1497 If you remove any term on the left and replace it with any term on the right, it results in a total which is further from 1800 than 1830 is. This proves that 1830 1800 or 61 60 is the closest to 1 you can get.
Paul said that we could probably solve this by programming some software to try every iteration.
Flynn from Parkside Primary School in Australia and Aidan from Sheldon School in England expressed all of the fractions as fractions with denominators of 100, which led to some very strange fractions (which mathematicians don't usually allow). Flynn got
62.5 100+16.6666666 100+15 100+4 100 or 1.02 (Rounded)
and Aidan got
3 5(60 100)+3 20(15 100)+4 15(26.7 100)(rounded up)=101.7 100
This is similar to what Saroja from India did using percentages:
3 5,3 20 and 4 15.
The first fraction is 60%.
The third one is a little more than 25%.
The total of these two is about 85%
Hence we have to choose a fraction which is about 15%.
Hence 3 20(equivalent to 15%) fits in well. The sum of the three fractions is 61 60 - nearly 1.
Zane from Shireland Collegiate Academy, Soham from Sutton Grammar School and Sheila, all in the UK, converted the fractions into decimals to solve the problem. Clickherehere to see Soham's complete solution, with explanation.
Teachers' Resources
The suggestions in these notes are adapted from Colin Foster's article,Sum Fractions.
Why do this problem?
Adding and subtracting fractions is a procedure which students often find very difficult to master. It is important to address the area without it feeling like an exact repetition of what they have done many times before.
One way to avoid the tedium of lots of repetitive practice is to embed practice in a bigger problem which students are trying to solve. This idea is explored in Colin Foster's article, Mathematical Etudes, and this problem is an example of a mathematical etude.
Possible approach
"What can you say about these six fractions?" 1 6 1 25 3 5 3 20 4 15 5 8
Students might note that they are all different, that they are all less than 1, that they are all positive, that they are all expressed in their simplest terms, that four are less than a half and two are greater than a half, that they are not in order of size, and so on.
Encourage students to say as many things as they can think of. Questions like this are a good way to encourage students to be mathematically observant.
"Which fraction do you think is the largest? Which is the smallest? Why?"
Since all of the fractions are expressed in their simplest terms, it is easy to see that none of them are equal. Students may compare fractions by making their denominators equal or converting them to decimals.
Encourage students to use 'informal' methods of comparing fractions, and only calculate when it becomes absolutely necessary. For example, 1 20 is bigger than 1 25, so 3 20 will certainly be bigger than 1 25.
"Write a fraction that is equal to 3 5. And another, and another..."
Students could write their fractions on mini-whiteboards.
They will probably list equivalent fractions such as 6 10,30 50 etc.
You could encourage a wider range of answers by introducing some constraints, for example... "Write down one with an odd denominator" or "Write down one where the numerator is a five-digit number that does not end in 0".
Then ask them to do the same with 5 8.
"How would you add 3 5 and 5 8 without a calculator?"
"The answer to 3 5+5 8 is a little bit more than 1. Is there any way that you could have predicted that the answer was going to be more than 1 without working it out exactly?"
Both fractions are more than 1 2, so their total must be more than 1.
This sort of reasoning can be very useful for estimating the size of an answer so that mistakes can be spotted. Estimation will be important in the main activity that follows.
Return to the original set of six fractions:
1 6 1 25 3 5 3 20 4 15 5 8
"Choose some of the fractions and add them together. You can use as many fractions as you like, but you can only use each fraction once."
"Can you get an answer that is close to 1?"
"What is the closest to 1 that you can get?"
Make it clear that calculators are not to be used!
If some students are unsure how to start, encourage them to talk to their partner.
Give students some time to work on the problem. This will be a good opportunity to circulate and see how students are getting on.
If students obtain an answer like 11 12 (from 1 6+3 5+3 20), they may think that they are as close as possible, as their answer is “only 1 away”, but because the “one” is “one twelfth” they are not really that close (1 12 is more than 8%), so they should aim to get even closer!
Allow plenty of time at the end of the lesson for students to share their approaches and reasoning.
Possible support
If students are not secure with equivalent fractions they could do some work with Fractional Wall.
Possible extension
Students could be asked to find the set of fractions which add up to as near to 1 2 as possible.
For other rich contexts that offer students an opportunity to practise manipulating fractions see Peaches Today, Peaches Tomorrow and Keep it Simple.
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NRICH is part of the family of activities in the Millennium Mathematics Project. |
13650 | https://mste.illinois.edu/dildine/tcd_files/metric/length.htm | Metric System Table
[Spreadsheet]
Conversion Tables and Methods to Use when converting
from Metric Units to Metric Units (Length)
Some terms to remember
Meter – LengthKilo – Thousand
Liter – VolumeMilli – Thousand
Gram – Mass/WeightCenti – Hundred
Celsius – TemperatureDeci – Ten
Measure of Lengths
10 millimeters (mm) =1 centimeter (cm)
10 centimeters =1 decimeter (dm) = 100 millimeters
100 centimeter =1 meter (m) = 1,000 millimeters
1000 meters =1 kilometer (km)
Operations for Length
Example: To get Centimeters from Meters you multiply the Meters by 100
You Have: 23 Meters
You want Centimters: So, Centimeters = 23 Meters
We Multiply the meters by 100:
You want Centimters: So, Centimeters = 23 Meters
There are 2300 centimeters in 23 meters or 100 centimeters for every meter.
If you have thisDo thisTo get this
millimeters (mm)Divide by 10 (mm/10)centimeter (cm)
centimeters (cm)Multiply by 10 (cm 10)Millimeters (mm)
Meters (m)Multiply by 100 (m 100)Centimeters (cm)
Centimeters (cm)Divide by 100 (cm/100)Meters (m)
Millimeters (mm)Divide by 1000 (mm/1000)Meters (m)
[page 1] [page 2] [page 3] [page 4] [page 5] [page 6] [page 7] [evaluation] |
13651 | https://www.barnesandnoble.com/w/recent-advances-in-geometric-inequalities-dragoslav-s-mitrinovic/1101004782 | Recent Advances in Geometric Inequalities by Dragoslav S. Mitrinovic, J. Pecaric, V. Volenec, Paperback | Barnes & Noble®
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Recent Advances in Geometric Inequalities 1101004782
Recent Advances in Geometric Inequalities
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by Dragoslav S. Mitrinovic, J. Pecaric, V. VolenecDragoslav S. Mitrinovic
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Paperback(Softcover reprint of the original 1st ed. 1989)
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Product Details
Table of Contents
Product Details
| ISBN-13: | 9789048184422 |
| Publisher: | Springer Netherlands |
| Publication date: | 09/17/2011 |
| Series: | Mathematics and its Applications , #28 |
| Edition description: | Softcover reprint of the original 1st ed. 1989 |
| Pages: | 710 |
| Product dimensions: | 6.10(w) x 9.25(h) x 0.06(d) |
Table of Contents
The Existence of a Triangle.- Duality between Geometric Inequalities and Inequalities for Positive Numbers.- Homogeneous Symmetric Polynomial Geometric Inequalities.- Duality between Different Triangle Inequalities and Triangle Inequalities with (R, r, s).- Transformations for the Angles of a Triangle.- Some Trigonometric Inequalities.- Some Other Transformations.- Convex Functions and Geometric Inequalities.- Miscellaneous Inequalities with Elements of a Triangle.- Special Triangles.- Triangle and Point.- Inequalities with Several Triangles.- The Möbius-Neuberg and the Möbius-Pompeiu Theorems.- Inequalities for Quadrilaterals.- Inequalities for Polygons.- Inequalities for a Circle.- Particular Inequalities in Plane Geometry.- Inequalities for Simplexes in En (n— 2).- Inequalities for Tetrahedra.- Other Inequalities in En (n— 2).
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Editorial Reviews
‘For the immediate future, however, this book should be (possibly chained!) in every university and college library, and, yes, in the library of every school which is intent on improving its mathematics teaching.'
The Americal Mathematical Monthly, December 1991
‘This book should make interesting reading for philosophers of mathematics, if they want to observe how mathematical ideas really develop.'
Advances in Mathematics, 86, 1991
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13652 | https://chemandy.com/calculators/return-loss-and-mismatch-calculator.htm | Return Loss and Mismatch Loss Calculator
Calculates the absolute load impedance, reflection coefficient, VSWR, return loss and mismatch loss of a load.
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The absolute value of the load impedance is calculated from the complex impedance (R + j).
Where R is the Real part of the impedance
and j is the Imaginary part of the impedance
The Load Reflection Coefficient ( Γ ) is calculated using the complex impedance of the load and the characteristic impedance of the source.
Where Zo is the Source Impedance
The VSWR is then calculated using the Reflection Coefficient
The Reflection Coefficient is used again to calculate the Return Loss
The Reflection Coefficient is used yet again to calculate the Mismatch Loss
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13653 | https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/Discrete_Structures/03%3A_Constructing_and_Writing_Proofs_in_Mathematics/3.04%3A_Using_Cases_in_Proofs | 3.4.1
3.4.1
n
n
3.4.1
3.4.1
a
b
ab=0
a=0
b=0
a=0
a≠0
a=0
a≠0
ab=0
dfrac1a
1a⋅ab=1a⋅0
b=0
a=0
b=0
a
b
ab=0
a=0
b=0
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3.4: Using Cases in Proofs
Last updated
: Sep 29, 2021
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3.3: Proof by Contradiction
3.5: The Division Algorithm and Congruence
Page ID
: 86112
Ted Sundstrom
Grand Valley State University via ScholarWorks @Grand Valley State University
( \newcommand{\kernel}{\mathrm{null}\,})
PREVIEW ACTIVITY 3.4.13.4.1: Using a Logical Equivalency
Complete a
truth table
to show that (P∨Q)→R(P∨Q)→R is logical equivalent to (P→R)∧(Q→R)(P→R)∧(Q→R).
2. Suppose that you are trying to prove a
statement
that is written in the form (P∨Q)→R(P∨Q)→R. Explain why you can complete this proof by writing separate and independent proofs of P→RP→R and Q→RQ→R.
3. Now consider the following proposition:
Proposition. For all integers xx and yy, if xyxy is odd, then xx is odd and yy is odd.
Write the
contrapositive
of this proposition.
4. Now prove that if xx is an even integer, then xyxy is an even integer. Also, prove that if yy is an even integer, then xyxy is an even integer.
5. Use the results proved in part (4) and the explanation in part (2) to explain why we have proved the
contrapositive
of the proposition in part (3).
Preview Activity 3.4.13.4.1: Using Cases in a Proof
The work in Preview Activity 3.4.13.4.1 was meant to introduce the idea of using cases in a proof. The method of using cases is often used when the hypothesis of the proposition is a
disjunction
. This is justified by the logical equivalency
[(P∨Q)→R]≡[(P→R)∧(Q→R)]
[(P∨Q)→R]≡(P→R)∧(Q→R)
See Theorem 2.8 on page 48 and Exercise (6) on page 50.
In some other situations when we are trying to prove a proposition or a theorem about an element xx in some
set
UU, we often run into the problem that there does not seem to be enough information about x to proceed. For example, consider the following proposition:
Proposition 1. If nn is an integer, then n2+nn2+n is an even integer.
If we were trying to write a
direct proof
of this proposition, the only thing we could assume is that nn is an integer. This is not much help. In a situation such as this, we will sometimes use cases to provide additional assumptions for the forward process of the proof. Cases are usually based on some common properties that the element xx may or may not possess. The cases must be chosen so that they exhaust all possibilities for the object xx in the hypothesis of the original proposition. For Proposition 1, we know that an integer must be even or it must be odd. We can thus use the following two cases for the integer nn:
The integer nn is an even integer;
The integer nn is an odd integer.
Complete the proof for the following proposition:
Proposition 2: If nn is an even integer, then n2+nn2+n is an even integer.
Proof. Let nn be an even integer. Then there exists an integer mm such that n=2mn=2m. Substituting this into the expression n2+nn2+n yields ...
2. Construct a proof for the following proposition:
Proposition 3: If nn is an odd integer, then n2+nn2+n is an even integer.
3. Explain why the proofs of Proposition 2 and Proposition 3 can be used to construct a proof of Proposition 1.
Some Common Situations to Use Cases
When using cases in a proof, the main rule is that the cases must be chosen so that they exhaust all possibilities for an object x in the hypothesis of the original proposition. Following are some common uses of cases in proofs.
| | |
--- |
| When the hypothesis is, "nn is an integer." | Case 1: nn is an even integer. Case 2: nn is an odd integer. |
| When the hypothesis is, "mm and nn are integers." | Case 1: mm and nn are even. Case 2: mm is even and nn is odd. Case 3: mm is odd and nn is even. Case 4: mm and nn are both odd. |
| When the hypothesis is, "xx is a real number." | Case 1: xx is rational. Case 2: xx is irrational. |
| When the hypothesis is, "xx is a real number." | Case 1: x=0x=0 OR Case 1: x>0x>0 Case 2: x≠0x≠0 Case 2: x=0x=0 Case 3: x<0x<0 |
| When the hypothesis is, "aa and bb are real numbers." | Case 1: a=ba=b OR Case 1: a>ba>b Case 2: a≠ba≠b Case 2: a=ba=b Case 3: a<ba<b |
Writing Guidelines for a Proof Using Cases
When writing a proof that uses cases, we use all the other writing guidelines. In addition, we make sure that it is clear where each case begins. This can be done by using a new paragraph with a label such as “Case 1,” or it can be done by starting a paragraph with a phrase such as, “In the case where . . . .”
Progress Check 3.21: Using Cases: nn Is Even or nn Is Odd
Complete the proof of the following proposition:
Proposition. For each integer nn, n2−5n+7n2−5n+7 is an odd integer.
Proof. Let nn be an integer. We will prove that n2−5n+7n2−5n+7 is an odd integer by examining the case where nn is even and the case where nn is odd.
Case 1. The integer nn is even. In this case, there exists an integer mm such that n=2mn=2m. Therefore, ....
Answer
: Add texts here. Do not delete this text first.
As another example of using cases, consider a situation where we know that aa and bb are real numbers and ab=0ab=0. If we want to make a conclusion about bb, the temptation might be to divide both sides of the equation by aa. However, we can only do this if a≠0a≠0. So, we consider two cases: one when a=0a=0 and the other when a≠0a≠0.
proposition 3.22
For all real numbers aa and bb, if ab=0ab=0, then a=0a=0 or b=0b=0.
Proof
: We let a and b be real numbers and assume that ab=0. We will prove that a=0 or b=0 by considering two cases: (1) a=0, and (2) a≠0.
In the case where a=0, the conclusion of the proposition is true and so there is nothing to prove.
In the case where a≠0, we can multiply both sides of the equation ab=0 by dfrac1a and obtain
1a⋅ab=1a⋅0
b=0
So in both cases, a=0 or b=0, and this proves that for all real numbers a and b, if ab=0, then a=0 or b=0.
Absolute Value
Most students by now have studied the concept of the absolute value of a real number. We use the notation |x||x| to stand for the absolute value of the real number xx. One way to think of the absolute value of xx is as the “distance” between xx and 0 on the number line. For example,
|-5| = 5 and |-7| = 7
Although this notion of absolute value is convenient for determining the absolute value of a specific number, if we want to prove properties about absolute value, we need a more careful and precise definition.
Definition: absolute value
For x∈Rx∈R, we define |x||x|, called the absolute value of xx, by
|x|={x if x ≥ 0;−x if x < 0.
Let’s first see if this definition is consistent with our intuitive notion of absolute value by looking at two specific examples.
Since 5 > 0, we see that |5| = 5, which should be no surprise.
Since -7 < 0, we see that |-7| = -(-7) = 7.
Notice that the definition of the absolute value of x is given in two parts, one for when x≥0 and the other for when x<0. This means that when attempting to prove something about absolute value, we often uses cases. This will be illustrated in Theorem 3.23.
Theorem 3.23
Let a be a positive real number. For each real number x,
|x|=a if and only if x=a or x=−a.
|−x|=|x|.
Proof
: The proof of Part (2) is part of Exercise (10). We will prove Part (1).
We let a be a positive real number and let x∈R. We will first prove that if |x|=a, then x=a or x=−a. So we assume that |x|=a. In the case where x≥0, we see that |x|=x, and since |x|=a, we can conclude that x=−a.
In the case where x<0, we see that |x|=−x. Since |x|=a, we can conclude that −x=a and hence that x=−a. These two cases prove that if |x|=a, then x=a or x=−a.
We will now prove that if x=a or x=−a, then |x|=a. We start by assuming that x=a or x=−a. Since the hypothesis of this
conditional statement
is a
disjunction
, we use two cases. When x=a, we see that
|x|=|a|=a since a>0.
When x=−a, we conclude that
|x|=|−a|=−(−a) since −a<0.
and hence, |x|=a. This proves that if x=a or x=−a, then |x|=a. Because we have proven both conditional statements, we have proven that |x|=a if and only if x=a or x=−a.
Progress Check 3.24: Equations Involving Absolute Values
What is |4,3| and what is |-π|?
Use the properties of absolute value in Proposition 3.23 to help solve the following equations for t, where t is a real number.
(a) |t|=12.
(b) |t+3|=5
(c) |t−4|=15.
(d) |3t−4|=8.
Answer
: Add texts here. Do not delete this text first.
Although solving equations involving absolute values may not seem to have anything to do with writing proofs, the point of Progress Check 3.24 is to emphasize the importance of using cases when dealing with absolute value. The following theorem provides some important properties of absolute value.
Theorem 3.25
Let a be a positive real number. For all real numbers x and y,
|x|<a if and only if −a<x<a.
|xy|=|x||y|.
|x+y|≤|x|+|y|. This is know as the Triangle Inequality.
Proof
: We will prove Part (1). The proof of Part (2) is included in Exercise (10), and the proof of Part (3) is Exercise (14). For Part (1), we will prove the
biconditional
proposition by proving the two associated conditional propositions.
So we let a be a positive real number and let x∈R and first assume that |x|<a. We will use two cases: either x≥0 or x<0.
In the case where x≥0, we know that |x|=x and so the inequality |x|<a implies that x<a. However, we also know that −a<0 and that x>0. Therefore, we conclude that −a<x and, hence, −a<x<a.
When x<0, we see that |x|=−x. Therefore, the inequality |x|<a implies that −x<a, which in turn implies that −a<x. In this case, we also know that x<a since x is negative and a is positive. Hence, −a<x<a
So in both cases, we have proven that −a<x<a and this proves that if |x|<a, then −a<x<a. We now assume that −a<x<a.
If x≥0, then |x|=x and hence, |x|<a.
If x<0, then |x|=−x and so |x|=−x. Thus, −a<−|x|. By multiplying both sides of the last inequality by -1, we conclude that |x|<a.
These two cases prove that if −a<x<a, then |x|<a. Hence, we have proven that |x|<a if and only if −a<x<a.
Exercises for section 3.4
In Preview Activity 3.4.2, we proved that if n is an integer, then n2+n is an even integer. We define two integers to be consecutive integers if one of the integers is one more than the other integer. This means that we can represent consecutive integers as m and m+1, where m is some integer.
Explain why the result proven in Preview Activity 3.4.2 can be used to prove that the product of any two consecutive integers is divisible by 2.
Prove that if u is an odd integer, then the equation x2+x−u=0 has no solution that is an integer.
Prove that if n is an odd integer, then n=4k+1 for some integer k or n=4k+3 for some integer k.
Prove the following proposition:
For each integer a, if a2=a, then a=0 or a=1.
5. (a) Prove the following proposition:
For all integers a, b, and d with d≠0, if d divides a or d divides b, then d divides the product ab.
Hint: Notice that the hypothesis is a
disjunction
. So use two cases.
(b) Write the
contrapositive
of the proposition in Exercise(5a).
(c) Write the of the proposition in Exercise (5a). Is the
converse
true or false? Justify your conclusion.
6. Are the following propositions true or false? Justify all your conclusions. If a
biconditional
statement
is found to be false, you should clearly determine if one of the conditional statements within it is true. In that case, you should state an appropriate theorem for this
conditional statement
and prove it. (a) For all integers m and n, m and n are consecutive integers if and only if 4 divides (m2+n2−1.
(b) For all integers m and n, 4 divides (m2−n2) if and only if m and n are both even or m and n are both odd.
7. Is the following proposition true or false? Justify your conclusion with a counterexample or a proof.
For each integer n, if n is odd, then 8|(n2−1).
8. Prove that there are no natural numbers a and n with n≥2 and a2+1=2n.
9. Are the following propositions true or false? Justify each conclusions with a counterexample or a proof.
(a) For all integers a and b with a≠0, the equation ax+b=0 has a rational number solution.
(b) For all integers a, b, and c, if a, b, and c are odd, then the equation ax2+bx+c=0 has no solution that is a rational number.
Hint: Do not use the quadratic formula. Use a
proof by contradiction
and recall that any rational number can be written in the form pq, where p and q are integers, q>0, and p and q have no common factor greater than 1.
(c) For all integers a, b, c, and d, if a, b, c, and d are odd, then the equation ax3+bx2+cx+d=0 has no solution that is a rational number.
10. (a) Prove Part (2) of Proposition 3.23.
For each x∈R, |−x|=|x|.
(b) Prove Part (2) of Proposition 3.25.
For all real numbers x and y, |xy|=|x||y|.
11. Let a be a positive real number. In Part (1) of Theorem 3.25, we proved that for each real number x, |x|<a if and only if −a<x<a. It is important to realize that the sentence −a<x<a is actually the of two inequalities. That is, −a<x<a means that −a<x and x<a.
(a) Complete the following
statement
: For each real number x, |x|≥a if and only if ....
(b) Prove that for each real number x, |x|≤a if and only if −a≤x≤a
(c) Complete the following
statement
: For each real number x, |x|>a if and only if ....
Prove each of the following:
(a) For each nonzero real number x, |x−1|=1|x|.
(b) For all real number x and y, |x−y|≥|x|−|y|.
Hint: An idea that is often used by mathematicians is to add 0 to an expression “intelligently”. In this case, we know that (−y)+y=0. Start by adding this “version” of 0 inside the absolute value sign of |x|.
(c) For all real number x and y, ||x|−|y||≤|x−y|.
13. Evaluation of proofs
See the instructions for Exercise (19) on page 100 from Section 3.1.
Theorem 3.4.1
The probabilities assigned to events by a distribution function on a sample space are given by.
Proof
: Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.
Theorem 3.4.1
The probabilities assigned to events by a distribution function on a sample space are given by.
Proof
: Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.
Explorations and Activities
14. Proof of the Triangle Inequality.
(a) Verify that the triangle inequality is true for several different real numbers x and y. Be sure to have some examples where the real numbers are negative.
(b) Explain why the following proposition is true: For each real number r, −|r|≤r≤|r|.
(c) Now let x and y be real numbers. Apply the result in Part (14b) to both x and y. Then add the corresponding parts of the two inequalities to obtain another inequality. Use this to prove that |x+y|≤|x|+|y|.
Answer
: Add texts here. Do not delete this text first.
3.3: Proof by Contradiction
3.5: The Division Algorithm and Congruence |
13654 | https://www.thelatinlibrary.com/101/IrregularVerbs.pdf | Irregular Verbs 1. sum, esse, fui (to be) Present Imperfect Future Perfect sum eram ero fui es eras eris fuisti est erat erit fuit sumus eramus erimus fuimus estis eratis eritis fuistis sunt erant erunt fuerunt 2. possum, posse, potui (to be able) possum poteram potero potui potes poteras poteris potuisti potest poterat poterit potuit possumus poteramus poterimus potuimus potestis poteratis poteritis potuistis possunt poterant poterunt potuerunt 3. fero, ferre, tuli, latum (to carry, bear) fero ferebam feram tuli fers ferebas feres tulisti fert ferebat feret tulit ferimus ferebamus feremus tulimus fertis ferebatis feretis tulistis ferunt ferebant ferent tulerunt 4. volo, velle, volui (to want, wish) volo volebam volam volui vis volebas voles voluisti vult volebat volet voluit volumus volebamus volemus voluimus vultis volebatis voletis voluistis volunt volebant volent voluerunt 5. nolo, nolle, nolui (to not want) nolo nolebam nolam nolui non vis nolebas noles noluisti non vult nolebat nolet noluit nolumus nolebamus nolemus noluimus non vultis nolebatis noletis noluistis nolunt nolebant nolent noluerunt 6. malo, malle, malui (to prefer) malo malebam malam malui mavis malebas males maluisti mavult malebat malet maluit malumus malebamus malemus maluimus mavultis malebatis maletis maluistis malunt malebant malent maluerunt |
13655 | https://www.youtube.com/watch?v=0f8OKc-hIHc | Mecánica de fluidos | Ecuación de Bernoulli | Ejemplo 4
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Posted: 20 Feb 2021
La figura muestra un chorro de agua que cae continuamente de la llave de agua de una cocina. En la boca de la llave el diámetro del chorro es 0.960 cm. El chorro llena un recipiente de 125 𝑐𝑚^3 en 16.3 s. Encuentre el diámetro del chorro a 13 cm abajo de la abertura de la llave
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Transcript:
la figura muestra un chorro de agua que cae continuamente de la llave de agua de una cocina en la boca de la llave el diámetro del chorro es punto 960 centímetros el chorro llena un recipiente de 125 centímetros cúbicos en 16.3 segundos encuentre el diámetro del chorro a 13 centímetros abajo de la abertura de la llave primeramente vamos a notar que el problema nos da el flujo o caudal del chorro que son 125 centímetros cúbicos cada 16 punto 3 segundos y recordando que el caudal corresponde al área por la velocidad y que es constante a lo largo de todo el chorro entonces nosotros si tenemos el área o la velocidad en cualquier punto podemos calcular el dato que nos falta asumiendo que el área de sección transversal del chorro es circular entonces el área sería igual a pi por radio al cuadrado y entonces el caudal es igual a pi por radio al cuadrado por velocidad con esto en consideración vamos a calcular entonces la velocidad que tiene el chorro al inicio es decir cuando sale de la llave para despejar la pasamos dividiendo pi y el radio al cuadrado y nos queda como si entonces el caudal es 125 sobre 16.3 el radio es de punto 480 centímetros que es la mitad del radio hacemos la operación y nos da una velocidad de 10.6 centímetros sobre segundos que si convertimos a metros sobre segundos va a ser punto 106 ahora por debajo de este punto a 13 centímetros tenemos que encontrar el diámetro del chorro y para esto utilizaremos la ecuación de werniul y dice que la presión 1 más la densidad por gravedad por la altura 1 más un medio de la densidad por la velocidad 1 al cuadrado es igual a la presión 2 más la densidad por la gravedad por la altura 2 más un medio de la densidad por la velocidad 2 al cuadrado vamos a considerar la altura 2 como 0 nuestro cero va a estar en ese punto y a 13 centímetros por encima está la llave entonces la altura 2 será 13 centímetros como el chorro está expuesto al aire entonces la presión va a ser igual a la presión atmosférica entonces la ecuación de berlín nos va a quedar como sigue sustituyendo tanto presión 1 como presión 2 por presión erika y eliminando el término de densidad por gravedad por altura 2 pues lo estamos considerando 0 nos queda como sigue como tenemos la presión atmosférica sumando ambos lados de la igualdad la podemos cancelar y a continuación como tenemos la densidad multiplicando en todos los términos también la podemos cancelar nos quedamos con la siguiente expresión ahora sólo tenemos que sustituir datos tenemos la gravedad que es 9.8 metros sobre segundos al cuadrado la altura vamos a ponerle en metros entonces sería punto 13 metros más un medio de la velocidad 1 al cuadrado lo habíamos calculado como punto 106 metros sobre segundos va a ser igual a un medio de la velocidad 2 al cuadrado hacemos las operaciones un medio de la velocidad 2 al cuadrado es igual a 1.28 y entonces la velocidad 2 va a ser igual a la raíz de 2 por 1 punto 28 que es 1.6 metros sobre segundos y para qué queremos la velocidad en ese punto recordemos que el caudal es igual al área por la velocidad que este valor es constante a lo largo de todo el short y recordemos además que el área es tipo radio al cuadrado entonces teniendo la velocidad podemos obtener el radio y después lo multiplicamos por 2 y nos del diámetro entonces vamos a despejar el radio va a ser igual a la raíz del caudal sobre pi por la velocidad 2 el caudal habíamos dicho que era 125 centímetros cúbicos en 16.3 segundos por pi por la velocidad 2 como estamos manejando centímetros vamos a ponerla en centímetros 160 centímetros por segundo y hacemos la operación y nos da un radio de punto 124 centímetros por lo tanto el diámetro que es dos veces el radio es punto doscientos cuarenta y ocho centímetros |
13656 | https://www.me.mtu.edu/~wjendres/ProductRealization1Course/DC_Motor_Calculations.pdf | Motor Calculations • Calculating Mechanical Power Requirements • Torque - Speed Curves • Numerical Calculation • Sample Calculation • Thermal Calculations Calculating Mechanical Power Requirements Physically, power is defined as the rate of doing work. For linear motion, power is the product of force multiplied by the distance per unit time. In the case of rotational motion, the analogous calculation for power is the product of torque multiplied by the rotational distance per unit time. ω × = M P rot Where: Prot = rotational mechanical power M = torque ω = angular velocity The most commonly used unit for angular velocity is rev/min (RPM). In calculating rotational power, it is necessary to convert the velocity to units of rad/sec. This is accomplished by simply multiplying the velocity in RPM by the constant (2 x π) /60: × = 60 2 sec / π ω ω rpm rad It is important to consider the units involved when making the power calculation. A reference that provides conversion tables is very helpful for this purpose. Such a reference is used to convert the torque-speed product to units of power (Watts). Conversion factors for commonly used torque and speed units are given in the following table. These factors include the conversion from RPM to rad/sec where applicable. Torque Units Speed Units Conversion Factor oz-in RPM 0.00074 oz-in rad/sec 0.0071 in-lb RPM 0.0118 in-lb rad/sec 0.1130 ft-lb RPM 0.1420 ft-lb rad/sec 1.3558 N-m RPM 0.1047 N-m rad/sec 1.0002 For example, assume that it is necessary to determine the power required to drive a torque load of 3 oz-in at a speed of 500 RPM. The product of the torque, speed, and the appropriate conversion factor from the table is: Watts rpm in oz 11 .
1 00074 .
0 500 3 = × × − Calculation of power requirements is often used as a preliminary step in motor or gearmotor selection. If the mechanical power required for a given application is known, then the maximum or continuous power ratings for various motors can be examined to determine which motors are possible candidates for use in the application. Torque - Speed Curves One commonly used method of displaying motor characteristics graphically is the use of torque - speed curves. While the use of torque - speed curves is much more common in technical literature for larger DC machines than it is for small, ironless core devices, the technique is applicable in either case. Torque - speed curves are generated by plotting motor speed, armature current, mechanical output power, and efficiency as functions of the motor torque. The following discussion will describe the construction of a set of torque - speed curves for a typical coreless DC motor from a series of raw data measurements. Motor 1624E009S is used as an example. Assume that we have a small motor that we know has a nominal voltage of 9 volts. With a few fundamental pieces of laboratory equipment, the torque - speed curves for the motor can be generated: Step One (measure basic parameters): Using a voltage supply set to 9 volts, run the motor unloaded and measure the rotational speed using a non-contacting tachometer (strobe, for instance). Measure the motor current under this no-load condition. A current probe is ideal for this measurement since it does not add resistance in series with the operating motor. Using an adjustable torque load such as a small particle brake coupled to the motor shaft, increase the torque load to the motor just to the point where stall occurs. At stall, measure the torque from the brake and the motor current. For the sake of this discussion, assume that the coupling adds no load to the motor and that the load from the brake does not include unknown frictional components. It is also useful at this point to measure the terminal resistance of the motor. Measure the resistance by contacting the mot or terminals. Then spin the motor shaft and take another measurement. The measurements should be very close in value. Continue to spin the shaft and take at least three measurements. This will ensure that the measurements were not taken at a point of minimum contact on the commutator. Now we have measured the: • n0= no-load speed • I0= no-load current • MH= stall torque • R= terminal resistance Step Two (plot current vs. torque and speed vs torque) Prepare a graph with motor torque on the horizontal axis, motor speed on the left vertical axis, and motor current on the right vertical axis. Scale the axes based on the measurements in step 1. Draw a straight line from the left origin of the graph (zero torque and zero current) to the stall current on the right v ertical axis (stall torque and stall current). This line represents a plot of the motor current as a function of the motor torque. The slope of this line is the proportionality constant for the relationship between motor current and motor torque (in units of current per unit torque). The reciprocal of this slope is the torque constant of the motor (in units of torque per unit current). For the resulting curves see Graph 1. Using the relationships between motor constants discussed earlier, calculate the velocity constant of the motor from the torque constant obtained above. By multiplying the velocity constant by the nominal motor voltage, obtain the theoretical no-load speed of the motor (zero torque and no-load speed) and plot it on the left vertical axis. Draw a straight line between this point and the stall torque and zero speed point on the graph. The slope of this line is the proportionality constant for the relationship between motor speed and motor torque (in units of speed per unit torque). The slope of the line is negative, indicating that motor speed decreases with increasing torque. This value is sometimes called the regulation constant of the motor. For the resulting curves see Graph 1. For the purpose of this discussion, it will be assumed that the motor has no internal friction. In practice, the motor friction torque is determined using the torque constant of the motor and the measured no-load current. The torque vs speed line and the torque vs current line are then started not at the left vertical axis but at an offset on the horizontal axis equal to the calculated friction torque. Step Three (plot power vs torque and efficiency vs torque) In most cases, two additional vertical axes are added for plotting power and efficiency as functions of torque. A second left vertical axis is usually used for efficiency and a second right vertical axis is used for power. For the sake of simplifying this discussion, efficiency vs. torque and power vs. torque will be plotted on a second graph separate from the speed vs. torque and current vs. torque plots. Construct a table of the motor mechanical power at various points from no-load to stall torque. Since mechanical power output is simply the product of torque and speed with a correction factor for units (see section on calculating mechanical power requirements), power can be calculated using the previously plotted line for speed vs. torque. A sample table of calculations for motor 1624E009S is shown in Table 1. Each calculated point is then plotted. The resulting curve is a parabolic curve as shown in Graph 1. The maximum mechanical power occurs at approximately one-half of the stall torque. The speed at this point is approximately one-half of the no-load speed. Construct a table of the motor efficiency at various points from no-load to stall torque. The voltage applied to the motor is given, and the current at various levels of torque has been plotted. The product of the motor current and the applied voltage is the power input to the motor. At each point selected for calculation, the efficiency of the motor is the mechanical power output divided by the electrical power input. Once again, a sample table for motor 1624E009S is shown in Table 1. and a sample curve in Graph 1. Maximum efficiency occurs at about 10% of the motor stall torque. Table 1. TORQUE SPEED CURRENT POWER EFFICIENCY (oz-in) (rpm) (mA) (Watts) (%) ----------- --------- ------------ ---------- ------------- 0.025 11247.65 0.024 0.208 0.10 0.05 10786.3 0.048 0.399 71.87 0.075 10324.95 0.072 0.573 75.27 0.1 9863.6 0.096 0.730 74.99 0.125 9402.25 0.120 0.870 73.25 0.15 8940.9 0.144 0.992 70.78 0.175 8479.55 0.168 1.098 67.89 0.2 8018.2 0.192 1.187 64.73 0.225 7556.85 0.217 1.258 61.40 0.25 7095.5 0.241 1.313 57.95 0.275 6634.15 0.265 1.350 54.41 0.3 6172.8 0.289 1.370 50.80 0.325 5711.45 0.313 1.374 47.14 0.35 5250.1 0.337 1.360 43.44 0.375 4788.75 0.361 1.329 39.71 0.4 4327.4 0.385 1.281 35.95 0.425 3866.05 0.409 1.216 32.17 0.45 3404.7 0.433 1.134 28.37 0.475 2943.35 0.457 1.035 24.56 0.5 2482.0 0.481 0.918 20.74 0.525 2020.65 0.505 0.785 16.90 0.55 1559.3 0.529 0.635 13.05 0.575 1097.95 0.553 0.467 9.20 0.6 636.6 0.577 0.283 5.34 0.625 175.25 0.602 0.081 1.47 Graph 1. 0 2000 4000 6000 8000 10000 12000 14000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Oz. In.
rpm 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Efficiency, Power, & Current Speed r/min Efficiency Current Output Power Numerical Calculation For an iron-less core, DC motor of relatively small size, the relationships that govern the behavior of the motor in various circumstances can be derived from physical laws and characteristics of the motors themselves. Kirchoff's voltage rule states, "The sum of the potential increases in a circuit loop must equal the sum of the potential decreases." When applied to a DC motor connected in series with a DC power source, Kirchoff's voltage rule can be expressed as "The nominal supply voltage from the power source must be equal in magnitude to the sum of the voltage drop across the resistance of the armature windings and the back EMF generated by the motor.": e o V R I V + × = ) ( Where: Vo = Power supply (Volts) I = Current (A) R = Terminal Resistance (Ohms) Ve = Back EMF (Volts) The back EMF generated by the motor is directly proportional to the angular velocity of the motor. The proportionality constant is the back EMF constant of the motor. e e k V × = ω Where: ω= angular velocity of the motor Ke = back EMF constant of the motor Therefore, by substitution: ) ( ) ( e o k R I V × + × = ω The back EMF constant of the motor is usually specified by the motor manufacturer in volts/RPM or mV/RPM. In order to arrive at a meaningful value for the back EMF, it is necessary to specify the motor velocity in units compatible with the specified back EMF constant. The motor constant is a function of the coil design and the strength and direction of the flux lines in the air gap. Although it can be shown that the three motor constants normally specified (back EMF constant, torque constant, and velocity constant) are equal if the proper units are used, calculation is facilitated by the specification of three constants in the commonly accepted units. The torque produced by the rotor is directly proportional to the current in the armature windings. The proportionality constant is the torque constant of the motor. M o k I M × = Where: Mo = torque developed at rotor kM = motor torque constant Substituting this relationship: ) ( ) ( e M k k R M V × + × = ω The torque developed at the rotor is equal to the friction torque of the motor plus the resisting torque due to external mechanical loading: f l o M M M + = Where: Mf = motor friction torque Ml = load torque Assuming that a constant voltage is applied to the motor terminals, the motor velocity will be directly proportional to sum of the friction torque and the load torque. The constant of proportionality is the slope of the torque-speed curve and can be calculated by: H M n M n / / 0 = ∆ ∆ Where: MH = stall torque n0= no-load speed An alternative approach to deriving this value is to solve for velocity, n: ) ( e m e o k k M k V n × − = Differentiating both sides with respect to M yields: ) ( e m k k R M n × − = ∆ ∆ Using dimensional analysis to check units, the result is: -Ohms/(oz-in/A) x (V/RPM) = -Ohm-A-RPM/V-oz-in = -RPM/oz-in It is a negative value describing loss of velocity as a function of increased torsional load. Sample Calculation Motor 1624T009S is to be operated with 9 volts applied to the motor terminals. The torque load is 0.2 oz-in. Find the resulting motor speed, motor current, efficiency, and mechanical power output. From the motor data sheet, it can be seen that the no-load speed of the motor at 9 volts is 11,700 rpm. If the torque load is not coupled to the motor shaft, the motor would run at this speed. The motor speed under load is simply the no-load speed less the reduction in speed due to the load. The proportionality constant for the relationship between motor speed and motor torque is the slope of the torque vs. speed curve, given by the motor no-load speed divided by the stall torque. In this example, the speed reduction caused by the 0.2 oz-in torque load is: 0.2 oz-in x (11,700 rpm/.634 oz-in) = -3690 rpm The motor speed under load must then be: 11,700 rpm - 3690 rpm = 8010 rpm The motor current under load is the sum of the no-load current and the current resulting from the load. The proportionality constant relating current to torque load is the torque constant (kM), in this case, 1.039 oz -in/A. In this case, the load torque is 0.2 oz-in, and the current resulting from the load must be: I = 0.2 oz-in x 1 amp/1.039 oz -in = 192 mA The total motor current must be the sum of this value and the motor no-load current. The data sheet lists the motor no-load current as 11 mA. Therefore, the total current is: 192 mA + 11 mA = 203 mA The mechanical power output of the motor is simply the product of the motor speed and the torque load with a correction factor for units (if required). Therefore, the mechanical power output of the motor in this application is: output power = 0.2 oz-in x 8010 rpm x .00074 = 1.18 Watts The mechanical power input to the motor is the product of the applied voltage and the total motor current in Amps. In this application: input power = 9 volts x .203 A = 1.82 Watts Since efficiency is simply power out divided by power in, the efficiency in this application is: efficiency = 1.18 Watts / 1.82 Watts = .65 = 65% Thermal Calculations A current I flowing through a resistance R results in a power loss as heat of I2R. In the case of a DC motor, the product of the square of the total motor current and the armature resistance is the power loss as heat in the armature windings. For example, if the total motor current was .203 A and the armature resistance 14.5 Ohms the power lost as heat in the windings is: power loss = .2032 x 14.5 = .59 Watts The heat resulting from I 2R losses in the coil is dissipated by conduction through motor components and airflow in the air gap. The ease with which this heat can be dissipated is a function of the motor type and construction. Motor manufacturers typically provide an indication of the motor’s ability to dissipate heat by providing thermal resistance values. Thermal resistance is a measure of the resistance to the passage of heat through a given thermal path. A large cross section aluminum plate would have a very low thermal resistance, for example, while the values for air or a vacuum would be considerably higher. In the case of DC motors, there is a thermal path from the motor windings to the motor case and a second between the motor case and the motor environment (ambient air, etc.). Some motor manufacturers specify a thermal resistance for each of the two thermal paths while others specify only the sum of the two as the total thermal resistance of the motor. Thermal resistance values are specified in temperature increase per unit power loss. The total I2R losses in the coil (the heat source) are multiplied by thermal resistances to determine the steady state armature temperature. The steady state temperature increase of the motor (T) is given by: ) ( 2 1 2 th th inc R R R I T + × = Where: Tinc = temperature increase I = current through motor windings R = resistance of motor windings Rth1 = thermal resistance from windings to case Rth2 = thermal resistance case to ambient For example, a 1624E009S motor running with a current of 0.203 Amps in the motor windings, with an armature resistance of 14.5 Ohms, a winding-to-case thermal resistance of 8 °C/Watt, and a case-to-ambient thermal resistance of 39 °C/Watt. The temperature increase of the windings is given by: T = .2032 x 14.5 x (8 + 39) = 28°C If it is assumed that the ambient air temperature is 22°C, then the final temperature of the motor windings is 50°C (22° + 28°). It is important to be certain that the final temperature of the windings does not exceed their rated value. In the example given above, the maximum permissible winding temperature is 100°C. Since the calculated winding temperature is only 50°C, thermal damage to the motor windings will not be a problem in this application. One could use similar calculations to answer a different kind of question. For example, an application may require that a motor run at its maximum torque without being damaged by heating. To continue with the example given above, suppose it is desired to run motor 1624E009S at the maximum possible torque with an ambient air temperature of 22°C. The designer wants to know how much torque the motor can safely provide without overheating. The data sheet for motor 1624E009S specifies a maximum winding temperature of 100°C. Since the ambient temperature is 22°C, a rotor temperature increase of 78°C is tolerable. The total thermal resistance for the motor is 47°C/Watt. By taking the reciprocal of the thermal resistance and multiplying this value by the acceptable temperature increase, the maximum power dissipation in the motor can be calculated: P = 78° x 1 Watt/47° = 1.66 Watts Setting I2R equal to the maximum power dissipation and solving for I yields the maximum continuous current allowable in the motor windings: I2R = 1.66 Watts I = .338 Amps The motor has a torque constant of 1.309 oz-in/A and a no-load current of 11 mA. Therefore, the maximum current available to produce useful torque is .327 Amps (.338 - .011), and the maximum usable torque available (M) is given by: M = .327 A x 1.309 oz-in/A = 0.428 oz-in The maximum allowable current through the motor windings could be increased by decreasing the thermal resistance of the motor. The rotor-to-case thermal resistance is primarily fixed by the motor design. The case-to-ambient thermal resistance can be decreased significantly by the addition of heat sinks. Motor thermal resistances for small DC motors are usually specified with the motor suspended in free air. Therefore, there is usually some heat sinking which results from simply mounting the motor into a framework or chassis. Some manufacturers of larger DC motors specify thermal resistance with the motor mounted into a metal plate of known dimensions and material. The preceding discussion does not take into account the change in resistance of the copper windings as a result of heating. While this change in resistance is important for larger machines, it is usually not significant for small, coreless motors and is often ignored for the sake of calculation. |
13657 | https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/solving-proportions/ | Solving Proportions | Mathematics for the Liberal Arts Corequisite
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Mathematics for the Liberal Arts Corequisite
Appendix A: Applications
Search for:
Solving Proportions
Learning Outcomes
Solve a proportion equation
Solve a proportion application
To solve a proportion containing a variable, we remember that the proportion is an equation. All of the techniques we have used so far to solve equations still apply. In the next example, we will solve a proportion by multiplying by the Least Common Denominator (LCD) using the Multiplication Property of Equality.
example
Solve: x 63=4 7 x 63=4 7
Solution
x 63=4 7 x 63=4 7
To isolate x x , multiply both sides by the LCD, 63 63.63(x 63)=63(4 7)63(x 63)=63(4 7)
Simplify.x=9⋅7⋅4 7 x=9⋅7⋅4 7
Divide the common factors.x=36 x=36
Check: To check our answer, we substitute into the original proportion.
x 63=4 7 x 63=4 7
Substitute x=36 x=36 36 63?=4 7 36 63=?4 7
Show common factors.4⋅9 7⋅9?=4 7 4⋅9 7⋅9=?4 7
Simplify.4 7=4 7 4 7=4 7
try it
In the next video we show another example of how to solve a proportion equation using the LCD.
When the variable is in a denominator, we’ll use the fact that the cross products of a proportion are equal to solve the proportions.
We can find the cross products of the proportion and then set them equal. Then we solve the resulting equation using our familiar techniques.
example
Solve: 144 a=9 4 144 a=9 4
Show Solution
Solution
Notice that the variable is in the denominator, so we will solve by finding the cross products and setting them equal.
Find the cross products and set them equal.4⋅144=a⋅9 4⋅144=a⋅9
Simplify.576=9 a 576=9 a
Divide both sides by 9 9.576 9=9 a 9 576 9=9 a 9
Simplify.64=a 64=a
Check your answer.
144 a=9 4 144 a=9 4
Substitute a=64 a=64 144 64?=9 4 144 64=?9 4
Show common factors..9⋅16 4⋅16?=9 4 9⋅16 4⋅16=?9 4
Simplify.9 4=9 4✓9 4=9 4✓
Another method to solve this would be to multiply both sides by the LCD, 4 a 4 a. Try it and verify that you get the same solution.
The following video shows an example of how to solve a similar problem by using the LCD.
try it
example
Solve: 52 91=−4 y 52 91=−4 y
Show Solution
Solution
Find the cross products and set them equal.
y⋅52=91(−4)y⋅52=91(−4)
Simplify.52 y=−364 52 y=−364
Divide both sides by 52 52.52 y 52=−364 52 52 y 52=−364 52
Simplify.y=−7 y=−7
Check:
52 91=−4 y 52 91=−4 y
Substitute y=−7 y=−7 52 91?=−4−7 52 91=?−4−7
Show common factors.13⋅4 13⋅4?=−4−7 13⋅4 13⋅4=?−4−7
Simplify.4 7=4 7✓4 7=4 7✓
try it
Solve Applications Using Proportions
The strategy for solving applications that we have used earlier in this chapter, also works for proportions, since proportions are equations. When we set up the proportion, we must make sure the units are correct—the units in the numerators match and the units in the denominators match.
example
When pediatricians prescribe acetaminophen to children, they prescribe 5 5 milliliters (ml) of acetaminophen for every 25 25 pounds of the child’s weight. If Zoe weighs 80 80 pounds, how many milliliters of acetaminophen will her doctor prescribe?
Show Solution
Solution
Identify what you are asked to find.How many ml of acetaminophen the doctor will prescribe
Choose a variable to represent it.Let a=a= ml of acetaminophen.
Write a sentence that gives the information to find it.If 5 5 ml is prescribed for every 25 25 pounds, how much will be prescribed for 80 80 pounds?
Translate into a proportion.
Substitute given values—be careful of the units.5 25=a 80 5 25=a 80
Multiply both sides by 80 80.80⋅5 25=80⋅a 80 80⋅5 25=80⋅a 80
Multiply and show common factors.16⋅5⋅5 5⋅5=80 a 80 16⋅5⋅5 5⋅5=80 a 80
Simplify.16=a 16=a
Check if the answer is reasonable.
Yes. Since 80 80 is about 3 3 times 25 25, the medicine should be about 3 3 times 5 5.
Write a complete sentence.The pediatrician would prescribe 16 16 ml of acetaminophen to Zoe.
You could also solve this proportion by setting the cross products equal.
try it
example
One brand of microwave popcorn has 120 120 calories per serving. A whole bag of this popcorn has 3.5 3.5 servings. How many calories are in a whole bag of this microwave popcorn?
Show Solution
Solution
Identify what you are asked to find.How many calories are in a whole bag of microwave popcorn?
Choose a variable to represent it.Let c=c= number of calories.
Write a sentence that gives the information to find it.If there are 120 120 calories per serving, how many calories are in a whole bag with 3.5 3.5 servings?
Translate into a proportion.
Substitute given values.120 1=c 3.5 120 1=c 3.5
Multiply both sides by 3.5 3.5.(3.5)(120 1)=(3.5)(c 3.5)(3.5)(120 1)=(3.5)(c 3.5)
Multiply.420=c 420=c
Check if the answer is reasonable.
Yes. Since 3.5 3.5 is between 3 3 and 4 4, the total calories should be between 360(3⋅120)360(3⋅120) and 480(4⋅120)480(4⋅120).
Write a complete sentence.The whole bag of microwave popcorn has 420 420 calories.
try it
example
Josiah went to Mexico for spring break and changed $325 325 dollars into Mexican pesos. At that time, the exchange rate had $1 1 U.S. is equal to 12.54 12.54 Mexican pesos. How many Mexican pesos did he get for his trip?
Show Solution
Solution
Identify what you are asked to find.How many Mexican pesos did Josiah get?
Choose a variable to represent it.Let p=p= number of pesos.
Write a sentence that gives the information to find it.If $1$1 U.S. is equal to 12.54 12.54 Mexican pesos, then $325$325 is how many pesos?
Translate into a proportion.
Substitute given values.1 12.54=325 p 1 12.54=325 p
The variable is in the denominator, so find the cross products and set them equal.p⋅1=12.54(325)p⋅1=12.54(325)
Simplify.c=4,075.5 c=4,075.5
Check if the answer is reasonable.
Yes, $100$100 would be $1,254$1,254 pesos. $325$325 is a little more than 3 3 times this amount.
Write a complete sentence.Josiah has 4075.5 4075.5 pesos for his spring break trip.
try it
In the following video we show another example of how to solve an application that involves proportions.
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Ex: Solve a Proportion by Clearing Fractions ((a/x=b/c, Fraction Solution). Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
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13658 | https://www.ucl.ac.uk/~ucahhwi/LTCC/sectionF-complex.pdf | F Laplace’s equation: Complex variables Let’s look at Laplace’s equation in 2D, using Cartesian coordinates: ∂2f ∂x2 + ∂2f ∂y2 = 0.
It has no real characteristics because its discriminant is negative (B2 −4AC = −4). But if we ignore this technicality and allow ourselves a complex change of variables, we can benefit from the same structure of solution that worked for the wave equation. Introduce η = x + iy x = (η + ξ)/2 ξ = x −iy y = (η −ξ)/2i.
Then the chain rule gives ∂ ∂x = ∂ ∂η + ∂ ∂ξ ∂ ∂y = i ∂ ∂η −∂ ∂ξ and the PDE becomes 4 ∂ ∂η ∂f ∂ξ = 0 whose solution is straightforward: f = p(η) + q(ξ) = p(x + iy) + q(x −iy).
Here p and q are differentiable complex functions; and assuming we wanted a real solution to the original (real) PDE, we have an additional constraint that the sum of the two functions must have no imaginary part.
We can formalise this in more standard notation: if we use the (x, y) plane to represent the complex plane in the usual way, we introduce the complex variable z = x + iy. Then its complex conjugate is z = x −iy and the solution we have just found is f = p(z) + q(z).
F.1 Cauchy-Riemann Equations Let’s look at our function p(η) = p(z), which forms half of our “characteristics”-style solution. It is obvious that ∂p ∂ξ = ∂p ∂z = 0 and using the chain rule, this tells us that 1 2 ∂p ∂x −1 2i ∂p ∂y = 0 ∂p ∂x = −i∂p ∂y .
Now if we divide the function into its real and imaginary parts: p(z) = u(x, y) + iv(x, y) 45 where u and v are real functions, we have ∂u ∂x + i∂v ∂x = −i∂u ∂y + ∂v ∂y This complex equation is equivalent to the pair of real equations: ∂u ∂x = ∂v ∂y ∂v ∂x = −∂u ∂y .
These are the Cauchy-Riemann equations, and are satisfied by the real and imaginary parts of any differentiable function of a complex variable z = x + iy.
In fact in a given domain, u and v (continuously differentiable) satisfy the Cauchy-Riemann equations if and only if p is an analytic function of z. We will not prove this here.
(Recall f(z) is analytic ≡holomorphic within a domain D if, in every circle |z −z1| < ρ lying in D, f can be represented as a power series in z −z1.) F.2 General solution of Laplace’s equation We had the solution f = p(z) + q(z) in which p(z) is analytic; but we can go further: remember that Laplace’s equation in 2D can be written in polar coordinates as ∇2f = 1 r ∂ ∂r r∂f ∂r + 1 r2 ∂2f ∂θ2 = 0 and we showed by separating variables that in the whole plane (except the origin) it has solutions f(r, θ) = A + B ln r + X n (an cos (nθ) + bn sin (nθ))(cnrn + dnr−n).
(In fact we also discarded some solutions which were not 2π-periodic in θ; these may be valid in a domain which does not encircle the origin.) Now in these variables, z = r exp [iθ] so we can also write the solution we found as f = Real A + B ln z + X n cn(an −ibn)zn + dn(an + ibn)z−n !
meaning that our solution is the real part of a function of z only: f = Real (g(z)).
Note that g(z) as given here is analytic in any simply connected domain that does not include the origin; if B = 0 it is analytic everywhere except the origin, and if additionally dn = 0, it is analytic everywhere.
We have shown that the real solution to Laplace’s equation we had found is the real part of an analytic function of z = x + iy in our domain; we can show 46 the converse very quickly from the Cauchy-Riemann equations. Consider an analytic function f(z) = u(x, y) + iv(x, y) Then the Cauchy-Riemann equations give ∂u ∂x = ∂v ∂y ∂v ∂x = −∂u ∂y .
Differentiating the first w.r.t. x and the second w.r.t. y gives: ∂2v ∂x∂y = ∂2u ∂x2 = −∂2u ∂y2 ∂2u ∂x2 + ∂2u ∂y2 = 0.
We can solve Laplace’s equation in any domain simply by taking the real part of any analytic function in that domain.
F.3 Composition of Analytic functions The composition of two analytic functions is analytic (providing, of course, the relevant domains are correctly specified): if f : D1 →D2 and g : D2 →D3 are both analytic, then the composed function g ◦f : D1 →D3 is also analytic on D1.
This has important ramifications for the solution of Laplace’s equation in odd-shaped domains or with boundary conditions which are unsuitable for separation of variables.
Suppose we are trying to find a real function u satisfying ∇2u = 0 in D1 with u = u(x, y) on ∂D1.
Of course this is equivalent to finding an analytic function f(z) on D1 whose real part satisfies the boundary condition on ∂D1.
If D1 is an awkward shape, and we can find an analytic function w(z) which maps it to a more helpful domain D2, then we can define f = g ◦w f(z) = g(w(z)) and we are now looking for an analytic function g defined on D2 such that Real (g(w(z))) = u(z) on ∂D1.
Real (g(w)) = u(z(w)) on ∂D2.
47 Example This is taken from an old UCL exam paper.
Find the solution to Laplace’s equation in the domain D1 given by the whole (x, y)-plane except for two semi-infinite plates |x| ≥1, y = 0. The boundary conditions on these two plates are u(x, 0) = 0 on x ≥1; u(x, 0) = 1 on x ≤−1.
The domain looks superficially suitable for separation of variables in Cartesian coordinates, but the boundary conditions are not suitable: we would need u(x, 0) to be prescribed for all x for separation to work.
Here we use the map w(z) = z + p (z2 −1). Note that the square root means this map is not analytic over the whole plane; we need a branch cut at each of z = 1, z = −1. Given the domain we are trying to transform, it makes sense to put the branch cuts on y = 0 (or z real) and |x| ≥1 (or |z| ≥1).
The point z = 0 maps to w = p (−1) and we can choose which of the possible values we take for the sign of the square root here: we choose w(0) = i. This choice, with the positioning of the branch cuts, determines w(z) everywhere in our domain – in the diagram I’ve marked the result of each of the square roots at points around it. So when z = x+iε and x > 1, both roots are positive; when z = x and |x| < 1, the root at z = 1 has argument i and the other is positive; when z = x −iε with x < −1, both roots have argument i so the product is negative, and so on: r r −i + + i i − In particular: w(−1) = −1 w(1) = 1 w(x) = x + i p (1 −x2) −1 < x < 1 w(x + iε) = x − p (x2 −1) x < −1 w(x −iε) = x + p (x2 −1) x < −1 w(x + iε) = x + p (x2 −1) x > 1 w(x −iε) = x − p (x2 −1) x > 1 Thus the branch cuts in the z-plane map onto the real line in the w-plane: the left-hand cut maps to (top side) w < −1 and (bottom side) −1 < w < 0, and the right-hand cut maps to (top side) w > 1 and (bottom side) 0 < w < 1. In the w-plane we now need to solve Laplace’s equation for a new function v with v(x, 0) = 1 on x ≤0 v(x, 0) = 0 on x ≥0.
This new problem is suitable for separation of variables in polar coordinates: the boundary conditions in terms of r and θ are v(r, 0) = 0 v(r, π) = 1.
Note that our domain now does not encircle the origin, so we must revisit our separable solution and include some terms we discarded earlier.
48 We look for the form v = R(r)T(θ) and derive the coupled ODEs r2R′′(r) + rR′(r) R(r) = A T ′′(θ) T(θ) = −A.
In the three cases A < 0, A > 0 and A = 0 respectively these yield: v = (Aµ exp [µθ] + Bµ exp [−µθ])(Cµ cos [µ ln r] + Dµ sin [µ ln r]) v = (aλ cos [λθ] + bλ sin [λθ])(cλrλ + dλr−λ) v = (α + β ln r)(γ + δθ).
Applying the boundary condition T(θ = 0) = 0 gives the three basis functions v = sinh µθ v = sin λθ v = θ(α + β ln r), and the condition that v must be well-behaved at r = 0 (since the origin is in our domain) fixes further: v = αθ + X λ cλrλ sin [λθ] The final boundary condition v(r, π) = 1 gives 1 = απ + X λ cλrλ sin [λπ] which is satisfied with α = 1/π and cλ = 0. Thus we have found v(r, θ) = θ π .
In order to convert this to a solution to our original problem, we first need to find the analytic function of which it is the real part. In this case the function is straightforward: v(r, θ) = θ π = −1 π Real (i[ln r + iθ]) = −1 π Real (i ln w) so the analytic function we need is g(w) = −i ln w π .
Finally we need to convert back to the original variables: f(z) = g ◦w(z) = −i π ln n z + p (z2 −1) o and the solution we need is the real part of this: u(x, y) = Real −i π ln n z + p (z2 −1) o = 1 π Imag ln n z + p (z2 −1) o .
In particular, on the “missing line” y = 0, −1 ≤x ≤1, we have u(x, 0) = 1 π Arg n x + i p (1 −x2) o = 1 π arctan p (1 −x2) x .
49 |
13659 | https://www.hydrocompinc.com/wp-content/uploads/documents/HC135-BladeAreaRatio.pdf | Blade Area Ratio Defined A HydroComp Technical Report Report 135 DEFINITIONS OVERVIEW The three graphics shown below illustrate the three different types of BAR for a common propeller. Blade area ratio, or BAR, is a parameter used to relate the size of a propeller blade to its diameter. It is critical to the control of cavitation and changes to BAR affect its efficiency and thrust-making performance. Projected Area Ratio (PAR) The Projected view is the one you see when you look down on the propeller. The Projected area is the area of the outline as projected onto a surface below. Projected area ratio is the smallest of the three. The generic term or “BAR”, however, does not sufficiently describe the blade area ratio. In fact, there are three types of BAR – Projected, Developed and Expanded. CALCULATION OF AREA RATIO Developed Area Ratio (DAR) Developed area is the area of the blade outline if it could be untwisted (i.e., as if the whole blade were unattached from the hub and brought to zero pitch). For all three types, the appropriate total blade area outside of the hub is divided by the propeller disk area (e.g., πR2) to derive the area ratio. Projected Developed Expanded Copyright © 2003, 2007 HydroComp, Inc. All Rights Reserved. www.hydrocompinc.com 1 Expanded Area Ratio (EAR) Expanded area is what if found if the Developed area could be flexibly unwrapped on a flat surface so that all sections were parallel. Expanded area is what is important to propeller designers, to treat the propeller blade like a wing. In other words, the Expanded view converts the propeller from its helix to a flat plane. Expanded area ratio is typically close in magnitude to Developed area ratio, and is often used interchangeably. CONVERSION You can use the the following formula to find approximate conversions between the three types of BAR. D = diameter P = pitch Z = number of blades D P DAR PAR × − = 229 .
0 067 .
1 + = Z DAR DAR EAR 75 .
2 34 .
0 For more technical articles like this one, visit the HydroComp Knowledge Library: www.hydrocompinc.com/knowledge/library.htm Copyright © 2003, 2007 HydroComp, Inc. All Rights Reserved. www.hydrocompinc.com 2 |
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Spiral similarity
Page made by vladimir.shelomovskii@gmail.com, vvsss
Contents
1 Definition
2 Simple problems
2.1 Explicit spiral similarity
2.2 Hidden spiral symilarity
2.3 Linearity of the spiral symilarity
2.4 Construction of a similar triangle
2.5 Center of the spiral symilarity for similar triangles
2.6 Spiral similarity in rectangle
2.7 Common point for 6 circles
2.8 Three spiral similarities
2.9 Superposition of two spiral similarities
2.10 Spiral similarity for circles
2.11 Remarkable point for spiral similarity
2.12 Remarkable point for pair of similar triangles
2.13 Remarkable point’s problems
2.13.1 Problem 1
2.13.2 Problem 2
2.13.3 Problem 3
2.13.4 Problem 4
2.13.5 Solutions
2.14 Japan Mathematical Olympiad Finals 2018 Q2
Definition
A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important.
Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation).
The transformation is linear and transforms any given object into an object homothetic to given.
On the complex plane, any spiral similarity can be expressed in the form where is a complex number. The magnitude is the dilation factor of the spiral similarity, and the argument is the angle of rotation.
The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane.
Let with corresponding complex numbers and so For any points and the center of the spiral similarity taking to point is also the center of a spiral similarity taking to This fact explain existance of Miquel point.
Case 1 Any line segment can be mapped into any other using the spiral similarity. Notation is shown on the diagram.
is circle is circle
is any point of is circle is the image under spiral symilarity centered at
is the dilation factor,
is the angle of rotation.
Case 2 Any line segment can be mapped into any other using the spiral similarity. Notation is shown on the diagram. is circle (so circle is tangent to is circle tangent to is any point of is circle is the image under spiral symilarity centered at is the dilation factor,
is the angle of rotation.
Simple problems
Explicit spiral similarity
Given two similar right triangles and Find and
Solution
The spiral similarity centered at with coefficient and the angle of rotation maps point to point and point to point
Therefore this similarity maps to
Hidden spiral symilarity
Let be an isosceles right triangle Let be a point on a circle with diameter The line is symmetrical to with respect to and intersects at Prove that
Proof
Denote Let cross perpendicular to in point at point
Then
Points and are simmetric with respect so
The spiral symilarity centered at with coefficient and the angle of rotation maps to and to point such that
Therefore
Linearity of the spiral symilarity
Points are outside
Prove that the centroids of triangles and are coinsite.
Proof
Let where be the spiral similarity with the rotation angle and
A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore We use the property of linearity and get Let be the centroid of so
is the centroid of the
Construction of a similar triangle
Let triangle and point on sideline be given. Construct where lies on sideline and lies on sideline
Solution
Let be the spiral symilarity centered at with the dilation factor and rotation angle
so image of any point lies on The spiral symilarity centered at with the dilation factor and rotation angle maps into and therefore the found triangle is the desired one.
Center of the spiral symilarity for similar triangles
Let triangle and point on sideline be given. where lies on sideline and lies on sideline The spiral symilarity maps into Prove
a)
b) Center of is the First Brocard point of triangles and
Proof
a) Let be the spiral symilarity centered at with the dilation factor and rotation angle
Denote Similarly
b) It is well known that the three circumcircles and have the common point (it is in the diagram).
Therefore is cyclic and
Similarly,
Similarly,
Therefore, is the First Brocard point of
is cyclic Similarly,
Therefore is the First Brocard point of and
Therefore the spiral symilarity maps into has the center the angle of the rotation
Spiral similarity in rectangle
Let rectangle be given. Let point
Let points and be the midpoints of segments and respectively.
Prove that
Proof
Let be the midpoint
is a parallelogram
and are corresponding medians of and
There is a spiral similarity centered at with rotation angle that maps to Therefore
Common point for 6 circles
Let and point on sideline be given. where lies on sideline and lies on sideline
Denote
Prove that circumcircles of triangles have the common point.
Proof
so there is the spiral symilarity taking to Denote the center of the center of is the secont crosspoint of circumcircles of and but this center is point so these circles contain point . Similarly for another circles.
Three spiral similarities
Let triangle be given. The triangle is constructed using a spiral similarity of with center , angle of rotation and coefficient
A point is centrally symmetrical to a point with respect to
Prove that the spiral similarity with center , angle of rotation and coefficient taking to
Proof
Corollary
Three spiral similarities centered on the images of the vertices of the given triangle and with rotation angles equal to the angles of take to centrally symmetric to with respect to
Superposition of two spiral similarities
Let be the spiral similarity centered at with angle and coefficient
Let be spiral similarity centered at with angle and coefficient
Let
Prove: a) is the crosspoint of bisectors and
b)
Algebraic proof
We use the complex plane Let Then Geometric proof
Denote
Then
Let be the midpoint be the point on bisector such that be the point on bisector such that Then
is the crosspoint of bisectors and
Corollary
There is another pair of the spiral similarities centered at and with angle coefficients and
In this case
Spiral similarity for circles
Let circle cross circle at points and Point lies on
Spiral similarity centered at maps into
Prove that points and are collinear.
Proof
Arcs
Corollary
Let points and be collinear.
Then exist the spiral similarity centered at such that
Let circle cross circle at points and
Points and lie on
Let be the tangent to be the tangent to
Prove that angle between tangents is equal angle between lines and
Proof
There is the spiral similarity centered at such that Therefore angles between these lines are the same.
Remarkable point for spiral similarity
Circles and centered at points and respectively intersect at points and Points and are collinear. Point is symmetrical to with respect to the midpoint point Prove:
a) b)
Proof
a) cross in midpoint
b) is parallelogram
Denote
Corollary
Let points and be collinear. Then
Therefore is the crosspoint of the bisectors and
Remarkable point for pair of similar triangles
Let
Let the points and be the circumcenters of and
Let point be the midpoint of
The point is symmetric to with respect point
Prove:
a) point be the crosspoint of the bisectors and
b)
Proof
is parallelogram Denote
Similarly,
The statement that was proved in the previous section.
Remarkable point’s problems
Problem 1
Let a convex quadrilateral be given,
Let and be the midpoints of and respectively.
Circumcircles and intersect a second time at point
Prove that points and are concyclic.
Problem 2
Let triangle be given.
Let point lies on sideline
Denote the circumcircle of the as , the circumcircle of the as .
Let be the circumcenter of
Let circle cross sideline at point
Let the circumcircle of the cross at point
Prove that
Problem 3
The circles and are crossed at points and points
Let be the tangent to be the tangent to
Point is symmetric with respect to
Prove that points and are concyclic.
Problem 4
The circles and are crossed at points and points
Let be the tangent to be the tangent to
Points and lye on bisector of the angle
Points and lye on external bisector of the angle
Prove that and bisector are tangent to the circle
bisector is tangent to the circle
Solutions
Solutions are clear from diagrams.
In each case we use remarcable point as the point of bisectors crossing.
Solution 1
We use bisectors and . The points and are concyclic.
Solution 2 We use bisectors of and
Solution 3
We use bisectors of and .
is the circumcenter of
Circle is symmetric with respect diameter
Point is symmetric to with respect diameter
Therefore
Solution 4
Let
Let be midpoint be midpoint .
We need prove that and
Denote
The angle between a chord and a tangent is half the arc belonging to the chord.
is tangent to
is diameter Similarly, is diameter is tangent to
Let be the spiral similarity centered at
Points and are collinear Points and are collinear
Therefore is tangent to
Similarly, is tangent to
Japan Mathematical Olympiad Finals 2018 Q2
Given a scalene let and be points on lines and respectively, so that
Let be the circumcircle of and the reflection of across
Lines and meet again at and respectively.
Prove that and intersect on
Proof
Let be the orthocenter of
Point is symmetrical to point with respect to height
Point is symmetrical to point with respect to height
is centered at
is symmetrical with respect to heightline
is symmetrical to point with respect to height
is symmetrical to point with respect to height
The isosceles triangles
a) are concyclic.
b) is the spiral center that maps to maps to
Therefore are concyclic and are concyclic.
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13661 | https://innovationspace.ansys.com/courses/wp-content/uploads/2020/12/Lesson4-Handout-Final.pdf | One Dimensional Waves Application of Shock-Expansion Theory – Lesson 4 • DECEMBER 2019 2 • So far, we have examined shock and expansion waves generated by flows over objects. These waves are stationary in the object’s reference frame, and steady-state analysis can be applied to derive their relations.
• In this lesson we will discuss propagation of transient shocks and expansion waves in one dimension.
• We will also briefly outline propagation of small amplitude acoustic waves.
• Finally, we will combine these elements of the transient wave theory in the classical Sod shock tube problem.
Intro Into 1D Wave Motion 3 Moving Normal Shock Waves • The governing equations of a moving normal shock can be readily derived from those obtained for the stationary shock using the following variables: ‐ 𝑊– velocity of the gas ahead of the shock relative to the shock, ‐ 𝑊−𝑢𝑝– velocity of the gas behind the shock relative to the shock Stationary shock 1 2 𝑢1 𝑢2 Gas motion downstream of shock Gas motion upstream of shock Shock moving with velocity 𝑊 1 2 𝑊 𝑢𝑝> 0 Motion induced by moving shock Stagnant gas ahead of shock 𝜌1𝑊= 𝜌2(𝑊−𝑢𝑝) 𝑝1 + 𝜌1𝑊2 = 𝑝2 + 𝜌2 𝑊−𝑢𝑝 2 ℎ2 + Τ 1 2 𝑣= ℎ1 + Τ 1 2 𝑊−𝑢𝑝 2 • The Mach number of the shock is given by: 𝑀1 = Τ 𝑊𝑎1 𝑎1 2 = ( Τ 𝑑𝑝𝑑𝜌)1 where the speed of sound is: 4 Moving Normal Shock Waves • The “jump” relations developed for stationary normal shocks can be easily rewritten for the moving shock using the transformation equation from the previous slide. For application to practical problems, it is more convenient to use the pressure ratio Τ 𝑝2 𝑝1 as the basic independent parameter. For an ideal gas, the moving shock relations become: 𝑊= 𝑀1𝑎1 = 𝑎1 𝛾−1 2𝛾 + 𝛾+ 1 2𝛾 𝑝2 𝑝1 1/2 shock velocity 𝑒2 −𝑒1 = 𝑝1 + 𝑝2 2 (𝑣1 −𝑣2) Hugoniot equation. This is identical in form to the stationary shock case since it relates thermodynamic variables independent of whether the shock is moving or stationary.
𝜌2 𝜌1 = 1 + 𝛾+ 1 𝛾−1 𝑝2 𝑝1 𝛾+ 1 𝛾−1 + 𝑝2 𝑝1 = 𝑢1 𝑢2 𝑇2 𝑇 1 = 𝑝2 𝑝1 𝛾+ 1 𝛾−1 + 𝑝2 𝑝1 1 + 𝛾+ 1 𝛾−1 𝑝2 𝑝1 • Relations for the density and temperature ratios can be obtained from the Hugoniot equation as: 5 Moving Normal Shock Waves (cont.) • The gas velocity behind the wave is: 𝑢𝑝= 𝑎1 𝛾 𝑝2 𝑝1 −1 2𝛾 𝛾+ 1 𝑝2 𝑝1 + 𝛾−1 𝛾+ 1 1/2 This is the velocity of moving gas behind the shock “experienced” by an observer in the laboratory frame of reference. Its magnitude can best be estimated in terms of the local Mach number: 𝑢𝑝 𝑎2 = 𝑢𝑝 𝑎1 𝑎1 𝑎2 = 𝑢𝑝 𝑎1 𝑇 1 𝑇2 1/2 ⟹ 𝑢𝑝 𝑎2 = 1 𝛾 𝑝2 𝑝1 −1 2𝛾 𝛾+ 1 𝑝2 𝑝1 + 𝛾−1 𝛾+ 1 1/2 1 + 𝛾+ 1 𝛾−1 𝑝2 𝑝1 𝛾+ 1 𝛾−1 𝑝2 𝑝1 + 𝑝2 𝑝1 2 1/2 ⟶ 𝑝2 𝑝1 →∞ 2 𝛾(𝛾−1) ⟹ In the limit of a very strong shock, 𝑢𝑝/𝑎2 →1.89 for air. Thus gas flow can be accelerated to supersonic speeds by strong shocks. • Finally, we should note that, unlike the case of stationary shocks, the total enthalpy ℎ0 is no longer constant for moving shocks: 𝜌𝐷ℎ0 𝐷𝑡= 𝜕𝑝 𝜕𝑡, 𝜕𝑝 𝜕𝑡≠0 ⟹ ℎ0 is not constant. 6 Acoustic Equations • Acoustic (sound) waves are small-amplitude disturbances propagating in gases (most notably in air). Given the assumption of smallness of acoustic fluctuations, governing questions of gas dynamics can be reduced to a set of linear acoustic equations. By applying a linearization technique and thermodynamic considerations, the equations reduce to the acoustic wave equation: 𝜕2𝜌 𝜕𝑡2 = 𝑎∞ 2 ∇𝜌 𝜕2𝑝 𝜕𝑡2 = 𝑎∞ 2 ∇p ⟺ Here 𝑎∞is the constant speed of sound. Note the equation can be written in the same form for density or pressure.
7 • The same form solution also holds for the velocity: Acoustic Equations (cont.) • For one-dimensional waves, the wave equation has a solution of the form: 𝜌(𝑥, 𝑡) = 𝐹𝑥−𝑎∞𝑡+ 𝐺(𝑥+ 𝑎∞𝑡) 𝑡 𝑥 𝑡2 𝑡1 0 characteristics 𝑡 𝑥 0 𝑑𝑥 𝑑𝑡= 𝑎∞ 𝑑𝑥 𝑑𝑡= −𝑎∞ 𝐹𝑥 𝐺𝑥 i.e., the solution is a superposition of left- and right-traveling simple waves propagating at the sound speed along characteristic lines with the slope Τ 𝑑𝑥𝑑𝑡= ±𝑎∞in the 𝑥−𝑡plane.
𝑢(𝑥, 𝑡) = 𝑓𝑥−𝑎∞𝑡+ 𝑔(𝑥+ 𝑎∞𝑡) • The relation between velocity, pressure and density is: 𝑢= ± Τ 𝑎∞𝜌𝜌∞= ± Τ 𝑝(𝜌∞𝑎∞) • The wave shape does not change with time in the acoustic approximation.
8 Isentropic Nonlinear Waves of Finite Amplitude • Previous analysis of small-amplitude acoustic waves is valid only if the wave amplitude is small. If such an assumption cannot be made, the wave can no longer be described by the linearized equations. Such waves are called finite waves. These waves have the following common characteristics: Fluctuation of density, temperature, velocity, etc., are large Any local part of the wave propagates at the local velocity 𝑢+ 𝑎relative to the laboratory frame The wave shape changes with time Full non-linear governing equations must be used to describe wave motion The finite wave approach is suitable for all real waves, linear or non-linear • The complete non-linear governing equations need to be solved to describe finite waves. Such a mathematical solution was obtained by Riemann and Earnshaw almost 160 years ago, and we will summarize its main steps here.
• Our derivation will be limited to 1D for clarity, but 3D generalization is also possible.
9 Isentropic Nonlinear Waves of Finite Amplitude (cont.) • From our discussion of thermodynamics, the density is a function of pressure and entropy, thus: 𝑑𝜌= 𝜕𝜌 𝜕𝑝 𝑠 𝑑𝑝+ 𝜕𝜌 𝜕𝑠 𝑑𝑝 𝑑𝑠= 𝜕𝜌 𝜕𝑝 𝑠 𝑑𝑝 = 0 for isentropic flows ⟹ 1/𝑎2 𝐷𝜌 𝐷𝑡= 1 𝑎2 𝐷𝑝 𝐷𝑡 ⟹ 1 𝑎2 𝜕𝑝 𝜕𝑡+ 𝑢𝜕𝑝 𝜕𝑥 + 𝜌𝜕𝑢 𝜕𝑥= 0 continuity equation • By adding and subtracting the new form of the continuity equation from the 1D momentum equation: 𝜕𝑢 𝜕𝑡+ (𝑢+ 𝑎) 𝜕𝑢 𝜕𝑥 + 1 𝜌𝑎 𝜕𝑝 𝜕𝑡+ (𝑢+ 𝑎) 𝜕𝑝 𝜕𝑥 = 0 𝜕𝑢 𝜕𝑡+ (𝑢−𝑎) 𝜕𝑢 𝜕𝑥 −1 𝜌𝑎 𝜕𝑝 𝜕𝑡+ (𝑢−𝑎) 𝜕𝑝 𝜕𝑥 = 0 𝑡 𝑥 𝑡1 • Considering these equations along two specific lines (characteristics) in the 𝑥−𝑡plane given by: 𝑑𝑢+ 𝑑𝑝 𝜌𝑎= 0 𝑑𝑢−𝑑𝑝 𝜌𝑎= 0 we obtain compatibility equations valid along characteristics: 𝑑𝑥= 𝑢+ 𝑎𝑑𝑡 𝑑𝑥= 𝑢−𝑎𝑑𝑡 𝑥1 𝑑𝑥 𝑑𝑡= 𝑢+ 𝑎 𝐶+ characteristic 𝐶−characteristic along Τ 𝑑𝑥𝑑𝑡= 𝑢+ 𝑎 along Τ 𝑑𝑥𝑑𝑡= 𝑢−𝑎 𝑑𝑥 𝑑𝑡= 𝑢−𝑎 10 Isentropic Nonlinear Waves of Finite Amplitude (cont.) • Integrating the two compatibility equations along respective characteristic lines gives rise to the Riemann invariants, which can be expressed for a calorically perfect gas as: 𝐽+ = 𝑢+ න𝑑𝑝 𝜌𝑎= 𝑢+ 2𝑎 𝛾−1 = const 𝐽−= 𝑢−න𝑑𝑝 𝜌𝑎= 𝑢− 2𝑎 𝛾−1 = const along 𝐶+ characteristic along 𝐶−characteristic ⟹ 𝑎= 𝛾−1 4 (𝐽+ −𝐽−) 𝑢= 1 2 (𝐽+ + 𝐽−) • If 𝐽+ and 𝐽−are known at a point in the 𝑥−𝑡plane, the velocity and speed of sound can be determined from the above relations.
11 Centered Expansion Wave • Let’s consider a tube filled with gas. A piston initially at rest is suddenly withdrawn to the right with the constant velocity 𝑢3. This will produce an expansion wave propagating to the left.
• As soon as the wave starts propagating, it begins to flatten. The velocity has a linear profile across the expansion wave and pressure has a corresponding profile. • From the analysis of Riemann invariants 𝐽+ and 𝐽−, and by considering: 1.
𝐽+ and 𝐽−are constant along respective characteristics 𝐶+ and 𝐶− 2.
Regions 3 and 4 are regions with constant flow properties 3 4 piston 𝑢3 𝑢3 𝑡 𝑥 Expansion front 𝑥= 𝑎4𝑡 Piston path 𝑥= −𝑢3𝑡 4 3 it can be shown 𝐽+ = const through the expansion wave ⟹ 𝑢+ 2𝑎 𝛾−1 = 𝑢4 + 2𝑎4 𝛾−1 = const = 0 any local point in expansion ⟹ 𝑎 𝑎4 = 1 −𝛾−1 2 𝑢 𝑎4 Relation between 𝑢and a in a simple expansion wave. 𝐶− 𝐶−𝐶− 𝑢3 𝑢4 𝑝4 𝑝3 𝑥 𝑥 𝑢 𝑝 12 Centered Expansion Wave (cont.) • Relations between the local gas velocity and temperature, pressure and density are obtained from the isentropic relations: 𝑇 𝑇 4 = 1 −𝛾−1 2 𝑢 𝑎4 2 𝑝 𝑝4 = 1 −𝛾−1 2 𝑢 𝑎4 2𝛾/(𝛾−1) 𝜌 𝜌4 = 1 −𝛾−1 2 𝑢 𝑎4 2/(𝛾−1) • Noting that the expansion wave shown in the sketch is the left-running wave expanding along the fan of 𝐶− characteristics, and considering characteristics are straight lines through the origin: 𝑑𝑥= 𝑢−𝑎𝑑𝑡 ⟹ 𝑥= 𝑢−𝑎𝑑𝑡 ⟹from the relation between 𝑢and a ⟹ 𝑢= 2 𝛾+ 1 𝑎4 + 𝑥 𝑡 Variation 𝑢as a function of 𝑥and 𝑡valid for the expansion region between its front and tail, −𝑎4 ≤ Τ 𝑥𝑡≤𝑢3 −𝑎3 • Similar relations can be obtained for the right-running expansion wave. 13 Sod Shock Tube Problem • The shock tube problem was first studied by Gary A. Sod in 1978.
• Let us consider a one-dimensional tube with a diaphragm initially separating regions of high pressure (left) and low pressure (right). The tube is closed on both ends.
• The gases on either side of the diaphragm can have different molecular weights and thermodynamic properties.
• All flow field properties depend on distance, 𝑥, and time, 𝑡.
• The basic parameter of the shock tube is the diaphragm pressure ratio, Τ 𝑝4 𝑝1.
𝑝1, 𝑇 1, 𝑎1, 𝛾1, 𝑅1 𝑝4, 𝑇 4, 𝑎4, 𝛾4, 𝑅4 1 4 0 𝑥 driver section driven section diaphragm 0 𝑥 𝑝1 𝑝4 𝑝 • At 𝑡= 0, the diaphragm is suddenly removed which initiates transient dynamics of gases in the tube.
14 Sod Shock Tube Problem (cont.) • At 𝑡> 0 the initial pressure jump splits into a shock wave propagating into the driven section and an expansion wave propagating into the driver section with the velocity 𝑊.
• Propagating normal shock increases the pressure behind it and induces a gas motion with velocity 𝑢𝑝.
• The interface between gases in the driver and driven section is called the contact surface, which can be thought of as a slip surface. The gases remain separated by this surface and the entropy changes across it are discontinuous.
• The pressure and velocity, however, do not change across the contact surface, 𝑝3 = 𝑝2 and 𝑢3 = 𝑢2 = 𝑢𝑝.
• The expansion wave propagates to the left, smoothly decreasing the pressure in regions 4.
1 4 𝑢𝑝 𝑊 2 3 Expansion propagating to the left Contact surface separating driver and driven gas and moving at the velocity of gas behind the shock Normal shock propagating to the right 15 Sod Shock Tube Problem (cont.) • The flow field inside the shock tube after the diaphragm is removed can be completely described by the diaphragm pressure ratio, Τ 𝑝4 𝑝1. All the relations follow from the previous analysis of moving shocks and expansions.
𝑢𝑝= 𝑢2 = 𝑎1 𝛾 𝑝2 𝑝1 −1 2𝛾 𝛾+ 1 𝑝2 𝑝1 + 𝛾−1 𝛾+ 1 1/2 Gas velocity induced by the shock 𝑝3 𝑝4 = 1 −𝛾4 −1 2 𝑢3 𝑎4 2𝛾4/(𝛾4−1) Pressure drop across the expansion ⟹ 𝑢3 = 2𝑎4 𝛾4 −1 1 −𝑝2 𝑝4 (𝛾4−1)/2𝛾4 𝑝3 = 𝑝2 ⟹ 𝑢3 = 𝑢2 𝑝4 𝑝1 = 𝑝2 𝑝1 1 − (𝛾4 −1)( Τ 𝑎1 𝑎4)( Τ 𝑝2 𝑝1 −1) 2𝛾1 2𝛾1 + (𝛾1 + 1))( Τ 𝑝2 𝑝1 −1 −2𝛾4/(𝛾4−1) Implicit relation for the shock strength Τ 𝑝2 𝑝1 as a function of the diaphragm pressure Τ 𝑝4 𝑝1 • The expansion strength is obtained as: 𝑝3 𝑝4 = 𝑝3 𝑝1 𝑝1 𝑝4 = Τ 𝑝2 𝑝1 Τ 𝑝4 𝑝1 • Once the shock strength is known, all other flow quantities can be found from the shock and expansion relations.
16 Sod Shock Tube Problem (cont.) Pressure Distribution along the shock tube ShockExpansion Wave 17 Summary • In this lesson we discussed unsteady wave motion in one dimension. • We covered moving normal shock, acoustic and finite nonlinear waves, and centered expansion wave .
• Finally, we combined this theory to describe the flow field inside a shock tube. |
13662 | https://askfilo.com/user-question-answers-smart-solutions/single-equivalent-discount-for-25-and-10-3334383430343134 | Question asked by Filo student
Determine the single equivalent discount when successive discounts of 25% and 10% are applied.
Views: 5,079 students
Updated on: Jun 8, 2025
Text SolutionText solutionverified iconVerified
Concepts
Successive Discount, Single Equivalent Discount, Percentage Calculation
Explanation
When two successive discounts are given, the net (equivalent) single discount can be calculated using the formula:
Single equivalent discount=d1+d2−100d1×d2
where d1 and d2 are the percentage rates of the successive discounts.
Step-By-Step Solution
Step 1
Let the marked price be Rs. 100 (for simplicity).
Step 2
Apply the first discount of 25%:
25% of Rs. 100=Rs.25
So, price after 1st discount:
Rs.100−Rs.25=Rs.75
Step 3
Apply the second discount of 10% on Rs. 75:
10% of Rs. 75=Rs.7.50
Price after 2nd discount:
Rs.75−Rs.7.50=Rs.67.50
Step 4
Calculate the equivalent single discount:
Original price = Rs. 100
Final price = Rs. 67.50
Discount = Rs. 100 - Rs. 67.50 = Rs. 32.50
Equivalent single discount percent:
=Rs.100Rs.32.50×100=32.5%
Step 5
Alternatively, using the formula for successive discounts:
Equivalent discount=25+10−10025×10
=35−2.5=32.5%
Final Answer
The single equivalent discount is 32.5%.
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| Question Text | Determine the single equivalent discount when successive discounts of 25% and 10% are applied. |
| Updated On | Jun 8, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 9 |
| Answer Type | Text solution:1 |
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13663 | https://stats.libretexts.org/Bookshelves/Introductory_Statistics/OpenIntro_Statistics_(Diez_et_al)./03%3A_Distributions_of_Random_Variables/3.04%3A_Binomial_Distribution_(Special_Topic) | Skip to main content
3.4: Binomial Distribution (Special Topic)
Last updated
: Apr 23, 2022
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3.3: Geometric Distribution (Special Topic)
3.5: More Discrete Distributions (Special Topic)
Page ID
: 272
David Diez, Christopher Barr, & Mine Çetinkaya-Rundel
OpenIntro Statistics
( \newcommand{\kernel}{\mathrm{null}\,})
Example 3.4.1: Shock Study
Suppose we randomly selected four individuals to participate in the "shock" study. What is the chance exactly one of them will be a success? Let's call the four people Allen (A), Brittany (B), Caroline (C), and Damian (D) for convenience. Also, suppose 35% of people are successes as in the previous version of this example.
Solution
Let's consider a scenario where one person refuses:
P(A=refuse;B=shock;C=shock;D=shock)=P(A=refuse)P(B=shock)P(C=shock)P(D=shock)=(0.35)(0.65)(0.65)(0.65)=(0.35)1(0.65)3=0.096
But there are three other scenarios: Brittany, Caroline, or Damian could have been the one to refuse. In each of these cases, the probability is again
P=(0.35)1(0.65)3.
These four scenarios exhaust all the possible ways that exactly one of these four people could refuse to administer the most severe shock, so the total probability is
4×(0.35)1(0.65)3=0.38.
Exercise 3.4.1
Verify that the scenario where Brittany is the only one to refuse to give the most severe shock has probability (0.35)1(0.65)3.
Answer
: P(A=shock;B=refuse;C=shock;D=shock)=(0.65)(0.35)(0.65)(0.65)=(0.35)1(0.65)3.
The Binomial Distribution
The scenario outlined in Example 3.4.1 is a special case of what is called the binomial distribution. The binomial distributiondescribes the probability of having exactly k successes in n independent Bernoulli trials with probability of a success p (in Example 3.4.1, n = 4, k = 1, p = 0.35). We would like to determine the probabilities associated with the binomial distribution more generally, i.e. we want a formula where we can use n, k, and p to obtain the probability. To do this, we reexamine each part of the example.
There were four individuals who could have been the one to refuse, and each of these four scenarios had the same probability. Thus, we could identify the nal probability as
[# of scenarios]×P(single scenario)(3.4.1)
The first component of this equation is the number of ways to arrange the k = 1 successes among the n = 4 trials. The second component is the probability of any of the four (equally probable) scenarios.
Consider P(single scenario) under the general case of k successes and n-k failures in the n trials. In any such scenario, we apply the Multiplication Rule for independent events:
pk(1−p)n−k(3.4.2)
This is our general formula for P(single scenario).
Secondly, we introduce a general formula for the number of ways to choose k successes in n trials, i.e. arrange k successes and n - k failures:
(nk)=n!k!(n−k)!(3.4.3)
The quantity (nk) is read n choose k.30 The exclamation point notation (e.g. k!) denotes a factorial expression.
0!1!2!3!4!n!=1=1=2×1=2=3×2×1=6=4×3×2×1=24⋮=n×(n−1)×⋯×3×2×1(3.4.4)
Substituting Equation 3.4.4 into Equation 3.4.3, we can compute the number of ways to choose k=1 successes in n=4 trials:
(41)=4!1!(4−1)!=4!1!3!=4×3×2×1(1)(3×2×1)=4(3.4.5)(3.4.6)(3.4.7)(3.4.8)
This result is exactly what we found by carefully thinking of each possible scenario in Example 3.4.1.
Other notations
Other notation for n choose k includes nCk, Ckn, and C(n,k).
Substituting n choose k for the number of scenarios and pk(1−p)n−k for the single scenario probability in Equation 3.4.1 yields the general binomial formula (Equation 3.4.9).
Definition: Binomial distribution
Suppose the probability of a single trial being a success is p. Then the probability of observing exactly k successes in n independent trials is given by
(nk)pk(1−p)n−k=n!k!(n−k)!pk(1−p)n−k(3.4.9)
Additionally, the mean, variance, and standard deviation of the number of observed successes are
μ=npσ2=np(1−p)σ=np(1−p)−−−−−−−−√(3.4.10)
TIP: Four conditions to check if it is binomial?
The trials independent.
The number of trials, n, is fixed.
Each trial outcome can be classified as a success or failure.
The probability of a success, p, is the same for each trial.
Example 3.4.2
What is the probability that 3 of 8 randomly selected students will refuse to administer the worst shock, i.e. 5 of 8 will?
Solution
We would like to apply the binomial model, so we check our conditions. The number of trials is fixed (n = 8) (condition 2) and each trial outcome can be classi ed as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4).
In the outcome of interest, there are k = 3 successes in n = 8 trials, and the probability of a success is p = 0.35. So the probability that 3 of 8 will refuse is given by
(83)(0.35)k(1−0.35)8−3=8!3!(8−3)!(0.35)k(1−0.35)8−3=8!3!5!(0.35)3(0.65)5
Dealing with the factorial part:
8!3!5!=8×7×6×5×4×3×2×1(3×2×1)(5×4×3×2×1)=8×7×63×2×1=56
Using (0.35)3(0.65)5≈0.005, the final probability is about 56 0.005 = 0.28.
TIP: computing binomial probabilities
The rst step in using the binomial model is to check that the model is appropriate. The second step is to identify n, p, and k. The final step is to apply the formulas and interpret the results.
TIP: computing n choose k
In general, it is useful to do some cancelation in the factorials immediately. Alternatively, many computer programs and calculators have built in functions to compute n choose k, factorials, and even entire binomial probabilities.
Exercise 3.4.2A
If you ran a study and randomly sampled 40 students, how many would you expect to refuse to administer the worst shock? What is the standard deviation of the number of people who would refuse? Equation 3.4.10 may be useful.
Answer
: We are asked to determine the expected number (the mean) and the standard deviation, both of which can be directly computed from the formulas in Equation 3.4.10:
μ=np=40×0.35=14
and
σ=np(1−p)−−−−−−−−√=40×0.35×0.65−−−−−−−−−−−−−√=0.02.
Because very roughly 95% of observations fall within 2 standard deviations of the mean (see Section 1.6.4), we would probably observe at least 8 but less than 20 individuals in our sample who would refuse to administer the shock.
Exercise 3.4.2B
The probability that a random smoker will develop a severe lung condition in his or her lifetime is about 0:3. If you have 4 friends who smoke, are the conditions for the binomial model satisfied?
Answer
: One possible answer: if the friends know each other, then the independence assumption is probably not satis ed. For example, acquaintances may have similar smoking habits.
Example 3.4.3
Suppose these four friends do not know each other and we can treat them as if they were a random sample from the population. Is the binomial model appropriate? What is the probability that
none of them will develop a severe lung condition?
One will develop a severe lung condition?
That no more than one will develop a severe lung condition?
Solution
To check if the binomial model is appropriate, we must verify the conditions. (i) Since we are supposing we can treat the friends as a random sample, they are independent. (ii) We have a fixed number of trials (n = 4). (iii) Each outcome is a success or failure. (iv) The probability of a success is the same for each trials since the individuals are like a random sample (p = 0.3 if we say a "success" is someone getting a lung condition, a morbid choice). Compute parts (a) and (b) from the binomial formula in Equation 3.4.9:
P(0)=(40)(0.3)0(0.7)4=1×1×0.74=0.2401
P(1)=(41)(0.3)1(0.7)3=0.4116.
Note: 0! = 1.
Part (c) can be computed as the sum of parts (a) and (b):
P(0)+P(1)=0.2401+0.4116=0.6517.(3.4.11)
That is, there is about a 65% chance that no more than one of your four smoking friends will develop a severe lung condition.
Exercise 3.4.3A
What is the probability that at least 2 of your 4 smoking friends will develop a severe lung condition in their lifetimes?
Answer
: The complement (no more than one will develop a severe lung condition) as computed in Example 3.4.3 as 0.6517, so we compute one minus this value: 0.3483.
Exercise 3.4.3B
Suppose you have 7 friends who are smokers and they can be treated as a random sample of smokers.
How many would you expect to develop a severe lung condition, i.e. what is the mean?
What is the probability that at most 2 of your 7 friends will develop a severe lung condition.
Answer a
: μ = 0.3 \times 7 = 2.1.
Answer b
: P(0, 1, or 2 develop severe lung condition) = P(k = 0)+P(k = 1)+P(k = 2) = 0:6471.
Below we consider the first term in the binomial probability, n choose k under some special scenarios.
Exercise 3.4.3C
Why is it true that (n0)=1 and (nn)=1 for any number n?
Solution
Frame these expressions into words. How many different ways are there to arrange 0 successes and n failures in n trials? (1 way.) How many different ways are there to arrange n successes and 0 failures in n trials? (1 way.)
Exercise 3.4.3D
How many ways can you arrange one success and n -1 failures in n trials? How many ways can you arrange n -1 successes and one failure in n trials?
Solution
One success and n - 1 failures: there are exactly n unique places we can put the success, so there are n ways to arrange one success and n - 1 failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations:
(n1)=n,(nn−1)=n(3.4.12)
Normal Approximation to the Binomial Distribution
The binomial formula is cumbersome when the sample size (n) is large, particularly when we consider a range of observations. In some cases we may use the normal distribution as an easier and faster way to estimate binomial probabilities.
Example 3.4.4
Approximately 20% of the US population smokes cigarettes. A local government believed their community had a lower smoker rate and commissioned a survey of 400 randomly selected individuals. The survey found that only 59 of the 400 participants smoke cigarettes. If the true proportion of smokers in the community was really 20%, what is the probability of observing 59 or fewer smokers in a sample of 400 people?
Solution
We leave the usual verification that the four conditions for the binomial model are valid as an exercise.
The question posed is equivalent to asking, what is the probability of observing k=0,1,…,58,or59 smokers in a sample of n = 400 when p = 0.20? We can compute these 60 different probabilities and add them together to nd the answer:
P(k=0ork=1or…ork=59)(3.4.13)
=P(k=0)+P(k=1)+⋯+P(k=59)(3.4.14)
=0.0041(3.4.15)
If the true proportion of smokers in the community is p = 0.20, then the probability of observing 59 or fewer smokers in a sample of n = 400 is less than 0.0041. The computations in Example 3.50 are tedious and long. In general, we should avoid such work if an alternative method exists that is faster, easier, and still accurate. Recall that calculating probabilities of a range of values is much easier in the normal model. We might wonder, is it reasonable to use the normal model in place of the binomial distribution? Surprisingly, yes, if certain conditions are met.
Exercise 3.4.4
Here we consider the binomial model when the probability of a success is p = 0.10. Figure 3.17 shows four hollow histograms for simulated samples from the binomial distribution using four different sample sizes: n = 10, 30, 100, 300. What happens to the shape of the distributions as the sample size increases? What distribution does the last hollow histogram resemble?
Solution
The distribution is transformed from a blocky and skewed distribution into one that rather resembles the normal distribution in last hollow histogram
Normal approximation of the binomial distribution
The binomial distribution with probability of success p is nearly normal when the sample size n is sufficiently large that np and n(1 - p) are both at least 10. The approximate normal distribution has parameters corresponding to the mean and standard deviation of the binomial distribution:
μ=npσ=np(1−p)−−−−−−−−√(3.4.16)
The normal approximation may be used when computing the range of many possible successes. For instance, we may apply the normal distribution to the setting of Example 3.50.
Example 3.4.5
How can we use the normal approximation to estimate the probability of observing 59 or fewer smokers in a sample of 400, if the true proportion of smokers is p = 0.20?
Solution
Showing that the binomial model is reasonable was a suggested exercise in Example 3.50. We also verify that both np and n(1- p) are at least 10:
np=400×0.20=80n(1−p)=400×0.8=320(3.4.17)
With these conditions checked, we may use the normal approximation in place of the binomial distribution using the mean and standard deviation from the binomial model:
μ=np=80σ=np(1−p)−−−−−−−−√=8(3.4.18)
We want to find the probability of observing fewer than 59 smokers using this model.
Exercise 3.4.5
Use the normal model N(μ=80,σ=8) to estimate the probability of observing fewer than 59 smokers. Your answer should be approximately equal to the solution of Example 3.50: 0.0041.
Answer
: Compute the Z score rst: Z = 59−808=−2.63. The corresponding left tail area is 0.0043.
Caution: The normal approximation may fail on small intervals
The normal approximation to the binomial distribution tends to perform poorly when estimating the probability of a small range of counts, even when the conditions are met.
The normal Approximation Breaks down on small intervals
Suppose we wanted to compute the probability of observing 69, 70, or 71 smokers in 400 when p = 0.20. With such a large sample, we might be tempted to apply the normal approximation and use the range 69 to 71. However, we would nd that the binomial solution and the normal approximation notably differ:
Binomial:0.0703Normal:0.0476(3.4.19)
We can identify the cause of this discrepancy using Figure 3.18, which shows the areas representing the binomial probability (outlined) and normal approximation (shaded). Notice that the width of the area under the normal distribution is 0.5 units too slim on both sides of the interval.
TIP: Improving the accuracy of the normal approximation to the binomial distribution
The normal approximation to the binomial distribution for intervals of values is usually improved if cutoff values are modified slightly. The cutoff values for the lower end of a shaded region should be reduced by 0.5, and the cutoff value for the upper end should be increased by 0.5.
The tip to add extra area when applying the normal approximation is most often useful when examining a range of observations. While it is possible to apply it when computing a tail area, the benefit of the modification usually disappears since the total interval is typically quite wide.
Contributors and Attributions
David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University)
3.3: Geometric Distribution (Special Topic)
3.5: More Discrete Distributions (Special Topic) |
13664 | https://planetmath.org/derivingthetrigonometricadditionformulaeusingareaandcosinerule1 | Deriving the trigonometric addition formulae using area and cosine rule
Deriving the trigonometric addition formulae using area and cosine rule
Abstract
The trigonometric addition formulae are very important and useful in Mathematics. They can be derived from geometric proof, relations from analytical geometry, Euler formula or relations from vectorial analysis. In this work, we prove the sine addition formula by considering area. Using cosine rule, we prove the cosine addition formula. Our proof is simpler because it requires the knowledge of area of triangles and cosine rule which are easily understood by students. From the addition formulae, we obtain the difference formulae by solving simultaneous equations.
Keywords
Trigonometric addition and difference formulae, area of triangles, cosine rule
Deriving the trigonometric addition formulae using area and cosine rule
Deriving the trigonometric addition formulae using area and cosine rule
1 Introduction
The sine and cosine addition and difference formulae are given by
sin(α±β)=sin α cos β±sin β cos α sin(α±β)=sinαcosβ±sinβcosα(1)
and
cos(α±β)=cos α cos β∓sin α sin β,cos(α±β)=cosαcosβ∓sinαsinβ,(2)
respectively. The sine and cosine addition and diference formulae can be obtained from geometric proofs , Euler’s formula, analytical geometry and vectorial analysis . Feldman uses the cosine rule and Pythagoras theorem to get the cosine difference formula. In this work, we consider the area of triangles to obtain the sine addition formula since the area of any triangle depends on the sine of the angle. Then, using the cosine rule twice, we obtain the cosine addition formula. Finally, we obtain the trigonometric difference formulae from addition formulae by solving two simultaneous equations.
2 Preliminaries
We use the important formulas and identities:
From Fig. (1), we have
Figure 1: Triangle XYZ
area of△XYZ=1 2 x y sin θ area of△XYZ=1 2xysinθ
and the cosine rule gives
z 2=x 2+y 2−2 x y cos θ.z 2=x 2+y 2-2xycosθ.
The simple trigonometric identities are given by
sin(180∘−θ)=sin θ,sin(180∘-θ)=sinθ,(3)
cos(180∘−θ)=−cos θ,cos(180∘-θ)=-cosθ,(4)
sin 2 θ+cos 2 θ=1.sin 2θ+cos 2θ=1.(5)
3 Derivation of Addition Trigonometric Formulae
Figure 2: Quadrilateral ABCD
Fig. (2) shows a quadrilateral ABCD which consists of two right-angled triangles A B C ABC and A C D ACD. In △A B C△ABC, C ˆ A B=α CA^B=α, A C=s AC=s. By simple trigonometry, we have
A B=s cos α,B C=s sin α.AB=scosα,BC=ssinα.(6)
In △A C D△ACD, C ˆ A D=β CA^D=β, A D=h AD=h. We have the following trigonometric equations:
cos β=s h,C D=h sin β.cosβ=s h,CD=hsinβ.(7)
We also note that B ˆ C D=180∘−α.BC^D=180∘-α.
We can express the area of quadrilateral ABCD as:
Area of△A B C+Area of△A C D=Area of△A B D+Area of△B C D Area of△ABC+Area of△ACD=Area of△ABD+Area of△BCD
1 2(A B)(B C)+1 2(A C)(C D)=1 2(A B)h sin(α+β)1 2(AB)(BC)+1 2(AC)(CD)=1 2(AB)hsin(α+β)
+1 2(C D)(B C)sin(180∘−α).+1 2(CD)(BC)sin(180∘-α).(8)
Substituting eqs. (3), (6) and (7) into eq. (3), we have, after simplifications,
1 2 s 2 sin α cos α+1 2 h s sin β=1 2 h s cos α sin(α+β)+1 2 h s sin β sin 2 α.1 2s 2sinαcosα+1 2hssinβ=1 2hscosαsin(α+β)+1 2hssinβsin 2α.(9)
Dividing eq. (9) by 1 2 h s,1 2hs, we obtain
s h sin α cos α+sin β=cos α sin(α+β)+sin β sin 2 α.s hsinαcosα+sinβ=cosαsin(α+β)+sinβsin 2α.
Simplifying and using eq. (5), we have
cos α sin(α+β)cosαsin(α+β)==s h sin α cos α+sin β(1−sin 2 α),s hsinαcosα+sinβ(1-sin 2α),(10)
==s h sin α cos α+sin β cos 2 α.s hsinαcosα+sinβcos 2α.
Dividing eq. (10) by cos α cosα and using eq. (7), we finally obtain the sine addition formula given by eq. (1) with the ++ sign.
We next consider △B C D△BCD. By using cosine rule and eqs. (4), (6) and (7), we have
B D 2 BD 2==C D 2+B C 2−2(C D)(B C)cos(180∘−α),CD 2+BC 2-2(CD)(BC)cos(180∘-α),(11)
==h 2 sin 2 β+s 2 sin 2 α+2 h s sin β sin α cos α.h 2sin 2β+s 2sin 2α+2hssinβsinαcosα.
Using cosine rule in △A B D△ABD and eq. (11), we obtain
cos(α+β)cos(α+β)==h 2+A B 2−B D 2 2 h(A B)h 2+AB 2-BD 2 2h(AB)(12)
==h 2(1−sin 2 β)+s 2(cos 2 α−sin 2 α)−2 h s sin β sin α cos α 2 h s cos α.h 2(1-sin 2β)+s 2(cos 2α-sin 2α)-2hssinβsinαcosα 2hscosα.
Using eq. (5) and s 2=h 2 cos 2 β s 2=h 2cos 2β, eq. (12) reduces to
cos(α+β)cos(α+β)==h 2 cos 2 β+s 2(cos 2 α−sin 2 α)−2 h s sin β sin α cos α 2 h s cos α h 2cos 2β+s 2(cos 2α-sin 2α)-2hssinβsinαcosα 2hscosα(13)
==s 2(1+cos 2 α−sin 2 α)−2 h s sin β sin α cos α 2 h s cos α s 2(1+cos 2α-sin 2α)-2hssinβsinαcosα 2hscosα
==2 s 2 cos 2 α−2 h s sin β sin α cos α 2 h s cos α 2s 2cos 2α-2hssinβsinαcosα 2hscosα
==s h cos α−sin α sin β.s hcosα-sinαsinβ.
Substituting eq. (7) into eq. (13), we finally obtain cosine addition formula given by eq. (2) with the ++ sign. We point out the proof of the cosine addition formula is a generalization to that of Feldman who used s=h=1.s=h=1.
4 Derivation of Trigonometric Difference Formulae
The difference formulas can be obtained by replacing β β by −β-β in eqs. (1) and (2).
But we adopt a different approach. We set γ=α+β γ=α+β so that β=γ−α.β=γ-α.
Then eqs. (1) and (2) reduce to
sin(γ)=sin α cos(γ−α)+cos α sin(γ−α)sin(γ)=sinαcos(γ-α)+cosαsin(γ-α)(14)
and
cos γ=cos α cos(γ−α)−sin α sin(γ−α),cosγ=cosαcos(γ-α)-sinαsin(γ-α),(15)
respectively. We obtain the trigonometric difference formulae by solving eqs. (14) and (15) simultaneously.
(Unknown node type: a)×cos α−(Unknown node type: a)×sin α(Unknown node type: a)×cosα-(Unknown node type: a)×sinα yields
sin γ cos α−cos γ sin α=(sin 2 α+cos 2 α)sin(γ−α)sinγcosα-cosγsinα=(sin 2α+cos 2α)sin(γ-α)
so that
sin(γ−α)=sin γ cos α−cos γ sin α sin(γ-α)=sinγcosα-cosγsinα(16)
using eq. (5).
Similarly, (Unknown node type: a)×sin α+(Unknown node type: a)×cos α(Unknown node type: a)×sinα+(Unknown node type: a)×cosα results in
cos(γ−α)=cos γ cos α+sin γ sin α.cos(γ-α)=cosγcosα+sinγsinα.(17)
References
| Title | Deriving the trigonometric addition formulae using area and cosine rule |
| Canonical name | DerivingTheTrigonometricAdditionFormulaeUsingAreaAndCosineRule1 |
| Date of creation | 2013-03-11 19:55:01 |
| Last modified on | 2013-03-11 19:55:01 |
| Owner | dkrbabajee (19083) |
| Last modified by | (0) |
| Numerical id | 1 |
| Author | dkrbabajee (0) |
| Entry type | Definition |
Generated on Fri Feb 9 21:12:49 2018 by LaTeXML |
13665 | https://books.google.com/books/about/Alice_in_Puzzle_land.html?id=JcRkt2A1_RoC | Alice in Puzzle-land: A Carrollian Tale for Children Under Eighty - Raymond M. Smullyan, Martin Gardner, Greer Fitting - Google Books
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Alice in Puzzle-land: A Carrollian Tale for Children Under Eighty ================================================================= Raymond M. Smullyan, Martin Gardner, Greer Fitting Courier Corporation, 2011 - Mathematics - 182 pages "A charming and challenging adventure." — Wordplay, the Crossword Blog of The New York Times Alice and her friends return for another romp through Wonderland and the Looking-Glass with these eighty-eight puzzles, paradoxes, and logic problems. Raymond M. Smullyan's characters speak and behave like the originals, and their puzzles abound in typical Carrollian word play, logic problems, and dark philosophical paradoxes. Isaac Asimov described this book as "amusing, entertaining, and surprisingly educational. And it might just send you back to reread Alice." Readers of all ages will delight in the charming stories and the wealth of ingenious puzzles. Written by a distinguished mathematician and creator of popular puzzle books, this volume requires no background in formal logic. The puzzles become progressively more complex, and complete solutions appear at the end. Puzzle authority Martin Gardner provides an Introduction to the text, which is enhanced by sixty charming illustrations. "An ingenious book," declared the Boston Globe, "magnificent for those who like conundrums, amusing for those who don't, and a tribute in itself to the genius of Lewis Carroll." More » Preview this book »
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Table of Contents
Common terms and phrases
accusedAlice thoughtAlice'sanswered yesasked Aliceawakebarberbelief is correctbelieve the statementBill the Lizardblack cardbrotherscarrying a blackcarrying a redCaterpillarCHAPTERCheshire CatclaimedconvictedCookcoursecried AliceDodoDormousedreamingDuchessEpimenidesexactlyfalse statementFish-FootmanFrog-FootmanGoatguiltyHearts ClubhenceinnocentJabberwockyJohnjudgeKing believesKing is asleepKnave of HeartsknewliedLooking-Glass logician believeslyingmiles an hourminutesMock TurtleMurdochneverpossibleQueen of HeartsquestionRaymond Smullyanred cardRed KingRed Queenrememberreplied Alicereplied Humpty Dumptyreplied the Gryphonreplied the Kingreplied the KnightROUNDsanesecond defendantshavesolved the problemstorySupposetellthingthird defendantthought Alicethree defendantstold the truthtrialtrue statementTweedledeeTweedledum and TweedledeeTweedledum carryingTypeWhite KnightWonderlandwrong
About the author(2011)
Raymond Smullyan received his PhD from Princeton University and has taught at Dartmouth, Princeton, Indiana University, and New York's Lehman College. Best known for his mathematical and creative logic puzzles and games, he is also a concert pianist and a magician.
Raymond Smullyan: The Merry PranksterRaymond Smullyan (1919– ), mathematician, logician, magician, creator of extraordinary puzzles, philosopher, pianist, and man of many parts. The first Dover book by Raymond Smullyan was First-Order Logic (1995). Recent years have brought a number of his magical books of logic and math puzzles: The Lady or the Tiger (2009); Satan, Cantor and Infinity (2009); an original, never-before-published collection, King Arthur in Search of His Dog and Other Curious Puzzles (2010); and Set Theory and the Continuum Problem (with Melvin Fitting, also reprinted by Dover in 2010). More will be coming in subsequent years.
In the Author's Own Words:"Recently, someone asked me if I believed in astrology. He seemed somewhat puzzled when I explained that the reason I don't is that I'm a Gemini."
"Some people are always critical of vague statements. I tend rather to be critical of precise statements: they are the only ones which can correctly be labeled 'wrong.'" — Raymond Smullyan
Critical Acclaim for The Lady or the Tiger:"Another scintillating collection of brilliant problems and paradoxes by the most entertaining logician and set theorist who ever lived." — Martin Gardner
Bibliographic information
Title Alice in Puzzle-land: A Carrollian Tale for Children Under Eighty
Dover Recreational Math
AuthorsRaymond M. Smullyan, Martin Gardner, Greer Fitting
Edition illustrated, reprint
Publisher Courier Corporation, 2011
ISBN 0486482006, 9780486482002
Length 182 pages
SubjectsMathematics
›
Recreations & Games
Games / Puzzles
Humor / General
Mathematics / Logic
Mathematics / Recreations & Games
Export CitationBiBTeXEndNoteRefMan
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13666 | https://www.xconvert.com/unit-converter/kilograms-to-pounds | Kilograms to Pounds | Convert kg To lb Online - XConvert
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Kilograms (kg) to Pounds (lb) conversion
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Kilograms to Pounds conversion table
| Kilograms (kg) | Pounds (lb) |
--- |
| 0 | 0 |
| 1 | 2.2046226218488 |
| 2 | 4.4092452436976 |
| 3 | 6.6138678655463 |
| 4 | 8.8184904873951 |
| 5 | 11.023113109244 |
| 6 | 13.227735731093 |
| 7 | 15.432358352941 |
| 8 | 17.63698097479 |
| 9 | 19.841603596639 |
| 10 | 22.046226218488 |
| 20 | 44.092452436976 |
| 30 | 66.138678655463 |
| 40 | 88.184904873951 |
| 50 | 110.23113109244 |
| 60 | 132.27735731093 |
| 70 | 154.32358352941 |
| 80 | 176.3698097479 |
| 90 | 198.41603596639 |
| 100 | 220.46226218488 |
| 1000 | 2204.6226218488 |
How to convert kilograms to pounds?
Converting between kilograms (kg) and pounds (lbs) is a common task, especially in contexts involving weight measurements in different regions. Here's how to perform these conversions accurately.
Kilograms to Pounds Conversion
The conversion factor between kilograms and pounds is based on the definition:
1 kg≈2.20462 lbs 1 \text{ kg} \approx 2.20462 \text{ lbs}
This means that one kilogram is approximately equal to 2.20462 pounds.
Step-by-Step Conversion: Kilograms to Pounds
To convert kilograms to pounds, multiply the number of kilograms by the conversion factor:
Pounds=Kilograms×2.20462\text{Pounds} = \text{Kilograms} \times 2.20462
For example, to convert 1 kg to pounds:
1 kg×2.20462=2.20462 lbs 1 \text{ kg} \times 2.20462 = 2.20462 \text{ lbs}
Pounds to Kilograms Conversion
To convert pounds to kilograms, you divide the number of pounds by the conversion factor:
Kilograms=Pounds 2.20462\text{Kilograms} = \frac{\text{Pounds}}{2.20462}
Step-by-Step Conversion: Pounds to Kilograms
For example, to convert 1 lb to kilograms:
1 lb 2.20462≈0.453592 kg\frac{1 \text{ lb}}{2.20462} \approx 0.453592 \text{ kg}
Law and Interesting Facts
International Prototype Kilogram (IPK): Until 2019, the kilogram was defined by a physical artifact known as the International Prototype Kilogram, a platinum-iridium cylinder stored at the International Bureau of Weights and Measures (BIPM) in France.
Redefinition of the Kilogram: On May 20, 2019, the kilogram was redefined based on fundamental constants of nature, specifically the Planck constant (h h), which provides a more stable and universally accessible definition. NIST - Kilogram: Introduction
Isaac Newton: Although not directly related to the kg-lbs conversion, Isaac Newton's work on gravity and mass laid the foundation for understanding these measurements. His laws of motion provide the basis for relating mass, force, and acceleration.
Real-World Examples
Shipping and Logistics:
Converting the weight of parcels or freight from kilograms to pounds is crucial for international shipping. For instance, a package weighing 15 kg would be approximately 33.07 lbs.
Cooking and Baking:
While many recipes use grams and ounces, converting between kilograms and pounds is essential in larger-scale food production. For instance, a recipe requiring 2 kg of flour translates to about 4.41 lbs.
Medical Field:
Patient weights are often recorded in both kilograms and pounds. If a patient weighs 70 kg, this is about 154.32 lbs.
Fitness and Exercise:
Weights for exercises are often listed in kilograms, while some fitness enthusiasts might prefer pounds. A 20 kg barbell is equivalent to approximately 44.09 lbs.
Purchasing Goods Internationally:
When buying products from other countries that list weights in kilograms, you might want to convert them to pounds for easier understanding. For example, buying 5 kg of coffee beans is approximately 11.02 lbs.
See below section for step by step unit conversion with formulas and explanations. Please refer to the table below for a list of all the Pounds to other unit conversions.
What is Kilograms?
Kilogram (kg) is the base unit of mass in the International System of Units (SI). It is a fundamental unit used to measure the amount of matter in an object. Unlike units like length or time, the kilogram's definition has historically been tied to a physical artifact.
Defining the Kilogram: From Artifact to Fundamental Constant
The IPK: A Piece of Platinum-Iridium
For over a century, the kilogram was defined by the International Prototype Kilogram (IPK), a platinum-iridium cylinder stored at the International Bureau of Weights and Measures (BIPM) in France. This meant that the mass of every kilogram in the world was traceable back to this single object.
The Problem with the Artifact
The IPK was not a perfect standard. Over time, it was observed to have changed mass slightly compared to its official copies. This instability, however small, was a major concern for scientific measurements that require utmost precision.
The New Definition: Based on Planck's Constant
On May 20, 2019, the kilogram was redefined based on fundamental constants of nature. The new definition is linked to the Planck constant (h h), a cornerstone of quantum mechanics. The Planck constant has a fixed numerical value when expressed in SI units:
h=6.62607015×10−34 k g⋅m 2⋅s−1 h = 6.62607015 × 10^{-34} kg⋅m^2⋅s^{-1}
By fixing the value of h h, scientists can realize the kilogram through experiments involving quantum phenomena. This change provides a far more stable and reproducible definition than relying on a physical object. The experiment that is commonly used to realize the kilogram based on Planck's constant is called a Kibble balance.
Mass vs. Weight
It's important to distinguish between mass and weight. Mass (measured in kilograms) is the amount of matter in an object, while weight is the force exerted on that object due to gravity. Weight is measured in Newtons (N). The relationship between mass and weight is:
W e i g h t=m a s s×g r a v i t y Weight = mass × gravity or W=m g W = mg
Where:
W W is weight (in Newtons)
m m is mass (in kilograms)
g g is the acceleration due to gravity (approximately 9.81 m/s 2 m/s^2 on Earth's surface)
Kilograms in Everyday Life: Examples
Food: Groceries are often sold by the kilogram, such as fruits, vegetables, and meat.
Human Body Weight: People often measure their body mass in kilograms.
Construction Materials: Cement, sand, and other building materials are often bought and sold by the kilogram or metric ton (1000 kg).
Shipping and Logistics: The weight of packages and cargo is a crucial factor in shipping costs and logistics.
Interesting Facts
Prefixes: Kilogram is unique in that it's the only SI base unit with a prefix already in its name ("kilo," meaning 1000).
The Kibble Balance: The Kibble balance (also known as a watt balance) is the instrument used to realize the new definition of the kilogram by linking mass to the Planck constant. For more information, you can read about the NIST's Kibble Balance.
What is Pounds?
The pound (lb) is a unit of mass used in the imperial and United States customary systems of measurement. It's widely used for measuring weight and mass in everyday applications. Let's delve into the details of what defines a pound.
Definition and History
The international avoirdupois pound, which is the standard pound used today, is defined as exactly 0.45359237 kilograms (kg).
1 lb=0.45359237 kg 1 \text{ lb} = 0.45359237 \text{ kg}
Historically, the pound has evolved through various forms and definitions. The avoirdupois pound, derived from the Old French "avoir de pois" (goods of weight), became the standard for general merchandise. The Troy pound, another historical variant, was used for precious metals and gemstones.
How Pounds are Formed
The current definition of the pound is directly tied to the kilogram, which is the base unit of mass in the International System of Units (SI). This means the pound's mass is derived from the kilogram's definition, ensuring a consistent standard. Previously, the pound was linked to a physical artifact which over time became degraded and inaccurate.
Notable Associations and Laws
While there isn't a specific "law" directly associated with the pound itself, the measurement is critical in physics, engineering, and commerce. Isaac Newton's laws of motion, for instance, use mass as a fundamental property. Pounds (or kilograms) are used to quantify this mass in calculations. Pounds are also used in various legal contexts to define maximum weight for vehicles or for packaging requirements of consumer products.
Real-World Examples
Body weight: People commonly measure their weight in pounds.
Food products: Packaged foods in the United States often list their weight in pounds or ounces (where 16 ounces = 1 pound).
Shipping and logistics: The weight of packages is crucial for determining shipping costs.
Construction materials: Materials like lumber, cement, and steel are often bought and sold based on weight in pounds.
Exercise Equipment: Weights used in gyms and fitness centers are typically measured in pounds.
For more information on the history of measurement check out NIST website.
Complete Kilograms conversion table
Enter # of Kilograms
| Convert 1 kg to other units | Result |
--- |
| Kilograms to Micrograms (kg to mcg) | 1000000000 |
| Kilograms to Milligrams (kg to mg) | 1000000 |
| Kilograms to Grams (kg to g) | 1000 |
| Kilograms to Metric Tonnes (kg to mt) | 0.001 |
| Kilograms to Ounces (kg to oz) | 35.27396194958 |
| Kilograms to Pounds (kg to lb) | 2.2046226218488 |
| Kilograms to Stones (kg to st) | 0.1574730444178 |
| Kilograms to Tons (kg to t) | 0.001102311310924 |
Mass conversions
Kilograms to Micrograms (kg to mcg)
Kilograms to Milligrams (kg to mg)
Kilograms to Grams (kg to g)
Kilograms to Metric Tonnes (kg to mt)
Kilograms to Ounces (kg to oz)
Kilograms to Pounds (kg to lb)
Kilograms to Stones (kg to st)
Kilograms to Tons (kg to t) |
13667 | https://www.youtube.com/watch?v=rhrqDAEDNsM | Converting ratios to percentage change
H. Gilbert Welch
1830 subscribers
58 likes
Description
4387 views
Posted: 9 Jun 2016
A primer for converting common medical ratios (relative risks, relative rates, hazard ratios) into percentage change (30% lower, 50% higher).
8 comments
Transcript:
Intro hi I'm dr. Gill Welch in this short take we'll talk about converting ratios to percentage change not that exciting a topic but in fact it's pretty useful thing to understand the reason is We love ratio measures in medicine because we love ratio measures in medicine we have relative risks relative rates hazard ratios standardized mortality ratios odds ratios all of these are ratios something about the event frequency in the group of interest over the event frequency in the comparison group event frequency may sound little odd and the most familiar measure of an event frequency is a probability so it may be the probability of heart attack in the group of interest over the probability of heart attack in the comparison group that's one ratio or the probability of cancer in one group versus another another ratio or the ratio of the probability of hip fracture in one group versus another or the ratio of the probability of death in one group versus another the general forum it's the probability of an event in one group versus another or some enough time you use the term the probability of an outcome now that is a ratio of two numbers x / wok and let's consider what that ratio means let's say that ratio is 1 what is What does a ratio of 1 mean? a ratio of 1 mean well it means the probability is the same in both groups the numerator in the denominator must be the same the probabilities must be the same how about a ratio of greater than 1 what is a ratio of greater than 1 me well it What does a ratio of greater than 1 mean? if it means the probability is higher in the group of interest than the comparison group it means the numerator is bigger than the denominator and how about a ratio of less than 1 what is a ratio of less than one mean What does a ratio of less than 1 mean? well it means the probability is lower in the group of interest than in the comparison group if numerator must be smaller than the denominator now probabilities can involve a lot of numbers things like let's see the five-year risk of heart attack is two per thousand that's a lot of numbers so let's work with a simpler quantity oh yeah money that's always easy let's consider my retirement account balance now versus my retirement account balance in 2000 let's make some real numbers here we got an x over a wide and that produces a ratio now ratio of 1 means my balance has stayed the same a ratio of greater than 1 means my current balance is higher than my 2000 balance and a ratio of less than 1 means my current balance is lower than my 2000 balance didn't have to get that small did it alright let's say I have 8,000 in my retirement account balance now it was $10,000 that produces a ratio of 0.8 what percent of my money have I lost well I lost two thousand from ten thousand so that's two thousand over ten thousand equals point two or twenty percent but that information is contained in the ratio of 0.8 a ratio point 8 always means that the numerator is 20 percent smaller than the denominator what's the algebra one minus point eight equals point two which equals 20 percent let's say my account balance now is $6,500 it was $10,000 that produces a ratio of 0.65 what percent of my money What percent of my money have I lost? have I lost well I've lost thirty five hundred dollars that's ten thousand minus sixty-five hundred dollars I lost it from ten thousand dollars so thirty five hundred dollars over ten thousand dollars is 0.35 or thirty five percent but that information is contained in the ratio of 0.65 0.65 always means the numerator is thirty-five percent smaller than the denominator what's the algebra one minus point six five equals point three five or thirty five percent what's the general lesson for ratios less than one the percentage decrease numerator relative to the denominator is one minus the ratio well these examples have been pretty painful for me I don't like to lose money okay if I make some let's say my retirement count balance now is $11,000 while it was $10,000 that produces a ratio of 1.1 what's my percentage game well I gained a thousand dollars from $10,000 or a thousand over 10,000 was 0.1 or 10% but that information is contained in the ratio of 1.1 a ratio of 1.1 always means that the numerator is 10 percent larger than the denominator what's the algebra it's one point one minus one equals point one or ten percent let's say my account balance now is $12,500 and I started with $10,000 that produces a ratio of 1.25 what's my percentage game well I gained $2,500 from $10,000 so 2500 over 10,000 is 0.25 or 25% but that information is contained in the ratio of 1.25 which always means the numerator is 25% larger than the denominator what's the algebra 1.25 minus 1 equals What's the algebra? 0.25 or 25% what's the general lesson for ratios greater than 1 the percentage increase numerator relative to the denominator is the ratio minus 1 note the subtle difference here I have a table and if the ratio is less than 1 that means the direction is a decrease numerator relative to the denominator and the percentage change is 1 minus the ratio if the ratio is greater than 1 that implies the direction is an increased numerator relative to the denominator and the percentage change is the ratio minus 1 well is it okay if I really make some money let's try that let's say my retirement balance now is $30,000 it used to be $10,000 that's a ratio of three what's my percentage gain well you What's my percentage gain? could say three minus one equals two and two is 200% increase but people have trouble once you talk about increases greater than a hundred percent what is it two hundred percent increase it has a way of confusing people wouldn't it be more understandable if I simply said I tripled my money I think so how about if my account balance now is $80,000 and I started with $10,000 that's a ratio of eight what's my percentage gain well you could say a seven hundred percent increase but wouldn't it be more understandable if I simply said my account increased eight full where does that leave us well I think there's a little bit more subtlety let's change that to ratios greater than one to less than two use the ratio minus one but once that ratio gets greater than two you've got a big increase forget the percentage change go times as high how does this all play out in medicine How does this all play out in medicine? here I'll give you three findings an RR point nine for heart attacks following aspirin versus placebo well the interpretation there is that patients taking aspirin have a 10% decreased risk of heart attack one minus point nine equals point one or 10% decrease or an RR one point three for breast cancer in women with early versus late menarchy that means women with early menarchy have a 30% increased risk of breast cancer relative to those with late menarchy that's 1.3 - one four point three and how about an RR of twenty four lung cancer and smokers versus non-smokers well we could say a 1900 percent increase but we wouldn't we'd say smokers have 20 times the risk of lung cancer as non-smokers these are all ratio measures and we should know that expressions of relative change like these tend to exaggerate both the perception of risk and the perception of benefit so I'm not necessarily advocating their use but you will see them a lot and they can be very misleading what you really want is to get the absolute numbers what you should know medicine is full of What you should know ratio measures big ratios to 3/10 are easy although we don't see them that often the numerator must be double triple or ten times the denominator ratios near one take a little bit more work 1.25 means the numerator is 25% higher that's the ratio minus one while the ratio of 0.75 means the numerator is 25% lower that's one minus the ratio well that's it you're now done I hope this helps Thanks |
13668 | https://www.cut-the-knot.org/Curriculum/Calculus/MVT.shtml | Rolle's and The Mean Value Theorems
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Rolle's and The Mean Value Theorems
The Mean Value Theorem (MVT, for short) is one of the most frequent subjects in mathematics education literature. It is one of important tools in the mathematician's arsenal, used to prove a host of other theorems in Differential and Integral Calculus. As a curiosity, it is most frequently derived as a consequence of its own special case -- Rolle's theorem. The latter is named after Michel Rolle (1652-1719), a French mathematician who established the now common symbol for the n th root and insisted that -a > -b, for positive a and b, a < b. The feat went against Descartes' teaching and laid the groundwork for the introduction of the ubiquitous number line.
Rolle's Theorem
Let f be continuous on a closed interval[a, b] and differentiable on the open interval(a, b). If f(a) = f(b), then there is at least one point c in (a, b) where f'(c) = 0.
(The tangent to a graph of f where the derivative vanishes is parallel to x-axis, and so is the line joining the two "end" points (a, f(a)) and (b, f(b)) on the graph. The line that joins to points on a curve -- a function graph in our context -- is often referred to as a secant. Thus Rolle's theorem claims the existence of a point at which the tangent to the graph is parallel to the secant, provided the latter is horizontal.)
Mean Value Theorem
Let f be continuous on a closed interval[a, b] and differentiable on the open interval(a, b). Then there is at least one point c in (a, b) where
(1)f'(c) = (f(b) - f(a)) / (b - a).
(The Mean Value Theorem claims the existence of a point at which the tangent is parallel to the secant joining (a, f(a)) and (b, f(b)). Rolle's theorem is clearly a particular case of the MVT in which f satisfies an additional condition, f(a) = f(b).)
The applet below illustrates the two theorems. It displays the graph of a function, two points on the graph that define a secant and a third point in-between to which a tangent to the graph is attached. The graph and the three points on it are draggable.
As of 2018, Java plugins are not supported by any browsers (find out more). This Wolfram Demonstration, Rolle's Theorem, shows an item of the same or similar topic, but is different from the original Java applet, named 'MVT'. The originally given instructions may no longer correspond precisely.
f(x)=- 2 x +2 x+3 y value1 solutions find show c0 -4 -2 2 4 x -4 -2 2 4 y f'(0)=2
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Proof of the Mean Value Theorem
Assume Rolle's theorem. The equation of the secant -- a straight line -- through points (a, f(a)) and (b, f(b)) is given by
g(x) = f(a) + (f(b) - f(a)) / (b - a).
The line is straight and, by inspection, g(a) = f(a) and g(b) = f(b). Because of this, the difference f - g satisfies the conditions of Rolle's theorem:
(f - g)(a) = f(a) - g(a) = 0 = f(b) - g(b) = (f - g)(b).
We are therefore guaranteed the existence of a point c in (a, b) such that (f - g)'(c) = 0. But
(f - g)'(x) = f'(x) - g'(x) = f'(x) - (f(b) - f(a)) / (b - a).
(f - g)'(c) = 0 is then the same as
f'(c) = (f(b) - f(a)) / (b - a).
Remark
The above is rather a standard proof of a standard formulation. The motivation for the choice of the auxiliary function g(x) is often questioned and even considered obscure. (f - g)(x) represents the vertical distance -- difference -- between the two graphs: that of f and the secant which is the graph of g. Other functions g can serve the same purpose.
Let A, B, X denote the points (a, f(a)), (b, f(b)), and (x, f(x)), respectively. Then the distance d(x) from X to AB can be easily computed and then differentiated. A more elegant approach depends on the observation that the product of d(x) and the length |AB| of AB equals twice the area S(x) of ΔABX. This area has a simple expression in determinants:
First of all observe that not unexpectedly S(a) = S(b) (= 0.) We are thus in a position to apply Rolle's theorem. The derivative S'(x) is given by
So that S'(c) = 0 immediately implies (1).
Both Rolle's and the Mean Value Theorem are statements of pure existence. Except for claiming that point c lies in the interval (a, b), neither provides more accurate information as to its location. For this reason, generations of students found the theorems perplexing. Following J. Dieudonné, R. P. Boas argued that the following form of the MVT is both more intuitive and no less useful:
Assume the derivative f' of function f is bounded on (a, b):
m ≤ f'(x) ≤ M.
Then
(b - a)m ≤ f(b) - f(a) ≤ M(b - a).
For a function, which is the integral of its derivative, this is the same as
which, in turn, is a form of what is known as the Mean Value Theorem for Integrals:
If f is continuous on a closed interval [a, b], then there is at least one point c such that
For further information and applications see superb pages by Timothy Gowers.
References
A Century of Calculus II, T. Apostol et al (eds), MAA, 1992
J. Dieudonné, Foundations of Modern Analysis, Academic Press, 1960
Related material
Read more...
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Mistrust Intuition of the Infinite
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Area of a Circle by Rabbi Abraham bar Hiyya Hanasi
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13669 | https://math.stackexchange.com/questions/2781790/functional-equation-fx2fy-yf2x | Skip to main content
Functional equation: f(x2+f(y))=y+f2(x).
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Functional Equation:
Solve for a given function f:R→R,
f(x2+f(y))=y+f2(x)(∗)
Note: f2(x)=f(x)⋅f(x).
Not sure if the proof is correct (I have made quite a few fake proofs):
My Attempt:
Fix x to show f is surjective, and assume f(y1)=f(y2). Then, sub y=y1 and y=y2 to show f is injective. Therefore, f is bijective.
One may find a k such that f(k)=0. Sub into (∗), giving
f(k2+f(y))=y.
Now, one may find m such that f(m)=−k2. Sub it in to obtain f(0)=m. Therefore, we have the following simultaneous equation:
f(f(0))and f(k)=−k2=0.
Set x=y=0, giving f(f(0))=f2(0). So we have f2(0)=−k2. Since the LHS≥0 and RHS≤0, then f2(0)=−k2=0 and thus, f(0)=0.
Now set x=0 to obtain that f(f(y))=y. In other words, f is an involution.
Now set y=0 to obtain f(x2)=f2(x). Thus, f2(−x)=f(x2)=f2(x). Since f is injective, then f(x)≠f(−x). Therefore, f(x)=−f(−x). In other words, f is odd.
Set y=f(x2) in (∗) to obtain that f(2x2)=2f(x2). Set x=1 and y=0 in (∗) to obtain f(1)=1. Set c=x2 to obtain f(2c)=2f(c). Therefore,
∀x>0, f(x)=x is the only solution.
Since f is an odd function, I claim that f(x)=x is certainly the only solution.
Edit: new last step
Sub x=f(z) and y=0 in (∗) gives
f(z2)=z2
Since f is an odd function, f(x)=x is the only solution.
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edited Jun 12, 2020 at 10:38
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asked May 15, 2018 at 4:44
abc...abc...
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There is a typo: Set y=f(x) in (∗) to obtain that f(2x2)=2f(x2). I guess you mean set y=f(x2).
– user260822
Commented
May 15, 2018 at 5:08
@user260822 c is a real number (which is always positive), and I fixed the typo.
– abc...
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May 15, 2018 at 5:09
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You correctly derived that f is odd, that f2(x)=f(x2), and that f is an involution.
From the latter, you have
f(x2+y)=f(y)+f2(x)
and therefore
z≥y⟹f(z)≥f(y).(1)
By induction (mimicking part of the work for Cauchy's equation),
f(nz)=nf(z)(2)
for n∈N and z≥0.
Indeed, this is clear for n=1, and with x=z√ and y=nz, we get the induction step: f(z+nz)=f(nz)+f2(z√)=nf(z)+f(z√2)=(n+1)f(z).
Using that f is odd, (2) also holds for z<0, as well as for all n∈Z. From 0≠f(1)=f(12)=f(1)2, we get f(1)=1, hence f(n)=n for all n∈Z.
Now let x be any real with mn≤x≤m+1n with m∈Z, n∈N. m≤nx<m+1 and by (1)
m=f(m)≤f(nx)=nf(x)≤f(m+1)=m+1
i.e.
mn≤f(x)≤m+1n.
As we can make the rational approximations arbitrarily good, we conclude that
f(x)=x
for all x.
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answered May 15, 2018 at 5:56
Hagen von EitzenHagen von Eitzen
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But isn't it possible to make rational approximations in Cauchy's function equation? So if it works here, why doesn't work in Cauchy's functional equations? Please explain as I am a bit confused.
– abc...
Commented
May 16, 2018 at 8:53
Since Hagen showed that the function is monotonic, then he found that f(any irrational number) is bounded by rational numbers. and then by the density of rational numbers, it has to be continuous.
– mastrok
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May 18, 2018 at 4:26
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Most of your proof are correct. However, your last claim is not justified.
The following is a simpler proof. Once you have proven that f2(x)=f(x2) and f(f(y))=y, you have
f(x2+f(f(y)))=f(x2+y)=f(y)+f(x2)
Then by replacing x2→x (but keeping that x>0 in mind) you have
f(x+y)=f(x)+f(y)
with x>0. Everything now is similar to Cauchy's functional equation.
Also with f(1)=1 and f(−x)=−f(x), you can be able to fix the freedom in Cauchy's solution and prove that f(x)=x for all rational x.
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edited May 15, 2018 at 5:48
answered May 15, 2018 at 5:33
mastrokmastrok
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But Cauchy has a lot more solutions ...
– Hagen von Eitzen
Commented
May 15, 2018 at 5:36
He has f(1)=1 and f(−x)=−f(x) therefore completely fix the freedom in Cauchy's problem.
– mastrok
Commented
May 15, 2018 at 5:37
Sorry, I still do not see where you pulled continuity from.
– Hagen von Eitzen
Commented
May 15, 2018 at 5:39
Oh, good point. There is no such assumption in the problem. Then my statement above only applies to rational number.
– mastrok
Commented
May 15, 2018 at 5:42
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How many distinct 2d arrangements of k k lines in general position are there?
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By 2d line arrangements in general position I mean arrangements as in the lazy caterer's sequence. By distinct I mean up to affine transformations. Also lines are allowed to be moved, but without crossing intersections of other lines.
My feeling is that this is a super hard question, but I am sure that people investigated this, so I am mostly interested in references. Also I would like to know if there is an OEIS sequence.
If I made no mistake, there should be a single isomorphism class for k=1,2,3,4 k=1,2,3,4. For k=5 k=5, I counted four arrangements:
It seems that I forgot at least one:
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euclidean-geometry
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edited Jan 13 at 19:04
Bipolar MindsBipolar Minds
asked Jan 13 at 8:38
Bipolar MindsBipolar Minds
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Not sure, but maybe OEIS A006248 and related sequences (e.g. OEIS A018242) might be relevant for your problem.Fabius Wiesner –Fabius Wiesner 2025-01-13 14:09:37 +00:00 Commented Jan 13 at 14:09
Thx! I am not sure either, according to this reference there is only one isomorphism class for k=5 k=5 so they might have a weaker notion of isomorphy Bipolar Minds –Bipolar Minds 2025-01-13 14:14:37 +00:00 Commented Jan 13 at 14:14
Is it oeis.org/A090338 ?Gerry Myerson –Gerry Myerson 2025-01-13 19:52:32 +00:00 Commented Jan 13 at 19:52
Yes, I also just found it :) Feel free to answer though Bipolar Minds –Bipolar Minds 2025-01-13 19:53:58 +00:00 Commented Jan 13 at 19:53
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is "Number of ways of arranging n n straight lines in general position in the (affine) plane." It appears that the exact number is only known for 0≤n≤8 0≤n≤8. Some references to the literature are given, but no formulas, and no estimates for how fast the number grows.
The numbers for 0≤n≤8 0≤n≤8 are 1,1,1,1,1,6,43,922,38609 1,1,1,1,1,6,43,922,38609.
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answered Jan 14 at 3:42
Gerry MyersonGerry Myerson
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13671 | https://users.sussex.ac.uk/~dienes/inference/Neyman%20Pearson.html | Neyman Pearson
Back to book
How to control your long run error rates
This page provides a few useful resources and web links for understanding and using standard (i.e. Neyman Pearson) statistics. You should read Chapter Three of Understanding psychology as a science first as an introduction to the issues. This webpage introduces some further technical details so you can apply the ideas to research more easily.
To test your intuitions concerning Bayesian versus Orthodox statistics try this quiz.
For more explication : Dienes, Z. (2011). Bayesian versus Orthodox statistics: Which side are you on?Perspectives on Psychological Sciences, 6(3), 274-290.
The aim in traditional (Neyman Pearson) statistics is to use decision procedures with known controlled long term error rates for accepting and rejecting hypotheses. Two hypotheses are formulated to pit against each other: the null and the alternative. The outcome of the decision procedure is to accept one hypothesis and reject the other. You can be in error by rejecting the null when it is actually true (Type I error) or accepting the null when it is actually false (Type II error). The error rates for both types of error must be controlled.
1. Controlling Type I errors.
Students are in general taught about controlling type I errors rates: Given the normal training of psychologists, it may seem the whole objective of any statistical procedure is to tell you whether p < .05 or not. It is this sentiment that has led to some of the frequent criticisms of significance testing. (See also Rozebooms' classic "the fallacy of the null hypothesis significance test".) To some extent these criticisms reflect the practice of only trying to control Type I errors and not employing the rest of the Neyman Pearson logic. Of course, many criticisms are not about the misuse of Neyman Pearson, but its very conceptual basis, as we discuss in the book.
When more than one significance test is conducted the question arises as to the long term error rate of the group ('family') of tests as a whole. The familywise error rate is the probability of falsely rejecting at least one null. Bonferroni is a generic way of controlling familywise error rate: If your family consists of k tests, use .05/k as the significance level for each individual test. For example if you were conducting three t-tests, you could reject any individual null only if its p< .05/3 = .017. With such a decision procedure, you would reject one or more of three true nulls no more than 5% of the time in the long run. Bonferonni controls familywise error rate to be no more than .05, but it does so more severely than needed: Type II error rates are higher than necessary. A decision procedure that controls familywise error just as well without increasing Type II as much as standard Bonferroni is this sequential Bonferroni procedure, which can be used for a family of independent significance tests (whether t-tests, correlations, chi squareds, etc). Take your k p-values and order them from smallest to largest. For example, if your p-values were .024, 001,.12, and .022, (so k = 4 in this example) you would order them:
p(1) = .001
p(2) = .022
p(3) = .024
p(4) = .12
Next, construct a threshold value for each p-value: p(1)'s threshold is.05/k, p(2)'s is 05/(k-1), p(3)'s is .05/(k-2). . . and so on, comparing p(k) with .05. So for our example:
Actual___Threshold
p(1)=0.001_____ .05/4 = .0124
p(2)=0.022_____.05/3 = .017
p(3)=0.024_____.05/2 = .025
p(4)=0.12______.05/1 = .05
Next, start at the bottom of the table and check if the last value there is smaller than its threshold. It is not in this case, p(4) = .12 is greater than .05, so this test is non-signfiicant. Move up to the next level and check. Here p(3)= .024 is less than .025 so it is significant and all p-values above it in the table are automatically signfiicant too whether they exceed their threshold or not . For example, p(2) is significant even though it is higher than its threshold (p(2) = .022 > .017). p(2) would not have been significant if the other tests below it had not been. (Of course, it is a general property of Neyman Pearson testing that the rejection or acceptance of a hypothesis depends on other testing of no evidential relation to the hypothesis under consideration.) This decision procedure is guaranteed to control familywise error rate at .05 for a set of independent tests, but you see it can result in more tests being declared significant than Bonferroni, so it has a lower Type II error rate. With Bonferroni testing, only p(1) would be declared significant. Note it follows from the sequential procedure that if you have a set of k tests and the largest p is still significant at the .05 level, then you can reject all nulls and still control familywise error rate at 5%.
With a group of tests, there is no logical reason why it has to be familywise error rate one controls. Why control specifically the probability of at least one false rejection of a null? Benjamini and Hochberg (1995) recommended instead controlling the expected proportion of erroneous rejections amongst all rejections which they called the false discovery rate (FDR). (See Chapter Three p. 63 for why uncorrected individual significance tests do not control this rate!) FDR can be controlled by ranking one's p-values as before from i = 1 . . .k. Then test each one against i.05/k, and, as for the sequential Bonferroni test described above, find the highest i such that p(i) is below its threshold. That p-value is significant as are all those smaller than it. So the procedure is the same as for the sequential Bonferroni above except the thresholds are different. For example, consider the four p-values: .001, .022, .031, .12. The thresholds are 1) 05/4, as for both standard and sequential Bonferroni procedures; 2) 2.05/4 = .025; 3) 3.05/4 = .0375; and 4) 4.05/4 = .05 as for the sequential Bonferroni. Notice that the the thresholds for p(2) and p(3) are higher than with the sequential Bonferroni procedure. That is, controlling FDR produces less Type II errors than controlling familywise error. The difference in sensitvity becomes even greater as the number of tests increases, which is why in situations where very large number of tests are employed, like brain imaging with fMRI where many brain regions are considered at once, FDR is often used to control long run Type I error rates for groups of tests. With the current example p(3) is below its threshold so p(1) to p(3) are significant, whereas with the sequential Bonferroni test only p(1) would be significant. You can decide to control FDR rather than familywise error rate in all multiple testing situations.
In Neyman Pearson you can change the testing procedure according to whether you strongly predicted a direction of an effect in advance. With two tailed tests one rejects the null if the outcome is extreme in either direction. For example, for a normally distributed z-score in the long run a true null would produce a z score greater than 1.96 2.5% of the time and also less than -1.96 2.5% of the time. Thus, one can reject the null if the obtained z score is greater than 1.96, whatever its sign, and hence control alpha at 5%. That is a two tailed test becuase you consider both tails of the distribution in setting up your rejection region. If one only considered it plausible there could be a difference greater than zero, you could declare you will reject the null if z > 1.64 and accept the null otherwise. 5% of the area of a normal lies beyond a z-score of 1.64. This is a classic one-tailed test because only one tail is considered, in this case the positive tail. Deciding to perform a one-tailed test amounts to declaring one would not reject the null no matter how strongly the results came out in the other direction. Would you really just accept the null if you obtained a z of e.g. -5.6? One response is to define an asymmetric rejection region, but not one as extreme as 0% of the area in one tail and 5% in the other.For example, to allow a sufficiently extreme outcome in the wrong direction legitimating the conclusion there is an effect in that direction one could use 0.1% of the aea in the negative tail and 4.9% in the positive tail; then one rejects the null if z is less than -3.75 or greater than 1.66. Or one could use 1% of the area in the negative tail and 4% in the positive tail, and still control overall alpha at 5%. In the latter case, one rejects the null if z is less than -2.32 or greater than 1.75. If you were looking at output for e.g. t-tests or correlations which gave you two-tailed p-values, halve the given p-value to get the one-tailed area. Thus a displayed two-tailed p-value less than .08 for an effect in the right direction would be significant with the latter rejection rule.
2. Controlling Type II errors.
As we discuss in the book, the most common conceptual error in applying classic statistics is ignoring sensitivity. Sensitivity can be determined by power and confidence intervals. To calculate power, it is easy to download and use Gpower.A tutorial on its use is here. Most statistical packages (e.g. SPSS) will report confidence intervals; a site for calculating confidence intervals is here.
Power calculations involve determing the minimally interesting effect size in standard units. For a t-test, effect size can be measured by Cohen's d. For a unrelated t-test, Cohen's d is the difference between the means of the groups divided by their pooled standard deviation, SDp. If the standard deviation in group one is SD1 and in group two is SD2, then SDp is defined by SDp squared= 0.5(SD1 squared + SD2 squared). For a related t-test, an equivalent measure (called dz in the related case) is the mean difference between conditions divded by the standard deviation of the difference scores. (For each subject find their difference in scores between the conditions. Make a column of these difference scores. Cohen's dz is the mean of this column divided by its standard deviation.) Note there are other measures of effect size, like correlation coefficients.
Based on his experience with psychology research at the time, Cohen (1988) gave rough arbitrary criteria for effect sizes, calling a d of 0.2 small, 0.5 medium, and 0.8 large for the betwen subjects case. Some papers report effect sizes as a matter of course for each test. If a paper has a between subjects design, it will generally report means and standard deviations so you can calculate effect size. (If it reports means and standard errors for each group, remember standard deviation = standard errorsquare root of number of subjects). A within subjects design can be more tricky, because you want the standard deviation of the differences, not the standard deviation of the score in each condition, and papers generally report only the latter. But if the authors of a paper report a t-test, t= dzsquare root of number of subjects, so you can get dz. From dz you can also recover the standard deviation of the differences given you know the mean difference.
The main point in looking at other papers and their effect size is help give you an idea of the typical standard deviations - subject variability - in a certain domain of investigation and the sort of mean differences various manipulations produce. These facts should feed your intuitions concerning what size of differences can be expected for other manipulations, and what size of differences can be expected on different theories. There is a danger that a measure of effect size, like Cohen's d, will be used to bypass deep thinking. You could follow a mechanical procedure: Set up power to detect a medium effect size, and then follow this rule for whatever experiment you run. While such a rule is a step up from current mechanical procedure used by many (in which power is ignored altogether), it is a short cut that should be used only in the genuine absence of good prior information. What you really should be doing is getting to know your literature so that you can make an informed estimate of the sort of mean difference and variability you could expect given a theory under test and a type of manipulation.
Power is something you decide on before running an experiment in order to determine subject number. Some statistical programs give automatic power calculations along with significance tests. These power calculations are worthless. They determine a power to detect the effect size actually measured, and this is a straightforward function of the p value. You get no information from these calculations. Power should be calculated for the effect size you are interested in detecting. Further, it is a property of your decision procedure, which is decided in advance. Baguley (2004) is a good summary of power and its misuse.
The best way to determine the sensitivity of your experiment after you have collected data is with confidence intervals. They tell you the set of values you reject as possible population means and the set you still hold under consideration. In fact, Chapter Three (page 73) describes ways you can use the confidence interval to decide when to stop running subjects while controlling error rates. Confidence intervals naturally allow you to assess if the effects allowed by the data are larger than some minimal interesting amount.
Rather than contrasting two specific values against each other as if they were the only possible values under consideration (the logic of hypothesis testing, which requires at least one specific value), confidence intervals reject and accept whole intervals of possible values, which in general makes more sense. Geoff Cumming provides free software for calculating confidence intervals.
For a collection of quotes by Ronald Fisher and Fisher's 1925 book online.
For tutors: A lecture on Neyman Pearson concepts I gave our undergraduates this year, as an introduction to chapter three of the book. Please feel free to adapt for your own purposes. I ask students to discuss each question with the person sitting next to them. Lecture lasts about an hour.
An essay I set students is the following: " For a paper published this year, consider the extent to which the authors strictly followed the demands of Neyman-Pearson hypothesis testing. Discuss whether or not substantial conclusions drawn from the data were compromised by either not adhering to the Neyman-Pearson approach, or adhering to it too strictly."
Guidance: "Did the authors set out their pre-set alpha and beta error rates in advance? Authors in general can be taken to be using an alpha rate of 5% by default, but authors rarely state their acceptable beta rate. Did they indicate what size of effect they would expect if their theory were true - and then use it to determine power? Is it clear what stopping rule was used? Widely varying sample sizes for effects just reaching significance may indicate numbers were topped up in some experiments to get the results significant. Was power (or confidence intervals) taken into account in interpreting null results? If a set of tests were addressing a single theme was there a correction for mutliple testing? It is somewhat arbitrary whether a set of tests counts as a "family" for which one should control familywise error rate, but at least when a set of tests is conducted where any one of them being significant means a "yes" answer to a single question (e.g. "Does the drug work at any time point?"), then the set is a family. Other mistakes include using p vallues to measure or compare sizes of effects or interpreting p values to mean probability of hypotheses."
See also this assessment of several topics from the book. |
13672 | https://content.byui.edu/file/b8b83119-9acc-4a7b-bc84-efacf9043998/1/Math-1-7-3.html | Simplifying Fractions
HomeMathWritingLanguageSkillsTutoring
Fractions:
Simplifying Fractions
To simplify a fraction, we do the prime factorization of the numerator and denominator and any numbers that are on the top and the bottom will “cancel out”, which means they divide to equal 1. We often cross out the numbers that cancel out and get rid of them, the following video will show why we can do this.
Video Source (07:01 mins) | Transcript
The following video goes through more examples of how to simplify fractions.
Video Source (04:42 mins) | Transcript
Remember, even if you cancel everything in the numerator or the denominator it doesn’t mean it is 0. There is still a 1 there. Anything multiplied to 1 is itself, so even when we divide out everything else, we will always have a 1 left.
Additional Resources
Khan Academy: Visualizing Equivalent Fractions (03:44 mins, Transcript)
Khan Academy: More on Equivalent Fractions (04:53 mins, Transcript)
Khan Academy: Introduction to Equivalent Fractions (additional links on the left of the webpage) (04:17 mins, Transcript)
Practice Problems
Simplify the following fractions to the lowest terms:
4 6
10 25
4 7
30 48
42 70
12 84
View Solutions
Solutions
4 6=2×2 3×2=2 3 (Written Solution) This fraction bar represents 4 out of 6 or 4/6. Four are shaded orange out of a total of six.
To simplify the fraction, divide the numerator and denominator by 2.
To simplify the fraction bar, group 2 rectangles into one bar. This is the same as to divide by 2.
The resulting fraction is 2 3; and 2 out of 3 bars are shaded orange.
10 25=2×5 5×5=2 5 (Written Solution)
We can simplify a fraction if the number in the numerator and the number in the denominator share at least one common factor.
We must first find the factorizations of these two numbers.
Prime factorization of 10:
10 is divisible by 2 since it is an even number.
2 × 5 = 10
Both the number 2 and the number 5 are prime numbers so the prime factorization of 10 is 2 × 5
Prime factorization of 25:
25 ends in a 5 and any number that ends in a 5 or a 0 is divisible by 5.
5 × 5 = 25
The number 5 is a prime number so the prime factorization of 25 is simply 5 × 5.
Now that we have the prime factorization of both the numerator and the denominator, we can rewrite the fraction like this: 10 25=2×5 5×5
Note:A dot between numbers represents multiplication. This dot is centered between the numbers and higher than a decimal point.
Since the numerator and the denominator both share one 5, and since 5 5=1 we can rewrite the fraction as 2×5 5×5=2 5×1
Anything multiplied by 1 is just itself, so 2 5×1=2 5.
There are no other common factors between 2 and 5 so this fraction is as simplified as it can be.
Our final solution: 2 5
4 7 is already simplified to lowest terms. (Written Solution) The 4 and 7 do not share a common factor other than 1, therefore this answer is written in its simplest or lowest terms.
30 48=2×3×5 2×2×2×2×3=5 8 (Solution Video | Transcript)
42 70=2×3×7 2×7×7=3 5 (Solution Video | Transcript)
12 84=2×2×3 2×2×3×7=1 7 (Written Solution)
If the numerator and denominator of a fraction have any common factors, the fraction can be simplified.
First, we will find the prime factorization of the numerator:
12 is divisible by 2 because it is an even number and all even numbers are divisible by 2.
2 × 6 = 12
6 is not prime so now we must factor 6 as well.
6 is also even so we can divide it by 2 as well.
2 × 3 = 6
The numbers 2 and 3 are both prime, so this is as far as we can factor our number.
The prime factorization of 12 is 2 × 2 × 3.
Next, we will find the prime factorization of the denominator:
84 is even so we will start by dividing it by 2.
2 × 42 = 84
The number 2 is prime but 42 is not so we still need to find the factors of 42.
42 is also even so we can divide it by 2 as well.
2 × 21 = 42
The number 2 is prime but 21 is not prime, so we need to factor 21 as well.
21 is divisible by 3.
3 × 7 = 21
Both 3 and 7 are prime, so this is as far as we can factor.
The prime factorization of 21 is 2 × 2 × 3 × 7.
Now we can use the prime factorizations to determine if there are any common factors in the numerator and the denominator.
Note:A dot between numbers represents multiplication. This dot is centered between the numbers and higher than a decimal point.
12 84=2⋅2⋅3 2⋅2⋅3⋅7 Here we see that both the numerator and the denominator have two 2s and a 3 in common. We can rewrite the fraction like this:
12 84=2×2×3 2×2×3×7=2 2×2 2×3 3×1 7
Note: you might think that there is only 0 left in the numerator after separating out the 2 × 2 × 3, but really, there is still a 1 because if there was a 0 in the numerator, the numerator would equal 0. There is always an invisible 1 being multiplied to everything. 2 × 2 × 3 × 1 = 12.
Anything divided by itself is equal to 1.
2 2⋅2 2⋅3 3⋅1 7=1⋅1⋅1⋅1 7
Anything multiplied by 1 is just itself, so we are just left with 1 7.
12 84 simplifies to 1 7. |
13673 | https://web.uvic.ca/~gmacgill/guide/GenFuncs.pdf | Generating Functions If you take f(x) = 1/(1 −x −x2) and expand it as a power series by long division, say, then you get f(x) = 1/(1−x−x2) = 1+x+2x2 +3x3 +5x4 +8x5 +13x6 +· · ·. It certainly seems as though the coefficient of xn in the power series is the n-th Fibonacci number.
This is not difficult to prove directly, or using techniques discussed below. You can think of the process of long division as generating the coefficients, so you might want to call 1/(1 −x −x2) the generating function for the Fibonacci numbers. The actual definition of generating function is a bit more general. Since the closed form and the power series represent the same function (within the circle of convergence), we will regard either one as being the generating function.
Definition of generating function. The generating function for the sequence a0, a1, . . .
is defined to be the function f(x) = ∞ n=0 anxn.
That is, the generating function for the sequence a0, a1, . . . is the function whose power series representation has an as the coefficient of xn.
We’ll call a0, a1, . . . the sequence generated by f(x). We will not be concerned with matters of convergence, and instead treat these as formal power series.
Perhaps “symbolic” would be a better word than formal.
When determining the sequence generated by a generating function, you will want to get a formula for the n-the term (that is, for the coefficient of xn), rather than just computing numerical values for the first few coefficients.
Useful facts.
• If x ̸= 1 then 1 + x + x2 + · · · + xr = xr+1−1 x−1 (although it is often best not to do anything with sums of only a few terms).
• 1 1−x = 1 + x + x2 + · · ·.
Both of these facts can be proved by letting S be the sum, calculating xS −S, and doing a little bit of algebra.
The second fact above says that 1 1−x is the generating function for the sequence 1, 1, 1, 1, . . ..
It also lets you determine the sequence generated by many other functions. For example: • 1 1−ax = 1 + a + a2x2 + a3x3 + · · · = ∞ n=0 anxn.
To see this, substitute y = ax into the series expansion of 1 1−y. Thus 1 1−ax is the generating function for the sequence a0, a1, a2, . . .. The coefficient of xn is an.
1 • 1 1+ax = 1 −a + a2x2 −a3x3 + · · · = ∞ n=0(−1)nanxn. To see this, substitute y = (−a)x into the series expansion of 1 1−y. Thus 1 1+ax is the generating function for the sequence a0, −a1, a2, . . .. The coefficient of xn is (−a)n = (−1)nxn.
More useful facts.
• 1 1−x k = ∞ n=0 n+k−1 n xn.
• 1 1−ax k = ∞ n=0 n+k−1 n anxn.
• 1 1+ax k = ∞ n=0(−1)nn+k−1 n anxn.
To see the first of these, consider the LHS, (1 + x + x2 + · · ·)k, multiplied out, but not simplified, Each term is a product of k (possibly different) powers of x (remembering that x0 = 1). By the rules of exponents, xn1xn2 · · · xnk = xn1+n2+···+nk. Thus, after simplifying, there is a term xn for every way of expressing n as a sum of k numbers each of which is one of 0, 1, 2, . . .. That is, the number of terms xn equals the number of integer solutions to n1 + n2 + · · · + nk = n, 0 ≤ni, ∀i. We know that this number is n+k−1 k−1 (count using “bars and stars”).
To see the second fact, let y = ax in 1 1−y k. (This is the same idea as before.) To see the third fact, let let y = −ax in 1 1−y k.
Multiplying a generating function by a constant multiplies every coefficient by that con-stant. For example, 3 1+5x = 3 1 1+5x = 3(1 −5x + 52x2 −53x3 + · · ·) = 3 −3 · 5x + 3 · 52x2 − 3 · 53x3 + · · · = ∞ n=0 3 · (−1)n5nxn. The coefficient of xn is 3 · (−1)n5n.
Multiplying a generating function by xk “shifts” the coefficients by k. This has the effect of introducing k zeros at the start of the sequence generated. For example, x 1+7x 4 = x4 1 1+7x 4 = x4 ∞ n=0(−1)nn+4−1 n 7nxn = ∞ n=0(−1)nn+4−1 3 7nxn+4 = ∞ t=4(−1)t−4t−1 3 7t−4xt.
(Let t = n + 4 in the second last sum. If n = 0 then t = 4, so that gives the lower limit for the sum. If n →∞, so does t, which gives the upper limit. And n = t −4.) Since (−1)t−4 = (−1)t, the last sum equals: ∞ t=4(−1)tt−1 3 7t−4xt. The coefficient of xt is zero if t < 4, and (−1)tt−1 3 7t−4 if t ≥4.
If you add two generating functions together, the coefficient of xn in the sum is what you would expect, the sum of the coefficients of xn in the summands. For example, adding the generating functions from the above two paragraphs gives: 2 3 1+5x + x 1+7x 4 = ∞ n=0 3(−1)n5nxn + ∞ n=4(−1)nn−1 3 7n−4xn = 3 −3(51)x + 3(52)x2 −3(53)x3 + ∞ n=4 3(−1)n5n + (−1)nn−1 3 7n−4 xn = 3 −3(51)x + 3(52)x2 −3(53)x3 + ∞ n=4(−1)n 3 · 5n + n−1 3 7n−4 xn.
Thus, if n ≤3, the coefficient of xn is 3(−1)n5n, and if n ≥4 it is (−1)n 3·5n+ n−1 3 7n−4 .
The above describes most of the basic tools you need. When trying to determine what sequence is generated by some generating function, your goal will be to write it as a sum of known generating functions, some of which may be multiplied by constants, or constants times some power of x.
Once you’ve done this, you can use the techniques above to determine the sequence.
Most of the time the known generating functions are among those described above. Occasionally you may need to directly compute the product of two generating functions: Cauchy Product.
∞ n=0 anxn∞ n=0 bnxn = a0b0 + (a0b1 + a1b0)x + (a0b2 + a1b1 + a2b0)x2 + · · · = ∞ n=0 n k=0 akbn−k xn To see this, carry out the multiplication and simplify.
The last tool you will need allows you to write a product of (known) generating functions as a sum of known generating functions, some of which may be multiplied by constants, or constants times some power of x. It is the method of partial fractions, which you should have learned while studying Calculus. We will do a couple of examples in the course of this section, but won’t give a comprehensive treatment of the method. For more info, look in (almost) any Calculus textbook.
Example 1. What sequence is generated by 2+x 1−x−8x2+12x3 ?
The first task is to factor the denominator. We want to write it as a product of terms of the form (1 −ax). Since the constant term is 1, if the a’s are all integers then they are among the divisors of the coefficient of the highest power of x, and their negatives. (This is similar to the hints for finding integer roots of polynomials discussed in the section on solving recurrences. To see it, imagine such a factorisation multiplied out and notice how the coefficient of the highest power of x arises.) Here, we find that 1 −x −8x2 + 12x3 = (1 + 3x)(1 −2x)2. Thus, we want to use partial fractions to write 2+x (1+3x)(1−2x)2 as a sum 3 of “known” generating functions. These arise from the factors in the denominator. Using partial fractions 2+x (1+3x)(1−2x)2 = A 1+3x + B 1−2x + C (1−2x)2 = A(1−2x)2+B(1+3x)(1−2x)+C(1+3x) (1+3x)(1−2x)2 Thus, 2 + x = A(1 −2x)2 + B(1 + 3x)(1 −2x) + C(1 + 3x). Expanding the RHS and equating like powers of x on the LHS and RHS gives A + B + C = 2, −4A + B + 3C = 1, 4A −6B = 0 The solution is A = 3/5, B = 2/5, C = 1. Therefore, 2+x (1+3x)(1−2x)2 = (3/5) 1 1+3x + (2/5) 1 1−2x + 1 1−2x 2.
The coefficient of xn on the RHS is (3/5)(−1)n3n + (2/5)2n + n+1 n 2n = (3/5)(−1)n3n + (2/5)2n + (n + 1)2n. Thus the sequence generated is (3/5)(−1)n3n + (2/5)2n + (n + 1)2n.
Deriving generating functions from recurrences. If you are given a sequence defined by a recurrence relation and initial conditions, you can use these to get a generating function for the sequence. Having done that, you can then apply the facts and methods above to get a formula for the n-th term of the sequence. This is another method of solving recurrences. We’ll illustrate the method first, and then try to give a description of it below. It would be wise to work through the example twice: once before reading the description of the method, and once after. That way you should be able to recognize (and understand) the major steps.
Example 2. Find the generating function for the sequence an defined by an = an−1 + 8an−2 −12an−3, n ≥3, with initial conditions a0 = 2, a1 = 3, and a3 = 19.
Let g(x) = ∞ n=0 anxn = 2 + 3x + 19x2 + ∞ n=3 anxn = 2 + 3x + 19x2 + ∞ n=3(an−1 + 8an−2 −12an−3)xn = 2 + 3x + 19x2 + ∞ n=3 an−1xn + ∞ n=3 8an−2xn −∞ n=3 12an−3xn = 2 + 3x + 19x2 + x ∞ n=3 an−1xn−1 + 8x2 ∞ n=3 an−2xn−2 −12x3 ∞ n=3 an−3xn−3 = 2 + 3x + 19x2 + x ∞ k=2 akxk + 8x2 ∞ k=1 akxk −12x3 ∞ k=0 akxk = 2+3x+19x2+ x ∞ k=0 akxk −x(2+3x)+8x2∞ k=0 akxk −8x2(2)−12x3 ∞ k=0 akxk = 2 + x + xg(x) + 8x2g(x) −12x3g(x).
Therefore g(x)−xg(x)−8x2g(x)−12xxg(x) = 2+x. That is, g(x)(1−x−8x2−12x3) = 2+x, so g(x) = 2+x 1−x−8x2−12x3 is the generating function. You can now apply the method in Example 1 (since this is the generating function from Example 1) to find that an = (3/5)(−1)n3n + (2/5)2n + (n + 1)2n.
4 The main idea is to let g(x) be the generating function, say g(x) = ∞ n=0 anxn, and then use the recurrence, initial conditions, manipulation of the sum(s), and algebra to convert the RHS into an expression which is a sum of terms some of which involve g(x) itself. The following steps are usually involved (and in this order): (1) Split the terms corresponding to the initial values from the sum on the RHS. The sum should then run from the first integer for which the recursive definition takes effect.
(2) Substitute the recursive definition of an into the summation.
(3) Split the sum into several sums, and in each one factor out the constant times a high enough power of x so that the subscript on the coefficient and the exponent of x are equal.
(4) Make a change of variable in each summation so each is the sum of anxn from some value to infinity. One of the sums should start at 0, another at one, and so on, until finally one starts at the subscript of the “last” initial value (assuming consecutive initial values are given). Also, the sum that starts at 0 should be multiplied by a constant (maybe 1), the one that starts at 1 should be multiplied by a constant (maybe 1) times x, and in general each sum that starts at t should be multiplied by a constant (maybe 1) times xt.
(5) By adding the “missing” (first few) terms, and subtracting them offoutside the sum (don’t forget to multiply by whatever is in front of the summation!), convert each sum so that it goes from 0 to infinity. That is, do algebra so that each sum involving a term of the recurrence equals g(x). If the recurrence is non-homogeneous, you may need to do the same sort of thing to convert the sum(s) arising from the non-homogeneous term(s) into known sums.
(6) Collect all terms involving g(x) on the LHS, factor g(x) out (if possible), then divide (or do whatever algebra is needed), to obtain a closed form for the generating function.
You might have noticed that the characteristic equation for the recurrence in Example 2 is x3 −x2 −8x + 12, while the denominator for the generating function is 1 −x − 8x2 −12x3.
It is always true that if the characteristic equation for a LHRRWCC is a0xk + a1xk−1 + · · · + ak−1, then the denominator for (some form of) the generating function is a0 +a1x+a2x2 +· · ·+ak−1xk. This follows from the general method described above.
Once you have the generating function, you can factor the denominator and (hopefully) use partial fractions to write it as a sum of multiples of known generating functions. By finding the coefficient of xn in each summand, you get a formula for an.
Example 3. Solve an = −3an−1 + 10an−2 + 3 · 2n, n ≥2 with initial conditions a0 = 0 5 and a1 = 6.
Let g(x) = ∞ n=0 anxn = a0 + a1x + ∞ n=2 anxn = 0 + 6x + ∞ n=2(−3an−1 + 10an−2 + 3 · 2n)xn = 6x −3x ∞ n=2 an−1xn−1 + 10x2 ∞ n=2 an−2xn−2 + 3 ∞ n=2 2nxn = 6x −3x ∞ k=1 akxk + 10x2 ∞ k=0 akxk + 3 ∞ n=2 2nxn = 6x −3x ∞ k=1 akxk −(−3x)(0) + 10x2 ∞ k=0 akxk + 3 ∞ n=2 2nxn = 6x −3xg(x) + 10x2g(x) + 3 ∞ n=0 2nxn −3(1 + 2x) Therefore, g(x)(1 + 3x −10x2) = −3 + 3 1−2x, so g(x) = −3 1+3x−10x2 + 3 (1−2x)(1+3x−10x2) = −3 1 (1+5x)(1−2x) + 3 1 (1−2x)2(1+5x) = 3 1−(1−2x) (1−2x)2(1+5x) = 6x (1−2x)2(1+5x).
We need to expand the RHS using partial fractions.
6x (1−2x)2(1+5x) = A 1−2x + B (1−2x)2 + C 1+5x = A(1−2x)(1+5x)+B(1+5x)+C(1−2x)2 (1−2x)2(1+5x) .
Thus 6x = A(1 −2x)(1 + 5x) + B(1 + 5x) + C(1 −2x)2. Expanding the RHS and then equating coefficients of like powers of x yields: A + B + C = 0, 3A + 5B −4C = 6, −10A + 4C = 0 The solution is A = −12/49, B = 6/7, C = −30/49. Thus, g(x) = (−12/49) 1 1−2x + (6/7) 1 (1−2x)2 + (−30/49) 1 1+5x, so an = (−12/49)2n + (6/7) n+1 1 2n −(30/49)(−1)n5n = (−12/49)2n + (6/7)(n + 1)2n + (30/49)(−1)n+15n.
Using generating functions to solve counting problems. In showing that (1 + x + x2 + · · ·)k = ∞ n=0 n+k−1 k−1 xn we argued that the coefficient of xn equals the number of ways of writing n as a sum of k integers, each of which occurs as an exponent in one of the factors on the LHS. Stated slightly differently, 1 1−x k is the generating function for the number of ways to write n as a sum of k non-negative integers.
Continuing this line of reasoning, the generating function for the number of ways to write n as a sum of five odd integers would be (x + x3 + x5 + · · ·)5 = [x(1 + x2 + x4 + · · ·)]5 = x 1−x2 5. (Think about what happens with the exponents when the LHS is multiplied out.) Similarly, the generating function for the number of ways to write n as a sum of two odd integers and an even integer would be (x + x3 + x5 + · · ·)2(1 + x2 + x4 + · · ·) = x2 (1−x)3 .
In general, suppose you want the generating function for the number of ways to write n as an (ordered) sum of k terms, some of which may be restricted (as in being odd, for example). Then the generating function will be a product of k factors each of which is a sum of powers of x. The exponents of x in the first factor will be the possibilities for 6 the first term in sum (that adds to n), the exponents of x in the second factor will be the possibilities for the second term in the sum, and so on until, finally, the exponents in the k-th term are the possibilities for the k-th term in the sum.
Remember that x0 = 1, so a factor has a 1 if and only if zero is one of the possibilities for the corresponding term in the sum. Similarly for x1 = x, and so on.
Example 4. Determine the generating function for the number of ways to distribute a large number of identical candies to four children so that the first two children receive an odd number of candies, the third child receives at least three candies, and the fourth child receives at most two candies.
Suppose the number of candies distributed is n. Then we are looking for the generating function for the number of ways to write n as a sum of four integers, the first two of which are odd, the third of which is at least 3, and the fourth of which is at most 2. Following the reasoning from above, the generating function is a product of four factors, and the exponents of x in each factor give the possible values for the corresponding term in the sum. Thus, the generating function is (x + x3 + x5 + · · ·)2(x3 + x4 + · · ·)(1 + x + x2) = x 1−x2 2 x3 1−x(1 + x + x2).
Once you have the generating function for the number of ways to do something, you can apply the methods you know (i.e. write the generating function as a sum of known generating functions) to determine the cooefficient of xn, and hence the number of ways.
Example 4 continued. (This gets a touch ugly at the end, but it is a good illustration of the methods, so hang with it.) Use the generating function to determine the number of ways.
The generating function is g(x) = x 1−x2 2 x3 1−x(1+x+x2) = (x2+x3+x4) 1 (1−x2)2(1−x) = (x2+x3+x4) 1 (1+x)2(1−x)3 .
An exercise in partial fractions shows that: 1 (1+x)2(1−x)3 = (1/8) 1 1−x 2 + (3/16) 1 1−x) + (1/4) 1 1+x 3 + (1/4) 1 1+x 2 + (3/16) 1 1+x .
Since these are all known generating functions, the coefficient of xn in this expression (which is not g(x)) is (1/8)+(3/16)(n+1)+(1/4)(−1)nn+3−1 3−1 +(1/4)(−1)n(n+1)+(3/16)(−1)n.
We need the coefficient of xn in g(x), which equals 1 (1+x)2(1−x)3 multiplied by (x2+x3+x4).
Thus, the coefficient of xn in g(x) is the sum of the coefficient of xn−2 in 1 (1+x)2(1−x)3 , the coefficient of xn−3 in 1 (1+x)2(1−x)3 , and the coefficient of xn−4 in 1 (1+x)2(1−x)3 . This equals a + b + c, where 7 a = (1/8) + (3/16)(n −1) + (1/4)(−1)n−2n 2 + (1/4)(−1)n−2(n −1) + (3/16)(−1)n−2, b = (1/8) + (3/16)(n −2) + (1/4)(−1)n−2n−1 2 + (1/4)(−1)n−3(n −2) + (3/16)(−1)n−3, c = (1/8) + (3/16)(n −3) + (1/4)(−1)n−4n−2 2 + (1/4)(−1)n−4(n −3) + (3/16)(−1)n−4.
No doubt this expression can be simplified.
Example 5. Use generating functions to find bn, the number of ways that n ≥0 identical candies can be distributed among 4 children and 1 adult so that each child receives an odd number of candies, and the adult receives 1 or 2 candies.
Following the method outlined above, the generating function is g(x) = (x + x3 + x5 + · · ·)4(x + x2) = x 1−x2 4(x + x2) = (x5 + x6) ∞ k=0 k+4−1 3 x2k = ∞ k=0 k+3 3 x2k+5 + ∞ k=0 k+3 3 x2k+6 The first sum contains all terms with odd exponents, and the second sum contains all terms with even exponents. Thus, if n ≤4, bn = 0. If n ≥5 and odd, bn = n+1/2 3 (let n = 2k + 5 in the first sum). If n ≥6 and even, bn = n/2 3 (let n = 2k + 6 in the second sum). These cases can all be described by the single expression bn = ⌈n/2⌉ 3 .
Example 6. In a certain game it is possible to score 1, 2, or 4 points on each turn. Find the generating functions for the number of ways to score n points in a game in which there are at least two turns where 4 points are scored.
Here we are looking for the number of ways to write n as an ordered sum of three terms.
The first term represents the number of points obtained from turns where 1 point was scored. The second term represents the number of points obtained from turns where 2 points were scored (and thus is a multiple of 2). The third term represents the number of points obtained from turns where 4 points were scored (and thus is a multiple of 4, and at least 8). Thus, the generating function will be a product of three factors, where the exponents in each factor correspond to the possibilities just discussed. Therefore, g(x) = (1 + x + x2 + · · ·)(1 + x2 + x4 + · · ·)(x8 + x12 + x16 + · · ·) = 1 1−x 1 1−x2 x8 1−x4 .
If you want an exercise in partial fractions, continue with Example 6 and determine the number of ways by writing g(x) as a sum of known generating functions and finding the coefficient of xn. This involves a 6×6 linear system (unless you makle an astute observation that is eluding me right now).
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13674 | https://accendoreliability.com/parallel-systems/ | Parallel Systems - Accendo Reliability
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Home » Articles » CRE Preparation Notes » Parallel Systems
by Fred Schenkelberg5 Comments
Parallel Systems
Speaking reliability-wise, parallel, means any of the elements in parallel structure permit the system to function. This does not mean they are physically parallel (in all cases), as capacitors in parallel provide a specific behavior in the circuit and if one capacitor fails that system might fail.
In this simple drawing, there are n components in parallel and any one component is needed for the system to function.
If component #2 fails, the others will permit the system to functions.
Simple. And, very useful. This construction permits the improvement of reliability overall, even above the reliability of the individual components. Unlike series system where the weakest component limits the reliability, here by adding redundancy the system reliability improves.
Consider a two component parallel system. If both components are both working, then the system is working. If either component 1 or 2 fails, the system is still working. If and only if both components fail, then the system fails. Unlike a series system where any one failure causes a system failure, in this simple example, two failure events have to occur before the system fails.
What is the chance of having two failures? The formula is based on the probability of component 1 or component 2 operating. Without doing the derivation, we can write the reliability of the 2 component parallel system as:
This get’s very complicated quickly with more than three components in parallel. Before exploring another way to calculate parallel systems, there is a special case situation to mention first.
When the components in parallel are the same (reliability wise), then the above simplifies to
Which should be obvious.
Often it’s easier to do parallel system calculations using the unreliability, or 1 – R(t). This is also the CDF or F(t). The math is now more like a series system with one correction. For the general case, the system reliability formula for a parallel system becomes
So, let’s say we have five components with Reliability at one year of use, R(1), at 90%, or 0.9. Calculate the system reliability.
Using the above formula and setting the reliability of each element at 0.9, we find
which is very reliable.
It’s expensive to add redundant parts to a system, yet in some cases, it is the right solution to create a system that meets the reliability requirements. Of course, actual systems have many variations and complications over simply setting components up in parallel. Load sharing, hot, warm, or cold standby, switching or voting systems, and many others can complicate the construction of a parallel system.
Using components in parallel is the only way to increase the system reliability over the limits of individual component reliability.
Related:
Reliability Block Diagrams Overview and Value(article)
Reliability Apportionment(article)
k out of n(article)
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Filed Under: Articles, CRE Preparation Notes, Reliability Modeling and PredictionsTagged With: Reliability Block Diagram (RBD)
About Fred Schenkelberg
I am the reliability expert at FMS Reliability, a reliability engineering and management consulting firm I founded in 2004. I left Hewlett Packard (HP)’s Reliability Team, where I helped create a culture of reliability across the corporation, to assist other organizations.
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Comments
Ahmet ÇETİN says
March 21, 2019 at 12:12 PM
Dear sir,
I want to calculate total breakdown time and breakdown number between A and B points.
There are 3 machines connected parallel. Reply
2. Alain Bensoussan says
July 17, 2019 at 4:19 AM
Hi Fred,
First of all I would like to thank you very much to share your knowledge.
I have a question regarding the failure definition of n elements in a parallel system reliability. As a failure can be either open circuit of short circuit the element failure in a parallel scheme will have different impact: if a short circuit on one element and all other sounds, then the system is in short so considered fail; while the open fail of one element will not affect the overal system reliability. Can you clarify this point ?
Complementary and second question: what if we are studying parametric drifts of a n parallel elements with a given failure criteria characterized by a WEIBULL law for each element (same law or different WEIBULL parameters for each one) ? Do you know some reference or book addressing this question for a serie or parallel systems reliability?
Thank you for your advise. Alain. Reply
Fred Schenkelbergsays
July 17, 2019 at 1:42 PM
Hi Alain,
Yep, if the items in parallel and one has a short that provides a short past the parallel structure, then it’s a single point failure causing a system failure. It would take a bit more work to model this situation, yet possible to estimate the system availability if one could define the probability of shorting…
And, I do not know of a good reference for the drift question – seems there may be good technical paper in that topic.
Cheers,
Fred Reply
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13675 | https://www.e-education.psu.edu/natureofgeoinfo/c2_p28.html | Map Projections | The Nature of Geographic Information
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The Nature of Geographic Information
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Latitude and longitude coordinates specify point locations within a coordinate system grid that is fitted to sphere or ellipsoid that approximates the Earth's shape and size. To display extensive geographic areas on a page or computer screen, as well as to calculate distances, areas, and other quantities most efficiently, it is necessary to flatten the Earth.
Figure 2.28.1 Map projections are mathematical equations that transform geographic coordinates (conventionally designated by the Greek symbols lambda for longitude and phi for latitude) into plane coordinates (x and y). If all the necessary parameters are known, inverse projection equations can be used to transform projected coordinates back into unprojected geographic coordinates.
Georeferenced plane coordinate systems like the Universal Transverse Mercator and State Plane Coordinates systems (examined elsewhere in this chapter) are created by first flattening the graticule, then superimposing a rectangular grid over the flattened graticule. The first step, transforming the geographic coordinate system grid from a more-or-less spherical shape to a flat surface, involves systems of equations called map projections.
Many different map projection methods exist. Although only a few are widely used in large-scale mapping, the projection parameters used vary greatly. Geographic information systems professionals are expected to be knowledgeable enough to select a map projection that is suitable for a particular mapping objective. Such professionals are expected to be able to recognize the type, amount, and distribution of geometric distortion associated with different map projections. Perhaps most important, they need to know about the parameters of map projections that must be matched in order to merge geographic data from different sources. The pages that follow introduce the key concepts. The topic is far too involved to master in one section of a single chapter, however. Indeed, Penn State offers an entire online course in Map Projections: Spatial Reference Systems in GIS(GEOG 861). If you are or plan to become, a GIS professional, you should own at least one good book on map projections. Several recommendations follow in the bibliography at the end of this chapter.
Students who successfully complete this section should be able to:
interpret distortion diagrams to identify geometric properties of the sphere that are preserved by a particular projection;
classify projected graticules by projection family.
‹ 26. SPC Zone Characteristicsup28. Geometric Properties Preserved and Distorted ›
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Author: David DiBiase, Senior Lecturer, John A. Dutton e-Education Institute, and Director of Education, Industry Solutions, Esri. Instructors and contributors: Jim Sloan, Senior Lecturer, John A. Dutton e-Education Institute; Ryan Baxter, Senior Research Assistant, John A. Dutton e-Education Institute, Beth King, Senior Lecturer, John A. Dutton e-Education Institute and Assistant Program Manager for Online Geospatial Education, and Adrienne Goldsberry, Senior Lecturer, John A. Dutton e-Education Institute; College of Earth and Mineral Sciences, The Pennsylvania State University.
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13676 | https://www.youtube.com/watch?v=yuqB-d5MjZA | Lagrange multipliers, using tangency to solve constrained optimization
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The Lagrange multiplier technique is how we take advantage of the observation made in the last video, that the solution to a constrained optimization problem occurs when the contour lines of the function being maximized are tangent to the constraint curve.
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In the last video, I introduced a constrained optimization problem where we were trying to maximize this function f ofxy = x^2 y but subject to a constraint that your values of x and y have to satisfy x^2 + y^2 = 1. And the way we were visualizing this was to look at the xy plane where this circle here represents our constraint. all of the points that make up this set x^2 + y^2= 1. And then this curvy line here is one of the contours of f. Meaning we're setting f ofxy equal to some constant. And then I was varying around that constant c. So for high values of c, the contour would look something like this. This is where the value of x^2 y is big. And then for small values of c, the contours would look like this. So all the points on this line would be f ofxy equals like 0.01 in this case something like that. Then the way to think about maximizing this function is to try to increase that value of c as much as you can without it falling off the circle. And the key observation is that happens when they're tangent. So you know you might kind of draw this out in a little sketch and say there's some curve representing your constraint which in this case would be you know kind of where our circle is. And then the curve representing the contour would just kiss that curve. Just barely touch it in some way. Now that's pretty, but in terms of solving the problem, we still have some work to do. And the main tool we're going to use here is the gradient. So let me go ahead and draw a lot more contour lines than there already are for x^2 y. So this is many of the contour lines. And I'll draw the gradient field, the gradient field of f. So, I've made a video about the relationship between the gradient and contour lines. And the upshot of it is that these gradient vectors, every time they pass through a contour line, they're perpendicular to it. And the basic reason for that is if you walk along the contour line, the function isn't changing value. So, if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change. But again, there's a whole video on that that's worth checking out if this feels unfamiliar. For our purposes, what it means is that when we're considering this point of tangency, the gradient of f at that point is going to be some vector perpendicular to both of the curves at that point. So that that little vector represents the gradient of f at this point on the plane. And we can do something very similar to understand the other curve. Right now I've just written it as a constraint x^2 + y^2= 1. But you know to give that function a name. Let's say that we've defined g ofxy to be x^2 + y^2 x^2 + y^2. In that case, this constraint is pretty much just one of the contour lines for the function g. And we can take a look at that. If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x^2 + y^2. And if we took a look at the gradient of g and we go over and ask about the gradient of g, it has that same property that every gradient vector if it passes through a contour line is perpendicular to it. So over on our drawing here, the gradient vector of G would also be perpendicular to both these curves. And you know, maybe in this case, it's not as long as the gradient of F or maybe it's longer. There's no reason that it would be the same length, but the important fact is that it's proportional. And the way that we're going to write this in formulas is to say that the gradient of f evaluated, let's see, evaluated at whatever the maximizing value of x and y are. So we should give that a name probably maybe you know x subm y subm the specific values of x and y that are going to be at this point that maximizes the function subject to our constraint. So that's going to be related to the gradient of g. It's not going to be quite equal so I'll leave some room here. Related to the gradient of g evaluated at that same point xm ym. And like I said they're not equal. They're proportional. So, we need to have some kind of proportionality constant in there. And you almost always use the variable lambda. And this guy has a fancy name. It's called a lrangege multiplier. Lrange. Lrange was one of those uh famous French mathematicians. I always get him confused with some of the other French mathematicians at the time like uh Leandre or Lelas. There's a whole bunch of things. Uh let's see. Multiplier. Distracting myself talking here. Um so, Lrange multiplier. So there there's a number of things in in multivariable calculus named after Lrange and this is one of the big ones. This this is a technique that he uh that he kind of developed or at the very least popularized and the core idea is to just set these gradients equal to each other because that represents when the contour line for one function is tangent to the contour line of another. So this this is something that we can actually work with. Um and let's let's start working it out, right? Let's see what this translates to in formulas. So I already have g written here. So let's go ahead and just evaluate what the gradient of g should be. Um and that's that's the gradient of x^2 + y^2. And the way that we take our gradient is it's going to be a vector whose components are all the partial derivatives. So the first component is the partial derivative with respect to x. So we treat x as a variable. y looks like a constant. The derivative is 2x. The second component the partial derivative with respect to y. So now we're treating y as the variable. x is the constant. So the derivative looks like 2y. Okay. So that's the the gradient of g. Then the gradient of f gradient of f is going to look like gradient of let's see what is x. What is f? It's x^2 y. So x^2 y. We do the same thing. First component partial derivative with respect to x. x looks like a variable. So its derivative is 2 x. And then that y looks like a constant when we're up here. But then partial derivative with respect to y, that y looks like a variable. That x squ just looks like a constant sitting in front of it. So that's what we get. And now if we kind of work out this lrangee multiplier expression using these two vectors what we have written what we're going to have written is that this vector 2xy x^2 is proportional with a proportionality constant lambda to the gradient vector for g to 2x 2 y. And if you want, you can think about this as two separate equations. I mean, right now it's one equation with vectors, but really what this is saying is, you've got two separate equations. 2 xy is equal to lambda. Got to change colors a lot here. lambda 2x going to be a stickler for color. Keep keep red all of the things associated with g. And then the second equation is that x^2 is equal to lambda 2 y. And this might seem like a problem because we have three unknowns, x, y, and this new lambda that we introduced. Kind of shot ourselves in the foot by giving ourselves a new variable to deal with. Uh, but we only have two equations. So in order to solve this, we're going to need three equations. And the third equation is something that we've known the whole time. It's been part of the original problem. It's the constraint itself. x^2 + y x^2 + y^2 excuse me equals 1. So that that third equation x^2 + y^2 is equal to 1. So these are the three equations that characterize our constrained optimization problem. The bottom one just tells you that we have to be on this this unit circle here. So let me just highlight it. We have to be on this unit circle. And then these top two tell us what's necessary in order for our contour lines, the contour of f and the contour of g to be perfectly tangent with each other. So uh in the next video I'll go ahead and solve this. At this point it's pretty much just algebra to deal with, but it's it's worth going through. Uh and then the next couple ones I'll talk about a way that you can encapsulate all three of these equations into one expression. Uh and also a little bit about the interpretation of this lambda that we introduced because it's not it's not actually just a dummy variable. has a pretty interesting meaning in physical context once you're actually dealing with a constrained optimization problem in practice. |
13677 | https://www.math.uchicago.edu/~lawler/srwbook.pdf | Random Walk: A Modern Introduction Gregory F. Lawler and Vlada Limic Contents Preface page 6 1 Introduction 9 1.1 Basic definitions 9 1.2 Continuous-time random walk 12 1.3 Other lattices 14 1.4 Other walks 16 1.5 Generator 17 1.6 Filtrations and strong Markov property 19 1.7 A word about constants 21 2 Local Central Limit Theorem 24 2.1 Introduction 24 2.2 Characteristic Functions and LCLT 27 2.2.1 Characteristic functions of random variables in Rd 27 2.2.2 Characteristic functions of random variables in Zd 29 2.3 LCLT — characteristic function approach 29 2.3.1 Exponential moments 42 2.4 Some corollaries of the LCLT 47 2.5 LCLT — combinatorial approach 51 2.5.1 Stirling’s formula and 1-d walks 52 2.5.2 LCLT for Poisson and continuous-time walks 56 3 Approximation by Brownian motion 63 3.1 Introduction 63 3.2 Construction of Brownian motion 64 3.3 Skorokhod embedding 68 3.4 Higher dimensions 71 3.5 An alternative formulation 72 4 Green’s Function 75 4.1 Recurrence and transience 75 4.2 Green’s generating function 76 4.3 Green’s function, transient case 81 4.3.1 Asymptotics under weaker assumptions 84 3 4 Contents 4.4 Potential kernel 86 4.4.1 Two dimensions 86 4.4.2 Asymptotics under weaker assumptions 90 4.4.3 One dimension 92 4.5 Fundamental solutions 95 4.6 Green’s function for a set 96 5 One-dimensional walks 103 5.1 Gambler’s ruin estimate 103 5.1.1 General case 106 5.2 One-dimensional killed walks 112 5.3 Hitting a half-line 115 6 Potential Theory 119 6.1 Introduction 119 6.2 Dirichlet problem 121 6.3 Difference estimates and Harnack inequality 125 6.4 Further estimates 132 6.5 Capacity, transient case 136 6.6 Capacity in two dimensions 144 6.7 Neumann problem 152 6.8 Beurling estimate 154 6.9 Eigenvalue of a set 157 7 Dyadic coupling 166 7.1 Introduction 166 7.2 Some estimates 167 7.3 Quantile coupling 170 7.4 The dyadic coupling 172 7.5 Proof of Theorem 7.1.1 175 7.6 Higher dimensions 176 7.7 Coupling the exit distributions 177 8 Addtional topics on Simple Random Walk 181 8.1 Poisson kernel 181 8.1.1 Half space 181 8.1.2 Cube 184 8.1.3 Strips and quadrants in Z2 188 8.2 Eigenvalues for rectangles 191 8.3 Approximating continuous harmonic functions 192 8.4 Estimates for the ball 193 9 Loop Measures 198 9.1 Introduction 198 9.2 Definitions and notations 198 9.2.1 Simple random walk on a graph 201 9.3 Generating functions and loop measures 201 9.4 Loop soup 206 Contents 5 9.5 Loop erasure 207 9.6 Boundary excursions 209 9.7 Wilson’s algorithm and spanning trees 214 9.8 Examples 216 9.8.1 Complete graph 216 9.8.2 Hypercube 217 9.8.3 Sierpinski graphs 220 9.9 Spanning trees of subsets of Z2 221 9.10 Gaussian free field 230 10 Intersection Probabilities for Random Walks 237 10.1 Long range estimate 237 10.2 Short range estimate 240 10.3 One-sided exponent 243 11 Loop-erased random walk 245 11.1 h-processes 245 11.2 Loop-erased random walk 248 11.3 LERW in Zd 250 11.3.1 d ≥3 250 11.3.2 d = 2 251 11.4 Rate of growth 254 11.5 Short-range intersections 257 12 Appendix 259 12.1 Some expansions 259 12.1.1 Riemann sums 259 12.1.2 Logarithm 260 12.2 Martingales 263 12.2.1 Optional Sampling Theorem 263 12.2.2 Maximal inequality 265 12.2.3 Continuous martingales 267 12.3 Joint normal distributions 267 12.4 Markov chains 269 12.4.1 Chains restricted to subsets 272 12.4.2 Maximal coupling of Markov chains 275 12.5 Some Tauberian theory 278 12.6 Second moment method 280 12.7 Subadditivity 281 References 285 Index of Symbols 286 Index 288 Preface Random walk – the stochastic process formed by successive summation of independent, identically distributed random variables – is one of the most basic and well-studied topics in probability theory.
For random walks on the integer lattice Zd, the main reference is the classic book by Spitzer .
This text considers only a subset of such walks, namely those corresponding to increment distributions with zero mean and finite variance. In this case, one can summarize the main result very quickly: the central limit theorem implies that under appropriate rescaling the limiting distribution is normal, and the functional central limit theorem implies that the distribution of the corresponding path-valued process (after standard rescaling of time and space) approaches that of Brownian motion.
Researchers who work with perturbations of random walks, or with particle systems and other models that use random walks as a basic ingredient, often need more precise information on random walk behavior than that provided by the central limit theorems. In particular, it is important to understand the size of the error resulting from the approximation of random walk by Brownian motion. For this reason, there is need for more detailed analysis. This book is an introduction to the random walk theory with an emphasis on the error estimates. Although “mean zero, finite variance” assumption is both necessary and sufficient for normal convergence, one typically needs to make stronger assumptions on the increments of the walk in order to get good bounds on the error terms.
This project embarked with an idea of writing a book on the simple, nearest neighbor random walk.
Symmetric, finite range random walks gradually became the central model of the text.
This class of walks, while being rich enough to require analysis by general techniques, can be studied without much additional difficulty. In addition, for some of the results, in particular, the local central limit theorem and the Green’s function estimates, we have extended the discussion to include other mean zero, finite variance walks, while indicating the way in which moment conditions influence the form of the error.
The first chapter is introductory and sets up the notation. In particular, there are three main classes of irreducible walks in the integer lattice Zd — Pd (symmetric, finite range), P′ d (aperiodic, mean zero, finite second moment), and P∗ d (aperiodic with no other assumptions).
Symmetric random walks on other integer lattices such as the triangular lattice can also be considered by taking a linear transformation of the lattice onto Zd.
The local central limit theorem (LCLT) is the topic for Chapter 2. Its proof, like the proof of the usual central limit theorem, is done by using Fourier analysis to express the probability of interest 6 Preface 7 in terms of an integral, and then estimating the integral. The error estimates depend strongly on the number of finite moments of the corresponding increment distribution. Some important corollaries are proved in Section 2.4; in particular, the fact that aperiodic random walks starting at different points can be coupled so that with probability 1 −O(n−1/2) they agree for all times greater than n is true for any aperiodic walk, without any finite moment assumptions. The chapter ends by a more classical, combinatorial derivation of LCLT for simple random walk using Stirling’s formula, while again keeping track of error terms.
Brownian motion is introduced in Chapter 3. Although we would expect a typical reader to already be familiar with Brownian motion, we give the construction via the dyadic splitting method.
The estimates for the modulus of continuity are given as well. We then describe the Skorokhod method of coupling a random walk and a Brownian motion on the same probability space, and give error estimates. The dyadic construction of Brownian motion is also important for the dyadic coupling algorithm of Chapter 7.
The Green’s function and its analog in the recurrent setting, the potential kernel, are studied in Chapter 4. One of the main tools in the potential theory of random walk is the analysis of martingales derived from these functions. Sharp asymptotics at infinity for the Green’s function are needed to take full advantage of the martingale technique. We use the sharp LCLT estimates of Chapter 2 to obtain the Green’s function estimates. We also discuss the number of finite moments needed for various error asymptotics.
Chapter 5 may seem somewhat out of place. It concerns a well-known estimate for one-dimensional walks called the gambler’s ruin estimate. Our motivation for providing a complete self-contained argument is twofold. Firstly, in order to apply this result to all one-dimensional projections of a higher dimensional walk simultaneously, it is important to shiw that this estimate holds for non-lattice walks uniformly in few parameters of the distribution (variance, probability of making an order 1 positive step). In addition, the argument introduces the reader to a fairly general technique for obtaining the overshoot estimates. The final two sections of this chapter concern variations of one-dimensional walk that arise naturally in the arguments for estimating probabilities of hitting (or avoiding) some special sets, for example, the half-line.
In Chapter 6, the classical potential theory of the random walk is covered in the spirit of and (and a number of other sources). The difference equations of our discrete space setting (that in turn become matrix equations on finite sets) are analogous to the standard linear partial differential equations of (continuous) potential theory. The closed form of the solutions is important, but we emphasize here the estimates on hitting probabilities that one can obtain using them. The martingales derived from the Green’s function are very important in this analysis, and again special care is given to error terms. For notational ease, the discussion is restricted here to symmetric walks.
In fact, most of the results of this chapter hold for nonsymmetric walks, but in this case one must distinguish between the “original” walk and the “reversed” walk, i.e. between an operator and its adjoint. An implicit exercise for a dedicated student would be to redo this entire chapter for nonsymmetric walks, changing the statements of the propositions as necessary. It would be more work to relax the finite range assumption, and the moment conditions would become a crucial component of the analysis in this general setting. Perhaps this will be a topic of some future book.
Chapter 7 discusses a tight coupling of a random walk (that has a finite exponential moment) and a Brownian motion, called the dyadic coupling or KMT or Hungarian coupling, originated in K´ omlos, Major, and Tusn´ ady [7, 8]. The idea of the coupling is very natural (once explained), but 8 Preface hard work is needed to prove the strong error estimate. The sharp LCLT estimates from Chapter 2 are one of the key points for this analysis.
In bounded rectangles with sides parallel to the coordinate directions, the rate of convergence of simple random walk to Brownian motion is very fast. Moreover, in this case, exact expressions are available in terms of finite Fourier sums. Several of these calculations are done in Chapter 8.
Chapter 9 is different from the rest of this book. It covers an area that includes both classical combinatorial ideas and topics of current research. As has been gradually discovered by a number of researchers in various disciplines (combinatorics, probability, statistical physics) several objects inherent to a graph or network are closely related: the number of spanning trees, the determinant of the Laplacian, various measures on loops on the trees, Gaussian free field, and loop-erased walks.
We give an introduction to this theory, using an approach that is focused on the (unrooted) random walk loop measure, and that uses Wilson’s algorithm for generating spanning trees.
The original outline of this book put much more emphasis on the path-intersection probabilities and the loop-erased walks. The final version offers only a general introduction to some of the main ideas, in the last two chapters. On the one hand, these topics were already discussed in more detail in , and on the other, discussing the more recent developments in the area would require familiarity with Schramm-Loewner evolution, and explaining this would take us too far from the main topic.
Most of the content of this text (the first eight chapters in particular) are well-known classical results. It would be very difficult, if not impossible, to give a detailed and complete list of refer-ences. In many cases, the results were obtained in several places at different occasions, as auxiliary (technical) lemmas needed for understanding some other model of interest, and were therefore not particularly noticed by the community. Attempting to give even a reasonably fair account of the development of this subject would have inhibited the conclusion of this project. The bibliography is therefore restricted to a few references that were used in the writing of this book. We refer the reader to for an extensive bibliography on random walk, and to for some additional references.
This book is intended for researchers and graduate students alike, and a considerable number of exercises is included for their benefit. The appendix consists of various results from probability theory, that are used in the first eleven chapters but are however not really linked to random walk behavior. It is assumed that the reader is familiar with the basics of measure-theoretic probability theory.
♣The book contains quite a few remarks that are separated from the rest of the text by this typeface. They are intended to be helpful heuristics for the reader, but are not used in the actual arguments.
A number of people have made useful comments on various drafts of this book including stu-dents at Cornell University and the University of Chicago. We thank Christian Beneˇ s, Juliana Freire, Michael Kozdron, Jos´ e Truillijo Ferreras, Robert Masson, Robin Pemantle, Mohammad Ab-bas Rezaei, Nicolas de Saxc´ e, Joel Spencer, Rongfeng Sun, John Thacker, Brigitta Vermesi, and Xinghua Zheng. The research of Greg Lawler is supported by the National Science Foundation.
1 Introduction 1.1 Basic definitions We will define the random walks that we consider in this book. We focus our attention on random walks in Zd that have bounded symmetric increment distributions although we occasionally discuss results for wider classes of walks. We also impose an irreducibility criterion to guarantee that all points in the lattice Zd can be reached.
Fig 1.1. The square lattice Z2 We start by setting some basic notation. We use x, y, z to denote points in the integer lattice Zd = {(x1, . . . , xd) : xj ∈Z}. We use superscripts to denote components, and we use subscripts to enumerate elements.
For example, x1, x2, . . . represents a sequence of points in Zd, and the point xj can be written in component form xj = (x1 j, . . . , xd j). We write e1 = (1, 0, . . . , 0), . . . , ed = (0, . . . , 0, 1) for the standard basis of unit vectors in Zd. The prototypical example is (discrete time) simple random walk starting at x ∈Zd. This process can be considered either as a sum of a sequence of independent, identically distributed random variables Sn = x + X1 + · · · + Xn 9 10 Introduction where P{Xj = ek} = P{Xj = −ek} = 1/(2d), k = 1, . . . , d, or it can be considered as a Markov chain with state space Zd and transition probabilities P{Sn+1 = z | Sn = y} = 1 2d, z −y ∈{±e1, . . . ± ed}.
We call V = {x1, . . . , xl} ⊂Zd \ {0} a (finite) generating set if each y ∈Zd can be written as k1x1 + · · · + klxl for some k1, . . . , kl ∈Z. We let G denote the collection of generating sets V with the property that if x = (x1, . . . , xd) ∈V then the first nonzero component of x is positive. An example of such a set is {e1, . . . , ed}. A (finite range, symmetric, irreducible) random walk is given by specifying a V = {x1, . . . , xl} ∈G and a function κ : V →(0, 1] with κ(x1) + · · · + κ(xl) ≤1.
Associated to this is the symmetric probability distribution on Zd p(xk) = p(−xk) = 1 2 κ(xk), p(0) = 1 − X x∈V κ(x).
We let Pd denote the set of such distributions p on Zd and P = ∪d≥1Pd. Given p the corresponding random walk Sn can be considered as the time-homogeneous Markov chain with state space Zd and transition probabilities p(y, z) := P{Sn+1 = z | Sn = y} = p(z −y).
We can also write Sn = S0 + X1 + · · · + Xn where X1, X2, . . . are independent random variables, independent of S0, with distribution p. (Most of the time we will choose S0 to have a trivial distribution.) We will use the phrase P-walk or Pd-walk for such a random walk. We will use the term simple random walk for the particular p with p(ej) = p(−ej) = 1 2d, j = 1, . . . , d.
We call p the increment distribution for the walk.
Given p ∈P, we write pn for the n-step distribution pn(x, y) = P{Sn = y | S0 = x} and pn(x) = pn(0, x). Note that pn(·) is the distribution of X1 + · · · + Xn where X1, . . . , Xn are independent with increment distribution p.
♣In many ways the main focus of this book is simple random walk, and a first-time reader might find it useful to consider this example throughout. We have chosen to generalize this slightly, because it does not complicate the arguments much and allows the results to be extended to other examples. One particular example is simple random walk on other regular lattices such as the planar triangular lattice. In Section 1.3, we show that walks on other d-dimensional lattices are isomorphic to p-walks on Zd.
If Sn = (S1 n, . . . , Sd n) is a P-walk with S0 = 0, then P{S2n = 0} > 0 for every even integer n; this follows from the easy estimate P{S2n = 0} ≥[P{S2 = 0}]n ≥p(x)2n for every x ∈Zd. We will call the walk bipartite if pn(0, 0) = 0 for every odd n, and we will call it aperiodic otherwise. In the 1.1 Basic definitions 11 latter case, pn(0, 0) > 0 for all n sufficiently large (in fact, for all n ≥k where k is the first odd integer with pk(0, 0) > 0). Simple random walk is an example of a bipartite walk since S1 n +· · ·+Sd n is odd for odd n and even for even n. If p is bipartite, then we can partition Zd = (Zd)e ∪(Zd)o where (Zd)e denotes the points that can be reached from the origin in an even number of steps and (Zd)o denotes the set of points that can be reached in an odd number of steps. In algebraic language, (Zd)e is an additive subgroup of Zd of index 2 and (Zd)o is the nontrivial coset. Note that if x ∈(Zd)o, then (Zd)o = x + (Zd)e.
♣It would suffice and would perhaps be more convenient to restrict our attention to aperiodic walks. Results about bipartite walks can easily be deduced from them. However, since our main example, simple random walk, is bipartite, we have chosen to allow such p.
If p ∈Pd and j1, . . . , jd are nonnegative integers, the (j1, . . . , jd) moment is given by E[(X1 1)j1 · · · (Xd 1)jd] = X x∈Zd (x1)j1 · · · (xd)jd p(x).
We let Γ denote the covariance matrix Γ = h E[Xj 1Xk 1 ] i 1≤j,k≤d .
The covariance matrix is symmetric and positive definite.
Since the random walk is truly d-dimensional, it is easy to verify (see Proposition 1.1.1 (a)) that the matrix Γ is invertible. There exists a symmetric positive definite matrix Λ such that Γ = Λ ΛT (see Section 12.3). There is a (not unique) orthonormal basis u1, . . . , ud of Rd such that we can write Γx = d X j=1 σ2 j (x · uj) uj, Λx = d X j=1 σj (x · uj) uj.
If X1 has covariance matrix Γ = Λ ΛT , then the random vector Λ−1 X1 has covariance matrix I.
For future use, we define norms J ∗, J by J ∗(x)2 = |x · Γ−1x| = |Λ−1x|2 = d X j=1 σ−2 j (x · uj)2, J (x) = d−1/2 J ∗(x).
(1.1) If p ∈Pd, E[J (X1)2] = 1 d E[J ∗(X1)2] = 1 d E |Λ−1X1|2 = 1.
For simple random walk in Zd, Γ = d−1 I, J ∗(x) = d1/2 |x|, J (x) = |x|.
We will use Bn to denote the discrete ball of radius n, Bn = {x ∈Zd : |x| < n}, and Cn to denote the discrete ball under the norm J , Cn = {x ∈Zd : J (x) < n} = {x ∈Zd : J ∗(x) < d1/2 n}.
12 Introduction We choose to use J in the definition of Cn so that for simple random walk, Cn = Bn. We will write R = Rp = max{|x| : p(x) > 0} and we will call R the range of p. The following is very easy, but it is important enough to state as a proposition.
Proposition 1.1.1 Suppose p ∈Pd.
(a) There exists an ǫ > 0 such that for every unit vector u ∈Rd, E[(X1 · u)2] ≥ǫ.
(b) If j1, . . . , jd are nonnegative integers with j1 + · · · + jd odd, then E[(X1 1)j1 · · · (Xd 1)jd] = 0.
(c) There exists a δ > 0 such that for all x, δ J (x) ≤|x| ≤δ−1 J (x).
In particular, Cδn ⊂Bn ⊂Cn/δ.
We note for later use that we can construct a random walk with increment distribution p ∈P from a collection of independent one-dimensional simple random walks and an independent multinomial process.
To be more precise, let V = {x1, . . . , xl} ∈G and let κ : V →(0, 1]l be as in the definition of P. Suppose that on the same probability space we have defined l independent one-dimensional simple random walks Sn,1, Sn,2, . . . , Sn,l and an independent multinomial process Ln = (L1 n, . . . , Ll n) with probabilities κ(x1), . . . , κ(xl). In other words, Ln = n X j=1 Yj, where Y1, Y2, . . . are independent Zl-valued random variables with P{Yk = (1, 0, . . . , 0)} = κ(x1), . . . , P{Yk = (0, 0, . . . , 1)} = κ(xl), and P{Yk = (0, 0, . . . , 0)} = 1 −[κ(x1) + · · · + κ(xl)]. It is easy to verify that the process Sn := x1 SL1 n,1 + x2 SL2 n,2 + · · · + xl SLl n,l (1.2) has the distribution of the random walk with increment distribution p. Essentially what we have done is to split the decision as to how to jump at time n into two decisions: first, to choose an element xj ∈{x1, . . . , xl} and then to decide whether to move by +xj or −xj.
1.2 Continuous-time random walk It is often more convenient to consider random walks in Zd indexed by positive real times. Given V, κ, p as in the previous section, the continuous-time random walk with increment distribution p is the continuous-time Markov chain ˜ St with rates p. In other words, for each x, y ∈Zd, P{ ˜ St+∆t = y | ˜ St = x} = p(y −x) ∆t + o(∆t), y ̸= x, 1.2 Continuous-time random walk 13 P{ ˜ St+∆t = x | ˜ St = x} = 1 − X y̸=x p(y −x) ∆t + o(∆t).
Let ˜ pt(x, y) = P{ ˜ St = y | ˜ S0 = x}, and ˜ pt(y) = ˜ pt(0, y) = ˜ pt(x, x + y). Then the expressions above imply d dt ˜ pt(x) = X y∈Zd p(y) [˜ pt(x −y) −˜ pt(x)].
There is a very close relationship between the discrete time and continuous time random walks with the same increment distribution. We state this as a proposition which we leave to the reader to verify.
Proposition 1.2.1 Suppose Sn is a (discrete-time) random walk with increment distribution p and Nt is an independent Poisson process with parameter 1. Then ˜ St := SNt has the distribution of a continuous-time random walk with increment distribution p.
There are various technical reasons why continuous-time random walks are sometimes easier to handle than discrete-time walks. One reason is that in the continuous setting there is no periodicity.
If p ∈Pd, then ˜ pt(x) > 0 for every t > 0 and x ∈Zd. Another advantage can be found in the following proposition which gives an analogous, but nicer, version of (1.2). We leave the proof to the reader.
Proposition 1.2.2 Suppose p ∈Pd with generating set V = {x1, . . . , xl} and suppose ˜ St,1, . . . , ˜ St,l are independent one-dimensional continuous-time random walks with increment distribution q1, . . . , ql where qj(±1) = p(xj). Then ˜ St := x1 ˜ St,1 + x2 ˜ St,2 + · · · + xl ˜ St,l (1.3) has the distribution of a continuous-time random walk with increment distribution p.
If p is the increment distribution for simple random walk, we call the corresponding walk ˜ St the continuous-time simple random walk in Zd. From the previous proposition, we see that the coordinates of the continuous-time simple random walk are independent — this is clearly not true for the discrete-time simple random walk. In fact, we get the following. Suppose ˜ St,1, . . . , ˜ St,d are independent one-dimensional continuous-time simple random walks. Then, ˜ St := ( ˜ St/d,1, . . . , ˜ St/d,d) is a continuous time simple random walk in Zd. In particular, if ˜ S0 = 0, then P{ ˜ St = (y1, . . . , yd)} = P{ ˜ St/d,1 = y1} · · · P{ ˜ St/d,l = yl}.
Remark. To verify that a discrete-time process Sn is a random walk with distribution p ∈Pd starting at the origin, it suffices to show for all positive integers j1 < j2 < · · · < jk and x1, . . . , xk ∈ Zd, P{Sj1 = x1, . . . , Sjk = xk} = pj1(x1) pj2−j1(x2 −x1) · · · pjk−jk−1(xk −xk−1).
To verify that a continuous-time process ˜ St is a continuous-time random walk with distribution p 14 Introduction starting at the origin, it suffices to show that the paths are right-continuous with probability one, and that for all real t1 < t2 < · · · < tk and x1, . . . , xk ∈Zd, P{ ˜ St1 = x1, . . . , ˜ Stk = xk} = ˜ pt1(x1) ˜ pt2−t1(x2 −x1) · · · ˜ ptk−tk−1(xk −xk−1).
1.3 Other lattices A lattice L is a discrete additive subgroup of Rd. The term discrete means that there is a real neighborhood of the origin whose intersection with L is just the origin. While this book will focus on the lattice Zd, we will show in this section that this also implies results for symmetric, bounded random walks on other lattices. We start by giving a proposition that classifies all lattices.
Proposition 1.3.1 If L is a lattice in Rd, then there exists an integer k ≤d and elements x1, . . . , xk ∈L that are linearly independent as vectors in Rd such that L = {j1 x1 + · · · + jk xk, j1, . . . , jk ∈Z}.
In this case we call L a k-dimensional lattice.
Proof Suppose first that L is contained in a one-dimensional subspace of Rd. Choose x1 ∈L \ {0} with minimal distance from the origin. Clearly {jx1 : j ∈Z} ⊂L. Also, if x ∈L, then jx1 ≤x < (j + 1)x1 for some j ∈Z, but if x > jx1, then x −jx1 would be closer to the origin than x1. Hence L = {jx1 : j ∈Z}.
More generally, suppose we have chosen linearly independent x1, . . . , xj such that the following holds: if Lj is the subgroup generated by x1, . . . , xj, and Vj is the real subspace of Rd generated by the vectors x1, . . . , xj, then L ∩Vj = Lj. If L = Lj, we stop. Otherwise, let w0 ∈L \ Lj and let U = {tw0 : t ∈R, tw0 + y0 ∈L for some y0 ∈Vj} = {tw0 : t ∈R, tw0 + t1x1 + · · · + tjxj ∈L for some t1, . . . , tj ∈[0, 1]}.
The second equality uses the fact that L is a subgroup. Using the first description, we can see that U is a subgroup of Rd (although not necessarily contained in L). We claim that the second description shows that there is a neighborhood of the origin whose intersection with U is exactly the origin. Indeed, the intersection of L with every bounded subset of Rd is finite (why?), and hence there are only a finite number of lattice points of the form tw0 + t1x1 + · · · + tjxj with 0 < t ≤1; and 0 ≤t1, . . . , tj ≤1. Hence there is an ǫ > 0 such that there are no such lattice points with 0 < |t| ≤ǫ. Therefore U is a one-dimensional lattice, and hence there is a w ∈U such that U = {kw : k ∈Z}. By definition, there exists a y1 ∈Vj (not unique, but we just choose one) such that xj+1 := w + y1 ∈L. Let Lj+1, Vj+1 be as above using x1, . . . , xj, xj+1. Note that Vj+1 is also the real subspace generated by {x1, . . . , xj, w0}. We claim that L ∩Vj+1 = Lj+1. Indeed, suppose that z ∈L ∩Vj+1, and write z = s0w0 + y2 where y2 ∈Vj. Then s0w0 ∈U, and hence s0w0 = lw for some integer l. Hence, we can write z = lxj+1 + y3 with y3 = y2 −ly1 ∈Vj. But, z −lxj+1 ∈Vj ∩L = Lj. Hence z ∈Lj+1.
1.3 Other lattices 15 ♣The proof above seems a little complicated. At first glance it seems that one might be able to simplify the argument as follows. Using the notation in the proof, we start by choosing x1 to be a nonzero point in L at minimal distance from the origin, and then inductively to choose xj+1 to be a nonzero point in L \ Lj at minimal distance from the origin. This selection method produces linearly independent x1, . . . , xk; however, it is not always the case that L = {j1x1 + · · · + jkxk : j1, . . . , jk ∈Z}.
As an example, suppose L is the 5-dimensional lattice generated by 2e1, 2e2, 2e3, 2e4, e1 + e2 + · · · + e5.
Note that 2e5 ∈L and the only nonzero points in L that are within distance two of the origin are ±2ej, j = 1, . . . , 5. Therefore this selection method would choose (in some order) ±2e1, . . . , ±2e5. But, e1 + · · · + e5 is not in the subgroup generated by these points.
It follows from the proposition that if k ≤d and L is a k-dimensional lattice in Rd, then we can find a linear transformation A : Rd →Rk that is an isomorphism of L onto Zk. Indeed, we define A by A(xj) = ej where x1, . . . , xk is a basis for L as in the proposition. If Sn is a bounded, symmetric, irreducible random walk taking values in L, then S∗ n := ASn is a random walk with increment distribution p ∈Pk. Hence, results about walks on Zk immediately translate to results about walks on L. If L is a k-dimensional lattice in Rd and A is the corresponding transformation, we will call | det A| the density of the lattice. The term comes from the fact that as r →∞, the cardinality of the intersection of the lattice and ball of radius r in Rd is asymptotically equal to | det A| rk times the volume of the unit ball in Rk. In particular, if j1, . . . , jk are positive integers, then (j1Z) × · · · × (jkZ) has density (j1 · · · jk)−1.
Examples.
• The triangular lattice, considered as a subset of C = R2 is the lattice generated by 1 and eiπ/3, LT = {k1 + k2 eiπ/3 : k1, k2 ∈Z}.
Note that e2iπ/3 = eiπ/3 −1 ∈LT. The triangular lattice is also considered as a graph with the above vertices and with edges connecting points that are Euclidean distance one apart. In this case, the origin has six nearest neighbors, the six sixth roots of unity. Simple random walk on the triangular lattice is the process that chooses among these six nearest neighbors equally likely.
Note that this is a symmetric walk with bounded increments. The matrix A = 1 −1/ √ 3 0 2/ √ 3 .
maps LT to Z2 sending {1, eiπ/3, e2iπ/3} to {e1, e2, e2 −e1}. The transformed random walk gives probability 1/6 to the following vectors: ±e1, ±e2, ±(e2 −e1). Note that our transformed walk has lost some of the symmetry of the original walk.
16 Introduction Fig 1.2. The triangular lattice LT and its transformation ALT • The hexagonal or honeycomb lattice is not a lattice in our sense but rather a dual graph to the triangular lattice. It can be constructed in a number of ways. One way is to start with the triangular lattice LT. The lattice partitions the plane into triangular regions, of which some point up and some point down. We add a vertex in the center of each triangle pointing down.
The edges of this graph are the line segments from the center points to the vertices of these triangles (see figure).
Fig 1.3. The hexagons within LT Simple random walk on this graph is the process that at each time step moves to one of the three nearest neighbors. This is not a random walk in our strict sense because the increment distribution depends on whether the current position is a “center” point or a “vertex” point.
However, if we start at a vertex in LT, the two-step distribution of this walk is the same as the walk on the triangular lattice with step distribution p(±1) = p(±eiπ/3) = p(±e2iπ/3) = 1/9; p(0) = 1/3.
When studying random walks on other lattices L, we can map the walk to another walk on Zd.
However, since this might lose useful symmetries of the walk, it is sometimes better to work on the original lattice.
1.4 Other walks Although we will focus primarily on p ∈P, there are times where we will want to look at more general walks. There are two classes of distributions we will be considering.
Definition 1.5 Generator 17 • P∗ d denotes the set of p that generate aperiodic, irreducible walks supported on Zd, i.e., the set of p such that for all x, y ∈Zd there exists an N such that pn(x, y) > 0 for n ≥N.
• P′ d denotes the set of p ∈P∗ d with mean zero and finite second moment.
We write P∗= ∪dP∗ d, P′ = ∪P′ d.
Note that under our definition P is not a subset of P′ since P contains bipartite walks. However, if p ∈P is aperiodic, then p ∈P′.
1.5 Generator If f : Zd →R is a function and x ∈Zd, we define the first and second difference operators in x by ∇xf(y) = f(y + x) −f(y), ∇2 xf(y) = 1 2 f(y + x) + 1 2 f(y −x) −f(y).
Note that ∇2 x = ∇2 −x. We will sometimes write just ∇j, ∇2 j for ∇ej, ∇2 ej. If p ∈Pd with generator set V , then the generator L = Lp is defined by Lf(y) = X x∈Zd p(x) ∇xf(y) = X x∈V κ(x) ∇2 xf(y) = −f(y) + X x∈Zd p(x) f(x + y).
In the case of simple random walk, the generator is often called the discrete Laplacian and we will represent it by ∆D, ∆Df(y) = 1 d d X j=1 ∇2 jf(y).
Remark. We have defined the discrete Laplacian in the standard way for probability. In graph theory, the discrete Laplacian of f is often defined to be 2d∆Df(y) = X |x−y|=1 [f(x) −f(y)].
♣We can define Lf(y) = X x∈Zd p(x) [f(x + y) −f(y)] for any p ∈P∗ d. If p is not symmetric, one often needs to consider LRf(y) = X x∈Zd p(−x) [f(x + y) −f(y)].
The R stands for “reversed”; this is the generator for the random walk obtained by looking at the walk with time reversed.
The generator of a random walk is very closely related to the walk. We will write Ex, Px to denote 18 Introduction expectations and probabilities for random walk (both discrete and continuous time) assuming that S0 = x or ˜ S0 = x. Then, it is easy to check that Lf(y) = Ey[f(S1)] −f(y) = d dtEy[f( ˜ St)] t=0 .
(In the continuous-time case, some restrictions on the growth of f at infinity are needed.) Also, the transition probabilities pn(x), ˜ pt(x) satisfy the following “heat equations”: pn+1(x) −pn(x) = Lpn(x), d dt ˜ pt(x) = L˜ pt(x).
The derivation of these equations uses the symmetry of p. For example to derive the first, we write pn+1(x) = X y∈Zd P{S1 = y; Sn+1 −S1 = x −y} = X y∈Zd p(y) pn(x −y) = X y∈Zd p(−y) pn(x −y) = pn(x) + Lpn(x).
The generator L is also closely related to a second order differential operator. If u ∈Rd is a unit vector, we write ∂2 u for the second partial derivative in the direction u. Let ˆ L be the operator ˆ Lf(y) = 1 2 X x∈V κ(x) |x|2 ∂2 x/|x|f(y).
In the case of simple random walk, ˆ L = (2d)−1 ∆, where ∆denotes the usual Laplacian, ∆f(x) = d X j=1 ∂xjxjf(y); Taylor’s theorem shows that there is a c such that if f : Rd →R is C4 and y ∈Zd, |Lf(y) −ˆ Lf(y)| ≤c R4 M4, (1.4) where R is the range of the walk and M4 = M4(f, y) is the maximal absolute value of a fourth derivative of f for |x −y| ≤R. If the covariance matrix Γ is diagonalized, Γx = d X j=1 σ2 j (x · uj) uj, where u1, . . . , ud is an orthonormal basis, then ˆ Lf(y) = 1 2 d X j=1 σ2 j ∂2 uj f(y).
For future reference, we note that if y ̸= 0, ˆ L[log J ∗(y)2] = ˆ L[log J (y)2] = ˆ L log d X j=1 σ−2 j (y · uj)2 = d −2 J ∗(y)2 = d −2 d J (y)2 .
(1.5) 1.6 Filtrations and strong Markov property 19 ♣The estimate (1.4) uses the symmetry of p. If p is mean zero and finite range, but not necessarily symmetric, we can relate its generator to a (purely) second order differential operator, but the error involves the third derivatives of f. This only requires f to be C3 and hence can be useful in the symmetric case as well.
1.6 Filtrations and strong Markov property The basic property of a random walk is that the increments are independent and identically dis-tributed. It is useful to set up a framework that allows more “information” at a particular time than just the value of the random walk. This will not affect the distribution of the random walk provided that this extra information is independent of the future increments of the walk.
A (discrete-time) filtration F0 ⊂F1 ⊂· · · is an increasing sequence of σ-algebras. If p ∈Pd, then we say that Sn is a random walk with increment distribution p with respect to {Fn} if: • for each n, Sn is Fn-measurable; • for each n > 0, Sn −Sn−1 is independent of Fn−1 and P{Sn −Sn−1 = x} = p(x).
Similarly, we define a (right continuous, continuous-time) filtration to be an increasing collection of σ-algebras Ft satisfying Ft = ∩ǫ>0Ft+ǫ. If p ∈Pd, then we say that ˜ St is a continuous-time random walk with increment distribution p with respect to {Ft} if: • for each t, ˜ St is Ft-measurable; • for each s < t, ˜ St −˜ Ss is independent of Fs and P{ ˜ St −˜ Ss = x} = ˜ pt−s(x).
We let F∞denote the σ-algebra generated by the union of the Ft for t > 0.
If Sn is a random walk with respect to Fn, and T is a random variable independent of F∞, then we can add information about T to the filtration and still retain the properties of the random walk. We will describe one example of this in detail here; later on, we will do similar adding of information without being explicit. Suppose T has an exponential distribution with parameter λ, i.e., P{T > λ} = e−λ. Let F′ n denote the σ-algebra generated by Fn and the events {T ≤t} for t ≤n. Then {F′ n} is a filtration, and Sn is a random walk with respect to F′ n. Also, given F′ n, then on the event {T > n}, the random variable T −n has an exponential distribution with parameter λ. We can do similarly for the continuous-time walk ˜ St.
We will discuss stopping times and the strong Markov property. We will only do the slightly more difficult continuous-time case leaving the discrete-time analogue to the reader. If {Ft} is a filtration, then a stopping time with respect to {Ft} is a [0, ∞]-valued random variable τ such that for each t, {τ ≤t} ∈Ft. Associated to the stopping time τ is a σ-algebra Fτ consisting of all events A such that for each t, A ∩{τ ≤t} ∈Ft. (It is straightforward to check that the set of such A is a σ-algebra.) Theorem 1.6.1 (Strong Markov Property) Suppose ˜ St is a continuous-time random walk with increment distribution p with respect to the filtration {Ft}. Suppose τ is a stopping time with respect to the process. Then on the event {τ < ∞} the process Yt = ˜ St+τ −˜ Sτ, is a continuous-time random walk with increment distribution p independent of Fτ.
20 Introduction Proof (sketch) We will assume for ease that P{τ < ∞} = 1. Note that with probability one Yt has right-continuous paths. We first suppose that there exists 0 = t0 < t1 < t2 < . . . such that with probability one τ ∈{t0, t1, . . .}. Then, the result can be derived immediately, by considering the countable collection of events {τ = tj}. For more general τ, let τn be the smallest dyadic rational l/2n that is greater than τ. Then, τn is a stopping time and the result holds for τn. But, Yt = lim n→∞ ˜ St+τn −˜ Sτn.
We will use the strong Markov property throughout this book often without being explicit about its use.
Proposition 1.6.2 (Reflection Principle.) Suppose Sn (resp., ˜ St) is a random walk (resp., continuous-time random walk) with increment distribution p ∈Pd starting at the origin.
(a) If u ∈Rd is a unit vector and b > 0, P{ max 0≤j≤n Sj · u ≥b} ≤2 P{Sn · u ≥b}, P{sup s≤t ˜ Ss · u ≥b} ≤2 P{ ˜ St · u ≥b}.
(b) If b > 0, P{ max 0≤j≤n |Sj| ≥b} ≤2 P{|Sn| ≥b}, P{ sup 0≤s≤t | ˜ St| ≥b} ≤2 P{| ˜ St| ≥b}.
Proof We will do the continuous-time case. To prove (a), fix t > 0 and a unit vector u and let An = An,t,b be the event An = max j=1,...,2n ˜ Sjt2−n · u ≥b .
The events An are increasing in n and right continuity implies that w.p.1, lim n→∞An = sup s≤t ˜ Ss · u ≥b .
Hence, it suffices to show that for each n, P(An) ≤2 P{ ˜ St · u ≥b}. Let τ = τn,t,b be the smallest j such that ˜ Sjt2−n · u ≥b. Note that 2n [ j=1 n τ = j; ( ˜ St −˜ Sjt2−n) · u ≥0 o ⊂{ ˜ St · u ≥b}.
Since p ∈P, symmetry implies that for all t, P{ ˜ St · u ≥0} ≥1/2. Therefore, using independence, P{τ = j; ( ˜ St −˜ Sjt2−n) · u ≥0} ≥(1/2) P{τ = j}, and hence P{ ˜ St · u ≥b} ≥ 2n X j=1 P n τ = j; ( ˜ St −˜ Sjt2−n) · u ≥0 o ≥1 2 2n X j=1 P{τ = j} = 1 2 P(An).
1.7 A word about constants 21 Part (b) is done similarly, by letting τ be the smallest j with {| ˜ Sjt2−n| ≥b} and writing 2n [ j=1 n τ = j; ( ˜ St −˜ Sjt2−n) · ˜ Sjt2−n ≥0 o ⊂{| ˜ St| ≥b}.
Remark. The only fact about the distribution p that is used in the proof is that it is symmetric about the origin.
1.7 A word about constants Throughout this book c will denote a positive constant that depends on the dimension d and the increment distribution p but does not depend on any other constants.
We write f(n, x) = g(n, x) + O(h(n)), to mean that there exists a constant c such that for all n, |f(n, x) −g(n, x)| ≤c |h(n)|.
Similarly, we write f(n, x) = g(n, x) + o(h(n)), if for every ǫ > 0 there is an N such that |f(n, x) −g(n, x)| ≤ǫ |h(n)|, n ≥N.
Note that implicit in the definition is the fact that c, N can be chosen uniformly for all x. If f, g are positive functions, we will write f(n, x) ≍g(n, x), n →∞, if there exists a c (again, independent of x) such that for all n, x, c−1 g(n, x) ≤f(n, x) ≤c g(n, x).
We will write similarly for asymptotics of f(t, x) as t →0.
As an example, let f(z) = log(1 −z), |z| < 1, where log denotes the branch of the complex logarithm function with log 1 = 0. Then f is analytic in the unit disk with Taylor series expansion log(1 −z) = − ∞ X j=1 zj j .
By the remainder estimate, for every ǫ > 0, log(1 −z) + k X j=1 zj j ≤ |z|k+1 ǫk (k + 1), |z| ≤1 −ǫ.
22 Introduction For a fixed value of k we can write this as log(1 −z) = − k X j=1 zj j + O(|z|k+1), |z| ≤1/2, (1.6) or log(1 −z) = − k X j=1 zj j + Oǫ(|z|k+1), |z| ≤1 −ǫ, (1.7) where we write Oǫ to indicate that the constant in the error term depends on ǫ.
Exercises Exercise 1.1 Show that there are exactly 2d −1 additive subgroups of Zd of index 2. Describe them and show that they all can arise from some p ∈P. (A subgroup G of Zd has index two if G ̸= Zd but G ∪(x + G) = Zd for some x ∈Zd.) Exercise 1.2 Show that if p ∈Pd, n is a positive integer, and x ∈Zd, then p2n(0) ≥p2n(x).
Exercise 1.3 Show that if p ∈P∗ d, then there exists a finite set {x1, . . . , xk} such that: • p(xj) > 0, j = 1, . . . , k, • For every y ∈Zd, there exist (strictly) positive integers n1, . . . , nk with n1 x1 + · · · + nk xk = y.
(1.8) (Hint: first write each unit vector ±ej in the above form with perhaps different sets {x1, . . . , xk}. Then add the equations together.) Use this to show that there exist ǫ > 0, q ∈P′ d, q′ ∈P∗ d such that q has finite support and p = ǫ q + (1 −ǫ) q′.
Note that (1.8) is used with y = 0 to guarantee that q has zero mean.
Exercise 1.4 Suppose that Sn = X1 + · · · + Xn where X1, X2, . . . are independent Rd-valued random variables with mean zero and covariance matrix Γ. Show that Mn := |Sn|2 −(trΓ) n is a martingale.
Exercise 1.5 Suppose that p ∈P′ d ∪Pd with covariance matrix Γ = ΛΛT and Sn is the corre-sponding random walk. Show that Mn := J (Sn)2 −n is a martingale.
1.7 A word about constants 23 Exercise 1.6 Let L be a 2-dimensional lattice contained in Rd and suppose x1, x2 ∈L are points such that |x1| = min{|x| : x ∈L \ {0}}, |x2| = min{|x| : x ∈L \ {jx1 : j ∈Z} }.
Show that L = {j1x1 + j2x2 : j1, j2 ∈Z}.
You may wish to compare this to the remark after Proposition 1.3.1.
Exercise 1.7 Let S1 n, S2 n be independent simple random walks in Z and let Yn = S1 n + S2 n 2 , S1 n −S2 n 2 , Show that Yn is a simple random walk in Z2.
Exercise 1.8 Suppose Sn is a random walk with increment distribution p ∈P∗∪P. Show that there exists an ǫ > 0 such that for every unit vector θ ∈Rd, P{S1 · θ ≥ǫ} ≥ǫ.
2 Local Central Limit Theorem 2.1 Introduction If X1, X2, . . . are independent, identically distributed random variables in R with mean zero and variance σ2, then the central limit theorem (CLT) states that the distribution of X1 + · · · + Xn √n (2.1) approaches that of a normal distribution with mean zero and variance σ2. In other words, for −∞< r < s < ∞, lim n→∞P r ≤X1 + · · · + Xn √n ≤s = Z s r 1 √ 2πσ2 e−y2 2σ2 dy.
If p ∈P1 is aperiodic with variance σ2, we can use this to motivate the following approximation: pn(k) = P{Sn = k} = P k √n ≤Sn √n < k + 1 √n ≈ Z (k+1)/√n k/√n 1 √ 2πσ2 e−y2 2σ2 dy ≈ 1 √ 2πσ2n exp −k2 2σ2n .
Similarly, if p ∈P1 is bipartite, we can conjecture that pn(k) + pn(k + 1) ≈ Z (k+2)/√n k/√n 1 √ 2πσ2 e−y2 2σ2 dy ≈ 2 √ 2πσ2n exp −k2 2σ2n .
The local central limit theorem (LCLT) justifies this approximation.
♣One gets a better approximation by writing P{Sn = k} = P k −1 2 √n ≤Sn √n < k + 1 2 √n ≈ Z (k+ 1 2 )/√n (k−1 2 )/√n 1 √ 2πσ2n e−y2 2σ2 dy.
If p ∈Pd with covariance matrix Γ = Λ ΛT = Λ2, then the normalized sums (2.1) approach a joint normal random variable with covariance matrix Γ, i.e., a random variable with density f(x) = 1 (2π)d/2 (det Λ) e−|Λ−1x|2/2 = 1 (2π)d/2 √ det Γ e−(x·Γ−1x)/2.
24 2.1 Introduction 25 (See Section 12.3 for a review of the joint normal distribution.) A similar heuristic argument can be given for pn(x). Recall from (1.1) that J ∗(x)2 = x · Γ−1x. Let pn(x) denote the estimate of pn(x) that one obtains by the central limit theorem argument, pn(x) = 1 (2πn)d/2 √ det Γ e−J ∗(x)2 2n = 1 (2π)d nd/2 Z Rd ei s·x √n e−s·Γs 2 dds.
(2.2) The second equality is a straightforward computation, see (12.14). We define pt(x) for real t > 0 in the same way. The LCLT states that for large n, pn(x) is approximately pn(x). To be more precise, we will say that an aperiodic p satisfies the LCLT if lim n→∞nd/2 sup x∈Zd |pn(x) −pn(x)| = 0.
A bipartite p satisfies the LCLT if lim n→∞nd/2 sup x∈Zd |pn(x) + pn+1(x) −2pn(x)| = 0.
In this weak form of the LCLT we have not made any estimate of the error term |pn(x) −pn(x)| other than that it goes to zero faster than n−d/2 uniformly in x. Note that pn(x) is bounded by c n−d/2 uniformly in x. This is the correct order of magnitude for |x| of order √n but pn(x) is much smaller for larger |x|. We will prove a LCLT for any mean zero distribution with finite second moment. However, the LCLT we state now for p ∈Pd includes error estimates that do not hold for all p ∈P′ d.
Theorem 2.1.1 (Local Central Limit Theorem) If p ∈Pd is aperiodic, and pn(x) is as defined in (2.2), then there is a c and for every integer k ≥4 there is a c(k) < ∞such that for all integers n > 0 and x ∈Zd the following hold where z = x/√n: |pn(x) −pn(x)| ≤ c(k) n(d+2)/2 (|z|k + 1) e−J ∗(z)2 2 + 1 n(k−3)/2 , (2.3) |pn(x) −pn(x)| ≤ c n(d+2)/2 |z|2 .
(2.4) We will prove this result in a number of steps in Section 2.3. Before doing so, let us consider what the theorem states. Plugging k = 4 into (2.3) implies that |pn(x) −pn(x)| ≤ c n(d+2)/2 .
(2.5) For “typical” x with |x| ≤√n, pn(x) ≍n−d/2. Hence (2.5) implies pn(x) = pn(x) 1 + O 1 n , |x| ≤√n.
The error term in (2.5) is uniform over x, but as |x| grows, the ratio between the error term and pn(x) grows. The inequalities (2.3) and (2.4) are improvements on the error term for |x| ≥√n.
Since pn(x) ≍n−d/2 e−J ∗(x)2/2n, (2.3) implies pn(x) = pn(x) 1 + Ok(|x/√n|k) n + Ok 1 n(d+k−1)/2 , |x| ≥√n, 26 Local Central Limit Theorem where we write Ok to emphasize that the constant in the error term depends on k.
An even better improvement is established in Section 2.3.1 where it is shown that pn(x) = pn(x) exp O 1 n + |x|4 n3 , |x| < ǫ n.
Although Theorem 2.1.1 is not as useful for atypical x, simple large deviation results as given in the next propositions often suffice to estimate probabilities.
Proposition 2.1.2 • Suppose p ∈P′ d and Sn is a p-walk starting at the origin. Suppose k is a positive integer such that E[|X1|2k] < ∞. There exists c < ∞such that for all s > 0 P max 0≤j≤n |Sj| ≥s √n ≤c s−2k.
(2.6) • Suppose p ∈Pd and Sn is a p-walk starting at the origin. There exist β > 0 and c < ∞such that for all n and all s > 0, P max 0≤j≤n |Sj| ≥s √n ≤c e−βs2.
(2.7) Proof It suffices to prove the results for one-dimensional walks. See Corollaries 12.2.6 and 12.2.7.
♣The statement of the LCLT given here is stronger than is needed for many applications. For example, to determine whether the random walk is recurrent or transient, we only need the following corollary. If p ∈Pd is aperiodic, then there exist 0 < c1 < c2 < ∞such that for all x, pn(x) ≤c2 n−d/2, and for |x| ≤√n, pn(x) ≥c1 n−d/2. The exponent d/2 is important to remember and can be understood easily. In n steps, the random walk tends to go distance √n. In Zd, there are of order nd/2 points within distance √n of the origin.
Therefore, the probability of being at a particular point should be of order n−d/2.
The proof of Theorem 2.1.1 in Section 2.2 will use the characteristic function. We discuss LCLTs for p ∈P′ d, where, as before, P′ d denotes the set of aperiodic, irreducible increment distributions p in Zd with mean zero and finite second moment. In the proof of Theorem 2.1.1, we will see that we do not need to assume that the increments are bounded. For fixed k ≥4, (2.3) holds for p ∈P′ d provided that E[|X|k+1] < ∞and the third moments of p vanish. The inequalities (2.5) and (2.4) need only finite fourth moments and vanishing third moments. If p ∈P′ d has finite third moments that are nonzero, we can prove a weaker version of (2.3). Suppose k ≥3, and E[|X1|k+1] < ∞.
There exists c(k) < ∞such that |pn(x) −pn(x)| ≤ c(k) n(d+1)/2 (|z|k + 1) e−J ∗(z)2 2 + 1 n(k−2)/2 .
Also, for any p ∈P′ d with E[|X1|3] < ∞, |pn(x) −pn(x)| ≤ c n(d+1)/2 , |pn(x) −pn(x)| ≤ c n(d−1)/2 |x|2 .
We focus our discussion in Section 2.2 on aperiodic, discrete-time walks, but the next theorem 2.2 Characteristic Functions and LCLT 27 shows that we can deduce the results for bipartite and continuous-time walks from LCLT for aperiodic, discrete-time walks. We state the analogue of (2.3); the analogue of (2.4) can be proved similarly.
Theorem 2.1.3 If p ∈Pd and pn(x) is as defined in (2.2), then for every k ≥4 there is a c = c(k) < ∞such that the follwing holds for all x ∈Zd.
• If n is a positive integer and z = x/√n, then |pn(x) + pn+1(x) −2 pn(x)| ≤ c n(d+2)/2 (|z|k + 1) e−J ∗(z)2/2 + 1 n(k−3)/2 .
(2.8) • If f t > 0 and z = x/ √ t, |˜ pt(x) −pt(x)| ≤ c t(d+2)/2 (|z|k + 1) e−J ∗(y)2/2 + 1 t(k−3)/2 .
(2.9) Proof (assuming Theorem 2.1.1) We only sketch the proof. If p ∈Pd is bipartite, then S∗ n := S2n is an aperiodic walk on the lattice Zd e. We can establish the result for S∗ n by mapping Zd e to Zd as described in Section 1.3. This gives the asymptotics for p2n(x), x ∈Zd e and for x ∈Zd o, we know that p2n+1(x) = X y∈Zd p2n(x −y) p(y).
The continuous-time walk viewed at integer times is the discrete-time walk with increment dis-tribution ˜ p = ˜ p1. Since ˜ p satisfies all the moment conditions, (2.3) holds for ˜ pn(x), n = 0, 1, 2, . . . .
If 0 < t < 1, we can write ˜ pn+t(x) = X y∈Zd ˜ pn(x −y) ˜ pt(y), and deduce the result for all t.
2.2 Characteristic Functions and LCLT 2.2.1 Characteristic functions of random variables in Rd One of the most useful tools for studying the distribution of the sums of independent random variables is the characteristic function. If X = (X1, . . . , Xd) is a random variable in Rd, then its characteristic function φ = φX is the function from Rd into C given by φ(θ) = E[exp{iθ · X}].
Proposition 2.2.1 Suppose X = (X1, . . . , Xd) is a random variable in Rd with characteristic function φ.
(a) φ is a uniformly continuous function with φ(0) = 1 and |φ(θ)| ≤1 for all θ ∈Rd.
(b) If θ ∈Rd then φX,θ(s) := φ(sθ) is the characteristic function of the one-dimensional random variable X · θ.
28 Local Central Limit Theorem (c) Suppose d = 1 and m is a positive integer with E[|X|m] < ∞. Then φ(s) is a Cm function of s; in fact, φ(m)(s) = im E[XmeisX].
(d) If m is a positive integer, E[|X|m] < ∞, and |u| = 1, then φ(su) − m−1 X j=0 ij E[(X · u)j] j!
sj ≤E[|X · u|m] m!
|s|m.
(e) If X1, X2, . . . , Xn are independent random variables in Rd, with characteristic functions φX1, . . . , φXn, then φX1+···+Xn(θ) = φX1(θ) · · · φXn(θ).
In particular, if X1, X2, . . . are independent, identically distributed with the same distribution as X, then the characteristic function of Sn = X1 + · · · + Xn is given by φSn(θ) = [φ(θ)]n.
Proof To prove uniform continuity, note that |φ(θ1 + θ) −φ(θ)| = |E[eiX(θ1+θ) −eiXθ]| ≤E[|eiXθ1 −1|], and the dominated convergence theorem implies that lim θ1→0 E[|eiXθ1 −1|] = 0.
The other statements in (a) and (b) are immediate.
Part (c) is derived by differentiating; the condition E[|X|m] < ∞is needed to justify the differentiation using the dominated convergence theorem (details omitted). Part (d) follows from (b), (c), and Taylor’s theorem with remainder.
Part (e) is immediate from the product rule for expectations of independent random variables.
We will write Pm(θ) for the m-th order Taylor series approximation of φ about the origin. Then the last proposition implies that if E[|X|m] < ∞, then φ(θ) = Pm(θ) + o(|θ|m), θ →0.
(2.10) Note that if E[X] = 0 and E[|X|2] < ∞, then P2(θ) = 1 −1 2 d X j=1 d X k=1 E[XjXk] θj θk = 1 −θ · Γθ 2 = 1 −E[(X · θ)2] 2 .
Here Γ denotes the covariance matrix for X. If E[|X|m] < ∞, we write Pm(θ) = 1 −θ · Γθ 2 + m X j=3 qj(θ), (2.11) where qj are homogeneous polynomials of degree j determined by the moments of X. If all the third moments of X exist and equal zero, q3 ≡0. If X has a symmetric distribution, then qj ≡0 for all odd j for which E[|X|j] < ∞.
2.3 LCLT — characteristic function approach 29 2.2.2 Characteristic functions of random variables in Zd If X = (X1, . . . , Xd) is a Zd-valued random variable, then its characteristic function has period 2π in each variable, i.e., if k1, . . . , kd are integers, φ(θ1, . . . , θd) = φ(θ1 + 2k1π, . . . , θd + 2kdπ).
The characteristic function determines the distribution of X; in fact, the next proposition gives a simple inversion formula. Here, and for the remainder of this section, we will write dθ for dθ1 · · · dθd.
Proposition 2.2.2 If X = (X1, . . . , Xd) is a Zd-valued random variable with characteristic func-tion φ, then for every x ∈Zd, P{X = x} = 1 (2π)d Z [−π,π]d φ(θ) e−ix·θ dθ.
Proof Since φ(θ) = E[eiX·θ] = X y∈Zd eiy·θ P{X = y}, we get Z [−π,π]d φ(θ) e−ix·θ dθ = X y∈Zd P{X = y} Z [−π,π]d ei(y−x)·θ dθ.
(The dominated convergence theorem justifies the interchange of the sum and the integral.) But, if x, y ∈Zd, Z [−π,π]d ei(y−x)·θ dθ = (2π)d, y = x 0, y ̸= x.
Corollary 2.2.3 Suppose X1, X2, . . . are independent, identically distributed random variables in Zd with characteristic function φ. Let Sn = X1 + · · · + Xn. Then, for all x ∈Zd, P{Sn = x} = 1 (2π)d Z [−π,π]d φn(θ) e−ix·θ dθ.
2.3 LCLT — characteristic function approach In some sense, Corollary 2.2.3 completely solves the problem of determining the distribution of a random walk at a particular time n. However, the integral is generally hard to evaluate and estimation of oscillatory integrals is tricky. Fortunately, we can use this corollary as a starting point for deriving the local central limit theorem. We will consider p ∈P′ in this section. Here, as before, we write pn(x) for the distribution of Sn = X1 +· · ·+Xn where X1, . . . , Xn are independent with distribution p. We also write ˜ St for a continuous time walk with rates p. We let φ denote the characteristic function of p, φ(θ) = X x∈Zd eiθ·x p(x).
30 Local Central Limit Theorem We have noted that the characteristic function of Sn is φn.
Lemma 2.3.1 The characteristic function of ˜ St is φ ˜ St(θ) = exp{t[φ(θ) −1]}.
Proof Since ˜ St has the same distribution as SNt where Nt is an independent Poisson process with parameter 1, we get φ ˜ St(θ) = E[eiθ· ˜ St] = ∞ X j=0 e−t tj j! E[eiθ·Sj] = ∞ X j=0 e−t tj j! φ(θ)j = exp{t[φ(θ) −1]}.
Corollary 2.2.3 gives the formulas pn(x) = 1 (2π)d Z [−π,π]d φn(θ) e−iθ·x dθ, (2.12) ˜ pt(x) = 1 (2π)d Z [−π,π]d et[φ(θ)−1] e−iθ·x dθ.
Lemma 2.3.2 Suppose p ∈P′ d.
(a) For every ǫ > 0, sup n |φ(θ)| : θ ∈[−π, π]d, |θ| ≥ǫ o < 1.
(b) There is a b > 0 such that for all θ ∈[−π, π]d, |φ(θ)| ≤1 −b|θ|2.
(2.13) In particular, for all θ ∈[−π, π]d, and r > 0, |φ(θ)|r ≤ 1 −b|θ|2r ≤exp −br|θ|2 .
(2.14) Proof By continuity and compactness, to prove (a) it suffices to prove that |φ(θ)| < 1 for all θ ∈[−π, π]d \ {0}. To see this, suppose that |φ(θ)| = 1. Then |φ(θ)n| = 1 for all positive integers n. Since φ(θ)n = X z∈Zd pn(z) eiz·θ, and for each fixed z, pn(z) > 0 for all sufficiently large n, we see that eiz·θ = 1 for all z ∈Zd.
(Here we use the fact that if w1, w2, . . . ∈C with |w1 + w2 + · · · | = 1 and |w1| + |w2| + · · · = 1, then there is a ψ such that wj = rjeiψ with rj ≥0.) The only θ ∈[−π, π]d that satisfies this is θ = 0. Using (a), it suffices to prove (2.13) in a neighborhood of the origin, and this follows from the second-order Taylor series expansion (2.10).
♣The last lemma does not hold for bipartite p. For example, for simple random walk φ(πi, πi, . . . , πi) = −1.
2.3 LCLT — characteristic function approach 31 In order to illustrate the proof of the local central limit theorem using the characteristic function, we will consider the one-dimensional case with p(1) = p(−1) = 1/4 and p(0) = 1/2. Note that this increment distribution is the same as the two-step distribution of (1/2 times) the usual simple random walk. The characteristic function for p is φ(θ) = 1 2 + 1 4 eiθ + 1 4 e−iθ = 1 2 + 1 2 cos θ = 1 −θ2 4 + O(θ4).
The inversion formula (2.12) tells us that pn(x) = 1 2π Z π −π e−ixθ φ(θ)n dθ = 1 2π√n Z π√n −π√n e−i(x/√n)s φ(s/√n)n ds.
The second equality follows from the substitution s = θ√n. For |s| ≤π √n, we can write φ s √n = 1 −s2 4n + O s4 n2 = 1 −(s2/4) + O(s4/n) n .
We can find δ > 0 such that if |s| ≤δ√n, s2 4 + O s4 n ≤n 2.
Therefore, using (12.3), if |s| ≤δ√n, φ s √n n = 1 −s2 4 n + O s4 n2 n = e−s2/4 eg(s,n), where |g(s, n)| ≤c s4 n .
If ǫ = min{δ, 1/ √ 8c} we also have |g(s, n)| ≤s2 8 , |s| ≤ǫ √n.
For ǫ √n < |s| ≤π √n, (2.13) shows that |e−i(x/√n)s φ(s/√n)n| ≤e−βn for some β > 0. Hence, up to an error that is exponentially small in n, pn(x) equals 1 2π√n Z ǫ√n −ǫ√n e−i(x/√n)s e−s2/4 eg(s,n) ds.
We now use |eg(s,n) −1| ≤ c s4/n, |s| ≤n1/4 es2/8, n1/4 < |s| ≤ǫ√n to bound the error term as follows: 1 2π√n Z ǫ√n −ǫ√n e−i(x/√n)s e−s2/4 [eg(s,n) −1] ds ≤ c √n Z ǫ√n −ǫ√n e−s2/4 |eg(s,n) −1| ds, Z n1/4 −n−1/4 e−s2/4 |eg(s,n) −1| ds ≤c n Z ∞ −∞ s4 e−s2/4 ds ≤c n, 32 Local Central Limit Theorem Z n1/4≤|s|≤ǫ√n e−s2/4 |eg(s,n) −1| ds ≤ Z |s|≥n1/4 e−s2/8 ds = o(n−1).
Hence we have pn(x) = O 1 n3/2 + 1 2π√n Z ǫ√n −ǫ√n e−i(x/√n)s e−s2/4 ds = O 1 n3/2 + 1 2π√n Z ∞ −∞ e−i(x/√n)s e−s2/4 ds.
The last term equals pn(x), see (2.2), and so we have shown that pn(x) = pn(x) + O 1 n3/2 .
We will follow this basic line of proof for theorems in this subsection. Before proceeding, it will be useful to outline the main steps.
• Expand log φ(θ) in a neighborhood |θ| < ǫ about the origin.
• Use this expansion to approximate [φ(θ/√n)]n, which is the characteristic function of Sn/√n.
• Use the inversion formula to get an exact expression for the probability and do a change of variables s = θ √n to yield an integral over [−π√n, π√n]d. Use Lemma 2.3.2 to show that the integral over |θ| ≥ǫ√n is exponentially small.
• Use the approximation of [φ(θ/√n)]n to compute the dominant term and to give an expression for the error term that needs to be estimated.
• Estimate the error term.
Our first lemma discusses the approximation of the characteristic function of Sn/√n by an exponential. We state the lemma for all p ∈P′ d, and then give examples to show how to get sharper results if one makes stronger assumptions on the moments.
Lemma 2.3.3 Suppose p ∈P′ d with covariance matrix Γ and characteristic function φ that we write as φ(θ) = 1 −θ · Γθ 2 + h(θ), where h(θ) = o(|θ|2) as θ →0. There exist ǫ > 0, c < ∞such that for all positive integers n and all |θ| ≤ǫ √n, we can write φ θ √n n = exp −θ · Γθ 2 + g(θ, n) = e−θ·Γθ 2 [1 + Fn(θ)] , (2.15) where Fn(θ) = eg(θ,n) −1 and |g(θ, n)| ≤min θ · Γθ 4 , n h θ √n + c|θ|4 n .
(2.16) In particular, |Fn(θ)| ≤e θ·Γθ 4 + 1.
2.3 LCLT — characteristic function approach 33 Proof Choose δ > 0 such that |φ(θ) −1| ≤1 2, |θ| ≤δ.
For |θ| ≤δ, we can write log φ(θ) = −θ · Γθ 2 + h(θ) −(θ · Γθ)2 8 + O(|h(θ)| |θ|2) + O(|θ|6).
(2.17) Define g(θ, n) by n log φ θ √n = −θ · Γθ 2 + g(θ, n), so that (2.15) holds. Note that |g(θ, n)| ≤n h θ √n + O |θ|4 n .
Since n h(θ/√n) = o(|θ|2), we can find 0 < ǫ ≤δ such that for |θ| ≤ǫ√n, |g(θ, n)| ≤θ · Γθ 4 .
♣The proofs will require estimates for Fn(θ). The inequality |ez −1| ≤O(|z|) is valid if z is restricted to a bounded set. Hence, the basic strategy is to find c1, r(n) ≤O(n1/4) such that n h θ √n ≤c1, |θ| ≤r(n) Since O(|θ|4/n) ≤O(1) for |θ| ≤n1/4, (2.16) implies |Fn(θ)| = eg(θ,n) −1 ≤c |g(θ, n)| ≤c n h θ √n + |θ|4 n2 , |θ| ≤r(n), |Fn(θ)| ≤e θ·Γθ 4 + 1, r(n) ≤|θ| ≤ǫ √n.
Examples We give some examples with different moment assumptions. In the discussion below, ǫ is as in Lemma 2.3.3 and θ is restricted to |θ| ≤ǫ √n.
• If E[|X1|4] < ∞, then by (2.11), h(θ) = q3(θ) + O(|θ|4), and log φ(θ) = −θ · Γθ 2 + f3(θ) + O(|θ|4), where f3 = q3 is a homogeneous polynomial of degree three. In this case, g(θ, n) = n f3 θ √n + O(|θ|4) n , (2.18) 34 Local Central Limit Theorem and there exists c < ∞such that |g(θ, n)| ≤min θ · Γθ 4 , c |θ|3 √n .
We use here and below the fact that |θ|3/√n ≥|θ|4/n for |θ| ≤ǫ√n.
• If E[X1|6] < ∞and all the third and fifth moments of X1 vanish, then h(θ) = q4(θ) + O(|θ|6), log φ(θ) = −θ · Γθ 2 + f4(θ) + O(|θ|6), where f4(θ) = q4(θ) −(θ · Γθ)2/8 is a homogeneous polynomial of degree four. In this case, g(θ, n) = n f4 θ √n + O(|θ|6) n2 , (2.19) and there exists c < ∞such that |g(θ, n)| ≤min θ · Γθ 4 , c |θ|4 n .
• More generally, suppose that k ≥3 is a positive integer such that E[|X1|k+1] < ∞. Then h(θ) = k X j=3 qj(θ) + O(|θ|k+1), log φ(θ) = −θ · Γθ 2 + k X j=3 fj(θ) + O(|θ|k+1), where fj are homogeneous polynomials of degree j that are determined by Γ, q3, . . . , qk. In this case, g(θ, n) = k X j=3 n fj θ √n + O(|θ|k+1) n(k−1)/2 , (2.20) Moreover, if j is odd and all the odd moments of X of degree less than or equal to j vanish, then fj ≡0. Also, |g(θ, n)| ≤min θ · Γθ 4 , c |θ|2+α nα/2 , where α = 2 if the third moments vanish and otherwise α = 1.
• Suppose E[eb·X] < ∞for all b in a real neighborhood of the origin. Then z 7→φ(z) = eiz·X1 is a holomorphic function from a neighborhood of the origin in Cn to C. Hence, we can choose ǫ so that log φ(z) is holomorphic for |z| < ǫ and hence z 7→g(z, n) and z 7→Fn(z) are holomorphic for |z| < ǫ √n.
The next lemma computes the dominant term and isolates the integral that needs to be estimated in order to obtain error bounds.
2.3 LCLT — characteristic function approach 35 Lemma 2.3.4 Suppose p ∈P′ d with covariance matrix Γ. Let φ, ǫ, Fn be as in Lemma 2.3.3. There exist c < ∞, ζ > 0 such that for all 0 ≤r ≤ǫ√n, if we define vn(x, r) by pn(x) = pn(x) + vn(x, r) + 1 (2π)d nd/2 Z |θ|≤r e−ix·θ √n e−θ·Γθ 2 Fn(θ) dθ, then |vn(x, r)| ≤c n−d/2 e−ζr2.
Proof The inversion formula (2.12) gives pn(x) = 1 (2π)d Z [−π,π]d φ(θ)n e−ix·θ dθ = 1 (2π)d nd/2 Z [−√nπ,√nπ]d φ s √n n e−iz·s ds, where z = x/√n. Lemma 2.3.2 implies that there is a β > 0 such that |φ(θ)| ≤e−β for |θ| ≥ǫ.
Therefore, 1 (2π)d nd/2 Z [−√nπ,√nπ]d φ s √n n e−iz·s ds = O(e−βn) + 1 (2π)d nd/2 Z |θ|≤ǫ√n φ s √n n e−iz·s ds.
For |s| ≤ǫ√n, we write φ s √n n = e−s·Γs 2 + e−s·Γs 2 Fn(s).
By (2.2) we have 1 (2π)d nd/2 Z Rd e−iz·s e−s·Γs 2 ds = pn(x).
(2.21) Also, 1 (2π)d nd/2 Z |s|≥ǫ√n e−iz·s e−s·Γs 2 ds ≤ 1 (2π)d nd/2 Z |s|≥ǫ√n e−s·Γs 2 ds ≤O(e−βn), for perhaps a different β. Therefore, pn(x) = pn(x) + O(e−βn) + 1 (2π)d nd/2 Z |θ|≤ǫ√n e −ix·θ √n e−θ·Γθ 2 Fn(θ) dθ.
This gives the result for r = ǫ√n. For other values of r, we use the estimate |Fn(θ)| ≤e θ·Γθ 4 + 1, to see that Z r≤|θ|≤ǫ√n e −ix·θ √n e−θ·Γθ 2 Fn(θ) dθ ≤2 Z |θ|≥r e−θ·Γθ 4 dθ = O(e−ζr2).
The next theorem establishes LCLTs for p ∈P′ with finite third moment and p ∈P′ with finite fourth moment and vanishing third moments. It gives an error term that is uniform over all x ∈Zd.
The estimate is good for typical x, but is not very sharp for atypically large x.
36 Local Central Limit Theorem Theorem 2.3.5 Suppose p ∈P′ with E[|X1|3] < ∞. Then there exists a c < ∞such for all n, x, |pn(x) −pn(x)| ≤ c n(d+1)/2 .
(2.22) If E[|X1|4] < ∞and all the third moments of X1 are zero, then there is a c such that for all n, x, |pn(x) −pn(x)| ≤ c n(d+2)/2 .
(2.23) Proof We use the notations of Lemmas 2.3.3 and 2.3.4. Letting r = n1/8 in Lemma 2.3.4, we see that, pn(x) = pn(x) + O(e−βn1/4) + 1 (2π)d nd/2 Z |θ|≤n1/8 e−ix·θ √n e−θ·Γθ 2 Fn(θ) dθ.
Note that |h(θ)| = O(|θ|2+α) where α = 1 under the weaker assumption and α = 2 under the stronger assumption. For |θ| ≤n1/8, |g(θ, n)| ≤c |θ|2+α/nα/2, and hence |Fn(θ)| ≤c |θ|2+α nα/2 .
This implies Z |θ|≤n1/8 e−ix·θ √n e−θ·Γθ 2 Fn(θ) dθ ≤ c nα/2 Z Rd |θ|2+α e−θ·Γθ 2 dθ ≤ c nα/2 .
♣The choice r = n1/8 in the proof above was somewhat arbitrary. The value r was chosen sufficiently large so that the error term vn(x, r) from Lemma 2.3.4 decays faster than any power of n but sufficiently small so that |g(θ, n)| is uniformly bounded for |θ| ≤r. We could just as well have chosen r(n) = nκ for any 0 < κ ≤1/8.
♣The constant c in (2.22) and (2.23) depends on the particular p. However, by careful examination of the proof, one can get uniform estimates for all p satisfying certain conditions. The error in the Taylor polynomial approximation of the characteristic function can be bounded in terms of the moments of p. One also needs a uniform bound such as (2.13) which guarantees that the walk is not too close to being a bipartite walk. Such uniform bounds on rates of convergence in CLT or LCLT are often called Berry-Esseen bounds. We will need one such result, see Proposition 2.3.13, but for most of this book, the walk p is fixed and we just allow constants to depend on p.
The estimate (2.5) is a special case of (2.23). We have shown that (2.5) holds for any symmetric p ∈P′ with E[|X1|4] < ∞. One can obtain a difference estimate for pn(x) from (2.5). However, we will give another proof below that requires only third moments of the increment distribution. This theorem also gives a uniform bound on the error term.
♣If α ̸= 0 and f(n) = nα + O(nα−1), (2.24) 2.3 LCLT — characteristic function approach 37 then f(n + 1) −f(n) = [(n + 1)α −nα] + [O((n + 1)α−1) −O(nα−1)].
This shows that f(n + 1) −f(n) = O(nα−1), but the best that we can write about the error terms is O((n + 1)α−1) −O(nα−1) = O(nα−1), which is as large as the dominant term. Hence an expression such as (2.24) is not sufficient to give good asymptotics on differences of f. One strategy for proving difference estimates is to go back to the derivation of (2.24) to see if the difference of the errors can be estimated. This is the approach used in the next theorem.
Theorem 2.3.6 Suppose p ∈P′ d with E[|X1|3] < ∞. Let ∇y denote the differences in the x variable, ∇ypn(x) = pn(x + y) −pn(x), ∇ypn(x) = pn(x + y) −pn(x), and ∇ j = ∇ej.
• There exists c < ∞such that for all x, n, y, |∇ypn(x) −∇ypn(x)| ≤ c |y| n(d+2)/2 .
• If E[|X1|4] < ∞and all the third moments of X1 vanish, there exists c < ∞such that for all x, n, y, |∇ypn(x) −∇ypn(x)| ≤ c |y| n(d+3)/2 .
Proof By the triangle inequality, it suffices to prove the result for y = ej, j = 1, . . . , d. Let α = 1 under the weaker assumptions and α = 2 under the stronger assumptions. As in the proof of Theorem 2.3.5, we see that ∇ jpn(x) = ∇ jpn(x) + O(e−βn1/4) + 1 (2π)d nd/2 Z |θ|≤n1/8 e− i(x+ej)·θ √n −e−ix·θ √n e−θ·Γθ 2 Fn(θ) dθ.
Note that e− i(x+ej)·θ √n −e−ix·θ √n = e− iej·θ √n −1 ≤|θ| √n, and hence Z |θ|≤n1/8 e− i(x+ej)·θ √n −e−ix·θ √n e−θ·Γθ 2 Fn(θ) dθ ≤ 1 √n Z |θ|≤n1/8 |θ| e−θ·Γθ 2 |Fn(θ)| dθ.
The estimate Z |θ|≤n1/8 |θ| e−θ·Γθ 2 |Fn(θ)| dθ ≤ c nα/2 , where α = 1 under the weaker assumption and α = 2 under the stronger assumption, is done as in the previous theorem.
38 Local Central Limit Theorem The next theorem improves the LCLT by giving a better error bound for larger x. The basic strategy is to write Fn(θ) as the sum of a dominant term and an error term. This requires a stronger moment condition. If E[|X1|j] < ∞, let fj be the homogeneous polynomial of degree j defined in (2.20). Let uj(z) = 1 (2π)d Z Rd e−is·z fj(s) e−s·Γs 2 ds.
(2.25) Using standard properties of Fourier transforms, we can see that uj(z) = f ∗ j (z) e−(z·Γ−1z)/2 = f ∗ j (z) e−J ∗(z)2 2 (2.26) for some jth degree polynomial f ∗ j that depends only on the distribution of X1.
Theorem 2.3.7 Suppose p ∈P′ d.
• If E[|X1|4] < ∞, there exists c < ∞such that pn(x) −pn(x) −u3(x/√n) n(d+1)/2 ≤ c n(d+2)/2 , (2.27) where u3 is a defined in (2.25).
• If E[|X1|5] < ∞and the third moments of X1 vanish there exists c < ∞such that pn(x) −pn(x) −u4(x/√n) n(d+2)/2 ≤ c n(d+3)/2 , (2.28) where u4 is a defined in (2.25).
If k ≥3 is a positive integer such that E[|X1|k] < ∞and uk is as defined in (2.25), then there is a c(k) such that |uk(z)| ≤c(k) (|z|k + 1) e−J ∗(z)2 2 .
Moreover, if j is a positive integer, there is a c(k, j) such that if Dj is a jth order derivative, |Djuk(z)| ≤c(k, j) |(|z|k+j + 1) e−J ∗(z)2 2 .
(2.29) Proof Let α = 1 under the weaker assumptions and α = 2 under the stronger assumptions. As in Theorem 2.3.5, nd/2 [pn(x) −pn(x)] = O(e−βn1/4) + 1 (2π)d Z |θ|≤n1/8 e−ix·θ √n e−θ·Γθ 2 Fn(θ) dθ.
(2.30) Recalling (2.20), we can see that for |θ| ≤n1/8, Fn(θ) = f2+α(θ) nα/2 + O(|θ|3+α) n(α+1)/2 .
Up to an error of O(e−βn1/4), the right-hand side of (2.30) equals 1 (2π)d Z Rd e−ix·θ √n e−θ·Γθ 2 f2+α(θ) n(α/2) dθ + 1 (2π)d Z |θ|≤n1/8 e−ix·θ √n e−θ·Γθ 2 Fn(θ) −f2+α(θ) nα/2 dθ.
2.3 LCLT — characteristic function approach 39 The second integral can be bounded as before Z |θ|≤n1/8 e−ix·θ √n e−θ·Γθ 2 Fn(θ) −f2+α(θ) nα/2 dθ ≤ c n(α+1)/2 Z Rd |θ|3+α e−θ·Γθ 2 dθ ≤ c n(α+1)/2 .
The estimates on uk and Djuk follows immediately from (2.26).
The next theorem is proved in the same way as Theorem 2.3.7 starting with (2.20), and we omit it. A special case of this theorem is (2.3). The theorem shows that (2.3) holds for all symmetric p ∈P′ d with E[|X1|6] < ∞. The results stated for n ≥|x|2 are just restatements of Theorem 2.3.5.
Theorem 2.3.8 Suppose p ∈P′ d and k ≥3 is a positive integer such that E[|X1|k+1] < ∞. There exists c = c(k) such that pn(x) −pn(x) − k X j=3 uj(x/√n) n(d+j−2)/2 ≤ c n(d+k−1)/2 , (2.31) where uj are as defined in (2.25).
In particular, if z = x/√n, |pn(x) −pn(x)| ≤ c n(d+1)/2 |z|k e−J ∗(z)2 2 + 1 n(k−2)/2 , n ≤|x|2, |pn(x) −pn(x)| ≤ c n(d+1)/2 , n ≥|x|2.
If the third moments of X1 vanish (e.g., if p is symmetric) then u3 ≡0 and |pn(x) −pn(x)| ≤ c n(d+2)/2 |z|k e−J ∗(z)2 2 + 1 n(k−3)/2 , n ≤|x|2, |pn(x) −pn(x)| ≤ c n(d+2)/2 , n ≥|x|2.
Remark. Theorem 2.3.8 gives improvements to Theorem 2.3.6. Assuming a sufficient number of moments on the increment distribution, one can estimate ∇ypn(x) up to an error of O(n−(d+k−1)/2) by taking ∇y of all the terms on the left-hand side of (2.31). These terms can be estimated using (2.29). This works for higher order differences as well.
The next theorem is the LCLT assuming only a finite second moment.
Theorem 2.3.9 Suppose p ∈P′ d. Then there exists a sequence δn →0 such that for all n, x, |pn(x) −pn(x)| ≤ δn nd/2 .
(2.32) Proof By Lemma 2.3.4, there exist c, ζ such that for all n, x and r > 0, nd/2 |pn(x) −pn(x)| ≤c e−ζr2 + Z θ≤r |Fn(θ)| dθ ≤c " e−ζr2 + rd sup θ≤r |Fn(θ)| # .
40 Local Central Limit Theorem We now refer to Lemma 2.3.3. Since h(θ) = o(|θ|2), lim n→∞sup |θ|≤r |g(θ, n)| = 0, and hence lim n→∞sup |θ|≤r |Fn(θ)| = 0.
In particular, for all n sufficiently large, nd/2 |pn(x) −pn(x)| ≤2 c e−ζr2.
The next theorem improves on this for |x| larger than √n. The proof uses an integration by parts.
One advantage of this theorem is that it does not need any extra moment conditions. However, if we impose extra moment conditions we get a stronger result.
Theorem 2.3.10 Suppose p ∈P′ d. Then there exists a sequence δn →0 such that for all n, x, |pn(x) −pn(x)| ≤ δn |x|2 n(d−2)/2 .
(2.33) Moreover, • If E[|X1|3] < ∞, then δn can be chosen O(n−1/2).
• If E[|X1|4] < ∞and the third moments of X1 vanish, then δn can be chosen O(n−1).
Proof If ψ1, ψ2 are C2 functions on Rd with period 2π in each component, then it follows from Green’s theorem (integration by parts) that Z [−π,π]d [∆ψ1(θ)] ψ2(θ) dθ = Z [−π,π]d ψ1(θ) [∆ψ2(θ)] dθ (the boundary terms disappear by periodicity). Since ∆[eix·θ] = −|x|2 e−x·θ, the inversion formula gives pn(−x) = 1 (2π)d Z [−π,π]d eix·θ ψ(θ) dθ = − 1 |x|2 (2π)d Z [−π,π]d eix·θ ∆ψ(θ) dθ, where ψ(θ) = φ(θ)n. Therefore, |x|2 n pn(−x) = − 1 (2π)d Z [−π,π]d eix·θ φ(θ)n−2 [(n −1) λ(θ) + φ(θ) ∆φ(θ)] dθ, where λ(θ) = d X j=1 [∂jφ(θ)]2 .
The first and second derivatives of φ are uniformly bounded. Hence by (2.14), we can see that φ(θ)n−2 [(n −1) λ(θ) + φ(θ) ∆φ(θ)] ≤c [1 + n|θ|2] e−n|θ|2b ≤c e−βn|θ|2, 2.3 LCLT — characteristic function approach 41 where 0 < β < b. Hence, we can write |x|2 n pn(−x) −O(e−βr2) = − 1 (2π)d Z |θ|≤r/√n eix·θ φ(θ)n−2 [(n −1) λ(θ) + φ(θ) ∆φ(θ)] dθ.
The usual change of variables shows that the right-hand side equals − 1 (2π)d nd/2 Z |θ|≤r eiz·θ φ θ √n n−2 (n −1) λ θ √n + φ θ √n ∆φ θ √n dθ, where z = x/√n.
Note that ∆ h e−θ·Γθ 2 i = e−θ·Γθ 2 [|Γθ|2 −tr(Γ)].
We define ˆ Fn(θ) by φ θ √n n−2 (n −1) λ θ √n + φ θ √n ∆φ θ √n = e−θ·Γθ 2 h |Γθ|2 −tr(Γ) −ˆ Fn(θ) i .
A straightforward calculation using Green’s theorem shows that pn(−x) = 1 (2π)d nd/2 Z Rd ei(x/√n)·θ e−θ·Γθ 2 dθ = − n |x|2 (2π)2 Z Rd ei(x/√n)·θ ∆[e−θ·Γθ 2 ] dθ.
Therefore (with perhaps a different β), |x|2 n pn(−x) = |x|2 n pn(−x) + O(e−βr2) − 1 (2π)d nd/2 Z |θ|≤r eiz·θ e−θ·Γθ 2 ˆ Fn(θ) dθ.
(2.34) The remaining task is to estimate ˆ Fn(θ). Recalling the definition of Fn(θ) from Lemma 2.3.3, we can see that φ θ √n n−2 (n −1) λ θ √n + φ θ √n ∆φ θ √n = e−θ·Γθ 2 [1 + Fn(θ)] (n −1) λ(θ/√n) φ(θ/√n)2 + ∆φ(θ/√n) φ(θ/√n) .
We make three possible assumptions: • p ∈P′ d.
• p ∈P′ d with E[|X1|3] < ∞.
• p ∈P′ d with E[|X1|4] < ∞and vanishing third moments.
We set α = 0, 1, 2, respectively, in these three cases. Then we can write φ(θ) = 1 −θ · Γθ 2 + q2+α(θ) + o(|θ|2+α), where q2 ≡0 and q3, q4 are homogeneous polynomials of degree 3 and 4 respectively. Because φ is C2+α and we know the values of the derivatives at the origin, we can write ∂jφ(θ) = −∂j θ · Γθ 2 + ∂jq2+α(θ) + o(|θ|1+α), 42 Local Central Limit Theorem ∂jjφ(θ) = −∂jj θ · Γθ 2 + ∂jjq2+α(θ) + o(|θ|α).
Using this, we see that Pd j=1 [∂jφ(θ)]2 φ(θ)2 = |Γθ|2 + ˜ q2+α(θ) + o(|θ|2+α), ∆φ(θ) φ(θ) = −tr(Γ) + ˆ qα(θ) + o(|θ|α), where ˜ q2+α is a homogeneous polyonomial of degree 2 + α with ˜ q2 ≡0, and ˆ qα is a homogeneous polynomial of degree α with ˆ q0 = 0. Therefore, for |θ| ≤n1/8, (n −1) λ(θ/√n) φ(θ/√n)2 + ∆φ(θ/√n) φ(θ/√n) = |Γθ|2 −tr(Γ) + ˜ q2+α(θ) + ˆ qα(θ) nα/2 + o |θ|α + |θ|α+2 nα/2 , which establishes that for α = 1, 2 | ˆ Fn(θ)| = O 1 + |θ|2+α nα/2 , |θ| ≤n1/16, and for α = 0, for each r < ∞, lim n→∞sup |θ|≤r | ˆ Fn(θ)| = 0.
The remainder of the argument follows the proofs of Theorem 2.3.5 and 2.3.9. For α = 1, 2 we can choose r = n7/16 in (2.34) while for α = 0 we choose r independent of n and then let r →∞.
2.3.1 Exponential moments The estimation of probabilities for atypical values can be done more accurately for random walks whose increment distribution has an exponential moment. In this section we prove the following.
Theorem 2.3.11 Suppose p ∈P′ d such that for some b > 0, E h eb|X1|i < ∞.
(2.35) Then there exists ρ > 0 such that for all n ≥0 and all x ∈Zd with |x| < ρ n, pn(x) = pn(x) exp O 1 √n + |x|3 n2 .
Moreover, if all the third moments of X1 vanish, pn(x) = pn(x) exp O 1 n + |x|4 n3 .
♣Note that the conclusion of the theorem can be written |pn(x) −pn(x)| ≤c pn(x) 1 nα/2 + |x|2+α n1+α , |x| ≤n 1+α 2+α , 2.3 LCLT — characteristic function approach 43 |pn(x) −pn(x)| ≤pn(x) exp O |x|2+α n1+α , |x| ≥n 1+α 2+α where α = 2 if the third moments vanish and α = 1 otherwise. In particular, if xn is a sequence of points in Zd, then as n →∞, pn(xn) ∼pn(xn) if |xn| = o(nβ), pn(xn) ≍pn(xn) if |xn| = O(nβ), where β = 2/3 if α = 1 and β = 3/4 if α = 2.
Theorem 2.3.11 will follow from a stronger result (Theorem 2.3.12). Before stating it we introduce some additional notation and make several preliminary observations. Let p ∈P′ d have characteristic function φ and covariance matrix Γ, and assume that p satisfies (2.35). If the third moments of p vanish, we let α = 2; otherwise, α = 1. Let M denote the moment generating function, M(b) = E[eb·X] = φ(−ib), which by (2.35) is well defined in a neighborhood of the origin in Cd.
Moreover, we can find C < ∞, ǫ > 0 such that E h |X|4e|b·X|i ≤C, |b| < ǫ.
(2.36) In particular, there is a uniform bound in this neighborhood on all the derivatives of M of order at most four. (A (finite) number of times in this section we will say that something holds for all b in a neighborhood of the origin. At the end, one should take the intersection of all such neighborhoods.) Let L(b) = log M(b), L(iθ) = log φ(θ). Then in a neighborhood of the origin we have M(b) = 1 + b · Γb 2 + O |b|α+2 , ∇M(b) = Γb + O |b|α+1 , L(b) = b · Γb 2 + O(|b|α+2), ∇L(b) = ∇M(b) M(b) = Γb + O(|b|α+1).
(2.37) For |b| < ǫ, let pb ∈P∗ d be the probability measure pb(x) = eb·x p(x) M(b) , (2.38) and let Pb, Eb denote probabilities and expectations associated to a random walk with increment distribution pb. Note that Pb{Sn = x} = eb·x M(b)−n P{Sn = x}.
(2.39) The mean of pb is equal to mb = E[X eb·X] E[eb·X] = ∇L(b).
A standard “large deviations” technique for understanding P{Sn = x} is to study Pb{Sn = x} where b is chosen so that mb = x/n. We will apply this technique in the current context. Since Γ is an invertible matrix, (2.37) implies that b 7→∇L(b) maps {|b| < ǫ} one-to-one and onto a 44 Local Central Limit Theorem neighborhood of the origin, where ǫ > 0 is sufficiently small. In particular, there is a ρ > 0 such that for all w ∈Rd with |w| < ρ, there is a unique |bw| < ǫ with ∇L(bw) = w.
♣One could think of the “tilting” procedure of (2.38) as “weighting by a martingale”. Indeed, it is easy to see that for |b| < ǫ, the process Nn = M(b)−n exp {bSn} is a martingale with respect to the filtration {Fn} of the random walk. The measure Pb is obtained by weighting by Nn. More precisely, if E is an Fn-measurable event, then Pb(E) = E [Nn 1E] .
The martingale property implies the consistency of this definition. Under the measure Pb, Sn has the distribution of a random walk with increment distribution pb and mean mb. For fixed n, x we choose mb = x/n so that x is a typical value for Sn under Pb. This construction is a random walk analogue of the Girsanov transformation for Brownian motion.
Let φb denote the characteristic function of pb which we can write as φb(θ) = Eb[eiθ·X] = M(iθ + b) M(b) .
(2.40) Then there is a neighborhood of the origin such that for all b, θ in the neighborhood, we can expand log φb as log φb(θ) = i mb · θ −θ · Γbθ 2 + f3,b(θ) + h4,b(θ).
(2.41) Here Γb is the covariance matrix for the increment distribution pb, and f3,b(θ) is the homogeneous polynomial of degree three f3,b(θ) = −i 6 Eb[(θ · X)3] + 2 (Eb[θ · X])3 .
Due to (2.36), the coefficients for the third order Taylor polynomial of log φb are all differential in b with bounded derivatives in the same neighborhood of the origin. In particular we conclude that |f3,b(θ)| ≤c |b|α−1 |θ|3, |b|, |θ| < ǫ.
To see this if α = 1 use the boundedness of the first and third moments. If α = 2, note that f3,0(θ) = 0, θ ∈Rd, and use the fact that the first and third moments have bounded derivatives as functions of b. Similarly, Γb = E[XXT eb·X] M(b) = Γ + O(|b|α), The error term h4,b is bounded by h4,b(θ) ≤c |θ|4, |b|, |θ| < ǫ.
Note that due to (2.37) (and invertibility of Γ) we have both |bw| = O(|w|) and |w| = O(|bw|).
Combining this with the above observations, we can conclude mb = Γb + O(|w|1+α), 2.3 LCLT — characteristic function approach 45 bw = Γ−1w + O(|w|1+α), (2.42) det Γbw = det Γ + O(|w|α) = det Γ + O(|bw|α), (2.43) L(bw) = bw · Γbw 2 + O(|bw|2+α) = w · Γ−1w 2 + O(|w|2+α).
(2.44) By examining the proof of (2.13) one can find a (perhaps different) δ > 0, such that for all |b| ≤ǫ and all θ ∈[−π, π]d, |e−imb·θ φb(θ)| ≤1 −δ |θ|2.
(For small θ use the expansion of φ near 0, otherwise consider maxθ,b |e−imb·θ φb(θ)| where the maximum is taken over all such θ ∈{z ∈[−π, π]d : |z| ≥ǫ} and all |b| ≤ǫ.) Theorem 2.3.12 Suppose p satisfies the assumptions of Theorem 2.3.11, and let L, bw be defined as above. Then there exists c < ∞and ρ > 0 such that the following holds. Suppose x ∈Zd with |x| ≤ρn and b = bx/n. Then (2π det Γb)d/2 nd/2 Pb{Sn = x} −1 ≤c (|x|α−1 + √n) n(α+1)/2 .
(2.45) In particular, pn(x) = P{Sn = x} = pn(x) exp O 1 nα/2 + |x|2+α n1+α .
(2.46) Proof [of (2.46) given (2.45)] We can write (2.45) as Pb{Sn = x} = 1 (2π det Γb)d/2 nd/2 1 + O |x|α−1 n(α+1)/2 + 1 nα/2 .
By (2.39), pn(x) = P{Sn = x} = M(b)n e−b·x Pb{Sn = x} = exp {n L(b) −b · x} Pb{Sn = x}.
From (2.43), we see that (det Γb)−d/2 = (det Γ)−d/2 1 + O |x|α nα , and due to (2.44), we have n L(b) −x · Γ−1x 2n ≤c |x|2+α n1+α .
Applying (2.42), we see that b · x = Γ−1 x n + O |x|α+1 nα+1 · x = x · Γ−1x n + O |x|α+2 nα+1 .
Therefore, exp {n L(b) −b · x} = exp −x · Γ−1x 2n exp O |x|2+α n1+α .
46 Local Central Limit Theorem Combining these and recalling the definition of pn(x) we get, pn(x) = pn(x) exp O |x|2+α n1+α .
Therefore, it suffices to prove (2.45). The argument uses an LCLT for probability distributions on Zd with non-zero mean. Suppose K < ∞and X is a random variable in Zd with mean m ∈Rd, covariance matrix Γ, and E[|X|4] ≤K. Let ψ be the characteristic function of X. Then there exist ǫ, C, depending only on K, such that for |θ| < ǫ, log ψ(θ) − im · θ + θ · Γθ 2 + f3(θ) ≤C |θ|4.
(2.47) where the term f3(θ) is a homogeneous polynomial of degree 3. Let us write K3 for the smallest number such that |f3(θ)| ≤K3|θ|3 for all θ. Note that there exist uniform bounds for m, Γ and K3 in terms of K. Moreover, if α = 2 and f3 corresponds to pb, then |K3| ≤c |b|. The next proposition is proved in the same was as Theorem 2.3.5 taking some extra care in obtaining uniform bounds.
The relation (2.45) then follows from this proposition and the bound K3 ≤c ( |x|/n)α−1.
Proposition 2.3.13 For every δ > 0, K < ∞, there is a c such that the following holds. Let p be a probability distribution on Zd with E[|X|4] ≤K. Let m, Γ, C, ǫ, ψ, K3 be as in the previous paragraph. Moreover, assume that |e−im·θ ψ(θ)| ≤1 −δ |θ|2, θ ∈[−π, π]d.
Suppose X1, X2, . . . are independent random variables with distribution p and Sn = X1 + · · · + Xn.
Then if nm ∈Zd, (2πn det Γ)d/2 P{Sn = nm} −1 ≤c[K3 √n + 1] n .
Remark. The error term indicates existence of two different regimes: K3 ≤n−1/2 and K3 ≥n−1/2.
Proof We fix δ, K and allow all constants in this proof to depend only on δ and K. Proposition 2.2.2 implies that P{Sn = nm} = 1 (2π)d Z [−π,π]d[e−im·θ ψ(θ)]n dθ.
The uniform upper bound on E[|X|4] implies uniform upper bounds on the lower moments. In particular, det Γ is uniformly bounded and hence it suffices to find n0 such that the result holds for n ≥n0. Also observe that (2.47) holds with a uniform C from which we conclude n log ψ θ √n −i √n (m · θ) −θ · Γθ 2 −n f3 θ √n ≤C |θ|4 n .
In addition we have |nf3(θ/√n)| ≤K3 |θ|3/√n. The proof proceeds as the proof of Theorem 2.3.5, the details are left to the reader.
2.4 Some corollaries of the LCLT 47 2.4 Some corollaries of the LCLT Proposition 2.4.1 If p ∈P′ with bounded support, there is a c such that X z∈Zd |pn(z) −pn(z + y)| ≤c |y| n−1/2.
Proof By the triangle inequality, it suffices to prove the result for y = e = ej. Let δ = 1/2d. By Theorem 2.3.6, pn(z + e) −pn(z) = ∇jpn(z) + O 1 n(d+2)/2 .
Also Corollary 12.2.7 shows that X |z|≥n(1/2)+δ |pn(z) −pn(z + e)| ≤ X |z|≥n(1/2)+δ [pn(z) + pn(z + e)] = o(n−1/2).
But, X |z|≤n(1/2)+δ |pn(z) −pn(z + e)| ≤ X z∈Zd |pn(z) −pn(z + e)| + X |z|≤n(1/2)+δ O 1 n(d+2)/2 ≤ O(n−1/2) + X z∈Zd |∇jpn(z)|.
A straightforward estimate which we omit gives X z∈Zd |∇jpn(z)| = O(n−1/2).
The last proposition holds with much weaker assumptions on the random walk. Recall that P∗ is the set of increment distributions p with the property that for each x ∈Zd, there is an Nx such that pn(x) > 0 for all n ≥Nx.
Proposition 2.4.2 If p ∈P∗, there is a c such that X z∈Zd |pn(z) −pn(z + y)| ≤c |y| n−1/2.
Proof In Exercise 1.3 it was shown that we can write p = ǫ q + (1 −ǫ)q′, where q ∈P′ with bounded support and q′ ∈P∗. By considering the process of first choosing q or q′ and then doing the jump, we can see that pn(x) = n X j=0 n j ǫj (1 −ǫ)n−j X z∈Zd qj(x −z) q′ n−j(z).
(2.48) 48 Local Central Limit Theorem Therefore, X x∈Zd |pn(x) −pn(x + y)| ≤ n X j=0 n j ǫj (1 −ǫ)n−j X x∈Zd q′ n−j(x) X z∈Zd |qj(x −z) −qj(x + y −z)|.
We split the first sum into the sum over j < (ǫ/2)n and j ≥(ǫ/2)n. Standard exponential estimates for the binomial (see Lemma 12.2.8) give X j<(ǫ/2)n n j ǫj (1 −ǫ)n−j X x∈Zd q′ n−j(x) X z∈Zd |qj(x −z) −qj(x + y −z)| ≤2 X j<(ǫ/2)n n j ǫj (1 −ǫ)n−j = O(e−αn), for some α = α(ǫ) > 0. By Proposition 2.4.1, X j≥(ǫ/2)n n j ǫj (1 −ǫ)n−j X x∈Zd q′ n−j(x) X z∈Zd |qj(x −z) −qj(x + y −z)| ≤c n−1/2 |y| X j≥(ǫ/2)n n j ǫj (1 −ǫ)n−j X x∈Zd q′ n−j(x) ≤c n−1/2 |y|.
The last proposition has the following useful lemma as a corollary. Since this is essentially a result about Markov chains in general, we leave the proof to the appendix, see Theorem 12.4.5.
Lemma 2.4.3 Suppose p ∈P∗ d. There is a c < ∞such that if x, y ∈Zd, we can define Sn, S∗ n on the same probability space such that: Sn has the distribution of a random walk with increment p with S0 = x; S∗ n has the distribution of a random walk with increment p with S0 = y; and such that for all n, P{Sm = S∗ m for all m ≥n} ≥1 −c |x −y| √n .
♣While the proof of this last lemma is somewhat messy to write out in detail, there really is not a lot of content to it once we have Proposition 2.4.2. Suppose that p, q are two probability distributions on Zd with 1 2 X z∈Zd |p(z) −q(z)| = ǫ.
Then there is an easy way to define random variables X, Y on the same probability space such that X has distribution p, Y has distribution q and P{X ̸= Y } = ǫ. Indeed, if we let f(z) = min{p(z), q(z)} we can let the probability space be Zd × Zd and define µ by µ(z, z) = f(z) 2.4 Some corollaries of the LCLT 49 and for x ̸= y, µ(x, y) = ǫ−1 [p(x) −f(x)] [q(y) −f(y)].
If we let X(x, y) = x, Y (x, y) = y, it is easy to check that the marginal of X is p, the marginal of Y is q and P{X = Y } = 1 −ǫ. The more general fact is not much more complicated than this.
Proposition 2.4.4 Suppose p ∈P∗ d. There is a c < ∞such that for all n, x, pn(x) ≤ c nd/2 .
(2.49) Proof If p ∈P′ d with bounded support this follows immediately from (2.22). For general p ∈P∗ d, write p = ǫ q + (1 −ǫ) q′ with q ∈P′ d, q′ ∈P∗ d as in the proof of Proposition 2.22. Then pn(x) is as in (2.48).
The sum over j < (ǫ/2)n is O(e−αn) and for j ≥(ǫ/2)n, we have the bound qj(x −z) ≤c n−d/2.
The central limit theorem implies that it takes O(n2) steps to go distance n. This proposition gives some bounds on large deviations for the number of steps.
Proposition 2.4.5 Suppose S is a random walk with increment distribution p ∈Pd and let τn = min{k : |Sk| ≥n}, ξn = min{k : J ∗(Sk) ≥n}.
There exist t > 0 and c < ∞such that for all n and all r > 0, P{τn ≤rn2} + P{ξn ≤rn2} ≤c e−t/r, (2.50) P{τn ≥rn2} + P{ξn ≥rn2} ≤c e−rt.
(2.51) Proof There exists a ˜ c such that ξ˜ cn ≤τn ≤ξn/˜ c so it suffices to prove the estimates for τn. It also suffices to prove the result for n sufficiently large. The central limit theorem implies that there is an integer k such that for all n sufficiently large, P{|Skn2| ≥2n} ≥1 2.
By the strong Markov property, this implies for all l P{τn > kn2 + l | τn > l} ≤1 2, and hence P{τn > jkn2} ≤(1/2)j = e−j log 2 = e−jk(log 2/k).
This gives (2.51). The estimate (2.50) on τn can be written as P max 1≤j≤rn2 |Sj| ≥n = P max 1≤j≤rn2 |Sj| ≥(1/√r) √ rn2 ≤c e−t/r, which follows from (2.7).
50 Local Central Limit Theorem ♣The upper bound (2.51) for τn does not need any assumptions on the distribution of the increments other than they be nontrivial, see Exercise 2.7.
Theorem 2.3.10 implies that for all p ∈P′ d, pn(x) ≤c n−d/2 (√n/|x|)2. The next proposition extends this to real r > 2 under the assumption that E[|X1|r] < ∞.
Proposition 2.4.6 Suppose p ∈P∗ d. There is a c such that for all n, x, pn(x) ≤ c nd/2 max 0≤j≤n P {|Sj| ≥|x|/2} .
In particular, if r > 2, p ∈P′ d and E[|X1|r] < ∞, then there exists c < ∞such that for all n, x, pn(x) ≤ c nd/2 √n |x| r .
(2.52) Proof Let m = n/2 if n is even and m = (n + 1)/2 if n is odd. Then, {Sn = x} = {Sn = x, |Sm| ≥|x|/2} ∪{Sn = x, |Sn −Sm| ≥|x|/2} .
Hence it suffices to estimate the probabilities of the events on the right-hand side. Using (2.49) we get P {Sn = x, |Sm| ≥|x|/2} = P {|Sm| ≥|x|/2} P {Sn = x | |Sm| ≥|x|/2} ≤ P {|Sm| ≥|x|/2} sup y pn−m(y, x) ≤ c n−d/2 P {|Sm| ≥|x|/2} .
The other probability can be estimated similarly since P {Sn = x, |Sn −Sm| ≥|x|/2} = P {Sn = x, |Sn−m| ≥|x|/2} .
We claim that if p ∈P′ d, r ≥2, and E[|X1|r] < ∞, then there is a c such that E[|Sn|r] ≤c nr/2.
Once we have this, the Chebyshev inequality gives for m ≤n, P{|Sm| ≥|x|} ≤c nr/2 |x|r .
The claim is easier when r is an even integer (for then we can estimate the expectation by expanding (X1 + · · · + Xn)r), but we give a proof for all r ≥2. Without loss of generality, assume d = 1. For a fixed n define T1 = ˜ T1 = min{j : |Sj| ≥c1 √n}, and for l > 1, ˜ Tl = min n j > ˜ Tl−1 : Sj −STl−1 ≥c1 √n o , Tl = ˜ Tl −˜ Tl−1, where c1 is chosen sufficiently large so that P{T1 > n} ≥1 2.
2.5 LCLT — combinatorial approach 51 The existence of such a c1 follows from (2.6) applied with k = 1.
Let Y1 = |ST1| and for l > 1, Yl = S ˜ Tl −S ˜ Tl−1 . Note that (T1, Y1), (T2, Y2), (T3, Y3), . . . are independent, identically distributed random variables taking values in {1, 2, . . .} × R. Let ξ be the smallest l ≥1 such that Tl > n. Then one can readily check from the triangle inequality that |Sn| ≤ Y1 + Y2 + · · · + Yξ−1 + c1 √n = c1 √n + ∞ X l=1 ˆ Yl, where ˆ Yl = Yl 1{Tl ≤n} 1{ξ > l −1}. Note that P{Y1 ≥c1 √n + t; T1 ≤n} ≤ P{|Xj| ≥t for some 1 ≤j ≤n} ≤ n P{|X1| ≥t}.
Letting Z = |X1|, we get E[ ˆ Y r 1 ] = E[Y r 1 ; Tl ≤n] = c Z ∞ 0 sr−1 P{Y1 ≥s; T1 ≤n} ds ≤ c " nr/2 + Z ∞ (c1+1)√n sr−1 n P{Z ≥s −c1 √n} ds # = c nr/2 + Z ∞ √n (s + √n)r−1 n P{Z ≥s} ds ≤ c nr/2 + 2r−1 Z ∞ √n sr−1 n P{Z ≥s} ds ≤ c h nr/2 + 2r−1 n E[Zr] i ≤c nr/2.
For l > 1, E[ ˆ Y r l ] ≤P{ξ > l −1} E[Y r l 1{Tl ≥n} | ξ > l −1] = 1 2 l−1 E[ ˆ Y r 1 ].
Therefore, E h ( ˆ Y1 + ˆ Y2 + · · · )ri = lim l→∞E h ( ˆ Y1 + · · · + ˆ Yl)ri ≤ lim l→∞ h E[ ˆ Y r 1 ]1/r + · · · + E[ ˆ Y r l ]1/rir = E[ ˆ Y r 1 ] " ∞ X l=1 1 2 (l−1)/r#r = c E[ ˆ Y r 1 ].
2.5 LCLT — combinatorial approach In this section, we give another proof of an LCLT with estimates for one-dimensional simple random walk, both discrete and continuous time, using an elementary combinatorial approach. Our results are no stronger than that derived earlier, and this section is not needed for the remainder of the 52 Local Central Limit Theorem book, but it is interesting to see how much can be derived by simple counting methods. While we focus on simple random walk, extensions to p ∈Pd are straightforward using (1.2). Although the arguments are relatively elementary, they do require a lot of calculation and estimation. Here is a basic outline: • Establish the result for discrete time random walk by exact counting of paths. Along the way we will prove Stirling’s formula.
• Prove an LCLT for Poisson random variables and use it to derive the result for one-dimensional continuous-time walks. (A result for d-dimensional continuous-time simple random walks follows immediately.) We could continue this approach and prove an LCLT for multinomial random variables and use it to derive the result for discrete-time d-dimensional simple random walk, but we have chosen to omit this.
2.5.1 Stirling’s formula and 1-d walks Suppose Sn is a simple one-dimensional random walk starting at the origin.
Determining the distribution of Sn reduces to an easy counting problem. In order for X1 + · · · + X2n to equal 2k, exactly n + k of the Xj must equal +1. Since all 2−2n sequences of ±1 are equally likely, p2n(2k) = P{S2n = 2k} = 2−2n 2n n + k = 2−2n (2n)!
(n + k)! (n −k)!.
(2.53) We will use Stirling’s formula, which we now derive, to estimate the factorial. In the proof, we will use some standard estimates about the logarithm, see Section 12.1.2.
Theorem 2.5.1 (Stirling’s formula) As n →∞, n! ∼ √ 2π nn+(1/2) e−n.
In fact, n!
√ 2π nn+(1/2) e−n = 1 + 1 12 n + O 1 n2 .
Proof Let bn = nn+(1/2)e−n/n!. Then, (12.5) and Taylor’s theorem imply bn+1 bn = 1 e 1 + 1 n n 1 + 1 n 1/2 = 1 −1 2n + 11 24n2 + O 1 n3 1 + 1 2n − 1 8n2 + O 1 n3 = 1 + 1 12n2 + O 1 n3 .
Therefore, lim m→∞ bm bn = ∞ Y l=n 1 + 1 12l2 + O 1 l3 = 1 + 1 12n + O 1 n2 .
2.5 LCLT — combinatorial approach 53 The second equality is obtained by log ∞ Y l=n 1 + 1 12l2 + O 1 l3 = ∞ X l=n log 1 + 1 12l2 + O 1 l3 = ∞ X l=n 1 12l2 + ∞ X l=n O 1 l3 = 1 12n + O 1 n2 .
This establishes that the limit C := h lim m→∞bm i−1 exists and bn = 1 C 1 − 1 12n + O 1 n2 , n! = C nn+(1/2) e−n 1 + 1 12n + O 1 n2 .
There are a number of ways to determine the constant C.
For example, if Sn denotes a one-dimensional simple random walk, then P{|S2n| ≤ √ 2n log n} = X |2k|≤ √ 2n log n 4−n 2n n + k = X |2k|≤ √ 2n log n 4−n (2n)!
(n + k)!(n −k)!.
Using (12.3), we see that as n →∞, if |2k| ≤ √ 2n log n, 4−n (2n)!
(n + k)!(n −k)! ∼ √ 2 C √n 1 + k n −(n+k) 1 −k n −(n−k) = √ 2 C √n 1 −k2/n n −n 1 + k2/n k −k 1 −k2/n k k ∼ √ 2 C √n e−k2/n.
Therefore, lim n→∞P{|S2n| ≤ √ 2n log n} = lim n→∞ X |k|≤√ n/2 log n √ 2 C √n e−k2/n = √ 2 C Z ∞ −∞ e−x2 dx = √ 2π C .
However, Chebyshev’s inequality shows that P{|S2n| ≥ √ 2n log n} ≤Var[S2n] 2n log2 n = 1 log2 n − →0.
Therefore, C = √ 2π.
♣By adapting this proof, it is easy to see that one can find r1 = 1/12, r2, r3, . . . such that for each positive integer k, n! = √ 2π nn+(1/2) e−n 1 + r1 n + r2 n2 + · · · + rk nk + O 1 nk+1 .
(2.54) 54 Local Central Limit Theorem We will now prove Theorem 2.1.1 and some difference estimates in the special case of simple random walk in one dimension by using (2.53) and Stirling’s formula. As a warmup, we start with the probability of being at the origin.
Proposition 2.5.2 For simple random walk in Z, if n is a positive integer, then P{S2n = 0} = 1 √πn 1 −1 8n + O 1 n2 .
Proof The probability is exactly 2−2n 2n n = (2n)!
4n (n!)2 .
By plugging into Stirling’s formula, we see that the right hand side equals 1 √πn 1 + (24n)−1 + O(n−2) [1 + (12n)−1 + O(n−2)]2 = 1 √πn 1 −1 8n + O 1 n2 .
In the last proof, we just plugged into Stirling’s formula and evaluated. We will now do the same thing to prove a version of the LCLT for one-dimensional simple random walk.
Proposition 2.5.3 For simple random walk in Z, if n is a positive integer and k is an integer with |k| ≤n, p2n(2k) = P{S2n = 2k} = 1 √πn e−k2/n exp O 1 n + k4 n3 .
In particular, if |k| ≤n3/4, then P{S2n = 2k} = 1 √πn e−k2/n 1 + O 1 n + k4 n3 .
♣Note that for one-dimensional simple random walk, 2 p2n(2k) = 2 1 p (2π) (2n) exp −(2k)2 2 (2n) = 1 √πn e−k2/n.
♣While the theorem is stated for all |k| ≤n, it is not a very strong statement when k is of order n. For example, for n/2 ≤|k| ≤n, we can rewrite the conclusion as p2n(2k) = 1 √πn e−k2/n eO(n) = eO(n), which only tells us that there exists α such that e−αn ≤p2n(2k) ≤eαn.
2.5 LCLT — combinatorial approach 55 In fact, 2p2n(2k) is not a very good approximation of p2n(2k) for large n. As an extreme example, note that p2n(2n) = 4−n, 2 p2n(2n) = 1 √πn e−n.
Proof If n/2 ≤|k| ≤n, the result is immediate using only the estimate 2−2n ≤P{S2n = 2k} ≤1.
Hence, we may assume that |k| ≤n/2. As noted before, P{S2n = 2k} = 2−2n 2n n + k = (2n)!
22n(n + k)!(n −k)!.
If we restrict to |k| ≤n/2, we can use Stirling’s formula (Lemma 2.5.1) to see that P{S2n = 2k} = 1 + O 1 n 1 √πn 1 −k2 n2 −1/2 1 −k2 n2 −n 1 − 2k n + k k .
The last two terms approach exponential functions. We need to be careful with the error terms.
Using (12.3) we get, 1 −k2 n2 n = e−k2/n exp O k4 n3 .
1 − 2k n + k k = e−2k2/(n+k) exp − 2k3 (n + k)2 + O k4 n3 = e−2k2/(n+k) exp −2k3 n2 + O k4 n3 , e−2k2/(n+k) = e−2k2/n exp 2k3 n2 + O k4 n3 .
Also, using k2/n2 ≤max{(1/n), (k4/n3)}, we can see that 1 −k2 n2 −1/2 = exp O 1 n + k4 n3 .
Combining all of this gives the theorem.
♣We could also prove “difference estimates” by using the equalities p2n(2k + 2) = n −k n + k + 1 p2n(2k), p2(n+1)(2k) = p2n(2k) 4−1 (2n + 1)(2n + 2) (n + k + 1)(n −k + 1).
56 Local Central Limit Theorem Corollary 2.5.4 If Sn is simple random walk, then for all positive integers n and all |k| < n, P{S2n+1 = 2k + 1} = 1 √πn exp ( −(k + 1 2)2 n ) exp O 1 n + k4 n3 .
(2.55) Proof Note that P{S2n+1 = 2k + 1} = 1 2 P{S2n = 2k} + 1 2 P{S2n = 2(k + 1)}.
Hence, P{S2n+1 = 2k + 1} = 1 2√πn[e−k2/n + e−(k+1)2/n] exp O 1 n + k4 n3 .
But, exp ( −(k + 1 2)2 n ) = e−k2/n 1 −k n + O k2 n2 , exp −(k + 1)2 n = e−k2/n 1 −2k n + O k2 n2 , which implies 1 2 h e−k2/n + e−(k+1)2/ni = exp ( −(k + 1 2)2 n ) 1 + O k2 n2 .
Using k2/n2 ≤max{(1/n), (k4/n3)}, we get (2.55).
♣One might think that we should replace n in (2.55) with n + (1/2). However, 1 n + (1/2) = 1 n 1 + O 1 n .
Hence, the same statement with n + (1/2) replacing n is also true.
2.5.2 LCLT for Poisson and continuous-time walks The next proposition establishes the strong LCLT for Poisson random variables. This will be used for comparing discrete-time and continuous-time random walks with the same p. If Nt is a Poisson random variable with parameter t, then E[Nt] = t, Var[Nt] = t. The central limit theorem implies that as t →∞, the distribution of (Nt −t)/ √ t approaches that of a standard normal. Hence, we might conjecture that P{Nt = m} = P m −t √ t ≤Nt −t √ t < m + 1 −t √ t ≈ Z (m+1−t)/ √ t (m−t)/ √ t 1 √ 2π e−x2 2 dx ≈ 1 √ 2πt e−(m−t)2 2t .
In the next proposition, we use a straightforward combinatorial argument to justify this approxi-mation.
2.5 LCLT — combinatorial approach 57 Proposition 2.5.5 Suppose Nt is a Poisson random variable with parameter t, and m is an integer with |m −t| ≤t/2. Then P{Nt = m} = 1 √ 2πt e−(m−t)2 2t exp O 1 √ t + |m −t|3 t2 .
Proof For notational ease, we will first consider the case where t = n is an integer, and we let m = n + k. Let q(n, k) = P{Nn = n + k} = e−n nn+k (n + k)!, and note the recursion formula q(n, k) = n n + k q(n, k −1).
Stirling’s formula (Theorem 2.5.1) gives q(n, 0) = e−n nn n!
= 1 √ 2πn 1 + O 1 n .
(2.56) By the recursion formula, if k ≤n/2, q(n, k) = q(n, 0) 1 + 1 n 1 + 2 n · · · 1 + k n −1 , and, log k Y j=1 1 + j n = k X j=1 log 1 + j n = k X j=1 j n + O j2 n2 = k2 2n + k 2n + O k3 n2 = k2 2n + O 1 √n + k3 n2 .
The last equality uses the inequality k n ≤max 1 √n, k3 n2 , which will also be used in other estimates in this proof. Using (2.56), we get log q(n, k) = −log √ 2πn −k2 2n + O 1 √n + k3 n2 , and the result for k ≥0 follows by exponentiating.
Similarly, q(n, −k) = q(n, 0) 1 −1 n 1 −2 n · · · 1 −k −1 n 58 Local Central Limit Theorem and log q(n, −k) = −log √ 2πn + log k−1 Y j=1 1 −j n = −log √ 2πn −k2 2n + O 1 √n + k3 n2 .
The proposition for integer n follows by exponentiating. For general t, let n = ⌊t⌋and note that P{Nt = n + k} = P{Nn = n + k} e−(t−n) 1 + t −n n n+k = P{Nn = n + k} 1 + t −n n k 1 + O 1 n = P{Nn = n + k} 1 + O |k| + 1 n = (2πn)−1/2 e−k2/(2n) exp O 1 √n + k3 n2 = (2πt)−1/2 e−(k+n−t)2/(2t) exp O 1 √ t + |n + k −t|3 t2 .
The last step uses the estimates 1 √ t = 1 √n 1 + O 1 t , e−k2 2t = e−k2 2n exp O k2 t2 .
We will use this to prove a version of the local central limit theorem for one-dimensional, continuous-time simple random walk.
Theorem 2.5.6 If ˜ St is continuous-time one-dimensional simple random walk, then if |x| ≤t/2, ˜ pt(x) = 1 √ 2πt e−x2 2t exp O 1 √ t + |x|3 t2 .
Proof We will assume that x = 2k is even; the odd case is done similarly. We know that ˜ pt(2k) = ∞ X m=0 P{Nt = 2m} p2m(2k).
Standard exponential estimates, see (12.12), show that for every ǫ > 0, there exist c, β such that P{|Nt −t| ≥ǫt} ≤c e−βt. Hence, ˜ pt(2k) = ∞ X m=0 P{Nt = 2m} p2m(2k) = O(e−βt) + X P{Nt = 2m} p2m(2k), (2.57) 2.5 LCLT — combinatorial approach 59 where here and for the remainder of this proof, we write just P to denote the sum over all integers m with |t −2m| ≤ǫt. We will show that there is an ǫ such that X P{Nt = 2m} p2m(2k) = 1 √ 2πt e−x2 2t exp O 1 √ t + |x|3 t2 .
A little thought shows that this and (2.57) imply the theorem.
By Proposition 2.5.3 we know that p2m(2k) = P{S2m = 2k} = 1 √πm e−k2 m exp O 1 m + k4 m3 , and by Proposition 2.5.5 we know that P{Nt = 2m} = 1 √ 2πt e−(2m−t)2 2t exp O 1 √ t + |2m −t|3 t2 .
Also, we have 1 2m = 1 t 1 + O |2m −t| t , 1 √ 2m = 1 √ t 1 + O |2m −t| t , which implies e−k2 m = e−2k2 t exp O k2 |2m −t| t2 .
Combining all of this, we can see that the sum in (2.57) can be written as 1 √ 2πt e−x2 2t exp O 1 √ t + |x|3 t2 X 2 √ 2πt e−(2m−t)2 2t exp O |2m −t|3 t2 .
We now choose ǫ so that |O(|2m −t|3/t2)| ≤(2m −t)2/(4t) for all |2m −t| ≤ǫt. We will now show that X 2 √ 2πt e−(2m−t)2 2t exp O |2m −t|3 t2 = 1 + O 1 √ t , which will complete the argument. Since e−(2m−t)2 2t exp O |2m −t|3 t2 ≤e−(2m−t)2 4t , is easy to see that the sum over |2m−t| > t2/3 decays faster than any power of t. For |2m−t| ≤t2/3 we write exp O |2m −t|3 t2 = 1 + O |2m −t|3 t2 .
The estimate X |2m−t|≤t2/3 2 √ 2πt e−(2m−t)2 2t = X |m|≤t2/3/2 2 √ 2πt e−2(m/ √ t)2 = O 1 √ t + 2 Z ∞ −∞ 1 √ 2π e−2y2 dy = 1 + O 1 √ t 60 Local Central Limit Theorem is a standard approximation of an integral by a sum. Similarly, X |2m−t|≤t2/3 O |2m −t|3 t2 2 √ 2πt e−(2m−t)2 2t ≤c √ t Z ∞ −∞ |y|3 √ 2πe−2y2 dy = O 1 √ t .
Exercises Exercise 2.1 Suppose p ∈P′ d, ǫ ∈(0, 1) and E[|X1|2+ǫ] < ∞. Show that the characteristic function has the expansion φ(θ) = 1 −θ · Γθ 2 + o(|θ|2+ǫ), θ →0.
Show that the δn in (2.32) can be chosen so that nǫ/2 δn →0.
Exercise 2.2 Show that if p ∈P∗ d, there exists a c such that for all x ∈Zd and all positive integers n, |pn(x) −pn(0)| ≤c |x| n(d+1)/2 .
(Hint: first show the estimate for p ∈P′ d with bounded support and then use (2.48). Alternatively, one can use Lemma 2.4.3 at time n/2, the Markov property, and (2.49). ) Exercise 2.3 Show that Lemma 2.3.2 holds for p ∈P∗.
Exercise 2.4 Suppose p ∈P′ d with E[|X|3] < ∞. Show that there is a c < ∞such that for all |y| = 1, |pn(0) −pn(y)| ≤ c n(d+2)/2 .
Exercise 2.5 Suppose p ∈P∗ d. Let A ⊂Zd and h(x) = Px{Sn ∈A i.o.}.
Show that if h(x) > 0 for some x ∈Zd, then h(x) = 1 for all x ∈Zd.
Exercise 2.6 Suppose Sn is a random walk with increment distribution p ∈Pd. Show that there exists a b > 0 such that sup n>1 E exp b|Sn|2 n < ∞.
Exercise 2.7 Suppose X1, X2, . . . are independent, identically distributed random variables in Zd with P{X1 = 0} < 1 and let Sn = X1 + · · · + Xn.
• Show that there exists an r such that for all n P{|Srn2| ≥n} ≥1 2.
2.5 LCLT — combinatorial approach 61 • Show that there exist c, t such that for all b > 0, P max 1≤j≤n2 |Sj| ≤bn ≤c e−t/b.
Exercise 2.8 Find r2, r3 in (2.54).
Exercise 2.9 Let Sn denote one-dimensional simple random walk. In this exercise we will prove without using Stirling’s formula that there exists a constant C such that p2n(0) = C √n 1 −1 8n + O 1 n2 .
a. Show that if n ≥1, p2(n+1) = 1 + 1 2n 1 + 1 n −1 p2n.
b. Let bn = √n p2n(0). Show that b1 = 1/2 and for n ≥1, bn+1 bn = 1 + 1 8n2 + O 1 n3 .
c. Use this to show that b∞= lim bn exists and is positive. Moreover, bn = b∞ 1 −1 8n + O 1 n2 .
Exercise 2.10 Show that if p ∈P′ d with E[|X1|3] < ∞, then ∇2 jpn(x) = ∇2 jpn(x) + O(n−(d+3)/2).
Exercise 2.11 Suppose q : Zd →R has finite support, and k is a positive integer such that for all l ∈{1, . . . , k −1} and all j1, . . . , jl ∈{1, . . . , d}, X x=(x1,...,xd)∈Zd xj1 xj2 . . . xjl q(x) = 0.
Then we call the operator Λf(x) := X y f(x + y) q(y) a difference operator of order (at least) k. The order of the operator is the largest k for which this is true. Suppose Λ is a difference operator of order k ≥1.
• Suppose g is a C∞function on Rd. Define gǫ on Zd by gǫ(x) = g(ǫx). Show that |Λgǫ(0)| = O(|ǫ|k), ǫ →0.
• Show that if p ∈P′ d with E[|X1|3] < ∞, then Λpn(x) = Λpn(x) + O(n−(d+1+k)/2).
62 Local Central Limit Theorem • Show that if p ∈P′ d is symmetric with E[|X1|4] < ∞, then Λpn(x) = Λpn(x) + O(n−(d+2+k)/2).
Exercise 2.12 Suppose p ∈P′ ∪P′ 2. Show that there is a c such that the following is true. Let Sn be a p-walk and let τn = inf{j : |Sj| ≥n}.
If y ∈Z2, let Vn(y) = τn−1 X j=0 1{Sj = y} denote the number of visits to y before time τn. Then, if 0 < |y| < n, E [Vk(y)] ≤c 1 + log n −log |y| n .
Hint: Show that there exist c1, β such that for each positive integer j, X jn2≤j<(j+1)n2 1{Sj = y; j < τn} ≤c1 e−βj n−1.
3 Approximation by Brownian motion 3.1 Introduction Suppose Sn = X1 + · · · + Xn is a one-dimensional simple random walk.
We make this into a (random) continuous function by linear interpolation, St = Sn + (t −n) [Sn+1 −Sn], n ≤t ≤n + 1.
For fixed integer n, the LCLT describes the distribution of Sn. A corollary of LCLT is the usual central limit theorem that states that the distribution of n−1/2 Sn converges to that of a standard normal random variable. A simple extension of this is the following: suppose 0 < t1 < t2 < . . . < tk = 1. Then as n →∞the distribution of n−1/2 (St1n, St2n, . . . , Stkn) converges to that of (Y1, Y1 + Y2, . . . , Y1 + Y2 + · · · Yk), where Y1, . . . , Yk are independent mean zero normal random variables with variances t1, t2 −t1, . . . , tk −tk−1, respectively.
The functional central limit theorem (also called the invariance principle or Donsker’s theorem) for random walk extends this result to the random function W (n) t := n−1/2 Stn.
(3.1) The functional central limit theorem states roughly that as n →∞, the distribution of this random function converges to the distribution of a random function t 7→Bt. From what we know about the simple random walk, here are some properties that would be expected of the random function Bt: • If s < t, the distribution of Bt −Bs is N(0, t −s).
• If 0 ≤t0 < t1 < . . . < tk, then Bt1 −Bt0, . . . , Btk −Btk−1 are independent random variables.
These two properties follow almost immediately from the central limit theorem. The third property is not as obvious.
• The function t 7→Bt is continuous.
Although this is not obvious, we can guess this from the heuristic argument: E[(Bt+∆t −Bt)2] ≈∆t, 63 64 Approximation by Brownian motion which indicates that |Bt+∆t −Bt| should be of order √ ∆t. A process satisfying these assumptions will be called a Brownian motion (we will define it more precisely in the next section).
There are a number of ways to make rigorous the idea that W (n) approaches a Brownian motion in the limit. For example, if we restrict to 0 ≤t ≤1, then W (n) and B are random variables taking values in the metric space C[0, 1] with the supremum norm. There is a well understood theory of convergence in distribution of random variables taking values in metric spaces.
We will take a different approach using a method that is often called strong approximation of random walk by Brownian motion. We start by defining a Brownian motion B on a probability space and then define the random walk Sn as a function of the Brownian motion, i.e., for each realization of random function Bt we associate a particular random walk path. We will do this in a way so that the random walk Sn has the distribution of simple random walk. We will then do some estimates to show that there exist positive numbers c, a such that if W (n) t is as defined in (3.1), then for all r ≤n1/4, P{∥B −W (n)∥≥r n−1/4 p log n } ≤c e−ra, (3.2) where ∥· ∥denotes the supremum norm on C[0, 1]. The convergence in distribution follows from the strong estimate (3.2).
♣There is a general approach here that is worth emphasizing. Suppose we have a discrete process and we want to show that it converges after some scaling to a continuous process. A good approach for proving such a result is to first study the conjectured limit process and then to show that the scaled discrete process is a small perturbation of the limit process.
We start by establishing (3.2) for one-dimensional simple random walk using Skorokhod embed-ding. We then extend this to continuous-time walks and all increment distributions p ∈P. The extension will not be difficult; the hard work is done in the one-dimensional case.
We will not handle the general case of p ∈P′ in this book. One can give strong approximations in this case to show that the random walk approaches Brownian motion. However, the rate of convergence depends on the moment assumptions. In particular, the estimate (3.2) will not hold assuming only mean zero and finite second moment.
3.2 Construction of Brownian motion A standard (one-dimensional) Brownian motion with respect to a filtration Ft is a collection of random variables Bt, t ≥0 satisfying the following: (a) B0 = 0; (b) if s < t, then Bt −Bs is an Ft-measurable random variable, independent of Fs, with a N(0, t −s) distribution; (c) with probability one, t 7→Bt is a continuous function.
If the filtration is not given explicitly, then it is assumed to be the natural filtration, Ft = σ{Bs : 0 ≤s ≤t}. In this section, we will construct a Brownian motion and derive an important estimate on the oscillations of the Brownian motion.
3.2 Construction of Brownian motion 65 We will show how to construct a Brownian motion. There are technical difficulties involved in defining a collection of random variables {Bt} indexed over an uncountable set. However, if we know a priori that the distribution should be supported on continuous functions, then we know that the random function t 7→Bt should be determined by its value on a countable, dense subset of times. This observation leads us to a method of constructing Brownian motion: define the process on a countable, dense set of times and then extend the process to all times by continuity.
Suppose (Ω, F, P) is any probability space that is large enough to contain a countable collection of independent N(0, 1) random variables which for ease we will index by Nn,k, n = 0, 1, . . . ; k = 0, 1, . . .
We will use these random variables to define a Brownian motion on (Ω, F, P). Let Dn = k 2n : k = 0, 1, . . .
, D = ∞ [ n=0 Dn denote the nonnegative dyadic rationals. Our strategy will be as follows: • define Bt for t in D satisfying conditions (a) and (b); • derive an estimate on the oscillation of Bt, t ∈D, that implies that with probability one the paths are uniformly continuous on compact intervals; • define Bt for other values of t by continuity.
The first step is straightforward using a basic property of normal random variables. Suppose X, Y are independent normal random variables, each mean 0 and variance 1/2. Then Z = X + Y is N(0, 1). Moreover, the conditional distribution of X given the value of Z is normal with mean Z/2 and variance 1/2. This can be checked directly using the density of the normals. Alternatively, one can check that if Z, N are independent N(0, 1) random variables then X := Z 2 + N 2 , Y := Z 2 −N 2 , (3.3) are independent N(0, 1/2) random variables. To verify this, one only notes that (X, Y ) has a joint normal distribution with E[X] = E[Y ] = 0, E[X2] = E[Y 2] = 1/2, E[XY ] = 0. (See Corollary 12.3.1.) This tells us that in order to define X, Y we can start with independent random variables N, Z and then use (3.3).
66 Approximation by Brownian motion Figure 3.1: The dyadic construction We start by defining Bt for t ∈D0 = N by B0 = 0 and Bj = N0,1 + · · · + N0,j.
We then continue recursively using (3.3). Suppose Bt has been defined for all t ∈Dn. Then we define Bt for t ∈Dn+1 \ Dn by B 2k+1 2n+1 = B k 2n + 1 2 h B k+1 2n −B k 2n i + 2−(n+2)/2 N2k+1,n+1.
By induction, one can check that for each n the collection of random variables Zk,n := Bk/2n − B(k−1)/2n are independent, each with a N(0, 2−n) distribution. Since this is true for each n, we can see that (a) and (b) hold (with the natural filtration) provided that we restrict to t ∈D. The scaling property for normal random variables shows that for each integer n, the random variables 2n/2 Bt/2n, t ∈D, have the same joint distribution as the random variables Bt, t ∈D.
We define the oscillation of Bt (restricted to t ∈D) by osc(B; δ, T) = sup{|Bt −Bs| : s, t ∈D; 0 ≤s, t ≤T; |s −t| ≤δ}.
For fixed δ, T, this is an FT -measurable random variable. We write osc(B; δ) for osc(B; δ, 1). Let Mn = max 0≤k<2n sup |Bt+k2−n −Bk2−n| : t ∈D; 0 ≤t ≤2−n .
The random variable Mn is similar to osc(B; 2−n) but is easier to analyze. Note that if r ≤2−n, osc(B; r) ≤osc(B; 2−n) ≤3 Mn.
(3.4) 3.2 Construction of Brownian motion 67 To see this, suppose δ ≤2−n, 0 < s < t ≤s + δ ≤1, and |Bs −Bt| ≥ǫ. Then there exists a k such that either k2−n ≤s < t ≤(k + 1)2−n or (k −1)2−n ≤s ≤k2−n < t ≤(k + 1)2−n. In either case, the triangle inequality tells us that Mn ≥ǫ/3. We will prove a proposition that bounds the probability of large values of osc(B; δ, T). We start with a lemma which gives a similar bound for Mn.
Lemma 3.2.1 For every integer n and every δ > 0, P{Mn > δ 2−n/2} ≤4 r 2 π 2n δ e−δ2/2.
Proof Note that P{Mn > δ 2−n/2} ≤2n P ( sup 0≤t≤2−n |Bt| > δ 2−n/2 ) = 2n P sup 0≤t≤1 |Bt| > δ .
Here the supremums are taken over t ∈D. Also note that P { sup{|Bt| : 0 ≤t ≤1, t ∈D} > δ } = lim n→∞P{ max{|Bk2−n| : k = 1, . . . , 2n} > δ } ≤ 2 lim n→∞P{ max{Bk2−n : k = 1, . . . , 2n} > δ }.
The reflection principle (see Proposition 1.6.2 and the remark following) shows that P{ max{Bk2−n : k = 1, . . . , 2n} > δ } ≤ 2 P{B1 > δ} = 2 Z ∞ δ 1 √ 2π e−x2/2 dx ≤ 2 Z ∞ δ 1 √ 2π e−xδ/2 dx = 2 r 2 π δ−1e−δ2/2.
Proposition 3.2.2 There exists a c > 0 such that for every 0 < δ ≤1, r ≥1, and positive integer T, P{osc(B; δ, T) > c r p δ log(1/δ)} ≤c T δr2.
Proof It suffices to prove the result for T = 1 since for general T we can estimate separately the oscillations over the 2T −1 intervals [0, 1], [1/2, 3/2], [1, 2], . . . , [T −1, T]. Also, it suffices to prove the result for δ ≤1/4. Suppose that 2−n−1 ≤δ ≤2−n. Using (3.4), we see that P{osc(B; δ) > c r p δ log(1/δ)} ≤P Mn > cr 3 √ 2 p 2−n log(1/δ) .
By Lemma 3.2.1, if c is chosen sufficiently large, the probability on the right-hand side is bounded by a constant times exp −1 4 c2r2 18 log(1/δ) , which for c large enough is bounded by a constant times δr2.
68 Approximation by Brownian motion Corollary 3.2.3 With probability one, for every integer T < ∞, the function t 7→Bt, t ∈D is uniformly continuous on [0, T].
Proof Uniform continuity on [0, T] is equivalent to saying that osc(B; 2−n, T) − →0 as n →∞. The previous proposition implies that there is a c1 such that P{osc(B; 2−n, T) > c1 2−n/2 √n} ≤c1 T 2−n.
In particular, ∞ X n=1 P{osc(B; 2−n, T) > c1 2−n/2 √n} < ∞, which implies by Borel-Cantelli that with probability one osc(B; 2−n, T) ≤c1 2−n/2 √n for all n sufficiently large.
Given the corollary, we can define Bt for t ̸∈D by continuity, i.e., Bt = lim tn→t Btn, where tn ∈D with tn →t. It is not difficult to show that this satisfies the definition of Brownian motion (we omit the details). Moreover, since Bt has continuous paths, we can write osc(B; δ, T) = sup{|Bt −Bs| : 0 ≤s, t ≤T; |s −t| ≤δ}.
We restate the estimate and include a fact about scaling of Brownian motion. Note that if Bt is a standard Brownian motion and a > 0, then Yt := a−1/2 Bat is also a standard Brownian motion.
Theorem 3.2.4 (Modulus of continuity of Brownian motion) There is a c < ∞such that if Bt is a standard Brownian motion, 0 < δ ≤1, r ≥c, T ≥1, P{osc(B; δ, T) > r p δ log(1/δ)} ≤c T δ(r/c)2.
Moreover, if T > 0, then osc(B; δ, T) has the same distribution as √ T osc(B, δ/T). In particular, if T ≥1, P{osc(B; 1, T) > c r p log T} = P{osc(B; 1/T) > r p (1/T) log T} ≤c T −(r/c)2.
(3.5) 3.3 Skorokhod embedding We will now define a procedure that takes a Brownian motion path Bt and produces a random walk Sn. The idea is straightforward. Start the Brownian motion and wait until it reaches +1 or −1.
If it hits +1 first we let S1 = 1; otherwise, we set S1 = −1. Now we wait until the new increment of the Brownian motion reaches +1 or −1 and we use this value for the increment of the random walk.
To be more precise, let Bt be a standard one-dimensional Brownian motion, and let τ = inf{t ≥0 : |Bt| = 1}.
Symmetry tells us that P{Bτ = 1} = P{Bτ = −1} = 1/2.
3.3 Skorokhod embedding 69 Lemma 3.3.1 E[τ] = 1 and there exists a b < ∞such that E[ebτ] < ∞.
Proof Note that for integer n P{τ > n} ≤P{τ > n −1, |Bn −Bn−1| ≤2} = P{τ > n −1} P{|Bn −Bn−1| ≤2}, which implies for integer n, P{τ > n} ≤P{|Bn −Bn−1| ≤2}n = e−ρn, with ρ > 0. This implies that E[ebτ] < ∞for b < ρ. If s < t, then E[B2 t −t | Fs] = B2 s −s (Exercise 3.1). This shows that B2 t −t is a continuous martingale. Also, E[|B2 t −t|; τ > t] ≤(t + 1) P{τ > t} − →0.
Therefore, we can use the optional sampling theorem (Theorem 12.2.9) to conclude that E[B2 τ −τ] = 0. Since E[B2 τ] = 1, this implies that E[τ] = 1.
More generally, let τ0 = 0 and τn = inf{t ≥τn−1 : |Bt −Bτn−1| = 1}.
Then Sn := Bτn is a simple one-dimensional random walk†. Let Tn = τn −τn−1. The random variables T1, T2, . . . are independent, identically distributed, with mean one satisfying E[ebTj] < ∞ for some b > 0. As before, we define St for noninteger t by linear interpolation. Let Θ(B, S; n) = max{|Bt −St| : 0 ≤t ≤n}.
In other words, Θ(B, S; n) is the distance between the continuous functions B and S in C[0, n] using the usual supremum norm. If j ≤t < j + 1 ≤n, then |Bt −St| ≤|Sj −St| + |Bj −Bt| + |Bj −Sj| ≤1 + osc(B; 1, n) + |Bj −Bτj|.
Hence for integer n, Θ(B, S; n) ≤1 + osc(B; 1, n) + max{|Bj −Bτj| : j = 1, . . . , n}.
(3.6) We can estimate the probabilities for the second term with (3.5). We will concentrate on the last term. Before doing the harder estimates, let us consider how large an error we should expect. Since T1, T2, . . . are i.i.d. random variables with mean 1 and finite variance, the central limit theorem says roughly that |τn −n| = n X j=1 [Tj −1] ≈√n.
Hence we would expect that |Bn −Bτn| ≈ p |τn −n| ≈n1/4.
From this reasoning, we can see that we expect Θ(B, S; n) to be at least of order n1/4. The next theorem shows that it is unlikely that the actual value is much greater than n1/4.
† We actually need the strong Markov property for Brownian motion to justify this and the next assertion. This is not difficult to prove, but we will not do it in this book.
70 Approximation by Brownian motion Theorem 3.3.2 There exist 0 < c1, a < ∞such that for all r ≤n1/4 and all integers n ≥3 P{Θ(B, S; n) > r n1/4 p log n} ≤c1 e−ar.
Proof It suffices to prove the theorem for r ≥9c2 ∗where c∗is the constant c from Theorem 3.2.4 (if we choose c1 ≥e9ac2 ∗, the result holds trivially for r ≤9c2 ∗). Suppose 9c2 ∗≤r ≤n1/4. If |Bn −Bτn| is large, then either |n −τn| is large or the oscillation of B is large. Using (3.6), we see that the event {Θ(B, S; n) ≥r n1/4 √log n} is contained in the union of the two events { osc(B; r√n, 2n) ≥(r/3) n1/4 p log n }, max 1≤j≤n |τj −j| ≥r √n .
Indeed, if osc(B; r√n, 2n) ≤(r/3) n1/4 √log n and |τj −j| ≤r √n for j = 1, . . . , n, then the three terms on the right-hand side of (3.6) are each bounded by (r/3) n1/4 √log n.
Note that Theorem 3.2.4 gives for 1 ≤r ≤n1/4, P{ osc(B; r√n, 2n) > (r/3) n1/4 p log n } ≤ 3 P{ osc(B; r√n, n) > (r/3) n1/4 p log n } = 3 P{ osc(B; r n−1/2) > (r/3) n−1/4 p log n} ≤ 3 P osc(B; r n−1/2) > (√r/3) q r n−1/2 log(n1/2/r) .
If √r/3 ≥c∗and r ≤n1/4, we can use Theorem 3.2.4 to conclude that there exist c, a such that P osc(B; r n−1/2) > (√r/3) q r n−1/2 log(n1/2/r) ≤c e−ar log n.
For the second event, consider the martingale Mj = τj −j.
Using (12.12) on Mj and −Mj, we see that there exist c, a such that P max 1≤j≤n |τj −j| ≥r √n ≤c e−ar2.
(3.7) ♣The proof actually gives the stronger upper bound of c [e−ar2 +e−ar log n] but we will not need this improve-ment.
Extending the Skorokhod approximation to continuous time simple random walk ˜ St is not difficult although in this case the path t 7→˜ St is not continuous. Let Nt be a Poisson process with parameter 1 defined on the same probability space and independent of the Brownian motion B. Then ˜ St := SNt has the distribution of the continuous-time simple random walk. Since Nt −t is a martingale, and 3.4 Higher dimensions 71 the Poisson distribution has exponential moments, another application of (12.12) shows that for r ≤t1/4, P max 0≤s≤t |Ns −s| ≥r √ t ≤c e−ar2.
Let Θ(B, ˜ S; n) = sup{|Bt −˜ St| : 0 ≤t ≤n}.
Then the following is proved similarly.
Theorem 3.3.3 There exist 0 < c, a < ∞such that for all 1 ≤r ≤n1/4 and all positive integers n P{Θ(B, ˜ S; n) ≥r n1/4 p log n} ≤c e−ar.
3.4 Higher dimensions It is not difficult to extend Theorems 3.3.2 and 3.3.3 to p ∈Pd for d > 1. A d-dimensional Brownian motion with covariance matrix Γ with respect to a filtration Ft is a collection of random variables Bt, t ≥0 satisfying the following: (a) B0 = 0; (b) if s < t, then Bt −Bs is an Ft-measurable random Rd-valued variable, independent of Fs, whose distribution is joint normal with mean zero and covariance matrix (t −s) Γ.
(c) with probability one, t 7→Bt is a continuous function.
Lemma 3.4.1 Suppose B(1), . . . , B(l) are independent one-dimensional standard Brownian motions and v1, . . . , vl ∈Rd. Then Bt := B(1) t v1 + · · · + B(l) t vl is a Brownian motion in Rd with covariance matrix Γ = AAT where A = [v1 v2 · · · vl].
Proof Straightforward and left to the reader.
In particular, a standard d-dimensional Brownian motion is of the form Bt = (B(1) t , . . . , B(d) t ) where B(1), . . . , B(d) are independent one-dimensional Brownian motions. Its covariance matrix is the identity.
The next theorem shows that one can define d-dimensional Brownian motions and d-dimensional random walks on the same probability space so that their paths are close to each other. Although the proof will use Skorokhod embedding, it is not true that the d-dimensional random walk is embedded into the d-dimensional Brownian motion. In fact, it is impossible to have an embedded walk since for d > 1 the probability that a d-dimensional Brownian motion Bt visits the countable set Zd after time 0 is zero.
72 Approximation by Brownian motion Theorem 3.4.2 Let p ∈Pd with covariance matrix Γ. There exist c, a and a probability space (Ω, F, P) on which are defined a Brownian motion B with covariance matrix Γ; a discrete-time random walk S with increment distribution p; and a continuous-time random walk ˜ S with increment distribution p such that for all positive integers n and all 1 ≤r ≤n1/4, P{Θ(B, S; n) ≥r n1/4 p log n} ≤c e−ar, P{Θ(B, ˜ S; n) ≥r n1/4 p log n} ≤c e−ar.
Proof Suppose v1, . . . , vl are the points such p(vj) = p(−vj) = qj/2 and p(z) = 0 for all other z ∈Zd \ {0}. Let Ln = (L1 n, . . . , Ll n) be a multinomial process with parameters q1, . . . , ql, and let B1, . . . , Bl be independent one-dimensional Brownian motions. Let S1, . . . , Sl be the random walks derived from B1, . . . , Bl by Skorokhod embedding. As was noted in (1.2), Sn := S1 L1 n v1 + . . . + Sl Ll n vl, has the distribution of a random walk with increment distribution p. Also, Bt := B1 t v1 + · · · + Bl t vl, is a Brownian motion with covariance matrix Γ. The proof now proceeds as in the previous cases.
One fact that is used is that the Lj n have a binomial distribution and hence we can get an exponential estimate P max 1≤j≤n |Li j −qi j| ≥a √n ≤c e−a.
3.5 An alternative formulation Here we give a slightly different, but equivalent, form of the strong approximation from which we get (3.2). We will illustrate this in the case of one-dimensional simple random walk. Suppose Bt is a standard Brownian motion defined on a probability space (Ω, F, P). For positive integer n, let B(n) t denote the Brownian motion B(n) t = n−1/2 Bnt.
Let S(n) denote the simple random walk derived from B(n) using the Skorokhod embedding. Then we know that for all positive integers T, P max 0≤t≤Tn |S(n) t −B(n) t | ≥c r (Tn)1/4 p log(Tn) ≤c e−ar.
If we let W (n) t = n−1/2 S(n) tn , then this becomes P max 0≤t≤T |W (n) t −Bt| ≥c r T 1/4 n−1/4 p log(Tn) ≤c e−ar.
3.5 An alternative formulation 73 In particular, if r = c1 log n where c1 = c1(T) is chosen sufficiently large, P max 0≤t≤T |W (n) t −Bt| ≥c1 n−1/4 log3/2 n ≤c1 n−2.
By the Borel-Cantelli lemma, with probability one max 0≤t≤T |W (n) t −Bt| ≤c1 n−1/4 log3/2 n for all n sufficiently large. In particular, with probability one W (n) converges to B in the metric space C[0, T].
By using a multinomial process (in the discrete-time case) or a Poisson process (in the continuous-time) case, we can prove the following.
Theorem 3.5.1 Suppose p ∈Pd with covariance matrix Γ. There exist c < ∞, a > 0 and a prob-ability space (Ω, F, P) on which are defined a d-dimensional Brownian motion Bt with covariance matrix Γ; an infinite sequence of discrete-time p-walks, S(1), S(2), . . .; and an infinite sequence of continuous time p-walks ˜ S(1), ˜ S(2), . . . such that the following holds for every r > 0, T ≥1. Let W (n) t = n−1/2 S(n) nt , ˜ W (n) t = n−1/2 ˜ S(n) nt .
Then, P max 0≤t≤T |W (n) t −Bt| ≥c r T 1/4 n−1/4 p log(Tn) ≤c e−ar.
P max 0≤t≤T | ˜ W (n) t −Bt| ≥c r T 1/4 n−1/4 p log(Tn) ≤c e−ar.
In particular, with probability one W (n) →B and ˜ W (n) →B in the metric space Cd[0, T].
Exercises Exercise 3.1 Show that if Bt is a standard Brownian motion with respect to the filtration Ft and s < t, then E[B2 t −t | Fs] = B2 s −s.
Exercise 3.2 Let X be an integer-valued random variable with P{X = 0} = 0 and E[X] = 0.
(a) Show that there exist numbers rj ∈(0, ∞], r1 ≤r2 ≤· · · , r−1 ≤r−2 ≤· · · , such that if Bt is a standard Brownian motion and T = inf{t : Bt ∈Z \ {0}, t ≤rBt}, then BT has the same distribution as X.
(b) Show that if X has bounded support, then there exists a b > 0 with E[ebT ] < ∞.
(c) Show that E[T] = E[X2].
(Hint: you may wish to consider first the cases where X is supported on {1, −1}, {1, 2, −1}, and {1, 2, −1, −2}, respectively.) 74 Approximation by Brownian motion Exercise 3.3 Show that there exist c < ∞, α > 0 such that the following is true.
Suppose Bt = (B1 t , B2 t ) is a standard two-dimensional Brownian motion and let TR = inf{t : |Bt| ≥R}. Let UR denote the unbounded component of the open set R2 \ B[0, TR]. Then, Px{0 ∈UR} ≤c(|x|/R)α.
(Hint: Show there is a ρ < 1 such that for all R and all |x| < R, Px{0 ∈U2R | 0 ∈UR} ≤ρ. ) Exercise 3.4 Show that there exist c < ∞, α > 0 such that the following is true. Suppose Sn is simple random walk in Z2 starting at x ̸= 0, and let ξR = min{n : |Sn| ≥R}. Then the probability that there is a nearest neighbor path starting at the origin and ending at {|z| ≥R} that does intersect {Sj : 0 ≤j ≤ξR} is no more than c(|x|/R)α. (Hint: follow the hint in Exercise 3.3, using the invariance principle to show the existence of a ρ.) 4 Green’s Function 4.1 Recurrence and transience A random walk Sn with increment distribution p ∈Pd ∪P∗ d is called recurrent if P{Sn = 0 i.o.} = 1.
If the walk is not recurrent it is called transient. We will also say that p is recurrent or transient.
It is easy to see using the Markov property that p is recurrent if and only if for each x ∈Zd, Px{Sn = 0 for some n ≥1} = 1, and p is transient if and only if the escape probability, q, is positive, where q is defined by q = P{Sn ̸= 0 for all n ≥1}.
Theorem 4.1.1 If p ∈P′ d with d = 1, 2, then p is recurrent. If p ∈P∗ d with d ≥3, then p is transient. For all p, q = " ∞ X n=0 pn(0) #−1 , (4.1) where the left-hand side equals zero if the sum is divergent.
Proof Let Y = P∞ n=0 1{Sn = 0} denote the number of visits to the origin and note that E[Y ] = ∞ X n=0 P{Sn = 0} = ∞ X n=0 pn(0).
If p ∈P′ d with d = 1, 2, the LCLT (see Theorem 2.1.1 and Theorem 2.3.9) implies that pn(0) ∼ c n−d/2 and the sum is infinite. If p ∈P∗ d with d ≥3, then (2.49) shows that pn(0) ≤c n−d/2 and hence E[Y ] < ∞. We can compute E(Y ) in terms of q. Indeed, the Markov property shows that, P{Y = j} = (1 −q)j−1 q. Therefore, if q > 0, E[Y ] = ∞ X j=0 j P{Y = j} = ∞ X j=0 j (1 −q)j−1 q = 1 q.
75 76 Green’s Function 4.2 Green’s generating function If p ∈P ∪P∗and x, y ∈Zd, we define the Green’s generating function to be the power series in ξ: G(x, y; ξ) = ∞ X n=0 ξn Px{Sn = y} = ∞ X n=0 ξn pn(y −x).
Note that the sum is absolutely convergent for |ξ| < 1. We write just G(y; ξ) for G(0, y, ξ). If p ∈P, then G(x; ξ) = G(−x; ξ).
The generating function is defined for complex ξ, but there is a particular interpretation of the sum for positive ξ ≤1. Suppose T is a random variable independent of the random walk S with a geometric distribution, P{T = j} = ξj−1 (1 −ξ), j = 1, 2, . . . , i.e., P{T > j} = ξj (if ξ = 1, then T ≡∞). We think of T as a “killing time” for the walk and we will refer to such T as a geometric random variable with killing rate 1 −ξ. At each time j, if the walker has not already been killed, the process is killed with probability 1 −ξ, where the killing is independent of the walk. If the random walk starts at the origin, then the expected number of visits to x before being killed is given by E X j j} = ∞ X j=0 P{Sj = x; T > j} = ∞ X j=0 pj(x) ξj = G(x; ξ).
Theorem 4.1.1 states that a random walk is transient if and only if G(0; 1) < ∞, in which case the escape probability is G(0; 1)−1. For a transient random walk, we define the Green’s function to be G(x, y) = G(x, y; 1) = ∞ X n=0 pn(y −x).
We write G(x) = G(0, x); if p ∈P, then G(x) = G(−x). The strong Markov property implies that G(0, x) = P{Sn = x for some n ≥0} G(0, 0).
(4.2) Similarly, we define ˜ G(x, y; ξ) = Z ∞ 0 ξt pt(x, y) dt.
For ξ ∈(0, 1) this is the expected amount of time spent at site y by a continuous-time random walk with increment distribution p before an independent “killing time” that has an exponential distribution with rate −log(1−ξ). We will now show that if we set ξ = 1, we get the same Green’s function as that induced by the discrete walk.
Proposition 4.2.1 If p ∈P∗ d is transient, then Z ∞ 0 ˜ pt(x) dt = G(x).
4.2 Green’s generating function 77 Proof Let Sn denote a discrete-time walk with distribution p, Nt an independent Poisson process with parameter 1, and let ˜ St denote the continuous-time walk ˜ St = SNt. Let Yx = ∞ X n=0 1{Sn = x}, ˜ Yx = Z ∞ 0 1{ ˜ St = x} dt, denote the amount of time spent at x by S and ˜ S, respectively. Then G(x) = E[Yx]. If we let Tn = inf{t : Nt = n}, then we can write ˜ Yx = ∞ X n=0 1{Sn = x} (Tn+1 −Tn).
Independence of S and N implies E[1{Sn = x} (Tn+1 −Tn)] = P{Sn = x} E[Tn+1 −Tn] = P{Sn = x}.
Hence E[ ˜ Yx] = E[Yx].
Remark. Suppose p is the increment distribution of a random walk in Zd. For ǫ > 0, let pǫ denote the increment of the “lazy walker” given by pǫ(x) = (1 −ǫ) p(x), x ̸= 0 ǫ + (1 −ǫ) p(0), x = 0 If p is irreducible and periodic on Zd, then for each 0 < ǫ < 1, pǫ is irreducible and aperiodic. Let L, φ denote the generator and characteristic function for p, respectively. Then the generator and characteristic function for pǫ are Lǫ = (1 −ǫ) L, φǫ(θ) = ǫ + (1 −ǫ) φ(θ).
(4.3) If p has mean zero and covariance matrix Γ, then pǫ has mean zero and covariance matrix Γǫ = (1 −ǫ) Γ, det Γǫ = (1 −ǫ)d det Γ.
(4.4) If p is transient, and G, Gǫ denote the Green’s function for p, pǫ, respectively, then similarly to the last proposition we can see that Gǫ(x) = 1 1 −ǫ G(x).
(4.5) For some proofs it is convenient to assume that the walk is aperiodic; results for periodic walks can then be derived using these relations.
If n ≥1, let fn(x, y) denote the probability that a random walk starting at x first visits y at time n (not counting time n = 0), i.e., fn(x, y) = Px{Sn = y; S1 ̸= y, . . . , Sn−1 ̸= y} = Px{τy = n}, where τy = min{j ≥1 : Sj = y}, τ y = min{j ≥0 : Sj = y}.
78 Green’s Function Let fn(x) = fn(0, x) and note that Px{τy < ∞} = ∞ X n=1 fn(x, y) = ∞ X n=1 fn(y −x) ≤1.
Define the first visit generating function by F(x, y; ξ) = F(y −x; ξ) = ∞ X n=1 ξnfn(y −x).
If ξ ∈(0, 1), then F(x, y; ξ) = Px{τy < Tξ}, where Tξ denotes an independent geometric random variable satisfying P{Tξ > n} = ξn.
Proposition 4.2.2 If n ≥1, pn(y) = n X j=1 fj(y) pn−j(0).
If ξ ∈C, G(y; ξ) = δ(y) + F(y; ξ) G(0; ξ), (4.6) where δ denotes the delta function. In particular, if |F(0, ξ)| < 1, G(0; ξ) = 1 1 −F(0; ξ).
(4.7) Proof The first equality follows from P{Sn = y} = n X j=1 P{τy = j; Sn −Sj = 0} = n X j=1 P{τy = j} pn−j(0).
The second equality uses ∞ X n=1 pn(x) ξn = " ∞ X n=1 fn(x) ξn # " ∞ X m=0 pm(0) ξm # , which follows from the first equality. For ξ ∈(0, 1], there is a probabilistic interpretation of (4.6).
If y ̸= 0, the expected number of visits to y (before time Tξ) is the product of the probability of reaching y and the expected number of visits to y given that y is reached before time Tξ. If y = 0, we have to add an extra 1 to account for p0(y).
♣If ξ ∈(0, 1), the identity (4.7) can be considered as a generalization of (4.1). Note that F(0; ξ) = ∞ X j=1 P{τ0 = j; Tξ > j} = P{τ0 < Tξ} 4.2 Green’s generating function 79 represents the probability that a random walk killed at rate 1 −ξ returns to the origin before being killed. Hence, the probability that the walker does not return to the origin before being killed is 1 −F(0; ξ) = G(0; ξ)−1.
(4.8) The right-hand side is the reciprocal of the expected number of visits before killing. If p is transient, we can plug ξ = 1 into this expression and get (4.1).
Proposition 4.2.3 Suppose p ∈Pd ∪P∗ d with characteristic function φ. Then if x ∈Zd, |ξ| < 1, G(x; ξ) = 1 (2π)d Z [−π,π]d 1 1 −ξφ(θ) e−ix·θ dθ.
If d ≥3, this holds for ξ = 1, i.e., G(x) = 1 (2π)d Z [−π,π]d 1 1 −φ(θ) e−ix·θ dθ.
Proof All of the integrals in this proof will be over [−π, π]d. The formal calculation, using Corollary 2.2.3, is G(x; ξ) = ∞ X n=0 ξn pn(x) = ∞ X n=0 ξn 1 (2π)d Z φ(θ)n e−ix·θ dθ = 1 (2π)d Z " ∞ X n=0 (ξ φ(θ))n # e−ix·θ dθ = 1 (2π)d Z 1 1 −ξ φ(θ) e−ix·θ dθ.
The interchange of the sum and the integral in the second equality is justified by the dominated convergence theorem as we now describe. For each N, N X n=0 ξnφ(θ)ne−ix·θ ≤ 1 1 −|ξ| |φ(θ)|.
If |ξ| < 1, then the right-hand side is bounded by 1/[1−|ξ|]. If p ∈P∗ d and ξ = 1, then (2.13) shows that the right-hand side is bounded by c |θ|−2 for some c. If d ≥3, |θ|−2 is integrable on [−π, π]d.
If p ∈Pd is bipartite, we can use (4.3) and (4.5).
Some results are easier to prove for geometrically killed random walks than for walks restricted to a fixed number of steps. This is because stopping time arguments work more nicely for such walks. Suppose that Sn is a random walk, τ is a stopping time for the random walk, and T is an independent geometric random variable. Then on the event {T > τ} the distribution of T −τ given Sn, n = 0, . . . , τ is the same as that of T. This “loss of memory” property for geometric and exponential random variables can be very useful. The next proposition gives an example of a result proved first for geometrically killed walks. The result for fixed length random walks can be deduced from the geometrically killed walk result by using Tauberian theorems. Tauberian theorems are one of the major tools for deriving facts about a sequence from its generating functions. We will only use some simple Tauberian theorems; see Section 12.5.
80 Green’s Function Proposition 4.2.4 Suppose p ∈Pd ∪P′ d, d = 1, 2. Let q(n) = P{Sj ̸= 0 : j = 1, . . . , n}.
Then as n →∞, q(n) ∼ r π−1 n−1/2, d = 1 r (log n)−1, d = 2.
where r = (2π)d/2 √ det Γ.
Proof We will assume p ∈P′ d; it is not difficult to extend this to bipartite p ∈Pd. We will establish the corresponding facts about the generating functions for q(n): as ξ →1−, ∞ X n=0 ξn q(n) ∼ r Γ(1/2) 1 √1 −ξ , d = 1, (4.9) ∞ X n=0 ξnq(n) ∼ r 1 −ξ log 1 1 −ξ −1 , d = 2.
(4.10) Here Γ denotes the Gamma function.† Since the sequence q(n) is monotone in n, Propositions 12.5.2 and 12.5.3 imply the proposition (recall that Γ(1/2) = √π).
Let T be a geometric random variable with killing rate 1 −ξ. Then (4.8) tells us that P{Sj ̸= 0 : j = 1, . . . , T −1} = G(0; ξ)−1.
Also, P{Sj ̸= 0 : j = 1, . . . , T −1} = ∞ X n=0 P{T = n + 1} q(n) = (1 −ξ) ∞ X n=0 ξn q(n).
Using (2.32) and Lemma 12.5.1, we can see that as ξ →1−, G(0; ξ) = ∞ X n=0 ξn pn(0) = ∞ X n=0 ξn 1 r nd/2 + o 1 nd/2 ∼1 r F 1 1 −ξ , where F(s) = Γ(1/2) √s, d = 1 log s, d = 2.
This gives (4.9) and (4.10).
Corollary 4.2.5 Suppose Sn is a random walk with increment distribution p ∈P′ d and τ = τ0 = min{j ≥1 : Sj = 0}.
Then E[τ] = ∞.
† We use the bold face Γ to denote the Gamma function to distinguish it from the covariance matrix Γ.
4.3 Green’s function, transient case 81 Proof If d ≥3, then transience implies that P{τ = ∞} > 0. For d = 1, 2, the result follows from the previous proposition which tells us P{τ > n} ≥ c n−1/2, d = 1, c (log n)−1, d = 2.
♣One of the basic ingredients of Proposition 4.2.4 is the fact that the random walk always starts afresh when it returns to the origin. This idea can be extended to returns of a random walk to a set if the set if sufficiently symmetric that it looks the same at all points. For an example, see Exercise 4.2.
4.3 Green’s function, transient case In this section, we will study the Green’s function for p ∈Pd, d ≥3.
The Green’s function G(x, y) = G(y, x) = G(y −x) is given by G(x) = ∞ X n=0 pn(x) = E " ∞ X n=0 1{Sn = x} # = Ex " ∞ X n=0 1{Sn = 0} # .
Note that G(x) = 1{x = 0} + X y p(x, y) Ey " ∞ X n=0 1{Sn = 0} # = δ(x) + X y p(x, y) G(y), In other words, LG(x) = −δ(x) = −1, x = 0, 0, x ̸= 0.
Recall from (4.2) that G(x) = P{τ x < ∞} G(0).
♣In the calculations above as well as throughout this section, we use the symmetry of the Green’s function, G(x, y) = G(y, x). For nonsymmetric random walks, one must be careful to distinguish between G(x, y) and G(y, x).
The next theorem gives the asymptotics of the Green’s function as |x| →∞. Recall that J ∗(x)2 = d J (x)2 = x · Γ−1x. Since Γ is nonsingular, J ∗(x) ≍J (x) ≍|x|.
Theorem 4.3.1 Suppose p ∈Pd with d ≥3. As |x| →∞, G(x) = C∗ d J ∗(x)d−2 + O 1 |x|d = Cd J (x)d−2 + O 1 |x|d , where C∗ d = d(d/2)−1 Cd = Γ(d−2 2 ) 2 πd/2 √ det Γ = Γ(d 2) (d −2) πd/2 √ det Γ .
82 Green’s Function Here Γ denotes the covariance matrix and Γ denotes the Gamma function.
In particular, for simple random walk, G(x) = d Γ(d 2) (d −2) πd/2 1 |x|d−2 + O 1 |x|d .
♣For simple random walk we can write Cd = 2d (d −2) ωd = 2 (d −2) Vd , where ωd denotes the surface area of unit (d −1)-dimensional sphere and Vd is the volume of the unit ball in Rd.
See Exercise 6.18 for a derivation of this relation. More generally, Cd = 2 (d −2) V (Γ) where V (Γ) denotes the volume of the ellipsoid {x ∈Rd : J (x) ≤1}.
The last statement of Theorem 4.3.1 follows from the first statement using Γ = d−1 I, J (x) = |x| for simple random walk.
It suffices to prove the first statement for aperiodic p; the proof for bipartite p follows using (4.4) and (4.5). The proof of the theorem will consist of two estimates: G(x) = ∞ X n=0 pn(x) = O 1 |x|d + ∞ X n=1 pn(x), (4.11) and ∞ X n=1 pn(x) = C∗ d J ∗(x)d−2 + o 1 |x|d .
The second estimate uses the next lemma.
Lemma 4.3.2 Let b > 1. Then as r →∞, ∞ X n=1 n−b e−r/n = Γ(b −1) rb−1 + O 1 rb+1 Proof The sum is a Riemann sum approximation of the integral Ir := Z ∞ 0 t−b e−r/t dt = 1 rb−1 Z ∞ 0 yb−2 e−y dy = Γ(b −1) rb−1 .
(4.12) If f : (0, ∞) →R is a C2 function and n is a positive integer, then Lemma 12.1.1 gives f(n) − Z n+(1/2) n−(1/2) f(s) ds ≤1 24 sup{|f ′′(t)t −n| ≤1/2}.
Choosing f(t) = t−b e−r/t, we get n−b e−r/n − Z n+(1/2) n−(1/2) t−b e−r/t dt ≤c 1 nb+2 1 + r2 n2 e−r/n, n ≥√r.
4.3 Green’s function, transient case 83 (The restriction n ≥√r is used to guarantee that e−r/(n+(1/2)) ≤c e−r/n.) Therefore, X n≥√r n−b e−r/n − Z n+(1/2) n−(1/2) t−b e−r/t dt ≤ c X n≥√r 1 nb+2 1 + r2 n2 e−r/n ≤ c Z ∞ 0 t−(b+2) 1 + r2 t2 e−r/t dt ≤ c r−(b+1) The last step uses (4.12). It is easy to check that the sum over n < √r and the integral over t < √r decay faster than any power of r.
Proof of Theorem 4.3.1. Using Lemma 4.3.2 with b = d/2, r = J ∗(x)2/2, we have ∞ X n=1 pn(x) = ∞ X n=1 1 (2πn)d/2 √ det Γ e−J ∗(x)2/(2n) = Γ(d−2 2 ) 2 πd/2 √ det Γ 1 J ∗(x)(d−2) + O 1 |x|d+2 .
Hence we only need to prove (4.11). A simple estimate shows that X n<|x| pn(x) as a function of x decays faster than any power of x. Similarly, using Proposition 2.1.2, X n<|x| pn(x) = o(|x|−d).
(4.13) Using (2.5), we see that X n>|x|2 |pn(x) −pn(x)| ≤c X n>|x|2 n−(d+2)/2 = O(|x|−d).
Let k = d + 3. For |x| ≤n ≤|x|2, (2.3) implies that there is an r such that |pn(x) −pn(x)| ≤c " |x| √n k e−r|x|2/n 1 n(d+2)/2 + 1 n(d+k−1)/2 # .
(4.14) Note that X n≥|x| n−(d+k−1)/2 = O(|x|−(d+k−3)/2) = O(|x|−d), and X n≥|x| |x| √n k e−r |x|2/n 1 n(d+2)/2 ≤c Z ∞ 0 |x| √ t k e−r|x|2/t ( √ t)d+2 dt ≤c |x|−d.
Remark. The error term in this theorem is very small. In order to prove that it is this small we need the sharp estimate (4.14) which uses the fact that the third moments of the increment distribution are zero.
If p ∈P′ d with bounded increments but with nonzero third moments, there exists a similar asymptotic expansion for the Green’s function except that the error term is O(|x|−(d−1)), see Theorem 4.3.5. We have used bounded increments (or at least the existence of sufficiently large 84 Green’s Function moments) in an important way in (4.13). Theorem 4.3.5 proves asymptotics under weaker moment assumptions; however, mean zero, finite variance is not sufficient to conclude that the Green’s function is asymptotic to c J ∗(x)2−d for d ≥4. See Exercise 4.5.
♣Often one does not use the full force of these asymptotics.
An important thing to remember is that G(x) ≍|x|2−d. There are a number of ways to remember the exponent 2 −d. For example, the central limit theorem implies that the random walk should visit on the order of R2 points in the ball of radius R. Since there are Rd points in this ball, the probability that a particular point is visited is of order R2−d. In the case of standard d-dimensional Brownian motion, the Green’s function is proportional to |x|2−d. This is the unique (up to multiplicative constant) harmonic, radially symmetric function on Rd \ {0} that goes to zero as |x| →∞(see Exercise 4.4).
Corollary 4.3.3 If p ∈Pd, then ∇jG(x) = ∇j Cd J ∗(x)d−2 + O(|x|−d).
In particular, ∇jG(x) = O(|x|−d+1). Also, ∇2 jG(x) = O(|x|−d).
Remark. We could also prove this corollary with improved error terms by using the difference estimates for the LCLT such as Theorem 2.3.6, but we will not need the sharper results in this book. If p ∈P′ d with bounded increments but nonzero third moments, we could also prove difference estimates for the Green’s function using Theorem 2.3.6. The starting point is to write ∇yG(x) = ∞ X n=0 ∇ypn(x) + ∞ X n=0 [∇ypn(x) −∇ypn(x)].
4.3.1 Asymptotics under weaker assumptions In this section we establish the asymptotics for G for certain p ∈P′ d, d ≥3. We will follow the basic outline of the proof of Theorem 4.3.1. Let G(x) = C∗ d/J ∗(x)d−2 denote the dominant term in the asymptotics. From that proof we see that G(x) = G(x) + o(|x|−d) + ∞ X n=0 [pn(x) −pn(x)].
In the discussion below, we let α ∈{0, 1, 2}. If E[|X1|4] < ∞and the third moments vanish, we set α = 2. If this is not the case, but E[|X1|3] < ∞, we set α = 1. Otherwise, we set α = 0. By Theorems 2.3.5 and 2.3.9 we can see that there exists a sequence δn →0 such that X n≥|x|2 |pn(x) −pn(x)| ≤c X n≥|x|2 δn + α |x|(d+α)/2 = o(|x|2−d), α = 0 O(|x|2−d−α), α = 1, 2.
This is the order of magnitude that we will try to show for the error term, so this estimate suffices 4.3 Green’s function, transient case 85 for this sum. The sum that is more difficult to handle and which in some cases requires additional moment conditions is X n<|x|2 [pn(x) −pn(x)].
Theorem 4.3.4 Suppose p ∈P′ 3. Then G(x) = G(x) + o 1 |x| .
If E[|X1|3] < ∞we can write G(x) = G(x) + O log |x| |x|2 .
If E[|X1|4] < ∞and the third moments vanish, then G(x) = G(x) + O 1 |x|2 .
Proof By Theorem 2.3.10, there exists δn →0 such that X n<|x|2 |pn(x) −pn(x)| ≤c X n<|x|2 δn + α |x|2 n(1+α)/2 .
The next theorem shows that if we assume enough moments of the distribution, then we get the asymptotics as in Theorem 4.3.1. Note that as d →∞, the number of moments assumed grows.
Theorem 4.3.5 Suppose p ∈P′ d, d ≥3.
• If E|X1|d+1] < ∞, then G(x) = G(x) + O(|x|1−d).
• If E|X1|d+3] < ∞and the third moments vanish, G(x) = G(x) + O(|x|−d).
Proof Let α = 1 under the weaker assumption and α = 2 under the stronger assumption, set k = d + 2α −1 so that E[|X1|k] < ∞. As mentioned above, it suffices to show that X n<|x|2 [pn(x) −pn(x)] = O(|x|2−d−α).
Let ǫ = 2(1 + α)/(1 + 2α). As before, X n<|x|ǫ pn(x) decays faster than any power of |x|. Using (2.52), we have X n<|x|ǫ pn(x) ≤ c |x|k X n<|x|ǫ n k−d 2 = O(|x|2−d−α).
86 Green’s Function (The value of ǫ was chosen as the largest value for which this holds.) For the range |x|ǫ ≤n < |x|2, we use the estimate from Theorem 2.3.8: |pn(x) −pn(x)| ≤ c n(d+α)/2 h |x/√n|k−1 e−r|x|2/n + n−(k−2−α)/2i .
As before, X n≥|x|ǫ |x/√n|k−1 n(d+α)/2 e−r|x|2/n ≤c Z ∞ 0 |x|k−1 ( √ t)k+d+α−1 e−r|x|2/t dt = O(|x|2−d−α).
Also, X n≥|x|ǫ 1 n d 2 + k 2 −1 = O(|x|−ǫ(d+α−5 2 )) ≤O(|x|2−d−α), provided that d + α −5 2 2(1 + α) 1 + 2α ≥d −2 + α, which can be readily checked for α = 1, 2 if d ≥3.
4.4 Potential kernel 4.4.1 Two dimensions If p ∈P∗ 2, the potential kernel is the function a(x) = ∞ X n=0 [pn(0) −pn(x)] = lim N→∞ " N X n=0 pn(0) − N X n=0 pn(x) # .
(4.15) Exercise 2.2 shows that |pn(0)−pn(x)| ≤c |x| n−3/2, so the first sum converges absolutely. However, since pn(0) ≍n−1, it is not true that a(x) = " ∞ X n=0 pn(0) # − " ∞ X n=0 pn(x) # .
(4.16) If p ∈P2 is bipartite, the potential kernel for x ∈(Z2)e is defined in the same way. If x ∈(Z2)o we can define a(x) by the second expression in (4.15). Many authors use the term Green’s function for a or −a. Note that a(0) = 0.
♣If p ∈P∗ d is transient, then (4.16) is valid, and a(x) = G(0) −G(x), where G is the Green’s function for p.
Since |pn(0) −pn(x)| ≤c |x| n−3/2 for all p ∈P∗ 2, the same argument shows that a exists for such p.
Proposition 4.4.1 If p ∈P′ 2, then 2 a(x) is the expected number of visits to x by a random walk starting at x before its first visit to the origin.
4.4 Potential kernel 87 Proof We delay this until the next section; see (4.31).
Remark. Using Proposition 4.4.1, we can see that if pǫ is defined as in (4.3), and aǫ denotes the potential kernel for pǫ then aǫ(x) = 1 1 −ǫ a(x).
(4.17) Proposition 4.4.2 If p ∈P2, La(x) = δ0(x) = 1, x = 0 0, x ̸= 0.
Proof Recall that L[pn(0) −pn(x)] = −Lpn(x) = pn(x) −pn+1(x).
For fixed x, the sequence pn(x) −pn+1(x) is absolutely convergent. Hence we can write La(x) = ∞ X n=0 L[pn(0) −pn(x)] = lim N→∞ N X n=0 L[pn(0) −pn(x)] = lim N→∞ N X n=0 [pn(x) −pn+1(x)] = lim N→∞[p0(x) −pN+1(x)] = p0(x) = δ0(x).
Proposition 4.4.3 If p ∈P∗ 2 ∪P2, then a(x) = 1 (2π)2 Z [−π,π]2 1 −e−ix·θ 1 −φ(θ) dθ.
Proof By the remark above, it suffices to consider p ∈P∗ 2. The formal calculation is a(x) = ∞ X n=0 [pn(0) −pn(x)] = ∞ X n=0 1 (2π)2 Z φ(θ)n [1 −e−ix·θ] dθ = 1 (2π)2 Z " ∞ X n=0 φ(θ)n # [1 −e−ix·θ] dθ = 1 (2π)2 Z 1 −e−ix·θ 1 −φ(θ) dθ.
All of the integrals are over [−π, π]2. To justify the interchange of the sum and the integral we use (2.13) to obtain the estimate N X n=0 φ(θ)n [1 −e−ix·θ] ≤|1 −e−ix·θ| 1 −|φ(θ)| ≤c |xθ| |θ|2 ≤c |x| |θ| .
88 Green’s Function Since |θ|−1 is an integrable function in [−π, π]2, the dominated convergence theorem may be applied.
Theorem 4.4.4 If p ∈P2, there exists a constant C = C(p) such that as |x| →∞, a(x) = 1 π √ det Γ log[J ∗(x)] + C + O(|x|−2).
For simple random walk, a(x) = 2 π log |x| + 2γ + log 8 π + O(|x|−2), where γ is Euler’s constant.
Proof We will assume that p is aperiodic; the bipartite case is done similarly. We write a(x) = X n≤J ∗(x)2 pn(0) − X n≤J ∗(x)2 pn(x) + X n>J ∗(x)2 [pn(0) −pn(x)].
We know from (2.23) that pn(0) = 1 2π √ det Γ 1 n + O 1 n2 .
We therefore get X n≤J ∗(x)2 pn(0) = 1 + O(|x|−2) + X 1≤n≤J ∗(x)2 1 2π √ det Γ 1 n + ∞ X n=1 pn(0) − 1 2π √ det Γ 1 n , where the last sum is absolutely convergent. Also, X 1≤n≤J ∗(x)2 1 n = 2 log[J ∗(x)] + γ + O(|x|−2), where γ is Euler’s constant (see Lemma 12.1.3). Hence, X n≤J ∗(x)2 pn(0) = 1 π √ det Γ log[J ∗(x)] + c′ + O(|x|−2) for some constant c′.
Proposition 2.1.2 shows that X n≤|x| pn(x) decays faster than any power of |x|. Theorem 2.3.8 implies that there exists c, r such that for n ≤J ∗(x)2, pn(x) − 1 2π n √ det Γ e−J ∗(x)2/2n ≤c " |x/√n|5 e−r|x|2/n n2 + 1 n3 # .
(4.18) 4.4 Potential kernel 89 Therefore, X |x|≤n≤J ∗(x)2 pn(x) − 1 2π n √ det Γ e−J ∗(x)2/2n ≤O(|x|−2) + c X |x| J ∗(x)2, we use Theorem 2.3.8 and Lemma 4.3.2 again to conclude that X n>J ∗(x)2 [pn(0) −pn(x)] = c Z 1 0 1 y h 1 −e−y/2i dy + O(|x|−2).
For simple random walk in two dimensions, it follows that a(x) = 2 π log |x| + k + O(|x|−2), for some constant k. To determine k, we use φ(θ1, θ2) = 1 −1 2 [cos θ1 + cos θ2].
Plugging this into Proposition 4.4.3 and doing the integral (details omitted, see Exercise 4.9), we get an exact expression for xn = (n, n) for integer n > 0 a(xn) = 4 π 1 + 1 3 + 1 5 + · · · + 1 2n −1 .
However, we also know that a(xn) = 2 π log n + 2 π log √ 2 + k + O(n−2).
Therefore, k = lim n→∞ −1 π log 2 −2 π log n + 4 π n X j=1 1 2j −1 .
90 Green’s Function Using Lemma 12.1.3 we can see that as n →∞, n X j=1 1 2j −1 = 2n X j=1 1 j − n X j=1 1 2j = 1 2 log n + log 2 + 1 2 γ + o(1).
Therefore, k = 3 π log 2 + 2 π γ.
♣Roughly speaking, a(x) is the difference between the expected number of visits to 0 and the expected number of visits to x by some large time N. Let us consider N >> |x|2. By time |x|2, the random walker has visited the origin about X n<|x|2 pn(0) ∼ X n<|x|2 c n ∼2 c log |x|, times where c = (2π √ det Γ)−1. It has visited x about O(1) times. From time |x|2 onward, pn(x) and pn(0) are roughly the same and the sum of the difference from then on is O(1). This shows why we expect a(x) = 2c log |x| + O(1).
Note that log |x| = log J ∗(x) + O(1).
♣Although we have included the exact value γ2 for simple random walk, we will never need to use this value.
Corollary 4.4.5 If p ∈P2, ∇ja(x) = ∇j 1 π √ det Γ log[J ∗(x)] + O(|x|−2).
In particular, ∇ja(x) = O(|x|−1). Also, ∇2 ja(x) = O(|x|−2).
Remark. One can give better estimates for the differences of the potential kernel by starting with Theorem 2.3.6 and then following the proof of Theorem 4.3.1. We give an example of this technique in Theorem 8.1.2.
4.4.2 Asymptotics under weaker assumptions We can prove asymptotics for the potential kernel under weaker assumptions. Let a(x) = [π √ det Γ]−1 log[J ∗(x)] denote the leading term in the asymptotics.
4.4 Potential kernel 91 Theorem 4.4.6 Suppose p ∈P′ 2. Then a(x) = a(x) + o(log |x|).
If E[|X1|3] < ∞, then there exists C < ∞such that a(x) = a(x) + C + O(|x|−1).
If E[|X1|6] < ∞and the third moments vanish, then a(x) = a(x) + C + O(|x|−2).
Proof Let α = 0, 1, 2 under the three possible assumptions, respectively. We start with α = 1, 2 for which we can write ∞ X n=0 [pn(0) −pn(x)] = ∞ X n=0 [pn(0) −pn(x)] + ∞ X n=0 [pn(0) −pn(0)] + ∞ X n=0 [pn(x) −pn(x)] (4.19) The estimate ∞ X n=0 [pn(0) −pn(x)] = a(x) + ˜ C + O(|x|−2) is done as in Theorem 4.4.4. Since |pn(0)−pn(0)| ≤c n−3/2, the second sum on the right-hand side of (4.19) converges, and we set C = ˜ C + ∞ X n=0 [pn(0) −pn(0)].
We write ∞ X n=0 [pn(x) −pn(x)] ≤ X n<|x|2 |pn(x) −pn(x)| + X n≥|x|2 |pn(x) −pn(x)|.
By Theorem 2.3.5 and and Theorem 2.3.9, X n≥|x|2 |pn(x) −pn(x)| ≤c X n≥|x|2 n−(2+α)/2 = O(|x|−α).
For α = 1, Theorem 2.3.10 gives X n<|x|2 |pn(x) −pn(x)| ≤ X n<|x|2 c |x|2 n1/2 = O(|x|−1).
If E[|X1|4] < ∞and the third moments vanish, a similar arugments shows that the sum on the left-hand side is bounded by O(|x|−2 log |x|) which is a little bigger than we want. However, if we also assume that E[|X1|6] < ∞, then we get an estimate as in (4.18), and we can show as in Theorem 4.4.4 that this sum is O(|x|−2).
If we only assume that p ∈P′ 2, then we cannot write (4.19) because the second sum on the right-hand side might diverge. Instead, we write ∞ X n=0 [pn(0) −pn(x)] = 92 Green’s Function X n≥|x|2 [pn(0) −pn(x)] + X n<|x|2 [pn(0) −pn(x)] + X n<|x|2 [pn(0) −pn(0)] + X n<|x|2 [pn(x) −pn(x)].
As before, X n<|x|2 [pn(0) −pn(x)] = a(x) + O(1).
Also, Exercise 2.2, Theorem 2.3.10, and (2.32), respectively, imply X n≥|x|2 |pn(0) −pn(x)| ≤ X n≥|x|2 c |x| n3/2 = O(1), X n<|x|2 |pn(x) −pn(x)| ≤ X n<|x|2 c |x|2 = O(1), X n<|x|2 |pn(0) −pn(0)| ≤ X n<|x|2 o 1 n = o(log |x|).
4.4.3 One dimension If p ∈P′ 1, the potential kernel is defined in the same way a(x) = lim N→∞ " N X n=0 pn(0) − N X n=0 pn(x) # .
In this case, the convergence is a little more subtle. We will restrict ourselves to walks satisfying E[|X|3] < ∞for which the proof of the next proposition shows that the sum converges absolutely.
Proposition 4.4.7 Suppose p ∈P′ 1 with E[|X|3] < ∞. Then there is a c such that for all x, a(x) σ2 −|x| ≤c log |x|.
If E[|X|4] < ∞and E[X3] = 0, then there is a C such that a(x) = |x| σ2 + C + O(|x|−1).
Proof Assume x > 0. Let α = 1 under the weaker assumption and α = 2 under the stronger assumption. Theorem 2.3.6 gives pn(0) −pn(x) = pn(0) −pn(x) + x O(n−(2+α)/2), which shows that X n≥x2 [pn(0) −pn(x)] −[pn(0) −pn(x)] ≤c x X n≥x2 n−(2+α)/2 ≤cx1−α.
4.4 Potential kernel 93 If α = 1, Theorem 2.3.5 gives X n<x2 [pn(0) −pn(0)] ≤c X n<x2 n−1 = O(log x).
If α = 2, Theorem 2.3.5 gives |pn(0) −pn(0)| = O(n−3/2) and hence X n<x2 [pn(0) −pn(0)] = C′ + O(|x|−1), C′ := ∞ X n=0 [pn(0) −pn(0)], In both cases, Theorem 2.3.10 gives X n<x2 [pn(x) −pn(x)] ≤c x2 X n<x2 n(1−α)/2 ≤c x1−α.
Therefore, a(x) = e(x) + ∞ X n=0 [pn(0) −pn(x)] = e(x) + ∞ X n=1 1 √ 2πσ2n 1 −e− x2 2σ2n , where e(x) = O(log x) if α = 1 and e(x) = C′ + O(x−1) if α = 2. Standard estimates (see Section 12.1.1), which we omit, show that there is a C′′ such that as x →∞, ∞ X n=1 1 √ 2πσ2n = C′′ + Z ∞ 0 1 √ 2πσ2t 1 −e−x2 2σ2t dt + o(x−1), and Z ∞ 0 1 √ 2πσ2t 1 −e−x2 2σ2t dt = 2x σ2 √ 2π Z ∞ 0 1 u2 1 −e−u2/2 du = x σ2 .
Since C = C′ + C′′ is independent of x, the result also holds for x < 0.
Theorem 4.4.8 If p ∈P1, and x > 0, a(x) = x σ2 + Ex a(ST ) −ST σ2 , (4.20) where T = min{n : Sn ≤0}. There exists β > 0 such that for x > 0, a(x) = x σ2 + C + O(e−βx), x →∞, (4.21) where C = lim y→∞Ey a(ST ) −ST σ2 .
In particular, for simple random walk, a(x) = |x|.
Proof Assume y > x, let Ty = min{n : Sn ≤0 or Sn ≥y}, and consider the bounded martingale Sn∧Ty. Then the optional sampling theorem implies that x = Ex[S0] = Ex[STy] = Ex[ST ; T ≤Ty] + Ex[STy; T > Ty].
94 Green’s Function If we let y →∞, we see that lim y→∞Ex[STy; T > Ty] = x −Ex[ST ].
Also, since E[STy | Ty < T] = y + O(1), we can see that lim y→∞y Px{Ty < T} = x −Ex[ST ].
We now consider the bounded martingale Mn = a(Sn∧Ty). Then the optional sampling theorem implies that a(x) = Ex[M0] = Ex[MTy] = Ex[a(ST ); T ≤Ty] + Ex[a(STy); T > Ty].
As y →∞, Ex[a(ST ); T < Ty] →Ex[a(ST )]. Also, as y →∞, Ex[a(STy); T > Ty] ∼Px{Ty < T} h y σ2 + O(1) i ∼x −Ex[ST ] σ2 .
This gives (4.20).
We will sketch the proof of (4.21); we leave it as an exercise (Exercise 4.12) to fill in the details.
We will show that there exists a β such that if 0 < x < y < ∞, then ∞ X j=0 |Px{ST = −j} −Py{ST = −j}| = O(e−βx).
(4.22) Even though we have written this as an infinite sum, the terms are nonzero only for j less than the range of the walk. Let ρz = min{n ≥0 : Sn ≤z}. Irreducibility and aperiodicity of the random walk can be used to see that there is an ǫ > 0 such that for all z > 0, Pz+1{Sρz = z} = P1{ρ0 = 0} > ǫ.
Let f(r) = f−j(r) = sup x,y≥r |Px{ST = −j} −Py{ST = −j}|.
Then if R denotes the range of the walk, we can see that f(r + 1) ≤(1 −ǫ) f(r −R).
Iteration of this inequality gives f(kR) ≤(1 −ǫ)k−1f(R) and this gives (4.22).
Remark. There is another (perhaps more efficient) proof of this result, see Exercise 4.13. One may note that the proof does not use the symmetry of the walk to establish a(x) = x σ2 + C + O(e−αx), x →∞.
Hence, this result holds for all mean zero walks with bounded increments. Applying the proof to negative x yields a(−x) = |x| σ2 + C−+ O(e−α|x|).
If the third moment of the increment distribution is nonzero, it is possible that C ̸= C−, see Exercise 4.14.
4.5 Fundamental solutions 95 ♣The potential kernel in one dimension is not as useful as the potential kernel or Green’s function in higher dimensions. For d ≥2, we use the fact that the potential kernel or Green’s function is harmonic on Zd \ {0} and that we have very good estimates for the asymptotics. For d = 1, similar arguments can be done with the function f(x) = x which is obviously harmonic.
4.5 Fundamental solutions If p ∈P, the Green’s function G for d ≥3 or the potential kernel a for d = 2 is often called the fundamental solution of the generator L since LG(x) = −δ(x), La(x) = δ(x).
(4.23) More generally, we write LxG(x, y) = LxG(y, x) = −δ(y −x), Lxa(x, y) = Lxa(y, x) = δ(y −x), where Lx denotes L applied to the x variable.
Remark.
Symmetry of walks in P is necessary to derive (4.23).
If p ∈P∗is transient, the Green’s function G does not satisfy (4.23). Instead it satisfies LRG(x) = −δ0(x) where LR denotes the generator of the “backwards random walk” with increment distribution pR(x) = p(−x). The function f(x) = G(−x) satisfies Lf(x) = −δ0(x) and is therefore the fundamental solution of the generator. Similarly, if p ∈P∗ 2, the fundamental solution of the generator is f(x) = a(−x).
Proposition 4.5.1 Suppose p ∈Pd with d ≥2, and f : Zd →R is a function satisfying f(0) = 0, f(x) = o(|x|) as x →∞, and Lf(x) = 0 for x ̸= 0. Then, there exists b ∈R such that f(x) = b [G(x) −G(0)], d ≥3, f(x) = b a(x), d = 2.
Proof See Propositions 6.4.6 and 6.4.8.
Remark. The assumption f(x) = o(|x|) is clearly needed since the function f(x1, . . . , xd) = x1 is harmonic.
Suppose d ≥3. If f : Zd →R is a function with finite support we define Gf(x) = X y∈Zd G(x, y) f(y) = X y∈Zd G(y −x) f(y).
(4.24) Note that if f is supported on A, then LGf(x) = 0 for x ̸∈A. Also if x ∈A, LGf(x) = Lx X y∈Zd G(x, y) f(y) = X y∈Zd LxG(x, y) f(y) = −f(x).
(4.25) In other words −G = L−1. For this reason the Green’s function is often called the inverse of the (negative of the) Laplacian. Similarly, if d = 2, and f has finite support, we define af(x) = X y∈Zd a(x, y) f(y) = X y∈Zd a(y −x) f(y).
(4.26) 96 Green’s Function In this case we get Laf(x) = f(x), i.e., a = L−1.
4.6 Green’s function for a set If A ⊂Zd and S is a random walk with increment distribution p, let τA = min{j ≥1 : Sj ̸∈A}, τ A = min{j ≥0 : Sj ̸∈A}.
(4.27) If A = Zd \ {x}, we write just τx, τ x, which is consistent with the definition of τx given earlier in this chapter. Note that τA, τ A agree if S0 ∈A, but are different if S0 ̸∈A. If p is transient or A is a proper subset of Zd we define GA(x, y) = Ex "τ A−1 X n=0 1{Sn = y} # = ∞ X n=0 Px{Sn = y; n < τ A}.
Lemma 4.6.1 Suppose p ∈Pd and A is a proper subset of Zd.
• GA(x, y) = 0 unless x, y ∈A.
• GA(x, y) = GA(y, x) for all x, y.
• For x ∈A, LxGA(x, y) = −δ(y −x). In particular if f(y) = GA(x, y), then f vanishes on Zd \ A and satisfies Lf(y) = −δ(y −x) on A.
• For each y ∈A, GA(y, y) = 1 Py{τA < τy} < ∞.
• If x, y ∈A, then GA(x, y) = Px{τ y < τA} GA(y, y).
• GA(x, y) = GA−x(0, y −x) where A −x = {z −x : z ∈A}.
Proof Easy and left to the reader. The second assertion may be surprising at first, but symmetry of the random walk implies that for x, y ∈A, Px{Sn = y; n < τA} = Py{Sn = x; n < τA}.
Indeed if zo = x, z1, z2, . . . , zn−1, zn = y ∈A, then Px{S1 = z1, S2 = z2, . . . , Sn = y} = Py{S1 = zn−1, S2 = zn−2, . . . , Sn = x}.
The next proposition gives an important relation between the Green’s function for a set and the Green’s function or the potential kernel.
4.6 Green’s function for a set 97 Proposition 4.6.2 Suppose p ∈Pd, A ⊂Zd, x, y ∈Zd.
(a) If d ≥3, GA(x, y) = G(x, y) −Ex[G(Sτ A, y); τ A < ∞] = G(x, y) − X z Px{Sτ A = z} G(z, y).
(b) If d = 1, 2 and A is finite, GA(x, y) = Ex[a(Sτ A, y)] −a(x, y) = "X z Px{Sτ A = z} a(z, y) # −a(x, y).
(4.28) Proof The result is trivial if x ̸∈A. We will assume x ∈A in which case τA = τ A.
If d ≥3, let Yy = P∞ n=0 1{Sn = y} denote the total number of visits to the point y. Then Yy = τA−1 X n=0 1{Sn = y} + ∞ X n=τA 1{Sn = y}.
If we assume S0 = x and take expectations of both sides, we get G(x, y) = GA(x, y) + Ex[G(SτA, y)].
The d = 1, 2 case could be done using a similar approach, but it is easier to use a different argument. If S0 = x and g is any function, then it is easy to check that Mn = g(Sn) − n−1 X j=0 Lg(Sj) is a martingale. We apply this to g(z) = a(z, y) for which Lg(z) = δ(z −y). Then, a(x, y) = Ex[M0] = Ex[Mn∧τA] = Ex[a(Sn∧τA, y)] −Ex (n∧τA)−1 X j=0 1{Sj = y} .
Since A is finite, the dominated convergence theorem implies that lim n→∞Ex[a(Sn∧τA, y)] = Ex[a(SτA, y)].
(4.29) The monotone convergence theorem implies lim n→∞Ex (n∧τA)−1 X j=0 1{Sj = y} = Ex τA−1 X j=0 1{Sj = y} = GA(x, y).
The finiteness assumption on A was used in (4.29). The next proposition generalizes this to all proper subsets A of Zd, d = 1, 2. Recall that Bn = {x ∈Zd : |x| < n}. Define a function FA by FA(x) = lim n→∞ log n π √ det Γ Px{τBn < τ A}, d = 2, FA(x) = lim n→∞ n σ2 Px{τBn < τ A}, d = 1.
98 Green’s Function The existence of these limits is established in the next proposition. Note that FA ≡0 on Zd \ A since Px{τ A = 0} = 1 for x ∈Zd \ A.
Proposition 4.6.3 Suppose p ∈Pd, d = 1, 2 and A is a proper subset of Zd. Then if x, y ∈Z2, GA(x, y) = Ex[a(Sτ A, y)] −a(x, y) + FA(x).
Proof The result is trivial if x ̸∈A so we will suppose that x ∈A. Choose n > |x|, |y| and let An = A ∩{|z| < n}. Using (4.28), we have GAn(x, y) = Ex[a(SτAn , y)] −a(x, y).
Note also that Ex[a(SτAn , y)] = Ex[a(SτA, y); τA ≤τBn] + Ex[a(SτBn, y); τA > τBn].
The monotone convergence theorem implies that as n →∞, GAn(x, y) − →GA(x, y), Ex[a(SτA, y); τA ≤τBn] − →Ex[a(SτA, y)].
Sincce GA(x, y) < ∞, this implies lim n→∞Ex[a(SτBn, y); τA > τBn] = GA(x, y) + a(x, y) −Ex[a(SτA, y)].
However, n ≤|SτBn| ≤n + R where R denotes the range of the increment distribution. Hence Theorems 4.4.4 and 4.4.8 show that as n →∞, Ex[a(SτBn, y); τA > τBn] ∼Px{τA > τBn} log n π √ det Γ , d = 2, Ex[a(SτBn, y); τA > τBn] ∼Px{τA > τBn} n σ2 , d = 1.
Remark. We proved that for d = 1, 2, FA(x) = GA(x, y) + a(x, y) −Ex[a(Sτ A, y)].
(4.30) This holds for all y. If we choose y ∈Zd \ A, then GA(x, y) = 0, and hence we can write FA(x) = a(x, y) −Ex[a(Sτ A, y)].
Using this expression it is easy to see that LFA(x) = 0, x ∈A.
Also, if Zd \ A is finite, FA(x) = a(x) + OA(1), x →∞.
In the particular case A = Zd \ {0}, y = 0, this gives FZd{0}(x) = a(x).
4.6 Green’s function for a set 99 Applying (4.30) with y = x, we get GZd{0}(x, x) = FZd{0}(x) + a(0, x) = 2 a(x).
(4.31) The next simple proposition relates Green’s functions to “escape probabilities” from sets. The proof uses a last-exit decomposition. Note that the last time a random walk visits a set is a random time that is not a stopping time. If A ⊂A′, the event {τ Zd\A < τ A′} is the event that the random walk visits A before leaving A′.
Proposition 4.6.4 (Last-Exit Decomposition) Suppose p ∈Pd and A ⊂Zd. Then, • If A′ is a proper subset of Zd with A ⊂A′, Px{τ Zd\A < τ A′} = X z∈A GA′(x, z) Pz{τZd\A > τA′}.
• If ξ ∈(0, 1) and Tξ is an independent geometric random variable with killing rate 1−ξ, then Px{τ Zd\A < Tξ} = X z∈A G(x, z; ξ) Pz{τZd\A ≥Tξ}.
• If d ≥3 and A is finite, Px{Sj ∈A for some j ≥0} = Px{τ Zd\A < ∞} = X z∈A G(x, z) Pz{τZd\A = ∞}.
Proof We will prove the first assertion; the other two are left as Exercise 4.11. We assume x ∈A′ (for otherwise the result is trivial). On the event {τ Zd\A < τ A′}, let σ denote the largest k < τ A′ such that Sk ∈A. Then, Px{τ Zd\A < τ A′} = ∞ X k=0 X z∈A Px{σ = k; Sσ = z} = X z∈A ∞ X k=0 Px{Sk = z; k < τA′; Sj ̸∈A, j = k + 1, . . . , τA′}.
The Markov property implies that Px{Sj ̸∈A, j = k + 1, . . . , τA′ | Sk = z; k < τA′} = Pz{τA′ < τZd\A}.
Therefore, Px{τ Zd\A < τ A′} = X z∈A ∞ X k=0 Px{Sk = z; k < τA′} Pz{τA′ < τZd\A} = X z∈A GA′(x, z) Pz{τA′ < τZd\A}.
100 Green’s Function The next proposition uses a last-exit decomposition to describe the distribution of a random walk conditioned to not return to its starting point before a killing time. The killing time is either geometric or the first exit time from a set.
Proposition 4.6.5 Suppose Sn is a p-walk with p ∈Pd; 0 ∈A ⊂Zd; and ξ ∈(0, 1). Let Tξ be a geometric random variable independent of the random walk with killing rate 1 −ξ. Let ρ = max{j ≥0 : j ≤τA, Sj = 0}, ρ∗= max{j ≥0 : j < Tξ, Sj = 0}.
• The distribution of {Sj : ρ ≤j ≤τA} is the same as the conditional distribution of {Sj : 0 ≤ j ≤τA} given ρ = 0.
• The distribution of {Sj : ρ∗≤j < Tξ} is the same as the conditional distribution of {Sj : 0 ≤j < Tξ} given ρ∗= 0.
Proof The usual Markov property implies that for any positive integer j, any x1, x2, . . . , xk−1 ∈ A \ {0} and any xk ∈Zd \ A, P{ρ = j, τA = j + k, Sj+1 = x1, . . . , Sj+k = xk} = P{Sj = 0, τA > j, Sj+1 = x1, . . . , Sj+k = xk} = P{Sj = 0, τA > j} P{S1 = x1, . . . , Sk = xk}.
The first assertion is obtained by summation over j, and the other equality is done similarly.
Exercises Exercise 4.1 Suppose p ∈Pd and Sn is a p-walk. Suppose A ⊂Zd and that Px{τ A = ∞} > 0 for some x ∈A. Show that for every ǫ > 0, there is a y with Py{τ A = ∞} > 1 −ǫ.
Exercise 4.2 Suppose p ∈Pd ∪P′ d, d ≥2 and let x ∈Zd \ {0}. Let T = min{n > 0 : Sn = jx for some j ∈Z}.
Show there exists c = c(x) such that as n →∞, P{T > n} ∼ c n−1/2, d = 2 c (log n)−1, d = 3 c, d ≥4.
Exercise 4.3 Suppose d = 1. Show that the only function satisfying the conditions of Proposition 4.5.1 is the zero function.
Exercise 4.4 Find all radially symmetric functions f in Rd \ {0} satisfying ∆f(x) = 0 for all x ∈Rd \ {0}.
Exercise 4.5 For each positive integer k find positive integer d and p ∈P′ d such that E[|X1|k] < ∞ and lim sup |x|→∞ |x|d−2 G(x) = ∞.
4.6 Green’s function for a set 101 (Hint: Consider a sequence of points z1, z2, . . . going to infinity and define P{X1 = zj} = qj. Note that G(zj) ≥qj. Make a good choice of z1, z2, . . . and q1, q2, . . .) Exercise 4.6 Suppose X1, X2, . . . are independent, identically distributed random variables in Z with mean zero. Let Sn = X1 + · · · + Xn denote the corresponding random walk and let Gn(x) = n X j=0 P{Sj = x} be the expected number of visits to x in the first n steps of the walk.
(i) Show that Gn(x) ≤Gn(0) for all n.
(ii) Use the law of large numbers to conclude that for all ǫ > 0 there is an Nǫ such that for n ≥Nǫ, X |x|≤ǫn Gn(x) ≥n 2.
(iii) Show that G(0) = lim n→∞Gn(0) = ∞ and conclude that the random walk is recurrent.
Exercise 4.7 Suppose A ⊂Zd and x, y ∈A. Show that GA(x, y) = lim n→∞GAn(x, y), where An = {z ∈A : |z| < n}.
Exercise 4.8 Let Sn denote simple random walk in Z2 starting at the origin and let ρ = min{j ≥ 1 : Sj = 0 or e1}. Show that P{Sρ = 0} = 1/2.
Exercise 4.9 Consider the random walk in Z2 that moves at each step to one of (1, 1), (1, −1), (−1, 1), (−1, −1) each with probability 1/4. Although this walk is not irreducible, many of the ideas of this chapter apply to this walk.
(i) Show that φ(θ1, θ2) = 1 −(cos θ1)(cos θ2).
(ii) Let a be the potential kernel for this random walk and ˆ a the potential kernel for simple random walk. Show that for every integer n, a((n, 0)) = ˆ a((n, n)). (see Exercise 1.7).
(iii) Use Proposition 4.4.3 (which is valid for this walk) to show that for all integers n > 0, a((n, 0)) −a((n −1, 0)) = 4 π(2n −1), a((n, 0)) = 4 π 1 + 1 3 + 1 5 + · · · + 1 2n −1 .
Exercise 4.10 Suppose p ∈P1 and let A = {1, 2, . . .}. Show that FA(x) = x −Ex[ST ] σ2 , 102 Green’s Function where T = min{j ≥0 : Sj ≤0} and FA is as in (4.30).
Exercise 4.11 Finish the details in Proposition 4.6.4.
Exercise 4.12 Finish the details in Theorem 4.4.8.
Exercise 4.13 Let Sj be a random walk in Z with increment distribution p satisfying r1 = min{j : p(j) > 0} < ∞, r2 = max{j : p(j) > 0} < ∞, and let r = r2 −r1.
(i) Show that if α ∈R and k is a nonnegative integer, then f(x) = αx xk satisfies Lf(x) = 0 for all x ∈R in and only if (s −α)k−1 divides the polynomial q(s) = E sX1 .
(ii) Show that the set of functions on {−r + 1, −r + 2, . . .} satisfying Lf(x) = 0 for x ≥1 is a vector space of dimension r.
(iii) Suppose that f is a function on {−r + 1, −r + 2, . . .} satisfying Lf(x) = 0 and f(x) ∼x as x →∞. Show that there exists c ∈R, c1, α > 0 such that |f(x) −x −c| ≤c1 e−αx.
Exercise 4.14 Find the potential kernel a(x) for the one-dimensional walk with p(−1) = p(−2) = 1 5, p(1) = 3 5.
5 One-dimensional walks 5.1 Gambler’s ruin estimate We will prove one of the basic estimates for one-dimensional random walks with zero mean and finite variance, often called the gambler’s ruin estimate.
We will not restrict to integer-valued random walks. For this section we assume that X1, X2, . . . are independent, identically distributed (one-dimensional) random variables with E[X1] = 0, E[X2 1] = σ2 > 0. We let Sn = S0+X1+· · ·+Xn be the corresponding random walk. If r > 0, we let ηr = min{n ≥0 : Sn ≤0 or Sn ≥r}, η = η∞= min{n ≥0 : Sn ≤0}.
We first consider simple random walk for which the gambler’s ruin estimates are identities.
Proposition 5.1.1 If Sn is one-dimensional simple random walk and j < k are positive integers, then Pj{Sηk = k} = j k.
Proof Since Mn := Sn∧ηk is a bounded martingale, the optional sampling theorem implies that j = Ej[M0] = Ej[Mηk] = k Pj{Sηk = k}.
Proposition 5.1.2 If Sn is one-dimensional simple random walk, then for positive integer n, P1{η > 2n} = P1{S2n > 0} −P1{S2n < 0} = P{S2n = 0} = 1 √π n + O 1 n3/2 .
Proof Symmetry and the Markov property tell us that each k < 2n and each positive integer x, P1{η = k, S2n = x} = P1{η = k} p2n−k(x) = P1{η = k, S2n = −x}.
Therefore, P1{η ≤2n, S2n = x} = P1{η ≤2n, S2n = −x}.
103 104 One-dimensional walks Symmetry also implies that for all x, P1{S2n = x + 2} = P1{S2n = −x}. Since P1{η > 2n, S2n = −x} = 0, for x ≥0, we have P1{η > 2n} = X x>0 P{η > 2n; S2n = x} = X x>0 [p2n(1, x) −p2n(1, −x)] = p2n(1, 1) + X x>0 [p2n(1, x + 2) −p2n(1, −x)] = p2n(0, 0) = 4−n 2n n = 1 √π n + O 1 n3/2 .
The proof of the gambler’s ruin estimate for more general walks follows the same idea as that in the proof of Proposition 5.1.1. However, there is a complication arising from the fact that we do not know the exact value of Sηk. Our first lemma shows that the application of the optional sampling theorem is valid. For this we do not need to assume that the variance is finite.
Lemma 5.1.3 If X1, X2, . . . are i.i.d. random variables in R with E(Xj) = 0 and P{Xj > 0} > 0, then for every 0 < r < ∞and every x ∈R, Ex[Sηr] = x.
(5.1) Proof We assume 0 < x < r for otherwise the result is trivial. We start by showing that Ex[|Sηr|] < ∞. Since P{Xj > 0} > 0, there exists an integer m and a δ > 0 such that P{X1 + · · · + Xm > r} ≥δ.
Therefore for all x and all positive integers j, Px{ηr > jm} ≤(1 −δ)m.
In particular, Ex[ηr] < ∞. By the Markov property, Px{|Sηr| ≥r + y; ηr = k} ≤Px{ηr > k −1; |Xk| ≥y} = Px{ηr > k −1} P{|Xk| ≥y}.
Summing over k gives Px{|Sηr| ≥r + y} ≤Ex[ηr] P{|Xk| ≥y}.
Hence Ex [|Sηr|] = Z ∞ 0 Px{|Sηr| ≥y} dy ≤ Ex[ηr] r + Z ∞ 0 P{|Xk| ≥y}dy = Ex[ηr] ( r + E [|Xj|] ) < ∞.
Since Ex[|Sηr|] < ∞, the martingale Mn := Sn∧ηr is dominated by the integrable random variable r+|Sηr|. Hence it is a uniformly integrable martingale, and (5.1) follows from the optional sampling theorem (Theorem 12.2.3).
5.1 Gambler’s ruin estimate 105 We now prove the estimates under the assumption of bounded range. We will take some care in showing how the constants in the estimate depend on the range.
Proposition 5.1.4 For every ǫ > 0 and K < ∞, there exist 0 < c1 < c2 < ∞such that if P{|X1| > K} = 0 and P{X1 ≥ǫ} ≥ǫ, then for all 0 < x < r, c1 x + 1 r ≤Px{Sηr ≥r} ≤c2 x + 1 r .
Proof We fix ǫ, K and allow constants in this proof to depend on ǫ, K. Let m be the smallest integer greater than K/ǫ. The assumption P{X1 ≥ǫ} ≥ǫ implies that for all x > 0, Px{SηK ≥K} ≥P{X1 ≥ǫ, . . . , Xm ≥ǫ} ≥ǫm.
Also note that if 0 ≤x ≤y ≤K then translation invariance and monotonicity give Px(Sηr ≥r) ≤ Py(Sηr ≥r). Therefore, for 0 < x ≤K, ǫm PK{Sηr ≥r} ≤Px{Sηr ≥r} ≤PK{Sηr ≥r}, (5.2) and hence it suffices to show for K ≤x ≤r that x r + K ≤Px{Sηr ≥r} ≤x + K r .
By the previous lemma, Ex[Sηr] = x. If Sηr ≥r, then r ≤Sηr ≤r + K.
If Sηr ≤0, then −K ≤Sηr ≤0. Therefore, x = Ex[Sηr] ≤Ex[Sηr; Sηr ≥r] ≤Px{Sηr ≥r} (r + K), and x = Ex[Sηr] ≥Ex[Sηr; Sηr ≥r] −K ≥r Px{Sηr ≥r} −K.
Proposition 5.1.5 For every ǫ > 0 and K < ∞, there exist 0 < c1 < c2 < ∞such that if P{|X1| > K} = 0 and P{X1 ≥ǫ} ≥ǫ, then for all x > 0, r > 1, c1 x + 1 r ≤Px{η ≥r2} ≤c2 x + 1 r .
Proof For the lower bound, we note that the maximal inequality for martingales (Theorem 12.2.5) implies P ( sup 1≤j≤n2 |X1 + · · · + Xj| ≥2Kn ) ≤E[S2 n2] 4K2n2 ≤1 4.
This tells us that if the random walk starts at z ≥3Kr, then the probability that it does not reach the origin in r2 steps is at least 3/4. Using this, the strong Markov property, and the last proposition, we get Px{η ≥r2} ≥3 4 Px{Sη3Kr ≥3Kr} ≥c1 (x + 1) r .
106 One-dimensional walks For the upper bound, we refer to Lemma 5.1.8 below. In this case, it is just as easy to give the argument for general mean zero, finite variance walks.
If p ∈Pd, d ≥2, then p induces an infinite family of one-dimensional non-lattice random walks Sn · θ where |θ| = 1. In Chapter 6, we will need gambler’s ruin estimates for these walks that are uniform over all θ. In particular, it will be important that the constant is uniform over all θ.
Proposition 5.1.6 Suppose Sn is a random walk with increment distribution p ∈Pd, d ≥2. There exist c1, c2 such that if θ ∈Rd with |θ| = 1 and Sn = Sn · θ, then the conclusions of Propositions 5.1.4 and 5.1.5 hold with c1, c2.
Proof Clearly there is a uniform bound on the range. The other condition is satisfied by noting the simple geometric fact that there is an ǫ > 0, independent of θ such that P{S1 · θ ≥ǫ} ≥ǫ, see Exercise 1.8.
5.1.1 General case We prove the gambler’s ruin estimate assuming only mean zero and finite variance. While we will not attempt to get the best values for the constants, we do show that the constants can be chosen uniformly over a wide class of distributions. In this section we fix K < ∞, δ, b > 0 and 0 < ρ < 1, and we let A(K, δ, b, ρ) be the collection of distributions on X1 with E[X1] = 0, E[X2 1] = σ2 ≤K2, P{X1 ≥1} ≥δ, inf n P{S1, . . . , Sn2 > −n} ≥b, ρ ≤inf n>0 P{Sn2 ≤−n}.
It is easy to check that for any mean zero, finite nonzero variance random walk Sn we can find a t > 0 and some K, δ, b, ρ such that the estimates above hold for tSn.
Theorem 5.1.7 (Gambler’s ruin) For every K, δ, b, ρ, there exist 0 < c1 < c2 < ∞such that if X1, X2, . . . are i.i.d. random variables whose distributions are in A(K, δ, b, ρ), then for all 0 < x < r, c1 x + 1 r ≤Px{η > r2} ≤c2 x + 1 r , c1 x + 1 r ≤Px{Sηr ≥r} ≤c2 x + 1 r .
Our argument consists of several steps. We start with the upper bound. Let η∗ r = min{n > 0 : Sn ≤0 or Sn ≥r}, η∗= η∗ ∞= min{n > 0 : Sn ≤0}.
Note that η∗ r differs from ηr in that the minimum is taken over n > 0 rather than n ≥0. As before we write P for P0.
5.1 Gambler’s ruin estimate 107 Lemma 5.1.8 P{η∗> n} ≤4K δ √n, P{η∗ n < η∗} ≤4K bδn.
Proof Let qn = P{η∗> n} = P{S1, . . . , Sn > 0}. Then P{S1, . . . , Sn ≥1} ≥δqn−1 ≥δqn.
Let Jk,n be the event Jk,n = {Sk+1, . . . , Sn ≥Sk + 1}.
We will also use Jk,n to denote the indicator function of this event. Let mn = min{Sj : 0 ≤j ≤n}, Mn = max{Sj : 0 ≤j ≤n}. For each real x ∈[mn, Mn], there is at most one integer k such that Sk ≤x and Sj > x, k < j ≤n. On the event Jk,n, the random set corresponding to the jump from Sk to Sk+1, {x : Sk ≤x and Sj > x, k < j ≤n}, contains an interval of length at least one.
In other words, there are P k Jk,n nonoverlapping intervals contained in [mn, Mn] each of length at least one. Therefore, n−1 X k=0 Jk,n ≤Mn −mn.
But, P(Jk,n) ≥δqn−k ≥δqn. Therefore, nδqn ≤E[Mn −mn] ≤2 E[ max{|Sj| : j ≤n} ].
Martingale maximal inequalities (Theorem 12.2.5) give P { max{|Sj| : j ≤n} ≥t } ≤E[S2 n] t2 ≤K2 n t2 .
Therefore, nδqn 2 ≤E[max{|Sj| : j ≤n}] = Z ∞ 0 P {max{|Sj| : j ≤n} ≥t} dt ≤ K √n + Z ∞ K√n K2 n t−2 dt = 2K √n.
This gives the first inequality. The strong Markov property implies P{η∗> n2 | η∗ n < η∗} ≥P{Sj −Sη∗ n > −n, 1 ≤j ≤n2 | η∗ n < η∗} ≥b.
Hence, b P{η∗ n < η∗} ≤P{η∗> n2}, (5.3) which gives the second inequality.
Lemma 5.1.9 (Overshoot lemma I) For all x > 0, Px{|Sη| ≥m} ≤1 ρ E[X2 1; |X1| ≥m].
(5.4) 108 One-dimensional walks Moreover if α > 0 and E[|X1|2+α] < ∞, then Ex [|Sη|α] ≤α ρ E[|X1|2+α].
♣Since Ex[η] = ∞, we cannot use the proof from Lemma 5.1.3.
Proof Fix ǫ > 0. For nonnegative integers k, let Yk = η X n=0 1{kǫ < Sn ≤(k + 1)ǫ} be the number of times the random walk visits (kǫ, (k + 1)ǫ] before hitting (−∞, 0], and let g(x, k) = Ex[Yk] = ∞ X n=0 Px{kǫ < Sn ≤(k + 1)ǫ; η > n}.
Note that if m, x > 0, Px{|Sη| ≥m} = ∞ X n=0 Px{|Sη| ≥m; η = n + 1} = ∞ X n=0 ∞ X k=0 Px{|Sη| ≥m; η = n + 1; kǫ < Sn ≤(k + 1)ǫ} ≤ ∞ X k=0 ∞ X n=0 Px{η > n; kǫ < Sn ≤(k + 1)ǫ; |Sn+1 −Sn| ≥m + kǫ} = ∞ X k=0 g(x, k) P{|X1| ≥m + kǫ} = ∞ X k=0 g(x, k) ∞ X l=k P{m + lǫ ≤|X1| < m + (l + 1)ǫ} = ∞ X l=0 P{m + lǫ ≤|X1| < m + (l + 1)ǫ} l X k=0 g(x, k).
Recall that P{Sn2 ≤−n} ≥ρ for each n. We claim that for all x, y, X 0≤k0 X 0≤k<⌊y/ǫ⌋ g(x, k).
5.1 Gambler’s ruin estimate 109 Note that the maximum is the same if we restrict to 0 < x ≤y. Then for any x ≤y, X 0≤k<⌊y/ǫ⌋ g(x, k) ≤y2 + Px{η ≥y2}E X n≥y2 1{Sn ≤y; n < η} η ≥y2 ≤y2 + (1 −ρ) Hy.
(5.6) By taking the supremum over x we get Hy ≤y2 + (1 −ρ)Hy which gives (5.5). We therefore have Px{|Sη| ≥m} ≤ 1 ρ ∞ X l=0 P{m + lǫ ≤|X1| < m + (l + 1)ǫ}(lǫ + ǫ)2 ≤ 1 ρ (E[(X1 −ǫ)2; X1 ≤−m] + E[(X1 + ǫ)2; X1 ≥m]).
Letting ǫ →0, we obtain (5.4).
To get the second estimate, let F denote the distribution function of |X1|. Then Ex[|Sη|α] = α Z ∞ 0 tα−1 Px{|Sη| ≥t} dt ≤ α ρ Z ∞ 0 tα−1 E[X2 1; |X1| ≥t] dt ≤ α ρ Z ∞ 0 E |X1|1+α; |X1| ≥t dt = α ρ Z ∞ 0 Z ∞ t x1+α dF(x) dt = α ρ Z ∞ 0 Z x 0 dt x1+α dF(x) = α ρ E[|X1|2+α].
♣The estimate (5.6) illustrates a useful way to prove upper bounds for Green’s functions of a set. If starting at any point y in a set V ⊂U, there is a probability q of leaving U within N steps, then the expected amount of time spent in V before leaving U starting at any x ∈U is bounded above by N + (1 −q) N + (1 −q)2 N + · · · = N q .
♣The lemma states that the overshoot random variable has two fewer moments than the increment distribution.
When the starting point is close to the origin, one might expect that the overshoot would be smaller since there are fewer chances for the last step before entering (−∞, 0] to be much larger than a typical step. The next lemma confirms this intuition and shows that one gains one moment if one starts near the origin.
Lemma 5.1.10 (Overshoot lemma II) Let c′ = 32 K bδ .
110 One-dimensional walks Then for all 0 < x ≤1, Px{|Sη| ≥m} ≤c′ ρ E[|X1|; |X1| ≥m].
Moreover if α > 0 and E[|X1|1+α] < ∞, then Ex[|Sη|α] ≤α c′ ρ E[|X1|1+α].
Proof The proof proceeds exactly as in Lemma 5.1.9 up to (5.5) which we replace with a stronger estimate that is valid for 0 < x ≤1: X 0≤kǫ<y g(x, k) ≤c′ y ρ .
(5.7) To derive this estimate we note that X 2j−1≤kǫ<2j g(x, k) equals the product of the probability of reaching a value above 2j−1 before hitting (−∞, 0] and the expected number of visits in this range given that event. Due to Lemma 5.1.8, the first probability is no more than 4K/(bδ2j−1) and the conditional expectation, as estimated in (5.5), is less than 22j/ρ. Therefore, X 0≤kǫ<2j g(x, k) ≤1 ρ j X l=1 4K bδ2l−1 22l ≤1 ρ 16K bδ 2j.
For general y we write 2j−1 < y ≤2j and obtain (5.7).
Given this, the same argument gives Px{|Sη| ≥m} ≤c′ ρ E[|X1|; |X1| ≥m], and E[|Sη|α] = α Z ∞ 0 tα−1 P{|Sη| ≥t} dt ≤ α c′ ρ Z ∞ 0 tα−1 E[|X1|; |X1| ≥t] dt ≤ α c′ ρ Z ∞ 0 E[Xα 1 ; |X1| ≥t] dt = α c′ ρ E[|X1|1+α].
♣The inequalities (5.5) and (5.7) imply that there exists a c < ∞such that for all y, Ey[η∗ n] < cn2, and Ex[η∗ n] ≤cn, 0 < x ≤1.
(5.8) 5.1 Gambler’s ruin estimate 111 Lemma 5.1.11 P{η∗ n < η∗} ≥c∗ n , where c∗= ρ δ 2 (ρ + 2c′K2), and c′ is as in Lemma 5.1.10. Also, P{η∗≥n} ≥b P{η∗ √n < η∗} ≥b c∗ √n.
Proof The last assertion follows immediately from the first one and the strong Markov property as in (5.3). Since P{η∗ n < η∗} ≥δ P1{η∗ n < η∗}, to establish the first assertion it suffices to prove that P1{η∗ n < η∗} ≥c∗ δ n.
Using (5.8), we have P1{|Sη∗ n| ≥s + n} ≤ ∞ X l=0 P1{η∗ n = l + 1; |Sη∗ n| ≥s + n} ≤ ∞ X l=0 P1{η∗ n > l; |Xl+1| ≥s} ≤ P{|X1| ≥s} E1[η∗ n] ≤ c′ n ρ P{|X1| ≥s}.
In particular, if t > 0, E1 |Sη∗ n|; |Sη∗ n| ≥(1 + t) n = Z ∞ tn P1{|Sη∗ n| ≥s + n} ds ≤ c′ n ρ Z ∞ tn P{|X1| ≥s} ds = c′ n ρ E[|X1|; |X1| ≥tn] ≤ c′ ρ t E |X1|2 ≤c′ K2 ρ t .
(5.9) Consider the martingale Mk = Sk∧η∗ n. Due to the optional stopping theorem we have 1 = E1[M0] = E1[M∞] ≤E1[Sη∗ n; Sη∗ n ≥n].
If we let t0 = 2c′K2/ρ in (5.9), we obtain E1[|Sη∗ n|; |Sη∗ n| ≥(1 + t0) n] ≤1 2, so it must be E1[Sη∗ n; n ≤Sη∗ n ≤(1 + t0)n] ≥1 2, 112 One-dimensional walks which implies P1{η∗ n < η∗} ≥P1{n ≤Sη∗ n ≤(1 + t0)n} ≥ 1 2(1 + t0)n.
Proof [of Theorem 5.1.7] Lemmas 5.1.8 and 5.1.11 prove the result for 0 < x ≤1. The result is easy if x ≥r/2 so we will assume 1 ≤x ≤r/2. As already noted, the function x 7→Px{Sηr ≥r} is nondecreasing in x. Therefore, P{Sη∗ r ≥r} = P{Sη∗ x ≥x} P{Sη∗ r ≥r | Sη∗ x ≥x} ≥P{Sη∗ x ≥x} Px{Sηr ≥r}.
Hence by Lemmas 5.1.8 and 5.1.11, Px{Sηr ≥r} ≤P{Sη∗ r ≥r} P{Sη∗ x ≥x} ≤4K c∗bδ x r .
For an inequality in the opposite direction, we first show that there is a c2 such that Ex[ηr] ≤c2 xr.
Recall from (5.8) that Ey[ηr] ≤cr for 0 < y ≤1. The strong Markov property and monotonicity can be used (Exercise 5.1) to see that Ex[ηr] ≤E1[ηr] + Ex−1[ηr].
(5.10) Hence we obtain the claimed bound for general x by induction. As in the previous lemma one can now see that Ex |Sη∗ r |; |Sη∗ r | ≥(1 + t) r ≤c2 K2 x t , and hence if t0 = 2c2K2, Ex Sη∗ r ; Sη∗ r ≥(1 + t0) r ≤x 2, Ex |Sη∗ r |; r ≤Sη∗ r ≤(1 + t0) r ≥x 2, so that Px r ≤Sη∗ r ≤(1 + t0) r ≥ x 2(1 + t0)r.
As we have already shown (see the beginning of the proof of Lemma 5.1.11), this implies Px{η∗≥r2} ≥b x 2(1 + t0)r.
5.2 One-dimensional killed walks A symmetric defective increment distribution (on Z) is a set of nonnegative numbers {pk : k ∈Z} with P pk < 1 and p−k = pk for all k. Given a symmetric defective increment distribution, we have the corresponding symmetric random walk with killing, that we again denote by S. More precisely, S is a Markov chain with state space Z ∪{∞}, where ∞is an absorbing state, and P{Sj+1 = k + l | Sj = k} = pl, P{Sj+1 = ∞| Sj = k} = p∞, 5.2 One-dimensional killed walks 113 where p∞= 1 −P pk. We let T = min{j : Sj = ∞} denote the killing time for the random walk. Note that P{T = j} = p∞(1 −p∞)j−1, j ∈{1, 2, . . .}.
Examples.
• Suppose p(j) is the increment distribution of a symmetric one-dimensional random walk and s ∈[0, 1). Then pj = s p(j) is a defective increment distribution corresponding to the random walk with killing rate 1 −s. Conversely, if pj is a symmetric defective increment distribution, and p(j) = pj/(1 −p∞), then p(j) is an increment distribution of a symmetric one-dimensional random walk (not necessarily aperiodic or irreducible). If we kill this walk at rate 1 −p∞, we get back pj.
• Suppose Sj is a symmetric random walk in Zd, d ≥2 which we write Sj = (Yj, Zj) where Yj is a random walk in Z and Zj is a random walk in Zd−1. Suppose the random walk is killed at rate 1 −s and let ˆ T denote the killing time. Let τ = min{j ≥1 : Zj = 0}, (5.11) pk = P{Yτ = k; τ < ˆ T}.
Note that pk = ∞ X j=1 P{τ = j; Yj = k; j < ˆ T} = ∞ X j=1 sj P{τ = j; Yj = k} = E[s ˆ T; Y ˆ T = k; ˆ T < ∞].
If Z is a transient random walk, then P{τ < ∞} < 1 and we can let s = 1.
• Suppose Sj = (Yj, Zj) and τ are as in the previous example and suppose A ⊂Zd−1 \ {0}. Let σA = min{j : Zj ∈A}, pk = P{Yτ = k; τ < σA}.
If P{Zj ∈A for some j} > 0, then {pk} is a defective increment distribution.
Given a symmetric defective increment distribution {pk} with corresponding walk Sj and killing time T, define the events V+ = {Sj > 0 : j = 1, . . . , T −1}, V + = {Sj ≥0 : j = 1, . . . , T −1}, V−= {Sj < 0 : j = 1, . . . , T −1}, V −= {Sj ≤0 : j = 1, . . . , T −1}.
Symmetry implies that P(V+) = P(V−), P(V +) = P(V −). Note that V+ ⊂V +, V−⊂V −and P(V+ ∩V −) = P(V + ∩V−) = P{T = 1} = p∞.
(5.12) Define a new defective increment distribution pk,−, which is supported on k = 0, −1, −2, . . ., by 114 One-dimensional walks setting pk,−equal to the probability that the first visit to {· · · , −2, −1, 0} after time 0 occurs at position k and this occurs before the killing time T, i.e., pk,−= ∞ X j=1 P{Sj = k ; j < T ; Sl > 0, l = 1, . . . , j −1}.
Define pk,+ similarly so that pk,+ = p−k,−. The strong Markov property implies P(V +) = P(V+) + p0,−P(V +), and hence P(V+) = (1 −p0,−) P(V +) = (1 −p0,+) P(V +).
(5.13) In the next proposition we prove a nonintuitive fact.
Proposition 5.2.1 The events V + and V−are independent. In particular, P(V−) = P(V+) = (1 −p0,+) P(V +) = q p∞(1 −p0,+).
(5.14) Proof Independence is equivalent to the statement P(V−∩V +) = P(V−) P(V +). We will prove the equivalent statement P(V−∩V c +) = P(V−) P(V c +). Note that V−∩V c + is the event that T > 1 but no point in {0, 1, 2 . . .} is visited during the times {1, . . . , T −1}. In particular, at least one point in {. . . , −2, −1} is visited before time T.
Let ρ = max{k ∈Z : Sj = k for some j = 1, . . . , T −1}, ξk = max{j ≥0 : Sj = k; j < T}.
In words, ρ is the rightmost point visited after time zero, and ξk is the last time that k is visited before the walk is killed. Then, P(V−∩V c +) = ∞ X k=1 P{ρ = −k} = ∞ X k=1 ∞ X j=1 P{ρ = −k; ξ−k = j}.
Note that the event {ρ = −k; ξ−k = j} is the same as the event {Sj = −k ; j < T ; Sl ≤−k, l = 1, . . . , j −1 ; Sl < −k, l = j + 1, . . . , T −1}.
Since, P{Sl < −k, l = j + 1, . . . , T −1 | Sj = −k ; j < T ; Sl ≤−k, l = 1, . . . , j −1} = P(V−), we have P{ρ = −k; ξ−k = j} = P{Sj = −k ; j < T ; Sl ≤−k, l = 1, . . . , j −1} P(V−).
Due to the symmetry of the random walk, the probability of the path [x0 = 0, x1, . . . , xj] is the same as the probability of the reversed path [xj −xj, xj−1 −xj, . . . , x0 −xj]. Note that if xj = −k 5.3 Hitting a half-line 115 and xl ≤−k, l = 1, . . . , j −1, then x0 −xj = k and Pl i=1(xj−i −xj−i+1) = xj−l −xj ≤0, for l = 1, . . . , j −1. Therefore we have P{Sj = −k ; j < T ; Sl ≤−k, l = 1, . . . , j −1} = P{η = j; j < T; Sj = k}, where η = min{j ≥1 : Sj > 0}.
Since ∞ X k=1 ∞ X j=1 P{η = j; j < T; Sj = k} = P{η < T} = P(V c −) = P(V c +), we obtain the stated independence. The equality (5.14) now follows from p∞= P(V−∩V +) = P(V−) P(V +) = P(V−) P(V+) 1 −p0,− = P(V+)2 1 −p0,+ .
5.3 Hitting a half-line We will give an application of Proposition 5.2.1 to walks in Zd. Suppose d ≥2 and Sn is a random walk with increment distribution p ∈Pd. We write Sn = (Yn, Zn) where Yn is a one-dimensional walk and Zn is a (d −1)-dimensional walk. Let Γ denote the covariance matrix for Sn and let Γ∗ denote the covariance matrix for Zn. Let T = min{j > 0 : Zj = 0} be the first time that the random walk returns to the line {(j, x) ∈Z × Zd−1 : x = 0}. Let T+, T + denote the corresponding quantities for the (nonpositive and negative) half-line T+ = min n n > 0 : Sn ∈{(j, x) ∈Z × Zd−1 : j ≤0, x = 0} o , T + = min n n > 0 : Sn ∈{(j, x) ∈Z × Zd−1 : j < 0, x = 0} o , and finally let p0,+ = P{YT+ = 0}.
Proposition 5.3.1 If p ∈Pd, d = 2, 3, there is a C such that as n →∞, (1 −p0,+) P{T + > n} ∼P{T+ > n} ∼ C n−1/4, d = 2, C (log n)−1/2, d = 3.
Proof We will prove the second asymptotic relation; a similar argument shows that the first and third terms are asympotic. Let σ = σξ denote a geometric random variable, independent of the random walk, with killing rate 1 −ξ, i.e., P{σ > k} = ξk. Let qn = P{T+ > n}, q(ξ) = P{T+ > σ}.
Then, q(ξ) = P{T+ > σ} = ∞ X n=1 P{σ = n; T+ > n} = ∞ X n=1 (1 −ξ) ξn−1 qn.
116 One-dimensional walks By Propositions 12.5.2 and 12.5.3, it suffices to show that q(ξ) ∼c (1 −ξ)1/4 if d = 2 and q(ξ) ∼ c [−log(1 −ξ)]−1/2 if d = 3.
This is the same situation as the second example of the last subsection (although (τ, T) there corresponds to (T, σ) here). Hence, Proposition 5.2.1 tells us that q(ξ) = q p∞(ξ) (1 −p0,+(ξ)), where p∞(ξ) = P{T > σ} and p0,+(ξ) = P{T+ ≤σ; YT+ = 0}. Clearly, as ξ →1−, 1 −p0,+(ξ) → 1 −p0,+ > 0. By applying (4.9) and (4.10) to the random walk Zn, we can see that P{T > σ} ∼c (1 −ξ)1/2, d = 2, P{T > σ} ∼c log 1 1 −ξ −1 , d = 3.
♣From the proof one can see that the constant C can be determined in terms of Γ∗and p0,+. We do not need the exact value and the proof is a little easier to follow if we do not try to keep track of this constant. It is generally hard to compute p0,+; for simple random walk, see Proposition 9.9.8.
♣The above proof uses the surprising fact that the events “avoid the positive x1-axis” and “avoid the negative x1- axis” are independent up to a multiplicative constant. This idea does not extend to other sets, for example the event “avoid the positive x1-axis” and “avoid the positive x2-axis” are not independent up to a multiplicative constant in two dimensions. However, they are in three dimensions (which is a nontrivial fact).
In Section 6.8 we will need some estimates for two-dimensional random walks avoiding a half-line. The argument given below uses the Harnack inequality (Theorem 6.3.9), which will be proved independently of this estimate. In the remainder of this section, let d = 2 and let Sn = (Yn, Zn) be the random walk. Let ζr = min {n > 0 : Yn ≥r} , ρr = min {n > 0 : Sn ̸∈(−r, r) × (−r, r)} , ρ∗ r = min {n > 0 : Sn ̸∈Z × (−r, r)} .
If |S0| < r, the event {ζr = ρr} occurs if and only if the first visit of the random walk to the complement of (−r, r) × (−r, r) is at a point (j, k) with j ≥r.
Proposition 5.3.2 If p ∈P2, then P{T+ > ρr} ≍r−1/2.
(5.15) Moreover, for all z ̸= 0, Pz{ρr < T+} ≤c |z|1/2 r−1/2 (5.16) 5.3 Hitting a half-line 117 In addition, there is a c < ∞such that if 1 ≤k ≤r and Ak = {je1 : j = −k, −k + 1, . . .} , then P{TAk > ρr} ≤c k−1/2 r−1/2.
(5.17) Proof It suffices to show that there exist c1, c2 with P{T+ > ρr} ≤c2 r−1/2, P{T+ > ρ∗ r} ≥c1 r−1/2.
The gambler’s ruin estimate applied to the second component implies that P{T > ρ∗ r} ≍r−1 and an application of Proposition 5.2.1 gives P{T+ > ρ∗ r} ≍r−1/2.
Using the invariance principle, it is not difficult to show that there is a c such that for r sufficiently large, P{ζr = ρr} ≥c. By translation invariance and monotonicity, one can see that for j ≥1, P−je1{ζr = ρr} ≤P{ζr = ρr}.
Hence the strong Markov property implies that P{ζr = ρr | T+ < ρr} ≤P{ζr = ρr}, therefore it has to be that P{ζr = ρr | T+ > ρr} ≥c and P{ρr < T+} ≤c P{ζr = ρr < T+}.
(5.18) Another application of the invariance principle shows that P{T+ > r2 | ζr = ρr < T+} ≥c, since this conditional probability is bounded below by the probability that a random walk goes no farther than distance r/2 in r2 steps. Hence, P{ρr < T+} ≤c P{ρr < T+, T+ > r2} ≤c P{T+ > r2} ≤c r−1/2.
This gives (5.15).
For the remaining results we will assume |z| is an integer greater than the range R of the walk, but one can easily adapt the argument to arbitrary z. Let hr(x) = Px{ρr < T+} and let M = M(r, |z|) be the maximum value of hr(x) over x ∈(−|z| −R, |z| + R) × (−|z| −R, |z| + R).
By translation invariance, this is maximized at a point with maximal first component and by the Harnack inequality (Theorem 6.3.9), c1 M ≤hr(x) ≤c2 M, x ∈(|z| −R, |z| + R) × (−|z| −R, |z| + R).
Together with strong Markov property this implies P{ρr < T+} ≤c M P{ρ|z| < T+}, and due to (5.18) P{ρr < T+} ≥c M P{ρ|z| = ζ|z| < T+} ≥c M P{ρ|z| < T+}.
Since P{ρr < T+} ≍r−1/2, we conclude that M ≍|z|1/2 r−1/2, implying (5.16). To prove (5.17), we write P{TAk > ρr} = P{TAk > ρk} P{TAk > ρr | TAk > ρk} ≤c P{TAk > ρk} (k/r)1/2.
So if suffices to show that P{TAk > ρk} ≤ck−1 This is very close to the gambler’s ruin estimate, but it is not exactly the form we have proved so far, so we will sketch a proof.
118 One-dimensional walks Let q(k) = P{TAk > ρk}.
Note that for all integers |j| < k, Pje1{TAk > ρk} ≥q(2k).
A last-exit decomposition focusing on the last visit to Ak before time ρk shows that 1 = X |j| ρk} ≥q(2k) X |j| n} ∼c n−1.
(ii) Show that there exists c1 such that P{T+ > n} ∼c1 n−1/2.
Establish the analog of Proposition 5.3.2 in this setting.
6 Potential Theory 6.1 Introduction There is a close relationship between random walks with increment distribution p and functions that are harmonic with respect to the generator L = Lp.
We start by setting some notation. We fix p ∈P. If A ⫋Zd, we let ∂A = {x ∈Zd \ A : p(y, x) > 0 for some y ∈A} denote the (outer) boundary of A and we let A = A ∪∂A be the discrete closure of A.
Note that the above definition of ∂A, A depends on the choice of p. We omit this dependence from the notation, and hope that this will not confuse the reader. In the case of simple random walk, ∂A = {x ∈Zd \ A : |y −x| = 1 for some y ∈A}.
Since p has finite range, if A is finite, then ∂A, A are finite. The inner boundary of A ⊂Zd is defined by ∂iA = ∂(Zd \ A) = {x ∈A : p(x, y) > 0 for some y ̸∈A}.
Figure 6.1: Suppose A is the set of lattice points “inside” the dashed curve. Then the points in A \ ∂iA, ∂iA and ∂A are marked by •, ◦and ×, respectively 119 120 Potential Theory A function f : A →R is harmonic (with respect to p) or p-harmonic in A if Lf(y) := X x p(x) [f(y + x) −f(y)] = 0 for every y ∈A. Note that we cannot define Lf(y) for all y ∈A, unless f is defined on A.
We say that A is connected (with respect to p) if for every x, y ∈A, there is a finite sequence x = z0, z1, . . . , zk = y of points in A with p(zj+1 −zj) > 0, j = 0, . . . , k −1.
♣This chapter contains a number of results about functions on subsets of Zd. These results have analogues in the continuous setting. The set A corresponds to an open set D ⊂Rd, the outer boundary ∂A corresponds to the usual topological boundary ∂D, and A corresponds to the closure D = D ∪∂D. The term domain is often used for open, connected subsets of Rd. Finiteness assumptions on A correspond to boundedness assumptions on D.
Proposition 6.1.1 Suppose Sn is a random walk with increment distribution p ∈Pd starting at x ∈Zd. Suppose f : Zd →R. Then Mn := f(Sn) − n−1 X j=0 Lf(Sj) is a martingale. In particular, if f is harmonic on A ⊂Zd, then Yn := f(Sn∧τA) is a martingale, where τ A is as defined in (4.27).
Proof Immediate from the definition.
Proposition 6.1.2 Suppose p ∈Pd and f : Zd →R is bounded and harmonic on Zd. Then f is constant.
Proof We may assume p is aperiodic; if not consider ˆ p = (1/2) p + (1/2)δ0 and note that f is p-harmonic if and only if it is ˆ p-harmonic. Let x, y ∈Zd. By Lemma 2.4.3 we can define random walks S, ˆ S on the same probability space so that S is a random walk starting at x; ˆ S is a random walk starting at y; and P{Sn ̸= ˆ Sn} ≤c |x −y| n−1/2.
In particular, |E[f(Sn)] −E[f( ˆ Sn)]| ≤2 c |x −y| n−1/2 ∥f∥∞− →0.
Proposition 6.1.1 implies that f(x) = E[f(Sn)], f(y) = E[f( ˆ Sn)].
♣The fact that all bounded harmonic functions are constant is closely related to the fact that a random walk eventually forgets its starting point. Lemma 2.4.3 gives a precise formulation of this loss of memory property.
The last proposition is not true for simple random walk on a regular tree.
6.2 Dirichlet problem 121 6.2 Dirichlet problem The standard Dirichlet problem for harmonic functions is to find a harmonic function on a region with specified values on the boundary.
Theorem 6.2.1 (Dirichlet problem I) Suppose p ∈Pd, and A ⊂Zd satisfies Px{τ A < ∞} = 1 for all x ∈A. Suppose F : ∂A →R is a bounded function. Then there is a unique bounded function f : A →R satisfying Lf(x) = 0, x ∈A, (6.1) f(x) = F(x), x ∈∂A.
(6.2) It is given by f(x) = Ex[F(Sτ A)].
(6.3) Proof A simple application of the Markov property shows that f defined by (6.3) satisfies (6.1) and (6.2). Now suppose f is a bounded function satisfying (6.1) and (6.2). Then Mn := f(Sn∧τ A) is a bounded martingale. Hence, the optional sampling theorem (Theorem 12.2.3) implies that f(x) = Ex[M0] = Ex[Mτ A] = Ex[F(Sτ A)].
Remark.
• If A is finite, then ∂A is also finite and all functions on A are bounded. Hence for each F on ∂A, there is a unique function satisfying (6.1) and (6.2). In this case we could prove existence and uniqueness using linear algebra since (6.1) and (6.2) give #(A) linear equations in #(A) unknowns. However, algebraic methods do not yield the nice probabilistic form (6.3).
• If A is infinite, there may well be more than one solution to the Dirichlet problem if we allow unbounded solutions. For example, if d = 1, p is simple random walk, A = {1, 2, 3, . . .}, and F(0) = 0, then there is an infinite number of solutions of the form fb(x) = bx. If b ̸= 0, fb is unbounded.
• Under the conditions of the theorem, it follows that any function f on A that is harmonic on A satisfies the maximum principle: sup x∈A |f(x)| = sup x∈∂A |f(x)|.
• If d = 1, 2 and A is a proper subset of Zd, then we know by recurrence that Px{τ A < ∞} = 1 for all x ∈A.
• If d ≥3 and Zd \ A is finite, then there are points x ∈A with Px{τ A = ∞} > 0. The function f(x) = Px{τ A = ∞} is a bounded function satisfying (6.1) and (6.2) with F ≡0 on ∂A.
Hence, the condition Px{τ A < ∞} = 1 is needed to guarantee uniqueness. However, as Proposition 6.2.2 below shows, all solutions with F ≡0 on ∂A are multiples of f.
122 Potential Theory Remark.
This theorem has a well-known continuous analogue. Suppose f : {|z| ∈Rd : |z| ≤ 1} →R is a continuous function with ∆f(x) = 0 for |x| < 1. Then f(x) = Ex[f(BT )], where B is a standard d-dimensional Brownian motion and T is the first time t that |Bt| = 1. If |x| < 1, the distribution of BT given B0 = x has a density with respect to surface measure on {|z| = 1}. This density h(x, z) = c (1 −|x|2)/|x −z|d is called the Poisson kernel and we can write f(x) = c Z |z|=1 f(z) 1 −|x|2 |x −z|d ds(z), (6.4) where s denotes surface measure. To verify that this is correct, one can check directly that f as defined above is harmonic in the ball and satisfies the boundary condition on the sphere. Two facts follow almost immediately from this integral formula: • Derivative estimates. For every k, there is a c = c(k) < ∞such that if f is harmonic in the unit ball and D denotes a kth order derivative, then |Df(0)| ≤ck ∥f∥∞.
• Harnack inequality.
For every r < 1, there is a c = cr < ∞such that if f is a positive harmonic function on the unit ball, then f(x) ≤c f(y) for |x|, |y| ≤r.
An important aspect of these estimates is the fact that the constants do not depend on f. We will prove the analogous results for random walk in Section 6.3.
Proposition 6.2.2 (Dirichlet problem II) Suppose p ∈Pd and A ⫋Zd. Suppose F : ∂A →R is a bounded function. Then the only bounded functions f : A →R satisfying (6.1) and (6.2) are of the form f(x) = Ex[F(Sτ A); τ A < ∞] + b Px{τ A = ∞}, (6.5) for some b ∈R.
Proof We may assume that p is aperiodic. We also assume that Px{τ A = ∞} > 0 for some x ∈A; if not, Theorem 6.2.1 applies. Assume that f is a bounded function satisfying (6.1) and (6.2). Since Mn := f(Sn∧τA) is a martingale, we know that f(x) = Ex[M0] = Ex[Mn] = Ex[f(Sn∧τA)] = Ex[f(Sn)] −Ex[f(Sn); τ A < n] + Ex[F(Sτ A); τ A < n].
Using Lemma 2.4.3, we can see that for all x, y, lim n→∞|Ex[f(Sn)] −Ey[f(Sn)]| = 0.
Therefore, |f(x) −f(y)| ≤2 ∥f∥∞[Px{τ A < ∞} + Py{τ A < ∞}].
Let Uǫ = {z ∈Zd : Pz{τ A = ∞} ≥1−ǫ}. Since Px{τ A = ∞} > 0 for some x, one can see (Exercise 4.1) that Uǫ is non-empty for each ǫ ∈(0, 1). Then, |f(x) −f(y)| ≤4 ǫ ∥f∥∞, x, y ∈Uǫ.
6.2 Dirichlet problem 123 Hence, there is a b such that |f(x) −b| ≤4 ǫ ∥f∥∞, x ∈Uǫ.
Let ρǫ be the minimum of τ A and the smallest n such that Sn ∈Uǫ. Then for every x ∈Zd, the optional sampling theorem implies f(x) = Ex[f(Sρǫ)] = Ex[F(Sτ A); τ A ≤ρǫ] + Ex[f(Sρǫ); τ A > ρǫ].
(Here we use the fact that Px{τA ∧ρǫ < ∞} = 1 which can be verified easily.) By the dominated convergence theorem, lim ǫ→0 Ex[F(Sτ A); τ A ≤ρǫ] = Ex[F(Sτ A); τ A < ∞].
Also, |Ex[f(Sρǫ); τ A > ρǫ] −b Px{τA > ρǫ}| ≤4ǫ∥f∥∞Px{τA > ρǫ}, and since ρǫ →∞as ǫ →0, lim ǫ→0 Ex[f(Sρǫ); τ A > ρǫ] = b Px{τA = ∞}.
This gives (6.5).
Remark. We can think of (6.5) as a generalization of (6.3) where we have added a boundary point at infinity. The constant b in the last proposition is the boundary value at infinity and can be written as F(∞). The fact that there is a single boundary value at infinity is closely related to Proposition 6.1.2.
Definition. If p ∈Pd and A ⊂Zd, then the Poisson kernel is the function H : A × ∂A →[0, 1] defined by HA(x, y) = Px{τ A < ∞; Sτ A = y}.
As a slight abuse of notation we will also write HA(x, ∞) = Px{τ A = ∞}.
Note that X y∈∂A HA(x, y) = Px{τ A < ∞}.
For fixed y ∈∂A, f(x) = HA(x, y) is a function on A that is harmonic on A and equals δ(· −y) on ∂A. If p is recurrent, there is a unique such function. If p is transient, f is the unique such function that tends to 0 as x tends to infinity. We can write (6.3) as f(x) = Ex[F(Sτ A)] = X y∈∂A HA(x, y) F(y), (6.6) and (6.5) as f(x) = Ex[F(Sτ A); τ A < ∞] + b Px{τ A = ∞} = X y∈∂A∪{∞} HA(x, y) F(y), 124 Potential Theory where F(∞) = b. The expression (6.6) is a random walk analogue of (6.4).
Proposition 6.2.3 Suppose p ∈Pd and A ⫋Zd. Let g : A →R be a function with finite support.
Then, the function f(x) = X y∈A GA(x, y) g(y) = Ex τ A−1 X j=0 g(Sj) , is the unique bounded function on A that vanishes on ∂A and satisfies Lf(x) = −g(x), x ∈A.
(6.7) Proof Since g has finite support, |f(x)| ≤ X y∈A GA(x, y) |g(y)| < ∞, and hence f is bounded. We have already noted in Lemma 4.6.1 that f satisfies (6.7). Now suppose f is a bounded function vanishing on ∂A satisfying (6.7). Then, Proposition 6.1.1 implies that Mn := f(Sn∧τA) + n∧τ A−1 X j=0 g(Sj), is a martingale. Note that |Mn| ≤∥f∥∞+ Y where Y = τ A−1 X j=0 |g(Sj)|, and that Ex[Y ] = X y GA(x, y) |g(y)| < ∞.
Hence Mn is dominated by an integrable random variable and we can use the optional sampling theorem (Theorem 12.2.3) to conclude that f(x) = Ex[M0] = Ex[Mτ A] = Ex τ A−1 X j=0 g(Sj) .
Remark. Suppose A ⊂Zd is finite with #(A) = m. Then GA = [GA(x, y)]x,y∈A is an m × m symmetric matrix with nonnegative entries. Let LA = [LA(x, y)]x,y∈A be the m × m symmetric matrix defined by LA(x, y) = p(x, y), x ̸= y; LA(x, x) = p(x, x) −1.
If g : A →R and x ∈A, then LAg(x) is the same as Lg(x) where g is extended to A by setting g ≡0 on ∂A. The last proposition can be rephrased as LA[GAg] = −g, or in other words, GA = −(LA)−1.
6.3 Difference estimates and Harnack inequality 125 Corollary 6.2.4 Suppose p ∈Pd and A ⊂Zd is finite. Let g : A →R, F : ∂A →R be given. Then, the function f(x) = Ex[F(Sτ A)] + Ex τ A−1 X j=0 g(Sj) = X z∈∂A HA(x, z) F(z) + X y∈A GA(x, y) g(y), (6.8) is the unique function on A that satisfies Lf(x) = −g(x), x ∈A.
f(x) = F(x), x ∈∂A.
In particular, for any f : A →R, x ∈A, f(x) = Ex[f(SτA)] −Ex τ A−1 X j=0 Lf(Sj) .
(6.9) Proof Use the fact that h(x) := f(x) −Ex[F(Sτ A)] satisfies the assumptions in the previous propo-sition.
Corollary 6.2.5 Suppose p ∈Pd and A ⊂Zd is finite. Then f(x) = Ex[τ A] = X y∈A GA(x, y) is the unique bounded function f : A →R that vanishes on ∂A and satisfies Lf(x) = −1, x ∈A.
Proof This is Proposition 6.2.3 with g ≡1A.
Proposition 6.2.6 Let ρn = τ Bn = inf{j ≥0 : |Sj| ≥n}. Then if p ∈Pd with range R and |x| < n, [n2 −|x|2] ≤(trΓ) Ex[ρn] ≤[(n + R)2 −|x|2].
Proof In Exercise 1.4 it was shown that Mj =: |Sj∧ρn|2 −(trΓ)(j ∧ρn) is a martingale.
Also, Ex[ρn] < ∞for each x, so Mj is dominated by the integrable random variable (n + R)2 + (trΓ) ρn.
Hence, |x|2 = Ex[M0] = Ex[Mρn] = Ex[|Sρn|2] −(trΓ) Ex[ρn].
Moreover, n ≤|Sρn| < (n + R).
6.3 Difference estimates and Harnack inequality In the next two sections we will prove useful results about random walk and harmonic functions.
The main tools in the proofs are the optional sampling theorem and the estimates for the Green’s function and the potential kernel. The basic idea in many of the proofs is to define a martingale 126 Potential Theory in terms of the Green’s function or potential kernel and then to stop it at a region at which that function is approximately constant. We recall that Bn = {z ∈Zd : |z| < n}, Cn = {z ∈Zd : J (z) < n}.
Also, there is a δ > 0 such that Cδn ⊂Bn ⊂Cn/δ.
We set ξn = τCn = min{j ≥1 : Sj ̸∈Cn}, ξ∗ n = τBn = min{j ≥1 : Sj ̸∈Bn}.
As the next proposition points out, the Green’s function and potential kernel are almost constant on ∂Cn. We recall that Theorems 4.3.1 and 4.4.4 imply that as x →∞, G(x) = Cd J (x)d−2 + O 1 |x|d , d ≥3, (6.10) a(x) = C2 log J (x) + γ2 + O 1 |x|2 .
(6.11) Here C2 = [π √ det Γ]−1 and γ2 = C + C2 log √ 2 where C is as in Theorem 4.4.4.
Proposition 6.3.1 If p ∈Pd, d ≥3 then for x ∈∂Cn ∪∂iCn, G(x) = Cd nd−2 + O(n1−d), d ≥2, a(x) = C2 log n + γ2 + O(n−1), d = 2, where Cd, C2, γ2 are as (6.10) and (6.11).
Proof This follows immediately from (6.10) and (6.11) and the estimate J (x) = n + O(1), x ∈∂Cn ∪∂iCn.
Note that the error term O(n1−d) comes from the estimates [n + O(1)]2−d = n2−d + O(n1−d), log[n + O(1)] = log n + O(n−1).
♣Many of the arguments in this section use Cn rather than Bn because we can then use Proposition 6.3.1.
We recall that for simple random walk Bn = Cn.
♣Proposition 6.3.1 requires the walk to have bounded increments.
If the walk does not have bounded increments, then many of the arguments in this chapter still hold. However, one needs to worry about “overshoot” estimates, i.e., giving upper bounds for the probability that the first visit of a random walk to the complement of Cn is far from Cn. These kinds of estimate can be done in a spirit similar to Lemmas 5.1.9 and 5.1.10, but they complicate the arguments. For this reason, we restrict our attention to walks with bounded increments.
6.3 Difference estimates and Harnack inequality 127 ♣Proposition 6.3.1 gives the estimates for the Green’s function or potential kernel on ∂Cn. In order to prove these estimates, it suffices for the error terms in (6.10) and (6.11) to be O(|x|1−d) rather than O(|x|−d). For this reason, many of the ideas of this section extend to random walks with bounded increments that are not necessarily symmetric (see Theorems 4.3.5 and 4.4.6). However, in this case we would need to deal with the Green’s functions for the reversed walk as well as the Green’s functions for the forward walk, and this complicates the notation in the arguments. For this reason, we restrict our attention to symmetric walks.
Proposition 6.3.2 If p ∈Pd, GCn(0, 0) = G(0, 0) −Cd nd−2 + O(n1−d), d ≥3, GCn(0, 0) = C2 log n + γ2 + O(n−1), d = 2.
(6.12) where Cd, γ2 are as defined in Proposition 6.3.1.
Proof Applying Proposition 4.6.2 at x = y = 0 gives GCn(0, 0) = G(0, 0) −E[G(SτCn, 0)], d ≥3, GCn(0, 0) = E[a(SτCn, 0)], d = 2.
We now apply Proposition 6.3.1.
♣It follows from Proposition 6.3.2 that GBn(0, 0) = G(0, 0) + O(n2−d), d ≥3, aBn(0, 0) = C2 log n + O(1), d = 2.
It can be shown that GBn(0, 0) = G(0, 0)−ˆ Cd n2−d +o(n1−d), aBn(0, 0) = C2 log n+ ˆ γ2 +O(n−1) where ˆ Cd, ˆ γ2 are different from Cd, γ2 but we will not need this in the sequel, hence omit the argument.
We will now prove difference estimates and a Harnack inequality for harmonic functions. There are different possible approaches to proving these results.
One would be to use the result for Brownian motion and approximate.
We will use a different approach where we start with the known difference estimates for the Green’s function G and the potential kernel a and work from there. We begin by proving a difference estimate for GA. We then use this to prove a result on probabilities that is closely related to the gambler’s ruin estimate for one-dimensional walks.
Lemma 6.3.3 If p ∈Pd, d ≥2, then for every ǫ > 0, r < ∞, there is a c such that if Bǫn ⊂A ⫋Zd, then for every |x| > ǫn and every |y| ≤r, |GA(0, x) −GA(y, x)| ≤ c nd−1 .
|2 GA(0, x) −GA(y, x) −GA(−y, x)| ≤c nd .
128 Potential Theory Proof It suffices to prove the result for finite A for we can approximate any A by finite sets (see Exercise 4.7). Assume that x ∈A, for otherwise the result is trivial. By symmetry GA(0, x) = GA(x, 0), GA(y, x) = GA(x, y). By Proposition 4.6.2, GA(x, 0) −GA(x, y) = G(x, 0) −G(x, y) − X z∈∂A HA(x, z) [G(z, 0) −G(z, y)], d ≥3, GA(x, y) −GA(x, 0) = a(x, 0) −a(x, y) − X z∈∂A HA(x, z) [a(z, 0) −a(z, y)], d = 2.
There are similar expressions for the second differences. The difference estimates for the Green’s function and the potential kernel (Corollaries 4.3.3 and 4.4.5) give, provided that |y| ≤r and |z| ≥(ǫ/2) n, |G(z) −G(z + y)| ≤cǫ n1−d, |2G(z) −G(z + y) −G(z −y)| ≤cǫ n−d for d ≥3 and |a(z) −a(z + y)| ≤cǫ n−1, |2a(z) −a(z + y) −a(z −y)| ≤cǫ n−2 for d = 2.
The next lemma is very closely related to the one-dimensional gambler’s ruin estimate. This lemma is particularly useful for x on or near the boundary of Cn. For x in Cn \ Cn/2 that are away from the boundary, there are sharper estimates. See Propositions 6.4.1 and 6.4.2.
Lemma 6.3.4 Suppose p ∈Pd, d ≥2. There exist c1, c2 such that for all n sufficiently large and all x ∈Cn \ Cn/2, Px{SτCn\Cn/2 ∈Cn/2} ≥c1 n−1, (6.13) and if x ∈∂Cn, Px{SτCn\Cn/2 ∈Cn/2} ≤c2 n−1.
(6.14) Proof We will do the proof for d ≥3; the proof for d = 2 is almost identical replacing the Green’s function with the potential kernel. It follows from (6.10) that there exist r, c such that for all n sufficiently large and all y ∈Cn−r, z ∈∂Cn, G(y) −G(z) ≥cn1−d.
(6.15) By choosing n sufficiently large, we can assure that ∂iCn/2 ∩Cn/4 = ∅.
Suppose that x ∈Cn−r and let T = τCn\Cn/2. Applying the optional sampling theorem to the bounded martingale G(Sj∧T ), we see that G(x) = Ex[G(ST )] ≤Ex[G(ST ); ST ∈Cn/2] + max z∈∂Cn G(z).
Therefore, (6.15) implies that Ex[G(ST ); ST ∈Cn/2] ≥c n1−d.
6.3 Difference estimates and Harnack inequality 129 For n sufficiently large, ST ̸∈Cn/4 and hence (6.10) gives Ex[G(ST ); ST ∈Cn/2] ≤c n2−d Px{τCn\Cn/2 < τCn}.
This establishes (6.13) for x ∈Cn−r.
To prove (6.13) for other x we note the following fact that holds for any p ∈Pd: there is an ǫ > 0 such that for all |x| ≥r, there is a y with p(y) ≥ǫ and J (x + y) ≤J (x) −ǫ. It follows that there is a δ > 0 such that for all n sufficiently large and all x ∈Cn, there is probability at least δ that a random walk starting at x reaches Cn−r before leaving Cn.
Since our random walk has finite range, it suffices to prove (6.14) for x ∈Cn \ Cn−r, and any finite r. For such x, G(x) = Cd n2−d + O(n1−d).
Also, Ex[G(ST ) | ST ∈Cn/2] = Cd 2d−2 n2−d + O(n1−d), Ex[G(ST ) | ST ∈∂Cn] = Cd n2−d + O(n1−d).
The optional sampling theorem gives G(x) = Ex[G(ST )] = Px{ST ∈Cn/2} Ex[G(ST ) | ST ∈Cn/2] + Px{ST ∈∂Cn} Ex[G(ST ) | ST ∈∂Cn].
The left-hand side equals Cd n2−d + O(n1−d) and the right-hand side equals Cd n2−d + O(n1−d) + Cd [2d−2 −1] n2−d Px{ST ∈Cn/2}.
Therefore Px{ST ∈Cn/2} = O(n−1).
Proposition 6.3.5 If p ∈Pd and x ∈Cn, GCn(0, x) = Cd h J (x)2−d −n2−di + O(|x|1−d), d ≥3, GCn(0, x) = C2 [log n −log J (x)] + O(|x|−1), d = 2.
In particular, for every 0 < ǫ < 1/2, there exist c1, c2 such that for all n sufficiently large, c1 n2−d ≤GCn(y, x) ≤c2 n2−d, y ∈Cǫn, x ∈∂iC2ǫn ∪∂C2ǫn.
Proof Symmetry and Lemma 4.6.2 tell us that GCn(0, x) = GCn(x, 0) = G(x, 0) −Ex[G(SτCn)], d ≥3, GCn(0, x) = GCn(x, 0) = Ex[a(SτCn)] −a(x), d = 2.
(6.16) Also, (6.10) and (6.11) give G(x) = Cd J (x)2−d + O(|x|−d), d ≥3, a(x) = C2 log[J (x)] + γ2 + O(|x|−2), d = 2, 130 Potential Theory and Proposition 6.3.1 implies that Ex[G(SτCn )] = Cd nd−2 + O(n1−d), d ≥3, Ex[a(SτCn)] = C2 log n + γ2 + O(n−1), d = 2.
Since |x| ≤c n, we can write O(|x|−d) + O(n1−d) ≤O(|x|1−d). To get the final assertion we use the estimate GC(1−ǫ)n(0, x −y) ≤GCn(y, x) ≤GC(1+ǫ)n(0, x −y).
We now focus on HCn, the distribution of the first visit of a random walker to the complement of Cn. Our first lemma uses the last-exit decomposition.
Lemma 6.3.6 If p ∈Pd, x ∈B ⊂A ⫋Zd, y ∈∂A, HA(x, y) = X z∈B GA(x, z) Pz{SτA\B = y} = X z∈B GA(z, x) Py{SτA\B = z}.
In particular, HA(x, y) = X z∈A GA(x, z) p(z, y) = X z∈∂iA GA(x, z) p(z, y).
Proof In the first display the first equality follows immediately from Proposition 4.6.4, and the second equality uses the symmetry of p. The second display is the particular case B = A.
Lemma 6.3.7 If p ∈Pd, there exist c1, c2 such that for all n sufficiently large and all x ∈Cn/4, y ∈ ∂Cn, c1 nd−1 ≤HCn(x, y) ≤ c2 nd−1 .
♣We think of ∂Cn as a (d −1)-dimensional subset of Zd that contains on the order of nd−1 points. This lemma states that the hitting measure is mutually absolutely continuous with respect to the uniform measure on ∂Cn (with a constant independent of n).
Proof By the previous lemma, HCn(x, y) = X z∈Cn/2 GCn(z, x) Py{SτCn\Cn/2 = z}.
Using Proposition 6.3.5 we see that for z ∈∂iCn/2, x ∈Cn/4, GCn(z, x) ≍n2−d. Also, Lemma 6.3.4 implies that X z∈Cn/2 Py{SτCn\Cn/2 = z} ≍n−1.
6.3 Difference estimates and Harnack inequality 131 Theorem 6.3.8 (Difference estimates) If p ∈Pd and r < ∞, there exists c such that the following holds for every n sufficiently large.
(a) If g : Bn →R is harmonic in Bn and |y| ≤r, |∇yg(0)| ≤c ∥g∥∞n−1, (6.17) |∇2 yg(0)| ≤c ∥g∥∞n−2.
(6.18) (b) If f : Bn →[0, ∞) is harmonic in Bn and |y| ≤r, then |∇yf(0)| ≤c f(0) n−1, (6.19) |∇2 yf(0)| ≤c f(0) n−2.
(6.20) Proof Choose ǫ > 0 such that C2ǫn ⊂Bn. Choose n sufficiently large so that Br ⊂C(ǫ/2)n and ∂iC2ǫn ∩Cǫn = ∅. Let H(x, z) = HC2ǫn(x, z). Then for |x| ≤r, g(x) = X z∈∂C2ǫn H(x, z) g(z), and similarly for f.
Hence to prove the theorem, it suffices to establish (6.19) and (6.20) for f(x) = H(x, z) (with c independent of n, z).
Let ρ = ρn,ǫ = τC2ǫn\Cǫn.
By Lemma 6.3.6, if x ∈C(ǫ/2)n, f(x) = X w∈∂iCǫn GC2ǫn(w, x) Pz{Sρ = w}.
Lemma 6.3.7 shows that f(x) ≍n1−d and in particular that f(z) ≤c f(w), z, w ∈C(ǫ/2)n.
(6.21) is a δ > 0 such that for n sufficiently large, |w| ≥δn for w ∈∂iCǫn. The estimates (6.19) and (6.20) now follow from Lemma 6.3.3 and Lemma 6.3.4.
Theorem 6.3.9 (Harnack inequality) Suppose p ∈Pd, U ⊂Rd is open and connected, and K is a compact subset of U. Then there exist c = c(K, U, p) < ∞and positive integer N = N(K, U, p) such that if n ≥N, Un = {x ∈Zd : n−1 x ∈U}, Kn = {x ∈Zd : n−1 x ∈K}, and f : Un →[0, ∞) is harmonic in Un, then f(x) ≤c f(y), x, y ∈Kn.
♣This is the discrete analogue of the Harnack principle for positive harmonic functions in Rd.
Suppose K ⊂U ⊂Rd where K is compact and U is open. Then there exists c(K, U) < ∞such that if f : U →(0, ∞) is harmonic, then f(x) ≤c(K, U) f(y), x, y ∈K.
132 Potential Theory Proof Without loss of generality we will assume that U is bounded. In (6.21) we showed that there exists δ > 0, c0 < ∞such that f(x) ≤c0 f(y) if |x −y| ≤δ dist(x, ∂Un).
(6.22) Let us call two points z, w in U adjacent if |z −w| < (δ/4) max{dist(z, ∂U), dist(w, ∂U)}. Let ρ denote the graph distance associated to this adjacency, i.e., ρ(z, w) is the minimum k such that there exists a sequence z = z0, z1, . . . , zk = w of points in U such that zj is adjacent to zj−1 for j = 1, . . . , k.
Fix z ∈U, and let Vk = {w ∈U : ρ(z, w) ≤k}, Vn,k = {x ∈Zd : n−1 x ∈Vk}.
For k ≥1, Vk is open, and connectedness of U implies that ∪Vk = U.
For n sufficiently large, if x, y ∈Vn,k, there is a sequence of points x = x0, x1, . . . , xk = y in Vn,k such that |xj −xj−1| < (δ/2) max{dist(xj, ∂U), dist(xj−1, ∂U)}. Repeated application of (6.22) gives f(x) ≤ck 0 f(y). Compactness of K implies that K ⊂Vk for some finite k, and hence Kn ⊂Vn,k.
6.4 Further estimates In this section we will collect some more facts about random walks in Pd restricted to the set Cn.
The first three propositions are similar to Lemma 6.3.4.
Proposition 6.4.1 If p ∈P2, m < n, T = τCn\Cm, then for x ∈Cn \ Cm, Px{ST ∈∂Cn} = log J (x) −log m + O(m−1) log n −log m .
Proof Let q = Px{ST ∈∂Cn}. The optional sampling theorem applied to the bounded martingale Mj = a(Sj∧T ) gives a(x) = Ex[a(ST )] = (1 −q) Ex[a(ST ) | ST ∈∂iCm] + q Ex[a(ST ) | ST ∈∂Cn].
From (6.11) and Proposition 6.3.1 we know that a(x) = C2 log J (x) + γ2 + O(|x|−2), Ex[a(ST ) | ST ∈∂iCm] = C2 log m + γ2 + O(m−1), Ex[a(ST ) | ST ∈∂Cn] = C2 log n + γ2 + O(n−1).
Solving for q gives the result.
Proposition 6.4.2 If p ∈Pd, d ≥3, T = τZd\Cm, then for x ∈Zd \ Cm, Px{T < ∞} = m J (x) d−2 1 + O(m−1) .
Proof Since G(y) is a bounded harmonic function on τZd\Cm with G(∞) = 0, (6.5) gives G(x) = Ex[G(ST ); T < ∞] = Px{T < ∞} Ex[G(ST ) | T < ∞].
6.4 Further estimates 133 But (6.10) gives G(x) = Cd J (x)2−d [1 + O(|x|−2)], Ex[G(ST ) | T < ∞] = Cd m2−d [1 + O(m−1)].
Proposition 6.4.3 If p ∈P2, n > 0, and T = τCn{0}, then for x ∈Cn, Px{ST = 0} = 1 −log J (x) + O(|x|−1) log n 1 + O 1 log n .
Proof Recall that Px{ST = 0} = GCn(x, 0)/GCn(0, 0). The estimate then follows immediately from Propositions 6.3.2 and 6.3.5. The O(|x|−1) term is superfluous except for x very close to ∂Cn.
Suppose m ≤n/2, x ∈Cm, z ∈∂Cn. By applying Theorem 6.3.8 O(m) times we can see that (for n sufficiently large) HCn(x, z) = Px{Sξn = z} = HCn(0, z) h 1 + O m n i .
(6.23) We will use this in the next two propositions to estimate some conditional probabilities.
Proposition 6.4.4 Suppose p ∈Pd, d ≥3, m < n/4, and Cn \ Cm ⊂A ⊂Cn. Suppose x ∈C2m with Px{SτA ∈∂Cn} > 0 and z ∈∂Cn. Then for n sufficiently large, Px{SτA = z | SτA ∈∂Cn} = HCn(0, z) h 1 + O m n i .
(6.24) Proof It is easy to check (using optional stopping) that it suffices to verify (6.24) for x ∈∂C2m.
Note that (6.23) gives Px{Sξn = z} = HCn(0, z) h 1 + O m n i , and since ∂A \ ∂Cn ⊂Cm, Px{Sξn = z | SτA ̸∈∂Cn} = HCn(0, z) h 1 + O m n i .
This implies Px{Sξn = z; SτA ̸∈∂Cn} = P{SτA ̸∈∂Cn} HCn(0, z) h 1 + O m n i .
The last estimate, combined with (6.24), yields Px{Sξn = z; SτA ∈∂Cn} = P{SτA ∈∂Cn} HCn(0, z) + HCn(0, z) O m n .
Using Proposition 6.4.2, we can see there is a c such that Px{SτA ∈∂Cn} ≥Px{Sj ̸∈Cm for all j} ≥c, x ∈∂C2m, which allows use to write the preceding expression as Px{Sξn = z; SτA ∈∂Cn} = P{SτA ∈∂Cn} HCn(0, z) h 1 + O m n i .
134 Potential Theory For d = 2 we get a similar result but with a slightly larger error term.
Proposition 6.4.5 Suppose p ∈P2, m < n/4, and Cn \ Cm ⊂A ⊂Cn. Suppose x ∈C2m with Px{SτA ∈∂Cn} > 0 and z ∈∂Cn. Then, for n sufficiently large, Px{SτA = z | SτA ∈∂Cn} = HCn(0, z) 1 + O m log(n/m) n .
(6.25) Proof The proof is essentially the same, except for the last step, where Proposition 6.4.1 gives us Px{SτA ∈∂Cn} ≥ c log(n/m), x ∈C2m, so that HCn(0, z) O m n can be written as Px{SτA ∈∂Cn} HCn(0, z) O m log(n/m) n .
The next proposition is a stronger version of Proposition 6.2.2. Here we show that the bounded-ness assumption of that proposition can be replaced with an assumption of sublinearity.
Proposition 6.4.6 Suppose p ∈Pd, d ≥3 and A ⊂Zd with Zd \ A finite. Suppose f : Zd →R is harmonic on A and satisfies f(x) = o(|x|) as x →∞. Then there exists b ∈R such that for all x, f(x) = Ex[f(Sτ A); τ A < ∞] + b Px{τ A = ∞}.
Proof Without loss of generality, we may assume that 0 ̸∈A. Also, we may assume that f ≡0 on Zd \ A; otherwise, we can consider ˆ f(x) = f(x) −Ex[f(SτA); τ A < ∞].
The assumptions imply that there is a sequence of real numbers ǫn decreasing to 0 such that |f(x)| ≤ǫn n for all x ∈Cn and hence |f(x) −f(y)| ≤2 ǫn n, x, y ∈∂Cn.
Since Lf ≡0 on A, (6.8) gives 0 = f(0) = E[f(Sξn)] − X y∈Zd\A GCn(0, y) Lf(y), (6.26) and since Zd \ A is finite, this implies that lim n→∞E[f(Sξn)] = b := X y∈Zd\A G(0, y) Lf(y).
6.4 Further estimates 135 If x ∈A ∩Cn, the optional sampling theorem implies that f(x) = Ex[f(SτA∧ξn)] = Ex[f(Sξn); τA > ξn] = Px{τA > ξn} Ex[f(Sξn) | τA > ξn].
For every w ∈∂Cn, we can write Ex[f(Sξn) | τA > ξn] −E[f(Sξn)] = X z∈∂Cn f(z) [Px{Sξn = z | τA > ξn} −HCn(0, z)] = X z∈∂Cn [f(z) −f(w)] [Px{Sξn = z | τA > ξn} −HCn(0, z)].
For n large, apply P w∈∂Cn and divide by |∂Cn| the above identity, and note that (6.24) now implies |Ex[f(Sξn) | τA > ξn] −E[f(Sξn)]| ≤c |x| n sup y,z∈∂Cn |f(z) −f(y)| ≤c |x| ǫn.
(6.27) Therefore, f(x) = lim n→∞Px{τA > ξn} Ex[f(Sξn) | τA > ξn] = Px{τA = ∞} lim n→∞E[f(Sξn)] = b Px{τA = ∞}.
Proposition 6.4.7 Suppose p ∈P2 and A is a finite subset of Z2 containing the origin.
Let T = TA = τ Z2\A = min{j ≥0 : Sj ∈A}. Then for each x ∈Z2 the limit gA(x) := lim n→∞C2 (log n) Px{ξn < T} (6.28) exists. Moreover, if y ∈A, gA(x) = a(x −y) −Ex[a(ST −y)].
(6.29) Proof If y ∈A and x ∈Cn \ A, the optional sampling theorem applied to the bounded martingale Mj = a(Sj∧T∧ξn −y) implies a(x −y) = Ex[a(ST∧ξn −y)] = Px{ξn < T} Ex[a(Sξn −y) | ξn < T] + Ex[a(ST −y)] −Px{ξn < T} Ex[a(ST −y) | ξn < T].
As n →∞, Ex[a(Sξn −y) | ξn < T] ∼C2 log n.
Letting n →∞, we obtain the result.
Remark. As mentioned before, it follows that the right-hand side of (6.29) is the same for all y ∈A. Also, since there exists δ such that Cδn ⊂Bn ⊂Cn/δ we can replace (6.28) with gA(x) := lim n→∞C2 (log n) Px{ξ∗ n < T}.
The astute reader will note that we already proved this proposition in Proposition 4.6.3.
Proposition 6.4.8 Suppose p ∈P2 and A is a finite subset of Z2. Suppose f : Z2 →R is harmonic on Z2 \ A; vanishes on A; and satisfies f(x) = o(|x|) as |x| →∞. Then f = b gA for some b ∈R.
136 Potential Theory Proof Without loss of generality, assume 0 ∈A and let T = TA be as in the previous proposition.
Using (6.8) and (6.12), we get E[f(Sξn)] = X y∈A GCn(0, y) Lf(y) = C2 log n X y∈A Lf(y) + O(1).
(Here and below the error terms may depend on A.) As in the argument deducing (6.27), we use (6.25) to see that |Ex[f(ST∧ξn) | ξn < T] −E[f(Sξn)]| ≤c |x| log n n sup y,z∈∂Cn |f(y) −f(z)| ≤c |x| ǫn log n, and combining the last two estimates we get f(x) = Ex[f(ST∧ξn)] = Px{ξn < T} Ex[f(STA∧ξn) | ξn < T] = Px{ξn < T} E[f(Sξn)] + |x| o(1) = b gA(x) + o(1), where b = P y∈A Lf(y).
6.5 Capacity, transient case If A is a finite subset of Zd, we let TA = τZd\A, T A = τ Zd\A, rad(A) = sup{|x| : x ∈A}.
If p ∈Pd, d ≥3, define EsA(x) = Px{TA = ∞}, gA(x) = Px{T A = ∞}.
Note that EsA(x) = 0 if x ∈A \ ∂iA. Furthermore, due to Proposition 6.4.6, gA is the unique function on Zd that is zero on A; harmonic on Zd \ A; and satisfies gA(x) ∼1 as |x| →∞. In particular, if x ∈A, LgA(x) = X y p(y) gA(x + y) = EsA(x).
Definition. If d ≥3, the capacity of a finite set A is given by cap(A) = X x∈A EsA(x) = X z∈∂iA EsA(z) = X x∈A LgA(x) = X z∈∂iA LgA(z).
♣The motivation for the above definition is given by the following property (stated as the next proposition): as z →∞, the probability that a random walk starting at z ever hits A is comparable to |z|2−d cap(A).
6.5 Capacity, transient case 137 Proposition 6.5.1 If p ∈Pd, d ≥3 and A ⊂Zd is finite, then Px{TA < ∞} = Cd cap(A) J (x)d−2 1 + O rad(A) |x| , |x| ≥2 rad(A).
Proof There is a δ such that Bn ⊂Cn/δ for all n. We will first prove the result for x ̸∈C2rad(A)/δ.
By the last-exit decomposition, Proposition 4.6.4, Px{T A < ∞} = X y∈A G(x, y) EsA(y).
For y ∈A, J (x −y) = J (x) + O(|y|). Therefore, G(x, y) = Cd J (x)2−d + O |y| |x|d−1 = Cd J (x)d−2 1 + O rad(A) |x| .
This gives the result for x ̸∈C2rad(A)/δ. We can extend this to |x| ≥2rad(A) by using the Harnack inequality (Theorem 6.3.9) on the set {z : 2 rad(A) ≤|z|; J (z) ≤(3/δ) rad(A)}.
Note that for x in this set, rad(A)/|x| is of order 1, so it suffices to show that there is a c such that, for any two points x, z in this set, Px{T A < ∞} ≤c Pz{T A < ∞}.
Proposition 6.5.2 If p ∈Pd, d ≥3, cap(Cn) = C−1 d nd−2 + O(nd−1).
Proof By Proposition 4.6.4, 1 = P{T Cn < ∞} = X y∈∂iCn G(0, y) EsCn(y), But for y ∈∂iCn, Proposition 6.3.1 gives G(0, y) = Cd n2−d [1 + O(n−1)].
Hence, 1 = Cd n2−d cap(Cn) 1 + O(n−1) .
Let TA,n = TA ∧ξn = inf{j ≥1 : Sj ∈A or Sj ̸∈Cn}.
If x ∈A ⊂Cn, Px{TA > ξn} = X y∈∂Cn Px{STA,n = y} = X y∈∂Cn Py{STA,n = x}.
138 Potential Theory The last equation uses symmetry of the walk. As a consequence, X x∈A Px{TA > ξn} = X x∈A X y∈∂Cn Py{STA,n = x} = X y∈∂Cn Py{TA < ξn}.
(6.30) Therefore, cap(A) = X x∈A EsA(x) = lim n→∞ X x∈A Px{TA > ξn} = lim n→∞ X y∈∂Cn Py{TA < ξn}.
(6.31) ♣The identities (6.30)–(6.31) relate, for a given finite set A, the probability that a random walker started uniformly in A “escapes” A and the probability that a random walker started uniformly on the boundary of a large ellipse, (far away from A) ever hits A. Formally, every path from A to infinity can also be considered as a path from infinity to A by reversal. This correspondence is manifested again in Proposition 6.5.4.
Proposition 6.5.3 If p ∈Pd, d ≥3, and A, B are finite subsets of Zd, then cap(A ∪B) ≤cap(A) + cap(B) −cap(A ∩B).
Proof Choose n such that A ∪B ⊂Cn. Then for y ∈∂Cn, Py{TA∪B < ξn} = Px{TA < ξn or TB < ξn} = Py{TA < ξn} + Py{TB < ξn} −Py{TA < ξn, TB < ξn} ≤ Py{TA < ξn} + Py{TB < ξn} −Py{TA∩B < ξn}.
The proposition then follows from (6.31).
Definition. If p ∈Pd, d ≥3, and A ⊂Zd is finite, the harmonic measure of A (from infinity) is defined by hmA(x) = EsA(x) cap(A), x ∈A.
Note that hmA is a probability measure supported on ∂iA. As the next proposition shows, it can be considered as the hitting measure of A by a random walk “started at infinity conditioned to hit A”.
Proposition 6.5.4 If p ∈Pd, d ≥3, and A ⊂Zd is finite, then for x ∈A, hmA(x) = lim |y|→∞Py{STA = x | TA < ∞}.
In fact, if A ⊂Cn/2 and y ̸∈Cn, then Py{STA = x | TA < ∞} = hmA(x) 1 + O rad(A) |y| .
(6.32) 6.5 Capacity, transient case 139 Proof If A ⊂Cn and y ̸∈Cn, the last-exit decomposition (Proposition 4.6.4) gives Py{STA = x} = X z∈∂Cn GZd\A(y, z) Pz{STA,n = x}, where, as before, TA,n = TA ∧ξn. By symmetry and (6.24), Pz{STA,n = x} = Px{STA,n = z} = Px{ξn < TA} HCn(0, z) 1 + O rad(A) n = EsA(x) HCn(0, z) 1 + O rad(A) n .
The last equality uses EsA(x) = Px{TA = ∞} = Px{TA > ξn} 1 + O rad(A)d−2 nd−2 , which follows from Proposition 6.4.2. Therefore, Py{STA = x} = EsA(x) 1 + O rad(A) n X z∈∂Cn GZd\A(y, z) HCn(0, z), and by summing over x, Py{TA < ∞} = cap(A) 1 + O rad(A) n X z∈∂Cn GZd\A(y, z) HCn(0, z).
We obtain (6.32) by dividing the last two expressions.
Proposition 6.5.5 If p ∈Pd, d ≥3, and A ⊂Zd is finite, then cap(A) = sup X x∈A f(x), (6.33) where the supremum is over all functions f ≥0 supported on A such that Gf(y) := X x∈Zd G(y, x) f(x) = X x∈A G(y, x) f(x) ≤1 for all y ∈Zd.
Proof Let ˆ f(x) = EsA(x). Note that Proposition 4.6.4 implies that for y ∈Zd, 1 ≥Py{T A < ∞} = X x∈A G(y, x) EsA(x).
Hence G ˆ f ≤1 and the supremum in (6.33) is at least as large as cap(A). Note also that G ˆ f is the unique bounded function on Zd that is harmonic on Zd \ A; equals 1 on A; and approaches 0 at infinity. Suppose f ≥0, f = 0 on Zd \ A, with Gf(y) ≤1 for all y ∈Zd. Then Gf is the unique bounded function on Zd that is harmonic on Zd \ A; equals Gf ≤1 on A; and approaches zero at 140 Potential Theory infinity. By the maximum principle, Gf(y) ≤G ˆ f(y) for all y. In particular, G( ˆ f −f) is harmonic on Zd \ A; is nonnegative on Zd; and approaches zero at infinity. We need to show that X x∈A f(x) ≤ X x∈A ˆ f(x).
If x, y ∈A, let KA(x, y) = Px{STA = y}.
Note that KA(x, y) = KA(y, x) and X y∈A KA(x, y) = 1 −EsA(x).
If h is a bounded function on Zd that is harmonic on Zd \ A and has h(∞) = 0, then h(z) = E[h(ST A); T A < ∞], z ∈Zd. Using this one can easily check that for x ∈A, Lh(x) = X y∈A KA(x, y) h(y) −h(x).
Also, if h ≥0, X x∈A X y∈A KA(x, y) h(y) = X y∈A h(y) X x∈A KA(y, x) = X y∈A h(y) [1 −EsA(y)] ≤ X y∈A h(y), which implies X x∈Zd Lh(x) = X x∈A Lh(x) ≤0.
Then, using (4.25), X x∈A f(x) = − X x∈A LGf ≤− X x∈A LGf − X x∈A LG( ˆ f −f) = X x∈A EsA(x).
Our definition of capacity depends on the random walk p. The next proposition shows that capacities for different p’s in the same dimension are comparable.
Proposition 6.5.6 Suppose p, q ∈Pd, d ≥3 and let capp, capq denote the corresponding capacities.
Then there is a δ = δ(p, q) > 0 such that for all finite A ⊂Zd, δ capp(A) ≤capq(A) ≤δ−1 capp(A).
Proof It follows from Theorem 4.3.1 that there exists δ such that δ Gp(x, y) ≤Gq(x, y) ≤δ−1 Gp(x, y), for all x, y. The proposition then follows from Proposition 6.5.5.
6.5 Capacity, transient case 141 Definition. If p ∈Pd, d ≥3, and A ⊂Zd, then A is transient if P{Sn ∈A i.o.} = 0.
Otherwise, the set is called recurrent.
Lemma 6.5.7 If p ∈Pd, d ≥3, then a subset A of Zd is recurrent if and only if for every x ∈Zd, Px{Sn ∈A i.o.} = 1.
Proof The if direction of the statement is trivial. To show the only if direction, let F(y) = Py{Sn ∈ A i.o.}, and note that F is a bounded harmonic function on Zd, so it must be constant by Propo-sition 6.1.2. Now if F(y) ≥ǫ > 0, y ∈Zd, then for each x there is an Nx such that Px{Sn ∈A for some n ≤Nx} ≥ǫ/2.
By iterating this we can see for all x, Px{Sn ∈A for some n < ∞} = 1, and the lemma follows easily.
♣Alternatively, {Sn ∈A i.o.} is an exchangeable event with respect to the i.i.d. steps of the random walk, and therefore Px(Sn ∈A i.o.) ∈{0, 1}.
Clearly, all finite sets are transient; in fact, finite unions of transient sets are transient. If A is a subset such that X x∈A G(x) < ∞, (6.34) then A is transient. To see this, let Sn be a random walk starting at the origin and let V denote the number of visits to A, VA = ∞ X j=0 1{Sn ∈A}.
Then (6.34) implies that E[VA] < ∞which implies that P{VA < ∞} = 1. In Exercise 6.3, it is shown that the converse is not true, i.e., there exist transient sets A with E[VA] = ∞.
Lemma 6.5.8 Suppose p ∈Pd, d ≥3, and A ⊂Zd. Then A is transient if and only if ∞ X k=1 P{T k < ∞} < ∞, (6.35) where T k = TAk and Ak = A ∩(C2k \ C2k−1).
Proof Let Ek be the event {T k < ∞}. Since the random walk is transient, A is transient if and only if P{Ek i.o.} = 0. Hence the Borel-Cantelli Lemma implies that any A satisfying (6.35) is transient.
142 Potential Theory Suppose ∞ X k=1 P{T k < ∞} = ∞.
Then either the sum over even k or the sum over odd k is infinite. We will assume the former; the argument if the latter holds is almost identical. Let Bk,+ = Ak ∩{(z1, . . . , zd) : z1 ≥0} and Bk,−= Ak ∩{(z1, . . . , zd) : z1 ≤0}. Since P{T 2k < ∞} ≤P{TB2k,+ < ∞} + P{TB2k,−< ∞}, we know that either ∞ X k=1 P{TB2k,+ < ∞} = ∞, (6.36) or the same equality with B2k,−replacing B2k,+. We will assume (6.36) holds and write σk = TB2k,+.
An application of the Harnack inequality (we leave the details as Exercise 6.11) shows that there is a c such that for all j ̸= k, P{σj < ∞| σj ∧σk = σk < ∞} ≤c P{σj < ∞}.
This implies P{σj < ∞, σk < ∞} ≤2c P{σj < ∞} P{σk < ∞}.
Using this and a special form of the Borel-Cantelli Lemma (Corollary 12.6.2) we can see that P{σj < ∞i.o.} > 0, which implies that A is not transient.
Corollary 6.5.9 (Wiener’s test) Suppose p ∈Pd, d ≥3, and A ⊂Zd. Then A is transient if and only if ∞ X k=1 2(2−d)k cap(Ak) < ∞ (6.37) where Ak = A ∩(C2k \ C2k−1). In particular, if A is transient for some p ∈Pd, then it is transient for all p ∈Pd.
Proof Due to Proposition 6.5.1, we have that P{T k < ∞} ≍2(2−d)k cap(Ak).
Theorem 6.5.10 Suppose d ≥3, p ∈Pd, and Sn is a p-walk. Let A be the set of points visited by the random walk, A = S[0, ∞) = {Sn : n = 0, 1, . . .}.
If d = 3, 4, then with probability one A is a recurrent set. If d ≥5, then with probability one A is a transient set.
Proof Since a set is transient if and only if all its translates are transient, we see that for each n, A is recurrent if and only if the set {Sm −Sn : m = n, n + 1, . . .} 6.5 Capacity, transient case 143 is recurrent. Hence the event {A is recurrent} is a tail event, and and the Kolmogorov 0-1 law now implies that it has probability 0 or 1.
Let Y denote the random variable that equals the expected number of visits to A by an inde-pendent random walker ˜ Sn starting at the origin. In other words, Y = X x∈A G(x) = X x∈Zd 1{x ∈A} G(x).
Then, E(Y ) = X x∈Zd P{x ∈A} G(x) = G(0)−1 X x∈Zd G(x)2.
Since G(x) ≍|x|2−d, we have G(x)2 ≍|x|4−2d. By examining the sum, we see that E(Y ) = ∞for d = 3, 4 and E(Y ) < ∞for d ≥5. If d ≥5, this gives Y < ∞with probability one which implies that A is transient with probability one.
We now focus on d = 4 (it is easy to see that if the result holds for d = 4 then it also holds for d = 3). It suffices to show that P{A is recurrent} > 0. Let S1, S2 be independent random walks with increment distribution p starting at the origin, and let σj k = min{n : Sj n ̸∈C2k}.
Let V j k = [C2k \ C2k−1] ∩Sj[0, σj k+1) = {x ∈C2k \ C2k−1 : Sj n = x for some n ≤σj k+1}.
Let Ek be the event {V 1 k ∩V 2 k ̸= ∅}. We will show that P{Ek i.o.} > 0 which will imply that with positive probability, {S1 n : n = 0, 1, . . .} is recurrent. Using Corollary 12.6.2, one can see that it suffices to show that ∞ X k=1 P(E3k) = ∞, (6.38) and that there exists a constant c < ∞such that for m < k, P(E3m ∩E3k) ≤c P(E3m) P(E3k).
(6.39) The event E3m depends only on the values of Sj n with σj 3m−1 ≤n ≤σj 3m+1. Hence, the Harnack inequality implies P(E3k | E3m) ≤c P(E3k) so (6.39) holds. To prove (6.38), let Jj(k, x) denote the indicator function of the event that Sj n = x for some n ≤σj k. Then, Zk := #(V 1 k ∩V 2 k ) = X x∈C2k \C2k−1 J1(k, x) J2(k, x).
There exist c1, c2 such that if x, y ∈C2k \ C2k−1, (recall d −2 = 2) E[Jj(k, x)] ≥c1 (2k)−2, E[Jj(k, x) Jj(k, y)] ≤c2 (2k)−2 [1 + |x −y|]−2.
(The latter inequality is obtained by noting that the probability that a random walker hits both x and y given that it hits at least one of them is bounded above by the probability that a random walker starting at the origin visits y −x.) Therefore, E[Zk] = X x∈C2k \C2k−1 E[J1(k, x)] E[J2(k, x)] ≥c X x∈C2k \C2k−1 (2k)−4 ≥c, 144 Potential Theory E[Z2 k] = X x,y∈C23k \C2k−1 E[J1(k, x) J1(k, y)] E[J2(k, x) J2(k, y)] ≤ c X x,y∈C2k\C2k−1 (2k)−4 1 [1 + |x −y|2]2 ≤ck, where for the last inequality note that there are O(24k) points in C2k \ C2k−1, and that for x ∈ C2k \ C2k−1 there are O(ℓ3) points y ∈C2k \ C2k−1 at distance ℓfrom from x. The second moment estimate, Lemma 12.6.1 now implies that P{Zk > 0} ≥c/k, hence (6.38) holds.
♣The central limit theorem implies that the number of points in Bn visited by a random walk is of order n2.
Roughly speaking, we can say that a random walk path is a “two-dimensional” set. Asking whether or not this is recurrent is asking whether or not two random two-dimensional sets intersect. Using the example of planes in Rd, one can guess that the critical dimension is four.
6.6 Capacity in two dimensions The theory of capacity in two dimensions is somewhat similar to that for d ≥3, but there are significant differences due to the fact that the random walk is recurrent. We start by recalling a few facts from Propositions 6.4.7 and 6.4.8. If p ∈P2 and 0 ∈A ⊂Z2 is finite, let gA(x) = a(x) −Ex[a(ST A)] = lim n→∞C2 (log n) Px{ξn < T A}.
(6.40) The function gA is the unique function on Z2 that vanishes on A; is harmonic on Z2\A; and satisfies gA(x) ∼C2 log J (x) ∼C2 log |x| as x →∞. If y ∈A, we can also write gA(x) = a(x −y) −Ex[a(ST A −y)].
To simplify notation we will mostly assume that 0 ∈A, and then a(x)−gA(x) is the unique bounded function on Z2 that is harmonic on Z2 \A and has boundary value a on A. We define the harmonic measure of A (from infinity) by hmA(x) = lim |y|→∞Py{STA = x}.
(6.41) Since Py{TA < ∞} = 1, this is the same as Py{STA = x | TA < ∞} and hence agrees with the definition of harmonic measure for d ≥3. It is not clear a priori that the limit exists, this fact is established in the next proposition.
Proposition 6.6.1 Suppose p ∈P2 and 0 ∈A ⊂Z2 is finite. Then the limit in (6.41) exists and equals LgA(x).
Proof Fix A and let rA = rad(A). Let n be sufficiently large so that A ⊂Cn/4. Using (6.25) on the set Z2 \ A, we see that if x ∈∂iA, y ∈∂Cn, Py{STA∧ξn = x} = Px{STA∧ξn = y} = Px{ξn < TA} HCn(0, y) 1 + O rA log n n .
6.6 Capacity in two dimensions 145 If z ∈Z2 \ Cn, the last-exit decomposition (Proposition 4.6.4) gives Pz{STA = x} = X y∈∂Cn GZ2\A(z, y) Py{STA∧ξn = x}.
Therefore, Pz{STA = x} = Px{ξn < TA} J(n, z) 1 + O rA log n n , (6.42) where J(n, z) = X y∈∂Cn HCn(0, y) GZ2\A(z, y).
If x ∈A, the definition of L, the optional sampling theorem, and the asymptotic expansion of gA respectively imply LgA(x) = Ex[gA(S1)] = Ex[gA(STA∧ξn)] = Ex[gA(Sξn); ξn < TA] = Px{ξn < TA} [C2 log n + OA(1)] .
(6.43) In particular, LgA(x) = lim n→∞C2 (log n) Px{ξn < TA}, x ∈A.
(6.44) (This is the d = 2 analogue of the relation LgA(x) = EsA(x) for d ≥3.) Note that (as in (6.30)) X x∈∂iA Px{ξn < TA} = X x∈∂iA X y∈∂Cn Px{Sξn∧TA = y} = X y∈∂Cn X x∈∂iA Py{Sξn∧TA = x} = X y∈∂Cn Py{TA < ξn}.
Proposition 6.4.3 shows that if x ∈A, then the probability that a random walk starting at x reaches ∂Cn before visiting the origin is bounded above by c log rA/ log n. Therefore, Py{TA < ξn} = Py{T{0} < ξn} 1 + O log rA log n .
As a consequence, X x∈∂iA Px{ξn < TA} = X y∈∂Cn Py{Sξn∧T{0} = 0} 1 + O log rA log n = P{ξn < T{0}} 1 + O log rA log n = [C2 log n]−1 1 + O log rA log n .
Combining this with (6.44) gives X x∈A LgA(x) = X x∈∂iA LgA(x) = 1.
(6.45) 146 Potential Theory Here we see a major difference between the recurrent and transient case. If d ≥3, the sum above equals cap(A) and increases in A, while it is constant in A if d = 2. (In particular, it would not be a very useful definition for a capacity!) Using (6.42) together with P x∈A Pz{STA = x} = 1, we see that J(n, z) X x∈A Px{ξn < TA} = 1 + O rA log n n , which by (6.43)– (6.45) implies that J(n, z) = C2 log n 1 + O rA log n n , uniformly in z ∈Z2 \ Cn, and the claim follows by (6.42).
We define the capacity of A by cap(A) := lim y→∞[a(y) −gA(y)] = X x∈A hmA(x) a(x −z), where z ∈A. The last proposition establishes the limit if z = 0 ∈A, and for other z use (6.29) and limy→∞a(x) −a(x −z) = 0. We have the expansion gA(x) = C2 log J (x) + γ2 −cap(A) + oA(1), |x| →∞.
It is easy to check from the definition that the capacity is translation invariant, that is, cap(A+y) = cap(A), y ∈Zd. Note that singleton sets have capacity zero.
Proposition 6.6.2 Suppose p ∈P2.
(a) If 0 ∈A ⊂B ⊂Zd are finite, then gA(x) ≥gB(x) for all x. In particular, cap(A) ≤cap(B).
(b) If A, B ⊂Zd are finite subsets containing the origin, then for all x gA∪B(x) ≥gA(x) + gB(x) −gA∩B(x).
(6.46) In particular, cap(A ∪B) ≤cap(A) + cap(B) −cap(A ∩B).
Proof The inequality gA(x) ≥gB(x) follows immediately from (6.40). The inequality (6.46) follows from (6.40) and the observation (recall also the argument for Proposition 6.5.3) Px{T A∪B < ξn} = Px{T A < ξn or T B < ξn} = Px{T A < ξn} + Px{T B < ξn} −Px{T A < ξn, T B < ξn} ≤ Px{T A < ξn} + Px{T B < ξn} −Px{T A∩B < ξn}, which implies Px{T A∪B > ξn} ≥Px{T A > ξn} + Px{T B > ξn} −Px{T A∩B > ξn}.
6.6 Capacity in two dimensions 147 We next derive an analogue of Proposition 6.5.5. If A is a finite set, let aA denote the #(A)×#(A) symmetric matrix with entries a(x, y). Let aA also denote the operator aA f(x) = X y∈A a(x, y) f(y) which is defined for all functions f : A →R and all x ∈Z2. Note that x 7→aAf(x) is harmonic on Z2 \ A.
Proposition 6.6.3 Suppose p ∈P2 and 0 ∈A ⊂Z2 is finite. Then cap(A) = sup X y∈A f(y) −1 , where the supremum is over all nonnegative functions f on A satisfying aAf(x) ≤1 for all x ∈A.
If A = {0} is a singleton set, the proposition is trivial since aAf(0) = 0 for all f and hence the supremum is infinity. A natural first guess for other A (which turns out to be correct) is that the supremum is obtained by a function f satisfying aAf(x) = 1 for all x ∈A. If {aA(x, y)}x,y∈A is invertible, there is a unique such function that can be written as f = a−1 A 1 (where 1 denotes the vector of all 1s). The main ingredient in the proof of Proposition 6.6.3 is the next lemma that shows this inverse is well defined assuming A has at least two points.
Lemma 6.6.4 Suppose p ∈P2 and 0 ∈A ⊂Z2 is finite with at least two points. Then a−1 A exists and a−1 A (x, y) = Px{STA = y} −δ(y −x) + LgA(x) LgA(y) cap(A) , x, y ∈A.
Proof We will first show that for all x ∈Z2.
X z∈A a(x, z) LgA(z) = cap(A) + gA(x).
(6.47) To prove this, we will need the following fact (see Exercise 6.7): lim n→∞[GCn(0, 0) −GCn(x, y)] = a(x, y).
(6.48) Consider the function h(x) = X z∈A a(x, z) LgA(z).
We first claim that h is constant on A. By a last-exit decomposition (Proposition 4.6.4), if x, y ∈A, 1 = Px{T A < ξn} = X z∈A GCn(x, z) Pz{ξn < TA} = X z∈A GCn(y, z) Pz{ξn < TA}.
Hence, (C2 log n) X z∈A [GCn(0, 0) −GCn(x, z)]Pz{ξn < TA} = 148 Potential Theory (C2 log n) X z∈A [GCn(0, 0) −GCn(y, z)]Pz{ξn < TA}.
Letting n →∞, and recalling that C2 (log n) Pz{ξn < TA} →LgA(z), we conclude that h(x) = h(y).
Theorem 4.4.4 and (6.45) imply that lim x→∞[a(x) −h(x)] = 0.
Hence, a(x) −h(x) is a bounded function that is harmonic in Z2 \ A and takes the value a −hA on A, where hA denotes the constant value of h on A. Now Theorem 6.2.1 implies that a(x) −h(x) = a(x) −gA(x) −hA. Therefore, hA = lim x→∞[a(x) −gA(x)] = cap(A).
This establishes (6.47).
An application of the optional sampling theorem gives for z ∈A GCn(x, z) = δ(z −x) + Ex[GCn(S1, z)] = δ(z −x) + X y∈A Px{STA∧ξn = y} GCn(y, z).
Hence, GCn(0, 0) −GCn(x, z) = −δ(z −x) + GCn(0, 0) Px{ξn < TA} + X y∈A Px{SτA∧ξn = y} [GCn(0, 0) −GCn(y, z)].
Letting n →∞and using (6.12) and (6.48), as well as Proposition 6.6.1, this gives δ(z −x) = −a(x, z) + LgA(x) + X y∈A Px{STA = y} a(y, z).
If x, z ∈A, we can use (6.47) to write the previous identity as δ(z −x) = X y∈A Px{STA = y} −δ(y −x) + LgA(x) LgA(y) cap(A) a(y, z), provided that cap(A) > 0.
Proof [of Proposition 6.6.3] Let ˆ f(x) = LgA(x)/cap(A). Applying (6.47) to x ∈A gives X y∈A a(x, y) ˆ f(y) = 1, x ∈A.
Suppose f satisfies the conditions in the statement of the proposition, and let h = aA ˆ f −aAf which is nonnegative in A. Then, using Lemma 6.6.4, X x∈A [ ˆ f(x) −f(x)] = X x∈A X y∈A a−1 A (x, y) h(y) ≥ X x∈A X y∈A Px{SτA = y} h(y) − X x∈A h(x) = X y∈A h(y) X x∈A Py{SτA = x} − X y∈A h(y) = 0.
6.6 Capacity in two dimensions 149 Proposition 6.6.5 If p ∈P2, cap(Cn) = C2 log n + γ2 + O(n−1), Proof Recall the asymptotic expansion for gCn. By definition of capacity we have, gCn(x) = C2 log J (x) + γ2 −cap(Cn) + o(1), x →∞.
But for x ̸∈Cn, gCn(x) = a(x) −Ex[a(STCn)] = C2 log J (x) + γ2 + O(|x|−2) −[C2 log n + γ2 + O(n−1)].
Lemma 6.6.6 If p ∈P2, and A ⊂B ⊂Z2 are finite, then cap(A) = cap(B) − X y∈B hmB(y) gA(y).
Proof gA −gB is a bounded function that is harmonic on Z2 \ B with boundary value gA on B.
Therefore, cap(B) −cap(A) = lim x→∞[gA(x) −gB(x)] = lim x→∞Ex[gA(ST B) −gB(ST B)] = X y∈B hmB(y) gA(y).
♣Proposition 6.6.5 tells us that the capacity of an ellipse of diameter n is C2 log n + O(1). The next lemma shows that this is also true for any connected set of diameter n. In particular, the capacities of the ball of radius n and a line of radius n are asymptotic as n →∞. This is not true for capacities in d ≥3.
Lemma 6.6.7 If p ∈P2, there exist c1, c2 such that the following holds. If A is a finite subset of Z2 with rad(A) < n satisfying #{x ∈A : k −1 ≤|x| < k} ≥1, k = 1, . . . , n, then (a) if x ∈∂C2n, Px{TA < ξ4n} ≥c1, (b) |cap(A) −C2 log n| ≤c2, (c) if x ∈∂C2n, m ≥4n, and An = A ∩Cn, then c1 ≤Px {TAn > ξm} log(m/n) ≤c2.
(6.49) 150 Potential Theory Proof (a) Let δ be such that Bδn ⊂Cn, and let B denote a subset of A contained in Bδn such that #{x ∈B : k −1 ≤|x| < k} = 1 for each positive integer k < δn. We will prove the estimate for B which will clearly imply the estimate for A. Let V = Vn,B denote the number of visits to B before leaving C4n, V = ξ4n−1 X j=0 1{Sj ∈B} = ∞ X j=0 X z∈B 1{Sj = z; j < ξ4n}.
The strong Markov property implies that if x ∈∂C2n, Ex[V ] = Px{TB < ξ4n} Ex[V | TB < ξ4n] ≤Px{TB < ξ4n} max z∈B Ez[V ].
Hence, we need only find a c1 such that Ex[V ] ≥c1 Ez[V ] for all x ∈∂C2n, z ∈B. Note that #(B) = δn + O(1). By Exercise 6.13, we can see that GC4n(x, z) ≥c for x, z ∈C2n. Therefore Ex[V ] ≥c n.
If z ∈B, there are at most 2k points w in B \ {z} satisfying |z −w| ≤k + 1, k = 1, . . . , δn. Using Proposition 6.3.5, we see that GC4n(z, w) ≤C2 [log n −log |z −w| + O(1)].
Therefore, Ez[V ] = X w∈B GC4n(z, w) ≤ δn X k=1 2 C2 [log n −log k + O(1)] ≤c n.
The last inequality uses the estimate n X k=1 log k = O(log n) + Z n 1 log x dx = n log n −n + O(log n).
(b) There exists a δ such that Bn ⊂Cn/δ for all n and hence cap(A) ≤cap(Bn) ≤cap(Cn/δ) ≤C2 log n + O(1).
Hence, we only need to give a lower bound on cap(A). By the previous lemma it suffices to find a uniform upper bound for gA on ∂C4n. For m > 4n, let rm = rm,n,A = max y∈C2n Py{ξm < TA}, r∗ m = rm,n,A = max y∈C4n Py{ξm < TA}.
Using part (a) and the strong Markov property, we see that there is a ρ < 1 such that rm ≤ρ r∗ m.
Also, if y ∈C4n Py{ξm < TA} = Py{ξm < TC2n} + Py{ξm > TC2n} Py{ξm < TA | ξm > TC2n} ≤ Py{ξm < TC2n} + ρ r∗ m.
Proposition 6.4.1 tells us that there is a c3 such that for y ∈C4n, Py{ξm < TC2n} ≤ c3 log m −log n + O(1).
6.6 Capacity in two dimensions 151 Therefore, gA(y) = lim m→∞C2 (log m) Py{ξm < TA} ≤C2 c3 1 −ρ.
(c) The lower bound for (6.49) follows from Proposition 6.4.1 and the observation Px{TAn > ξm} ≥Px{TCn > ξm}.
For the upper bound let u = un = max x∈∂C2n Px{TAn > ξm}.
Consider a random walk starting at y ∈∂C2n and consider TCn ∧ξm. Clearly, Py{TAn > ξm} = Py{ξm < TCn} + Py{ξm > TCn; ξm < TAn}.
By Proposition 6.4.1, for all y ∈∂C2n Py {ξm < TCn} ≤ c log(m/n).
Let σ = σn = min{j ≥TCn : Sj ∈∂C2n}. Then, by the Markov property, Py{ξm > TCn, ξm ≤TAn} ≤u Py{S[0, σ] ∩An = ∅}.
Part (a) shows that there is a ρ < 1 such that Py{S[0, σ] ∩An = ∅} ≤ρ and hence, we get Py{TAn > ξm} ≤ c log(m/n) + ρ u.
Since this holds, for all y ∈∂C2n, this implies u ≤ c log(m/n) + ρ u, which gives us the upper bound.
♣A major example of a set satisfying the condition of the theorem is a connected (with respect to simple random walk) subset of Z2 with radius between n −1 and n. In the case of simple random walk, there is another proof of part (a) based on the observation that the simple random walk starting anywhere on ∂C2n makes a closed loop about the origin contained in Cn with a probability uniformly bounded away from 0. One can justify this rigorously by using an approximation by Brownian motion. If the random walk makes a closed loop, then it must intersect any connected set. Unfortunately, it is not easy to modify this argument for random walks that take non-nearest neighbor steps.
152 Potential Theory 6.7 Neumann problem We will consider the following “Neumann problem”. Suppose p ∈Pd and A ⊂Zd with nonempty boundary ∂A. If f : A →R is a function, we define its normal derivative at y ∈∂A by Df(y) = X x∈A p(y, x) [f(x) −f(y)].
Given D∗: ∂A →A, the Neumann problem is to find a function f : A →R such that Lf(x) = 0, x ∈A, (6.50) Df(y) = D∗(y), y ∈∂A.
(6.51) ♣The term normal derivative is motivated by the case of simple random walk and a point y ∈∂A such that there is a unique x ∈A with |y −x| = 1. Then Df(y) = [f(x) −f(y)]/2d, which is a discrete analogue of the normal derivative.
A solution to (6.50)–(6.51) will not always exist. The next lemma which is a form of Green’s theorem shows that if A is finite, a necessary condition for existence is X y∈∂A D∗(y) = 0.
(6.52) Lemma 6.7.1 Suppose p ∈Pd, A is a finite subset of Zd and f : A →R is a function. Then X x∈A Lf(x) = − X y∈∂A Df(y).
Proof X x∈A Lf(x) = X x∈A X y∈A p(x, y) [f(y) −f(x)] = X x∈A X y∈A p(x, y) [f(y) −f(x)] + X x∈A X y∈∂A p(x, y) [f(y) −f(x)] However, X x∈A X y∈A p(x, y) [f(y) −f(x)] = 0, since p(x, y) [f(y) −f(x)] + p(y, x) [f(x) −f(y)] = 0 for all x, y ∈A. Therefore, X x∈A Lf(x) = X y∈∂A X x∈A p(x, y) [f(y) −f(x)] = − X y∈∂A Df(y).
Given A, the excursion Poisson kernel is the function H∂A : ∂A × ∂A − →[0, 1], 6.7 Neumann problem 153 defined by H∂A(y, z) = Py {S1 ∈A, SτA = z} = X x∈A p(y, x) HA(x, z), where HA : A × ∂A →[0, 1] is the Poisson kernel. If z ∈∂A and H(x) = HA(x, z), then DH(y) = H∂A(y, z), y ∈∂A \ {z}, DH(z) = H∂A(z, z) −Pz{S1 ∈A}.
More generally, if f : A →R is harmonic in A, then f(y) = P z∈∂A f(z)HA(y, z) so that Df(y) = X z∈∂A H∂A(y, z) [f(z) −f(y)].
(6.53) Note that if y ∈∂A then X z∈∂A H∂A(y, z) = Py{S1 ∈A} ≤1.
It is sometimes useful to consider the Markov transition probabilities ˆ H∂A where ˆ H∂A(y, z) = H∂A(y, z) for y ̸= z, and ˆ H∂A(y, y) is chosen so that X z∈∂A ˆ H∂A(y, z) = 1.
Note that again (compare with (6.53)) Df(y) = X z∈∂A ˆ H∂A(y, z) [f(z) −f(y)], which we can write in matrix form Df = [ ˆ H∂A −I] f.
If A is finite, then the #(∂A)×#(∂A) matrix ˆ H∂A−I is sometimes called the Dirichlet-to-Neumann map because it takes the boundary values f (Dirichlet conditions) of a harmonic function to the derivatives Df (Neumann conditions). The matrix is not invertible since constant functions f are mapped to zero derivatives. We also know that the image of the map is contained in the subspace of functions D∗satisfying (6.52). The next proposition shows that the rank of the matrix is #(∂A)−1.
It will be useful to define random walk “reflected off∂A”. There are several natural ways to do this. We define this to be the Markov chain with state space A and transition probabilities q where q(x, y) = p(x, y) if x ∈A or y ∈A; q(x, y) = 0 if x, y ∈∂A are distinct; and q(y, y) is defined for y ∈∂A so that P z∈A q(y, z) = 1. In words, this chain moves like random walk with transition probability p while in A, and whenever its current position y is in ∂A, the only moves allowed are those into A ∪{y}. While the original walk could step out of A ∪{y} with some probability ˜ p(y) = ˜ p(y, A, p), the modified walk stays at y with probability p(y, y) + ˜ p(y).
Proposition 6.7.2 Suppose p ∈Pd, A is a finite, connected subset of Zd, and D∗: ∂A →R is a 154 Potential Theory function satisfying (6.52). Then there is a function f : A →R satisfying (6.50) and (6.51). The function f is unique up to an additive constant. One such function is given by f(x) = −lim n→∞Ex n X j=0 D∗(Yj) 1{Yj ∈∂A} , (6.54) where Yj is a Markov chain with transition probabilities q as defined in the previous paragraph.
Proof It suffices to show that f as defined in (6.54) is well defined and satisfies (6.50) and (6.51).
Indeed, if this is true then f + c also satisfies it. Since the image of the matrix ˆ H∂A −I contains the set of functions satisfying (6.52) and this is a subspace of dimension #(∂A) −1, we get the uniqueness.
Note that q is an irreducible, symmetric Markov chain and hence has the uniform measure as the invariant measure π(y) = 1/m where m = #(A).
Because the chain also has points with q(y, y) > 0, it is aperiodic. Also, Ex n X j=0 D∗(Yj) 1{Yj ∈∂A} = n X j=0 X z∈∂A qj(x, z) D∗(z) = n X j=0 X z∈∂A qj(x, z) −1 m D∗(z).
By standard results about Markov chains (see Section 12.4), we know that qj(x, z) −1 m ≤c e−αj, for some positive constants c, α. Hence the sum is convergent. It is then straightforward to check that it satisfies (6.50) and (6.51).
6.8 Beurling estimate The Beurling estimate is an important tool for estimating hitting (avoiding) probabilities of sets in two dimensions. The Beurling estimate is a discrete analogue of what is known as the Beurling projection theorem for Brownian motion in R2.
Recall that a set A ⊂Zd is connected (for simple random walk) if any two points in A can be connected by a nearest neighbor path of points in A.
Theorem 6.8.1 (Beurling estimate) If p ∈P2, there exists a constant c such that if A is an infinte connected subset of Zd containing the origin and S is simple random walk, then P{ξn < TA} ≤ c n1/2 , d = 2.
(6.55) We prove the result for simple random walk, and then we describe the extension to more general walks.
Definition. Let Ad denote the collection of infinite subsets of Zd with the property that for each positive integer j, #{z ∈A : (j −1) ≤|z| < j} = 1.
6.8 Beurling estimate 155 One important example of a set in Ad is the half-infinite line L = {je1 : j = 0, 1, . . .}.
We state two immediate facts about Ad.
• If A′ is an infinite connected subset of Zd containing the origin, then there exists a (not necessarily connected) A ∈Ad with A ⊂A′.
• If z ∈A ∈Ad, then for every real r > 0.
{w ∈A : |z −w| ≤r} ≤#{w ∈A : |z| −r ≤|w| ≤|z| + r} ≤2r + 1.
(6.56) Theorem 6.8.1 for simple random walk is implied by the following stronger result.
Theorem 6.8.2 For simple random walk in Z2 there is a c such that if A ∈Ad, then P{ξn < TA} ≤ c n1/2 .
(6.57) Proof We fix n and let V = Vn = {y1, . . . , yn} where yj denotes the unique point in A with j ≤|yj| < j + 1.
We let K = Kn = {x1, . . . , xn} where xj = je1. Let Gn = GBn, B = Bn3, G = GB, ξ = ξn3. Let v(z) = Pz{ξ < TVn}, q(z) = Pz{ξ < TKn}.
By (6.49), there exist c1, c2 such that for z ∈∂B2n, c1 log n ≤v(z) ≤ c2 log n, c1 log n ≤q(z) ≤ c2 log n.
We will establish v(0) ≤ c n1/2 log n and then the Markov property will imply that (6.57) holds. Indeed, note that v(0) ≥P(ξ2n < TV2n)P(ξ < ξn|ξ2n < TV2n).
By (5.17) and (6.49), we know that there is a c such that for j = 1, . . . , n, q(xj) ≤c n−1/2 h j−1/2 + (n −j + 1)−1/2i [log n]−1; (6.58) In particular, q(0) ≤c/(n1/2 log n) and hence it suffices to prove that v(0) −q(0) ≤ c n1/2 log n.
(6.59) If |x|, |y| ≤n, then Gn3−2n(0, y −x) ≤Gn3(x, y) ≤Gn3+2n(0, y −x), and hence (4.28) and Theorem 4.4.4 imply G(x, y) = 2 π log n3 + γ2 −a(x, y) + O 1 n2 , |x|, |y| ≤n.
(6.60) 156 Potential Theory Using Proposition 4.6.4, we write v(0) −q(0) = P{ξ < TV } −P{ξ < TK} = P{ξ > TK} −P{ξ > TV } = n X j=1 G(0, xj) q(xj) − n X j=1 G(0, yj) v(yj) = n X j=1 [G(0, xj) −G(0, yj)] q(xj) + n X j=1 G(0, yj) [q(xj) −v(yj)].
Using (6.58) and (6.60), we get (log n) n X j=1 [G(0, xj) −G(0, yj)] q(xj) ≤O(n−1) + c n X j=1 |a(xj) −a(yj)| (j−1/2 + (n −j)−1/2) n−1/2.
Since |xj| = j, |yj| = j + O(1), (4.4.4) implies that |a(xj) −a(yj)| ≤c j , and hence (log n) n X j=1 [G(0, xj) −G(0, yj)] q(xj) ≤O(n−1) + c n X j=1 1 j3/2 n1/2 ≤ c n1/2 .
For the last estimate we note that n X j=1 1 j(n −j)1/2 ≤ n X j=1 1 j3/2 .
In fact, if a, b ∈Rn are two vectors such that a has non-decreasing components (that is, a1 ≤a2 ≤ . . . ≤an) then a · b ≤a · b∗where b∗= (bπ(1), . . . , bπ(n)) and π is any permutation that makes bπ(1) ≤bπ(2) ≤. . . ≤bπ(n).
Therefore, to establish (6.59), it suffices to show that n X j=1 G(0, yj) [q(xj) −v(yj)] ≤ c n1/2 log n.
(6.61) Note that we are not taking absolute values on the left-hand side. Consider the function F(z) = n X j=1 G(z, yj) [q(xj) −v(yj)], and note that F is harmonic on B \ V . Since F ≡0 on ∂B, either F ≤0 everywhere (in which case (6.61) is trivial) or it takes its maximum on V . Therefore, it suffices to find a c such that for all k = 1, . . . , n, n X j=1 G(yk, yj) [q(xj) −v(yj)] ≤ c n1/2 log n.
6.9 Eigenvalue of a set 157 By using Proposition 4.6.4 once again, we get n X j=1 G(yk, yj) v(yj) = Pyk{T V ≤ξ} = 1 = Pxk{T K ≤ξ} = n X j=1 G(xk, xj) q(xj).
Plugging in, we get n X j=1 G(yk, yj) [q(xj) −v(yj)] = n X j=1 [G(yk, yj) −G(xk, xj)] q(xj).
We will now bound the right-hand side. Note that |xk −xj| = |k −j| and |yk −yj| ≥|k −j| −1.
Hence, using (6.60), G(yk, yj) −G(xk, xj) ≤ c |k −j| + 1 and therefore for each k = 1, . . . , n n X j=1 [G(yk, yj) −G(xk, xj)] q(xj) ≤c n X j=1 1 (|k −j| + 1) j1/2 log n ≤ c n1/2 log n.
One can now generalize this result.
Definition. If p ∈P2 and k is a positive integer, let A∗= A∗ 2,k,p denote the collection of infinite subsets of Z2 with the property that for each positive integer j, #{z ∈A : (j −1)k ≤J (z) < jk} ≥1, and let A denote the collection of subsets with #{z ∈A : (j −1)k ≤J (z) < jk} = 1.
If A ∈A∗then A contains a subset in A.
Theorem 6.8.3 If p ∈P2 and k is a positive integer, there is a c such that if A ∈A∗, then P{ξn < TA} ≤ c n1/2 .
The proof is done similarly to that of the last theorem. We let K = {x1, . . . , xn} where xj = jle1 and l is chosen sufficiently large so that J (le1) > k, and set V = {y1, . . . , yn} where yj ∈A with jJ ∗(le1) ≤|yj| < (j + 1)J ∗(le1). See Exercise 5.2.
6.9 Eigenvalue of a set Suppose p ∈Pd and A ⊂Zd is finite and connected (with respect to p) with #(A) = m. The (first) eigenvalue of A is defined to be the number αA = e−λA such that for each x ∈A, as n →∞, Px{τA > n} ≍αn A = e−λAn.
Let P A denote the m × m matrix [p(x, y)]x,y∈A and, as before, let LA = P A −I. Note that (P A)n is the matrix [pA n (x, y)] where pA n (x, y) = Px{Sn = y; n < τA}. We will say that p ∈Pd is aperiodic 158 Potential Theory restricted to A if there exists an n such that (P A)n has all entries strictly positive; otherwise, we say that p is bipartite restricted to A. In order for p to be aperiodic restricted to A, p must be aperiodic. However, it is possible for p to be aperiodic but for p to be bipartite restricted to A (Exercise 6.16). The next two propositions show that αD is the largest eigenvalue for the matrix P A, or, equivalently, 1 −αA is the smallest eigenvalue for the matrix LA.
Proposition 6.9.1 If p ∈Pd, A ⊂Zd is finite and connected, and p restricted to A is aperiodic, then there exist numbers 0 < β = βA < α = αA < 1 such that if x, y ∈A, pA n (x, y) = αn gA(x) gA(y) + OA(βn).
(6.62) Here gA : A →R is the unique positive function satisfying P AgA(x) = αA gA(x), x ∈A, X x∈A gA(x)2 = 1.
In particular, Px{τA > n} = ˜ gA(x) αn + OA(βn), where ˜ gA(x) = gA(x) X y∈A gA(y), We write OA to indicate that the implicit constant in the error term depends on A.
Proof This is a general fact about irreducible Markov chains, see Proposition 12.4.3. In the notation of that proposition v = w = g. Note that Px{τA > n} = X y∈A pA n(x, y).
.
Proposition 6.9.2 If p ∈Pd, A ⊂Zd is finite and connected, and p is bipartite restricted to A, then there exist numbers 0 < β = βA < α = αA < 1 such that if x, y ∈A for all n sufficiently large, pA n(x, y) + pA n+1(x, y) = 2 αn gA(x) gA(y) + OA(βn).
Here gA : A →R is the unique positive function satisfying X x∈A gA(x)2 = 1, P AgA(x) = α gA(x), x ∈A.
Proof This can be proved similarly using Markov chains. We omit the proof.
Proposition 6.9.3 Suppose p ∈Pd; ǫ ∈(0, 1), and pǫ = ǫ δ0 + (1 −ǫ) p is the corresponding lazy walker. Suppose A is a finite, connected subset of Zd and let α, αǫ, g, gǫ be the eigenvalues and eigenfunctions for A using p, pǫ, respectively. Then 1 −αǫ = (1 −ǫ) (1 −α) and gǫ = g.
6.9 Eigenvalue of a set 159 Proof Let P A, P A ǫ be the corresponding matrices. Then P A ǫ = (1 −ǫ) P A + ǫ I and hence P A ǫ gA = [(1 −ǫ) α + ǫ] gA.
♣A standard problem is to estimate λA or αA as A gets large and αA →1, λA →0. In these cases it usually suffices to consider the eigenvalue of the lazy walker with ǫ = 1/2. Indeed let ˜ λA be the eigenvalue for the lazy walker. Since, λA = 1 −αA + O((1 −αA)2), αA →1 −.
we get ˜ λA = 1 2 λA + O(λ2 A), λA →0.
Proposition 6.9.1 gives no bounds for the β. The optimal β is the maximum of the absolute values of the eigenvalues other than α. In general, it is hard to estimate β, and it is possible for βto be very close to α. We will show that in the case of the nice set Cn there is an upper bound for β independent of n. We fix p ∈Pd with p(x, x) > 0 and let e−λm = αCm, gm = gCm, and pm n (x, y) = pCm n (x, y). For x ∈Cm we let ρm(x) = dist(x, ∂Cm) + 1 m , and we set ρm ≡0 on Zd \ Cm.
Proposition 6.9.4 There exist c1, c2 such that for all m sufficiently large and all x, y ∈Cm, c1 ρm(x) ρm(y) ≤md pm m2(x, y) ≤c2 ρm(x) ρm(y).
(6.63) Also, there exist c3, c4 such that for every n ≥m2, and all x, y ∈Cm, c3 ρm(y) m−d ≤Px{Sn = y | τCm > n} ≤c4 ρm(y) m−d.
♣This proposition is an example of a parabolic boundary Harnack principle. At any time larger than rad2(Cm), the position of the random walker, given that it has stayed in Cm up to the current time, is independent of the initial state up to a multiplicative constant.
Proof For notational ease, we will restrict to the case where m is even. (If m is odd, essentially the same proof works except m2/4 must be replaced with ⌊m2/4⌋, etc.) We write ρ = ρm. Note that pm m2(x, y) = X z,w pm m2/4(x, z) pm m2/2(z, w) pm m2/4(w, y).
(6.64) The local central limit theorem implies that there is a c such that for all z, w, pm m2/2(z, w) ≤ pm2/2(z, w) ≤c m−d. Therefore, pm m2(x, y) ≤m−d Px{τCm > m2/4} Py{τCm > m2/4}.
160 Potential Theory Gambler’s ruin (see Proposition 5.1.6) implies that Px{τCm > m2/4} ≤c ρm(x). This gives the upper bound for (6.63).
For the lower bound, we first note that there is an ǫ > 0 such that pm ⌊ǫm2⌋(z, w) ≥ǫ m−d, |z|, |w| ≤ǫ m.
Indeed, the Markov property implies that pm ⌊ǫm2⌋(z, w) ≥p⌊ǫm2⌋(z, w) −max{pk(˜ z, w) : k ≤⌊ǫm2⌋, ˜ z ∈Zd \ Cm}, (6.65) and the local central limit theorem establishes the estimate. Using this estimate and the invariance principle, one can see that for every ǫ > 0, there is a c such that for z, w ∈C(1−ǫ)m, pm2/2(z, w) ≥c m−d.
Indeed, in order to estimate pm2/2(z, w), we split the path into three pieces: the first m2/8 steps, the middle m2/4 steps; and the final m2/8 steps (here we are assuming m2/8 is an integer for notational ease). We estimate both the probability that the walk starting at z has not left Cm and is in the ball of radius ǫm at time m2/8 and corresponding probability for the walk in reverse time starting at w using the invariance principle. There is a positive probability for this, where the probability depends on ǫ. For the middle piece we use (6.65), and then we “connect” the paths to obtain the lower bound on the probability.
Using (6.64), we can then see that it suffices to find ǫ > 0 and c > 0 such that X z∈C(1−ǫ)m pm m2/4(x, z) ≥c ρ(x).
(6.66) Let T = τCm \ τCm/2 as in Lemma 6.3.4 and let Tm = T ∧(m2/4). Using that lemma and Theorem 5.1.7, we can see that Px{STm ∈Cm} ≤c1 ρ(x).
Propositions 6.4.1 and 6.4.2 can be used to see that Px{ST ∈Cm/2} ≥c2 ρ(x).
We can write Px{ST ∈Cm/2} = X z Px{STm = z} Px{ST ∈Cm/2 | STm = z}.
The conditional expectation can be estimated again by Lemma 6.3.4; in particular, we can find an ǫ such that Pz{ST ∈Cm/2} ≤c2 2c1 , z ̸∈C(1−ǫ)m.
This implies, X z∈C(1−ǫ)m Px{STm = z} ≥ X z∈C(1−ǫ)m Px{STm = z} Px{ST ∈Cm/2 | STm = z} ≥c2 2 ρ(x).
A final appeal to the central limit theorem shows that if ǫ ≤1/4, X z∈C(1−ǫ)m pm m2/4(x, z) ≥c X z∈C(1−ǫ)m Px{STm = z}.
6.9 Eigenvalue of a set 161 The last assertion follows for n = m2 by noting that Px{Sm2 = y | τCm > m2} = pm m2(x, y) P z pm m2(x, z) and X z pm m2(x, z) ≍ρm(x) m−d X z ρm(z) ≍ρm(x).
For n > m2, we can argue similarly by conditioning on the walk at time n −m2.
Corollary 6.9.5 There exists c1, c2 such that c1 ≤m2λm ≤c2.
Proof See exercise 6.10.
Corollary 6.9.6 There exists c1, c2 such that for all m and all x ∈Cm/2, c1 e−λmn ≤Px{ξm > n} ≤c2e−λmn.
Proof Using the previous corollary, it suffices to prove the estimates for n = km2, k ∈{1, 2, . . .}. Let βk(x) = βk(x, m) = Px{ξm > km2} and let βk = maxx∈Cm βk(x). Using the previous proposition, we see there is a c1 such that βk ≥βk(x) ≥c1 βk, x ∈Cm/2.
Due to the same estimates, Px{Sξm ∈Cm/2 | ξm > km2} ≥c2.
Therefore, there is a c3 such that c3βj βk ≤βj+k ≤βj βk, which implies (see Corollary 12.7.2) e−λmm2k ≤βk ≤c−1 3 e−λmm2k, and hence for x ∈Cm/2, c1 e−λmm2k ≤βk(x) ≤c−1 3 e−λmm2k.
Exercises Exercise 6.1 Show that Proposition 6.1.2 holds for p ∈P∗.
Exercise 6.2 162 Potential Theory (i) Show that if p ∈Pd and x ∈Cn, Ex[ξn] = X y∈Cn GCn(x, y) = n2 −J (x) + O(n).
(Hint: see Exercise 1.5.) (ii) Show that if p ∈P′ d and x ∈Cn, Ex[ξn] = X y∈Cn GCn(x, y) = n2 −J (x) + o(n2).
Exercise 6.3 In this exercise we construct a transient subset A of Z3 with X y∈A G(0, y) = ∞.
(6.67) Here G denotes the Green’s function for simple random walk. Our set will be of the form A = ∞ [ k=1 Ak, Ak = {z ∈Z3 : |z −2k e1| ≤ǫk 2k}.
for some ǫk →0.
(i) Show that (6.67) holds if and only if P∞ k=1 ǫ3 k 22k = ∞.
(ii) Show that A is transient if and only if P∞ k=1 ǫk < ∞.
(iii) Find a transient A satisfying (6.67).
Exercise 6.4 Show that there is a c < ∞such that the following holds. Suppose Sn is simple random walk in Z2 and let V = Vn,N be the event that the path S[0, ξN] does not disconnect the origin from ∂Bn. Then if x ∈B2n, Px(V ) ≤ c log(N/n).
(Hint: There is a ρ > 0 such that the probability that a walk starting at ∂Bn/2 disconnects the origin before reaching ∂Bn is at least ρ, see Exercise 3.4.) Exercise 6.5 Suppose p ∈Pd, d ≥3. Show that there exists a sequence Kn →∞such that if A ⊂Zd is a finite set with at least n points, then cap(A) ≥Kn.
Exercise 6.6 Suppose p ∈Pd and r < 1. Show there exists c = cr < ∞such that the following holds.
(i) If |e| = 1, and x ∈Crn, X y∈Cn |GCn(x + e, y) −GCn(x, y)| ≤c n, (ii) Suppose f, g, F are as in Corollary 6.2.4 with A = Cn. Then if x ∈Crn, |∇jf(x)| ≤c n ∥F∥∞+ n2 ∥g∥∞ .
6.9 Eigenvalue of a set 163 Exercise 6.7 Show that if p ∈P2 and r > 0, lim n→∞[GCn+r(0, 0) −GCn(0, 0)] = 0.
Use this and (6.16) to conclude that for all x, y, lim n→∞[GCn(0, 0) −GCn(x, y)] = a(x, y).
Exercise 6.8 Suppose p ∈Pd and A ⊂Zd is finite. Define QA(f, g) = X x,y∈A p(x, y) [f(y) −f(x)] [g(y) −g(x)].
and QA(f) = QA(f, f). Let F : ∂A →R be given. Show that the infimum of QA(f) restricted to functions f : A →R with f ≡F on ∂A is obtained by the unique harmonic function with boundary value F.
Exercise 6.9 Write the two-dimensional integer lattice in complex form, Z2 = Z + iZ and let A be the upper half plane A = {j + ik ∈Z2 : k > 0}. Show that for simple random walk GA(x, y) = a(x, y) −a(x, y), x, y ∈A, HA(x, j) = 1 4 [a(x, j −i) −a(x, j + i)] + δ(x −j), x ∈A, j ∈Z.
where j + ik = j −ik denotes complex conjugate. Find lim k→∞k HA(ik, j).
Exercise 6.10 Prove Corollary 6.9.5.
Exercise 6.11 Provide the details of the Harnack inequality argument in Lemma 6.5.8 and Theorem 6.5.10.
Exercise 6.12 Suppose p ∈Pd.
(i) Show that there is a c < ∞such that if x ∈A ⊂Cn and z ∈C2n, Px{Sξ2n = z | ξ2n < TA} ≤c n1−d Px{ξn < TA} Px{ξ2n < TA}.
(ii) Let A be the line {je1 : j ∈Z}. Show that there is an ǫ > 0 such that for all n sufficiently large, P{dist(Sξn, A) ≥ǫn | ξn < TA} ≥ǫ.
(Hint: you can use the gambler’s ruin estimate to estimate Px{ξn/2 < TA}/Px{ξn < TA}.) Exercise 6.13 Show that for each p ∈P2 and each r ∈(0, 1), there is a c such that for all n sufficiently large, GCn(x, y) ≥c, x, y ∈Crn, 164 Potential Theory GBn(x, y) ≥c, x, y ∈Brn.
Exercise 6.14 Suppose p ∈P2 and let A = {x1, x2} be a two-point set.
(i) Prove that hmA(x1) = 1/2.
(ii) Show that there is a c < ∞such that if A ⊂Cn, then for y ∈Z2 \ C2n, Py{STA = x1} −1 2 ≤ c log n.
(Hint: Suppose Py{STA = xj} ≥1/2 and let V be the set of z such that Pz{STA = xj} ≤1/2.
Let σ = min{j : Sj ∈V }. Then it suffices to prove that Py{TA < σ} ≤c/ log n.) (iii) Show that there is a c < ∞such that if A = Z2 \ {x} with x ̸= 0, then GA(0, 0) −4 π log |x| ≤c.
Exercise 6.15 Suppose p ∈P2. Show that there exist c1, c2 > 0 such that the following holds.
(i) If n is sufficiently large, A is a set as in Lemma 6.6.7, and An = A ∩{|z| ≥n/2}, then for x ∈∂Bn/2, Px{TA < ξ∗ n} ≥c.
(ii) If x ∈∂Bn/2, GZ2\A(x, 0) ≤c.
(iii) If A′ is a set with Bn/2 ⊂A′ ⊂Z2 \ An, GA′(0, 0) −2 π log n ≤c.
Exercise 6.16 Give an example of an aperiodic p ∈Pd and a finite connected (with respect to p) set A for which p is bipartite restricted to A.
Exercise 6.17 Suppose Sn is simple random walk in Zd so that ξn = ξ∗ n. If |x| < n, let u(x, n) = Ex [|Sξn| −n] and note that 0 ≤u(x, n) ≤1.
(i) Show that n2 −|x|2 + 2n u(x, n) ≤Ex[ξn] ≤n2 −|x|2 + (2n + 1) u(x, n).
(ii) Show that if d = 2, π 2 GBn(0, x) = log n −log |x| + u(x, n) n + O(|x|−2).
(iii) Show that if d ≥3, C−1 d GBn(0, x) = 1 |x|d−2 − 1 nd−2 + (d −2) u(x, n) nd−1 + O(|x|−d).
6.9 Eigenvalue of a set 165 Exercise 6.18 Suppose Sn is simple random walk in Zd with d ≥3. For this exercise assume that we know that G(x) ∼ Cd |x|d−2 , |x| →∞ for some constant Cd but no further information on the asymptotics. The purpose of this exercise is to find Cd. Let Vd be the volume of the unit ball in Rd and ωd = d Vd the surface area of the boundary of the unit ball.
(i) Show that as n →∞, X x∈Bn G(0, x) ∼Cd ωd n2 2 = Cd d Vd n2 2 .
(ii) Show that as n →∞, X x∈Bn [G(0, x) −GBn(0, x)] ∼Cd Vd n2.
(iii) Show that as n →∞, X x∈Bn GBn(0, x) ∼n2.
(iv) Conclude that Cd Vd d 2 −1 = 1.
7 Dyadic coupling 7.1 Introduction In this chapter we will study the dyadic or KMT coupling which is a coupling of Brownian motion and random walk for which the paths are significantly closer to each other than in the Skorokhod embedding. Recall that if (Sn, Bn) are coupled by the Skorokhod embedding, then typically one expects |Sn −Bn| to be of order n1/4. In the dyadic coupling, |Sn −Bn| will be of order log n. We mainly restrict our consideration to one dimension, although we discuss some higher dimensional versions in Section 7.6.
Suppose p ∈P′ 1 and Sn = X1 + · · · + Xn is a p-walk. Suppose that there exists b > 0 such that E[X2 1] = σ2, E[eb|X1|] < ∞.
(7.1) Then by Theorem 2.3.11, there exist N, c, ǫ such that if we define δ(n, x) by pn(x) := P{Sn = x} = 1 √ 2πσ2n e− x2 2σ2n exp{δ(n, x)}, then for all n ≥N and |x| ≤ǫn, |δ(n, x)| ≤c 1 √n + |x|3 n2 .
(7.2) Theorem 7.1.1 Suppose p ∈P′ d satisfies (7.1) and (7.2).
Then one can define on the same probability space (Ω, F, P), a Brownian motion Bt with variance parameter σ2 and a random walk with increment distribution p such that the following holds. For each α < ∞, there is a cα such that P max 1≤j≤n |Sj −Bj| ≥cα log n ≤cα n−α.
(7.3) Remark.
From the theorem it is easy to conclude the corresponding result for bipartite or continuous-time walks with p ∈P1. In particular, the result holds for discrete-time and continuous-time simple random walk.
166 7.2 Some estimates 167 ♣We will describe the dyadic coupling formally in Section 7.4, but we will give a basic idea here. Suppose that n = 2m. One starts by defining S2m as closely to B2m as possible. Using the local central limit theorem, we can do this in a way so that with very high probability |S2m −B2m| is of order 1. We then define S2m−1 using the values of B2m, B2m−1, and again get an error of order 1. We keep subdividing intervals using binary splitting, and every time we construct the value of S at the middle point of a new interval. If at each subdivision we get an error of order 1, the total error should be at most of order m, the number of subdivisions needed. (Typically it might be less because of cancellation.) ♣The assumption E[eb|X1|] < ∞for some b > 0 is necessary for (7.3) to hold at j = 1. Suppose p ∈P′ 1 such that for each n there is a coupling with P{|S1 −B1| ≥ˆ c log n} ≤ˆ c n−1.
It is not difficult to show that as n →∞, P{|B1| ≥ˆ c log n} = o(n−1), and hence P{|S1| ≥2ˆ c log n} ≤P{|S1 −B1| ≥ˆ c log n} + P{|B1| ≥ˆ c log n} ≤2 ˆ c n−1 for n sufficiently large. If we let x = 2ˆ c log n, this becomes P{|X1| ≥x} ≤2ˆ c e−x/(2ˆ c), for all x sufficiently large which implies E[eb|X1|] < ∞for b < (2ˆ c)−1.
Some preliminary estimates and definitions are given in Sections 7.2 and 7.3, the coupling is defined in Section 7.4, and we show that it satisfies (7.3) in Section 7.5. The proof is essentially the same for all values of σ2. For ease of notation we will assume that σ2 = 1. It also suffices to prove the result for n = 2m and we will assume this in Sections 7.4 and 7.5.
For the remainder of this chapter, we fix b, ǫ, c0, N and assume that p is an increment distribution satisfying E[eb|X1|] < ∞, (7.4) and pn(x) = 1 √ 2πn e−x2 2n exp{δ(n, x)}, where |δ(n, x)| ≤c0 1 √n + |x|3 n2 , n ≥N, |x| ≤ǫn.
(7.5) 7.2 Some estimates In this section we collect a few lemmas about random walk that will be used in establishing (7.3).
The reader may wish to skip this section at first reading and come back to the estimates as they are needed.
168 Dyadic coupling Lemma 7.2.1 Suppose Sn is a random walk with increment distribution p satisfying (7.4) and (7.5). Define δ∗ n(n, x, y) by P{Sn = x | S2n = y} = 1 √πn exp −(x −(y/2))2 n exp{δ∗(n, x, y)}.
Then if n ≥N, |x|, |y| ≤ǫn/2, |δ∗(n, x, y)| ≤9 c0 1 √n + |x|3 n2 + |y|3 n2 .
♣Without the conditioning, Sn is approximately normal with mean zero and variance n. Conditioned on the event S2n = y, Sn is approximately normal with mean y/2 and variance n/2. Note that specifying the value at time 2n reduces the variance of Sn.
Proof Note that P{Sn = x | S2n = y} = P{Sn = x, S2n −Sn = y −x} P{S2n = y} = pn(x) pn(y −x) p2n(y) .
Since |x|, |y|, |x −y| ≤ǫn, we can apply (7.5). Note that |δ∗(n, x, y)| ≤|δ(n, x)| + |δ(n, y −x)| + |δ(2n, y)|.
We use the simple estimate |y −x|3 ≤8(|x|3 + |y|3).
Lemma 7.2.2 If x1, x2, . . . , xn ∈R, then n X j=1 (x1 + · · · + xj)2 2j ≤2 √ 2 + 1 √ 2 −1 n X j=1 x2 j 2j .
(7.6) Proof Due to homogeneity of (7.6) we may assume that P 2−j x2 j = 1. Let yj = 2−j/2 xj, y = (y1, . . . , yn). Then n X i=1 (x1 + · · · + xi)2 2i = n X i=1 X 1≤j,k≤i xjxk 2i = n X j=1 n X k=1 xjxk n X i=j∨k 2−i ≤ 2 n X j=1 n X k=1 2−(j∨k) xjxk = 2 n X j=1 n X k=1 2−|k−j|/2 yjyk = 2⟨Ay, y⟩≤2λ ∥y∥2 = 2λ, 7.2 Some estimates 169 where A = An is the n × n symmetric matrix with entries a(j, k) = 2−|k−j|/2 and λ = λn denotes the largest eigenvalue of A. Since λ is bounded by the maximum of the row sums, λ ≤1 + 2 ∞ X j=1 2−j/2 = √ 2 + 1 √ 2 −1.
♣We will use the fact that the left-hand side of (7.6) is bounded by a constant times the term in brackets on the right-hand side. The exact constant is not important.
Lemma 7.2.3 Suppose Sn is a random walk with increment distribution satisfying (7.4) and (7.5).
Then for every α there exists a c = c(α) such that P X log2 n 0, see Exercise 7.3. To overcome this difficulty, we use a striaghtforward truncation argument.
Proof We fix α > 0 and allow constants in this proof to depend on α. Using (7.4), we see that there is a β such that P{|Sn| ≥n} ≤e−βn.
Hence, we can find c1 such that X log2 n<j≤n [ P{|S2j| ≥c12j} + P{|S2j −S2j−1| ≥c12j} ] = O(e−αn).
Fix this c1, and let j0 = ⌊log2 n + 1⌋be the smallest integer greater than log2 n. Let Yj = 0 for j < j0; Yj0 = S2j0; and for j > j0, let Yj = S2j −S2j−1. Then, except for an event of probability O(e−αn), |Yj| ≤c12j for j ≥j0 and hence P n X j=j0 Y 2 j 2j ̸= n X j=j0 Y 2 j 2j 1{|Yj| ≤c12j} ≤O(e−αn).
Note that X log2 n 0 such that for each n, E exp t S2 n n ; |Sn| ≤c1n ≤e (see Exercise 7.2). Therefore, E exp t n X j=1 Y 2 j 2j 1{|Yj| ≤c12j} ≤en, which implies P n X j=1 Y 2 j 2j 1{|Yj| ≤c12j} ≥t−1 (α + 1) n ≤e−αn.
7.3 Quantile coupling In this section we consider the simpler problem of coupling Sn and Bn for a fixed n. The following is a general definition of quantile coupling. We will only use quantile coupling in a particular case where F is supported on Z or on (1/2)Z.
Definition. Suppose F is the distribution function of a discrete random variable supported on the locally finite set · · · < a−1 < a0 < a1 < · · · , and Z is a random variable with a continuous, strictly increasing distribution function G. Let rk be defined by G(rk) = F(ak), i.e., if F(ak) > F(ak−), G(rk) −G(rk−1) = F(ak) −F(ak−).
Let f be the step function f(z) = ak if rk−1 < z ≤rk, and let X be the random variable f(Z). We call X the quantile coupling of F with Z, and f the quantile coupling function of F and G.
Note that the event {X = ak} is the same as the event {rk−1 < Z ≤rk}. Hence, P{X = ak} = P{rk−1 < Z ≤rk} = G(rk) −G(rk−1) = F(ak) −F(ak−), and X has distribution function F. Also, if G(ak −t) ≤F(ak−1) < F(ak) ≤G(ak + t), (7.8) 7.3 Quantile coupling 171 then it is immediate from the above definitions that {X = ak} ⊂{|X −Z| = |ak −Z| ≤t}. Hence, if we wish to prove that |X −Z| ≤t on the event {X = ak}, it suffices to establish (7.8).
As an intermediate step in the construction of the dyadic coupling, we study the quantile coupling of the random walk distribution with normal random variable that has the same mean and variance.
Let Φ denote the standard normal distribution function, and let Φβ (where β > 0) denote the distribution function of a mean zero normal random variable with variance β.
Proposition 7.3.1 For every ǫ, b, c0, N there exist c, δ such that if Sn is a random walk with increment distribution p satisfying (7.4) and (7.5) the following holds for n ≥N. Let Fn denote the distribution function of Sn, and suppose Z has distribution function Φn. Let (X, Z) be the quantile coupling of Fn with Z. Then, |X −Z| ≤c 1 + X2 n , |X| ≤δn.
Proposition 7.3.2 For every ǫ, b, c0, N there exist c, δ such that the following holds for n ≥N.
Suppose Sn is a random walk with increment distribution p satisfying (7.4) and (7.5). Suppose |y| ≤δn with P{S2n = y} > 0. Let Fn,y denote the conditional distribution function of Sn −(S2n/2) given S2n = y, and suppose Z has distribution function Φn/2. Let (X, Z) be the quantile coupling of Fn,y with Z. Then, |X −Z| ≤c 1 + X2 n + y2 n , |X|, |y| ≤δn.
Using (7.8), we see that in order to prove the above propositions, it suffices to show the following estimate for the corresponding distribution functions.
Lemma 7.3.3 For every ǫ, b, c0, N there exist c, δ such that if Sn is a random walk with increment distribution p satisfying (7.4) and (7.5) the following holds for n ≥N. Let Fn, Fn,y be as in the propositions above. Then for y ∈Z, |x|, |y| ≤δn, Φn x −c 1 + x2 n ≤Fn(x −1) ≤Fn(x) ≤Φn x + c 1 + x2 n , (7.9) Φn/2 x −c 1 + x2 n + y2 n ≤Fn,y(x −1) ≤Fn,y(x) ≤Φn/2 x + c 1 + x2 n + y2 n , Proof It suffices to establish the inequalities in the case where x is a non-negative integer. Implicit constants in this proof are allowed to depend on ǫ, b, c0 and we assume n ≥N. If F is a distribution function, we write F = 1 −F. Since for t > 0, (t + 1)2 2n = t2 n + O 1 √n + t3 n2 , (consider t ≤√n and t ≥√n), we can see that (7.4) and (7.5) imply that we can write pn(x) = Z x+1 x 1 √ 2πn e−t2/(2n) exp O 1 √n + t3 n2 dt, |x| ≤ǫn.
172 Dyadic coupling Hence, using (12.12), for some a and all |x| ≤ǫn, F n(x) = P{Sn ≥ǫn} + P{x ≤Sn < ǫn} = O(e−an) + Z ǫn x 1 √ 2πn e−t2/(2n) exp O 1 √n + t3 n2 dt.
From this we can conclude that for |x| ≤ǫn, F n(x) = Φn(x) exp O 1 √n + x3 n2 , (7.10) and from this we can conclude (7.9). The second inequality is done similarly by using Lemma 7.2.1 to derive F n,y(x) = Φn/2(x) exp O 1 √n + x3 n2 + y3 n2 , for |x|, |y| ≤δn. Details are left as Exercise 7.4.
♣To derive Propositions 7.3.1 and Proposition 7.3.2 we use only estimates on the distribution functions Fn, Fn,y and not pointwise estimates (local central limit theorem). However, the pointwise estimate (7.5) is used in the proof of Lemma 7.2.1 which is used in turn to estimate Fn,y.
7.4 The dyadic coupling In this section we define the dyadic coupling. Fix n = 2m and assume that we are given a stan-dard Brownian motion defined on some probability space. We will define the random variables S1, S2, . . . , S2m as functions of the random variables B1, B2, . . . , B2m so that S1, . . . , S2m has the distribution of a random walk with increment distribution p.
In Chapter 3, we constructed a Brownian motion from a collection of independent normal random variables by a dyadic construction. Here we reverse the process, starting with the Brownian motion, Bt, and obtaining the independent normals. We will only use the random variables B1, B2, . . . , B2m.
Define Γk,j by Γk,j = Bk2m−j −B(k−1)2m−j, j = 0, 1, . . . , m; k = 1, 2, 3, . . . , 2j.
For each j, {Γk,j : k = 1, 2, 3, . . . , 2j} are independent normal random variables with mean zero and variance 2m−j. Let Z1,0 = B2m and define Z2k+1,j, j = 1, . . . , m, k = 0, 1, . . . , 2j−1 −1, recursively by Γ2k+1,j = 1 2 Γk+1,j−1 + Z2k+1,j, (7.11) so that also Γ2k+2,j = 1 2 Γk+1,j−1 −Z2k+1,j.
7.4 The dyadic coupling 173 One can check (see Corollary 12.3.1) that the random variables {Z2k+1,j : j = 0, . . . , 2m, k = 0, 1, . . . , 2m−1 −1} are independent, mean zero, normal random variables with E[Z2 1,0] = 2m and E[Z2 2k+1,j] = 2m−j−1 for j ≥1. We can rewrite (7.11) as B(2k+1)2m−j = 1 2 Bk2m−j+1 + B(k+1)2m−j+1 + Z2k+1,j.
(7.12) Let fm(·) denote the quantile coupling function for the distribution functions of S2m and B2m.
If y ∈Z, let fj(·, y) denote the quantile coupling function for the conditional distribution of S2j −1 2S2j+1 given S2j+1 = y and a normal random variable with mean zero and variance 2j−1. This is well defined as long as P{S2j+1 = y} > 0. Note that the range of fj(·, y) is contained in (1/2)Z. This conditional distribution is symmetric about the origin (see Exercise 7.1), so fj(−z, y) = −fj(z, y).
We can now define the dyadic coupling.
• Let S2m = fm(B2m).
• Suppose the values of Sl2m−j+1, l = 1, . . . , 2j−1 are known. Let ∆k,i = Sk2m−i −S(k−1)2m−i.
Then we let S(2k−1)2m−j = 1 2 [S(k−1)2m−j+1 + Sk2m−j+1] + fm−j(Z2k−1,j, ∆k,j−1), so that ∆2k−1,j = 1 2∆k,j−1 + fm−j(Z2k−1,j, ∆k,j−1), ∆2k,j = 1 2∆k,j−1 −fm−j(Z2k−1,j, ∆k,j−1).
It follows immediately from the definition that (S1, S2, . . . , S2m) has the distribution of the ran-dom walk with increment p.
Also Exercise 7.1 shows that ∆2k−1,j and ∆2k,j have the same conditional distribution given ∆k,j−1.
It is convenient to rephrase this definition in terms of random variables indexed by dyadic inter-vals. Let Ik,j denote the interval Ik,j = [(k −1)2m−j, k2m−j], j = 0, . . . , m; k = 1, . . . , 2j.
We write Z(I) for the normal random variable associated to the midpoint of I, Z(Ik,j) = Z2k−1,j+1 Then the Z(I) are independent mean zero normal random variables indexed by the dyadic intervals with variance |I|/4 where | · | denotes length. We also write Γ(Ik,j) = Γk,j, ∆(Ik,j) = ∆k,j.
Then the definition can be given as follows.
• Let Γ(I1,0) = B2m, ∆(I1,0) = fm(B2m).
174 Dyadic coupling • Suppose I is a dyadic interval of length 2m−j+1 that is the union of consecutive dyadic intervals I1, I2 of length 2m−j. Then Γ(I1) = 1 2 Γ(I) + Z(I), Γ(I2) = 1 2 Γ(I) −Z(I) (7.13) ∆(I1) = 1 2 ∆(I) + fj(Z(I), ∆(I)), ∆(I2) = 1 2 ∆(I) −fj(Z(I), ∆(I)).
(7.14) • Note that if j ≥1 and k ∈{1, . . . , 2j}, then Bk2m−j = X i≤k Γ([(i −1)2m−j, i2m−j]), Sk2m−j = X i≤k ∆([(i −1)2m−j, i2m−j]).
(7.15) We next note a few important properties of the coupling.
• If I = I1 ∪I2 as above, then Γ(I1), Γ(I2), ∆(I1), ∆(I2) are deterministic functions of Γ(I), ∆(I), Z(I). The conditional distributions of (Γ(I1), ∆(I1)) and (Γ(I2), ∆(I2)) given (Γ(I), ∆(I)) are the same.
• By iterating this we get the following. For each interval Ik,j consider the joint distribution random variables (Γ(Il,i), ∆(Il,i)), i = 0, . . . , j, where l = l(i, k, j) is chosen so that Ik,j ⊂Il,i.
Then this distribution is the same for all k = 1, 2, . . . , 2j. In particular, if Rk,j = j X i=0 |Γ(Il,i) −∆(Il,i)|, then the random variables R1,j, . . . , R2j,j are identically distributed. (They are not independent.) • For k = 1, Γ(I1,j) −∆(I1,j) = 1 2 [Γ(I1,j−1) −∆(I1,j−1)] + [Z1,j −fj(Z1,j, S2m−j+1)] By iterating this, we get R1,j ≤|S2m −B2m| + 2 j X l=1 |fm−l(Z1,l, S2m−l+1) −Z1,l|.
(7.16) • Define Θ(I1,0) = |B2m −S2m| = |Γ(I1,0) −∆(I1,0)|. Suppse j ≥1 and Ik,j is an interval with “parent” interval I′. Define Θ(Ik,j) to be the maximum of |Bt −St| where the maximum is over three values of t: the left endpoint, midpoint, and right endpoint of I. We claim that Θ(Ik,j) ≤Θ(I′) + |Γ(Ik,j) −∆(Ik,j)|.
Since the endpoints of Ik,j are either endpoints or midpoints of I′, it suffices to show that |Bt −St| ≤max |Bs−−Ss−|, |Bs+ −Ss+|| + |Γ(Ik,j) −∆(Ik,j)|, 7.5 Proof of Theorem 7.1.1 175 where t, s−, s+ denote the midpoint, left endpoint, and right endpoint of Ik,j, respectively. But using (7.13), (7.14), and (7.15), we see that Bt −St = 1 2 (Bs−−Ss−) + (Bs+ −Ss+) + |Γ(Ik,j) −∆(Ik,j)|, and hence the claim follows from the simple inequality |x + y| ≤2 max{|x|, |y|}.
Hence, by induction, we see that Θ(Ik,j) ≤Rk,j.
(7.17) 7.5 Proof of Theorem 7.1.1 Recall that n = 2m. It suffices to show that for each α there is a cα such that for each integer j, P {|Si −Bi| ≥cα log n} ≤cα n−α.
(7.18) Indeed if the above holds, then P max 1≤i≤n |Si −Bi| ≥cα log n ≤ n X i=1 P {|Si −Bi| ≥cα log n} ≤cα n−α+1.
We claim in fact, that it suffices to find a sequence 0 = i0 < i1 < · · · < il = n such that |ik −ik−1| ≤cα log n and such that (7.18) holds for these indices. Indeed, if we prove this and |j −ik| ≤cα log n, then exponential estimates show that there is a c′ α such that P{|Sj −Sik| ≥c′ α log n} + P{|Bj −Bik| ≥c′ α log n} ≤c′ α n−α, and hence the triangle inequality gives (7.18) (with a different constant).
For the remainder of this section we fix α and allow constants to depend on α. By the reasoning of the previous paragraph and (7.17), it suffices to find a c such that for log2 m + c ≤j ≤m, and k = 1, . . . , 2m−j, P {Rk,j ≥cα m} ≤cα e−αm, and as pointed out in the previous section, it suffices to consider the case k = 1, and show P {R1,j ≥cα m} ≤cα e−αm, for j = log2 m + c, . . . , m.
(7.19) Let δ be the minimum of the two values given in Propositions 7.3.1 and 7.3.2, and recall that there is a β = β(δ) such that P{|S2j| ≥δ2j} ≤exp{−β2j} In particular, we can find a c3 such that X log2 m+c3≤j≤m P{|S2j| ≥δ2j} ≤O(e−αm).
Proposition 7.3.1 tells us that on the event {|S2m| ≤δ2m}, |S2m −B2m| ≤c 1 + S2 2m 2m .
176 Dyadic coupling Similarly, Proposition 7.3.2 tells us that on the event {max{|S2m−l|, |S2m−l+1|} ≤δ2m−l}, we have |Z1,l −fm−l(Z1,l, S2m−l+1)| ≤c " 1 + S2 2m−l+1 + S2 2m−l 2m−l # .
Hence, by (7.16), we see that on the same event, simultaneously for all j ∈[log2 m + c3, m], |S2m−j −B2m−j| ≤R1,j + |S2m −B2m| ≤c m + X log2 m−c3≤i≤m S2 2i 2i .
We now use (7.7) (due to the extra term −c3 in the lower limit of the sum, one may have to apply (7.7) twice) to conclude (7.19) for j ≥log2 m + c3.
7.6 Higher dimensions Without trying to extend the result of the previous section to to the general (bounded exponential moment) walks in higher dimensions, we indicate two immediate consequences.
Theorem 7.6.1 One can define on the same probability space (Ω, F, P), a Brownian motion Bt in R2 with covariance matrix (1/2) I and a simple random walk in Z2. such that the following holds.
For each α < ∞, there is a cα such that P max 1≤j≤n |Sj −Bj| ≥cα log n ≤cα n−α.
Proof We use the trick from Exercise 1.7.
Let (Sn,1, Bn,1), (Sn,2, Bn,2) be independent dyadic couplings of one-dimensional simple random walk and Brownian motion. Let Sn = Sn,1 + Sn,2 2 , Sn,1 −Sn,2 2 , Bn = Bn,1 + Bn,2 2 , Bn,1 −Bn,2 2 .
Theorem 7.6.2 If p ∈Pd, one can define on the same probability space (Ω, F, P), a Brownian motion Bt in Rd with covariance matrix Γ and a continuous-time random walk ˜ St with increment distribution p such that the following holds. For each α < ∞, there is a cα such that P max 1≤j≤n | ˜ Sj −Bj| ≥cα log n ≤cα n−α.
(7.20) Proof Recall from (1.3) that we can write any such ˜ St as ˜ St = ˜ S1 q1t x1 + · · · + ˜ Sl qlt xl, 7.7 Coupling the exit distributions 177 where q1, . . . , ql > 0; x1, . . . , xl ∈Zd; and ˜ S1, . . . , ˜ Sl are independent one-dimensional simple continuous-time random walks. Choose l independent couplings as in Theorem 7.1.1, (S1 t , B1 t ), (S2 t , B2 t ), . . . , (Sl t, Bl t), where B1, . . . , Bl are standard Brownian motions. Let Bt = B1 q1t x1 + · · · + Bl qlt xl.
This satisfies (7.20).
7.7 Coupling the exit distributions Proposition 7.7.1 Suppose p ∈Pd. Then one can define on the same probability space a (discrete-time) random walk Sn with increment distribution p; a continuous-time random walk ˜ St with in-crement distribution p; and a Brownian motion Bt with covariance matrix Γ such that for each n, r > 0, P |Sξn −Bξ′ n| ≥r log n = P n | ˜ S˜ ξn −Bξ′ n| ≥r log n o ≤c r, where ξn = min{j : J (Sj) ≥n}, ˜ ξn = min{t : J ( ˜ St) ≥n}, ξ′ n = min{t : J (Bt) = n}.
♣We advise caution when using the dyadic coupling to prove results about random walk. If (Sn, Bt) are coupled as in the dyadic coupling, then Sn and Bt are Markov processes, but the joint process (Sn, Bn) is not Markov.
Proof It suffices to prove the result for ˜ St, Bt, for then we can define Sj to be the discrete-time “skeleton” walk obtained by sampling ˜ St at times of its jumps. We may also assume r ≤n; indeed, since | ˜ S˜ ξn| + |Bξ′ n| ≤O(n), for all n sufficiently large P n | ˜ S˜ ξn −Bξ′ n| ≥n log n o = 0.
By Theorem 7.6.2 we can define ˜ S, B on the same probability space such that except for an event of probability O(n−4), | ˜ St −Bt| ≤c1 log n, 0 ≤t ≤n3.
We claim that P{˜ ξn > n3} decay exponentially in n. Indeed, the central limit theorem shows that there is a c > 0 such that for n sufficiently large and |x| < n , Px{˜ ξn ≤n2} ≥c. Iterating this gives Px{˜ ξn > n3} ≤(1−c)n. Similarly, P{ξ′ n > n3} decays exponentially. Therefore, except on an event of probability O(n−4), | ˜ St −Bt| ≤c1 log n, 0 ≤t ≤max{˜ ξn, ξ′ n}.
(7.21) Note that the estimate (7.21) is not sufficient to directly yield the claim, since it is possible that one of the two paths (say ˜ S) first exits Cn at some point y, then moves far away from y (while 178 Dyadic coupling staying close to ∂Cn) and that only then the other path exits Cn, while all along the two paths stay close. The rest of the argument shows that such event has small probability. Let ˜ σn(c1) = min{t : dist( ˜ St, Zd \ Cn) ≤c1 log n} and σ′ n(c1) = min{t : dist(Bt, Zd \ Cn) ≤c1 log n}, and define ρn := σn(c1) ∧σ′ n(c1).
Since ρn ≤max{˜ ξn, ξ′ n}, we conclude as in (7.21) that with an overwhelming (larger than 1−O(n−4)) probability, | ˜ St −Bt| ≤c1 log n, 0 ≤t ≤ρn, and in particular that | ˜ Sρn −Bρn| ≤c1 log n.
(7.22) On the event in (7.22) we have max{dist( ˜ Sρn, Zd \ Cn), dist(Bρn, Zd \ Cn)} ≤2c1 log n, by triangle inequality, so in particular max{˜ σn(2c1), σ′ n(2c1)} ≤ρn.
(7.23) Using the gambler’s ruin estimate (see Exercise 7.5) and strong Markov property for each process separately (recall, they are not jointly Markov) P{| ˜ S˜ σn(2c1) −˜ Sj| ≤r log n for all j ∈[˜ σn(2c1), ˜ ξn]} ≥1 −c2 r , (7.24) and also P{|Bσ′ n(2c1) −Bt| ≤r log n for all t ∈[σ′ n(2c1), ξ′ n]} ≥1 −c2 r .
(7.25) Applying the triangle inequality to ˜ S˜ ξn −Bξ′ n = ( ˜ S˜ ξn −˜ Sρn) + ( ˜ Sρn −Bρn) + (Bρn −Bξ′ n), on the intersection of the four events from (7.22)–(7.25), yields ˜ S˜ ξn −Bξ′ n ≤(2r + c1) log n, and the complement has probability bounded by O(1/r).
Definition. A finite subset A of Zd is simply connected if both A and Zd \ A are connected. If x ∈Zd, let Sx denote the closed cube in Rd of side length one, centered at x, with sides parallel to the coordinate axes. If A ⊂Zd, let DA be the domain defined as the interior of ∪x∈ASx. The inradius of A is defined by inrad(A) = min{|y| : y ∈Zd \ A}.
Proposition 7.7.2 Suppose p ∈P2. Then one can define on the same probability space a (discrete-time) random walk Sn with increment distribution p and a Brownian motion Bt with covariance 7.7 Coupling the exit distributions 179 matrix Γ such that the following holds. If A is a finite, simply connected set containing the origin and ρA = inf{t : Bt ̸∈DA}, then each if r > 0, P{|SτA −BρA| ≥r log[inrad(A)]} ≤ c √r.
Proof Similar to the last proposition except that the gambler’s ruin estimate is replaced with the Beurling estimate.
Exercises Exercise 7.1 Suppose Sn = X1+· · ·+Xn where X1, X2, . . . are independent, identically distributed random variables. Suppose P{S2n = 2y} > 0 for some y ∈R. Show that the conditional distribution of Sn −y conditioned on {S2n = 2y} is symmetric about the origin.
Exercise 7.2 Suppose Sn is a random walk in Z whose increment distribution satisfies (7.4) and (7.5) and let C < ∞. Show that there exists a t = t(b, ǫ, c0, C) > 0 such that for all n, E exp t S2 n n ; |Sn| ≤Cn ≤e.
Exercise 7.3 Suppose ˜ St is continuous-time simple random walk in Z.
(i) Show that there is a c < ∞such that for all positive integers n, P{ ˜ Sn = n2} ≥c−1 exp{−cn2 log n}.
(Hint: consider the event that the walk makes exactly n2 moves by time n, each of them in the positive direction.) (ii) Show that if t > 0, E " exp ( t ˜ S2 n n )# = ∞, Exercise 7.4 Let Φ be the standard normal distribution function, and let Φ = 1 −Φ.
(i) Show that as x →∞, ¯ Φ(x) ∼e−x2/2 √ 2π Z ∞ 0 e−xt dt = 1 x √ 2π e−x2/2.
(ii) Prove (7.10).
(iii) Show that for all 0 ≤t ≤x, Φ(x + t) ≤e−tx e−t2/2 Φ(x) ≤etx e−t2/2 Φ(x) ≤Φ(x −t).
180 Dyadic coupling (iv) For positive integer n, let Φn(x) = Φ(x/√n) denote the distribution function of a mean zero normal random variable with variance n, and Φn = 1−Φn. Show that for every b > 0, there exist δ > 0 and 0 < c < ∞such that if 0 ≤x ≤δn, Φn x + c 1 + x2 n exp 2b 1 √n + x3 n2 ≤ Φn(x) exp b 1 √n + x3 n2 ≤ Φn x −c 1 + x2 n .
(v) Prove (7.9).
Exercise 7.5 In this exercise we prove the following version of the gambler’s ruin estimate. Suppose p ∈Pd, d ≥2. Then there exists c such that the following is true. If θ ∈Rd with |θ| = 1 and r ≥0, P {Sj · θ ≥−r, 0 ≤j ≤ξ∗ n} ≤c(r + 1) n .
(7.26) Here ξ∗ n is as defined in Section 6.3.
(i) Let q(x, n, θ) = Px{Sj · θ > 0, 1 ≤j ≤ξ∗ n}.
Show that there is a c1 > 0 such that for all n sufficiently large and all θ ∈Rd with |θ| = 1, the cardinality of the set of x ∈Zd with |x| ≤n/2 and q(x, n, θ) ≥c1 q(0, 2n, θ) is at least c1 nd−1.
(ii) Use a last-exit decomposition to conclude X x∈Cn GBn(0, x) q(x, n, θ) ≤1, and use this to conclude the result for r = 0.
(iii) Use Lemma 5.1.6 and the invariance principle to show that there is a c2 > 0 such that for all |θ| = 1, q(0, n, θ) ≥c2 n .
(iv) Prove (7.26) for all r ≥0.
8 Addtional topics on Simple Random Walk In this chapter we only consider simple random walk on Zd. In particular, S will always denote a simple random walk in Zd. If d ≥3, G denotes the corresponding Green’s function, and we simplify the notation by setting G(z) = −a(z), d = 1, 2, where a is the potential kernel. Note that then the equation LG(z) = −δ(z) holds for all d ≥1.
8.1 Poisson kernel Recall that if A ⊂Zd and τA = min{j ≥0 : Sj ̸∈A}, τ A = min{j ≥1 : Sj ̸∈A}, then the Poisson kernel is defined for x ∈A, y ∈∂A by HA(x, y) = Px{SτA = y}.
For simple random walk, we would expect the Poisson kernel to be very close to that of Brownian motion. If D ⊂Rd is a domain with sufficiently smooth boundary, we let hD(x, y) denote the Poisson kernel for Brownian motion. This means that, for each x ∈D, hD(x, ·) is the density with respect to surface measure on ∂D of the distribution of the point at which the Brownian motion visits ∂D for the first time. For sets A that are rectangles with sides perpendicular to the coordinate axes (with finite or infinite length), explicit expressions can be obtained for the Poisson kernel and one can show convergence to the Brownian quantities with relatively small error terms. We give some of these formulas in this section.
8.1.1 Half space If d ≥2, we define the discrete upper half space H = Hd by H = {(x, y) ∈Zd−1 × Z : y > 0}, with boundary ∂H = Zd−1 × {0} and “closure” H = H ∪∂H. Let T = τ H, and let HH denote the Poisson kernel, which for convenience we will write as a function HH : H × Zd−1 →[0, 1], HH(z, x) = HH(z, (x, 0)) = Pz{ST = (x, 0)}.
If z = (x, y) ∈Zd−1 × Z, we write z for its “conjugate”, z = (x, −y). If z ∈H, then z ∈−H. If z ∈∂H, then z = z. Recall the Green’s function for a set defined in Section 4.6.
181 182 Addtional topics on Simple Random Walk Proposition 8.1.1 For simple random walk in Zd, d ≥2, if z, w ∈H, GH(z, w) = G(z −w) −G(z −w), HH(z, 0) = 1 2d [G(z −ed) −G(z + ed)] .
(8.1) Proof To establish the first relation, note that for w ∈H, the function f(z) = G(z−w)−G(z−w) = G(z−w)−G(z−w) is bounded on H, Lf(z) = −δw(z), and f ≡0 on ∂H. Hence f(z) = GH(z, w) by the characterization of Proposition 6.2.3. For the second relation, we use a last-exit decomposition (focusing on the last visit to ed before leaving H) to see that HH(z, 0) = 1 2d GH(z, ed).
The Poisson kernel for Brownian motion in the upper half space H = Hd = {(x, y) ∈Rd−1 × (0, ∞)} is given by hH((x, y), 0) = hH((x + z, y), z) = 2y ωd |(x, y)|d , where ωd = 2πd/2/Γ(d/2) is the surface area of the (d −1)-dimensional sphere of radius 1 in Rd.
The next theorem shows that this is also the asymptotic value for the Poisson kernel for the random walk in H = Hd, and that the error term is small.
Theorem 8.1.2 If d ≥2 and z = (x, y) ∈Zd−1 × {1, 2, . . .}, then HH(z, 0) = 2y ωd|z|d 1 + O |y| |z|2 + O 1 |z|d+1 .
(8.2) Proof We use (8.1). If we did not need to worry about the error terms, we would naively estimate 1 2d [G(z −ed) −G(z + ed)] by Cd 2d h |z −ed|2−d −|z + ed|2−di , d ≥3, (8.3) C2 4 log |z −ed| |z + ed|, d = 2.
(8.4) Using Taylor series expansion, one can check that the quantities in (8.3) and (8.4) equal 2y ωd|z|d + O |y|2 |z|d+2 .
However, the error term in the expansion of the Green’s function or potential kernel is O(|z|−d), so we need to do more work to show that the error term in (8.2) is of order O(|y|2/|z|d+2)+O(|z|−(d+1)).
8.1 Poisson kernel 183 Assume without loss of generality that |z| > 1. We need to estimate G(z −ed) −G(z + ed) = ∞ X n=1 [pn(z −ed) −pn(z + ed)] = ∞ X n=1 [pn(z −ed) −pn(z + ed)] − ∞ X n=1 [pn(z −ed) −pn(z −ed) −pn(z + ed) + pn(z + ed)] .
Note that z −ed and z + ed have the same “parity” so the above series converge absolutely even if d = 2. We will now show that 1 2d ∞ X n=1 [pn(z −ed) −pn(z + ed)] = 2y ωd|z|d + O y2 |z|d+2 .
(8.5) Indeed, pn(z −ed) −pn(z + ed) = dd/2 (2π)d/2 1 nd/2 e−|x|2+(y−1)2 2n/d 1 −e− 4y 2n/d .
For n ≥y, we can use a Taylor approximation for 1 −e− 4y 2n/d . The terms with n < y, do not contribute much. More specifically, the left-hand side of (8.5) equals 2dd/2+1 (2π)d/2 ∞ X n=1 y n1+d/2 e−|x|2+(y−1)2 2n/d + O ∞ X n=1 y2 n2+d/2 e−|x|2+(y−1)2 2n/d !
+ O(y2e−|z|).
Lemma 4.3.2 then gives ∞ X n=1 [pn(z −ed) −pn(z + ed)] = 2 d Γ(d/2) πd/2 y (|x|2 + (y −1)2)d/2 + O y2 |z|d+2 = 2 d Γ(d/2) πd/2 y |z|d + O y2 |z|d+2 = 4d ωd y |z|d + O y2 |z|d+2 .
The remaining work is to show that ∞ X n=1 [pn(z −ed) −pn(z −ed) −pn(z + ed) + pn(z + ed)] = O(|z|−(d+1)).
We mimic the argument used for (4.11), some details are left to the reader.
Again the sum over n < |z| is negligible. Due to the second (stronger) estimate in Theorem 2.3.6, the sum over n > |z|2 is bounded by X n>|z|2 c n(d+3)/2 = O 1 |z|d+1 .
For n ∈[|z|, |z|2], apply Theorem 2.3.8 with k = d + 5 (for the case of symmetric increment 184 Addtional topics on Simple Random Walk distribution) to give pn(w) = pn(w) + d+5 X j=3 uj(w/√n) n(d+j−2)/2 + O 1 n(d+k−1)/2 , where w = z ± ed. As remarked after Theorem 2.3.8, we then can estimate |pn(z −ed) −pn(z −ed) −pn(z −ed) + pn(z −ed)| up to an error of O(n(−d+k−1)/2) by I3,d+5(n, z) := d+5 X j=3 1 n(d+j−2)/2 uj z + ed √n −uj z −ed √n .
Finally, due to Taylor expansion and the uniform estimate (2.29), one can obtain a bound on the sum P n∈[|z|,|z|2] I3,d+5(n, z) by imitating the final estimate in the proof of Theorem 4.3.1. We leave this to the reader.
In Section 8.1.3 we give an exact expression for the Poisson kernel in H2 in terms of an integral. To motivate it, consider a random walk in Z2 starting at e2 stopped when it first reaches {xe1 : x ∈Z}.
Then the distribution of the first coordinate of the stopping position gives a probability distribution on Z. In Corollary 8.1.7, we show that the characteristic function of this distribution is φ(θ) = 2 −cos θ − p (2 −cos θ)2 −1.
Using this and Proposition 2.2.2, we see that the probability that the first visit is to xe1 is 1 2π Z π −π e−ixθ φ(θ) dθ = 1 2π Z π −π cos(xθ) φ(θ) dθ.
If instead the walk starts from ye2, then the position of its first visit to the origin can be considered as the sum of y independent random variables each with characteristic function φ. The sum has characteristic function φy, and hence HH(ye2, xe1) = 1 2π Z π −π cos(xθ) φ(θ)y dθ.
8.1.2 Cube In this subsection we give an explicit form for the Poisson kernel on a finite cube in Zd.
Let Kn = Kn,d be the cube Kn = {(x1, . . . , xd) ∈Zd : 1 ≤xj ≤n −1}.
Note that #(Kn) = (n −1)d and ∂Kn consists of 2d copies of Kn,d−1. Let Sj denote simple random walk and τ = τn = min{j ≥0 : Sj ̸∈Kn}. Let Hn = HKn denote the Poisson kernel Hn(x, y) = Px{Sτn = y}.
If d = 1, the gambler’s ruin estimate gives Hn(x, n) = x n, x = 0, 1, . . . , n, 8.1 Poisson kernel 185 so we will restrict our consideration to d ≥2. By symmetry, it suffices to determine Hn(x, y) for y in one of the (d −1)-dimensional sub-cubes of ∂Kn. We will consider y ∈∂1 n := {(n, ˜ y) ∈Zd : ˜ y ∈Kn,d−1}.
The set of functions on Kn that are harmonic in Kn and equal zero on ∂Kn \ ∂1 n is a vector space of dimension #(∂1 n), and one of its bases is {Hn(·, y) : y ∈∂1 n}. In the next proposition we will use another basis which is more explicit.
♣The proposition below uses a discrete analogue of a technique from partial differential equations called separation of variables.
We will then compare this to the Poisson kernel for Brownian motion that can be computed using the usual separation of variables.
Proposition 8.1.3 If x = (x1, . . . , xd) ∈Kn,d and y = (y2, . . . , yd) ∈Kn−1,d, then Hn(x, (n, y)) equals 2 n d−1 X z∈Kn,d−1 sinh(αzx1π/n) sinh(αzπ) sin z2x2π n · · · sin zdxdπ n sin z2y2π n · · · sin zdydπ n , where z = (z2, . . . , zd) and αz = αz,n is the unique nonnegative number satisfying cosh αzπ n + d X j=2 cos zjπ n = d.
(8.6) Proof If z = (z2, . . . , zd) ∈Rd−1, let fz denote the function on Zd, fz(x1, . . . , xd) = sinh αzx1π n sin z2x2π n · · · sin zdxdπ n , where αz satisfies (8.6). It is straightforward to check that for any z, fz is a discrete harmonic function on Zd with fz(x) = 0, x ∈∂Kn \ ∂1 n.
We now restrict our consideration to z ∈Kn,d−1. Let ˆ fz = 2(d−1)/2 n(d−1)/2 sinh(αzπ) fz, z ∈Kn,d−1, and let ˆ f ∗ z denote the restriction of ˆ fz to ∂1 n, considered as a function on Kn,d−1, ˆ f ∗ z (x) = ˆ fz((n, x)) = 2 n (d−1)/2 sin z2x2π n · · · sin zdxdπ n , x = (x2, . . . , xd) ∈Kn,d−1.
For integers 1 ≤j, k ≤n −1, one can see (via the representation of sin in terms of exponentials) that n−1 X l=1 sin jlπ n sin klπ n = 0 j ̸= k n/2 j = k.
(8.7) 186 Addtional topics on Simple Random Walk Therefore, { ˆ f ∗ z : z ∈Kn,d−1} forms an orthonormal basis for the set of functions on Kn,d−1, in symbols, X x∈Kn,d−1 ˆ f ∗ z (x) ˆ f ∗ ˆ z (x) = 0, z ̸= ˆ z 1, z = ˆ z .
Hence any function g on Kn,d−1 can be written as g(x) = X z∈Kn,d−1 C(g, z) ˆ f ∗ z (x), where C(g, z) = X y∈Kn,d−1 ˆ f ∗ z (y) g(y).
In particular, if y ∈Kn,d−1, δy(x) = X z∈Kn,d−1 ˆ f ∗ z (y) ˆ f ∗ z (x).
Therefore, for each y = (y2, . . . , yn) the function x 7→ X z∈Kn,d−1 sinh(αzx1π/n) sinh(αzπ) ˆ fz(y) ˆ fz((n, x2, . . . , xn)), is a harmonic function in Kn,d whose value on ∂Kn,d is δ(n,y) and hence it must equal to x 7→ Hn(x, (n, y)).
To simplify the notation, we will consider only the case d = 2 (but most of what we write extends to d ≥3). If d = 2, HKn((x1, x2), (n, y)) = 2 n n X k=1 sinh(akx1π/n) sinh(akπ) sin kx2π n sin kyπ n , (8.8) where ak = ak,n is the unique positive solution to cosh akπ n + cos kπ n = 2.
Alternatively, we can write ak = n πr kπ n , (8.9) where r is the even function r(t) = cosh−1(2 −cos t).
(8.10) Using cosh−1(1 + x) = √ 2x + O(x3/2) as x →0+, we get r(t) = |t| + O(|t|3), t ∈[−1, 1].
Now (8.9)–(8.10) imply ak = k + O k3 n2 .
(8.11) 8.1 Poisson kernel 187 Since ak increases with k, (8.11) implies that there is an ǫ > 0 such that ak ≥ǫ k, 1 ≤k ≤n −1.
(8.12) We will consider the scaling limit.
Let Bt denote a two-dimensional Brownian motion.
Let K = (0, 1)2 and let T = inf{t : Bt ̸∈K}.
The corresponding Poisson kernel hK can be computed exactly in terms of an infinite series using the continuous analogue of the procedure above giving h((x1, x2), (1, y)) = 2 ∞ X k=0 sinh(kx1π) sinh(kπ) sin(kx2π) sin(kyπ) (8.13) (see Exercise 8.2).
Roughly speaking, we expect HKn((nx1, nx2), (n, ny)) ≈1 nh((x1, x2), (1, y2)), and the next proposition gives a precise formulation of this.
Proposition 8.1.4 There exists c < ∞such that if 1 ≤j1, j2, l ≤n−1 are integers, xi = ji/n, y = l/n, n HKn((j1, j2), (n, l)) −h((x1, x2), (1, y)) ≤ c (1 −x1)6 n2 sin(x2π) sin(yπ).
♣A surprising fact about this proposition is how small the error term is. For fixed x1 < 1, the error is O(n−2) where one might only expect O(n−1).
Proof Let ρ = x1. Given k ∈N, note that | sin(kt)| ≤k sin t for 0 < t < π.
(To see this, t 7→k sin t ± sin(kt) is increasing on [0, tk] where tk ∈(0, π/2) solves sin(tk) = 1/k, and sin(·) continues to increase up to π/2, while sin(k ·) stays bounded by 1. For π/2 < t < π consider π −t instead, details are left to the reader.) Therefore, 1 sin(x2π) sin(yπ) ∞ X k≥n2/3 sinh(kx1π) sinh(kπ) sin(kx2π) sin(kyπ) ≤ ∞ X k≥n2/3 k2 sinh(kρπ) sinh(kπ) ≤ 1 n2 ∞ X k≥n2/3 k5 sinh(kρπ) sinh(kπ) ≤ c n2 ∞ X k≥n2/3 k5 e−k(1−ρ)π.
(8.14) 188 Addtional topics on Simple Random Walk Similarly, using (8.12), 1 sin(x2π) sin(yπ) ∞ X k≥n2/3 sinh(akx1π) sinh(akπ) sin(kx2π) sin(kyπ) ≤1 n2 X k≥n2/3 k5 e−kǫ(1−ρ)π.
For 0 ≤x ≤1 and k < n2/3, sinh(xakπ) = sinh(xkπ) 1 + O k3 n2 .
Therefore, 1 sin(x2π) sin(yπ) X k<n2/3 sinh(kx1π) sinh(kπ) −sinh(akx1π) sinh(akπ) sin(kx2π) sin(kyπ) ≤ c n2 X k≤n2/3 k3 e−k(1−ρ)π ≤c n2 X k≤n2/3 k5 e−k(1−ρ)π, where the second to last inequality is obtained as in (8.14). Combining this with (8.8) and (8.13), we see that |n HKn((j1, j2), (n, nl)) −h((x1, x2), (1, y))| sin(x2π) sin(yπ) ≤c n2 ∞ X k=1 k5 e−ǫ(1−ρ)k ≤ c (1 −ρ)6 n2 .
♣The error term in the last proposition is very good except for x1 near 1. For x1 close to 1, one can give good estimates for the Poisson kernel by using the Poisson kernel for a half plane (if x2 is not near 0 or 1) or by a quadrant (if x2 is near 0 or 1). These Poisson kernels are discussed in the next subsection.
8.1.3 Strips and quadrants in Z2 In the continuing discussion we think of Z2 as Z + iZ, and we will use complex numbers notation in this section. Recall r defined in (8.10) and note that er(t) = 2 −cos t + p (2 −cos t)2 −1, e−r(t) = 2 −cos t − p (2 −cos t)2 −1.
For each t ≥0, the function ft(x + iy) = exr(t) sin(yt), ˆ ft(x + iy) = e−xr(t) sin(yt), is harmonic for simple random walk, and so is the function sinh(xr(t))·sin(yt). The next proposition is an immediate generalization of Proposition 8.1.3 to rectangles that are not squares. The proof is the same and we omit it. We then take limits as the side lengths go to infinity to get expressions for other “rectangular” subsets of Z2.
8.1 Poisson kernel 189 Proposition 8.1.5 If m, n are positive integers, let Am,n = {x + iy ∈Z × iZ : 1 ≤x ≤m −1, 1 ≤y ≤n −1}.
Then HAm,n(x + iy, iy1) = HAm,n((m −x) + iy, m + iy1) = 2 n n−1 X j=1 sinh(r(jπ n )(m −x)) sinh(r(jπ n )m) sin jπy n sin jπy1 n .
(8.15) Corollary 8.1.6 If n is a positive integer, let A∞,n = {x + iy ∈Z × iZ : 1 ≤x < ∞, 1 ≤y ≤n −1}.
Then HA∞,n(x + iy, iy1) = 2 n n−1 X j=1 exp −r jπ n x sin jπy n sin jπy1 n , (8.16) and HA∞,n(x + iy, x1) = 2 π Z π 0 sinh(r(t)(n −y)) sinh(r(t)n) sin(tx) sin(tx1) dt.
(8.17) Proof Note that HA∞,n(x + iy, iy1) = lim m→∞HAm,n(x + iy, iy1), and lim m→∞ sinh(r(jπ n )(m −x)) sinh(r(jπ n )m) = exp −r jπ n x .
This combined with (8.15) gives the first identity. For the second we write HA∞,n(x + iy, x1) = lim m→∞HAm,n(x + iy, x1) = lim m→∞HAn,m(y + ix, ix1) = lim m→∞ 2 m m−1 X j=1 sinh(r(jπ m )(n −y)) sinh(r(jπ m )n) sin jπx m sin jπx1 m = 2 π Z π 0 sinh(r(t)(n −y)) sinh(r(t)n) sin(tx) sin(tx1) dt.
♣We derived (8.16) as a limit of (8.15). We could also have derived it directly by considering the collection of harmonic functions exp −r jπ n x sin jπy n , j = 1, . . . , n −1.
190 Addtional topics on Simple Random Walk Corollary 8.1.7 Let A+ = {x + iy ∈Z × iZ : x > 0}.
Then HA+(x + iy, 0) = 1 2π Z π −π e−xr(t) cos(yt) dt.
Remark. If H denotes the discrete upper half plane, then this corollary implies HH(iy, x) = HH(−x + iy, 0) = 1 2π Z π −π e−yr(t) cos(xt) dt.
Proof Note that HA+(x + iy, 0) = lim n→∞HA∞,2n(x + i(n + y), in) = lim n→∞ 1 n 2n−1 X j=1 exp −r jπ 2n x sin jπ 2 sin jπ(n + y) 2n .
Note that sin(jπ/2) = 0 if j is even. For odd j, we have sin2(jπ/2) = 1 and cos(jπ/2) = 0, hence sin jπ 2 sin jπ(n + y) 2n = cos jπy 2n .
Therefore, HA+(x + iy, 0) = lim n→∞ 1 n n X j=1 exp −r (2j −1)π 2n x cos (2j −1)πy 2n = 1 π Z π 0 e−xr(t) cos(yt) dt.
Remark. As already mentioned, using the above expression for (HA+(i, x), x ∈Z), one can read offthe characteristic function of the stopping position of simple random walk started from e2 and stopped at its first visit to the Z × {0} (see also Exercise 8.1).
Corollary 8.1.8 Let A∞,∞= {x + iy ∈Z + iZ : x, y > 0}.
Then, HA∞,∞(x + iy, x1) = 2 π Z π 0 e−r(t)y sin(tx) sin(tx1) dt.
8.2 Eigenvalues for rectangles 191 Proof Using (8.17), HA∞,∞(x + iy, x1) = lim n→∞HA∞,n(x + iy, x1) = lim n→∞ 2 π Z π 0 sinh(r(t)(n −y)) sinh(r(t)n) sin(tx) sin(tx1) dt = 2 π Z π 0 e−r(t)y sin(tx) sin(tx1) dt.
8.2 Eigenvalues for rectangles In general it is hard to compute the eigenfunctions and eigenvectors for a finite subset A of Zd with respect to simple random walk. One feasible case is that of a rectangle A = R(N1, . . . , Nd) := {(x1, . . . , xd) ∈Zd : 0 < xj < Nj}.
If k = (k1, . . . , kd) ∈Zd with 1 ≤kj < Nj, let fk(x1, . . . , xd) = fk,N1,...,Nd(x1, . . . , xd) = d Y j=1 sin xj kj π Nj .
Note that fk ≡0 on ∂R(N1, . . . , Nd). A straightforward computation shows that Lfk(x1, . . . , xd) = α(k) fk.
where α(k) = α(k; N1, . . . , Nd) = 1 d d X j=1 cos kj π Nj −1 .
Using (8.7) we can see that the functions {fk : 1 ≤kj ≤Nj −1} , form an orthogonal basis for the set of functions on R(N1, . . . , Nd) that vanish on ∂R(N1, . . . , Nd).
Hence this gives a complete set of eigenvalues and eigenvectors.
We conclude that the (first) eigenvalue α of R(N1, . . . , Nd), defined in Section 6.9, is given by α = 1 d d X j=1 cos π Nj .
In particular, as n →∞, the eigenvalue for Kn = Kn,d = R(n, . . . , n), is given by αKn = cos π n = 1 −π2 2n2 + O 1 n4 .
192 Addtional topics on Simple Random Walk 8.3 Approximating continuous harmonic functions It is natural to expect that discrete harmonic functions in Zd, when appropriately scaled, converge to (continuous) harmonic functions in Rd. In this section we discuss some versions of this principle.
We let U = Ud = {x ∈Rd : |x| < 1} denote the unit ball in Rd.
Proposition 8.3.1 There exists c < ∞such that the following is true for all positive integers n, m. Suppose f : (n + m)U →R is a harmonic function. Then there is a function ˆ f on Bn with L ˆ f(x) = 0, x ∈Bn and such that |f(x) −ˆ f(x)| ≤c ∥f∥∞ m2 , x ∈Bn.
(8.18) In fact, one can choose (recall ξn from Section 6.3) ˆ f(x) = Ex[f(Sξn)].
Proof Without loss of generality, assume ∥f∥∞= 1. Since f is defined on (n + 1)Ud, ˆ f is well defined. By definition we know that L ˆ f(x) = 0, x ∈Bn. We need to prove (8.18). By (6.9), if x ∈Bn, f(x) = Ex f(Sξn) − ξn−1 X j=0 Lf(Sj) = ˆ f(x) −φ(x), where φ(x) = X z∈Bn GBn(x, z) Lf(z).
In Section 6.2, we observed that there is a c such that all 4th order derivatives of f at x are bounded above by c (n + m −|x|)−4. By expanding in a Taylor series, using the fact that f is harmonic, and also using the symmetry of the random walk, this implies |Lf(x)| ≤ c (n + m −|x|)4 .
Therefore, we have |φ(x)| ≤c n−1 X k=0 X k≤|z|<k+1 GBn(x, z) (n + m −k)4 .
(8.19) We claim that there is a c1 such that for all x, X n−l≤|z|≤n−1 GBn(x, z) ≤c l2.
Indeed, the proof of this is essentially the same as the proof of (5.5). Once we have the last estimate, summing by parts the right-hand side of (8.19) gives that |φ(x)| ≤c/m2.
The next proposition can be considered as a converse of the last proposition. If f : Zd →R is a function, we will also write f for the piecewise constant function on Rd defined as follows. For 8.4 Estimates for the ball 193 each x = (x1, . . . , xj) ∈Zd, let □x denote the cube of side length 1 centered at x, □x = (y1, . . . , yd) ∈Rd : −1 2 ≤yj −xj < 1 2 .
The sets {□x : x ∈Zd} partition Rd.
Proposition 8.3.2 Suppose fn is a sequence of functions on Zd satisfying Lfn(x) = 0, x ∈Bn and supx |fn(x)| ≤1. Let gn : Rd →R be defined by gn(y) = fn(ny). Then there exists a subsequence nj and a function g that is harmonic on U such that gnj − →g uniformly on every compact K ⊂U.
Proof Let J be a countable dense subset of U. For each y ∈J, the sequence gn(y) is bounded and hence has a subsequential limit. By a standard diagonalization procedure, we can find a function g on J such that gnj(y) − →g(y), y ∈J.
For notational convenience, for the rest of this proof we will assume that in fact gn(y) − →g(y), but the proof works equally well if there is only a subsequence.
Given r < 1, let rU = {y ∈U : |y| < r}. Using Theorem 6.3.8, we can see that there is a cr < ∞such that for all n, |gn(y1) −gn(y2)| ≤cr [|y1 −y2| + n−1] for y1, y2 ∈rU. In particular, |g(y1) −g(y2)| ≤cr |y1 −y2| for y1, y2 ∈J ∩rU. Hence, we can extend g continuously to rU such that |g(y1) −g(y2)| ≤cr |y1 −y2|, y1, y2 ∈rU, (8.20) and a standard 3ǫ-argument shows that gn converges to g uniformly on rU.
Since g is continuous, in order to show that g is harmonic, it suffices to show that it has the spherical mean value property, i.e., if y ∈U and |y| + ǫ < 1, Z |z−y|=ǫ g(z) dsǫ(z) = g(y).
Here sǫ denotes surface measure normalized to have measure one. This can be established from the discrete mean value property for the functions fn, using Proposition 7.7.1 and (8.20). We omit the details.
8.4 Estimates for the ball One is often interested in comparing quantities for the simple random walk on the discrete ball Bn with corresponding quantities for Brownian motion. Since the Brownian motion is rotationally invariant, balls are very natural domains to consider. However, the lattice effects at the boundary mean that it is harder to control the rate of convergence of the simple random walk. This section presents some basic comparison estimates.
We first consider the Green’s function GBn(x, y). If x = 0, Proposition 6.3.5 gives sharp estimates.
It is trickier to estimate this for other x, y. We will let g denote the Green’s function for Brownian motion in Rd with covariance matrix d−1I, g(x, y) = Cd |x −y|2−d, d ≥3, 194 Addtional topics on Simple Random Walk g(x, y) = −Cd log |x −y|, d = 2, and define gn(x, y) by gn(x, y) = g(x, y) −Ex[g(BTn, y)] .
Here B is a d-dimensional Brownian motion with covariance d−1I and Tn = inf{t : |Bt| = n}.
The reader should compare the formula for gn(x, y) to corresponding formulas for GBn(x, y) in Proposition 4.6.2. The Green’s function for standard Brownian motion is gn(x, y)/d.
Proposition 8.4.1 If d ≥2 and x, y ∈Bn, GBn(x, y) = gn(x, y) + O 1 |x −y|d + O log2 n (n −|y|)d−1 .
♣This estimate is not optimal, but improvements will not be studied in this book. Note that if follows that for every ǫ > 0, if |x|, |y| ≤(1 −ǫ)n, GBn(x, y) = gn(x, y) 1 + O 1 |x −y| 2 + Oǫ log2 n n , where we write Oǫ to indicate that the implicit constants depend on ǫ but are uniform in x, y, n. In particular, we have uniform convergence on compact subsets of the open unit ball.
Proof We will do the d ≥3 case; the d = 2 case is done similarly. By Proposition 4.6.2, GBn(x, y) = G(x, y) −Ex [G(Sξn, y)] .
Therefore, |gn(x, y) −GBn(x, y)| ≤|g(x, y) −G(x, y)| + |Ex[g(BTn, y)] −Ex[G(Sξn, y)]| .
By Theorem 4.3.1, |g(x, y) −G(x, y)| ≤ c |x −y|d G(Sξn, y) = Cd |Sξn −y|d−2 + O 1 (1 + |Sξn −y|)d , Note that Ex[(1 + |Sξn −y|)−d] ≤ |n + 1 −|y||−2 Ex[G(Sξn, y)] ≤ c |n + 1 −|y||−2 G(x, y) ≤ c |n + 1 −|y||−2 |x −y|2−d ≤ c h |x −y|−d + (n + 1 −|y|)1−di .
We can define a Brownian motion B and a simple random walk S on the same probability space such that for each r, P {|BTn −Sξn| ≥r log n} ≤c r, 8.4 Estimates for the ball 195 see Proposition 7.7.1. Since |BTn −Sξn| ≤c n, we see that E [|BTn −Sξn|] ≤ cn X k=1 P(|BTn −Sξn| ≥k) ≤c log2 n.
Also, Cd |x −y|d−2 − Cd |z −y|d−2 ≤ c |x −z| [n −|y|]d−1 .
Let αBn denote the eigenvalue for the ball as in Section 6.9 and define λn by αBn = e−λn. Let λ = λ(d) be the eigenvalue of the unit ball for a standard d-dimensional Brownian motion Bt, i.e., P{|Bs| < 1, 0 ≤s ≤t} ∼c e−λt, t →∞.
Since the random walk suitably normalized converges to Brownian motion, one would conjecture that dn2λn is approximately λ for large n. The next proposition establishes this but again not with the optimal error bound.
Proposition 8.4.2 λn = λ dn2 1 + O 1 log n .
Proof By Theorem 7.1.1, we can find a b > 0 such that a simple random walk Sn and a standard Brownian motion Bt can be defined on the same probability space so that P max 0≤j≤n3 |Sj −Bj/d| ≥b log n ≤b n−1.
By Corollary 6.9.6, there is a c1 such that for all n and all j, P{|Sj| < n, j ≤kn2} ≥c1 e−λnk n2.
(8.21) For Brownian motion, we know there is a c2 such that P{|Bt| < 1, 0 ≤t ≤k} ≤c2 e−λk.
By the coupling, we know that for all n sufficiently large P{|Sj| < n, j ≤d λ−1n2 log n} ≤P{|Bt| < n + b log n, t ≤λ−1 n2 log n} + b n−1, and due to Brownian scaling, we obtain c1 exp −dn2λn λ log n ≤ c2 exp − n2 log n (n + b log n)2 + b n−1 ≤ c3 exp −log n + O log2 n n .
Taking logarithms, we get dn2λn λ ≥1 −O 1 log n .
196 Addtional topics on Simple Random Walk A similar argument, reversing the roles of the Brownian motion and the random walk, gives dn2λn λ ≤1 + O 1 log n .
Exercises Exercise 8.1 Suppose Sn is simple random walk in Z2 started at the origin and T = min {j ≥1 : Sj ∈{xe1 : x ∈Z}} .
Let X denote the first component of ST . Show that the characteristic function of X is φ(t) = 1 − p (2 −cos t)2 −1.
Exercise 8.2 Let V = {(x, y) ∈R2 : 0 < x, y < 1} and let ∂1V = {(1, y) : 0 ≤y ≤1}. Suppose g : ∂V →R is a continuous function that vanishes on ∂V \ ∂1V . Show that the unique continuous function on V that is harmonic in V and agrees with g on ∂V is f(x, y) = 2 ∞ X k=1 ck sinh(kxπ) sinh(kπ) sin(kyπ), where ck = Z 1 0 sin(tkπ) g(1, t) dt.
Use this to derive (8.13).
Exercise 8.3 Let A∞,∞be as in Corollary 8.1.8.
Suppose xn, yn, kn are sequences of positive integers with lim n→∞ xn n = x, lim n→∞ yn n = y, lim n→∞ kn n = k, where x, y, k are positive real numbers. Find lim n→∞n HA∞,∞(xn + iyn, kn).
Exercise 8.4 Let fn be the eigenfunction associated to the d-dimensional simple random walk in Zd on Bn, i.e., Lfn(x) = (1 −e−λn) fn(x), x ∈Bn, with fn ≡0 on Zd \ Bn or equivalently, fn(x) = (1 −e−λn) X y∈Bn GBn(x, y) f(y).
This defines the function up to a multiplicative constant; fix the constant by asserting fn(0) = 1.
Extend fn to be a function of Rd as in Section 8.3 and let Fn(x) = fn(nx).
8.4 Estimates for the ball 197 The goal of this problem is to show that the limit F(x) = lim n→∞Fn(x), exists and satisfies F(x) = λ Z |y|≤1 g(x, y) F(y) ddy, (8.22) where g is the Green’s function for Brownian motion with constant chosen as in Section 8.4. In other words, F is the eigenfunction for Brownian motion. (The eigenfunction is the same whether we choose covariance matrix I or d−1I.) Useful tools for this exercise are Proposition 6.9.4, Exercise 6.6, Proposition 8.4.1, and Proposition 8.4.2. In particular, (i) Show that there exist c1, c2 such that c1[1 −|x|] ≤Fn(x) ≤c2[1 −|x| + n−1].
(ii) Use a diagonalization argument to find a subsequence nj such that for all x with rational coordinates the limit F(x) = lim j→∞Fnj(x) exists.
(iii) Show that for every r < 1, there is a cr such that for |x|, |y| ≤r, |Fn(x) −Fn(y)| ≤cr [|x −y] + n−1].
(iv) Show that F is uniformly continuous on the set of points in the unit ball with rational coordinates and hence can be defined uniquely on {|z| ≤1} by continuity.
(v) Show that if |x|, |y| < cr, then |F(x) −F(y)| ≤cr |x −y|.
(vi) Show that F satisfies (8.22).
(vii) You may take it as a given that there is a unique solution to (8.22) with F(0) = 1 and F(x) = 0 for |x| = 1. Use this to show that Fn converges to F uniformly.
9 Loop Measures 9.1 Introduction Problems in random walks are closely related to problems on loop measures, spanning trees, and determinants of Laplacians. In this chapter we will gives some of the relations. Our basic viewpoint will be different from that normally taken in probability. Instead of concentrating on probability measures, we consider arbitrary (positive) measures on paths and loops.
Considering measures on paths or loops that are not probability measures is standard in statistical physics. Typically one consider weights on paths of the form e−βE where β is a parameter and E is the “energy” of a configuration. If the total mass is finite (say, if the there are only a finite number of configurations) such weights can be made into probability measures by normalizing.
There are times where it is more useful to think of the probability measures and other times where the unnormalized measure is important. In this chapter we take the configurational view.
9.2 Definitions and notations Throughout this chapter we will assume that X = {x0, x1, . . . , xn−1}, or X = {x0, x1, . . .} is a finite or countably infinite set of points or vertices with a distinguished vertex x0 called the root.
• A finite sequence of points ω = [ω0, ω1, . . . , ωk] in X is called a path of length k. We write |ω| for the length of ω.
• A path is called a cycle if ω0 = ωk. If ω0 = x, we call the cycle an x-cycle and call x the root of the cycle.
We allow the trivial cycles of length zero consisting of a single point.
• If x ∈X, we write x ∈ω if x = ωj for some j = 0, . . . , |ω|.
• If A ⊂X, we write ω ⊂A if all the vertices of ω are in A.
• A weight q is a nonnegative function q : X × X →[0, ∞) that induces a weight on paths q(ω) = |ω| Y j=1 q(ωj−1, ωj).
198 9.2 Definitions and notations 199 By convention q(ω) = 1 if |ω| = 0.
• q is symmetric if q(x, y) = q(y, x) for all x, y ♣Although we are doing this in generality, one good example to have in mind is X = Zd or X equal to a finite subset of Zd containing the origin with x0 = 0. The weight q is that obtained from simple random walk, i.e., q(x, y) = 1/2d if |x −y| = 1 and q ≡0 otherwise.
• We say that X is q-connected if for every x, y ∈X there exists a path ω = [ω0, ω1, . . . , ωk] with ω0 = x, ωk = y and q(ω) > 0.
• q is called a (Markov) transition probability if for each x X y q(x, y) = 1.
In this case q(ω) denotes the probability that the chain starting at ω0 enters states ω1, . . . , ωk in that order. If X is q-connected, q is called irreducible.
• q is called a subMarkov transition probability if for each x X y q(x, y) ≤1, and it is called strictly subMarkov if the sum is strictly less than one for at least one x. Again, q is called irreducible if X is q-connected. A subMarkov transition probability q on X can be made into a transition probability on X ∪{∆} by setting q(∆, ∆) = 1 and q(x, ∆) = 1 − X y q(x, y).
The first time that this Markov chain reaches ∆is called the killing time for the subMarkov chain.
♣If q is the weight corresponding to simple random walk in Zd, then q is a transition probability if X = Zd and q is a strictly subMarkov transition probability if X is a proper subset of Zd.
• If q is a transition probability on X, two important ways to get subMarkov transition probabilities are: – Take A ⊂X and consider q(x, y) restricted to A. This corresponds to the Markov chain killed when it leaves A.
– Let 0 < λ < 1 and consider λq. This corresponds to the Markov chain killed at geometric rate (1 −λ).
• The rooted loop measure m = mq is the measure on cycles defined by m(ω) = 0 if |ω| = 0 and m(ω) = mq(ω) = q(ω) |ω| , |ω| ≥1.
200 Loop Measures • An unrooted loop or cycle is an equivalence class of cycles under the equivalence [ω0, ω1, . . . , ωk] ∼[ωj, ωj+1, . . . , ωk, ω1 . . . , ωj].
(9.1) We denote unrooted loops by ω and write ω ∼ω if ω is a cycle that produces the unrooted loop ω.
The lengths and weights of all representatives of ω are the same, so it makes sense to write |ω| and q(ω).
• If ω is an unrooted loop, let K(ω) = #{ω : ω ∼ω} be the number of representatives of the equivalence class. The reader can easily check that K(ω) divides |ω| but can be smaller. For example, if ω is the unrooted loop corresponding to a rooted loop ω = [x, y, x, y, x] with distinct vertices x, y, then |ω| = 4 but K(ω) = 2.
• The unrooted loop measure is the measure m = mq obtained from m by “forgetting the root”, i.e., m(ω) = X ω∼ω q(ω) |ω| = K(ω) q(ω) |ω| .
• A weight q generates a directed graph with vertices X and directed edges = {(x, y) ∈X × X : q(x, y) > 0}. Note that this allows “self-loops” of the form (x, x). If q is symmetric, then this is an undirected graph. In this chapter graph will mean undirected graph.
• If #(X) = n < ∞, a spanning tree T (of the complete graph) on vertices X is a collection of n −1 edges in X such that X with these edges is a connected graph.
• Given q, the weight of a tree T (with respect to root x0) is q(T ; x0) = Y (x,x′)∈T q(x, x′), where the product is over all directed edges (x, x′) ∈T and the direction is chosen so that the unique self-avoiding path from x to x0 in T goes through x′.
• If q is symmetric, then q(T ; x0) is independent of the choice of the root x0 and we will write q(T ) for (T ; x0). Any tree with positive weight is a subgraph of the graph generated by q.
• If q is a weight and λ > 0, we write qλ for the weight λq. Note that qλ(ω) = λ|ω| q(ω), qλ(T ) = λn−1 q(T ). If q is a subMarkov transition probability and λ ≤1, then qλ is also a subMarkov transition probability for a chain moving as q with an additional geometric killing.
• Let Lj denote the set of (rooted) cycles of length j and Lj(A) = {ω ∈Lj : ω ⊂A}, Lx j (A) = {ω ∈Lj(A) : x ∈ω}, Lx j = Lx j (X).
L = ∞ [ j=0 Lj, L(A) = ∞ [ j=0 Lj(A), Lx(A) = ∞ [ j=0 Lx j (A), Lx = ∞ [ j=0 Lx j .
9.3 Generating functions and loop measures 201 We also write Lj, Lj(A), etc., for the analogous sets of unrooted cycles.
9.2.1 Simple random walk on a graph An important example is simple random walk on a graph. There are two different definitions that we will use. Suppose X is the set of the vertices of an (undirected) graph. We write x ∼y if x is adjacent to y, i.e., if {x, y} is an edge. Let deg(x) = #{y : x ∼y} be the degree of x. We assume that the graph is connected.
• Simple random walk on the graph is the Markov chain with transition probability q(x, y) = 1 deg(x), x ∼y.
If X is finite, the invariant probability measure for this Markov chain is proportional to d(x).
• Suppose d = sup x∈X deg(x) < ∞.
The lazy (simple random) walk on the graph, is the Markov chain with symmetric transition probability q(x, y) = 1 d, x ∼y, q(x, x) = d −deg(x) d .
We can also consider this as simple random walk on the augmented graph that has added d − deg(x) self-loops at each vertex x. If X is finite, the invariant probability measure for this Markov chain is uniform.
• A graph is regular (or d-regular) if deg(x) = d for all x. For regular graphs, the lazy walk is the same as the simple random walk.
• A graph is transitive if “all the vertices look the same”, i.e., if for each x, y ∈X there is a graph isomorphism that takes x to y. Any transitive graph is regular.
9.3 Generating functions and loop measures In this section, we fix a set of vertices X and a weight q on X.
• If x ∈X, the x-cycle generating function is given by g(λ; x) = X ω∈L,ω0=x λ|ω| q(ω) = X ω∈L,ω0=x qλ(ω).
If q is a subMarkov transition probability and λ ≤1, then g(λ; x) denotes the expected number of visits of the chain to x before being killed for a subMarkov chain with weight qλ started at x.
202 Loop Measures • For any cycle ω we define d(ω) = #{j : 1 ≤j ≤|ω|, ωj = ω0}, and we call ω an irreducible cycle if d(ω) = 1.
• The first return to x generating function is defined by f(λ; x) = X ω∈L,|ω|≥1,ω0=x,d(ω)=1 q(ω) λ|ω|.
If λq is a subMarkov transition probability, then f(λ; x) is the probability that the chain starting at x returns to x before being killed.
One can check as in (4.6), that g(λ; x) = 1 + f(λ; x) g(λ, x), which yields g(λ; x) = 1 1 −f(λ; x).
(9.2) • If X is finite, the cycle generating function is g(λ) = X x∈X g(λ; x) = X ω∈L λ|ω| q(ω) = X ω∈L qλ(ω).
Since each x ∈X has a unique cycle of length 0 rooted at x, g(0; x) = 1, g(0) = #(X).
• If X is finite, the loop measure generating function is Φ(λ) = X ω∈L λ|ω| mq(ω) = X ω∈L λ|ω| mq(ω) = X ω∈L, |ω|≥1 λ|ω| |ω| q(ω).
Note that if #(X) = n < ∞, Φ(0) = 0, g(λ) = λ Φ′(λ) + n, Φ(λ) = Z λ 0 g(s) −n s ds.
• If A ⊂X is finite, we write F(A; λ) = exp X ω∈L(A),|ω|≥1 q(ω) λ|ω| |ω| = exp X ω∈L(A),|ω|≥1 q(ω) K(ω) λ|ω| |ω| .
In other words, log F(A; λ) is the loop measure (with weight qλ) of the set of loops in A. In particular, F(X; λ) = eΦ(λ).
• If x ∈A (A not necessarily finite), let log Fx(A; λ) denote the loop measure (with weight qλ) of the set of loops in A that include x, i.e., Fx(A; λ) = exp X ω∈Lx(A),|ω|≥1 q(ω) λ|ω| |ω| = exp X ω∈L x(A),|ω|≥1 q(ω) K(ω) λ|ω| |ω| .
9.3 Generating functions and loop measures 203 More generally, if V ⊂A, log FV (A; λ) denotes the loop measure of loops in A that intersect V , FV (A; λ) = exp X ω∈L(A),|ω|≥1,V ∩ω̸=∅ q(ω) λ|ω| |ω| = exp X ω∈L(A),|ω|≥1,V ∩ω̸=∅ q(ω) K(ω) λ|ω| |ω| .
If η is a path, we write Fη for FV where V denotes the vertices in η. Note that F(A; λ) = FA(A; λ).
• We write F(A) = F(A; 1), Fx(A) = Fx(A; 1).
Proposition 9.3.1 If A = {y1, . . . , yk}, then F(A; λ) = FA(A; λ) = Fy1(A; λ) Fy2(A1; λ) · · · Fyk(Ak−1; λ), (9.3) where Ai = A \ {y1, . . . , yi}. More generally, if V = {y1, . . . , yj} ⊂A then FV (A; λ) = Fy1(A; λ) Fy2(A1; λ) · · · Fyj(Aj−1; λ).
(9.4) In particular, the products on the right-hand side of (9.3) and (9.4) are independent of the ordering of the vertices.
Proof This follows from the definition and the observation that the collection of loops that intersect V can be partitioned into those that intersect y1, those that do not intersect y1 but intersect y2, etc.
The next lemma is an important relationship between one generating function and the exponential of another generating function.
Lemma 9.3.2 Suppose x ∈A ⊂X. Let gA(λ; x) = X ω∈L(A),ω0=x q(ω) λ|ω|.
Then, Fx(A; λ) = gA(λ; x).
Remark.
If λ = 1 and q is a transition probability, then gA(1; x) (and hence by the lemma Fx(A) = Fx(A; 1)) is the expected number of visits to x by a random walk starting at x before its first visit to X \ A. In other words, Fx(A)−1 is the probability that a random walk starting at x reaches X \ A before its first return to x. Using this interpretation for Fx(A), the fact that the product on the right-hand side of (9.3) is independent of the ordering is not so obvious. See Exercise 9.2 for a more direct proof in this case.
Proof Suppose ω ∈L x(A). Let dx(ω) be the number of times that a representative ω of ω visits x (this is the same for all representatives ω). For representatives ω with ω0 = x, d(ω) = dx(ω). It is easy to verify that the number of representatives ω of ω with ω0 = x is K(ω)dx(ω)/|ω|. From this we see that m(ω) = X ω∼ω q(ω) |ω| = K(ω) q(ω) |ω| = X ω∼ω,ω0=x q(ω) d(ω).
204 Loop Measures Therefore, X ω∈Lx(A) m(ω) λ|ω| = X ω∈L(A),ω0=x q(ω) λ|ω| d(ω) = ∞ X j=1 1 j X ω∈L(A),ω0=x,d(ω)=j q(ω) λ|ω|.
An x-cycle ω with dx(ω) = j can be considered as a concatenation of j x-cycles ω′ with d(ω′) = 1.
Using this we can see that X ω∈L(A),ω0=x,d(ω)=j q(ω) λ|ω| = X ω∈L(A),ω0=x,d(ω)=1 q(ω) λ|ω| j = fA(λ; x)j.
Therefore, log Fx(A; λ) = ∞ X j=1 fA(x; λ)j j = −log[1 −fA(λ; x)] = log gA(λ; x).
The last equality uses (9.2).
Proposition 9.3.3 Suppose #(X) = n < ∞and λ > 0 satisfies F(X; λ) < ∞. Then F(X; λ) = 1 det[I −λQ], where Q denotes the n × n matrix [q(x, y)]x,y∈X .
Proof Without loss of generality we may assume λ = 1. We prove by induction on n. If n = 1 and Q = (r), then F(X; 1) = 1/(1 −r). To do the inductive step, suppose n > 1 and x ∈X, then g(1; x) = ∞ X j=0 Qj x,x = (I −Q)−1 x,x = det[I −Qx] det[I −Q] , where Qx denotes the matrix Q with the row and column corresponding to x removed. The last equality follows from the adjoint form of the inverse. Using (9.3) and the inductive hypothesis on X \ {x}, we get the result.
Remark. The matrix I −λQ is often called the (negative of the) Laplacian. The last proposition and others below relate the determinant of the Laplacian to loop measures and trees.
• Let λ0,x denote the radius of convergence of g(λ; x).
– If X is q-connected, then λ0,x is independent of x and we write just λ0.
– If X is q-connected and finite, then λ0 is also the radius of convergence for g(λ) and F(X; λ) and 1/λ0 is the largest eigenvalue for the matrix Q = (q(x, y)). If q is a transition probability, λ0 = 1. If q is an irreducible, strictly subMarkov transition probability, then λ0 > 1.
If X is q-connected and finite, then g(λ0) = g(λ0; x) = F(X; λ0) = ∞. However, one can show easily that F(X \ {x}; λ0) < ∞. The next proposition shows how to compute the last quantity from the generating functions.
9.3 Generating functions and loop measures 205 Proposition 9.3.4 Let λ0 be the radius of convergence of g. Then if x ∈X, log F(X \ {x}; λ0) = lim λ→λ0−[log F(X; λ) −log g(λ; x)].
Proof log F(X \ {x}; λ0) = lim λ→λ0−log F(X \ {x}; λ) = lim λ→λ0−[log F(X; λ) −log Fx(X; λ)].
If #(X) < ∞and a subset E of edges is given, then simple random walk on the graph (X, E) is the Markov chain corresponding to q(x, y) = [#{z : (x, z) ∈E}]−1 , (x, y) ∈E.
If #(X) = n < ∞and (X, E) is transitive, then g(λ; x) = n−1 g(λ), and hence we can write Proposition 9.3.4 as log F(X \ {x}, λ0) = log n + lim λ→λ0−[Φ(λ) −log g(λ)].
Proposition 9.3.5 Suppose X is finite and q is an irreducible, transition probability, reversible with respect to the invariant probability π. Let α1 = 1, α2, . . . , αn denote the eigenvalues of Q = [q(x, y)].
Then for every x ∈X, 1 F(X \ {x}) = π(x) n Y j=2 (1 −αj) Proof Since the eigenvalues of I −λQ are 1 −λ α1, . . . , 1 −λ αn, we see that lim λ→1− det[I −λQ] 1 −λ = n Y j=2 (1 −αj).
If λ < 1, then Proposition 9.3.3 states F(X; λ) = 1 det[I −λQ].
Also, as λ →1−, g(λ; x) ∼π(x) (1 −λ)−1, where π denotes the invariant probability. (This can be seen by recalling that g(λ; x) is the number of visits to x by a chain starting at x before a geometric killing time with rate (1−λ). The expected number of steps before killing is 1/(1 −λ), and since the killing is independent of the chain, the 206 Loop Measures expected number of visits to x before begin killed is asymptotic to π(x)/(1 −λ).) Therefore, using Proposition 9.3.4, log F(X \ {x}) = lim λ→1−[log F(X; λ) −log g(λ; x)] = −log π(x) − n X j=2 log(1 −αj).
9.4 Loop soup • If V is a countable set and ν : V →[0, ∞) is a measure, then a (Poisson) soup from ν is a collection of independent Poisson processes N x t , x ∈V, where N x t has parameter ν(x). A soup realization is the corresponding collection of multi-sets† At where the number of times that x appears in At is N x t . This can be considered as a stochastic process taking values in multi-sets of elements of V .
• The rooted loop soup Ct is a soup realization from m.
• The unrooted loop soup Ct is a soup realization from m.
From the definitions of m and m we can see that we can obtain an unrooted loop soup Ct from a rooted loop soup Ct by “forgetting the roots” of the loops in Ct. To obtain Ct from Ct, we need to add some randomness. More specifically, if ω is a loop in an unrooted loop soup Ct, we choose a rooted loop ω by choosing uniformly among the K(ω) representatives of ω. It is not hard to show that with probability one, for each t, there is at most one loop in Ct \ "[ s<t Cs # .
Hence we can order the loops in Ct (or Ct) according to the “time” at which they were created; we call this the chronological order.
Proposition 9.4.1 Suppose x ∈A ⊂X with Fx(A) < ∞. Let Ct(A; x) denote an unrooted loop soup Ct restricted to ¯ Lx(A). Then with probability one, C1(A; x) contains a finite number of loops which we can write in chronological order ω1, . . . , ωk.
Suppose that independently for each unrooted loop ωj, a rooted loop ωj with ω0 = x is chosen uni-formly among the K(ω) dx(ω)/|ω| representatives of ω rooted at x, and these loops are concatenated to form a single loop η = ω1 ⊕ω2 ⊕· · · ⊕ωk.
† A multi-set is a generalization of a set where elements can appear multiple times in the collection.
9.5 Loop erasure 207 Then for any loop η′ ⊂A rooted at x, P{η = η′} = q(η) Fx(A).
Proof We first note that ω1, . . . , ωk as given above is the realization of the loops in a Poissonian realization corresponding to the measure m∗ x(ω) = q(ω) dx(ω), up to time 1 listed in chronological order, restricted to loops ω ∈L(A) with ω0 = x. Using the argument of Lemma 9.3.2, the probability that no loop appears is exp − X ω∈L(A);ω0=x m∗ x(ω) = exp − X ω∈L(A);ω0=x q(ω) dx(ω) = 1 Fx(A).
More generally, suppose η′ ∈L(A) is given with η′ 0 = x and d(η′) = k. For any choice of positive j1, . . . , jr integers summing to k, we have a decomposition of η′, η′ = ω1 ⊕· · · ⊕ωr where ωi is a loop rooted at x with dx(ωi) = ji. The probability that ω1, . . . , ωr (and no other loops) appear in the realization up to time 1 in this order is exp n −P ω∈L(A);ω0=x m∗ x(ω) o r!
m∗ x(ω1) · · · m∗ x(ωr) = 1 r! Fx(A) q(ω1) · · · q(ωr) j1 · · · jr = q(η′) Fx(A) 1 r! (j1 · · · jr).
The proposition then follows from the following combinatorial fact that we leave as Exercise 9.1: X j1+···+jr=k 1 r! (j1 · · · jr) = 1.
9.5 Loop erasure • A path ω = [ω0, . . . , ωn] is self-avoiding if ωj ̸= ωk for 0 ≤j < k ≤n.
Given a path ω = [ω0, . . . , ωn] there are a number of ways to obtain a self-avoiding subpath of ω that goes from ω0 to ωn. The next definition gives one way.
• If ω = [ω0, . . . , ωn] is a path, LE(ω) denotes its (chronological) loop-erasure defined as follows.
– Let σ0 = max{j ≤n : ωj = 0}. Set η0 = ω0 = ωσ0.
– Suppose σi < n. Let σi+1 = max{j ≤n : ωj = ωσi+1}. Set ηi+1 = ωσi+1 = ωσi+1.
– If iω = min{i : σi = n} = min{i : ηi = ωn}, then LE(ω) = [η0, . . . , ηi].
208 Loop Measures A weight q on paths induces a new weight ˆ qA on self-avoiding paths by specifying that the weight of a self-avoiding path η is the sum of the weights of all the paths ω in A for which η = L(ω). The next proposition describes this weight.
Proposition 9.5.1 Suppose A ⊂X, i ≥1 and η = [η0, . . . , ηi] is a self-avoiding path whose vertices are in A. Then, ˆ qA(η) := X ω∈L(A);LE(ω)=η q(ω) = q(η) Fη(A), (9.5) where, as before, Fη(A) = exp X ω∈L(A),|ω|≥1,η∩ω̸=∅ q(ω) |ω| .
Proof Let A−1 = A, Aj = A \ {η0, . . . , ηj}. Given any ω with LE(ω) = η, we can decompose ω as ω = ω0 ⊕[η0, η1] ⊕ω1 ⊕[η1, η2] ⊕· · · ⊕[ηi−1, ηi] ⊕ωi, where ωj denotes the loop [ωσj−1+1, . . . , ωσj] (here σ−1 = −1). The loop ωj can be any loop rooted at ηj contained in Aj−1. The total measure of such loops is Fηj(Aj−1), see Lemma 9.3.2. The result then follows from (9.4).
In particular, ˆ qA(η) depends on A. The next proposition discusses the “Radon-Nikodym deriva-tive” of ˆ qA1 with respect to ˆ qA for A1 ⊂A.
• If V1, V2 ⊂A, Let FV1,V2(A) = exp X ω∈L(A),ω∩V1̸=∅,ω∩V2̸=∅ q(ω) |ω| .
Proposition 9.5.2 Suppose A1 ⊂A and η = [η0, . . . , ηi] is a self-avoiding path whose vertices are in A1. Then ˆ qA(η) = ˆ qA1(η) Fη,A\A1(A).
Proof This follows immediately from the relation Fη(A) = Fη(A1) Fη,A\A1(A).
The “inverse” of loop erasing is loop addition. Suppose η = [η0, . . . , ηk] is a self-avoiding path.
We define a random variable Zη taking values in the set of paths ω with LE(ω) = η as follow. Let Ct be a realization of the unrooted loop soup in A as in Proposition 9.4.1. For each 0 ≤j ≤k, let ω1,j, ω2,j, . . . , ωsj,j, denote the loops in C1 that intersect ηj but do not intersect {η0, . . . , ηj−1}. These loops are listed in the order that they appear in the soup. For each such loop ωi,j, choose a representative ωi,j 9.6 Boundary excursions 209 roooted at ηj; if there is more than one choice for the representative, choose it uniformly. We then concatenate these loops to give ˜ ωj = ω1,j ⊕· · · ⊕ωsj,j.
If sj = 0, define ˜ ωj to be the trivial loop [ηj]. We then concatenate again to define ω = Z(η) = ˜ ω0 ⊕[η0, η1] ⊕˜ ω1 ⊕[η1, η2] ⊕· · · ⊕[ηk−1, ηk] ⊕˜ ωk.
Proposition 9.4.1 tells us that there is another way to construct a random variable with the distribu-tion of Zη. Suppose ˜ ω0, . . . , ˜ ωk are chosen independently (given η) with ˜ ωj having the distribution of a cycle in A \ {η0, . . . , ηj−1} rooted at ηj. In other words if ω′ ∈L(A \ {η0, . . . , ηj−1}) with ω′ 0 = ηj, then the probability that ˜ ωj = ω′ is q(ω′)/Fηj(A \ {η0, . . . , ηj−1}).
9.6 Boundary excursions Boundary excursions in a set A are paths that begin and end on the boundary and otherwise stay in A. Suppose X, q are given. If A ⊂X we define ∂A = (∂A)q = {y ∈X \ A : q(x, y) + q(y, x) > 0 for some x ∈A}.
• A (boundary) excursion in A is a path ω = [ω0, . . . , ωn] with n ≥2 such that ω0, ωn ∈∂A and ω1, . . . , ωn−1 ∈A.
• The set of boundary excursions with ω0 = x and ω|ω| = y is denoted EA(x, y), and EA = [ x.y∈∂A EA(x, y).
• Let ˆ EA(x, y), ˆ EA denote the subsets of EA(x, y), EA, respectively, consisting of the self-avoiding paths. If x = y, the set ˆ EA(x, y) is empty.
• The measure q restricted to EA is called excursion measure on A.
• The measure q restricted to ˆ EA is called the self-avoiding excursion measure on A.
• The loop-erased excursion measure on A, is the measure on ˆ EA given by ˆ q(η) = q{ω ∈EA : LE(ω) = η}.
As in (9.5), we can see that ˆ q(η) = q(η) Fη(A).
(9.6) • If x, y ∈∂A, q, ˆ q can also be considered as measures on EA(x, y) or ˆ EA(x, y) by restricting to those paths that begin at x and end at y. If x = y, these measures are trivial for the self-avoiding and loop-erased excursion measures.
♣If µ is a measure on a set K and K1 ⊂K, then the restriction of µ to K1 is the measure ν defined by ν(V ) = µ(V ∩K1). If µ is a probability measure, this is related to but not the same as the conditional measure given K1; the conditional measure normalizes to make the measure a probability measure. A family of measures µA, indexed by subsets A, supported on EA (or EA(x, y)) is said to have the restriction property if whenever A1 ⊂A, then µA1 is µA restricted to EA1 (EA1(x, y)). The excursion measure and the self-avoiding excursion 210 Loop Measures measure have the restriction property. However, the loop-erased excursion measure does not have the restriction property. This can be seen from (9.6) since it is possible that Fη(A1) ̸= Fη(A).
The loop-erased excursion measure ˆ q is obtained from the excursion measure q by a deterministic function on paths (loop erasure). Since this function is not one-to-one, we cannot obtain q from ˆ q without adding some extra randomness. However, one can obtain q from ˆ q by adding random loops as described at the end of Section 9.5.
The next definition is a generalization of the boundary Poisson kernel defined in Section 6.7.
• The boundary Poisson kernel is the function H∂A : ∂A × ∂A →[0, ∞) given by H∂A(x, y) = X ω∈EA(x,y) q(ω).
Note that if ω ∈EA(x, y), then LE(ω) ∈ˆ EA(x, y). In particular, if x ̸= y, H∂A(x, y) = X η∈ˆ EA(x,y) ˆ q(η).
Suppose k is a positive integer and x1, . . . , xk, y1, . . . , yk are distinct points in ∂A. We write x = (x1, . . . , xk), y = (y1, . . . , yk). We let EA(x, y) = EA(x1, y1) × · · · × EA(xk, yk), and we write [ω] = (ω1, . . . , ωk) for an element of EA(x, y) and q([ω]) = (q × · · · × q)([ω]) = q(ω1) q(ω2) · · · q(ωk).
We can consider q × · · · × q as a measure on EA(x, y). We define ˆ EA(x, y) similarly.
• The nonintersecting excursion measure qA(x, y) at (x, y) is the restriction of the measure q×· · ·×q to the set of [ω] ∈EA(x, y) that do not intersect, i.e., ωi ∩ωj = ∅, 1 ≤i < j ≤k.
• The nonintersecting self-avoiding excursion measure at (x, y) is the restriction of the measure q × · · · × q to the set of [ω] ∈ˆ EA(x, y) that do not intersect. Equivalently, it is the restriction of the nonintersecting excursion measure to ˆ EA(x, y).
There are several ways to define the nonintersecting loop-erased excursion measure. It turns out that the most obvious way (restricting the loop-erased excursion measure to k-tuples of walks that do not intersect) is neither the most important nor the most natural. To motivate our definition, let us consider the nonintersecting excursion measure with k = 2. This is the measure on pairs of excursions (ω1, ω2). that gives measure q(ω1) q(ω2) to each (ω1, ω2) satisfying ω1∩ω2 = ∅. Another way of saying this is the following.
• Given ω1, the measure on ω2 is q restricted to those excursions ω ∈EA(x2, y2) such that ω∩ω1 = ∅.
In other words, the measure is q restricted to EA\ω1(x2, y2).
More generally, if k ≥2 and 1 ≤j ≤k −1, the following holds.
• Given ω1, . . . , ωj, the measure on ωj+1 is q restricted to excursions in EA(xj+1, yj+1) that do not intersect ω1 ∪· · · ∪ωj. In other words, the measure is q restricted to EA(ω1∪···∪ωj)(xj+1, yj+1).
9.6 Boundary excursions 211 The nonintersecting self-avoiding excursion measure satisfies the analogous property. We will use this as the basis for our definition of the nonintersecting loop-erased measure ˆ qA(x, y) at (x, y). We want our definition to satisfy the following.
• Given η1, . . . , ηj, the measure on ηj+1 is the same as ˆ qA(η1∪···∪ηj)(xj+1, yj+1).
This leads to the following definition.
• The measure ˆ qA(x, y) is the measure on ˆ EA(x, y) obtained by restricting qA(x, y) to the set V of k-tuples [ω] ∈EA(x, y) that satisfy ωj+1 ∩[η1 ∪· · · ∪ηj] = ∅, j = 1, . . . , k −1, (9.7) where ηj = LE(ωj), and then considering it as a measure on the loop erasures. In other words, ˆ qA(η1, . . . , ηk) = q{(ω1, . . . , ωk) ∈V : LE(ωj) = ηj, j = 1, . . . , k, satisfying (9.7)}.
This definition may look unnatural because it seems that it might depend on the order of the pairs of vertices. However, the next proposition shows that this is not the case.
Proposition 9.6.1 The ˆ qA(x, y)-measure of a k-tuple (η1, . . . , ηk) is k Y j=1 ˆ qA(ηj) 1{ηi ∩ηj ̸= ∅, 1 ≤i < j ≤n}Fη1,...,ηk(A)−1, where Fη1,...,ηk(A) = exp X ω∈L(A) q(ω) |ω| J(ω; η1, . . . , ηk) , and J(ω; η1, . . . , ηk) = max{0, s −1}, where s is the number of paths η1, . . . , ηk intersected by ω.
Proof Proposition 9.5.1 implies k Y j=1 ˆ qA(ηj) = k Y j=1 q(ηj) k Y j=1 exp X ω∈L(A),|ω|≥1,ω∩ηj̸=∅ q(ω) |ω| .
(9.8) However, assuming that ηi ∩ηj = ∅for i ̸= j, ˆ qA(η1, . . . , ηj) = k Y j=1 q(ηj) k Y j=1 exp X ω∈L(A(η1∪···ηj−1),|ω|≥1,ω∩ηj̸=∅ q(ω) |ω| .
(9.9) If a loop ω intersects s of the ηj, where s ≥2, then it appears s times in (9.8) but only one time in (9.9).
• Let ˆ H∂A(x, y) denote the total mass of the measure ˆ qA(x, y).
212 Loop Measures If k = 1, we know that ˆ H∂A(x, y) = H∂A(x, y). The next proposition shows that for k > 1, we can describe ˆ H∂A(x, y) in terms of the quantities H∂A(xi, yj). The identity is a generalization of a result of Karlin and McGregor on Markov chains (see Exercise 9.3). If π is a permutation of {1, . . . , k}, we also write π(y) for (yπ(1), . . . , yπ(k)).
Proposition 9.6.2 (Fomin’s identity) X π (−1)sgnπ ˆ H∂A(x, π(y)) = det [H∂A(xi, yj)]1≤i,j≤k .
(9.10) Remark. If A is a simply connected subset of Z2 and q comes from simple random walk, then topological considerations tell us that ˆ H∂A(x, π(y)) is nonzero for at most one permutation π. If we order the vertices so that this permutation is the identity, Fomin’s identity becomes ˆ H∂A(x, y) = det [H∂A(xi, yj)]1≤i,j≤k .
Proof We will say that [ω] is nonintersecting if (9.7) holds and otherwise we call it intersecting.
Let E∗= [ π EA(x, π(y)), (9.11) let E∗ NI be the set of nonintersecting [ω] ∈E∗, and let E∗ I = E∗\ E∗ NI be the set of intersecting [ω].
We will define a function φ : E∗→E∗with the following properties.
• φ is the identity on E∗ NI.
• q([ω]) = q(φ([ω])).
• If [ω] ∈E∗ I ∩EA(x, π(y)), then φ([ω]) ∈EA(x, π1(y)) where sgnπ1 = −sgnπ. In fact, π1 is the composition of π and a transposition.
• φ ◦φ is the identity. In particular, φ is a bijection.
To show that existence of such a φ proves the proposition, first note that det [H∂A(xi, yj)]1≤i,j≤k = X π (−1)sgnπ k Y i=1 H∂A(xi, yπ(i)).
Also, H∂A(xi, yπ(i)) = X ω∈EA(xi,yπ(i)) q(ω).
Therefore, by expanding the product, we have det [H∂A(xi, yj)]1≤i,j≤k = X [ω]∈E∗ (−1)sgnπ q([ω]) = X [ω]∈E∗ (−1)sgnπ1 q([φ(ω)]).
In the first summation the permutation π is as in (9.11). Hence the sum of all the terms that come from ω ∈E∗ I is zero, and det [H∂A(xi, yj)]1≤i,j≤k = X [ω]∈E∗ NI (−1)sgnπ q([ω]).
9.6 Boundary excursions 213 But the right-hand side is the same as the left-hand side of (9.10). We define φ to be the identity on E∗ NI, and we now proceed to define the bijection φ on E∗ I .
Let us first consider the k = 2 case. Let [ω] ∈E∗ I , ξ = ω1 = [ξ0, . . . , ξm] ∈EA(x1, y), ω = ω2 = [ω0, . . . , ωn] ∈EA(x2, y′), η = [η0, . . . , ηl] = LE(ξ), where y = y1, y′ = y2 or y = y2, y′ = y1. Since [ω] ∈E∗ I , we know that η ∩ω = L(ξ) ∩ω ̸= ∅.
Define s = min{l : ηl ∈ω}, t = max{l : ξl = ηs}, u = max{l : ωl = ηs}.
Then we can write ξ = ξ−⊕ξ+, ω = ω−⊕ω+ where ξ−= [ξ0, . . . , ξt], ξ+ = [ξt, . . . , ξm], ω−= [ω0, . . . , ωu], ω+ = [ωu, . . . , ωn].
We define φ([ω]) = φ((ξ−⊕ξ+, ω−⊕ω+)) = (ξ−⊕ω+, ω−⊕ξ+).
Note that ξ−⊕ω+ ∈EA(x1, y′), ω−⊕ξ+ ∈EA(x2, y), and q(φ([ω])) = q([ω]). A straightforward check shows that φ ◦φ is the identity.
x1 x2 y y′ ξ− ξ+ ω− ω+ ξt = ηs = wu Figure 0-a 214 Loop Measures Suppose k > 2 and [ω] ∈E∗ I . We will change two paths as in the k = 2 case and leave the others fixed being careful in our choice of the paths to make sure that φ(φ([ω])) = [ω]. Let ηi = LE(ωi).
We define r = min{i : ηi ∩ωj ̸= ∅for some j > i}, s = min{l : ηr l ∈ωi+1 ∪· · · ∪ωk}, b = min{j > r : ηr s ∈ωj}, t = max{l : ωr l = ηr s}, u = max{l : ωb l = ηr s}.
We make the interchange (ωr,−⊕ωr,+, ωb,−⊕ωb,+) ← →(ωr,−⊕ωb,+, ωb,−⊕ωr,+) as in the previous paragraph (with (ωr, ωb) = (ξ, ω)) leaving the other paths fixed. This defines φ, and it is then straightforward to check that φ ◦φ is the identity.
9.7 Wilson’s algorithm and spanning trees Kirchhoffwas the first to relate the number of spanning trees of a graph to a determinant. Here we derive a number of these results. We use a more recent technique, Wilson’s algorithm, to establish the results.
This algorithm is an efficient method to produce spanning trees from the uniform distribution using loop-erased random walk. We describe it in the proof of the next proposition.
The basic reason why this algorithm works is that the product on the right-hand side of (9.3) is independent of the ordering of the vertices.
Proposition 9.7.1 Suppose #(X) = n < ∞and q are transition probabilities for an irreducible Markov chain on X. Then X T q(T ; x0) = 1 F(X \ {x0}).
(9.12) Proof We will describe an algorithm due to David Wilson that chooses a spanning tree at random.
Let X = {x0, . . . , xn−1}.
• Start the Markov chain at x1 and let it run until it reaches x0. Take the loop-erasure of the set of points visited, [η0 = x1, η1, . . . , ηi = x0]. Add the edges [η0, η1], [η1, η2], . . . , [ηi−1, ηi] to the tree.
• If the edges form a spanning tree we stop. Otherwise, we let j be the smallest index such that xj is not a vertex in the tree. Start a random walk at xj and let it run until it reaches one of the vertices that has already been added. Perform loop-erasure on this path and add the edges in the loop-erasure to the tree.
• Continue until all vertices have been added to the tree.
We claim that for any tree T , the probability that T is output in this algorithm is q(T ; x0) F(X \ {x0}).
(9.13) 9.7 Wilson’s algorithm and spanning trees 215 The result (9.12) follows immediately. To prove (9.13), suppose that a spanning tree T is given.
Then this gives a collection of self-avoiding paths: η1 = [y1,1 = x1, y1,2, . . . , y1,k1, z1 = x0] η2 = [y2,1, y2,2, . . . , y2,k2, z2] .
.
.
ηm = [ym,1, ym,2, . . . , ym,km, zm] .
Here η1 is the unique self-avoiding path in the tree from x1 to x0; for j > 1, yj,1 is the vertex of smallest index (using the ordering x0, x1, . . . , xn−1) that has not been listed so far; and ηj is the unique self-avoiding path from yj,1 to a vertex zj in η1 ∪· · · ∪ηj−1. Then the probability that T is chosen is exactly the product of the probabilities that • if a random walk starting at x1 is stopped at x0, the loop-erasure is η1; • if a random walk starting at y2,1 is stopped at η1, then the loop-erasure is η2 .
.
.
• if a random walk starting at ym,1 is stopped at η1 ∪· · · ∪ηm−1, then the loop-erasure is ηm.
With this decomposition, we can now use (9.5) and (9.3), we obtain (9.13).
Corollary 9.7.2 If Cn denotes the number of spanning trees of a connected graph with vertices {x0, x1, . . . , xn−1}, then log Cn = n−1 X j=1 log d(xj) −log F(X \ {x0}) = n−1 X j=1 log d(xj) + lim λ→1−[log g(λ; x0) −Φ(λ)].
Here the implicit q is the transition probability for simple random walk on the graph and d(xj) denotes the degree of xj. If Cn is a transitive graph of degree d, log Cn = (n −1) log d −log n + lim λ→1−[log g(λ) −Φ(λ)].
(9.14) Proof For simple random walk on the graph, for all T , q(T ; x0) = n−1 Y j=1 d(xj) −1 .
In particular, it is the same for all trees, and (9.12) implies that the number of spanning trees is [q(T ; x0) F(X \ {x0})]−1 = n−1 Y j=1 d(xj) F(X \ {x0})−1.
216 Loop Measures The second equality follows from Proposition 9.3.4 and the relation Φ(λ) = F(x; λ). If the graph is transitive, then g(λ) = n g(λ; x0), from which (9.14) follows.
♣If we take a connected graph and add any number of self-loops at vertices, this does not change the number of spanning trees. The last corollary holds regardless of how many self-loops are added. Note that adding self-loops affects both the value of the degree and the value of F(X \ {x0}).
Proposition 9.7.3 Suppose X is a finite, connected graph with n vertices and maximal degree d, and P is the transition matrix for the lazy random walk on X as in Section 9.2.1. Suppose the eigenvalues of P are α1 = 1, α2, . . . , αn.
Then the number of spanning trees of X is dn−1 n−1 n Y j=2 (1 −αj).
Proof Since the invariant probability is π ≡1/n, Proposition 9.3.5 tells us that for each x ∈X, 1 F(X \ {x}) = n−1 n Y j=2 (1 −αj).
♣The values 1 −αj are the eigenvalues for the (negative of the) Laplacian I −Q for simple random walk on the graph. In graph theory, it is more common to define the Laplacian to be ±d(I −Q). When looking at formulas, it is important to know which definition of the Laplacian is being used.
9.8 Examples 9.8.1 Complete graph The complete graph on a collection of vertices is the graph with all (distinct) vertices adjacent.
Proposition 9.8.1 The number of spanning trees of the complete graph on X = {x0, . . . , xn−1} is nn−2.
Proof Consider the Markov chain with transition probabilities q(x, y) = 1/n for all x, y.
Let Aj = {xj, . . . , xn−1}. The probability that the chain starting at xj has its first visit (after time zero) to {x0, . . . , xj} at xj is 1/(j + 1) since each vertex is equally likely to be the first one visited.
Using the interpretation of Fxj(Aj) as the reciprocal of the probability that the chain starting at xj visits {x0, . . . , xj−1} before returning to xj we see that Fxj(Aj) = j + 1 j , j = 1, . . . , n −1 9.8 Examples 217 and hence (9.3) gives F(X \ {x0}) = n.
With the self-loops, each vertex has degree n and hence for each spanning tree q(T ; x0) = n−(n−1).
Therefore the number of spanning trees is [q(T ; x0) F(X \ {x0})]−1 = nn−2.
9.8.2 Hypercube The hypercube Xn is the graph whose vertices are {0, 1}n with vertices adjacent if they agree in all but one component.
Proposition 9.8.2 If Cn denotes the number of spanning trees of the hypercube Xn := {0, 1}n, then log Cn := (2n −n −1) log 2 + n X k=1 n k log k.
By (9.14), Proposition 9.8.2 is equivalent to lim λ→1−[log g(λ) −Φ(λ)] = −(2n −1) log n + (2n −1) log 2 + n X k=1 n k log k.
where g is the cycle generating function for simple random walk on Xn. The next proposition computes g.
Proposition 9.8.3 Let g be the cycle generating function for simple random walk on the hypercube Xn. Then g(λ) = n X j=0 n j n n −λ(n −2j) = 2n + n X j=0 n j λ(n −2j) n −λ(n −2j).
Proof [of Proposition 9.8.2 given Proposition 9.8.3] Note that Φ(λ) = Z λ 0 g(s) −2n s ds = Z λ 0 n X j=0 n j n −2j n −s(n −2j) ds = (2n −1) log n −log(1 −λ) − n X j=1 n j log[n −λ(n −2j)].
218 Loop Measures Let us write λ = 1 −ǫ so that log g(λ) = log n X j=0 n j n ǫ n + 2j(1 −ǫ) , Φ(λ) = (2n −1) log n −log ǫ − n X j=1 n j log[ǫn + (1 −ǫ) 2j].
As ǫ →0+, log g(λ) = log(1/ǫ) + log n X j=0 n j n n + 2j 1−ǫ ǫ = −log(ǫ) + o(1), Φ(λ) = (2n −1) log n −log ǫ − n X j=1 n j log(2j) + o(1) and hence lim λ→1−[log g(λ) −Φ(λ)] = (1 −2n) log n + (2n −1) log 2 + n X j=1 n j log j, which is what we needed to show.
The remainder of this subsection will be devoted to proving Proposition 9.8.3.
Let L(n, 2k) denote the number of cycles of length 2k in Xn. By definition L(n, 0) = 2n. Let gn denote the generating function on Xn using weights 1 (instead of 1/n) on the edges on the graph and zero otherwise, gn(λ) = X ω λ−|ω| = ∞ X k=0 L(n, 2k) λ2k.
Then if g is as in Proposition 9.8.3, g(λ) = gn(λ/n). Then Proposition 9.8.3 is equivalent to gn(λ) = n X j=0 n j 1 1 −λ(n −2j), (9.15) which is what we will prove. By convention we set L(0, 0) = 1; L(0, k) = 0 for k > 0, and hence g0(λ) = ∞ X k=0 L(0, 2k) λ2k = 1, which is consistent with (9.15).
Lemma 9.8.4 If n, k ≥0, L(n + 1, 2k) = 2 k X j=0 2k 2j L(n, 2j).
9.8 Examples 219 Proof This is immediate for n = 0 since L(1, 2k) = 2 for every k ≥0. For n ≥1, consider any cycle in Xn+1 of length 2k. Assume that there are 2j steps that change one of the first n components and 2(k −j) that change the last component. There are 2k 2j ways to choose which 2j steps make changes in the first n components. Given this choice, there are L(n, 2j) ways of moving in the first n components. The movement in the last component is determined once the initial value of the (n + 1)-component is chosen; the 2 represents the fact that this initial value can equal 0 or 1.
Lemma 9.8.5 For all n ≥0, gn+1(λ) = 1 1 −λ gn λ 1 −λ + 1 1 + λ gn λ 1 + λ .
Proof gn+1(λ) = ∞ X k=0 L(n + 1, 2k) λ2k = 2 ∞ X k=0 k X j=0 2k 2j L(n, 2j) λ2k = 2 ∞ X j=0 L(n, 2j) ∞ X k=0 2j + 2k 2j λ2j+2k = 2 1 −λ ∞ X j=0 L(n, 2j) λ 1 −λ 2j ∞ X k=0 2j + 2k 2j (1 −λ)2j+1 λ2k Using the identity (see Exercise 9.4).
∞ X k=0 2j + 2k 2j p2j+1 (1 −p)2k = 1 2 + 1 2 p 2 −p 2j+1 , (9.16) we see that gn+1(λ) equals 1 1 −λ ∞ X j=0 L(n, 2j) λ 1 −λ 2j + 1 1 + λ ∞ X j=0 L(n, 2j) λ 1 + λ 2j , which gives the result.
Proof [Proof of Proposition 9.8.3] Setting λ = (n + β)−1, we see that it suffices to show that gn 1 n + β = n X j=0 n j β + n β + 2j .
(9.17) This clearly holds for n = 0. Let Hn(λ) = λ gn(λ). Then the previous lemma gives the recursion relation Hn+1 1 n + 1 + β = Hn 1 n + β + Hn 1 n + 2 + β .
220 Loop Measures Hence by induction we see that Hn 1 n + β = n X j=0 n j 1 β + 2j .
9.8.3 Sierpinski graphs In this subsection we consider the Sierpinski graphs which is a sequence of graphs V0, V1, . . . defined as follows. V0 is a triangle, i.e., a complete graph on three vertices. For n > 0, Vn will be a graph with 3 vertices of degree 2 (which we call the corner vertices) and [3n+1 −3]/2 vertices of degree 4. We define the graph inductively. Suppose we are given three copies of Vn−1, V (1) n−1, V (2) n−1, V (3) n−1, with corner vertices x(1) 1 , x(1) 2 , x(1) 3 , . . . , x(3) 1 , x(3) 2 , x(3) 3 . Then Vn is obtained from these three copies by identifying the vertex x(k) j with the vertex x(j) k . We call the graphs Vn the Sierpinski graphs.
Proposition 9.8.6 Let Cn denote the number of spanning trees of the Sierpinski graph Vn. Then Cn satisfies the recursive equation Cn+1 = 2 (5/3)n C3 n.
(9.18) Hence, Cn = (3/20)1/4 (3/5)n/2 (540)3n/4.
(9.19) Proof It is clear that C0 = 3, and a simple induction argument shows that the solution to (9.18) with C0 = 3 is given by (9.19). Hence we need to show the recursive equation Cn+1 = 2 (5/3)n C3 n.
x0 x1 x2 x3 x4 x5 For n ≥1, we will write Vn = {x0, x1, x2, x3, x4, x5, . . . , xMn} where Mn = [3n + 3]/2, x0, x1, x2 are corner vertices of Vn and x3, x4, x5 are the other vertices that are corner vertices for the three copies of Vn−1. They are chosen so that x3 lies between x0, x1; x4 between x1, x2; x5 between x2, x0.
9.9 Spanning trees of subsets of Z2 221 Using Corollary 9.7.2 and Lemma 9.3.2, we can write Cn = Ψn Jn where Ψn = Mn Y j=1 d(xj), Jn = Mn Y j=1 pj,n.
Here pj,n denotes the probability that simple random walk in Vn started at xj returns to xj before visiting {x0, . . . , xj−1}. Note that Ψn = 22 4(3n+1−3)/2, and hence Ψn+1 = 4Ψ3 n. Hence we need to show that Jn+1 = (1/2) (5/3)n J3 n.
(9.20) We can write Jn+1 = p1,n+1 p2,n+1 J∗ n+1 where J∗ n+1 denotes the product over all the other vertices (the non-corner vertices). From this, we see that Jn+1 = p1,n+1 p2,n+1 · · · p5,n+1 (J∗ n)3 = p1,n+1 p2,n+1 · · · p5,n+1 p3 1,n p3 2,n J3 n.
The computations of pj,n are straightforward computations familiar to those who study random walks on the Sierpinski gasket and are easy exercises in Markov chains. We give the answers here, leaving the details to the reader. By induction on n one can show that p2,n+1 = (3/5) p2,n and from this one can see that p2,n+1 = 3 5 n+1 , p1,n = 3 4 3 5 n+1 .
Also, p5,n+1 = p2,n = 3 5 n , p4,n+1 = 15 16 3 5 n , p3,n+1 = 5 6 3 5 n .
This gives (9.20).
9.9 Spanning trees of subsets of Z2 Suppose A ⊂Z2 is finite, and let e(A) denote the set of edges with at least one vertex in A. We write e(A) = ∂eA ∪eo(A) where ∂eA denotes the “boundary edges” with one vertex in ∂A and eo(A) = e(A) \ ∂eA, the “interior edges”. There will be two types of spanning trees of A, we will consider.
• Free. A collection of #(A)−1 edges from e0(A) such that the corresponding graph is connected.
• Wired. The set of vertices is A ∪{∆} where ∆denotes the boundary. The edges of the graph are the same as e(A) except that each edge in ∂eA is replaced with an edge connecting the point in A to ∆. (There can be more than one edge connecting a vertex in A to ∆.) A wired spanning tree is a collection of edges from e(A) such that the corresponding subgraph of A ∪{∆} is a spanning tree. Such a tree has #(A) edges.
222 Loop Measures In both cases, we will find the number of trees by considering the Markov chain given by simple random walk in Z2. The different spanning trees correspond to different “boundary conditions” for the random walks.
• Free. The lazy walker on A as described in Section 9.2.1, i.e., q(x, y) = 1 4, x, y ∈A, |x −y| = 1, and q(x, x) = 1 −P y q(x, y).
• Wired. Simple random walk on A killed when it leaves A, i.e., q(x, y) = 1 4, x, y ∈A, |x −y| = 1, and q(x, x) = 0. Equivalently, we can consider this as the Markov chain on A ∪{∆} where ∆is an absorbing point and q(x, ∆) = 1 − X y∈A q(x, y).
♣In other words, free spanning trees correspond to reflecting or Neumann boundary conditions and wired spanning trees correspond to Dirichlet boundary conditions.
We let F(A) denote the quantity for the wired case. This is the same as F(A) for simple random walk in Z2. If x ∈A, we write F ∗(A{x}) for the corresponding quantity for the lazy walker. (The lazy walker is a Markov chain on A and hence F ∗(A) = ∞. In order to get a finite quantity, we need to remove a point x.) The following are immediate corollaries of results in Section 9.7.
Proposition 9.9.1 If A ⊂Z2 is connected with #(A) = n < ∞, then the number of wired spanning trees of A is 4n F(A)−1 = 4n n Y j=1 (1 −βj), where β1, . . . , βn denote the eigenvalues of QA = [q(x, y)]x,y∈A.
Proof This is a particular case of Corollary 9.7.2 using the graph A ∪{∆} and x0 = ∆. See also Proposition 9.3.3.
Proposition 9.9.2 Suppose α1 = 1, . . . , αn are the eigenvalues of the transition matrix for the lazy walker on a finite, connected A ⊂Z2 of cardinality n. Then the number of spanning trees of A is 4n−1n−1 n Y j=2 (1 −αj).
Proof This is a particular case of Proposition 9.7.3.
9.9 Spanning trees of subsets of Z2 223 Recall that log F(A) = X ω∈L(A),|ω|≥1 1 4|ω| |ω| = X x∈A ∞ X n=1 1 2n Px{S2n = 0; Sj ∈A, j = 1, . . . , 2n}.
(9.21) The first order term in an expansion of log F(A) is X x∈A ∞ X n=1 1 2n Px{S2n = 0}, which ignores the restriction that Sj ∈A, j = 1, . . . , 2n. The actual value involves a well known constant Ccat called Catalan’s constant. There are many equivalent definitions of this constant. For our purposes we can use the following Ccat = π 2 log 2 −π 4 ∞ X n=1 1 2n 4−2n 2n n 2 = .91596 · · · .
Proposition 9.9.3 If S = (S1, S2) is simple random walk in Z2, then ∞ X n=1 1 2n P{S2n = 0} = log 4 −4 π Ccat, where Ccat denotes Catalan’s constant. In particular, if A ⊂Z2 is finite, log F(A) = [log 4 −(4/π) Ccat] #(A) − X x∈A ψ(x; A), (9.22) where ψ(x; A) = ∞ X n=1 1 2nPx{S2n = 0; Sj ̸∈A for some 0 ≤j ≤2n}.
Proof Using Exercise 1.7, we get ∞ X n=1 1 2n P{S2n = 0} = ∞ X n=1 1 2n [P{S1 2n = 0}]2 = ∞ X n=1 1 2n 4−2n 2n n 2 .
Since P{S(2n) = 0} ∼c n−1, we can see that the sum is finite. The exact value follows from our (conveniently chosen) definition of Ccat. The last assertion then follows from (9.21).
Lemma 9.9.4 There exists c < ∞such that if A ⊂Z2, x ∈A, and ψ(x; A) is defined as in Proposition 9.9.3, then ψ(x; A) ≤ c dist(x, ∂A)2 .
Proof We only sketch the argument leaving the details as Exercise 9.5. Let r = dist(x, ∂A). Since it takes about r2 steps to reach ∂A, the loops with fewer than that many steps rooted at x tend not to leave A. Hence ψ(x; A) is at most of the order of X n≥r2 1 2n P{S2n = 0} ≍ X n≥r2 n−2 ≍r−2.
224 Loop Measures Using this and (9.22) we immediately get the following.
Proposition 9.9.5 Suppose An is a sequence of finite, connected subsets of Z2 satisfying the following condition (that roughly means “measure of the boundary goes to zero”). For every r > 0, lim n→∞ #{x ∈An : dist(x, ∂An) ≤r} #(An) = 0.
Then, lim n→∞ log F(An) #(An) = log 4 −4Ccat π .
Suppose Am,n is the (m −1) × (n −1) discrete rectangle, Am,n = {x + iy : 1 ≤x ≤m −1, 1 ≤y ≤n −1}.
Note that #(Am,n) = (m −1) (n −1), #(∂Am,n) = 2 (m −1) + 2 (n −1).
Theorem 9.9.6 4(m−1)(n−1) F(Am,n) ≍e4Ccatmn/π ( √ 2 −1)m+n n−1/2.
(9.23) More precisely, for every b ∈(0, ∞) there is a cb < ∞such that if b−1 ≤m/n ≤b then both sides of (9.23) are bounded above by cb times the other side. In particular, if Cm,n denotes the number of wired spanning trees of Am,n, log Cmn = 4Ccat π mn + log( √ 2 −1) (m + n) −1 2 log n + O(1) = 4Ccat π #(Am,n) + 2Ccat π + 1 2 log( √ 2 −1) #(∂Am,n) −1 2 log n + O(1).
♣Although our proof will use the exact values of the eigenvalues, it is useful to consider the result in terms of (9.22). The dominant term is already given by (9.22). The correction comes from loops rooted in Am,n that leave A. The biggest contribution to these comes from points near the boundary. It is not surprising then that the second term is proportional to the number of points on the boundary. The next correction to this comes from the corners of the rectangle. This turns out to contribute a logarithmic term and after that all other correction terms are O(1). We arbitrarily write log n rather than log m; note that log m = log n + O(1).
Proof The expansion for log Cm,n follows immediately from Proposition 9.9.1 and (9.23), so we only need to establish (9.23). The eigenvalues of I −QA can be given explicitly (see Section 8.2), 1 −1 2 cos jπ m + cos kπ n , j = 1, . . . , m −1; k = 1, . . . , n −1, 9.9 Spanning trees of subsets of Z2 225 with corresponding eigenfunctions f(x, y) = sin jπx m sin kπy n , where the eigenfunctions have been chosen so that f ≡0 on ∂Am,n. Therefore, −log F(Am,n) = log det[I −QA] = m−1 X j=1 n−1 X k=1 log 1 −1 2 cos jπ m + cos kπ n .
Let g(x, y) = log 1 −cos(x) + cos(y) 2 .
Then (mn)−1 log det[I −QA] is a Riemann sum approximation of 1 π2 Z π 0 Z π 0 g(x, y) dx dy.
To be more precise, Let V (j, k) = Vm,n(j, k) denote the rectangle of side lengths π/m and π/n centered at (jπ/m) + i(kπ/n). Then we will consider J(j, k) := 1 mn g jπ m , kπ n = 1 mn 1 −1 2 cos jπ m + cos kπ n as an approximation to 1 π2 Z V (j,k) g(x, y) dx dy.
Note that V = m−1 [ j=1 n−1 [ k=1 V (j, k) = x + iy : π 2m ≤x ≤π 1 −1 2m , π 2n ≤y ≤π 1 −1 2n .
One can show (using ideas as in Section 12.1.1, details omitted), log det[I −QA] = mn Z V g(x, y) dx dy + O(1).
Therefore, log det[I −QA] = mn "Z [0,π]2 g(x, y) dx dy − Z [0,π]2\V g(x, y) dx dy # + O(1).
The result will follow if we show that mn Z [0,π]2\V g(x, y) dx dy = (m + n) log 4 −(m + n) log(1 − √ 2) + 1 2 log n + O(1).
We now estimate the integral over [0, π]2 \ V which we write as the sum of integrals over four thin strips minus the integrals over the “corners” that are doubly counted. One can check (using an integral table, e.g.) that 1 π Z π 0 log 1 −cos x + cos y 2 dy = −2 log 2 + log[2 −cos x + p 2(1 −cos x) + (1 −cos x)2].
226 Loop Measures Then, 1 π2 Z ǫ 0 Z π 0 log 1 −cos x + cos y 2 dy dx = −2ǫ π log 2 + ǫ2 2π + O(ǫ3).
If we choose ǫ = π/(2m) or ǫ = π/2n, this gives mn π2 Z π/(2m) 0 Z π 0 log 1 −cos x + cos y 2 dy dx = m log 2 + O(1).
mn π2 Z π 0 Z π/(2n) 0 log 1 −cos x + cos y 2 dy dx = −n log 2 + O(1).
Similarly, 1 π2 Z π π−ǫ Z π 0 log 1 −cos x + cos y 2 dy dx = −2ǫ π log 2 + ǫ π log[3 + 2 √ 2] + O(ǫ3) = −2ǫ π log 2 −2ǫ π log[ √ 2 −1] + O(ǫ3), which gives mn π2 Z π π−π 2m Z π 0 log 1 −cos x + cos y 2 dy dx = −n log 2 −n log[ √ 2 −1] + O(n−1), mn π2 Z π 0 Z π π−π 2n log 1 −cos x + cos y 2 dy dx = −m log 2 −m log[ √ 2 −1] + O(n−1).
The only nontrivial “corner” term comes from Z ǫ 0 Z δ 0 log 1 −cos x + cos y 2 dxdy = 2 ǫ δ log(ǫ) + O(ǫ δ).
Therefore, mn π2 Z π 2m 0 Z π 2n 0 log 1 −cos x + cos y 2 dxdy = −1 2 log n + O(1).
All of the other corners give O(1) terms.
Combining it all, we get m−1 X j=1 n−1 X k=1 log 1 −1 2 cos jπ m + cos kπ n equals Imn + (m + n) log 4 + (m + n) log[1 − √ 2] −1 2 log n + O(1)., where I = 1 π2 Z π 0 Z π 0 log 1 −cos x + cos y 2 dydx.
9.9 Spanning trees of subsets of Z2 227 Proposition 9.9.5 tells us that I = 4 Ccat π −log 4.
Theorem 9.9.6 allows us to derive some constants for simple random walk that are hard to show directly. Write (9.23) as log F(Am,n) = B1 mn + B2 (m + n) + 1 2 log n + O(1), (9.24) where B1 = log 4 −4Ccat π , B2 = log( √ 2 + 1) −log 4.
The constant B1 was obtained by considering the rooted loop measure and B2 was obtained from the exact value of the eigenvalues. Recall from (9.3) that if we enumerate Am,n, Am,n = {x1, x2, . . . , xK}, K = (m −1) (n −1), then log F(Am,n) = K X j=1 log Fxj(Am,n \ {x1, . . . , xj−1}), and Fx(V ) is the expected number of visits to x for a simple random walk starting at x before leaving V . We will define the lexicographic order of Z + iZ by x + iy ≺x1 + iy1 if x < x1 or x = x1 and y < y1.
Proposition 9.9.7 If V = {x + iy : y > 0} ∪{0, 1, 2, . . . , }, then F0(V ) = 4 e−4Ccat/π.
Proof Choose the lexicographic order for An,n. Then one can show that Fxj(An,n \ {x1, . . . , xj−1}) = F0(V ) [1 + error] , where the error term is small for points away from the boundary. Hence log F(An,n) = #(An,n) log F0(V ) [1 + o(1)].
which implies log F0(V ) = B1 as in (9.24).
Proposition 9.9.8 Let V ⊂Z × iZ be the subset V = (Z × iZ) \ {· · · , −2, −1}, 228 Loop Measures Then, F0(V ) = 4 ( √ 2 −1). In other words, the probability that the first return to {· · · , −2, −1, 0} by a simple random walk starting at the origin is at the origin equals 1 − 1 F0(V ) = 3 − √ 2 4 .
Proof Consider A = An = {x + iy : x = 1, . . . , n −1; −(n −1) ≤y ≤n −1}.
Then A is a translation of A2n,n and hence (9.24) gives log F(A) = 2 B1 n2 + 3 B2 n + 1 2 log n + O(1).
Order A so that the first n −1 vertices of A are 1, 2, . . . , n −1 in order. Then, we can see that log F(A) = n−1 X j=1 log Fj(A \ {1, . . . , j −1}) + 2 log F(An,n).
Using (9.24) again, we see that 2 log F(An,n) = 2 B1 n2 + 4 B2 n + log n + O(1), and hence n−1 X j=1 log Fj(A \ {1, . . . , j −1}) = −B2 n −1 2 log n + O(1).
Now we use the fact that log Fj(A \ {1, . . . , j −1}) = log F0(V ) [1 + error] , where the error term is small for points away from the boundary to conclude that F0(V ) = e−B2.
Let ˜ Am,n be the m × n rectangle ˜ Am,n = {x + iy : 0 ≤x ≤m −1, 0 ≤y ≤n −1}.
Note that #( ˜ Am,n) = mn, #(∂˜ Am,n) = 2(m + n).
Let ˜ Cm,n denote the number of (free) spanning trees of ˜ Am,n.
Theorem 9.9.9 ˜ Cm,n ≍e4Ccatmn/π ( √ 2 −1)m+n n−1/2.
More precisely, for every b ∈(0, ∞) there is a cb < ∞such that if b−1 ≤m/n ≤b then both sides of (9.23) are bounded above by cb times the other side.
9.9 Spanning trees of subsets of Z2 229 Proof We claim that the eigenvalues for the lazy walker Markov chain on ˜ Am,n are: 1 −1 2 cos jπ m + cos kπ n , j = 0, . . . , m −1; k = 0, . . . , n −1, with corresponding eigenfunctions f(x, y) = cos jπ(x + 1 2) m !
cos kπ(y + 1 2) n !
.
Indeed, these are eigenvalues and eigenfunctions for the usual discrete Laplacian, but the eigen-functions have been chosen to have boundary conditions f(0, y) = f(−1, y), f(m −1, y) = f(m, y), f(x, 0) = f(x, −1), f(x, n −1) = f(x, n).
For these reason we can see that they are also eigenvalues and eigenvalues for the lazy walker.
Using Proposition 9.9.2, we have ˜ Cmn = 4mn−1 mn Y (j,k)̸=(0,0) 1 −1 2 cos jπ m + cos kπ n .
Recall that if F(An,m) is as in Theorem 9.9.6, then 1 F( ˜ Am,n) = Y 1≤j≤m−1,1≤k≤n−1 1 −1 2 cos jπ m + cos kπ n .
Therefore, ˜ Cmn = 4(m−1)(n−1) F(Am,n) 4m+n−1 mn n−1 Y j=1 1 2 −1 2 cos jπ n m−1 Y j=1 1 2 −1 2 cos jπ m .
Using (9.23), we see that it suffices to prove that 4n n n−1 Y j=1 1 2 −1 2 cos jπ n ≍1, or equivalently, n−1 X j=1 log 1 −cos jπ n = −n log 2 + log n + O(1).
(9.25) To establish (9.25), note that 1 n n−1 X j=1 log 1 −cos jπ n is a Riemann sum approximation of 1 π Z π 0 f(x) dx 230 Loop Measures where f(x) = log[1 −cos x]. Note that f ′(x) = sin x 1 −cos x, f ′′(x) = − 1 1 −cos x.
In particular |f ′′(x)| ≤c x−2. Using this we can see that 1 n 1 −cos jπ n = 1 n O(j−2) + 1 π Z jπ n + π 2n jπ n −π 2n f(x) dx.
Therefore, 1 n n−1 X j=1 log 1 −cos jπ n = O(n−1) + 1 π Z π−π 2n π 2n f(x) dx = O(n−1) + 1 π Z π 0 f(x) dx −1 π Z π 2n 0 f(x) dx = O(n−1) −log 2 + 1 n log n.
9.10 Gaussian free field We introduce the Gaussian free field.
In this section we assume that q is a symmetric transi-tion probability on the space X. Some of the definitions below are straightforward extensions of definitions for random walk on Zd.
• We say e = {x, y} is an edge if q(e) := q(x, y) > 0.
• If A ⊂X, let e(A) denote the set of edges with at least one vertex in A. We write e(A) = ∂eA ∪eo(A) where ∂eA are the edges with one vertex in ∂A and eo(A) are the edges with both vertices in A.
• We let ∂A = {y ∈X \ A : q(x, y) > 0 for some x ∈A}, A = A ∪∂A.
• If f : A →R and x ∈A, then ∆f(x) = X y q(x, y) [f(y) −f(x)].
We say that f is harmonic at x if ∆f(x) = 0, and f is harmonic on A if ∆f(x) = 0 for all x ∈A.
• If e ∈e(A), we set ∇ef = f(y) −f(x) where e = {x, y}. This defines ∇ef up to a sign. Note that ∇ef ∇eg is well defined.
Throughout this section we assume that A ⊂X with #(A) < ∞.
9.10 Gaussian free field 231 • If f, g : A →R are functions, then we define the energy or Dirichlet form E to be the quadratic form EA(f, g) = X e∈e(A) q(e) ∇ef ∇eg We let EA(f) = EA(f, f).
Lemma 9.10.1 (Green’s formula) Suppose f, h : A →R. Then, EA(f, h) = − X x∈A f(x) ∆h(x) + X x∈∂A X y∈A f(x) [h(x) −h(y)] q(x, y).
(9.26) • If h is harmonic in A, EA(f, h) = X x∈∂A X y∈A f(x) [h(x) −h(y)] q(x, y).
(9.27) • If f ≡0 on ∂A, EA(f, h) = − X x∈A f(x) ∆h(x).
(9.28) • If h is harmonic in A and f ≡0 on ∂A, then EA(f, h) = 0 and hence EA(f + h) = EA(f) + EA(h).
(9.29) Proof EA(f, h) = X e∈e(A) q(e) ∇ef ∇eh = 1 2 X x,y∈A q(x, y) [f(y) −f(x)] [h(y) −h(x)] + X x∈A X y∈∂A q(x, y) [f(y) −f(x)] [h(y) −h(x)] = − X x∈A X y∈A q(x, y) f(x) [h(y) −h(x)] − X x∈A X y∈∂A q(x, y) f(x) [h(y) −h(x)] + X x∈A X y∈∂A q(x, y) f(y) [h(y) −h(x)] = − X x∈A f(x) ∆h(x) + X y∈∂A X x∈A q(x, y) f(y) [h(y) −h(x)].
This gives (9.26) and the final three assertions follow immediately.
Suppose x ∈∂A and let hx denote the function that is harmonic on A with boundary value δx on ∂A. Then it follows from (9.27) that EA(hx) = X y∈A [1 −hx(y)] q(x, y).
We extend hx to X by setting hx ≡0 on X \ A.
232 Loop Measures Lemma 9.10.2 Let Yj be a Markov chain on X with transition probability q. Let Tx = min{j ≥1 : Yj = x}, τA = min{j ≥1 : Yj ̸∈A}.
If A ⊂X, x ∈∂A, A′ = A ∪{x}, EA′(hx) = Px{Tx ≥τA′} = 1 Fx(A′).
(9.30) Proof If y ∈A, then hx(y) = Py{Tx = τA}. Note that Px{Tx < τA′} = q(x, x) + X y∈A q(x, y) Py{Tx = τA} = q(x, x) + X y∈A q(x, y) hx(y).
Therefore, Px{Tx ≥τA′} = 1 −Px{Tx < τA′} = X z̸∈A′ q(x, z) + X y∈A q(x, y) [1 −hx(y)] = −∆hx(x) = − X y∈A′ hx(y) ∆hx(y) = EA′(hx).
The last equality uses (9.28). The second equality in (9.30) follows from Lemma 9.3.2.
• If v : X \ A →R, f : A →R, we write EA(f; v) for EA(fv) where fv ≡f on A and fv ≡v on ∂A.
If v is omitted, then v ≡0 is assumed.
• The Gaussian free field on A with boundary condition v is the measure on functions f : A →R whose density with respect to Lebesgue measure on RA is (2π)−#(A)/2 e−EA(f;v)/2.
• If v ≡0, we call this the field with Dirichlet boundary conditions.
• If A ⊂X is finite and v : X \ A →R, define the partition function C(A; v) = Z (2π)−#(A)/2 e−EA(f;v)/2 d f, where d f indicates that this is an integral with respect to Lebesgue measure on RA. If v ≡0, we write just C(A). By convention, we set C(∅; v) = 1.
We will give two proofs of the next fact.
Proposition 9.10.3 For any A ⊂X with #(A) < ∞, C(A) = p F(A) = exp 1 2 X ω∈L(A) m(ω) .
(9.31) 9.10 Gaussian free field 233 Proof We prove this inductively on the cardinality of A. If A = ∅, the result is immediate. From (9.3), we can see that it suffices to show that if A ⊂X is finite, x ̸∈A, and A′ = A ∪{x}, C(A′) = C(A) p Fx(A′).
Suppose f : A′ →R and extend f to X by setting f ≡0 on X \ A′. We can write f = g + t h where g vanishes on X \ A; t = f(x); and h is the function that is harmonic on A with h(x) = 1 and h ≡0 on X \ A′. The edges in e(A′) are the edges in e(A) plus those edges of the form {x, z} with z ∈X \ A. Using this, we can see that EA′(f) = EA(f) + X y̸∈A′ q(x, y) t2.
(9.32) Also, by (9.29), EA(f) = EA(g) + EA(th) = EA(g) + t2 EA(h), which combined with (9.32) gives exp −1 2EA′(f) = exp −1 2EA(g) exp −t2 2 EA′(h) .
Integrating over A first, we get C(A′) = C(A) Z ∞ −∞ 1 √ 2π e−t2EA′(h)/2 dt = C(A) Z ∞ −∞ 1 √ 2π e−t2/[2Fx(A′)] dt = C(A) p Fx(A′).
The second equality uses (9.30).
Let Q = QA as above and denote the entries of Qn by qn(x, y). The Green’s function on A is the matrix G = (I −Q)−1; in other words, the expected number of visits to y by the chain starting at x equals ∞ X n=0 qn(x, y) which is the (x, y) entry of (I −Q)−1.
Since Q is strictly subMarkov, (I −Q) is symmetric, strictly positive definite, and (I −Q)−1 is well defined. The next proposition uses the joint normal distribution as discussed in Section 12.3.
Proposition 9.10.4 Suppose the random variables {Zx : x ∈A} have a (mean zero) joint normal distribution with covariance matrix G = (I −Q)−1. Then the distribution of the random function f(x) = Zx is the same as the Gaussian free field on A with Dirichlet boundary conditions.
234 Loop Measures Proof Plugging Γ = G = (I −Q)−1 into (12.14) , we see that the joint density of {Zx} is given by (2π)−#(A) [det(I −Q)]1/2 exp −f · (I −Q)f 2 .
But (9.28) implies that f · (I −Q)f = EA(f). Since this is a probability density this shows that C(A) = s 1 det(I −Q), and hence (9.31) follows from Proposition 9.3.3.
♣The scaling limit of the Gaussian free field for random walk in Zd is the Gaussian free field in Rd. There are technical subtleties required in the definition. For example if d = 2 and U is a bounded open set, we would like to define the Gaussian free field {Zz : z ∈U} with Dirichlet boundary conditions to be the collection of random variables such that each finite collection (Zz1, . . . , Zzk) has a joint normal distribution with covariance matrix [GU(zi, zj)]. Here GU denotes the Green’s function for Brownian motion in the domain. However, the Green’s function GU(z, w) blows up as w approaches z, so this gives an infinite variance for the random variable Zz. These problems can be overcome, but the collection {Zz} is not a collection of random variables in the usual sense.
♣The proof of Proposition 9.10.3 is not really needed given the quick proof in Proposition 9.10.4. However, we choose to include it since it uses more directly the loop measure interpretation of F(A) rather than the interpretation as a determinant. Many computations with the loop measure have interpretations in the scaling limit.
Exercises Exercise 9.1 Show that for all positive integers k X j1+···+jr=k 1 r! (j1 · · · jr) = 1.
Here are two possible approaches.
• Show that the number of permutations of k elements with exactly r cycles is X j1+···+jr=k k!
r! j1j2 · · · jr .
• Consider the equation 1 1 −t = exp{−log(1 −t)}, expand both sides in power series in t, and compare coefficients.
9.10 Gaussian free field 235 Exercise 9.2 Suppose Xn is an irreducible Markov chain on a countable state space X and A = {x1, . . . , xk} is a proper subset of X. Let A0 = A, Aj = A \ {x1, . . . , xj}. If z ∈V ⊂X, let gV (z) denote the expected number of visits to z by the chain starting at z before leaving V .
(i) Show that gA(x1) gA{x1}(x2) = gA{x2}(x1) gA(x2).
(9.33) (ii) By iterating (9.33) show that the quantity k Y j=1 gAj−1(xj) is independent of the ordering of x1, . . . , xk.
Exercise 9.3 [Karlin-McGregor] Suppose X1 n, . . . , Xk n are independent realizations from a Markov chain with transition probability q on a finite state space X. Assume x1, . . . , xk, y1, . . . , yj ∈X.
Consider the event V = Vn(y1, . . . , yk) = Xi m ̸= Xj m, m = 0, . . . , n; Xj n = yj, 1 ≤j ≤n .
Show that P{V | X1 0 = x1, . . . , X1 n = xn} = det [qn(xi, yj)]1≤i,j≤k , where qn(xi, yj) = P X1 n = yj | X1 0 = xi .
Exercise 9.4 Suppose Bernoulli trials are performed with probability p of success. Let Yn denote the number of failures before the nth success, and let r(n) be the probability that Yn is even. By definition, r(0) = 1. Give a recursive equation for r(n) and use it to find r(n). Use this to verify (9.16).
Exercise 9.5 Give the details of Lemma 9.9.4.
Exercise 9.6 Suppose q is the weight arising from simple random walk in Zd. Suppose A1, A2 are disjoint subsets of Zd and x ∈Zd. Let p(x, A1, A2) denote the probability that a random walk starting at x enters A2 and subsequently returns to x all without entering A1. Let g(x, A1) denote the expected number of visits to x before entering A1 for a random walk starting at x. Show that the unrooted loop measure of the set of loops in Zd \ A1 that intersect both x and A2 is bounded above by p(x, A1, A2) g(x, A1). Hint: for each unroooted loop that intersects both x and A2 choose a (not necessarily unique) representative that is rooted at x and enters A2 before its first return to x.
Exercise 9.7 We continue the notation of Exercise 9.6 with d ≥3. Choose an enumeration of Zd = {x0, x1, . . .} such that j < k implies |xj| ≤|xk|.
(i) Show there exists c < ∞such that if r > 0, u ≥2, and |xj| ≤r, p(xj, Aj−1, Zd \ Bur) ≤c1 |xj|−2 (ur)2−d.
236 Loop Measures (Hint: Consider a path that starts at xj, leaves Bur and then returns to xj without visiting Aj−1. Split such a curve into three pieces: the “beginning” up to the first visit to Zd \ Bur; the “end” which (with time reversed) is a walk from xj to the first (last) visit to Zd \B3|xj|/2; and the “middle” which ties these walks together.) (ii) Show that there exists c1 < ∞, such that if r > 0 and u ≥2, then the (unrooted) loop measure of the set of loops that intersect both Br and Zd \ Bur is bounded above by c1 u2−d.
10 Intersection Probabilities for Random Walks ‘ 10.1 Long range estimate In this section we prove a fundamental inequality concerning the probability of intersection of the paths of two random walks. If Sn is a random walk, we write S[n1, n2] = {Sn : n1 ≤n ≤n2}.
Proposition 10.1.1 If p ∈Pd, there exist c1, c2 such that for all n ≥2, c1 φ(n) ≤ P{S[0, n] ∩S[2n, 3n] ̸= ∅} ≤ P{S[0, n] ∩S[2n, ∞) ̸= ∅} ≤c2 φ(n), where φ(n) = 1, d < 4, (log n)−1, d = 4, n(4−d)/2, d > 4.
♣As n →∞, we get a result about Brownian motions. If B is a standard Brownian motion in Rd, then P{B[0, 1] ∩B[2, 3] ̸= ∅} > 0, d ≤3 = 0, d = 4. .
Four is the critical dimension in which Brownian paths just barely avoid each other.
Proof The upper bound is trivial for d ≤3, and the lower bound for d ≤2 follows from the lower bound for d = 3. Hence we can assume that d ≥3. We will assume the walk is aperiodic (only a trivial modification is needed for the bipartite case). The basic strategy is to consider the number of intersections of the paths, Jn = n X j=0 3n X k=2n 1{Sj = Sk}, Kn = n X j=0 ∞ X k=2n 1{Sj = Sk}.
237 238 Intersection Probabilities for Random Walks Note that P{S[0, n] ∩S[2n, 3n] ̸= ∅} = P{Jn ≥1}, P{S[0, n] ∩S[2n, ∞) ̸= ∅} = P{Kn ≥1}.
We will derive the following inequalities for d ≥3, c1 n(4−d)/2 ≤E(Jn) ≤E(Kn) ≤c2 n(4−d)/2, (10.1) E(J2 n) ≤ c n, d = 3, c log n, d = 4, c n(4−d)/2, d ≥5.
(10.2) Once these are established, the lower bound follows by the second moment lemma (Lemma 12.6.1), P{Jn > 0} ≥E(Jn)2 4 E(J2 n).
Let us write p(n) for P{Sn = 0}. Then, E(Jn) = n X j=0 3n X k=2n p(k −j), and similarly for E(Kn). Since p(k −j) ≍(k −j)−d/2, we get E(Jn) ≍ n X j=0 3n X k=2n 1 (k −j)d/2 ≍ n X j=0 3n X k=2n 1 (k −n)d/2 ≍ n X j=0 n1−(d/2) ≍n2−(d/2), and similarly for E(Kn). This gives (10.1). To bound the second moments, note that E(J2 n) = X 0≤j,i≤n X 2n≤k,m≤3n P{Sj = Sk, Si = Sm} ≤ 2 X 0≤j≤i≤n X 2n≤k≤m≤3n [P{Sj = Sk, Si = Sm} + P{Sj = Sm, Si = Sk}].
If 0 ≤i, j ≤n and 2n ≤k ≤m ≤3n, then P{Sj = Sk, Si = Sm} ≤ max l≥n,x∈Zd P{Sl = x} max x∈Zd P{Sm−k = x} ≤ c nd/2 (m −k + 1)d/2 .
The last inequality uses the local central limit theorem. Therefore, E(J2 n) ≤c n2 X 2n≤k≤m≤3n 1 nd/2 (m −k + 1)d/2 ≤c n2−(d/2) X 0≤k≤m≤n 1 (m −k + 1)d/2 .
This yields (10.2).
The upper bound is trivial for d = 3 and for d ≥5 it follows from (10.1) and the inequality P{Kn ≥1} ≤E[Kn]. Assume d = 4. We will consider E[Kn | Kn ≥1]. On the event {Kn ≥1}, let k be the smallest integer ≥2n such that Sk ∈S[0, n]. Let j be the smallest index such that 10.1 Long range estimate 239 Sk = Sj. Then by the Markov property, given [S0, . . . , Sk] and Sk = Sj, the expected value of K2n is n X i=0 ∞ X l=k P{Sl = Si | Sk = Sj} = n X i=0 G(Si −Sj).
Define a random variable, depending on S0, . . . , Sn, Yn = min j=0,...,n n X i=0 G(Si −Sj).
For any r > 0, we have that E[Kn | Kn ≥1, Yn ≥r log n] ≥r log n.
Note that for each r, P{Yn < r log n} ≤(n + 1) P X i≤n/2 G(Si) < r log n .
Using Lemma 10.1.2 below, we can find an r such that P{Yn < r log n} = o(1/ log n) But, c ≥E[Kn] ≥ P{Kn ≥1; Yn ≥r log n}E[Kn | Kn ≥1, Yn ≥r log n] ≥ P{Kn ≥1; Yn ≥r log n}[r log n].
Therefore, P{Kn ≥1} ≤P{Yn < r log n} + P{Kn ≥1; Yn ≥r log n} ≤ c log n.
This finishes the proof except for the one lemma that we will now prove.
Lemma 10.1.2 Let p ∈P4.
(a) For every α > 0, there exist c, r such that for all n sufficiently large, P ξn−1 X j=0 G(Sj) ≤r log n ≤c n−α.
(b) For every α > 0, there exist c, r such that for all n sufficiently large, P n X j=0 G(Sj) ≤r log n ≤c n−α.
Proof It suffices to prove (a) when n = 2l for some integer l, and we write ξk = ξ2k.
Since G(x) ≥c/(|x| + 1)2, we have ξl−1 X j=0 G(Sj) ≥ l X k=1 ξk−1 X j=ξk−1 G(Sj) ≥c l X k=1 2−2k [ξk −ξk−1].
240 Intersection Probabilities for Random Walks The reflection principle (Proposition 1.6.2) and the central limit theorem show that for every ǫ > 0, there is a δ > 0 such that if n is sufficiently large, and x ∈Cn/2, then Px{ξn ≤δ n2} ≤ǫ. Let Ik denote the indicator function of the event {ξk −ξk−1 ≤δ 22k}. Then we know that P(Ik = 1 | S0, . . . , Sξk−1) ≤ǫ.
Therefore, Jl := Pl k=1 Ik is stochastically bounded by a binomial random variable with parameters l and ǫ. By exponential estimates for binomial random variables (see Lemma 12.2.8), we can find an α such that P{Jl ≥l/2} ≤c 2−αl.
But on the event {Jl < l/2} we know that G(Sj) ≥c(l/2) δ ≥r log n, where the r depends on α.
For part (b) we need only note that P{n < ξn1/4} decays faster than any power of n and P n X j=0 G(Sj) ≤r 4 log n ≤P ξn1/4 X j=0 G(Sj) ≤r log n1/4 + P{n < ξn1/4}.
♣The proof of the upper bound for d = 4 in Proposition 10.1.1 can be compared to the proof of an easier estimate P{0 ∈S[n, ∞)} ≤c n1−d 2 , d ≥3.
To prove this, one uses the local central limit theorem to show that the expected number of visits to the origin is O(n1−d 2 ).
On the event that 0 ∈S[n, ∞), we consider the smallest j ≥n such that Sj = 0.
Then using the strong Markov property, one shows that the expected number of visits given at least one visit is G(0, 0) < ∞. In Proposition 10.1.1 we consider the event that S[0, n] ∩S[2n, ∞) ̸= ∅and try to take the “first” (j, k) ∈[0, n] × [2n, ∞) such that Sj = Sk. This is not well defined since if (i, l) is another pair it might be the case that i < j and l > k. To be specific, we choose the smallest k and then the smallest j with Sj = Sk.
We then say that the expected number of intersections after this time is the expected number of intersections of S[k, ∞) with S[0, n]. Since Sk = Sj this is like the number of intersections of two random walks starting at the origin. In d = 4, this is of order log n. However, because Sk, Sj have been chosen specifically, we cannot use a simple strong Markov property argument to assert this. This is why the extra lemma is needed.
10.2 Short range estimate We are interested in the probability that the paths of two random walks starting at the origin do not intersect up to some finite time. We discuss only the interesting dimensions d ≤4. Let S, S1, S2, . . .
be independent random walks starting at the origin with distribution p ∈Pd. If 0 < λ < 1, let Tλ, T 1 λ, T 2 λ, . . . denote independent geometric random variables with killing rate 1 −λ and we write λn = 1 −1 n. We would like to estimate P{S(0, n] ∩S1[0, n] = ∅}, 10.2 Short range estimate 241 or P S(0, Tλn] ∩S1[0, T 2 λn] = ∅ .
The next proposition uses the long range estimate to bound a different probability, P S(0, Tλn] ∩(S1[0, T 1 λn] ∪S2[0, T 2 λn]) = ∅ .
Let Q(λ) = X y∈Zd P0,y{S[0, Tλ] ∩S1[0, T 1 λ] ̸= ∅} = (1 −λ)2 X y∈Zd ∞ X j=0 ∞ X k=0 λj+k P0,y{S[0, j] ∩S1[0, k] ̸= ∅}.
Here we write Px,y to denote probabilities assuming S0 = x, S1 0 = y. Using Proposition 10.1.1, one can show that as n →∞(we omit the details), Q(λn) ≍ nd/2, d < 4 n2 [log n]−1, d = 4.
Proposition 10.2.1 Suppose S, S1, S2 are independent random walks starting at the origin with increment p ∈Pd. Let Vλ be the event that 0 ̸∈S1(0, T 1 λ]. Then, P Vλ ∩{S(0, Tλ] ∩(S1(0, T 1 λ] ∪S2(0, T 2 λ]) = ∅} = (1 −λ)2 Q(λ).
(10.3) Proof Suppose ω = [ω0 = 0, . . . , ωn], η = [η0, . . . , ηm] are paths in Zd with p(ω) := n Y j=1 p(ωj−1, ωj) > 0 p(η) :== m Y j=1 p(ηj−1, ηj) > 0.
Then we can write Q(λ) = (1 −λ)2 ∞ X n=0 ∞ X m=0 X ω,η λn+m p(ω) p(η), where the last sum is over all paths ω, η with |ω| = n, |η| = m, ω0 = 0 and ω ∩η ̸= ∅. For each such pair (ω, η) we define a 4-tuple of paths starting at the origin (ω−, ω+, η−, η+) as follows. Let s = min{j : ωj ∈η}, t = min{k : ηk = ωs}.
ω−= [ωs −ωs, ωs−1 −ωs, . . . , ω0 −ωs], ω+ = [ωs −ωs, ωs+1 −ωs, . . . , ωn −ωs], η−= [ηt −ηt, ηt−1 −ηt, . . . , η0 −ηt], η+ = [ηt −ηt, ηt+1 −ηt, . . . , ηm −ηt].
Note that p(ω) = p(ω−) p(ω+), p(η) = p(η−) p(η+). Also, 0 ̸∈[η− 1 , . . . , η− t ], [ω− 1 , . . . , ω− s ] ∩[η−∪η+] = ∅.
(10.4) 242 Intersection Probabilities for Random Walks Conversely, for each 4-tuple (ω−, ω+, η−, η+) of paths starting at the origin satisfying (10.4), we can find a corresponding (ω, η) with ω0 = 0 by inverting this procedure. Therefore, Q(λ) = (1 −λ)2 X 0≤n−,n+,m−,m+ X ω,ω+,η−,η+ λn−+n++m−+m+ p(ω−)p(ω+)p(η−)p(η+), where the last sum is over all (ω−, ω+, η−, η+) with |ω−| = n−, |ω+| = n+, |η−| = m−, |η+| = m+ satisfying (10.4). Note that there is no restriction on the path ω+. Hence we can sum over n+ and ω+ to get Q(λ) = (1 −λ) X 0≤n,m−,m+ X ω,η−,η+ λn+m−+m+ p(ω)p(η−)p(η+), But it is easy to check that the left-hand side of (10.3) equals (1 −λ)3 X 0≤n,m−,m+ X ω,η−,η+ λn+m−+m+ p(ω)p(η−)p(η+).
Corollary 10.2.2 For d = 2, 3, 4, P{S(0, n] ∩(S1(0, n] ∪S2[0, n]) = ∅} ≍ P{S(0, Tλn] ∩(S1(0, T 1 λn] ∪S2[0, T 2 λn]) = ∅} ≍ (1 −λn)2 Q(λn) ≍ ( n d−4 2 , d = 2, 3 (log n)−1, d = 4 Proof [Sketch] We have already noted the last relation. The previous proposition almost proves the second relation. It gives a lower bound. Since P{Tλn = 0} = 1/n, the upper bound will follow if we show that P[Vλn | S(0, Tλn] ∩(S1(0, T 1 λn] ∪S2[0, T 2 λn]) = ∅, Tλn > 0] ≥c > 0.
(10.5) We leave this as an exercise (Exercise 10.1).
One direction of the first relation can be proved by considering the event {Tλn, T 1 λn, T 2 λn ≤n} which is independent of the random walks and whose probability is bounded below by a c > 0 uniformly in n. This shows P{S(0, Tλn] ∩(S1(0, T 1 λn] ∪S2[0, T 2 λn]) = ∅} ≥c P{S(0, n] ∩(S1(0, n] ∪S2[0, n]) = ∅}.
For the other direction, it suffices to show that P{S(0, Tλn] ∩(S1(0, T 1 λn] ∪S2[0, T 2 λn]) = ∅; Tλn, T 1 λn, T 2 λn ≥n} ≥c (1 −λn)2 Q(λn).
This can be established by going through the construction in proof of Proposition 10.2.1. We leave this to the interested reader.
10.3 One-sided exponent 243 10.3 One-sided exponent Let q(n) = P{S(0, n] ∩S1(0, n] = ∅}.
This is not an easy quantity to estimate. If we let Yn = P S(0, n] ∩S1(0, n] = ∅| S(0, n] , then we can write q(n) = E[Yn].
Note that if S, S1, S2 are independent, then E[Y 2 n ] = P S(0, n] ∩(S1(0, n] ∪S2(0, n]) = ∅ .
Hence, we see that E[Y 2 n ] ≍ ( (log n)−1, d = 4 n d−4 2 , d < 4 (10.6) Since 0 ≤Yn ≤1, we know that E[Y 2 n ] ≤E[Yn] ≤ p E[Y 2 n ].
(10.7) If it were true that (E[Yn])2 ≍E[Y 2 n ] we would know how E[Yn] behaves. Unfortunately, this is not true for small d.
As an example, consider simple random walk on Z. In order for S(0, n] to avoid S1[0, n], ei-ther S(0, n] ⊂{1, 2, . . .} and S1[0, n] ⊂{0, −1, −2, . . .} or S(0, n) ⊂{−1, −2, . . .} and S1[0, n] ⊂ {0, 1, 2, . . .}.
The gambler’s ruin estimate shows that the probability of each of these events is comparable to n−1/2 and hence E[Yn] ≍n−1, E[Y 2 n ] ≍n−3/2.
Another way of saying this is P{S(0, n] ∩S2(0, n] = ∅} ≍n−1, P{S(0, n] ∩S2(0, n] = ∅| S(0, n] ∩S1(0, n] = ∅} ≍n−1/2.
For d = 4, it is true that (E[Yn])2 ≍E[Y 2 n ]. For d < 4, the relation (E[Yn])2 ≍E[Y 2 n ] does not hold. The intersection exponent ζ = ζd is defined by saying E[Yn] ≍n−ζ. One can show the existence of the exponent by first showing the existence of a similar exponent for Brownian motion (this is fairly easy) and then showing that the random walk has the same exponent (this takes more work, see ). This argument does not establish the value of ζ. For d = 2, it is known that ζ = 5/8. The techniques of the proof use conformal invariance of Brownian motion and a process called the Schramm-Loewner evolution (SLE). For d = 3, the exact value is not known (and perhaps will never be known). Corollary 10.2.2 and (10.7) imply that 1/4 ≤ζ ≤1/2, and it has been proved that both inequalities are actually strict. Numerical simulations suggest a value of about .29.
♣The relation E[Y 2 n ] ≈E[Yn]2 or equivalently P{S(0, n] ∩S2(0, n] = ∅} ≍P{S(0, n] ∩S2(0, n] = ∅| S(0, n] ∩S1(0, n] = ∅} 244 Intersection Probabilities for Random Walks is sometimes called mean-field behavior. Many systems in statistical physics have mean-field behavior above a critical dimension and also exhibit such behavior at the critical dimension with a logarithmic correction. Below the critical dimension they do not have mean-field behavior. The study of the exponents E[Y r n ] ≍n−ζ(r) sometimes goes under the name of multifractal analysis. The function ζ2(r) is known for all r ≥0, see .
Exercise 10.1 Prove (10.5).
11 Loop-erased random walk Loop-erased walks were introduced in Chapter 9 as a measure on self-avoiding paths obtained by starting with a (not necessarily probability) measure on intersecting paths and erasing loops chronologically. In this chapter we will study the corresponding process obtained when one starts with a probability measure on paths; we call this the loop-erased random walk (LERW) or the Laplacian random walk. We will consider only LERW derived from simple random walk in Zd, but many of the ideas can be extended to loop-erased random walks obtained from Markov chains.
♣The terms loop-erased walk and loop-erased random walk tend to be used synonymously in the literature.
We will make a distinction in this book, reserving LERW for a stochastic process associated to a probability measure on paths.
Throughout this section Sn will denote a simple random walk in Zd. We write S1 n, S2 n, . . . for independent realizations of the walk. We let τA, τ A be defined as in (4.27) and we use τ j A, τ j A to be the corresponding quantities for Sj. We let pA n (x, y) = pA n (y, x) = Px{Sn = y : τ A > n}.
If x ̸∈A, then pA n (x, y) = 0 for all n, y.
11.1 h-processes We will see that the loop-erased random walk looks like a random walk conditioned to avoid its past. As the LERW grows, the “past” of the walk also grows; this is an example of what is called a “moving boundary”. In this section we consider the process obtained by conditioning random walk to avoid a fixed set. This is a special case of an h-process.
Suppose A ⊂Zd and h : Zd →[0, ∞) is a strictly positive and harmonic function on A that vanishes on Zd \ A. Let (∂A)+ = (∂A)+,h = {y ∈∂A : h(y) > 0} = {y ∈Zd \ A : h(y) > 0}.
The (Doob) h-process (with reflecting boundary) is the Markov chain on A with transition prob-ability ˜ q = ˜ qA,h defined as follows.
245 246 Loop-erased random walk • If x ∈A and |x −y| = 1, ˜ q(x, y) = h(y) P |z−x|=1 h(z) = h(y) 2d h(x).
(11.1) • If x ∈∂A and |x −y| = 1, ˜ q(x, y) = h(y) P |z−x|=1 h(z).
The second equality in (11.1) follows by the fact that Lh(x) = 0. The definition of ˜ q(x, y) for x ∈∂A is the same as that for x ∈A, but we write it separately to emphasize that the second equality in (11.1) does not necessarily hold for x ∈∂A. The h-process stopped at (∂A)+ is the chain with transition probability q = qA,h which equals ˜ q except for • q(x, x) = 1, x ∈(∂A)+.
Note that if x ∈∂A \ (∂A)+, then q(y, x) = ˜ q(y, x) = 0 for all y ∈A. In other words, the chain can start in x ∈∂A \ (∂A)+, but it cannot visit there at positive times. Let ˜ qn = ˜ qA,h n , qn = qA,h n denote the usual n-step transition probabilities for the Markov chains.
Proposition 11.1.1 If x, y ∈A, qA,h n (x, y) = pA n (x, y) h(y) h(x).
In particular, qA,h n (x, x) = pA n (x, x).
Proof Let ω = [ω0 = x, ω1, . . . , ωn = y] be a nearest neighbor path with ωj ∈A for all j. Then the probability that first n points of the h-process starting at x are ω1, . . . , ωn in order is n Y j=1 h(ωj) 2d h(ωj−1) = (2d)−n h(y) h(x).
By summing over all paths ω, we get the proposition.
♣If we consider qA,h and pA as measures on finite paths ω = [ω0, . . . , ωn] in A, then we can rephrase the proposition as dqA,h dpA (ω) = h(ωn) h(ω0) .
Formulations like this in terms of Radon-Nikodym derivatives of measures can be extended to measures on continuous paths such as Brownian motion.
♣The h-process can be considered as the random walk “weighted by the function h”. One can define this 11.1 h-processes 247 for any positive function on A, even if h is not harmonic, using the first equality in (11.1). However, Proposition 11.1.1 will not hold if h is not harmonic.
Examples • If A ⊂Zd and V ⊂∂A, let hV,A(x) = Px{Sτ A ∈V } = X y∈V HA(x, y).
Assume hV,A(x) > 0 for all x ∈A. By definition, hV,A ≡1 on V and hV,A ≡0 on Zd \ (A ∪V ).
The hV,A-process corresponds to simple random walk conditioned to leave A at V . We usually consider the version stopped at V = (∂A)+.
– Suppose x ∈∂A \ V and H∂A(x, V ) := Px{S1 ∈A; SτA ∈V } = X y∈A H∂A(x, y) > 0.
If |x −y| > 1 for all y ∈V , then the excursion measure as defined in Section 9.6 corresponding to paths from x to V in A normalized to be a probability measure is the hV,A-process. If there is a y ∈V with |x −y| = 1, the hV,A-process allows an immediate transition to y while the normalized excursion measure does not.
• Let A = H = {x + iy ∈Z × iZ : y > 0} and h(z) = Im(z). Then h is a harmonic function on H that vanishes on ∂A. This h-process corresponds to simple random walk conditioned never to leave H and is sometimes called an H-excursion. With probability one this process never leaves H. Also, if q = qH,h and x + iy ∈H, q(x + iy, (x ± 1) + iy) = 1 4, q(x + iy, x + i(y + 1)) = y + 1 4y , q(x + iy, +i(y −1)) = y −1 4y .
• Suppose A is a proper subset of Zd and V = ∂A. Then the hV,A-process is simple random walk conditioned to leave A. If d = 1, 2 or Zd \ A is a recurrent subset of Zd, then hV,A ≡1 and the hV,A-process is the same as simple random walk.
• Suppose A is a connected subset of Zd, d ≥3 such that h∞,A(x) := Px{τ A = ∞} > 0.
Then the h∞,A-process is simple random walk conditioned to stay in A.
• Let A be a connected subset of Z2 such that Z2 \ A is finite and nonempty, and let h(x) = a(x) −Ex [a(Sτ A)] , be the unique function that is harmonic on A; vanishes on ∂A; and satisfies h(x) ∼(2/π) log |x| as x →∞, see (6.40). Then the h-process is simple random walk conditioned to stay in A. Note that this “conditioning” is on an event of probability zero. Using (6.40), we can see that this is the limit as n →∞of the hVn,An processes where An = A ∩{|z| < n}, Vn = ∂An ∩{|z| ≥n}.
248 Loop-erased random walk Note that for large n, Vn = ∂Bn.
11.2 Loop-erased random walk Suppose A ⊂Zd, V ⊂∂A, and x ∈A with that hV,A(x) > 0. The loop-erased random walk (LERW) from x to V in A is the probability measure on paths obtained by taking the hV,A-process stopped at V and erasing loops. We can define the walk equivalently as follows.
• Take a simple random walk Sn started at x and stopped when it reaches ∂A. Condition on the event (of positive probability) {SτA ∈V }. The conditional probability gives a probability measure on (finite) paths ω = [S0 = x, S1, . . . , Sn = SτA] .
• Erase loops from each ω which produces a self-avoiding path η = L(ω) = [ ˆ S0 = x, ˆ S1, . . . , ˆ Sm = SτA], with ˆ S1, . . . , ˆ Sm−1 ∈A. We now have a probability measure on self-avoiding paths from x to V , and this is the LERW.
Similarly, if x ∈∂A \ V with Px{SτA ∈V } > 0, we define LERW from x to V in A by erasing loops from the hV,A-process started at x stopped at V . If x ∈V , we define LERW from x to V to be the trivial path of length zero.
We write the LERW as ˆ S0, ˆ S1, . . . , ˆ Sρ.
Here ρ is the length of the loop-erasure of the h-process.
The LERW gives a probability measure on paths which we give explicitly in the next proposition.
We will use the results and notations from Chapter 9 where the weight q from that chapter is the weight associated to simple random walk, q(x, y) = 1/2d if |x −y| = 1.
Proposition 11.2.1 Suppose V ⊂∂A, x ∈A \ V and ˆ S0, ˆ S1, . . . , ˆ Sρ is LERW from x to V in A.
Suppose η = [η0, . . . , ηn] is a self-avoiding path with η0 = x ∈A, ηn ∈V , and ηj ∈A for 0 < j < n.
Then P{ρ = n; [ ˆ S0, . . . , ˆ Sn] = η} = 1 (2d)n Px{SτA ∈V } Fη(A).
Proof This is proved in the same way as Proposition 9.5.1. The extra term Px{SτA ∈V } comes from the normalization to be a probability measure.
If ω = [ω0, ω1, . . . , ωm] and ωR denotes the reversed path [ωm, ωm−1, . . . , ω0], it is not necessarily true that L(ωR) = [L(ω)]R (the reader might want to find an example). However, the last proposi-tion shows that for any self-avoiding path η with appropriate endpoints, the probability that LERW produces η depends only on the set {η1, . . . , ηn−1}. For this reason we have the following corollary which shows that the distribution of LERW is reversible.
Corollary 11.2.2 (Reversibility of LERW) Suppose x, y ∈∂A and ˆ S0, ˆ S1, . . . , ˆ Sρ is LERW from x to y in A. Then the distribution of ˆ Sρ, ˆ Sρ−1, . . . , ˆ S0 is that of LERW from y to x.
11.2 Loop-erased random walk 249 Proposition 11.2.3 If x ∈A with hV,A(x) > 0, then the distribution of LERW from x to V in A stopped at V is the same as that of LERW from x to V in A \ {x} stopped at V .
Proof Let X0, X1, . . . denote an hV,A-process started at x, and let τ = τA be the first time that the walk leaves A, which with probability one is the first time that the walk visits V . Let σ = max{m < τA : Xm = x}.
Then using last-exit decomposition ideas (see Proposition 4.6.5) and Proposition 11.1.1, the distri-bution of [Xσ, Xσ+1, . . . , XτA] is the same as that of an hV,A-process stopped at V conditioned not to return to x. This is the same as an hV,A{x}-process.
If x ∈∂A \ V , then the first step ˆ S1 of the LERW from x to V in A has the same distribution as the first step of the hV,A-process from x to V . Hence, Px{ ˆ S1 = y} = hV,A(y) P |z−x|=1 hV,A(z).
Proposition 11.2.4 Suppose x ∈A \ V and ˆ S0, . . . , ˆ Sρ denotes LERW from x to V in A. Suppose η = [η0, . . . , ηm] is a self-avoiding path with η0 = x and η1, . . . , ηm ∈A. Then Px{ρ > m; [ ˆ S0, . . . , ˆ Sm] = η} = Pηm{SτA\η ∈V } (2d)m Px{SτA ∈V } Fη(A).
Proof Let ω = [ω0, . . . , ωn] be a nearest neighbor path with ω0 = x, ωn ∈V and ω0, . . . , ωn−1 ∈A such that the length of LE(ω) is greater than m and the first m steps of LE(ω) agrees with η. Let s = max{j : ωj = ηm} and write ω = ω−⊕ω+ where ω−= [ω0, ω1, . . . , ωs], ω+ = [ωs, ωs+1, . . . , ωn].
Then L(ω−) = η and ω+ is a nearest neighbor path from ηm to V with ωs = ηm, {ωs+1, . . . , ωn−1} ∈A \ η, ωn ∈V.
(11.2) Every such ω can be obtained by concatenating an ω−in A with L(ω−) = η with an ω+ satisfying (11.2). The total measure of the set of ω−is given by (2d)−m Fη(A) and the total measure of the set of ω+ is given by Pηm{SτA\η ∈V }. Again, the term Px{SτA ∈V } comes from the normalization to make the LERW a probability measure.
The LERW is not a Markov process. However, we can consider the LERW from x to V in A as a Markov chain on a different state space. Fix V , and consider the state space X of ordered pairs (x, A) with x ∈Zd, A ⊂Zd \ (V ∪{x}) and either x ∈V or Px{SτA ∈V } > 0. The states (x, A), x ∈V are absorbing states. For other states, the probability of the transition (x, A) − →(y, A \ {y}) 250 Loop-erased random walk is the same as the probability that an hV,A-process starting at x takes its first step to y. The fact that this is a Markov chain is sometimes called the domain Markov property for LERW.
11.3 LERW in Zd The loop-erased random walk in Zd is the process obtained by erasing the loops from the path of a d-dimensional simple random walk. The d = 1 case is trivial, so we will focus on d ≥2. We will use the term self-avoiding path for a nearest neighbor path that is self-avoiding.
11.3.1 d ≥3 The definition of LERW is easier in the transient case d ≥3 for then we can take the infinite path [S0, S1, S2, . . .] and erase loops chronologically to obtain the path [ ˆ S0, ˆ S1, ˆ S2, . . .].
To be precise, we let σ0 = max{j ≥0 : Sj = 0}, and for k > 0, σk = max{j > σk−1 : Sj = Sσk−1+1}, and then [ ˆ S0, ˆ S1, ˆ S2, . . .] = [Sσ0, Sσ1, Sσ2, . . .].
♣It is convenient to define chronological erasing as above by considering the last visit to a point. It is not difficult to see that this gives the same path as obtained by “nonanticipating” loop erasure, i.e., every time one visits a point that is on the path one erases all the points in between.
The following properties follow from the previous sections in this chapter and we omit the proofs.
• Given ˆ S0, . . . , ˆ Sm, the distribution of ˆ Sm+1 is that of the h∞,Am-process starting at ˆ Sm where Am = Zd \ { ˆ S0, . . . , ˆ Sm}. Indeed, P{ ˆ Sm+1 = x | [ ˆ S0, . . . , ˆ Sm]} = h∞,Am(x) P |y−ˆ Sm|=1 h∞,Am(y), |x −ˆ Sm| = 1.
• If η = [η0, . . . , ηm] is a self-avoiding path with η0 = 0, P n [ ˆ S0, . . . , ˆ Sm] = η o = EsAm(ηm) (2d)m Fη(Zd) = EsAm(ηm) (2d)m m Y j=0 GAj−1(ηj, ηj).
Here A−1 = Zd.
11.3 LERW in Zd 251 • Suppose Zd \ A is finite, Ar = A ∩{|z| < r}, V r = ∂Ar ∩{|z| ≥r}, and ˆ S(r) 0 , . . . , ˆ S(r) m denotes (the first m steps of) a LERW from 0 to V r in Ar. Then for every self-avoiding path η, P n [ ˆ S0, . . . , ˆ Sm] = η o = lim r→∞P n [ ˆ S(r) 0 , . . . , ˆ S(r) m ] = η o .
11.3.2 d = 2 There are a number of ways to define LERW in Z2; all the reasonable ones give the same answer.
One possibility (see Exercise 11.2) is to take simple random walk conditioned not to return to the origin and erase loops. We take a different approach in this section and define it as the limit as N →∞of the measure obtained by erasing loops from simple random walk stopped when it reaches ∂BN. This approach has the advantage that we obtain an error estimate on the rate of convergence.
Let Sn denote simple random walk starting at the origin in Z2.
Let ˆ S0,N, . . . , ˆ SρN,N denote LERW from 0 to ∂BN in BN. A This can be obtained by erasing loops from [S0, S1, . . . , SξN ].
As noted in Section 11.2, if we condition on the event that τ0 > ξN, we get the same distribution on the LERW. Let ΞN denote the set of self-avoiding paths η = [0, η1, . . . , ηk] with η1, . . . , ηk−1 ∈BN, and ηN ∈∂BN and let νN denote the corresponding probability measure on ΞN, νN(η) = P{[ ˆ S0,N, . . . , ˆ Sn,N] = η}.
If n < N, we can also consider νN as a probability measure on Ξn, by considering the path η up to the first time it visits ∂Bn and removing the rest of the path. The goal of this subsection is to prove the following result.
Proposition 11.3.1 Suppose d = 2 and n < ∞. For each N ≥n, consider νN as a probability measure on Ξn. Then the limit ν = lim N→∞νN, exists. Moreover, for every η ∈Ξn.
νN(η) = ν(η) 1 + O 1 log(N/n) , N ≥2n.
(11.3) ♣To be more specific, (11.3) means that there is a c such that for all N ≥2n and all η ∈Ξn, νN(η) ν(η) −1 ≤ c log(N/n).
The proof of this proposition will require an estimate on the loop measure as defined in Chapter 252 Loop-erased random walk 9. We start by stating the following proposition which is an immediate application of Proposition 11.2.4 to our situation.
Proposition 11.3.2 If n ≤N and η = [η0, . . . , ηk] ∈Ξn, νN(η) = Pηk n ξN < τZd\η o (2d)|η| Fη(BN) = Pηk n ξN < τZd\η o (2d)|η| P{ξN < τ0} Fη(BN \ {0}).
♣Since 0 ∈η, Fη(BN) = GBN (0, 0) Fη(BN \ {0}) = P{ξN < τ0}−1 Fη(BN \ {0}), which shows the second equality in the proposition.
We will say that a loop disconnects the origin from a set A if there is no self-avoiding path starting at the origin ending at A that does not intersect the loop; in particular, loops that intersect the origin disconnect the origin from all sets. Let m denote the unrooted loop measure for simple random walk as defined in Chapter 9.
Lemma 11.3.3 There exists c < ∞such that the following holds for simple random walk in Z2.
For every n < N/2 < ∞consider the set U = U(n, N) of unrooted loops ω satisfying ω ∩Bn ̸= ∅, ω ∩(Zd \ BN) ̸= ∅ and such that ω does not disconnect the origin from ∂Bn. Then m(U) ≤ c log(N/n).
Proof Order Z2 = {x0 = 0, x1, x2, . . .} so that j < k implies |xj| ≤|xk|.
Let Ak = Z2 \ {x0, . . . , xk−1}. For each unrooted loop ω, let k be the smallest index with xk ∈ω and, as be-fore, let dxk(ω) denote the number of times that ω visits xk. By choosing the root uniformly among the dxk(ω) visits to xk, we can see that m(U) = ∞ X k=1 X ω∈˜ Uk 1 (2d)|ω| dxk(ω) ≤ ∞ X k=1 X ω∈˜ Uk 1 (2d)|ω| , where ˜ Uk = ˜ Uk(n, N) denotes the set of (rooted) loops rooted at xk satisfying the following three properties: • ω ∩{x0, . . . , xk−1} = ∅, • ω ∩(Zd \ BN) ̸= ∅, • ω does not disconnect the origin from ∂Bn.
We now give an upper bound for the measure of ˜ Uk for xk ∈Bn. Suppose ω = [ω0, . . . , ω2l] ∈˜ Uk.
Let s0 = ω0, s5 = 2l and define s1, . . . , s4 as follows.
• Let s1 be the smallest index s such that |ωs| ≥2|xk|.
11.3 LERW in Zd 253 • Let s2 be the smallest index s ≥s1 such that |ωs| ≥n • Let s3 be the smallest index s ≥s2 such that |ωs| ≥N.
• Let s4 be the largest index s ≤2l such that |ωs| ≥2|xk|.
Then we can decompose ω = ω1 ⊕ω2 ⊕ω3 ⊕ω4 ⊕ω5, where ωj = [ωsj−1, . . . , ωsj]. We can use this decomposition to estimate the probability of ˜ Uk.
• ω1 is a path from xk to ∂B2|xk| that does not hit {x0, . . . , xk−1}. Using gambler’s ruin (or a similar estimate), the probability of such a path is bounded above by c/|xk|.
• ω2 is a path from ∂B2|xk| to ∂Bn that does not disconnect the origin from ∂Bn.
There exists c, α such that the probability of reaching distance n without disconnecting the origin is bounded above by c (|xk|/n)α (see Exercise 3.4).
• ω3 is a path from ∂Bn to ∂BN that does not disconnect the origin from ∂Bn. The probability of such paths is bounded above by c/ log(N/n), see Exercise 6.4.
• The reverse of ω5 is a path like ω1 and has probability c/|xk|.
• Given ω3 and ω5, ω4 is a path from ωs3 ∈∂BN to ωs4 ∈∂B2|xk| that does not enter {x0, . . . , xk−1}. The expected number such paths is O(1).
Combining all these estimates we see that the measure of ˜ Uk is bounded above by a constant times 1 |xk|2−α nα 1 log(N/n).
By summing over xk ∈Bn, we get the proposition.
♣Being able to verify all the estimates in the last proof is a good test that one has absorbed a lot of material from this book!
Proof [of Proposition 11.3.1] Let ǫr = 1/ log r. We will show that for 2n ≤N ≤M and η = [η0, . . . , ηk] ∈Ξn, νM(η) = νN(η) [1 + O(ǫN/n)].
(11.4) Standard arguments using Cauchy sequences then show the existence of ν satisfying (11.3). Propo-sition 11.3.2 implies νM(η) = νN(η) Fη(BM) Fη(BN) Pηk ξM < τZ2\η | ξN < τZ2\η .
The set of loops contributing to the term Fη(BM)/Fη(BN) are of two types: those that disconnect the origin from ∂Bn and those that do not. Loops that disconnect the origin from ∂Bn intersect every ˜ η ∈Ξn and hence contribute a factor C(n, N, M) that is independent of η. Hence, using Lemma 11.3.3, we see that Fη(BM) Fη(BN) = C(n, N, M) [1 + O(ǫN/n)], (11.5) 254 Loop-erased random walk Using Proposition 6.4.1, we can see that for every x ∈∂BN, Px{ξM < τZ2\∂Bn} = log(N/n) log(M/n) 1 + O(ǫN/n) (actually the error is of smaller order than this). Using (6.49), if x ∈Bn, Px{ξN < τZ2\η} ≤ c log(N/n).
We therefore get Pηk ξM < τZ2\η | ξN < τZ2\η = log(N/n) log(M/n) 1 + O(ǫN/n) .
Combining this with (11.5) we get νM(η) = νN(η) C(n, N, M) log(N/n) log(M/n) 1 + O(ǫN/n) , where we emphasize that the error term is bounded uniformly in η ∈Ξn. However, both νN and νM are probability measures. By summing over η ∈Ξn on both sides, we get C(n, N, M) log(N/n) log(M/n) = 1 + O(ǫN/n), which gives (11.4).
The following is proved similarly (see Exercise 9.7).
Proposition 11.3.4 Suppose d ≥3 and n < ∞. For each N ≥n, consider νN as a probability measure on Ξn. Then the limit ν = lim N→∞νN, exists and is the same as that given by the infinite LERW. Moreover, for every η ∈Ξn.
νN(η) = ν(η) h 1 + O (n/N)d−2i , N ≥2n.
(11.6) 11.4 Rate of growth If ˆ S0, ˆ S1, . . . , denotes LERW in Zd, d ≥2, we let ˆ ξn = min{j : | ˆ Sj| ≥n}.
Let ˆ F(n) = ˆ Fd(n) = E[ˆ ξn].
In other words it takes about ˆ F(n) steps for the LERW to go distance n. Recall that for simple random walk, E[ξn] ∼n2. Note that ˆ F(n) = X x∈Bn P{ ˆ Sj = x for some j < ˆ ξn}.
11.4 Rate of growth 255 By Propositions 11.3.1 and 11.3.4, we know that if x ∈Bn, P{ ˆ Sj = x for some j < ˆ ξn} ≍ P{x ∈LE(S[0, ξ2n])} = ∞ X j=0 P{j < ξ2n; Sj = x; LE(S[0, j]) ∩S[j + 1, ξ2n] = ∅}.
If S, S1 are independent random walks, let ˆ Q(λ) = (1 −λ)2 X x∈Zd ∞ X n=0 ∞ X m=0 λn+m P0,x{LE(S[0, n]) ∩S1[0, m] = ∅}.
In Proposition 10.2.1, a probability of nonintersection of random walks starting at the origin was computed in terms of a “long-range” intersection quantity Q(λ).
We do something similar for LERW using the quantity ˆ Q(λ). The proof of Proposition 10.2.1 used a path decomposition: given two intersecting paths, the proof focused on the first intersection (using the time scale of one of the paths) and then translating to make that the origin. The proof of the next proposition is similar given a simple random walk that intersects a loop-erased walk.
However, we get two different results depending on whether we focus on the first intersection on the time scale of the simple walk or on the time scale of the loop-erased walk.
Proposition 11.4.1 Let S, S1, S2, S3 be independent simple random walks starting at the origin in Zd with independent geometric killing times Tλ, T 1 λ, . . . , T 3 λ.
(i) Let V 1 = V 1 λ be the event that Sj[1, T 1 λ] ∩LE(S[0, Tλ]) = ∅, j = 1, 2, and S3[1, T 3 λ] ∩[LE(S[0, Tλ]) \ {0}] = ∅.
Then P(V 1) = (1 −λ)2 ˆ Q(λ).
(11.7) (ii) Let V 2 = V 2 λ be the event that S1[1, T 1 λ] ∩LE(S[0, Tλ]) = ∅, and S2[1, T 2 λ] ∩ LE(S[0, Tλ]) ∪LE(S1[0, T 1 λ]) = ∅ .
Then P(V 2) = (1 −λ)2 ˆ Q(λ).
(11.8) Proof We use some of the notation from the proof of Proposition 10.2.1. Note that ˆ Q(λ) = (1 −λ)2 ∞ X n=0 ∞ X m=0 X ω,η λn+m p(ω) p(η), 256 Loop-erased random walk where the last sum is over all ω, η with |ω| = n, |η| = m, ω0 = 0 and L(ω) ∩η ̸= ∅. We write ˆ ω = L(ω) = [ˆ ω0, . . . , ˆ ωl].
To prove (11.7), on the event ˆ ω ∩η ̸= ∅, we let u = min{j : ˆ ωj ∈η}, s = max{j : ωj = ˆ ωu}, t = min{k : ηk = ˆ ωu}.
We define the paths ω−, ω+, η−, η+ as in the proof of Proposition 10.2.1 using these values of s, t.
Our definition of s, t implies for j > 0, ω+ j ̸∈LER(ω−), η− j ̸∈LER(ω−), η+ j ̸∈LER(ω−) \ {0}.
(11.9) Here we write LER to indicate that one traverses the path in the reverse direction, erases loops, and then reverses the path again — this is not necessarily the same as LE(ω−). Conversely, for any 4-tuple (ω−, ω+, η−, η+) satisfying (11.9), we get a corresponding (ω, η) satisfying L(ω) ∩η ̸= ∅.
Therefore, ˆ Q(λ) = (1 −λ)2 X 0≤n−,n+,m−,m+ X ω,ω+,η−,η+ λn−+n++m−+m+ p(ω−)p(ω+)p(η−)p(η+), where the last sum is over all (ω−, ω+, η−, η+) with |ω−| = n−, |ω+| = n+, |η−| = m−, |η+| = m+ satisfying (11.9). Using Corollary 11.2.2, we see that the sum is the same if replace (11.9) with: for j > 0, ω+ j ̸∈LE(ω−), η− j ̸∈LE(ω−), η+ j ̸∈LE(ω−) \ {0}.
To prove (11.8), on the event ˆ ω ∩η ̸= ∅, we let t = min{k : ηk ∈ˆ ω}, s = max{j : ωj = ηt}, and define (ω−, ω+, η−, η+) as before. The conditions now become for j > 0, ω+ j ̸∈LER(ω−), η− j ̸∈[LER(ω−) ∪LE(ω+)], It is harder to estimate ˆ Q(λ) then Q(λ).
We do not give a proof here but we state that if λn = 1 −1 n, then as n →∞, ˆ Q(λn) ≍ nd/2, d < 4 n2 [log n]−1, d = 4.
This is the same behavior as for Q(λn). Roughly speaking, if two random walks of length n start distance √n away, then the probability that one walk intersects the loop erasure of the other is of order 1 for d ≤3 and of order 1/ log n for d = 4. For d = 1, 2, this is almost obvious for topological reasons. The hard cases are d = 3, 4. For d = 3 the set of “cut points” (i.e., points Sj such that S[0, j] ∩S[j + 1, n] = ∅) has a “fractal dimension” strictly greater than one and hence tends to be hit by a (roughly two-dimensional) simple random walk path. For d = 4, one can also show that the probability of hitting the cut points is of order 1/ log n. Since all cut points are retained in loop 11.5 Short-range intersections 257 erasure this gives a bound on the probability of hitting the loop-erasure. This estimate of ˆ Q(λ) yields P(V 1 λn) ≍P(V 2 λn) ≍ ( n d−4 2 , d < 4 1 log n, d = 4.
To compute the growth rate we would like to know the asymptotic behavior of P{LE(S[0, λn]) ∩S1[1, Tλn] = ∅} = E[Yn], where Yn = P{LE(S[0, λn]) ∩S1[1, Tλn] = ∅| S[0, λn]}.
Note that P(V 1 λn) ≍E[Y 3 n ].
11.5 Short-range intersections Studying the growth rate for LERW leads one to try to estimate probabilities such as P{LE(S[0, n]) ∩S[n + 1, 2n] = ∅}, which by Corollary 11.2.2 is the same as ˆ qn =: P{LE(S[0, n]) ∩S1[1, n] = ∅}, where S, S1 are independent walks starting at the origin. If d ≥5, ˆ qn ≥P{S[0, n] ∩S1[1, n] = ∅} ≥c > 0, so we will restrict our discussion to d ≤4. Let ˆ Yn = P{LE(S[0, n]) ∩S[n + 1, 2n] = ∅| S[0, n]}.
Using ideas similar to those leading up to (11.7), one can show that E[ ˆ Y 3 n ] ≍ ( (log n)−1, d = 4 n d−4 2 , d = 1, 2, 3.
(11.10) This can be compared to (10.6) where the second moment for an analogous quantity is given. We also know that E[ ˆ Y 3 n ] ≤E[ ˆ Yn] ≤ E[ ˆ Yn] 1/3 .
(11.11) In the “mean-field” case d = 4, it can be shown that E[ ˆ Y 3 n ] ≍ E[ ˆ Yn] 3 . and hence that qn ≍(log n)−1/3.
Moreover, if we appropriately scale the process, the LERW converges to a Brownian motion.
258 Loop-erased random walk For d = 2, 3, we do not expect E[ ˆ Y 3 n ] ≍ E[ ˆ Yn] 3 . Let us define an exponent α = αd roughly as qn ≈n−α. The relations (11.10) and (11.11) imply that α ≥4 −d 6 .
It has been shown that for d = 2 the exponent α exists with α = 3/8; in other words, the expected number of points in LE(S[0, n]) is of order n5/8 (see and also ). This suggests that if we scale appropriately, there should be a limit process whose paths have fractal dimension 5/4. In fact, this has been proved. The limit process is the Schramm-Loewner evolution (SLE) with parameter κ = 2 .
For d = 3, we get the bound α ≥1/6 which states that the number of points in LE(S[0, n]) should be no more than n5/6. We also expect that this bound is not sharp. The value of α3 is any open problem; in fact, the existence of an exponent satisfying qn ≈n−α has not been established.
However, the existence of a scaling limit has been shown in Exercises Exercise 11.1 Suppose d ≥3 and Xn is simple random walk in Zd conditioned to return to the origin. This is the h-process with h(x) = Px{Sn = 0 for some n ≥0}.
Prove that (i) For all d ≥3, Xn is a recurrent Markov chain.
(ii) Assume X0 = 0 and let T = min{j > 0 : Xj = 0}. Show that there is a c = cd > 0 such that P{T = 2n} ≍n−d/2, n →∞.
In particular, E[T] = ∞, d ≤4, < ∞, d ≥5.
Exercise 11.2 Suppose Xn is simple random walk in Z2 conditioned not to return to the origin.
This is the h-process with h(x) = a(x).
(i) Prove that Xn is a transient Markov chain.
(ii) Show that if loops are erased chronologically from this chain, then one gets LERW in Z2.
12 Appendix 12.1 Some expansions 12.1.1 Riemann sums In this book we often approximate sums by integrals. Here we give bounds on the size of the error in such approximation.
Lemma 12.1.1 If f : (0, ∞) →R is a C2 function, and bn is defined by bn = f(n) − Z n+(1/2) n−(1/2) f(s) ds, then |bn| ≤1 24 sup |f ′′(r)n −r| ≤1 2 .
(12.1) If P |bn| < ∞, let C = ∞ X n=1 bn, Bn = ∞ X j=n+1 |bn|.
Then n X j=1 f(j) = Z n+(1/2) 1/2 f(s) ds + C + O(Bn).
Also, for all m < n n X j=m f(j) − Z n+(1/2) m−(1/2) f(s) ds ≤Bm.
Proof Taylor’s theorem shows that for |s −n| ≤1/2, f(s) = f(n) + (s −n) f ′(n) + 1 2 f ′′(rs) (rs −n)2, for some |n −rs| < 1/2. Hence, for such s, |f(s) + f(−s) −2f(n)| ≤s2 2 sup{|f ′′(r)n −r| ≤s}.
259 260 Appendix Integrating gives (12.1). The rest is straightforward.
Example. Suppose α < 1, β ∈R and f(n) = nα logβ n.
Note that for t ≥2, |f ′′(t)| ≤c tα−2 logβ t.
Therefore, there is a C(α, β) such that n X j=2 nα logβ n = Z n+(1/2) 2 tα logβ t dt + C(α, β) + O(nα−1 logβ n) = Z n 2 tα logβ t dt + 1 2 nα logβ n + C(α, β) + O(nα−1 logβ n) (12.2) 12.1.2 Logarithm Let log denote the branch of the complex logarithm on {z ∈C; Re(z) > 0} with log 1 = 0. Using the power series log(1 + z) = k X j=1 (−1)j+1zj j + Oǫ(|z|k+1), |z| ≤1 −ǫ.
we see that if r ∈(0, 1) and |ξ| ≤rt, log 1 + ξ t t = ξ −ξ2 2t + ξ3 3t2 + · · · + (−1)k+1 ξk ktk−1 + Or |ξ|k+1 tk , 1 + ξ t t = eξ exp −ξ2 2t + ξ3 3t2 + · · · + (−1)k+1 ξk ktk−1 + Or |ξ|k+1 tk .
(12.3) If |ξ|2/t is not too big, we can expand the exponential in a Taylor series. Recall that for fixed R < ∞, we can write ez = 1 + z + z2 2! + · · · + zk k! + OR(|z|k+1), |z| ≤R.
Therefore, if r ∈(0, 1), R < ∞, |ξ| ≤rt, |ξ|2 ≤Rt, we can write 1 + ξ t t = eξ 1 −ξ2 2t + 8ξ3 + 3ξ4 24t2 + · · · + fk(ξ) tk−1 + O |ξ|2k tk , (12.4) where fk is a polynomial of degree 2(k −1) and the implicit constant in the O(·) term depends only on r, R and k. In particular, 1 + 1 n n = e 1 −1 2n + 11 24 n2 + · · · + bk nk + O 1 nk+1 .
(12.5) 12.1 Some expansions 261 Lemma 12.1.2 For every positive integer k, there exist constants c(k, l), l = k + 1, k + 2, . . ., such that for each m > k, ∞ X j=n k jk+1 = 1 nk + m X l=k+1 c(k, l) nl + O 1 nm+1 .
(12.6) Proof If n > 1, n−k = ∞ X j=n [j−k −(j + 1)−k] = ∞ X j=n j−k [1 −(1 + j−1)−k] = ∞ X l=k b(k, l) ∞ X j=n l jl+1 , with b(k, k) = 1 (the other constants can be given explicitly but we do not need to). In particular, n−k = m X l=k b(k, j) ∞ X j=n l jl+1 + O 1 nm+1 .
The expression (12.6) can be obtained by inverting this expression; we omit the details.
Lemma 12.1.3 There exists a constant γ (called Euler’s constant) and b2, b3, . . . such that for every integer k ≥2, n X j=1 1 j = log n + γ + 1 2n + k X l=2 bl nl + O 1 nk+1 .
In fact, γ = lim n→∞ n X j=1 1 j −log n = Z 1 0 (1 −e−t) 1 t dt − Z ∞ 1 e−t 1 t dt.
(12.7) Proof Note that n X j=1 1 j = log n + 1 2 + log 2 + n X j=1 βj, where βj = 1 j −log j + 1 2 + log j −1 2 = − ∞ X k=1 2 (2k + 1) (2j)2k+1 .
In particular, βj = O(j−3), and hence P βj < ∞. We can write n X j=1 1 j = log n + 1 2 + γ − ∞ X j=n+1 βj = log n + γ + ∞ X l=1 (−1)l+1 2l nl − ∞ X j=n+1 βj, where γ is the constant γ = log 2 + ∞ X j=1 βj.
262 Appendix Using (12.6), we can write ∞ X j=n+1 βj = k X l=3 al nl + O 1 nk+1 , for some constants al.
We will sketch the proof of (12.7) leaving the details to the reader. By Taylor’s series, we know that log n = −log 1 − 1 −1 n = ∞ X j=1 1 −1 n j 1 j .
Therefore, γ = lim n→∞ n X j=1 1 j −log n = lim n→∞ n X j=1 " 1 − 1 −1 n j# 1 j − lim n→∞ ∞ X j=n+1 1 −1 n j 1 j .
We now use the approximation (1 −n−1)n ∼e−1 to get lim n→∞ n X j=1 " 1 − 1 −1 n j# 1 j = lim n→∞ n X j=1 1 n (1 −e−j/n) 1 j/n = Z 1 0 (1 −e−t) 1 t dt, lim n→∞ ∞ X j=n+1 1 −1 n j 1 j = lim n→∞ ∞ X j=n+1 1 n e−j/n 1 j/n = Z ∞ 1 e−t 1 t dt.
Lemma 12.1.4 Suppose α ∈R and m is a positive integer. There exist constants r0, r1, . . ., such that if k is a positive integer and n ≥m, n Y j=m 1 −α j = r0 n−α 1 + r1 n + · · · + rk nk + O 1 nk+1 .
Proof Without loss of generality we assume that |α| ≤2m; if this does not hold we can factor out the first few terms of the product and then analyze the remaining terms. Note that log n Y j=m 1 −α j = n X j=m log 1 −α j = − n X j=m ∞ X l=1 αl l jl = − ∞ X l=1 n X j=m αl l jl .
For the l = 1 term we have n X j=m α j = − m−1 X j=1 α j + α " log n + γ + 1 2n + k X l=2 bl nl + O 1 nk+1 # .
12.2 Martingales 263 All of the other terms can be written in powers of (1/n). Therefore, we can write log n Y j=m 1 −α j = −α log n + k X l=0 Cl nl + O 1 nk+1 .
The lemma is then obtained by exponentiating both sides.
12.2 Martingales A filtration F0 ⊂F1 ⊂· · · is an increasing sequence of σ-algebras.
Definition.
A sequence of integrable random variables M0, M1, . . . is called a martingale with respect to the filtration {Fn} if each Mn is Fn-measurable and for each m ≤n, E[Mn | Fm] = Mm.
(12.8) If (12.8) is replaced with E[Mn | Fm] ≥Mm the sequence is called a submartingale. If (12.8) is replaced with E[Mn | Fm] ≤Mm the sequence is called a supermartingale.
Using properties of conditional expectation, it is easy to see that to verify (12.8) it suffices to show for each n that E[Mn+1 | Fn] = Mn. This equality only needs to hold up to an event of probability zero; in fact, the conditional expectation is only defined up to events of probability zero. If the filtration is not specified, then the assumption is that Fn is the σ-algebra generated by M0, . . . , Mn. If M0, X1, X2, . . . are independent random variables with E[|M0|] < ∞and E[Xj] = 0 for j ≥1, and Mn = M0 + X1 + · · · + Xn, then M0, M1, . . . is a martingale. We omit the proof of the next lemma which is the conditional expectation version of Jensen’s inequality.
Lemma 12.2.1 (Jensen’s inequality) If X is an integrable random variable; f : R →R is convex with E[|f(X)|] < ∞; and F is a σ-algebra, then E[f(X) | F] ≥f(E[X | F]). In particular, if M0, M1, . . . is a martingale; f : R →R is convex with E[|f(Mn)|] < ∞for all n; and Yn = f(Mn); then Y0, Y1, . . . is a submartingale.
In particular, if M0, M1, . . . is a martingale then • if α ≥1, Yn := |Mn|α is a submartingale; • if b ∈R, then Yn := ebMn is a submartingale.
In both cases, this is assuming that E[Yn] < ∞.
12.2.1 Optional Sampling Theorem A stopping time with respect to a filtration {Fn} is a {0, 1, . . .} ∪{∞}-valued random variable T such that for each n, {T ≤n} is Fn-measurable. If T is a stopping time, and n is a positive integer, then Tn := T ∧n is a stopping time satisfying Tn ≤n.
264 Appendix Proposition 12.2.2 Suppose M0, M1, . . . is a martingale and T is a stopping time each with respect to the filtration Fn. Then Yn := MTn is a martingale with respect to Fn. In particular, E[M0] = E[MTn].
Proof Note that Yn+1 = MTn 1{T ≤n} + Mn+1 1{T ≥n + 1}.
The event {T ≥n + 1} is the complement of the event {T ≤n} and hence is Fn-measurable.
Therefore, by properties of conditional expectation, E[Mn+1 1{T ≥n + 1} | Fn] = 1{T ≥n + 1} E[Mn+1 | Fn] = 1{T ≥n + 1} Mn.
Therefore, E[Yn+1 | Fn] = MTn 1{T ≤n} + Mn 1{T ≥n + 1} = Yn.
The optional sampling theorem states that under certain conditions, if P{T < ∞} = 1, then E[M0] = E[MT ]. However, this does not hold without some further assumptions. For example, if Mn is one-dimensional simple random walk starting at the origin and T is the first n such that Mn = 1, then P{T < ∞} = 1, MT = 1, and hence E[M0] ̸= E[MT ]. In the next theorem we list a number of sufficient conditions under which we can conclude that E[M0] = E[MT ].
Theorem 12.2.3 (Optional Sampling Theorem) Suppose M0, M1, . . . is a martingale and T is a stopping time with respect to the filtration {Fn}. Suppose that P{T < ∞} = 1. Suppose also that at least one of the following conditions holds: • There is a K < ∞such that P{T ≤K} = 1.
• There exists an integrable random variable Y such that for all n, |MTn| ≤Y .
• E [|MT |] < ∞and limn→∞E[|Mn|; T > n] = 0.
• The random variables M0, M1, . . . are uniformly integrable, i.e., for every ǫ > 0 there is a Kǫ < ∞such that for all n, E[|Mn|; |Mn| > Kǫ] < ǫ.
• There exists an α > 1 and a K < ∞such that for all n, E[|Mn|α] ≤K.
Then E[M0] = E[MT ].
Proof We will consider the conditions in order. The sufficiency of the first follows immediately from Proposition 12.2.2. We know that MTn →MT with probability one. Proposition 12.2.2 gives E[MTn] = E[M0]. Hence we need to show that lim n→∞E[MTn] = E[MT ].
(12.9) If the second condition holds, then this limit is justified by the dominated convergence theorem.
Now assume the third condition. Note that MT = MTn + MT 1{T > n} −Mn 1{T > n}.
12.2 Martingales 265 Since P{T > n} →0, and E[|MT |] < ∞, it follows from the dominated convergence theorem that lim n→∞E[MT 1{T > n}] = 0.
Hence if E[Mn 1{T > n}] →0, we have (12.9). Standard exercises show that the fourth implies the third and the fifth condition implies the fourth, so either the fourth or fifth condition is sufficient.
12.2.2 Maximal inequality Theorem 12.2.4 (Maximal inequality) Suppose M0, M1, . . . is a nonnegative submartingale with respect to {Fn} and λ > 0. Then P max 0≤j≤n Mj ≥λ ≤E[Mn] λ .
Proof Let T = min{j ≥0 : Mj ≥λ}. Then, P max 0≤j≤n Mj ≥λ = n X j=0 P{T = j}, E[Mn] ≥E[Mn; T ≤n] = n X j=0 E[Mn; T = j].
Since Mn is a submartingale and {T = j} is Fj-measurable, E[Mn; T = j] = E[E[Mn | Fj]; T = j] ≥E[Mj; T = j] ≥λ P{T = j}.
Combining these estimates gives the theorem.
Combining Theorem 12.2.4 with Lemma 12.2.1 gives the following theorem.
Theorem 12.2.5 (Martingale maximal inequalities) Suppose M0, M1, . . . is a martingale with respect to {Fn} and λ > 0. Then if α ≥1, b ≥0, P max 0≤j≤n |Mj| ≥λ ≤E[|Mn|α] λα , (12.10) P max 0≤j≤n Mj ≥λ ≤E[ebMn] ebλ .
Corollary 12.2.6 Let X1, X2, . . . be independent, identically distributed random variables in R with mean zero, and let k be a positive integer for which E[|X1|2k] < ∞. There exists c < ∞such that for all λ > 0, P max 0≤j≤n |Sj| ≥λ √n ≤c λ−2k.
(12.11) 266 Appendix Proof Fix k and allow constants to depend on k. Note that E[S2k n ] = X E[Xj1 · · · Xj2k], where the sum is over all (j1, . . . , j2k) ∈{1, . . . , n}2k. If there exists an l such that ji ̸= jl for i ̸= l, then we can use independence and E[Xjl] = 0 to see that E[Xj1 · · · Xj2k] = 0. Hence E[S2k n ] = X E[Xj1 · · · Xj2k], where the sum is over all (2k)-tuples such that if l ∈{j1, . . . , j2k}, then l appears at least twice.
The number of such (2k)-tuples is O(nk) and hence we can see that E " Sn √n 2k# ≤c.
Hence we can apply (12.10) to the martingale Mj = Sj/√n.
Corollary 12.2.7 Let X1, X2, . . . be independent, identically distributed random variables in R with mean zero, variance σ2, and such that for some δ > 0, the moment generating function ψ(t) = E[etXj] exists for |t| < δ. Let Sn = X1 + · · · + Xn. Then for all 0 ≤r ≤δ√n/2, P max 0≤j≤n Sj ≥r σ √n ≤e−r2/2 exp O r3 √n .
(12.12) If P{X1 ≥R} = 0 for some R, this holds for all r > 0.
Proof Without loss of generality, we may assume σ2 = 1. The moment generating function of Sn = X1 + · · · + Xn is ψ(t)n. Letting t = r/√n, we get P max 0≤j≤n Sj ≥r √n ≤e−r2 ψ(r/√n)n.
Using the expansion for ψ(t) at zero, ψ(t) = 1 + t2 2 + O(t3), |t| ≤δ 2, we see that for 0 ≤r ≤δ√n/2, ψ(r/√n)n = 1 + r2 2n + O r3 n3/2 n ≤er2/2 exp O r3 √n .
This gives (12.12). If P{X1 > R} = 0, then (12.12) holds for r > R√n trivially, and we can choose δ = 2R.
Remark. From the last corollary, we also get the following modulus of continuity result for random walk. Let X1, X2, . . . and Sn be as in the previous lemma. There exist c, b such that for every m ≤n and every 0 ≤r ≤δ√m/2 P{ max 0≤j≤n max 1≤k≤m |Sk+j −Sj| ≥r √m} ≤c n e−br2.
(12.13) This next lemma is not about martingales, but it does concern exponential estimates for proba-bilities so we will include it here.
12.3 Joint normal distributions 267 Lemma 12.2.8 If 0 < α < ∞, 0 < r < 1 and Xn is a binomial random variable with parameters n and α e−2α/r, then P{Xn ≥rn} ≤e−αn.
Proof P{Xn ≥rn} ≤e−2αn E[e(2α/r)Xn] ≤e−2αn [1 + α]n ≤e−αn.
12.2.3 Continuous martingales A process Mt adapted to a filtration Ft is called a continuous martingale if for each s < t, E[Mt | Ms] = Ms and with probability one the function t 7→Mt is continuous. If Mt is a continuous martingale, and δ > 0, then M(δ) n := Mδn is a discrete time martingale. Using this, we can extend results about discrete time martingales to continuous martingales. We state one such result here.
Theorem 12.2.9 (Optional Sampling Theorem) Suppose Mt is a uniformly integrable con-tinuous martingale and τ is a stopping time with P{τ < ∞} = 1 and E[|Mτ|] < ∞. Suppose that lim t→∞E[|Mt|; τ > t] = 0.
Then E[MT ] = E[M0].
12.3 Joint normal distributions A random vector Z = (Z1, . . . , Zd) ∈Rd is said to have a (mean zero) joint normal distribution if there exist independent (one-dimensional) mean zero, variance one normal random variables N1, . . . , Nn and scalars ajk such that Zj = aj1 N1 + · · · + ajn Nn, j = 1, . . . , d, or in matrix form Z = AN.
Here A = (ajk) is a d × n matrix and Z, N are column vectors. Note that E(ZjZk) = n X m=1 ajm akm.
In other words, the covariance matrix Γ = [ E(ZjZk) ] is the d × d symmetric matrix Γ = AAT .
268 Appendix We say Z has a nondegenerate distribution if Γ is invertible.
The characteristic function of Z can be computed using the known formula for the characteristic function of Nk, E[eitNk] = e−t2/2, E[exp{iθ · Z}] = E exp i d X j=1 θj n X k=1 ajk Nk = E exp i n X k=1 Nk d X j=1 θjajk = n Y k=1 E exp iNk d X j=1 θjajk = n Y k=1 exp −1 2 d X j=1 θjajk 2 = exp −1 2 n X k=1 d X j=1 d X l=1 θj θl ajk alk = exp −1 2 θAATθT = exp −1 2 θΓθT .
Since the characteristic function determines the distribution, we see that the distribution of Z depends only on Γ.
The matrix Γ is symmetric and nonnegative definite. Hence we can find an orthogonal basis u1, . . . , ud of unit vectors in Rd that are eigenvectors of Γ with nonnegative eigenvalues α1, . . . , αd.
The random variable Z = √α1 N1 u1 + · · · + √αd Nd ud has a joint normal distribution with covariance matrix Γ. In matrix language, we have written Γ = Λ ΛT = Λ2 for a d × d nonnegative definite symmetric matrix Λ. The distribution is nondegenerate if and only if all of the αj are strictly positive.
♣Although we allow the matrix A to have n columns, what we have shown is that there is a symmetric, positive definite d × d matrix Λ which gives the same distribution. Hence joint normal distribution in Rd can be described as linear combinations of d independent one-dimensional normals. Moreover, if we choose the correct orthogonal basis for Rd, the components of Z with respect to that basis are independent normals.
If Γ is invertible, then Z has a density f(z1, . . . , zd) with respect to Lebesgue measure that can 12.4 Markov chains 269 be computed using the inversion formula f(z1, . . . , zd) = 1 (2π)d Z e−iθ·z E[exp{iθ · Z}] dθ = 1 (2π)d Z exp −iθ · z −1 2 θΓθT dθ.
(Here and for the remainder of this paragraph the integrals are over Rd and dθ represents ddθ.) To evaluate the integral, we start with the substitution θ1 = Λθ which gives Z exp −iθ · z −1 2 θΓθT dθ = 1 det Λ Z e−|θ1|2/2 e−i(θ1·Λ−1z) dθ1.
By completing the square we see that the right-hand side equals e−|Λ−1z|2/2 det Λ Z exp 1 2 (iθ1 −Λ−1z) · (iθ1 −Λ−1z) dθ1.
The substitution θ2 = θ1 −iΛ−1z gives Z exp 1 2 (iθ1 −Λ−1z) · (iθ1 −Λ−1z) dθ1 = Z e−|θ2|2/2 dθ2 = (2π)d/2.
Hence, the density of Z is f(z) = 1 (2π)d/2 √ det Γ e−|Λ−1z|2/2 = 1 (2π)d/2 √ det Γ e−(z·Γ−1z)/2.
(12.14) Corollary 12.3.1 Suppose Z = (Z1, . . . , Zd) has a mean zero, joint normal distribution such that E[ZjZk] = 0 for all j ̸= k. Then Z1, . . . , Zd are independent.
Proof Suppose E[ZjZk] = 0 for all j ̸= k. Then Z has the same distribution as (b1N1, . . . , bdNd), where bj = q E[Z2 j ]. In this representation, the components are obviously independent.
♣If Z1, . . . , Zd are mean zero random variables satisfying E[ZjZk] = 0 for all j ̸= k, they are called orthogonal. Independence implies orthogonality but the converse is not always true. However, the corollary tells us that the converse is true in the case of joint normal random variables. Orthogonality is often easier to verify than independence.
12.4 Markov chains A (time-homogeneous) Markov chain on a countable state space D is a process Xn taking values in D whose transitions satisfy P{Xn+1 = xn+1 | X0 = x0, . . . , Xn = xn} = p(xn, xn+1) 270 Appendix where p : D × D →[0, 1] is the transition function satisfying P y∈D p(x, y) = 1 for each x. If A is finite, we call the transition function the transition matrix P = [p(x, y)]x,y∈A. The n-step transitions are given by the matrix P n. In other words, if pn(x, y) is defined to be P{Xn = y | X0 = x}, then pn(x, y) = X z∈D p(x, z) pn−1(z, y) = X z∈D pn−1(x, z) p(z, y).
A Markov chain is called irreducible if for each x, y ∈A, there exists an n = n(x, y) ≥0 with pn(x, y) > 0. The chain is aperiodic if for each x there is an Nx such that for n ≥Nx, pn(x, x) > 0.
If D is finite, then the chain is irreducible and aperiodic if and only if there exists an n such that P n has strictly positive entries.
Theorem 12.4.1 [Perron-Froebenius Theorem] If P is an m×m matrix such that for some positive integer n, P n has all entries strictly positive, then there exists α > 0 and vectors v, w, with strictly positive entries such that v P = α v, P w = α w.
This eigenvalue is simple and all other eigenvalues of P have absolute value strictly less than α. In particular, if P is the transition matrix for an irreducible aperiodic Markov chain there is a unique invariant probability π satisfying X x∈D π(x) = 1, π(x) = X y∈D π(y) P(y, x).
Proof We first assume that P has all strictly positive entries. It suffices to find a right eigenvector, since the left eigenvector can be handled by considering the transpose of P. We write w1 ≥w2 if every component of w1 is greater than or equal to the corresponding component of w2. Similarly, we write w1 > w2 if all the components of w1 are strictly greater than the corresponding components of w2. We let 0 denote the zero vector and ej the vector whose jth component is 1 and whose other components are 0. If w ≥0, let λw = sup{λ : Pw ≥λw}.
Clearly λw < ∞, and since P has strictly positive entries, λw > 0 for all w > 0. Let α = sup{λw : w ≥0, m X j=1 [w]j = 1}.
By compactness and continuity arguments we can see that there exists a w with w ≥0, P j[w]j = 1 and λw = α. We claim that Pw = αw. Indeed, if [Pw]j > α[w]j for some j, one can check that there exist positive ǫ, ρ such that P[w + ǫej] ≥(α + ρ) [w + ǫej], which contradicts the maximality of α. If v is a vector with both positive and negative component, then for each j, |[Pv]j| < [P|v|]j ≤α [|v|]j.
Here we write |v| for the vector whose components are the absolute values of the components of v.
Hence any eigenvector with both positive and negative values has an eigenvalue with absolute value strictly less than α. Also, if w1, w2 are positive eigenvectors with eigenvalue α, then w1 −tw2 is an eigenvector for each t. If w1 is not a multiple of w2 then there is some value of t such that w1 −tw2 12.4 Markov chains 271 has both positive and negative values. Since this is impossible, we conclude that the eigenvector w is unique. If w ≥0 is an eigenvector, then the eigenvalue must be positive. Therefore, α has a unique eigenvector (up to constant), and all other eigenvalues have absolute value strictly less than α. Note that if v > 0, then Pv has all entries strictly positive; hence the eigenvector w must have all entries strictly positive.
We claim, in fact, that α is a simple eigenvalue. To see this, one can use the argument as in the previous paragraph to all submatrices of the matrix to conclude that all eigenvalues of all submatrices of the matrix are strictly less than α in absolute value. Using this (details omitted), one can see that the derivative of the function f(λ) = det(λI −P) is nonzero at λ = α which shows that the eigenvalue is simple.
If P is a matrix such that P n has all entries strictly positive, and w is an eigenvector of P with eigenvalue α, then w is an eigenvector for P n with eigenvalue αn. Using this, we can conclude the result for P. The final assertion follows by noting that the vector of all 1s is a right eigenvector for a stochastic matrix.
♣A different derivation of the Perron-Froebenius Theorem which generalizes to some chains on infinite state spaces is done in Exercise 12.4.
If P is the transition matrix for a irreducible, aperiodic Markov chain, then pn(x, y) →π(y) as n →∞. In fact, this holds for countable state space provided the chain is positive recurrent, i.e., if there exists an invariant probability measure. The next proposition gives a simple, quantitative version of this fact provided the chain satisfies a certain condition which always holds for the finite irreducible, aperiodic case.
Proposition 12.4.2 Suppose p : D×D →[0, 1] is the transition probability for a positive recurrent, irreducible, aperiodic Markov chain on a countable state space D.
Let π denote the invariant probability measure. Suppose there exist ǫ > 0 and a positive integer k such that for all x, y ∈D, 1 2 X z∈D |pk(x, z) −pk(y, z)| ≤1 −ǫ.
(12.15) Then for all positive integers j and all x ∈A, 1 2 X z∈D |pj(x, z) −π(z)| ≤c e−βj, where c = (1 −ǫ)−1 and e−β = (1 −ǫ)1/k.
Proof If ν is any probability distribution on D, let νj(x) = X y∈D ν(y)pj(y, x).
Then (12.15) implies that for every ν, 1 2 X z∈D |νk(z) −π(z)| ≤1 −ǫ.
272 Appendix In other words we can write νk = ǫ π + (1 −ǫ) ν(1) for some probability measure ν(1). By iterating (12.15), we can see that for every integer i ≥1 we can write νik = (1 −ǫ)i ν(i) + [1 −(1 −ǫ)i]π for some probability measure ν(i). This establishes the result for j = ki (with c = 1 for these values of j) and for other j we find i with ik ≤j < (i + 1)k.
12.4.1 Chains restricted to subsets We will now consider Markov chains restricted to a subset of the original state space. If Xn is an irreducible, aperiodic Markov chain with state space D and A is a finite proper subset of D, we write PA = [p(x, y)]x,y∈A. Note that (PA)n = [pA n (x, y)]x,y∈A where pA n (x, y) = P{Xn = y; X0, . . . , Xn ∈A | X0 = x} = Px{Xn = y, τA > n}, (12.16) where τA = inf{n : Xn ̸∈A}. Note that Px{τA > n} = X y∈A pA n(x, y).
We call A connected and aperiodic (with respect to P) if for each x, y ∈A, there is an N such that for n ≥N, pA n(x, y) > 0. If A is finite, then A is connected and aperiodic if and only if there exists an n such that (PA)n has all entries strictly positive. In this case all of the row sums of PA are less than or equal to one and (since A is a proper subset) there is at least one row whose sum is strictly less than one.
Suppose Xn is an irreducible, aperiodic Markov chain with state space D and A is a finite, connected, aperiodic proper subset of D. Let α be as in the Perron-Froebenius Theorem for the matrix PA. Then 0 < α < 1. Let v, w be the corresponding positive eigenvectors which we write as functions, X x∈A v(x) p(x, y) = α v(y), X y∈A w(y) p(x, y) = α w(x).
We normalize the functions so that X x∈A v(x) = 1, X x∈A v(x) w(x) = 1, and we let π(x) = v(x) w(x). Let qA(x, y) = α−1 p(x, y) w(y) w(x).
Note that X y∈A qA(x, y) = P y∈A p(x, y) w(y) α w(x) = 1.
In other words, QA := [qA(x, y)]x,y∈A is the transition matrix for a Markov chain which we will denote by Yn. Note that (QA)n = [qA n (x, y)]x,y∈A where qA n (x, y) = α−n pA n (x, y) w(y) w(x) 12.4 Markov chains 273 and pA n (x, y) is as in (12.16). From this we see that the chain is irreducible and aperiodic. Since X x∈A π(x) qA(x, y) = X x∈A v(x) w(x) α−1 p(x, y) w(y) w(x) = π(y), we see that π is the invariant probability for this chain.
Proposition 12.4.3 Under the assumptions above, there exist c, β such that for all n, |α−n pA n(x, y) −w(x) v(y)| ≤c e−βn.
In particular, P{X0, . . . , Xn ∈A | X0 = x} = w(x) αn [1 + O(e−βn)].
Proof Consider the Markov chain with transition matrix QA. Choose positive integer k and ǫ > 0 such that qA k (x, y) ≥ǫ π(y) for all x, y ∈A. Proposition 12.4.2 gives |qA n (x, y) −π(y)| ≤c e−βn, for some c, β.
Since π(y) = v(y) w(y) and qA n (x, y) = α−n pA n (x, y) w(y)/w(x), we get the first assertion, using the fact that A is finite so that inf v > 0. The second assertion follows from the first using P y v(y) = 1 and P{X0, . . . , Xn ∈A | X0 = x} = X y∈A pA n(x, y).
If the Markov chain is symmetric (p(y, x) = p(x, y)), then w(x) = c v(x), π(x) = c v(x)2. The function g(x) = √c v(x) can be characterized by the fact that g is strictly positive and satisfies P Ag(x) = α g(x), X x∈A g(x)2 = 1.
♣The chain Yn can be considered the chain derived from Xn by conditioning the chain to “stay in A forever”.
The probability measures v, π are both “invariant” (sometimes the word quasi-invariant is used) probability measures but with different interpretations.
Roughly speaking, the three measures v, w, π can be described as follows.
• Suppose the chain Xn is observed at a large time n and it known that the chain has stayed in A for all times up to n. Then the conditional distribution on Xn given this information approaches v.
• For x ∈A, the probability that the chain stays in A up to time n is asymptotic to w(x) αn.
• Suppose the chain Xn is observed at a large time n and it is known that the chain has stayed in A and will stay in A for all times up to N where N ≫n. Then the conditional distribution on Xn given this information approaches π. We can think of the first term of the product v(x) w(x) as the conditional probability of being at x given that the walk has stayed in A up to time n and the second part of the product is the conditional probability given this that the walk stays in A for times between n and N.
The next proposition gives a criterion for determining the rate of convergence to the invariant distribution v. Let us write ˆ pA n (x, y) = pA n (x, y) P z∈A pA n(x, z) = Px{Xn = y | τA > n}.
274 Appendix Proposition 12.4.4 Suppose Xn is an irreducible, aperiodic Markov chain on the countable state space D. Suppose A is a finite, proper subset of D and A′ ⊂A. Suppose there exist ǫ > 0 and integer k > 1 such that the following is true.
• If x ∈A, X y∈A′ ˆ pk(x, y) ≥ǫ.
(12.17) • If x, x′ ∈A, X y∈A′ [ˆ pA k (x, y) ∧ˆ pA k (x′, y)] ≥ǫ.
(12.18) • If x ∈A, y ∈A′ and n is a positive integer Py{τA > n} ≥ǫ Px{τA > n}.
(12.19) Then there exists δ > 0, depending only on ǫ, such that for all x, z ∈A and all integers m ≥0, 1 2 X y∈A ˆ pA km(x, y) −ˆ pA km(z, y) ≤(1 −δ)m.
Proof We fix ǫ and allow all constants in this proof to depend on ǫ. Let qn = maxy∈A Py{τA > n}.
Then (12.19) implies that for all y ∈A′ and all n, Py{τA > n} ≥ǫ qn. Combining this with (12.17) gives for all positive integers k, n, c qn Px{τA > k} ≤Px{τA > k + n} ≤qn Px{τA > k}.
(12.20) Let m be a positive integer and let Y0, Y1, Y2, . . . Ym be the process corresponding to X0, Xk, X2k, . . . , Xmk conditioned so that τA > mk. This is a time inhomogeneous Markov chain with transition probabilities P{Yj = y | Yj−1 = x} = pA k (x, y) Py{τA > (m −j)k} Px{τA > (m −j + 1)k} , j = 1, 2, . . . , m.
Note that (12.20) implies that for all y ∈A, P{Yj = y | Yj−1 = x} ≤c2 ˆ pA k (x, y), and if y ∈A′, P{Yj = y | Yj−1 = x} ≥c1 ˆ pA k (x, y).
Using this and (12.18), we can see that there is a δ > 0 such that if x, z ∈A and j ≤m, 1 2 X y∈A |P{Yj = y | Yj−1 = x} −P{Yj = y | Yj−1 = z}| ≤1 −δ, and using an argument as in the proof of Proposition 12.4.2 we can see that 1 2 X y∈A |P{Ym = y | Y0 = x} −P{Ym = y | Y0 = z}| ≤(1 −δ)m.
12.4 Markov chains 275 12.4.2 Maximal coupling of Markov chains Here we will describe the maximal coupling of a Markov chain. Suppose that p : D × D →[0, 1] is the transition probability function for an irreducible, aperiodic Markov chain with countable state space D. Assume that g1 0, g2 0 are two initial probability distributions on D. Let gj n denote the corresponding distribution at time n, given recursively by gj n(x) = X z∈D gj n−1(z) p(z, x).
Let ∥· ∥denote the total variation distance, ∥g1 n −g2 n∥= 1 2 X x∈D |g1 n(x) −g2 n(x)| = 1 − X x∈D [g1 n(x) ∧g2 n(x)].
Suppose X1 0, X1 1, X1 2, . . . , X2 0, X2 1, X2 2, . . .
are defined on the same probability space such that for each j, {Xj n : n = 0, 1, . . .} has the distribution of the Markov chain with initial distribution gj 0. Then it is clear that P{X1 n = X2 n} ≤1 −∥g1 n −g2 n∥= X x∈D g1 n(x) ∧g2 n(x).
(12.21) The following theorem shows that there is a way to define the chains on the same probability space so that equality is obtained in (12.21). This theorem gives one example of the powerful probabilistic technique called coupling. Coupling refers to the defining of two or more processes on the same probability space in a way so that each individual process has a certain distribution but the joint distribution has some particularly nice properties. Often, as in this case, the two processes are equal except for an event of small probability.
Theorem 12.4.5 Suppose p, g1 n, g2 n are as defined in the previous paragraph.
We can define (X1 n, X2 n), n = 0, 1, 2, . . . on the same probability space such that: • for each j, Xj 0, Xj 1, . . . has the distribution of the Markov chain with initial distribution gj 0; • for each integer n ≥0, P{X1 m = X2 m for all m ≥n} = 1 −∥g1 n −g2 n∥.
♣Before doing this proof, let us consider the easier problem of defining (X1, X2) on the same probability space so that Xj has distribution gj 0 and P{X1 = X2} = 1 −∥g1 0 −g2 0∥.
Assume 0 < ∥g1 0 −g2 0∥< 1. Let f j(x) = gj 0(x) −[g1 0(x) ∧g2 0(x)].
• Suppose that J, X, W 1, W 2 are independent random variables with the following distributions.
P{J = 0} = 1 −P{J = 1} = ∥g1 0 −g2 0∥.
P{X = x} = g1(x) ∧g2(x) 1 −∥g1 0 −g2 0∥, x ∈D 276 Appendix P{W j = x} = f j(x) ∥g1 0 −g2 0∥, x ∈D.
• Let Xj = 1{J = 1} X + 1{J = 0}W j.
It is easy to check that this construction works.
Proof For ease, we will assume that ∥g1 0 −g2 0∥= 1 and ∥g1 n −g2 n∥→0 as n →∞; the adjustment needed if this does not hold is left to the reader. Let (Z1 n, Z2 n) be independent Markov chains with the appropriate distributions. Let f j n(x) = gj n(x)−[g1 n(x)∧g2 n(x)] and define hj n by hj 0(x) = gj 0(x) = f j 0(x) and for n > 1, hj n(x) = X z∈S f j n−1(z) p(z, x).
Note that f j n+1(x) = hj n+1(x) −[h1 n(x) ∧h2 n(x)]. Let ρj n(x) = h1 n(x) ∧h2 n(x) hj n(x) if hj n(x) ̸= 0.
We set ρj n(x) = 0 if hj n(x) = 0. We let {Y j(n, x) : j = 1, 2; n = 1, 2, . . . ; x ∈D} be independent 0-1 random variables, independent of (Z1 n, Z2 n), with P{Y j(n, x) = 1} = ρj n(x).
We now define 0-1 random variables Jj n as follows: • Jj 0 ≡0 • If Jj n = 1, then Jj m = 1 for all m ≥n.
• If Jj n = 0, then Jj n+1 = Y j(n + 1, Zj n+1).
We claim that P{Jj n = 0; Zj n = x} = f j n(x).
For n = 0, this follows immediately from the definition. Also, P{Jj n+1 = 0; Zj n+1 = x} = X z∈D P{Jj n = 0; Zj n = z}P{Zj n+1 = x, Y j(n + 1, x) = 0 | Jj n = 0; Zj n = z}.
The random variable Y j(n+1, x) is independent of the Markov chain, and the event {Jj n = 0; Zj n = z} depends only on the chain up to time n and the values of {Y j(k, y) : k ≤n}. Therefore, P{Zj n+1 = x, Y j(n + 1, x) = 0 | Jj n = 0; Zj n = z} = p(z, x) [1 −ρj n+1(x)].
Therefore, we have the inductive argument P{Jj n+1 = 0; Zj n+1 = x} = X z∈D f j n(z) p(z, x) [1 −ρj n+1(x)] = hj n+1(x) [1 −ρj n+1(x)] = hj n+1(x) −[h1 n+1(x) ∧h2 n+1(x)] = f j n+1(x), which establishes the claim.
12.4 Markov chains 277 Let Kj denote the smallest n such that Jj n = 1. The condition ∥g1 0 −g2 0∥→0 implies that Kj < ∞with probability one. A key fact is that for each n and each x, P{K1 = n + 1; Z1 n = x} = P{K2 = n + 1; Z2 n = x} = h1 n+1(x) ∧h2 n+1(x).
This is immediate for n = 0 and for n > 0, P{Kj = n + 1; Zj n+1 = x} = X z∈D P{Jn = 0; Zj n = z} P{Y j(n + 1, x) = 1; Zj n+1 = x | Jn = 0; Zj n = z} = X z∈D f j n(z) p(z, x) ρj n+1(x) = hj n+1(x) ρj n+1(x) = h1 n+1(x) ∧h2 n+1(x).
The last important observation is that the distribution of Wm := Xj m−n given the event {Kj = n; Xj n = x} is that of a Markov chain with transition probability p starting at x.
The reader may note that for each j, the process (Zj n, Jj n) is a time-inhomogeneous Markov chain with transition probabilities P{(Zj n+1, Jj n+1) = (y, 1) | (Zj n, Jj n) = (x, 1)} = p(x, y), P{(Zj n+1, Jn+1) = (y, 0) | (Zj n, Jj n) = (x, 0)} = p(x, y) [1 −ρj n+1(y)], P{(Zj n+1, Jn+1) = (y, 1) | (Zj n, Jj n) = (x, 0)} = p(x, y) ρj n+1(y).
The chains (Z1 n, J1 n) and (Z2 n, J2 n) are independent. However, the transition probabilities for these chains depend on both initial distributions and p.
We are now ready to make our construction of (X1 n, X2 n).
• Define for each (n, x) a process {W n,x m : m = 0, 1, 2, . . .} that has the distribution of the Markov chain with initial point x. Assume that all these processes are independent.
• Choose (n, x) according to the probability distribution h1 n+1(x) ∧h2 n+1(x) = P{Kj = n; Zj n = x}.
Set Jj m = 1 for m ≥n, Jj m = 0 for m < n, and K1 = K2 = n. Note that Kj is the smallest n such that Jj n = 1.
• Given (n, x), choose X1 0, . . . , X1 n from the conditional distribution of the Markov chain with initial distribution g1 0 conditioned on the event {K1 = n; Z1 n = x}.
• Given (n, x), choose X2 0, . . . , X2 n (conditionally) independent of X1 0, . . . , X1 n from the condi-tional distribution of the Markov chain with initial distribution g2 0 conditioned on the event {K2 = n; Z2 n = x}.
• Let Xj m = W n,x m−n, m = n, n + 1, . . . .
The two conditional distributions above are not easy to express explicitly; fortunately, we do not need to do so.
To finish the proof, we need only check that the above construction satisfies the conditions. For 278 Appendix fixed j, the fact that Xj 0, Xj 1, . . . has the distribution of the chain with initial distribution gj 0 is im-mediate from construction and the earlier observation that the distribution of {Xj n, Xj n+1, . . .} given {Kj = n; Zj n = x} is that of the Markov chain starting at x. Also, the construction immediately gives X1 m = X2 m if m ≥K1 = K2. Also, P{Jj n = 0} = X x∈D f j n(z) = ∥gn 1 −gn 2 ∥.
Remark. A review of the proof of Theorem 12.4.5 shows that we do not need to assume that the Markov chain is time-homogeneous. However, time-homogeneity makes the notation a little simpler and we use the result only for time-homogenous chains.
12.5 Some Tauberian theory Lemma 12.5.1 Suppose α > 0. Then as ξ →1−, ∞ X n=2 ξn nα−1 ∼ Γ(α) (1 −ξ)α .
Proof Let ǫ = 1 −ξ. First note that X n≥ǫ−2 ξn nα−1 = X n≥ǫ−2 [(1 −ǫ)1/ǫ]nǫ nα−1 ≤ X n≥ǫ−2 e−nǫ nα−1, and the right-hand side decays faster than every power of ǫ. For n ≤ǫ−2 we can do the asymptotics ξn = exp{n log(1 −ǫ)} = exp{n(−ǫ −O(ǫ2))} = e−nǫ[1 + O(nǫ2)].
Hence, X n≤ǫ−2 ξn nα−1 = ǫ−α X n≤ǫ−2 ǫ e−nǫ(nǫ)α−1 [1 + (nǫ) O(ǫ)].
Using Riemann sum approximations we see that lim ǫ→0+ ∞ X n=1 ǫ e−nǫ(nǫ)α−1 = Z ∞ 0 e−t tα−1 dt = Γ(α).
Proposition 12.5.2 Suppose un is a sequence of nonnegative real numbers. If α > 0, the following two statements are equivalent: ∞ X n=0 ξn un ∼ Γ(α) (1 −ξ)α , ξ →1−, (12.22) N X n=1 un ∼α−1 N α, N →∞.
(12.23) 12.5 Some Tauberian theory 279 Moreover, if the sequence is monotone, either of these statements implies un ∼nα−1, n →∞.
Proof Let Un = P j≤n uj where U−1 = 0. Note that ∞ X n=0 ξn un = ∞ X n=0 ξn [Un −Un−1] = (1 −ξ) ∞ X n=0 ξn Un.
(12.24) If (12.23) holds, then by the previous lemma ∞ X n=0 ξn un ∼(1 −ξ) ∞ X n=0 ξn α−1 nα ∼Γ(α + 1) α (1 −ξ)α = Γ(α) (1 −ξ)α .
Now suppose (12.22) holds. We first give an upper bound on Un. Using 1 −ξ = 1/n, we can see as n →∞, Un ≤ n−1 1 −1 n −2n 2n−1 X j=n 1 −1 n j Uj ≤ n−1 1 −1 n −2n ∞ X j=0 1 −1 n j Uj ∼e2 Γ(α) nα.
The last relation uses (12.24). Let ν(j) denote the measure on [0, ∞) that gives measure j−α un to the point n/j. Then the last estimate shows that the total mass of ν(j) is uniformly bounded on each compact interval and hence there is a subsequence that converges weakly to a measure ν that is finite on each compact interval. Using (12.22) we can see that that for each λ > 0, Z ∞ 0 e−λx ν(dx) = Z ∞ 0 e−λx xα−1 dx.
This implies that ν is xα−1 dx. Since the limit is independent of the subsequence, we can conclude that ν(j) →ν and this implies (12.23).
The fact that (12.23) implies the last assertion if un is monotone is straightforward using Un(1+ǫ) −Un ∼α−1 [(n(1 + ǫ))α −nα] , n →∞, The following is proved similarly.
Proposition 12.5.3 Suppose un is a sequence of nonnegative real numbers. If α ∈R, the following two statements are equivalent: ∞ X n=0 ξn un = 1 1 −ξ logα 1 1 −ξ , ξ →1−, (12.25) N X n=1 un ∼N logα N N →∞.
(12.26) 280 Appendix Moreover, if the sequence is monotone, either of these statements implies un ∼logα n, n →∞.
12.6 Second moment method Lemma 12.6.1 Suppose X is a nonnegative random variable with E[X2] < ∞and 0 < r < 1.
Then P {X ≥rE(X)} ≥(1 −r)2 E(X)2 E(X2) .
Proof Without loss of generality, we may assume that E(X) = 1. Since E[X; X < r] ≤r, we know that E[X; X ≥r] ≥(1 −r). Then, E(X2) ≥E[X2; X ≥r] = P{X ≥r} E[X2 | X ≥r] ≥ P{X ≥r} (E[X | X ≥r])2 ≥ E[X; X ≥r]2 P{X ≥r} ≥ (1 −r)2 P{X ≥r}.
Corollary 12.6.2 Suppose E1, E2, . . . is a collection of events with P P(En) = ∞. Suppose there is a K < ∞such that for all j ̸= k, P(Ej ∩Ek) ≤K P(Ej) P(Ek). Then P{Ek i.o.} ≥1 K .
Proof Let Vn = Pn k=1 1Ek. Then the assumptions imply that lim n→∞E(Vn) = ∞, and E(V 2 n ) ≤ k X j=1 P(Ej) + X j̸=k K P(Ej) P(Ek) ≤E(Vn) + K E(Vn)2 = 1 E(Vn) + K E(Vn)2.
By Lemma 12.6.1, for every r > 0, P{Vn ≥rE(Vn)} ≥ (1 −r)2 K + E(Vn)−1 .
Since E(Vn) →∞, this implies P{V∞= ∞} ≥(1 −r)2 K .
Since this holds for every r > 0, we get the result.
12.7 Subadditivity 281 12.7 Subadditivity Lemma 12.7.1 (Subadditivity lemma) Suppose f : {1, 2, . . .} →R is subadditive, i.e., for all n, m, f(n + m) ≤f(n) + f(m). Then, lim n→∞ f(n) n = inf n>0 f(n) n .
Proof Fix integer N > 0. We can write any integer n as jN + k where j is a nonnegative integer and k ∈{1, . . . , N}. Let bN = max{f(1), . . . , f(N)}. Then subadditivity implies f(n) n ≤jf(N) + f(k) jN ≤f(N) N + bN jN .
Therefore, lim sup n→∞ f(n) n ≤f(N) N .
Since this is true for every N, we get the lemma.
Corollary 12.7.2 Suppose rn is a sequence of positive numbers and b1, b2 > 0 such that for every n, m, b1 rn rm ≤rn+m ≤b2 rn rm.
(12.27) Then there exists α > 0 such that for all n, b−1 2 αn ≤rn ≤b−1 1 αn.
Proof Let f(n) = log rn + log b2. Then f is subadditive and hence lim n→∞ f(n) n = f(n) n := α.
This shows that rn ≥αn/b2. Similarly, by considering the subadditive function g(n) = −log rn − log b1, we get rn ≤b−1 1 αn.
Remark. Note that if rn satisfies (12.27), then so does βnrn for each β > 0. Therefore, we cannot determine the value of α from (12.27).
Exercises Exercise 12.1 Find f3(ξ), f4(ξ) in (12.4).
Exercise 12.2 Go through the proof of Lemma 12.5.1 carefully and estimate the size of the error term in the asymptotics.
282 Appendix Exercise 12.3 Suppose E1 ⊃E2 ⊃· · · is a decreasing sequence of events with P(En) > 0 for each n. Suppose there exist α > 0 such that ∞ X n=1 |P(En | En−1) −(1 −αn−1)| < ∞.
Show there exists c such that P(En) ∼c n−α.
(12.28) (Hint: use Lemma 12.1.4.) Exercise 12.4 In this exercise we will consider an alternative approach to the Perron-Froebenius Theorem. Suppose q : {1, 2, . . .} × {1, 2, . . .} →[0, ∞), is a function such that for each x > 0, q(x) := X y q(x, y) ≤1.
Define qn(x, y) by matrix multiplication as usual, that is, q1(x, y) = q(x, y) and qn(x, y) = X z qn−1(x, z) q(z, y).
Assume for each x, qn(x, 1) > 0 for all n sufficiently large. Define qn(x) = X y qn(x, y), pn(x, y) = qn(x, y) qn(x) , qn = sup x qn(x), q(x) = inf n qn(x) qn .
Assume there is a function F : {1, 2, . . .} →[0, 1] and a positive integer m such that pm(x, y) ≥F(y), 1 ≤x, y < ∞, and such that ρ := X y F(y) q(y) > 0.
(i) Show there exists 0 < α ≤1 such that lim n→∞q1/n n = α.
Moreover, qn ≥αn. (Hint: qn+m ≤qn qm.) (ii) Show that pn+k(x, y) = X z νn,k(x, z) pk(z, y), where νn,k(x, z) = pn(x, z) qk(z) P w pn(x, w) qk(w) = qn(x, z) qk(z) P w qn(x, w) qk(w).
12.7 Subadditivity 283 (iii) Show that if k, x are positive integers and n ≥km, 1 2 X y |pkm(1, y) −pn(x, y)| ≤(1 −ρ)k.
(iv) Show that the limit v(y) = lim n→∞pn(1, y) exists and if k, x are positive integers and n ≥km, 1 2 X y |v(y) −pn(x, y)| ≤(1 −ρ)k.
(v) Show that v(y) = α X x v(x) q(x, y).
(vi) Show that for each x, the limit w(x) = lim n→∞α−n qn(x) exists, is positive, and w satisfies w(x) = α X y q(x, y) w(y).
(Hint: consider qn+1(x)/qn(x).) (vii) Show that there is a C = C(ρ, α) < ∞such that if ǫn(x) is defined by qn(x) = w(x) αn [1 + ǫn(x)] , then |ǫn(x)| ≤C e−βn, where β = −log(1 −ρ)/m.
(viii) Show that there is a C = C(ρ, α) < ∞such that if ǫn(x, y) is defined by qn(x, y) = w(x) αn [v(y) + ǫn(x, y)] , then |ǫn(x, y)| ≤C e−βn.
(ix) Suppose that Q is an N × N matrix with nonnegative entries such that Qm has all positive entries.
Suppose that the row sums of Q are bounded by K.
For 1 ≤j, k ≤N, let q(j, k) = K−1 q(j, k); set q(j, k) = 0 if k > N; and q(k, j) = δj,1 if k > N. Show that the conditions are satisfied (and hence we get the Perron-Froebenius Theorem).
Exercise 12.5 In the previous exercise, let q(x, 1) = 1/2 for all k, q(2, 2) = 1/2 and q(x, y) = 0 for all other x, y. Show that there is no F such that ρ > 0.
284 Appendix Exercise 12.6 Suppose X1, X2, . . . are i.i.d. random variables in R with mean zero, variance one, and such that for some t > 0, β := 1 + E X2 1etX1; X1 ≥0 < ∞.
Let Sn = X1 + · · · + Xn.
(i) Show that for all n, E etSn ≤eβnt2/2.
(Hint: expand the moment generating function for X1 about s = 0.) (ii) Show that if r ≤tβn, P{Sn ≥r} ≤exp −r2 2βn .
Exercise 12.7 Suppose X1, X2, . . . are i.i.d. random variables in R with mean zero, variance one, and such that for some t > 0 and 0 < α < 1, β := 1 + E X2 1etXα 1 ; X1 ≥0 < ∞.
Let Sn = X1 + · · · + Xn. Suppose r > 0 and n is a positive integer. Let K = nβt r 1 1−α , ˜ Xj = Xj 1{Xj ≤K}, ˜ Sn = ˜ X1 + · · · + ˜ Xn.
(i) Show that P{Xj ̸= ˜ Xj} ≤(β −1) K−2 e−tKα.
(ii) Show that E h etKα−1 ˜ Sni ≤eβnt2K2(α−1)/2.
(iii) Show that P{Sn ≥r} ≤exp −r2 2βn + n (β −1) K−2 e−tKα.
References Bhattacharya, R. and Rao R. (1976). Normal Approximation and Asymptotic Expansions, John Wiley & Sons.
Bousquet-M´ elou, M. and Schaeffer, G. (2002). Walks on the slit plane, Prob. Theor. Rel. Fields 124, 305–344.
Duplantier, B. and David, F. (1988). Exact partition functions and correlation functions of multiple Hamilton walks on the Manhattan lattice, J. Stat. Phys. 51, 327–434.
Fomin, S. (2001). Loop-erased walks and total positivity, Trans. AMS 353, 3563–3583.
Fukai, Y. and Uchiyama, K. (1996). Potential kernel for two-dimensional random walk, Annals of Prob. 24, 1979–1992.
Kenyon, R. (2000). The asymptotic determinant of the dsicrete Laplacian, Acta Math. 185, 239–286.
Koml¨ os, J., Major, P., and Tusn´ ady, G. (1975). An approximation of partial sums if independent rv’s and the sample df I, Z. Warsch. verw. Geb. 32, 111-131.
Koml¨ os, J., Major, P., and Tusn´ ady, G. (1975). An approximation of partial sums if independent rv’s and the sample df II, Z. Warsch. verw. Geb. 34, 33-58.
Kozma, G. (2007). The scaling limit of loop-erased random walk in three dimensions, Acta Math. 191, 29-152.
Lawler, G. (1996). Intersections of Random Walks, Birkh¨ auser.
Lawler, G., Puckette, E. (2000). The intersection exponent for simple random walk, Combin., Probab., and Comp. 9, 441–464.
Lawler, G., Schramm, O., and Werner, W. (2001). Values of Brownian intersection exponents II: plane exponents, Acta Math. 187, 275–308.
Lawler, G., Schramm, O., and Werner, W. (2004). Conformal invariance of planar loop-erased random walk and uniform spanning trees, Annals of Probab. 32 , 939–995.
Lawler, G. and Trujillo Ferreras, J. A., Random walk loop soup, Trans. Amer. Math. Soc. 359, 767–787.
Masson, R. (2009), The growth exponent for loop-erased walk, Elect. J Prob. 14, 1012–1073 Spitzer, F. (1976). Principles of Random Walk, Springer-Verlag.
Teufl, E. and Wagner, S. (2006). The number of spanning trees of finite Sierpinski graphs, Fourth Colloquium on Mathematics and Computer Science, DMTCS proc. AG, 411-414 Wilson, D. (1996). Gnerating random spanning trees more quickly than the cover time, Proc. STOC96, 296–303.
285 286 References Index of Symbols If an entry is followed by a chapter number, then that notation is used only in that chap-ter. Otherwise, the notation may appear through-out the book.
a(x), 86 aA, 147 a(x), 90 Am,n [Chap. 8], 225 Ad, 154 b [Chap. 5], 106 Bt, 64 Bn, 11 cap, 136, 146 C2, 126 Cd (d ≥3), 82 C∗ d, 82 Cn, 11 C(A; v), C(A) [Chap. 8], 233 Ct, Ct [Chap. 8], 207 Df(y), 152 d(ω) [Chap. 8], 203 deg [Chap. 8], 202 ∆k,i [Chap. 7], 173 ej, 9 EsA(x), 136 EA(f, g), EA(f) [Chap. 8], 231 EA(x, y) [Chap. 8], 211 EA(x, y), EA [Chap. 8], 210 ˆ EA(x, y) [Chap. 8], 210 f(λ; x) [Chap. 8], 203 F(A; λ), Fx(A; λ), FV (A; λ), F(A) [Chap.
8], 203 F(x, y; ξ) [Chap. 4], 78 Fn(θ) [Chap. 2], 32 FV1,V2(A) [Chap. 8], 209 g(λ; x), g(λ) [Chap. 8], 203 g(θ, n) [Chap. 2], 32 gA(x), 135 G(x), 76 G(x, y), 76 G(x, y; ξ), 76 G(x; ξ), 76 GA(x, y), 96 G(x), 84 γ2, 126 Γ, 11 Γk,j [Chap. 7], 172 Γ, 80 hD(x, y), 181 hmA, 138 hmA(x), 144 HA(x, y), 123 H∂A(x, y), 153, 211 ˆ H∂A(x, y) [Chap. 8], 212 inrad(A), 178 J , 11 J ∗, 11 K [Chap. 5], 106 K(ω) [Chap. 8], 200 LE(ω) [Chap. 8], 208 L, 17 LA, 157 L, Lj, Lx j , Lx [Chap. 8], 201 L, Lj, L x j , L x [Chap. 8], 201 m = mq [Chap. 8], 200 m(ω) [Chap. 8], 201 o(·), 21 osc, 66 O(·), 21 [ω] [Chap. 8], 211 ω [Chap. 8], 200 pn(x, y), 10 pn(x), 25 ˜ pt(x, y), 13 P A, 157 P, Pd, 10 P′, P′ d, 17 INDEX OF SYMBOLS 287 P∗, P∗ d, 17 q(ω) [Chap. 8], 199 q(T ; x0) [Chap. 8], 201 qλ [Chap. 8], 201 ˆ qA [Chap. 8], 209 ˆ qA(x, y) [Chap. 8], 212 r(t) [Chap. 8], 187 rad(A), 136 Rk,j [Chap. 7], 174 ρ [Chap. 5], 106 Sn, 10 ˜ S, 13 TA [Chap. 6], 136 T A [Chap. 6], 136 T [Chap. 8], 201 τA, 96 τ A, 96 X, 199 Zd, 9 Θ(B, S; n), 69 φ(θ), 27 Φ(λ) [Chap. 8], 203 Φ, Φσ2, 171 η, ηr [Chap. 5], 103 η∗, η∗ r [Chap. 5], 106 ξn, 126 ξ∗ n, 126 ∇x, 17 ∇2 x, 17 ∂A, 119 ∂iA, 119 x ∼y [Chap. 8], 202 288 References Index h-process, 245 adjacent, 201 aperiodic, 10, 158 Berry-Esseen bounds, 36 Beurling estimate, 154 bipartite, 10, 158 boundary, inner, 119 boundary, outer, 119 Brownian motion, standard, 64, 71, 172 capacity d ≥3, 136 capacity, two dimensions, 146 central limit theorem (CLT), 24 characteristic function, 27 closure, discrete, 119 connected (with respect to p), 120 coupling, 48 covariance matrix, 11 cycle, 198 unrooted, 199 defective increment distribution, 112 degree, 201 determinant of the Laplacian, 204 difference operators, 17 Dirichlet form, 230 Dirichlet problem, 121, 122 Dirichlet-to-Neumann map, 153 domain Markov property, 250 dyadic coupling, 166, 172 eigenvalue, 157, 191 excursion measure, 209 loop-erased, 209 nonintersecting, 210 self-avoiding, 209 excursions (boundary), 209 filtration, 19, 263 Fomin’s identity, 212 functional central limit theorem, 63 fundamental solution, 95 gambler’s ruin, 103, 106 Gaussian free field, 230, 232 generating function, 202 cycle, 202 loop measure, 202 generating function, first visit, 78 generating set, 10 generator, 17 Girsanov transformation, 44 Green’s function, 76, 96 half-line, 115, 155 harmonic (with respect to p), 120 harmonic function difference estimates, 125, 131 harmonic functions, 192 harmonic measure, 138, 144 Harnack inequality, 122, 131 hexagonal lattice, 16 honeycomb lattice, 16 increment distribution, 10 inradius, 178 invariance principle, 63 irreducible cycle, 202 joint normal distribution, 24, 267 killing rate, 76 killing time, 76, 113, 199 KMT coupling, 166 Laplacian, discrete, 17 last-exit decomposition, 99, 130 lattice, 14 lazy walker, 77, 201 length (of a path or cycle), 198 local central limit theorem (LCLT), 24, 25 loop unrooted, 199 loop measure rooted, 199 INDEX 289 unrooted, 200 loop soup, 206 loop-erased random walk (LERW), 248 loop-erasure(chronological), 207 martingale, 120, 263 maximal inequality, 265 maximal coupling, 275 maximal inequality, 265 maximum principle, 121 modulus of continuity of Brownian motion, 68 Neumann problem, 152 normal derivative, 152 optional sampling theorem, 125, 263, 267 oscillation, 66 overshoot, 108, 110, 126 Perron-Froebenius theorem, 270 Poisson kernel, 122, 123, 181 boundary, 210 excursion, 152 potential kernel, 86 quantile coupling, 170 function, 170 range, 12 recurrent, 75 recurrent set, 141 reflected random walk, 153 reflection principle, 20 regular graph, 201 restriction property, 209 root, 198 Schramm-Loewner evolution, 243, 258 second moment method, 280 simple random walk, 9 on a graph, 201 simply connected, 178 Skorokhod embedding, 68, 166 spanning tree, 200 complete graph, 216 counting number of trees, 215 free, 221 hypercube, 217 rectangle, 224 Sierpinski graphs, 220 uniform, 214 Stirling’s formula, 52 stopping time, 19, 263 strong approximation, 64 strong Markov property, 19 subMarkov, 199 strictly, 199 Tauberian theorems, 79, 278 transient, 75 transient set, 141 transitive graph, 201 triangular lattice, 15 weight, 198 symmetric, 199 Wiener’s test, 142 Wilson’s algorithm, 214 |
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Evaluating the integral $\int_{0}^{\infty} \frac{\cos ax}{1+x^2}dx$ [duplicate]
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I encountered this problem while calculating the Fourier Cosine Integral for the function $f(x) = \frac{1}{1+x^2}$. I know that $\int \frac{1}{1+x^2} = \arctan x + c$. I just can't figure out what to do with $cosax$ term combined. If I do integration by parts, I get: $$\left [ \cos ax \times \tan^{-1}x \right ]_{0}^{\infty} +\int_{0}^{\infty}a \sin ax \times \tan^{-1}xdx$$ I don't know how to proceed after that. Perhaps a hint could help.
integration
definite-integrals
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edited Oct 24, 2017 at 13:01
bikalpabikalpa
asked Oct 24, 2017 at 12:37
bikalpabikalpa
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$\begingroup$ the $a\sin(ax)$ should be "inside" the integral, multiplied with the $\tan^{-1}(x)$. $\endgroup$
Diffusion
– Diffusion
2017-10-24 12:51:16 +00:00
Commented Oct 24, 2017 at 12:51
1
$\begingroup$ This is quite easy using a contour integral - do you know enough complex analysis to do these? $\endgroup$
Joppy
– Joppy
2017-10-24 12:59:35 +00:00
Commented Oct 24, 2017 at 12:59
$\begingroup$ @Messney Sorry my bad! Thanks for pointing out. $\endgroup$
bikalpa
– bikalpa
2017-10-24 13:01:45 +00:00
Commented Oct 24, 2017 at 13:01
$\begingroup$ @Joppy I am quite new to complex analysis, but I do know the basics of Cauchy's theorems and Residue theorem. How do you propose the solultioin using complex analysis? $\endgroup$
bikalpa
– bikalpa
2017-10-24 13:02:54 +00:00
Commented Oct 24, 2017 at 13:02
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$\begingroup$ See also: math.stackexchange.com/q/35396/9464 $\endgroup$
user9464
– user9464
2017-10-24 13:42:29 +00:00
Commented Oct 24, 2017 at 13:42
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Your idea of using integration by parts can be used to evaluate the integral; you'll just have to couple that with Feynman's Trick. Making the substitution $u=1/(1+x^2)$ and $dv=\cos ax\, dx$ gives$$\begin{align}I(a)=\int\limits_0^{\infty}\frac {\cos ax}{1+x^2}\, dx & =\frac {\sin ax}{a(1+x^2)}\,\Biggr\rvert_{0}^{\infty}+\frac 2a\int\limits_{0}^{\infty}\frac {x\sin ax}{(1+x^2)^2}\, dx\ & =\frac 2a\int\limits_0^{\infty}\frac {x\sin ax}{\left(1+x^2\right)^2}\, dx\end{align}$$ Multiplying both sides by $a$ and differentiating with respect to $a$, we see that$$\begin{align}a\cdot I'(a)+I(a) & =2\int\limits_0^{\infty}\frac {x^2\cos ax}{(1+x^2)^2}\, dx\ & =2I(a)-2\int\limits_0^{\infty}\frac {\cos ax}{\left(1+x^2\right)^2}\, dx\end{align}$$So$$a\cdot I'(a)-I(a)=-2\int\limits_0^{\infty}\frac {\cos ax}{\left(1+x^2\right)^2}\, dx$$Differentiating a second time and solving the little differential equation that follows gives$$I(a)=C_1e^a+C_2e^{-a}$$To find the constants, set $a=0$ and $a\to\infty$. Evaluating, we find that$$(C_1,C_2)=\left(0,\tfrac {\pi}2\right)$$Hence$$\int\limits_{0}^{\infty}\frac {\cos ax}{1+x^2}\, dx=\color{blue}{\frac {\pi}{2e^{|a|}}}$$With the absolute value there because $I(a)=I(-a)$ from the cosine function.
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edited Oct 24, 2017 at 18:08
answered Oct 24, 2017 at 13:36
CrescendoCrescendo
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$\begingroup$ that suspicious the left function is even in a-variable but the right is not $\endgroup$
Guy Fsone
– Guy Fsone
2017-10-24 13:40:17 +00:00
Commented Oct 24, 2017 at 13:40
$\begingroup$ how is $C_1 + C_2 = \pi $? Shouldn't it be $\frac \pi 2$? $\endgroup$
bikalpa
– bikalpa
2017-10-24 14:09:44 +00:00
Commented Oct 24, 2017 at 14:09
$\begingroup$ @bikalpa Yes that too. When I solved it, I rewrote the limits as from $-\infty$ to $\infty$. $\endgroup$
Crescendo
– Crescendo
2017-10-24 14:12:41 +00:00
Commented Oct 24, 2017 at 14:12
$\begingroup$ All clear except this line: "As a tends to $\infty$, $I(a)=0$". Could you make that point clear? $\endgroup$
bikalpa
– bikalpa
2017-10-24 14:18:24 +00:00
Commented Oct 24, 2017 at 14:18
$\begingroup$ I think I got it now. I guess it is from $I(a)=\frac 2a\int\limits_{0}^{\infty}\frac {x\sin ax}{(1+x^2)^2}\, dx$ $\endgroup$
bikalpa
– bikalpa
2017-10-24 14:26:18 +00:00
Commented Oct 24, 2017 at 14:26
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I'll give a purely probabilistic solution.
Let $X%$ be a random variable that takes the value of $a$ and $-a$ each with probability $1/2$. The characteristic function of $X$ is then $\phi_X(t)=E[e^{itX}]=\frac{1}{2}(e^{iatx}+e^{-iatx})=cos(at)$. Now let $Y$ be a random variable with standard cauchy distribution.
$\int_{-\infty}^{\infty}\frac{cos(at)}{1+t^2} dt=\pi\int_{-\infty}^{\infty}\phi_X(t)f_y(t) dt$
Using Parseval's relation you have that
$\pi\int_{-\infty}^{\infty}\phi_X(t)f_y(t) dt=\pi\int_{-\infty}^{\infty}\phi_Y(t)dFx(t) dt$
Where $\phi_Y(t)$ is the characteristic function of the standard cauchy distribution, $\phi_Y(t)=e^{-|t|}$. Now, you should note that the distribution of $X$ takes only two values $a$ and $-a$ with probability $\frac{1}{2}$ hence:
$\pi\int_{-\infty}^{\infty}e^{-|t|}dFx(t) dt=\frac{\pi}{2}(e^{ -|a|}+e^{ -|-a|})=\frac{\pi}{e^{|a|}}$.
Now this is the value for the full integral from $(-\infty, \infty)$ in your case you have $(0,\infty)$ but since $cos(x)$ is an even function and the denominator $(1+t^2)$ is the same for negative and positive numbers, you have:
$\int_{-\infty}^{\infty}\frac{cos(at)}{1+t^2} dt=2\int_{0}^{\infty}\frac{cos(at)}{1+t^2} dt$
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answered Oct 24, 2017 at 13:18
Daniel OrdoñezDaniel Ordoñez
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Recall that, if we consider the Fourier transform $$\mathcal Ff (a) =\int_\Bbb R e^{-ia x}f(x)dx$$ then its Fourier inverse is defined as $$\mathcal F^{-1}f (x) =\frac{1}{2\pi}\int_\Bbb R e^{it x}f(t)dt.$$
But we have, \begin{split} \mathcal F(e^{-|t|})(x) = \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt &=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\ &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\ &=&\frac{1}{1-ix}+\frac{1}{1+ix}\ &=&\frac{2}{x^2+1}. \end{split} Then, $$ \begin{align} e^{-|a|}=\mathcal F^{-1}\left( \frac{2}{x^2+1}\right)(a) &=\frac{1}{2\pi}\int_\Bbb R \frac{2}{x^2+1}e^{ix a}\,dx = \frac{1}{\pi}\int_\Bbb R\frac{e^{ix a}}{x^2+1}\,dx \&=\frac{1}{\pi}\int_\Bbb R\frac{\cos a x}{x^2+1}\,dx = \frac{2}{\pi}\int_0^\infty\frac{\cos ax}{x^2+1}\,dx \end{align} $$ Given that, as $x\mapsto\sin ax $ is an old function we have, $$\int_\Bbb R \frac{\sin{a x}}{x^2+1}dx= 0.$$
Thus we have, $$ \int_0^\infty\frac{\cos ax}{x^2+1}\,dx =\frac{\pi}{2}e^{-|a|} $$
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edited Nov 7, 2017 at 12:56
answered Oct 24, 2017 at 13:11
Guy FsoneGuy Fsone
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Electrocardiographic Monitoring
Chapter Outline
ELECTROCARDIOGRAPHY
Normal Electrical Activity
P Wave
PR Interval
QRS Complex
ST Segment and T Wave
Lead Systems
Standard Limb and Precordial Leads
Electrocardiogram Monitoring Systems
Three-Electrode System
Modified Three-Electrode System
Five-Electrode System
Invasive Electrocardiography
DISPLAY, RECORDING, AND INTERPRETATION
Basic Requirements
Oscilloscope Displays
Standard Electrocardiogram Recordings
Artifacts
Patient
Electrodes and Leads
Operating Room Environment
Monitoring System
INDICATIONS
Diagnosis of Arrhythmias
Diagnosis of Ischemia
Diagnosis of Conduction Defects
AUTOMATED RECORDING
Computer-Assisted Electrocardiogram Interpretation
Arrhythmias
Myocardial Ischemia
ELECTROCARDIOGRAPHIC MONITORING DURING MAGNETIC RESONANCE IMAGING
Electrocardiography
The intraoperative use of the electrocardiogram (ECG) has developed markedly over the past several decades. Originally, this monitor was primarily used during anesthesia for the detection of arrhythmias in high-risk patients. In recent years, however, it has become a standard perioperative monitor used during the administration of all anesthetics. Beyond its usefulness for the intraoperative recognition of arrhythmias, one of the major indications for ECG monitoring is the intraoperative diagnosis of myocardial ischemia. ECG monitoring for ischemia is inexpensive and noninvasive. Most modern operating room (OR) monitors provide automated ST-segment monitoring, which can be set to alarm if changes are detected.
Normal Electrical Activity
Figure 13-1shows the segments and intervals of the normal ECG. These elements are explained in the following subsections.
FIGURE 13-1
Segments and intervals of the normal electrocardiogram.
P Wave
Under normal circumstances, the sinoatrial (SA) node has the most rapid rate of spontaneous depolarization and therefore is the dominant cardiac pacemaker. From the SA node, the impulse spreads through the right and left atria. Specialized tracts can conduct the impulse to the atrioventricular (AV) node, but they are not essential. On the ECG, depolarization of the atria is represented by the P wave. The initial depolarization primarily involves the right atrium and predominantly occurs in an anterior, inferior, and leftward direction. Subsequently, it proceeds to the left atrium, which is located in a more posterior position.
PR Interval
Once the wave of depolarization has reached the AV node, a delay is observed. The delay permits contraction of the atria and allows supplemental filling of the ventricular chambers. On the ECG, this delay is represented by the PR interval.
QRS Complex
After passing through the AV node, the electrical impulse is conducted along the ventricular conduction pathways, consisting of the common bundle of His, the left and right bundle branches, the distal bundle branches, and the Purkinje fibers. The QRS complex represents the progress of the depolarization wave through this conduction system. After terminal depolarization, the ECG normally returns to baseline.
ST Segment and T Wave
Repolarization of the ventricles, which begins at the end of the QRS complex, consists of the ST segment and T wave. Ventricular depolarization occurs along established conducting pathways, but ventricular repolarization is a prolonged process that occurs independently in every cell. The T wave represents the uncanceled potential differences of ventricular repolarization. The junction of the QRS complex and the ST segment is called the J junction. The T wave is sometimes followed by a small U wave, the origin of which is unclear. Prominent U waves are characteristic of hypokalemia (as well as hypothermia, hypomagnesemia, and hypocalcemia) and sometimes also observed after cerebrovascular accidents. Very prominent U waves may be seen in patients taking medications such as sotalol or one of the phenothiazines. Negative U waves may appear with positive T waves, an abnormal finding that has been noted in left ventricular hypertrophy and myocardial ischemia.
Lead Systems
Standard Limb and Precordial Leads
The small electric currents produced by the electrical activity of the heart spread throughout the body, which behaves as a volume conductor, allowing the surface ECG to be recorded at any site. The standard leads are bipolar leads because they measure differences in potential between two electrodes. The electrodes are placed on the right arm, the left arm, and the left leg. The leads are formed by the imaginary lines connecting the electrodes, and the polarities correspond to the conventions of Einthoven’s triangle. They are labeled leads I, II, and III. By convention, lead I is formed by connecting the right arm and left arm electrodes, with the left arm being positive; lead II is formed by connecting the right arm and left leg, with the left leg being positive; lead III is formed by connecting the left arm and left leg, with the left leg being positive. If the three electrodes of the standard leads are connected through resistances of 5000 ohms each, a common central terminal with zero potential is obtained. When this common electrode is used with another active electrode, the potential difference between the two represents the actual potential. On a standard 12-lead ECG, three unipolar limb leads usually are recorded: aVR, aVL, and aVF. The a indicates that the limb leads are augmented and were obtained via Goldberger’s modification, in which the resistors are removed from the lead wires and the exploring electrode is disconnected from the central terminal. Goldberger’s modification produces larger voltage deflections on the ECG.
Additional information on the heart’s electrical activity is obtained when electrodes are placed closer to the heart or around the thorax. In the precordial lead system ( Fig. 13-2), the neutral electrode is formed by the standard leads, and an exploring electrode is placed on the chest wall. The ECG normally is recorded with the exploring electrode in one or more of six precordial positions. Each lead is indicated by the letter V followed by a subscript numeral from 1 to 6, which indicates the location of the electrode on the chest wall.
FIGURE 13-2
Multiple-lead electrocardiographic system consisting of four extremity electrodes: right arm ( RA), left arm ( LA), right leg ( RL), left leg ( LL), and V 5 leads.
Electrocardiogram Monitoring Systems
Three-Electrode System
As the name implies, the three-electrode system uses only three electrodes to record the ECG. In such a system, the ECG is observed along one bipolar lead between two of the electrodes; the third electrode serves as a neutral lead, and a selector switch allows the user to alter the designation of the electrodes. Three ECG leads can be examined in sequence without changing the location of the electrodes. Although the three-electrode system has the advantage of simplicity, its use is limited in the detection of myocardial ischemia because it provides a narrow picture of myocardial electrical activity.
Modified Three-Electrode System
Numerous modifications of the standard bipolar limb lead system have been developed. Some of these are displayed in Figure 13-3. They are used in an attempt to maximize P-wave amplitude for the diagnosis of atrial arrhythmias or to increase the sensitivity of the ECG for the detection of anterior myocardial ischemia. In clinical studies, these modified three-electrode systems have been shown to be at least as sensitive as the standard V 5 lead system for the intraoperative diagnosis of ischemia.
FIGURE 13-3
Modified precordial lead arrangements. RA, right arm; LA, left arm; LL, left leg.
(From Thys DM, Kaplan JA, editors: The ECG in anesthesia and critical care. New York, 1987, Churchill Livingstone.)
Central Subclavicular Lead
The central subclavicular (CS 5 ) lead (see Fig. 13-3) is particularly well suited for the detection of anterior wall myocardial ischemia. The right arm (RA) electrode is placed under the right clavicle, the left arm (LA) electrode is placed in the V 5 position, and the left leg electrode is in its usual position to serve as a neutral lead. Lead I is selected for detection of anterior wall ischemia, and lead II can be selected either for monitoring inferior wall ischemia or for the detection of arrhythmias. If a unipolar precordial electrode is unavailable, this CS 5 bipolar lead is the best and easiest alternative to a true V 5 lead for monitoring myocardial ischemia.
Central Back Lead
The central back (CB 5 ) lead is useful for the detection of ischemia and supraventricular arrhythmias, as demonstrated in a study that compared CB 5 and V 5 in patients with closed and open chests. The P wave was 90% larger in lead CB 5 than in lead V 5 , and a good association between ventricular deflections of CB 5 and V 5 leads was noted. CB 5 is obtained by placing the RA electrode over the center of the right scapula and placing the LA electrode in the V 5 position. The lead selector switch should be set to lead I. The CB 5 lead may be useful in patients with ischemic heart disease who are susceptible to the development of perioperative arrhythmias.
When modified bipolar limb leads are used, the user should be aware that in certain aspects, they differ significantly from true unipolar precordial leads. The modified precordial leads usually show a greater R-wave amplitude than standard precordial leads, which can result in amplification of the ST-segment response. The criteria for diagnosing myocardial ischemia may therefore need to be adjusted when modified bipolar leads are used. It has been shown during exercise stress testing that normalization of the degree of ST-segment depression to the height of the R wave increases the sensitivity and specificity of the ECG for the recognition of myocardial ischemia. Although similar corrections have not yet been tested during intraoperative monitoring, their possible importance should be kept in mind when intraoperative ECG recordings are examined.
Five-Electrode System
The use of five electrodes permits the recording of the six standard limb leads—I, II, III, aVR, aVL, and aVF—as well as one precordial unipolar lead. In general, the unipolar lead is placed in the V 5 position, along the anterior axillary line in the fifth intercostal space. With the addition of only two electrodes to the ECG system, up to seven different leads can be monitored simultaneously. This allows several areas of the myocardium to be monitored for ischemia, which is helpful in the establishment of a differential diagnosis between atrial and ventricular arrhythmias.
In 1976, Kaplan and King suggested monitoring lead V 5 as the best choice for the detection of intraoperative ischemia. London, and colleagues demonstrated that in high-risk patients undergoing noncardiac surgery, when a single lead was used the greatest sensitivity was obtained with lead V 5 (75%), followed by lead V 4 (61%). Combining leads V 4 and V 5 increased the sensitivity to 90%, whereas the standard combination of leads II and V 5 produced a sensitivity of only 80%. They also suggested that if three leads (II, V 4 , and V 5 ) could be examined simultaneously, the sensitivity would increase to 98%.
Recently, Landesberg and colleagues investigated the usefulness of analyzing data obtained from continuous online 12-lead ECG monitoring for the detection of myocardial ischemia. During 11,132 patient-hours of monitoring, the investigators reported that V 4 was most sensitive to ischemia (83.3%), followed by V 3 and V 5 (75% each). Combining two precordial leads increased the sensitivity for detecting ischemia (97.4% for V 3 + V 5 and 92.1% for either V 4 + V 5 or V 3 + V 4 ) and infarction (100% for V 4 + V 5 or V 3 + V 5 and 83.3% for V 3 + V 4 ). They found that the baseline preanesthesia ST segment was above isoelectric in V 1 through V 3 and below isoelectric in V 5 through V 6 . Lead V 4 was closest to the isoelectric level on the baseline ECG, rendering it most suitable for ischemia monitoring. The authors recommended that two or more precordial leads were necessary to approach a sensitivity of greater than 95% for the detection of perioperative ischemia and infarction.
However, in an editorial entitled “Multilead Precordial ST-segment Monitoring: ‘The Next Generation?’” London concluded that multilead precordial ST-segment monitoring is not practical. It presents a significant logistical problem, particularly if a patient needs to be mobilized quickly. It is also likely to result in a high rate of false-positive responses and artifacts. Instead, London recommended the use of true V 4 or V 5 leads, control of heart rate and pain, and the use of β-blockers as tolerated for all patients at risk (diabetics or patients with left ventricular hypertrophy).
Invasive Electrocardiography
The electrical potentials of the heart can be measured not only from a surface ECG but also from body cavities adjacent to the heart (esophagus or trachea) or from within the heart itself.
Esophageal Electrocardiography
The concept of esophageal ECG is not new, and numerous studies have demonstrated the usefulness of this approach in the diagnosis of complicated arrhythmias. A prominent P wave usually is displayed in the presence of atrial depolarization, and its relationship to the ventricular electrical activity can be examined. The esophageal electrodes are incorporated into an esophageal stethoscope and are welded to conventional ECG wires ( Fig. 13-4). To record a bipolar esophageal ECG, the electrodes are connected to the right and left arm terminals, and lead I is selected on the monitor. In one study of 20 cardiac patients, 100% of atrial arrhythmias were correctly diagnosed with the esophageal lead (intracavitary ECG was used as the standard); lead II led to a correct diagnosis in 54% of the cases, and V 5 led to a correct diagnosis in 42% of the cases. In addition, the esophageal ECG may be helpful in the detection of posterior wall ischemia because of its proximity to the posterior aspect of the left ventricle. Jain described the use of esophageal ECG and compared it with surface ECG (SECG) in patients undergoing coronary bypass grafting. He found that the recognition and measurement of all the PQRST waves could be improved and automated by simultaneous use of esophageal ECG and SECG. The P-wave amplitude was greater in esophageal ECG than in SECG, which might facilitate the identification of supraventricular versus ventricular arrhythmias. ST-segment deviation in the unipolar esophageal ECG was not suitable for the routine detection of ischemia because of excessive noise.
FIGURE 13-4
The cardioesophagoscope. The esophageal leads are embedded in plastic. The electrocardiogram wires are connected to the right arm and left arm leads, and lead I is selected on the monitor.
To minimize the risk of esophageal burn injury, an electrocautery protection filter capable of filtering radio frequencies greater than 20 kHz should be inserted between the ECG cable and the esophageal lead.
Intracardiac Electrocardiography
For many years, long central venous catheters filled with saline have been used to record the intracardiac ECG (IC-ECG). To best illustrate an IC-ECG, 1) attach a plastic adapter to the hub of the central venous catheter (CVC), whose most distal port is at the low superior vena cava or superior region of the right atrium; 2) instill saline via a syringe needle inserted through the rubber head of the adapter, ensuring by previous aspiration that no blood clots are present and air bubbles are cleared; 3) connect the needle of the syringe with one of the six precordial cables of a standard ECG machine using an alligator clamp; and 4) record a standard ECG, which will show the IC-ECG substituting for the selected V lead. Lead V 1 is used most often, but any of leads V 1 through V 6 can be used.
Chatterjee and colleagues described the use of a modified balloon-tipped flotation catheter for recording intracavitary electrograms. The multipurpose pulmonary artery catheter presently available has all the features of a standard pulmonary artery catheter. In addition, three atrial and two ventricular electrodes have been incorporated into the catheter ( Fig. 13-5). These electrodes allow the recording of intracavitary ECGs and the establishment of atrial or AV pacing. The diagnostic capabilities of this catheter are great because atrial, ventricular, and AV nodal arrhythmias and conduction blocks can be demonstrated. The large voltages obtained from the intracardiac electrodes are relatively insensitive to electrocautery interference and are therefore useful for intraaortic balloon pump triggering. Other pulmonary artery catheters have ventricular and atrial ports that allow passage of pacing wires. These catheters also can be used for diagnostic purposes or for therapeutic interventions (pacing).
FIGURE 13-5
The multipurpose pacing pulmonary artery catheter. Three atrial and two ventricular electrodes are shown.
Benzadon and colleagues compared the amplitude of the P wave obtained by IC-ECG with those of the P waves obtained by esophageal ECG and SECG. They found that IC-ECG and esophageal ECG made it possible to register P waves larger than those registered by SECG but reported no difference between IC-ECG and esophageal ECG.
Tracheal Electrocardiography
Tracheal ECG allows monitoring when it is impractical or impossible to monitor the SECG. The tracheal ECG consists of a standard tracheal tube in which electrodes have been embedded. In a recent report, a tracheal tube was described with two coiled-wire stainless steel electrodes embedded in the tube’s cuff ( Fig. 13-6). The same safety precautions as for esophageal ECG should be followed for tracheal ECG.
FIGURE 13-6
Schematic representation of an endotracheal tube with cuff electrodes, spiral reference electrode, and electrocardiograph ( ECG) amplifier connector.
(From Hayes JK, Peters JL, Smith KW, et al: Monitoring normal and aberrant electrocardiographic activity from an endotracheal tube: comparison of the surface, esophageal, and tracheal electrocardiograms. J Clin Monit 1994;10:81-90.)
Display, Recording, and Interpretation
The American Heart Association (AHA) has published instrumentation and practice standards for ECG monitoring in special care units. Because many of the principles enumerated in these standards also are applicable to intraoperative monitoring, they are often referred to in this chapter.
Basic Requirements
The function of the ECG monitor is to detect, amplify, display, and record the ECG signal. The ECG signal usually is displayed on an oscilloscope, and most monitors now offer nonfade storage oscilloscopes to facilitate wave recognition. All ECG monitors for use in patients with cardiac disease also should have paper recording capabilities. The recorder is needed to make accurate diagnoses of complex arrhythmias and allow careful analysis of all the ECG waveforms. In addition, the recorder allows differentiation of real ECG changes from oscilloscope artifacts. The AHA special report defines a number of requirements that should be met by ECG monitoring equipment ( Boxes 13-1 to 13-3).
BOX 13-1
Electrocardiographic Monitoring Performance Requirements
From Mirvis DM, Berson AS, Goldberger AL, et al: Instrumentation and practice standards for electrocardiographic monitoring in special care units. Circulation 1989;79:464-471.
1.
Protection from overload. Protection should be adequate (no damage) for 1 V (peak to peak), 60 Hz, applied for 10 seconds to any electrode connection. The device should recover within 8 seconds after a defibrillation shock of at least 5000 V, with a delivered energy of ≥360 J.
2.
Isolated patient connection. The system should include isolated patient connections to meet standards defined in American National Standards for Safe Current Limits for Electromedical Apparatus.
3.
QRS detection. Monitors should detect QRS complexes with amplitudes of 0.5-5.0 mV, slopes of 6-300 mV/sec, and durations of 70-140 ms for adult use or 40-120 ms for pediatric use. The system should not respond to signals with an amplitude of ≤0.15 mV or a duration of ≤10 ms.
Accuracy of heart rate meter. The rate meter should be accurate to within the lesser of ±10% or ±5 beats/min over the range of 30-200 beats/min for adult use or 30-250 beats/min for pediatric use.
5.
Alarm range and accuracy. Alarm rates should be accurate to within the lesser of ±10% or ±5 beats/min over the range of 30-100 beats/min for the lower limit and 100-200 beats/min (adult) or 100-250 beats/min (pediatric) for the upper limit. Time to alarm after exceeding rate limits should not exceed 10 seconds.
6.
Noise tolerance. Heart rate meters should remain accurate during application of a 60-Hz signal, 100 mV peak to peak, minimum. Accuracy should not be affected when a triangular wave of 4 mV at 0.1 Hz is superimposed on a train of QRS signals of 0.5-mV amplitude and 100-ms duration.
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13680 | https://en.wikipedia.org/wiki/Filled_Julia_set | Filled Julia set - Wikipedia
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1 Formal definition
2 Relation to the Fatou set
3 Relation between Julia, filled-in Julia set and attractive basin of infinity
4 Spine
5 Images
6 Names
7 Notes
8 References
Filled Julia set
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From Wikipedia, the free encyclopedia
The filled-in Julia setK(f){\displaystyle K(f)} of a polynomial f{\displaystyle f} is a Julia set and its interior, non-escaping set.
Formal definition
[edit]
The filled-in Julia setK(f){\displaystyle K(f)} of a polynomial f{\displaystyle f} is defined as the set of all points z{\displaystyle z} of the dynamical plane that have boundedorbit with respect to f{\displaystyle f}K(f)=d e f{z∈C:f(k)(z)↛∞as k→∞}{\displaystyle K(f){\overset {\mathrm {def} }{{}={}}}\left{z\in \mathbb {C} :f^{(k)}(z)\not \to \infty ~{\text{as}}~k\to \infty \right}} where:
C{\displaystyle \mathbb {C} } is the set of complex numbers
f(k)(z){\displaystyle f^{(k)}(z)} is the k{\displaystyle k} -fold composition of f{\displaystyle f} with itself = iteration of functionf{\displaystyle f}
Relation to the Fatou set
[edit]
The filled-in Julia set is the (absolute) complement of the attractive basin of infinity. K(f)=C∖A f(∞){\displaystyle K(f)=\mathbb {C} \setminus A_{f}(\infty )}
The attractive basin of infinity is one of the components of the Fatou set. A f(∞)=F∞{\displaystyle A_{f}(\infty )=F_{\infty }}
In other words, the filled-in Julia set is the complement of the unbounded Fatou component: K(f)=F∞C.{\displaystyle K(f)=F_{\infty }^{C}.}
Relation between Julia, filled-in Julia set and attractive basin of infinity
[edit]
Wikibooks has a book on the topic of: Fractals
The Julia set is the common boundary of the filled-in Julia set and the attractive basin of infinityJ(f)=∂K(f)=∂A f(∞){\displaystyle J(f)=\partial K(f)=\partial A_{f}(\infty )} where: A f(∞){\displaystyle A_{f}(\infty )} denotes the attractive basin of infinity = exterior of filled-in Julia set = set of escaping points for f{\displaystyle f}
A f(∞)=d e f{z∈C:f(k)(z)→∞a s k→∞}.{\displaystyle A_{f}(\infty )\ {\overset {\underset {\mathrm {def} }{}}{=}}\ {z\in \mathbb {C} :f^{(k)}(z)\to \infty \ as\ k\to \infty }.}
If the filled-in Julia set has no interior then the Julia set coincides with the filled-in Julia set. This happens when all the critical points of f{\displaystyle f} are pre-periodic. Such critical points are often called Misiurewicz points.
Spine
[edit]
Rabbit Julia set with spine
Basilica Julia set with spine
The most studied polynomials are probably those of the formf(z)=z 2+c{\displaystyle f(z)=z^{2}+c}, which are often denoted by f c{\displaystyle f_{c}}, where c{\displaystyle c} is any complex number. In this case, the spine S c{\displaystyle S_{c}} of the filled Julia set K{\displaystyle K} is defined as arc between β{\displaystyle \beta }-fixed point and −β{\displaystyle -\beta }, S c=[−β,β]{\displaystyle S_{c}=\left[-\beta ,\beta \right]} with such properties:
spine lies inside K{\displaystyle K}. This makes sense when K{\displaystyle K} is connected and full
spine is invariant under 180 degree rotation,
spine is a finite topological tree,
Critical pointz c r=0{\displaystyle z_{cr}=0} always belongs to the spine.
β{\displaystyle \beta }-fixed point is a landing point of external ray of angle zero R 0 K{\displaystyle {\mathcal {R}}_{0}^{K}},
−β{\displaystyle -\beta } is landing point of external rayR 1/2 K{\displaystyle {\mathcal {R}}_{1/2}^{K}}.
Algorithms for constructing the spine:
detailed version is described by A. Douady
Simplified version of algorithm:
connect −β{\displaystyle -\beta } and β{\displaystyle \beta } within K{\displaystyle K} by an arc,
when K{\displaystyle K} has empty interior then arc is unique,
otherwise take the shortest way that contains 0{\displaystyle 0}.
Curve R{\displaystyle R}: R=d e f R 1/2∪S c∪R 0{\displaystyle R{\overset {\mathrm {def} }{{}={}}}R_{1/2}\cup S_{c}\cup R_{0}} divides dynamical plane into two components.
Images
[edit]
Filled Julia set for f c, c=1−φ=−0.618033988749…, where φ is the Golden ratio
Filled Julia with no interior = Julia set. It is for c=i.
Filled Julia set for c=−1+0.1i. Here Julia set is the boundary of filled-in Julia set.
Douady rabbit
Filled Julia set for c = −0.8 + 0.156i.
Filled Julia set for c = 0.285 + 0.01i.
Filled Julia set for c = −1.476.
Names
[edit]
airplane
Douady rabbit
dragon
basilica or San Marco fractal or San Marco dragon
cauliflower
dendrite
Siegel disc
Notes
[edit]
^Douglas C. Ravenel: External angles in the Mandelbrot set: the work of Douady and Hubbard. University of RochesterArchived 2012-02-08 at the Wayback Machine
^John Milnor: Pasting Together Julia Sets: A Worked Out Example of Mating. Experimental Mathematics Volume 13 (2004)
^Saaed Zakeri: Biaccessiblility in quadratic Julia sets I: The locally-connected case
^A. Douady, “Algorithms for computing angles in the Mandelbrot set,” in Chaotic Dynamics and Fractals, M. Barnsley and S. G. Demko, Eds., vol. 2 of Notes and Reports in Mathematics in Science and Engineering, pp. 155–168, Academic Press, Atlanta, Georgia, USA, 1986.
^K M. Brucks, H Bruin: Topics from One-Dimensional Dynamics Series: London Mathematical Society Student Texts (No. 62) page 257
^The Mandelbrot Set And Its Associated Julia Sets by Hermann Karcher
References
[edit]
Peitgen Heinz-Otto, Richter, P.H.: The beauty of fractals: Images of Complex Dynamical Systems. Springer-Verlag 1986. ISBN978-0-387-15851-8.
Bodil Branner: Holomorphic dynamical systems in the complex plane. Department of Mathematics Technical University of Denmark, MAT-Report no. 1996-42.
| hide v t e Fractals |
| Characteristics | Fractal dimensions Assouad Box-counting Higuchi Correlation Hausdorff Packing Topological Recursion Self-similarity |
| Iterated function system | Barnsley fern Cantor set Koch snowflake Menger sponge Sierpiński carpet Sierpiński triangle Apollonian gasket Fibonacci word Space-filling curve Blancmange curve De Rham curve Minkowski Dragon curve Hilbert curve Koch curve Lévy C curve Moore curve Peano curve Sierpiński curve Z-order curve String T-square n-flake Vicsek fractal Gosper curve Pythagoras tree Weierstrass function |
| Strange attractor | Multifractal system |
| L-system | Fractal canopy Space-filling curve H tree |
| Escape-time fractals | Burning Ship fractal Julia set Filled Newton fractal Douady rabbit Lyapunov fractal Mandelbrot set Misiurewicz point Multibrot set Newton fractal Tricorn Mandelbox Mandelbulb |
| Rendering techniques | Buddhabrot Orbit trap Pickover stalk |
| Random fractals | Brownian motion Brownian tree Brownian motor Fractal landscape Lévy flight Percolation theory Self-avoiding walk |
| People | Michael Barnsley Georg Cantor Bill Gosper Felix Hausdorff Desmond Paul Henry Gaston Julia Niels Fabian Helge von Koch Paul Lévy Aleksandr Lyapunov Benoit Mandelbrot Hamid Naderi Yeganeh Lewis Fry Richardson Wacław Sierpiński |
| Other | Coastline paradox Fractal art List of fractals by Hausdorff dimension The Fractal Geometry of Nature (1982 book) The Beauty of Fractals (1986 book) Chaos: Making a New Science (1987 book) Kaleidoscope Chaos theory |
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Choosing a Drug to Prevent Malaria
Health Care Providers
July 18, 2024
At a glance
Drugs to prevent malaria differ by country of travel.
Counsel patients to use personal protective measures along with malaria chemoprophylaxis.
Consider contraindications and drug-drug interactions when prescribing a malaria prophylaxis.
Consider the timing of when to start and stop chemoprophylaxis. Some medications require patients take them weeks in advance of travel and continued after leaving the malaria-endemic area.
Considerations
Considerations when choosing a drug for malaria prophylaxis:
Recommendations for drugs to prevent malaria differ by country of travel and can be found in CDC's Yellow Book chapter on Malaria Prevention Information, by Country. Recommended drugs for each country are listed in alphabetical order and have comparable efficacy in that country.
When used correctly malaria chemoprophylaxis is very effective. Using multiple prevention strategies together offer additional protection. Counsel patients on the use of personal protective measures, (i.e., insect repellent, long sleeves, long pants, sleeping in a mosquito-free setting or using an insecticide-treated bed net) along with malaria chemoprophylaxis.
For all medicines, also consider the possibility of drug-drug interactions with other medicines that the person might be taking as well as other medical contraindications, such as drug allergies.
When several different drugs are recommended for an area, the following table might help in the decision process.
CDC has replaced the Malaria Prevention Country Tables with the Yellow Book 2024 chapter on Malaria Prevention Information, by Country. You can find the same information regarding chemoprophylaxis by country or region.
Drug/Dosage
Reasons that might make you consider using this drug
Reasons that might make you avoid using this drug
Atovaquone/Proguanil (Malarone)
Begin 1 – 2 days before travel, daily during travel, and for 7 days after leaving.
Adults: 1 adult tablet daily.
Children: 5-8 kg: ½ pediatric tablet daily.
8-10 kg: ¾ pediatric tablet daily.
10-20 kg: 1 pediatric tablet daily.
20-30 kg: 2 pediatric tablets daily.
30-40 kg 3 pediatric tablets daily.
40 kg and over: 1 adult tablet daily.
Good for last-minute travelers because the drug is started 1-2 days before traveling to an area where malaria transmission occurs
Some people prefer to take a daily medicine
Good choice for shorter trips because you only have to take the medicine for 7 days after traveling rather than 4 weeks
Very well tolerated medicine – side effects uncommon
Pediatric tablets are available and may be more convenient
Cannot be used by women who are pregnant or breastfeeding a child less than 5 kg
Cannot be taken by people with severe renal impairment
Tends to be more expensive than some of the other options (especially for trips of long duration)
Some people (including children) would rather not take a medicine every day
Chloroquine
Begin 1 – 2 weeks before travel, once/week during travel, and for 4 weeks after leaving.
Adults: 300 mg base (500 mg salt), once/week.
Children: 5 mg/kg base (8.3 mg/kg salt) (maximum is adult dose), once/week.
Some people would rather take medicine weekly
Good choice for long trips because it is taken only weekly
Some people are already taking hydroxychloroquine chronically for rheumatologic conditions. In those instances, they may not have to take an additional medicine
Can be used in all trimesters of pregnancy
Cannot be used in areas with chloroquine or mefloquine resistance
May exacerbate psoriasis
Some people would rather not take a weekly medication
For trips of short duration, some people would rather not take medication for 4 weeks after travel
Not a good choice for last-minute travelers because drug needs to be started 1-2 weeks prior to travel
Doxycycline
Begin 1 – 2 days before travel, daily during travel, and for 4 weeks after leaving.
Adults: 100 mg daily.
Children: ≥8 years old: 2.2 mg/kg (maximum is adult dose) daily.
Some people prefer to take a daily medicine
Good for last-minute travelers because the drug is started 1-2 days before traveling to an area where malaria transmission occurs
Tends to be the least expensive antimalarial
Some people are already taking doxycycline chronically for prevention of acne. In those instances, they do not have to take an additional medicine
Doxycycline also can prevent some additional infections (e.g., Rickettsiae and leptospirosis) and so it may be preferred by people planning to do lots of hiking, camping, and wading and swimming in fresh water
Cannot be used by pregnant women and children <8 years old
Some people would rather not take a medicine every day
For trips of short duration, some people would rather not take medication for 4 weeks after travel
Women prone to getting vaginal yeast infections when taking antibiotics may prefer taking a different medicine
Persons planning on considerable sun exposure may want to avoid the increased risk of sun sensitivity
Some people are concerned about the potential of getting an upset stomach from doxycycline
Mefloquine
Begin 1 – 2 weeks before travel, weekly during travel, and for 4 weeks after leaving.
Adults: 228 mg base (250 mg salt), weekly.
Children: ≤9 kg: 4.6 mg/kg base (5 mg/kg salt), weekly.
9-19 kg: ¼ tablet weekly. >19-30 kg: ½ tablet weekly.
30-45 kg: ¾ tablet weekly.
45 kg: 1 tablet weekly.
Some people would rather take medicine weekly
Good choice for long trips because it is taken only weekly
Can be used during pregnancy
Cannot be used in areas with mefloquine resistance
Cannot be used in patients with certain psychiatric conditions
Cannot be used in patients with a seizure disorder
Not recommended for persons with cardiac conduction abnormalities
Not a good choice for last-minute travelers because drug needs to be started at least 2 weeks prior to travel
Some people would rather not take a weekly medication
For trips of short duration, some people would rather not take medication for 4 weeks after travel
Primaquine
Begin 1 – 2 days prior to travel, daily during travel, and for 7 days after leaving
Adults: 30 mg base (52.6 mg salt), daily
Children: 0.5 mg/kg base (0.8 mg/kg salt) up to adult dose daily
It is one of the most effective medicines for preventing P. vivax and so it is a good choice for travel to places with > 90% P. vivax
Good choice for shorter trips because you only have to take the medicine for 7 days after traveling rather than 4 weeks
Good for last-minute travelers because the drug is started 1-2 days before traveling to an area where malaria transmission occurs
Some people prefer to take a daily medicine
Cannot be used in patients with glucose-6-phosphatase dehydrogenase (G6PD) deficiency
Cannot be used in patients who have not been tested for G6PD deficiency
There are costs and delays associated with getting a G6PD test done; however, it only has to be done once. Once a normal G6PD level is verified and documented, the test does not have to be repeated the next time primaquine is considered
Cannot be used by pregnant women
Cannot be used by women who are breastfeeding unless the infant has also been tested for G6PD deficiency
Some people (including children) would rather not take a medicine every day
Some people are concerned about the potential of getting an upset stomach from primaquine
Tafenoquine (Arakoda TM)
Begin daily for 3 days prior to travel, weekly during travel, and for 1 week after leaving.
Adults only: 200 mg per dose.
One of the most effective drugs for prevention of P. vivax malaria, but also prevents P. falciparum
Good choice for shorter trips because you only have to take the medicine once, 1 week after traveling rather than 4 weeks
Good for last-minute travelers because the drug is started 3 days before traveling to an area where malaria transmission occurs
Cannot be used in patients with glucose-6-phosphatase dehydrogenase (G6PD) deficiency
Cannot be used in patients who have not been tested for G6PD deficiency
There are costs and delays associated with getting a G6PD test done; however, it only has to be done once. Once a normal G6PD level is verified and documented, the test does not have to be repeated the next time tafenoquine is considered
Cannot be used by children
Cannot be used by pregnant women
Cannot be used by women who are breastfeeding unless the infant has also been tested for G6PD deficiency
Not recommended in those with psychotic disorders
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13682 | https://www.sciencedirect.com/topics/computer-science/boundary-condition | Skip to Main content
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Mathematical preliminaries
2.6 Boundary condition terms
If a boundary condition involves a non-zero value then we must extend the assumed approximate solution to include additional constants to be used to satisfy the essential boundary conditions. Usually these conditions are invoked prior to or during the solution of the corresponding algebraic equations. Of course, since we are going to satisfy both the differential equation and the boundary conditions th total number of algebraic equations developed must be equal to the number of unknown parameters. To illustrate the algebraic procedure that is usually used we will solve the ODE in Eq. 2.15 with an approximation that allows any value to be assigned as the boundary condition at x = 0, say u(0) = Φ1. To apply the boundary condition after we have selected an approximate solution we will use Eq. 2.16 and pick g(x) = (1 − x) so that only the boundary condition at x = 1 is satisfied in advance. We can add another constant to f(x) to allow any boundary condition at x = 0: f(x) = Φ1 + Φ2 x + Φ3 x2. Then the residual error is
Since we now have three unknown degrees of freedom, Φ, we must have three weighted residual equations. For simplicity we will choose the collocation method and pick three equally spaced collocation points. Evaluating the residual at the quarter points and multiplying by the common denominator gives the three equations
Note that since we know u at x = 0 these unknowns are not independent. Substituting x = 0 into our approximate solution and equating it to the assign boundary value there gives u(0) = Φ1. We call this an essential boundary condition on Φ1 There are an infinite number of possible boundary conditions and we gain flexibility by allowing extra constants to satisfy them. Since Φ1 will be a known number only the last two rows are independent for determining the remaining terms in Φ. Note that the first column of numbers, in the last two rows, is now multiplied by a known value and thus they can be carried to the right hand side to give the reduced algebraic system for the independent Φ:
An equivalent matrix modification routine in the MODEL code deletes the redundant coefficients, but keeps the matrix the same size to avoid re-ordering all the coefficients as done above. For the common original boundary condition of u(0) = 0, we have Φ1 = 0 and the changes to the right hand side (RHS) are not necessary. But the above form also allows us the option of specifying any non-zero boundary condition we need. Using the zero value gives a solution of Φ2 = 0. 2058 and Φ3 = 0. 1598. The resulting values at the interior quarter points are 0. 046, 0.071, and 0.061, respectively. These compare well with the previous results.
If the second boundary condition had been applied other than at x = 0 then we would have a more complicated relation between the Φ. For example, assume we move the boundary condition to x = 0.5. Then evaluating the approximate solution there yields
which is called a linear constraint equation on Φ, or a multipoint constraint (MPC). In other words we would have to solve the weighted residual algebraic system subject to a linear constraint. This is a fairly common situation in practical design problems and adaptive analysis procedures. The computational details for enforcing the above essential boundary conditions are discussed in detail later.
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Book2005, Finite Element Analysis with Error EstimatorsJ.E. Akin
Chapter
Boundary Conditions
2005, Computational Fluid Dynamics: Principles and Applications (Second Edition)J. Blazek
The following types of boundary conditions are in general encountered in the numerical solution of the Euler and the Navier-Stokes equations:
•
: solid wall,
•
: farfield in external flows,
•
: inflow and outflow in internal flows,
•
: injection boundary,
•
: symmetry,
•
: coordinate cut and periodic boundary,
•
: boundary between blocks.
The numerical treatment of these boundary conditions is described in detail in the following sections. For literature on further boundary conditions like heat radiation on walls or like free surfaces, the reader is referred to Section 3.4.
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Book2005, Computational Fluid Dynamics: Principles and Applications (Second Edition)J. Blazek
Chapter
Computer-Aided Design
2003, Encyclopedia of Information SystemsGeorge Gustav Savii
IX.C. Imposing Boundary Conditions
To solve the set of partial differential equations, some consistent boundary conditions must be specified. For example, in a stress/strain problem, the conditions are in the form of applied forces/pressures and/or displacements. In a temperature field problem, the boundary conditions can be heat fluxes and/or temperatures.
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Reference work2003, Encyclopedia of Information SystemsGeorge Gustav Savii
Chapter
Quantitative Systems Pharmacology
2018, Computer Aided Chemical EngineeringDiego Caccavo, ... Anette Larsson
3.6 Initial and boundary conditions
Assuming the stress-free state:
(48)
(49)
Finally, a boundary condition has to be introduced in the form of an implicit nonlinear Dirichlet condition (Lucantonio et al., 2013):
(50)
according to which the chemical potential internal to the system is equal to the chemical potential of the external solvent (μ1, ext).
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Book series2018, Computer Aided Chemical EngineeringDiego Caccavo, ... Anette Larsson
Chapter
Batch Processing
2003, Encyclopedia of Physical Science and Technology (Third Edition)Narses Barona
VIII.A.1 Constraints
Problem definition requires specification of the initial state of the system and boundary conditions, which are mathematical constraints describing the physical situation at the boundaries. These may be thermal energy, momentum, or other types of restrictions at the geometric boundaries. The system is determined when one boundary condition is known for each first partial derivative, two boundary conditions for each second partial derivative, and so on. In a plate heated from ambient temperature to 1200 °F, the temperature distribution in the plate is determined by the heat equation ∂T/∂t = α∇2T. The initial condition is T = 60 °F at t = 0, all over the plate. The boundary conditions indicate how heat is applied to the plate at the various edges: y = 0, 0 < x < a, ∂T/∂y = 0; y = b, 0 < x < a, ∂T/∂y = 0; x = 0, 0 < y < b, ∂T/∂x = 0; x = a, 0 < y < b, −k(∂T/∂x) = h(T − TA). The first three conditions indicate that the plate is insulated while it is heated by convection along the fourth edge, x = a, from an environment at temperature TA.
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Reference work2003, Encyclopedia of Physical Science and Technology (Third Edition)Narses Barona
Review article
2021 Review Issue
2021, The Journal of Strategic Information SystemsGerit Wagner, ... Guy Paré
Boundary condition 1: Digitality
The first boundary condition refers to the digitality of the matching, contracting, and executing processes and the mediation by a digital platform. It excludes research on nondigital platforms (de Reuver et al. 2018), such as traditional, offline knowledge work and labor markets, as well as matchmaking platforms mediating between a supply-side and a demand-side of work without getting involved in the contracting and execution processes. Thus, DPKW can be considered as a digital successor of traditional matchmaking organizations and contracting agencies (Barley and Kunda 2006; Kunda et al. 2002). Digital mediation of the entire value chain from posting a project to rating the work upon completion enables a more comprehensive scope of data-driven operation and governance compared to platforms restricted to individual macro processes. We contend that considering the constituent (macro) processes and their mediation by the digital platform offers a meaningful criterion for distinguishing DPKW in the light of increasingly complex service ecosystems that cover selected processes, combine different technological components, and integrate with complementary platforms.
Technical consequences of boundary condition 1 pertain to the extensibility and adaptability of the platform artifact, which aligns with the broader literature on digital platforms (e.g., de Reuver et al. 2018; Tiwana et al. 2010). Extensibility refers to the modification of the platform core by the platform operator, the integration of knowledge-intensive services into existing workflows through boundary resources (e.g., APIs), as well as to practices of “circumtechventing” (cf. Sison and Lavilles 2018). These practices refer to “the use of technology to find ways around perceived obstacles to one’s main concern” (Sison and Lavilles 2018, p.12).
Digitalizing knowledge work services has been associated with significant economic implications. Consistent with economics of electronic (service) markets (e.g., Kauffman and Walden 2001; Malone et al. 1987), transaction cost theory (Williamson 1979; Williamson 1991), and the literature on outsourcing (Lacity et al. 2009), digitalizing (knowledge work) services and mediating them on a platform results in lower search, transaction, and coordination costs (Gefen and Carmel 2008; Gong et al. 2018).
From a macroeconomic perspective, digitalization of labor markets that are not rooted in the physical world (Huang et al. 2020) further implies unrestricted access of workers on a global scale (Chan and Wang 2014). In this context, Clemens (2011) insightfully contends that digitizing work arrangements could – by circumventing legal barriers to physical migration – result in global GDP gains “one or two orders of magnitude larger than the gains from dropping all remaining restrictions on international flows of goods and capital” (p.83). Enabled by lower barriers to labor mobility, the success of platforms like DPKW is partly driven by vast global imbalances in the cost of labor. The corresponding potential for global labor arbitrage has been considered to induce client preferences for workers from low-income countries over workers from high-income countries and to trigger migration of workers from high-income countries to low-cost destinations (cf. Roach 2003; Schlagwein et al. 2019). These implications are primarily driven by digitality but also enabled by the departure from traditional value-chain logic and organizational-level outsourcing arrangements for knowledge work.
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Journal2021, The Journal of Strategic Information SystemsGerit Wagner, ... Guy Paré
Chapter
Disease Modelling and Public Health, Part B
2017, Handbook of StatisticsChristina Kuttler
3.3 Initial and Boundary Conditions
For a concrete applied problem and a concrete mathematical solution, one needs to define how the system has to be started, the so-called initial condition, typically formulated as
If the area of interest is not bounded, no boundary conditions are needed. But in most situation in nature or also in laboratory experiments, space is limited, e.g., for bacteria which are in a Petri dish. For such problems, realistic boundary conditions are needed. Also from a more mathematical point of view, well-suited boundary conditions are needed; without these, uniqueness of the solution cannot be guaranteed.
Let us consider a typical reaction–diffusion equation on a bounded domain , i.e., .
We choose the diffusion constant D = 1, w.l.o.g. In general, boundary conditions may have the form
This implicit notation looks a bit confusing at first glance, but in this way boundary conditions are also formulated in some software packages like MATLAB®. There are some typical types of boundary conditions, formulated more explicitly.
: Dirichlet boundary condition:
where the function b is prescribed. In case of b = 0, it is the so-called homogenous Dirichlet boundary condition. That means that everything which reaches the boundary disappears.
: Neumann boundary condition:
n denotes the outer normal to Ω at x ∈ ∂Ω. In case of b = 0 it is called the homogeneous Neumann boundary condition. That corresponds to the “no flux condition,” as no individuals or particles can leave or enter the domain Ω via the boundary ∂Ω.
: Robin boundary condition:
This boundary condition connects the flux to the concentration/density of the considered substance or population.
Please note, that the Robin boundary condition is also called “mixed boundary condition.”
Of course, also different types of boundary conditions on different parts of the boundary can be combined. One just has to make sure that they do not contradict each other, i.e., are compatible. The required compatibility also concerns initial and boundary conditions, where they meet. The choice of properly posed boundary conditions and initial conditions is essential for the existence of solutions and their smoothness properties.
Even though nonlinear boundary conditions are possible in principle, linear conditions in u are used in nearly all relevant cases, as shown above in the three classical types.
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Handbook2017, Handbook of StatisticsChristina Kuttler
Chapter
23 European Symposium on Computer Aided Process Engineering
2013, Computer Aided Chemical EngineeringMateusz Korpyś, ... Janusz Wójcik
3.2 Boundary conditions
The boundary conditions related to the fluid dynamics and heat transfer in the CFD code were defined as follows:
•
: At the inlet, the mass flow rate of the fluid was set as a constant.
•
: The flow direction was defined normal to the boundary.
•
: The constant heat flux 115 W was specified at the bottom wall of the heat exchanger for all cases.
•
: At the outlet, a fully developed flow (pressure outlet) condition was imposed.
•
: At all the walls, a no-slip boundary condition was proposed.
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Book series2013, Computer Aided Chemical EngineeringMateusz Korpyś, ... Janusz Wójcik
Chapter
Boundary Conformal Field Theory
2006, Encyclopedia of Mathematical PhysicsJ. Cardy
Boundary CFT
In any field theory in a domain with a boundary, one needs to consider how to impose a set of consistent boundary conditions. Since CFT is formulated independently of a particular set of fundamental fields and a Lagrangian, this must be done in a more general manner. A natural requirement is that the off-diagonal component T∥⊥ of the stress tensor parallel/perpendicular to the boundary should vanish. This is called the conformal boundary condition. If the boundary is parallel to the time axis, it implies that there is no momentum flow across the boundary. Moreover, it can be argued that, under the RG, any uniform boundary condition will flow into a conformally invariant one. For a given bulk CFT, however, there may be many possible distinct such boundary conditions, and it is one task of BCFT to classify these.
To begin with, take the domain to be the upper-half plane, so that the boundary is the real axis. The conformal boundary condition then implies that when z is on the real axis. This has the immediate consequence that correlators of are those of T, analytically continued into the lower-half plane. The conformal Ward identity, cf. , now reads
In radial quantization, in order that the Hilbert spaces defined on different hypersurfaces be equivalent, one must choose semicircles centered on some point on the boundary, conventionally the origin. The dilatation operator is now
where S is a semicircle. Using the conformal boundary condition, this can also be written as
where C is a complete circle around the origin. As before, one may similarly define the , and they satisfy a Virasoro algebra.
Note that there is now only one Virasoro algebra. This is related to the fact that conformal mappings which preserve the real axis correspond to real analytic functions. The eigenstates of correspond to boundary operators acting on the vacuum state | 0〉. It is well known that in a renormalizable QFT operators at the boundary require a different renormalization from those in the bulk, and this will in general lead to a different set of conformal weights. It is one of the tasks of BCFT to determine these, for a given allowed boundary condition.
However, there is one feature unique to boundary CFT in two dimensions. Radial quantization also makes sense, leading to the same form for the dilation operator, if the boundary conditions on the negative and positive real axes are different. As far as the structure of BCFT goes, correlation functions with this mixed boundary condition behave as though a local scaling field were inserted at the origin. This has led to the term “boundary condition changing (bcc) operator,” but it must be stressed that these are not local operators in the conventional sense.
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Reference work2006, Encyclopedia of Mathematical PhysicsJ. Cardy
Review article
Effect of ventilator configuration on the distributed climate of greenhouses: A review of experimental and CFD studies
2010, Computers and Electronics in AgriculturePierre-Emmanuel Bournet, Thierry Boulard
6.1.3 Boundary conditions and source terms
Boundary conditions are generally inferred from experimental data and, consequently, poor input data quality can affect the quality of the simulations. These experimental data are generally not directly used but adapted instead, like in the case of the logarithmic wind velocity profile inferred from several measurements at given heights. Consequently, uncertainties about the parameter settings of these functions may cause discrepancies in the results. Likewise, the formulation of the source terms that account for specific processes, such as the pressure drop caused by insect screens or crop drag, also involves empirical parameters that are poorly estimated. A compromise must therefore be found between the accuracy of the modelling approach and the information necessary for establishing the appropriate boundary conditions and the model parameters.
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Journal2010, Computers and Electronics in AgriculturePierre-Emmanuel Bournet, Thierry Boulard
Related terms:
Approximation (Algorithm)
Boundary Value Problems
Finite Element Method
Genetic Algorithm
Partial Differential Equation
Computational Domain
Decision Variable
Discretization
Eigenvalue
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13683 | https://ghomi.math.gatech.edu/LectureNotes/LectureNotes9G.pdf | Math 598 Mar 2, 20051 Geometry and Topology II Spring 2005, PSU Lecture Notes 9 3 Some Topics from Differential Topology 3.1 Regular points and values; Fundamental Theorem of Algebra Let f : M →N be a smooth map. We say that p ∈M is a regular point of f provided that rank(d fp) = dim(N); otherwise we say that p is a critical point or a singular point. We say q ∈N is a regular value of f provided that every p ∈f −1(q) is a regular point. If q is not a regular value of f, then we say that it is a critical value or singular value.
Exercise 1. Show that if f : M m →N n has rank k at some point p ∈M, then it has rank ≥k on a neighborhood of p.
Proposition 2. If f : M m →N n is a smooth map, q is a regular value of f, and f −1(q) ̸= ∅, then f −1(q) is a smooth (m −n)-dimensional submanifold of M.
Proof. By definition, every p ∈f −1(q) is a regular point. Thus, f has con-stant rank n on f −1(q).
This implies that f has constant rank n on an open neighborhood of U of f −1(q). Since f : U →N has constant rank n, it now follows, as we had proved earlier, that f −1(q) is an (m −n)-dimensional submanifold of U, and therefore of M.
Exercise 3. Show that if f : M →N is a smooth map, dim(M) = dim(N), M is compact, and q is a regular value of f, then f −1(q) consists of a finite number of points.
Further, show that if we denote the number of these points by #f −1(q), then #f −1(q) is locally constant, i.e., there is an open neighborhood U of q in N, such that #f −1(q′) = #f −1(q), for all q′ ∈U.
1Last revised: September 23, 2022 1 The following proof demonstrates the elegant and powerful utility of basic techniques of Differential topology.
Theorem 4 (Fundamental Theorem of Algebra). Every nonconstant complex polynomial has a zero.
Proof(After Milnor). Let P : R2 →R2 be a complex polynomial, and π+ : S2− {(0, 0, 1)} →R2 be the stereographic projection from the north pole. Define f : S2 →S2 by f((0, 0, 1)) := (0, 0, 1), and f(p) := (π+)−1 ◦P ◦π+(p), if p ̸= (0, 0, 1).
We claim that f is smooth. This is obvious on S2 −{(0, 0, 1)} where f is the composition of three smooth functions. To see that f is smooth in a neighborhood of (0, 0, 1) as well, let π−: S2 −{(0, 0, −1)} →R2 be the stere-ographic projection from the south pole, U ⊂R2 be a small neighborhood of the origin o ∈R2, and define Q: U →R2 by Q(z) := π−◦f ◦(π−)−1(z).
Note that if U is sufficiently small f ◦(π−)−1(z) is close to f ◦(π−)−1(o) = (0, 0, 1) for all z ∈U. In particular, if U is sufficiently small, f ◦(π−)−1(z) ̸= (0, 0, −1) for all z ∈U. Thus Q is well defined. Secondly, note that π+ ◦ (π−)−1 is inversion with respect to the unit circle in R2, i.e., π+◦(π−)−1(z) = z/∥z∥2 = 1/z. This yields that, if P(z) = Pn i=0 aizi, with an ̸= o, then Q(z) = π−◦(π+)−1 ◦P ◦π+ ◦(π−)−1(z) = ((π+) ◦(π−)−1)−1 ◦P(1/z) = 1 P(1/z) = zn Pn i=0 aizn−i.
Thus Q is smooth on U, which yields that f is smooth near (0, 0, 1).
Next note that d fp is singular, if and only of π+(p) is a root of the complex polynomial P ′(z) = Pn i=1 aiizi−1 (see the next exercise).
But a complex polynomial has only finitely many roots unless it is identically zero.
So, since by assumption an ̸= o, we conclude that the set of regular values of f 2 consists of S2 minus a finite number of points. In particular, the set of regular values of f is connected and open. So the locally constant function #f −1 is constant on the set of regular values of f. Since the number of singular points of f are finite, #f −1 cannot be zero everywhere, so it is zero nowhere on the set of regular values of f. This yields that f is onto. In particular, there exists p ∈S2 such that f(p) = (0, 0, −1). So P(π+(p)) = π+(f(p)) = o.
Exercise 5. Let P : R2 →R2 be a complex polynomial. Show that dPz(w) = P ′(z)w.
More precisely if θz : TzR2 →R2 denotes the standard isomorphism, then dPz(w) = θ−1 z (P ′(z)θz(w)). In particular z is a singular point of P if and only if z is a root of P ′(z) 3.2 Manifolds with boundary Next we generalize the last proposition, concerning the inverse image of reg-ular values, to manifolds with boundary. Recall that M is a manifold with boundary if every point of M has an open neighborhood which is homeomor-phic to an open subset of the half space Hn := {(x1, . . . , xn) ∈Rn | xn ≥0}.
We define the interior, int Hn, and boundary, ∂Hn, as the subsets of Hn where xn > 0 and xn = 0 respectively. We say that p is an interior point of M if an open neighborhood of p is homeomorphic to an open neighborhood in int Hn; otherwise, we say that p is a boundary point.
Exercise 6. Show that if M n is an n-manifold with ∂M ̸= ∅, then ∂M is an (n −1)-manifold without boundary.
Similar to the case of manifolds (without boundary), we say that a man-ifold with boundary is smooth if it admits an atlas such that all the local charts in that atlas are C∞compatible, i.e., if (U, φ) and (V, ψ) are local charts of M then φ ◦ψ−1 : ψ(U ∩V ) →Rn is smooth. Further, it is im-portant to recall that if A ⊂M is any subset, then f : A →N is smooth provided that for every p ∈A there exists an open neighborhood U of p in M and a smooth function fp : U →N such that f|U∩A = fp. For any p ∈M, we may define TpM as the equivalence class of all the half curves 3 which originate or end at p. More explicitly, define CurvespM as the set of curves α: [0, ϵ) →M and α: (−ϵ, 0] →M such α(0) = p. We say that a pair of curves α, β ∈CurvespM are equivalent provided that there exists a local chart (U, φ) centered at p such that (φ ◦α)′(0) = (ψ ◦β)′(0). Then TpM is defined as the collection of equivalence classes CurvespM/ ∼.
Exercise 7. Check that the above definition for TpM coincides with the one we had given earlier whenever p ∈int M.
Exercise 8. Show that if M n is a manifold with boundary, p ∈M, and (U, φ) is any local chart centered at p, then the mapping TpM ∋[α] 7− →(φ ◦α)′(0) ∈Rn is a bijection. Thus we may use this map to endow TpM with the structure of an n-dimensional vector space.
The differential of any smooth map f : M →N, where M is a manifold with boundary, is defined as before.
Exercise 9. Let f : Hm →M be a smooth map and p ∈∂H. Show that if U is any open neighborhood of p in Rn and ˜ f : U →M is any smooth map such that ˜ f = f on U ∩Hm, then d fp = d ˜ fp.
Exercise 10. Let M m be a smooth manifold with boundary and f : M m → R be a smooth map which has 0 as a regular value. Then f −1([0, ∞)) is a smooth m-manifold with boundary and ∂f −1([0, ∞)) = f −1(0).
Our main aim in this section is to prove: Theorem 11. Let M m be a manifold with boundary and f : M m →N n be a smooth map. Suppose that q ∈N is a regular value of both f and f|∂M, and f −1(q) ̸= ∅, then f −1(q) is a smooth m −n manifold with boundary, and ∂ f −1(q) = f −1(q) ∩∂M.
To prove the above theorem we need a couple of more exercises: Exercise 12. Show that to prove the above theorem it is enough to consider the case of M m = Hm.
4 Exercise 13. Show that if M is a manifold (without boundary) and f : M → N is a smooth map which has q as a regular value, then, for every p ∈f −1(q), Tpf −1(q) is the null space of d fp.
Proof of Theorem 11. We know that f −1(q) ∩int M is a smooth manifold, because int M is a manifold (without boundary) and q is a regular value of f|int M. Thus it remains to consider points p ∈f −1(q) ∩∂M.
We may suppose M = Hm. Then, by definition of smoothness, there exists, for every p ∈∂M = ∂H, an open neighborhood V of p in Rm and a smooth function ˜ f : V →N such that ˜ f = f on V ∩Hm. Since, by a previous exercise, d ˜ fp = d fp, it follows that p is a regular point of ˜ f. Thus, after replacing V by a smaller neighborhood of p, which we again denote by V , we may assume that ˜ f as no critical points in V . In particular, it follows that q is a regular value of ˜ f. So ˜ f −1(q) is a smooth submanifold of V . Now define g: ˜ f −1(q) →R by g(x1, . . . , xn) = xn. Then g(p) = 0 and V ∩f −1(q) = H ∩˜ f −1(q) = g−1([0, ∞)). Thus to complete the proof it suffices to show that 0 is a regular value of g. Suppose not. Then Tp ˜ f −1(q) is equal to the null space of dgp. But the null space of dgp is a subset of Tp∂H. Thus, since Tp ˜ f −1(q) and Tp∂H have the same dimension, it follows that Tp ˜ f −1(q) = Tp∂H. But Tp ˜ f −1(q) is the null space of d ˜ fp which is equal to the null space of d fp. So the null space of d fp is equal to Tp∂H, which contradicts the assumption that p is a regular point of f|∂H.
3.3 Sard’s Theorem, and Brouwer’s Fixed Point The-orem Let f : M →N be a smooth map. Sard’s theorem states that almost ev-ery point q ∈N is a regular value of f, where “almost every”, or “almost all”, means except for a set of measure zero. This theorem, whose proof we postpone for the time being, has great many applications, including the Brouwer’s fixed point theorem which we prove below.
Exercise 14. Let Γ ⊂R2 be a smooth simple closed curve, i.e., the image of a smooth embedding of S1. For any unit vector u ∈S1, the height function hu : Γ →R is defined as hu(p) = ⟨p, u⟩. Use Sard’s theorem to show that, for almost all u ∈S2, hu has a finite number of critical points (Hint: It is enough to show that for almost every u ∈S1 there exist only finitely many tangent lines of Γ which are orthogonal to u).
5 Our main aim in this section is to use Sard’s theorem to show that Theorem 15 (Brouwer). For n ≥2, any continuous map f : Bn →Bn has a fixed point, where Bn denotes the n-dimensional closed unit ball in Rn.
The proof is by contradiction and requires the following lemmas: Lemma 16. If there exists a continuous map f : Bn →Bn without fixed points, then there exists a smooth map ˜ f : Bn →Bn without fixed points.
Proof. If f has no fixed points, then, since Bn is compact, there exists an ϵ > 0 such that ∥f(p)−p∥> ϵ for all p ∈Bn. By Wierstrauss approximation theorem, there exists a smooth map f : Bn →Rn such that ∥f(p) −f(p)∥< ϵ/2 for all p ∈Bn. Let ˜ f := 1 1 + ϵ/2f.
Then ˜ f : Bn →Bn, because, by the triangle inequality, ∥f(p)∥≤∥f(p)∥+ ∥f(p) −f(p)∥< 1 + ϵ/2, which yields that ∥˜ f∥≤1. Further note that, again by the triangle inequality, ∥˜ f(p) −f(p)∥≤∥˜ f(p) −f(p)∥+ ∥f(p) −f(p)∥< ϵ/2 + ϵ/2 = ϵ.
Thus, since ∥f(p) −p∥> ϵ, it follows that ∥˜ f(p) −p∥≥∥f(p) −p∥−∥˜ f(p) −f(p)∥> 0.
So ˜ f has no fixed points.
Lemma 17. If there exists a smooth map f : Bn →Bn without fixed points, then there exists a smooth map r: Bn →Sn−1 such that r(p) = p for all p ∈Sn−1.
Proof. Consider the ray which starts from f(p) and passes through p. Let r(p) be the intersection of this ray with Sn−1. To see that r is smooth, note that the ray may be parametrized by ℓ(t) := f(p) + t p −f(p) , where t ≥0. Solving ∥ℓ(t)∥= 1 for t and substituting the solution back in ℓ(t) gives an explicit expression for r(p), which one may check to be smooth.
6 Exercise 18. Find the explicit expression for r in the above lemma and show that r is smooth.
Now we are ready to prove the main result of this section: Proof of Brouwer’s Theorem. Let r be as in the previous lemma. By Sard’s theorem, r has a regular value p ∈Sn−1. Then r−1(p) is a 1-dimensional manifold with boundary which contains p, since r(p) = p. Further recall that ∂r−1(p) = r−1(p) ∩Sn−1. But r−1(p) ∩Sn−1 = {p} because r is one-to-one on Sn−1. Thus ∂r−1(p) = {p}. But r−1(p), being the closed subset of a compact space, is compact, and each component of a compact 1-dimensional manifold with boundary must have either zero or two boundary points. So we have a contradiction.
Note that the above proof uses the fact that every compact connected one dimensional manifold with boundary is either homeomorphic to S1 or the interval [0, 1]. Can you prove this fact?
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13684 | https://ntrs.nasa.gov/api/citations/20200006182/downloads/Introduction%20to%20Numerical%20Methods%20in%20Heat%20Transfer.pdf | 1 Introduction to Numerical Methods in Heat TransferPresented by:Steven L. RickmanNASA Technical Fellow for Passive ThermalNASA Engineering and Safety CenterThermal and Fluids Analysis Workshop 2016Cleveland, Ohio August 2024 NASA Photo: ISS016E009207 Introduction Heat transfer is best understood through theory and application of principles in thermal analysis; Modern thermal analysis leverages the power of computers and numerical methods to simulate heat transfer in networks representing a physical system; This lesson is an introduction to numerical methods in heat transfer.
2 Overview 3 Students may have experience with numerical methods in college level courses; Once in the workplace, however, they often use commercially available modeling tools; Unless graduate study is pursued, students must create opportunities to understand the fundamentals behind these methods; This course is intended to provide such focus.
Scope of this Lesson 4 What is heat transfer?
Governing differential equation; Finite difference; Finite element; Thermal radiation in heat transfer analysis.
Lesson Roadmap 5 What is Heat Transfer?
Finite Difference Governing Differential Equation for 1-D Heat Transfer Finite Element Calculus of Variations Thermal Modeling Basics Fourier's Heat Conduction Law Solution for 1-D Heat Transfer Establishing the Functional Equation Example 4: Steady State Analysis - Distributed Heating Example 1: Solving the Transient Heat Equation (1-D) Example 3: Adding Thermal Radiation Example 2: Transient Thermal Analysis Example 5: Steady State Analysis Example 6: Transient Analysis Implicit Formulation Numerical Methods What is Heat Transfer? (Ref. 1) Heat transfer is energy transfer due to a temperature difference and can only be measured at the boundary of a system. Conduction - Heat transfer from one substance to another by direct contact.
Convection - Heat transfer via movement of fluids.
Radiation – Heat transfer via electromagnetic waves. 6 Thermal Modeling Basics 7 Engineers use thermal models as a tool to aid in design and to understand the performance of thermal systems; A thermal model is an abstraction of a physical system.
Thermal Modeling Basics 8 Closed-form solutions do not exist for all physical systems or geometries -- so they are discretized into smaller pieces, called elements.
Note: Geometry and mesh created using MSC Patran® Geometry Geometry Discretized into a Finite Element Mesh Thermal systems can also be represented by representations analogous to electrical resistance-capacitance (RC) networks.
Geometry that is discretized is represented by nodes. A node is isothermal and has a constant temperature throughout the entire volume. Heat flow between nodes is modeled with conductors (the inverse of thermal resistance) and heat storage is modeled using a capacitor.
Thermal Modeling Basics 9 Thermal Modeling Basics 10 Nodes come in the following varieties: Diffusion -- a node representing a finite mass with finite capacitance; Arithmetic -- a massless node representing zero capacitance; Boundary -- a node representing a source or sink with infinite capacitance.
Thermal Modeling Basics 11 Conductors come in the following varieties: Conduction -- heat transfer between solid objects (or a solid and a gas/fluid); Convection -- heat transfer between a solid object and a convecting liquid or gas; Radiation -- heat transfer via electromagnetic radiation between objects.
Thermal Modeling Basics 12 Heat transfer via conduction and convection is proportional to DT: But, heat transfer via radiation takes this form: Thermal Modeling Basics 13 The network representation for this configuration might look something like this: Consider this geometry Some Terminology Heat rate, denoted by , is [energy/time]; Heat flux, denoted by , is [energy/area/time]; Volumetric heat rate, denoted by , is [energy/volume/time]; Heat per unit length, denoted by , is [energy/length/time].
14 Fourier Heat Conduction Law Fourier assumed that conduction heat transfer is directly proportional to the temperature difference across a solid: 15 A A L k T1 T2 Conservation of Energy In heat transfer calculations, energy must be conserved: where... is energy per unit time entering; is energy per unit time generated; is energy per unit time leaving; is time rate of change of energy stored.
16 17 x x+dx Deriving the Heat Equation in One Dimension (Ref. 2) The energy per unit time stored in a mass can be expressed as: where... is the mass [mass]; is the specific heat [energy/mass/temperature]; is the material density [mass/volume]; is the material volume [volume]; is the differential element cross sectional area [area]; is the differential element length [length]; is the mass' time rate change of temperature [temperature/time].
18 Deriving the Heat Equation in One Dimension (Ref. 2) The heat entering the mass per unit time at location x is: where... is the thermal conductivity [energy/time/length/temperature]; is the differential element cross sectional area [area]; is temperature gradient at x [temperature/length]. 19 Deriving the Heat Equation in One Dimension (Ref. 2) The heat leaving the mass per unit time at location x+Dx is: where... is the thermal conductivity [energy/time/length/temperature]; is the differential element cross sectional area [area]; is temperature gradient at x+dx [temperature/length]. 20 Deriving the Heat Equation in One Dimension (Ref. 2) But we recognize that the heat transfer at x+dx may be expressed as: where, the first term is simply the heat entering the left face per unit time and the second term is the change in heat transfer rate with respect to x over the control volume times the distance to the other end of the control volume.
21 Deriving the Heat Equation in One Dimension (Ref. 2) We can now substitute our new expressions for the original terms: This reduces to: 22 Deriving the Heat Equation in One Dimension (Ref. 2) Rearranging and simplifying yields: Or, for a constant thermal conductivity: This is known as the heat equation in one dimension expressed per unit volume.
23 Deriving the Heat Equation in One Dimension (Ref. 2) Extension to 2-D and 3-D More generally, the heat equation for isotropic thermal conductivity can be expanded into additional dimensions: or, using the Laplacian: 24 One Dimensional Heat Equation: Steady State Assuming the heat generation term is constant, the only time dependency appears on the right hand side of the equation: At steady state, the time rate of change of temperature, ∂T/ ∂ t = 0; We see that steady state behavior is independent of density, r, and specific heat, Cp .
25 0 For a case with no internal heat generation, we have: We recognize this to be a linear differential equation of second order in the spatial dimension, x and first order in time, t.
Fortunately, there exists a solution.
26 One Dimensional Heat Equation: Transient 0 Consider the following example: The rod is initially at a uniform temperature, T0. At time, t = 0, the temperature at both ends (x=0 and x=2L) is raised to T1; How does the temperature profile in the rod change with time?
Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) 27 r , Cp , k x x=0 x=2L T0 T1 T1 The governing differential equation is the previously derived heat equation: We recast the equation noting that a = k/rCp and q=T-T1: 28 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) Using this new form, we examine our boundary conditions for q(x,t): For 0 < x < 2L and t = 0: q(x,0) = T0 - T1 For x = 0 and t > 0: q(0,t) = 0 For x = 2L and t > 0: q(2L,t) = 0 29 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) To solve this differential equation, we can use separation of variables by assuming: where...
X(x) is a function of only x, and; Y(t) is a function of only t.
30 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) Forming the required derivatives of q(x, t): So our transformed differential equation becomes: 31 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) We can rearrange the previous equation to read: To facilitate solution, we can set both equations equal to a constant, where l is called the separation constant: 32 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) We can now write two, independent linear differential equations: 33 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) We recognize that the first equation: has a solution of the form: 34 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) The second equation: has a solution of the form: 35 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) Our expression for q(x, t), then, becomes: We can simplify this expression by letting: 36 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) The expression becomes: The previously defined boundary conditions are now used to solve for constants, C1 and C2: For 0 < x < 2L and t = 0: q(x,0) = T0 - T1 For x = 0 and t > 0: q(0,t) = 0 For x = 2L and t > 0: q(2L,t) = 0 37 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) When we apply the second boundary condition (i.e., for x = 0 and t > 0): From this, we see that C1 must be zero.
We note that C2 cannot be zero -- if it was, q(x, t) = 0 everywhere.
38 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) When we apply the third boundary condition (i.e., x = 2L and t > 0): Since C2 ≠ 0, we see that the only way for this to happen is when: 39 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) This happens when: So the solution may be expressed in the form of a series given by: 40 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) Cn may be determined by integrating considering the initial conditions (for n = 1, 3, 5, ...): The overall solution becomes (for n = 1, 3, 5, ...): 41 Example 1: Solving the Transient Heat Equation (Adapted from Ref. 2) 42 Fraction of Rod Length (x/2L) Temperature (C) n = 1 n = 3 n = 5 n = 21 Time = 0.1 seconds T0 = 100 C T1 = 200 C Example 1: Solving the Transient Heat Equation 43 Fraction of Rod Length (x/2L) Temperature (C) t = 1 sec t = 2 sec t = 5 sec t = 10 sec t = 20 sec t = 100 sec n = 199 T0 = 100 C T1 = 200 C Example 1: Solving the Transient Heat Equation Numerical Methods We'll focus on two different numerical methods: Finite Difference -- uses the differential formulation -- i.e., equations are formulated using the governing differential equation -- where we replace the partial derivatives by approximations obtained by Taylor expansions near the point of interest; Finite Element -- uses a variational formulation -- i.e., equations are formulated from an integral formulation arising from the Calculus of Variations.
44 Formulation of the Finite Difference for 1-D Heat Transfer Finite difference relies on a differential formulation -- that is, a description of the heat transfer using derivatives; For our one-dimensional heat transfer case, recall the governing differential equation is: 45 Formulation of the Finite Difference for 1-D Heat Transfer This equation involves the second derivative of temperature with respect to a spatial dimension (i.e., x) and the first derivative of temperature with respect to time; Let's take a closer look at how these derivatives are formed for a numerical solution.
46 Formulation of the Finite Difference for 1-D Heat Transfer Consider this temperature distribution about x: 47 x T(x) x x +Dx x -Dx Dx Dx Formulation of the Finite Difference for 1-D Heat Transfer (Ref. 3) At a given instant in time, the first derivative, taken to the "right" of x, is given by: Similarly, the first derivative, taken to the "left" of x, is given by: Each expression yields an approximation of the slope in the specified region about x.
48 Formulation of the Finite Difference for 1-D Heat Transfer (Ref. 3) The second derivative is just the slope of the slope over the region, or, the slope of the first derivatives: which becomes...
49 Formulation of the Finite Difference for 1-D Heat Transfer (Ref. 3) Further simplification and rearranging yields: Hence we have an expression for the second derivative in terms of temperatures at specific locations.
But this derivation assumed the distance between nodes was Dx on either side of the node. What if it isn't?
50 Formulation of the Finite Difference for 1-D Heat Transfer (Adapted from Ref. 3) Consider the case where the distance between the last node and the boundary is only Dx/2: If we go through the same process, our expression for the second derivative at the left end and right end, respectively becomes: 51 Formulation of the Finite Difference for 1-D Heat Transfer (Ref. 3) Forming the first derivative of temperature, T with respect to time, t for a specific node is considerably easier: Again, we have an expression for a derivative in terms of temperature and time.
52 Subscripting and Superscripting We're going to be working with both time and spatial dimension; We'll need to do some bookkeeping; A subscripting and superscripting scheme is shown here and will be used in the example.
53 Subscripting and Superscripting 54 For the node of interest, i, at time, n, we have: Time T of Node i-1 T of Node i T of Node i+1 n n+1 i i+1 i-1 Review We have...
derived expressions for the second derivative of temperature with respect to distance and for the derivative of temperature with respect to time, and; established a subscript and superscript convention to aid our analysis; We're now ready to present an example problem.
55 Example 2: Explicit Finite Difference Consider the following configuration: For an initial rod temperature of 20 C, determine the system transient response.
56 20 C (const.) L = 0.1 m 100 W (for t > 0) r = 0.01 m aluminum 20 C (const.) Example 2: Explicit Finite Difference 57 L = 0.1 m 100 W 0.02 m 0.01 m 20 C (const.) 20 C (const.) 1 2 3 4 5 We'll model the system with five diffusion ( ) nodes: We also establish two boundary ( ) nodes at either end of the rod.
Consider the following simplified geometry; We wish to write the difference equation for node i accounting for heat transfer via conduction to node i+1 and the addition of heat to node i.
Example 2: The Explicit Finite Difference Solution 58 The overall energy balance for a segment of the rod is: where each term represents energy per unit time per unit volume; This leads to the governing differential equation: Example 2: The Explicit Finite Difference Solution 59 We arrive at the difference equation describing our system: which becomes...
For an explicit scheme, note that the temperature at time, n+1 is completely determined by parameters known at time n.
Example 2: The Explicit Finite Difference Solution 60 Example 2: The Explicit Finite Difference Solution 61 Temperatures at time n are known and are used to calculate temperatures at time n+1 -- the n+1 solution becomes the n solution for the next iteration.
Example 2: Explicit Finite Difference 62 20 30 40 50 60 70 0 5 10 15 20 25 30 35 40 x = 0.00 m x = 0.01 m x = 0.03 m x = 0.05 m x = 0.07 m x = 0.09 m x = 0.10 m Temperature (C) Time (s) Example 2: Explicit Finite Difference 63 X Location (m) Temperature (C) 10 20 30 40 50 60 70 0.00 0.02 0.04 0.06 0.08 0.10 t=0 s t = 5 s t = 10 s t = 20 s t = 150 s Time Step and Time Constant Finite difference temperature solution is calculated at discrete time steps; Accuracy and stability of the solution is influenced by the time step; Commercial software will auto calculate the time step; Upper limit to the time step required for solution stability- function of the system's time constant, t.
64 Time Step and Time Constant For transient solutions, the time constant, t, is determined by: In other words, the minimum of the quotient of node capacitance divided by the sum of the conductances to that node is the limiting rate at which marching in time can occur.
65 Example 2: Effect of the Time Step 66 Time Step = 1.0 s Time Step = 2.5 s Time Step = 3.5 s Time Step, t < 1.93 s, required for stability Adding Radiation 67 Conduction and convection heat transfer are linear functions of the temperature difference; Radiation heat transfer is nonlinear and is proportional to difference of the fourth power of absolute temperatures.
Adding Radiation (Ref. 5) 68 We seek to express the radiation heat transfer between the two nodes of interest in terms of DT.
Linearized Radiation Conductor Consider the following simplified geometry; We wish to write the difference equation for node i including radiation to the environment.
Adding Radiation 69 Adding Radiation We can perform an energy balance on node i (adapted from Ref. 6): This is an expression of energy transfer per unit time per unit volume; 70 Adding Radiation Expanding and assuming nodes are of equal length, Dx: Add nodal subscripting and linearization: 71 Adding Radiation Next, we add accommodation for node-specific heating and a means of indexing the equation in time (i.e., n, n+1): With some additional manipulation, we arrive at our desired result: 72 Example 3: Explicit Finite Difference with Radiation Added Consider the following configuration: Assume a surface e = 0.8 and solve for two different Tenv: 20 C and -250 C.
73 20 C (const.) L = 0.1 m 100 W r = 0.01 m aluminum 20 C (const.) e= 0.8 Tenv = 20 C, -250 C 74 20 30 40 50 60 70 0.00 0.02 0.04 0.06 0.08 0.10 No Radiation With Radiation (20 C Boundary) With Radiation (-250 C Boundary) X Location (m) Steady State Temperature (C) Example 3: Explicit Finite Difference with Radiation Added The Implicit Finite Difference Solution The explicit technique shown allows determination of temperatures at time n+1 in terms of known temperatures at time n; However, different formulations of the differencing equation exist; Common Implicit formulations include the Backward difference and the Central difference (Crank-Nicolson) methods.
75 The Implicit Finite Difference Solution (Adapted from Ref. 3) Crank-Nicolson takes the form: Note that since the n+1 superscript appears on both sides of the equation, we can't express temperatures at that time as explicit functions of n; Hence, this is referred to as an implicit technique.
76 Solution Accuracy (Adapted from Ref. 4) An assessment of the solution accuracy can be made by observing the order of the terms truncated in the Taylor series approximation; For a first derivative formed as: The truncated terms are: 77 A second derivative with respect to x, though, is inherently first order accurate in x because it is formed using first derivatives that are accurate to the first order in x.
Techniques exist to boost Forward and Backward differences to higher accuracy.
78 Solution Accuracy (Adapted from Ref. 4) Both Forward and Backward differencing are inherently first order accurate in x; A Central difference, such as Crank-Nicolson,, however, is inherently: 79 Solution Accuracy (Adapted from Ref. 4) Finite Difference Wrap-Up Formulated the finite difference; Demonstrated explicit solution technique; Brief discussion of time step and time constant; Added effects of radiation; Implicit techniques; Solution accuracy.
80 Finite Element Strategy The finite element formulation centers around minimization of a functional; But what is a functional?
To understand this, we need some background on the Calculus of Variations.
81 Calculus of Variations Finite differencing relies on a differential formulation of the heat equation; Finite element relies on a variational formulation; For steady state, one-dimensional heat transfer, we seek to minimize the following integral, called a functional: 82 Calculus of Variations (Adapted from Ref. 7) But where does the functional come from and what is the theoretical basis for establishing the functional?
The functional is established using the Calculus of Variations; In general, we seek a function to minimize the integral: 83 Calculus of Variations (Adapted from Ref. 7) 84 u(x) is the function we want to minimize.
ũ(x,e) is the set of all functions that can minimize the integral, I.
ũ(x,e) u(x) e(x) x = 0 x = L Calculus of Variations (Adapted from Ref. 7) The set of all functions to minimize I can be expressed as: where...
u(x) is the function we seek to minimize; (x) is an arbitrary function constrained by: 85 Calculus of Variations (Adapted from Ref. 7) This ensures that ũ(x,e) is correct at the boundaries: This works if e = 0 so that: or...
86 0 Calculus of Variations (Adapted from Ref. 7) Let's substitute our expression for ũ(x,e) into the original equation but note we've added the variable, e : and note that when e = 0, we return to the original expression.
87 Calculus of Variations (Adapted from Ref. 7) So, we want our function I(e) to be a minimum when e = 0; We can accomplish this by differentiating the expression with respect to e; 88 Calculus of Variations (Adapted from Ref. 7) By chain rule, the expression becomes: This looks messy, but we can clean it up.
89 Calculus of Variations (Adapted from Ref. 7) Remember: So that...
and...
90 Calculus of Variations (Adapted from Ref. 7) The integral expression becomes: But note that: can be integrated by parts.
91 w dv Calculus of Variations (Adapted from Ref. 7) Recall integration by parts: So the entire integral expression becomes: 92 Calculus of Variations (Adapted from Ref. 7) But, recall that (0) = 0 and (L) = 0, so: So the expression simplifies to: 93 0 Calculus of Variations (Adapted from Ref. 7) Consider the quantity inside the brackets: We force e to be zero so that: 94 Calculus of Variations (Adapted from Ref. 7) Since (x) is arbitrary, this forces us to conclude that the bracketed quantity: Note that since e = 0, the ũ and ũ' become u and u'.
This is called the Euler-Lagrange equation and we will use it to establish the functional for heat transfer.
95 Forming the Functional Recall the governing differential equation for steady state, one-dimensional conduction: Area, A, cancels out and we can re-write this as: where 96 Forming the Functional Next, we note that T' replaces u' and which means that T replaces u in the Euler-Lagrange equation: So the equation becomes: 97 Forming the Functional The heat equation can be rewritten as: In comparing it with our expression: We see that: 98 Forming the Functional And since: We can integrate this expression with respect to T' to get: 99 0 Forming the Functional Finally, we arrive at the desired expression: 100 Finite Element Method Strategy We will employ the following strategy to solve a sample problem: Form the variational statement; Formulate relations at the element level; Minimize the integral at the element level; Assemble equations into a system for solution; Apply boundary conditions; Solve for nodal temperatures.
101 Example 4: One-Dimensional Steady State Finite Element Consider, the following configuration: What is the steady state temperature distribution in the rod?
102 20 C 20 C L = 0.1 m 5000 W/m L = 0.02 m This is a steady state conduction problem with heat application/generation and fixed boundary temperatures; The governing differential equation is: but for steady state, the right hand side is equal to zero.
Example 4: One-Dimensional Steady State Finite Element 103 0 Example 4: One-Dimensional Steady State Finite Element First, we form the variational relationship; From the Euler-Lagrange equation, we have: And note that: 104 Example 4: One-Dimensional Steady State Finite Element So the Euler-Lagrange equation becomes: We can rewrite our governing differential equation in this form: 105 Example 4: One-Dimensional Steady State Finite Element By comparing terms, we see that: and...
106 Example 4: One-Dimensional Steady State Finite Element We can solve for F by integrating the first expression: But, by integrating the second expression (i.e., the one with the T' term), we get another expression for F.
107 Example 4: One-Dimensional Steady State Finite Element But both of these expressions must be equal to one another because they both represent F; So we conclude that...
And the integral we seek to minimize is: 108 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) Now that we have our variational statement, we need to apply it to our problem; Most geometries are quite complex and we cannot apply this expression directly; So, we discretize the system into finite elements and apply the expression to each element: 109 Example 4: One-Dimensional Steady State Finite Element Let's consider the following element breakdown; Note that for finite elements, the node points are at the boundary of each element; For finite difference, the nodes were centered in the element.
110 20 C 20 C 5000 W/m (1) (2) (3) (4) (5) 1 3 2 4 5 6 Example 4: One-Dimensional Transient Finite Element (Adapted from Ref. 7) In this case, we seek to minimize an integral expression considering the effects of conduction (Ik), and applied (or internally generated) heating (Iq): 111 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) For an individual element, our integral becomes: where xi and xj represent the bounds of the element, A(e) is the element cross sectional area and is the heating per unit length per unit time.
112 (e) i j xj xi x Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) Let's assume a linear temperature profile across each element At the element boundaries, we have: But we are left with unknown constants, c1 and c2.
113 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) But, with two equations and two unknowns, we can solve for constants c1 and c2.
114 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) But, we can express the unknown constants in terms of known variables from the Ti and Tj expressions to give us an expression for T(e) in terms of, either, known or to-be-determined quantities: 115 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) We'll also need an expression for the derivative of T with respect to x; Differentiating the previous expression yields: 116 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 8) Similarly, we must establish an expression for the distribution of heating across the element: As with our temperature calculation, this leads to: 117 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) So, starting with the integral expression for an element, we have: 118 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) And substituting our expressions for the element temperature and gradient, we arrive at: 119 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) Before we integrate, we differentiate this previous expression with respect to Ti : 120 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) And we arrive at...
121 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) We can now integrate the expression with respect to x: 122 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) This simplifies to: But, for constant heating across an element: So the expression becomes: 123 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) Remember, the overall integral, I, is a function of all of the m temperatures in the network: To minimize the overall integral, we'll need to find the derivative of each element integral with respect to every temperature and set the sum equal to zero: 124 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) For our linear, one-dimensional elements, though, I(e) is a function of only two temperatures; The expression reduces to: 125 m m+1 m-1 (a) (b) Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) To simplify matters for our example, let's assume all elements are the same size, so: First, for conduction only, the expression reduces to: 126 m m+1 m-1 (a) (b) Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) The expression is different when a boundary is considered; For our example, both the left and right boundary temperatures are the same, Tbound : 127 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) We can perform a similar operation for heating and we arrive at: As a result, half of the heating is applied to one node and the remaining half to the other.
128 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) The system of equations in matrix form is given by: 129 Not used Not used Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) But, before solving, we must apply the boundary conditions 130 Example 4: One-Dimensional Steady State Finite Element (Adapted from Ref. 7) Moving the terms associated with the boundary temperatures over to the right hand side yields a reduced matrix: 131 Example 4: One-Dimensional Steady State Finite Element We see this is of the form: where...
132 Example 4: One-Dimensional Steady State Finite Element 133 For our sample problem, consider 5 equally sized elements with 6 equally spaced nodes: L = 0.1 m r = 0.01 m k = 167 W/mC Tboundary = 20 C qL = 5000 W/m .
Example 4: One-Dimensional Steady State Finite Element 134 This leads to the following derived quantities: Dx = 0.02 m A = 3.1415910-4 m2 qLDx = 100 W kA/Dx = 2.623 W/C Tb1 = Tb6 = 20 C .
Example 4: One-Dimensional Steady State Finite Element Filling in numbers: we get...
135 Example 4: One-Dimensional Steady State Finite Element To solve for temperatures: 136 0 10 20 30 40 50 60 70 0 0.02 0.04 0.06 0.08 0.1 Example 4: Finite Element Solution 137 X Location (m) Temperature (C) Poor discretization artificially flattens the temperature profile Example 4: Finite Element Solution Revisited 138 To fix the problem, adjust discretization; We seek a calculation point somewhere within the heated region to resolve local temperature differences; Note: Discretization was increased across the entire rod to maintain a constant element size.
20 C 20 C 5000 W/m (1) (2) (3) (4) (5) 1 3 2 4 5 6 7 8 9 10 11 (7) (8) (6) (9) (10) Example 4: Finite Element Solution Revisited 139 0 10 20 30 40 50 60 70 0 0.02 0.04 0.06 0.08 0.1 X Location (m) Temperature (C) Improved discretization allows temperature variation to be resolved Example 5: One-Dimensional Steady State Finite Element Let's solve the Example 2 geometry using the finite element method; Instead of a distributed heat load (as was the case in Example 4), all heating is applied at the center (i.e., Node 3).
140 20 C 20 C 100 W (1) (2) (3) (4) 1 3 2 4 5 Example 5: One-Dimensional Steady State Finite Element The system of equations in matrix form is given by: 141 Not used Not used Example 5: One-Dimensional Steady State Finite Element But, before solving, we must apply the boundary conditions 142 Not used Not used Example 5: One-Dimensional Steady State Finite Element Moving the terms associated with the boundary temperatures over to the right hand side yields the reduced matrix: 143 Example 5: One-Dimensional Steady State Finite Element 144 Consider 4 equally spaced elements with 5 nodes: L = 0.1 m r = 0.01 m k = 167 W/mC Tboundary = 20 C Q = 100 W .
Example 5: One-Dimensional Steady State Finite Element 145 This leads to the following derived quantities: Dx = 0.025 m A = 3.1415910-4 m2 Q = 100 W kA/Dx = 2.099 W/C Tb1 = Tb5 = 20 C .
Example 5: One-Dimensional Steady State Finite Element Filling in numbers: we get...
146 Example 5: One-Dimensional Steady State Finite Element To solve for temperatures: 147 Example 5: Finite Element Solution 148 X Location (m) Temperature (C) 0 10 20 30 40 50 60 70 0 0.02 0.04 0.06 0.08 0.1 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) Let's look at the same problem geometry but, this time, we'll model the transient response.
In this case, we seek to minimize an integral expression considering the effects of conduction (Ik), applied heating (Iq) and capacitance (Ic): 149 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) The individual components are: where is heating rate and is time.
150 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) And the overall integral to be minimized is: We already have expressions for the conduction and heating so let's focus on the capacitance term.
151 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) Recall, from Example 4, our expression for temperature as a function of location in an element: And because we know the temperatures at the ends, we solved for constants c1 and c2: 152 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) So, for an individual element, the integral becomes: After substituting in expressions for c1 and c2, the expression inside the integral is a function of, only, Ti, Tj, xi, xj and x. 153 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) We differentiate the expression with respect to Ti and then integrate over x to obtain: Similarly, we perform the differentiation with respect to Tj and integrate, we obtain: 154 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) So, for an element, we see that...
155 m m+1 m-1 (a) (b) Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) But...
and over a specified time interval, Dq, gives us... 156 Example 6: One-Dimensional Transient Finite Element (Adapted from Ref. 7) The previous expressions form the basis for our element capacitance matrix: 157 Example 6: One-Dimensional Transient Finite Element But we can assemble the element matrices into a global capacitance matrix through superposition; For similarly sized elements, the matrix becomes: 158 Example 6: One-Dimensional Transient Finite Element With the previously derived conduction matrix, we can now assemble our system of equations: Becomes...
159 Example 6: One-Dimensional Transient Finite Element (Ref. 9) This equation can be expressed in a more general form : so when...
f = 0, we have the Forward difference f = 1/2, we have Crank-Nicolson f = 2/3, we have Galerkin f = 1, we have the Backward difference 160 Example 6: One-Dimensional Transient Finite Element (Ref. 9) Let's use Backward differencing to solve this problem, i.e., f = 1; When we substitute into the general equation, we arrive at: Our unknowns are the temperatures at time n+1, shown in blue.
161 Example 6: One-Dimensional Transient Finite Element (Ref. 9) We recognize this equation is of the form: Inverting [A] and multiplying it with {B} yields: But we must remember to apply the boundary conditions as we did before.
162 Example 6: One-Dimensional Transient Finite Element 163 Time (s) Temperature (C) 10 20 30 40 50 60 70 0 50 100 150 200 250 300 T1 T2 T3 T4 T5 Comparing Finite Difference, Finite Element with Exact Solution 164 0 10 20 30 40 50 60 70 0 0.02 0.04 0.06 0.08 0.1 Exact FD FE X Location (m) Temperature (C) Conclusion 165 An overview of numerical methods in heat transfer has been presented; The governing differential equation was formulated from first principles; The finite difference was developed and demonstrated through examples; Finite element was developed and demonstrated through examples.
Acknowledgements 166 Dr. Vitali Volovoi of the NESC Statistics Team and Dr. Jeff Rickman of Lehigh University are acknowledged for their assistance and guidance with some key calculations.
Ms. Carol Mosier, Mrs. Melissa Flores, Mr. Richard Wear, Ms. Ruth Amundsen, Mr. Arturo Avila, Mr. David Gilmore, Dr. Bruce Drolen, Ms. Robin Beck, Mr. Hai Nguyen, Mr. John Sharp, Mr. Jack Ercol and Mr. Hume Peabody are acknowledged for their helpful review of the draft lesson and for their comments.
References/Credits 1. Van Wylen, G. J. and Sonntag, R. E., Fundamentals of Classical Thermodynamics, SI Version 2e, New York, John Wiley and Sons, 1978.
2. Holman, J. P., Heat Transfer (Fifth Edition), New York, McGraw Hill Book Company, 1981.
3. Anderson, John D., Jr., Computational Fluid Dynamics, The Basics with Applications, New York, McGraw-Hill, Inc., 1995.
4. Hornbeck, R., Numerical Methods, Englewood Cliffs, Prentice-Hall, Inc., 1975.
5. McMurchy, R., Thermal Network Modeling Handbook, 14690-H003-R0-00, NASA Contract 9-10435, January 1972.
167 References/Credits 6. Jiji, L. M., Heat Conduction, Heidelberg, Springer-Verlag, 2009.
7. Myers, G. E., Analytical Methods in Conduction Heat Transfer, Schenectady, Genium Publishing Corporation, 1987.
8. Akin, J. E., Finite Element for Undergraduates, Academic Press, 1985.
9. Ramamurty, G., Applied Finite Element Analysis, New Dehli, I.K. International Publishing Houst Pvt. Ltd., 2009.
Microsoft® Clip Art was used in this presentation.
Wolfram Mathematica® was used for some calculations in this presentation.
168 For Additional Information Variational Principles Liu, G., Exact Variational Principle for 3-D Unsteady Heat Conduction with Second Sound, Journal of Thermal Science, Vol. 15, No. 4, December 2006.
Vujanovic, B. D. and Jones, S. E., Variational Methods in Nonconservative Phenomena, Boston, Academic Press, Inc., 1989.
Shih, T. M. (editor), Numerical Properties and Methodologies in Heat Transfer, Proceedings of the Second National Symposium, Washington, Hemisphere Publishing Corporation, 1983.
169 To Contact the Author 170 Address: Steven L. Rickman NASA Engineering and Safety Center (NESC) NASA - Lyndon B. Johnson Space Center 2101 NASA Parkway Mail Code: WE Houston, TX 77058 Phone: 281-483-8867 Email: steven.l.rickman@nasa.gov |
13685 | https://app.oncoursesystems.com/curriculum/12346/9654728/unit/9731876 | Unit 2 - Ratios and Proportional Relationships
Content Area: Math
Course(s): Sample Course
Time Period: OctNov
Length: 6 Weeks & Grade 6
Status: Published
Title Section
Department of Curriculum and Instruction
Belleville Public Schools
Curriculum Guide
MATHEMATICS
GRADE 6 - UNIT 2
Belleville Board of Education
102 Passaic Avenue
Belleville, NJ 07109
Prepared by: Margaret M. Sabino Dr. Richard Tomko, Superintendent of Schools
Mr. Thomas D’Elia, Director of Curriculum and Instruction
Ms. Diana Kelleher, District Supervisor of ELA/Social Studies
Mr. George Droste, District Supervisor of Math/Science
Board Approved: August 24, 2015
Board Approved Revisions: August 22, 2016
Unit Overview
Students should expect to learn from this unit:
• How to express ratios in different mathematical notations
• Understand ratios and ratio relationships between two quantities
• Reason about ratios through the use of tables, double number lines and graphs
• Create equivalent ratios
• Create and use unit rates to compare and solve real world applications
• Extend ratios to include percents
• Estimate and use benchmark percents
• Create and evaluate equivalent fractions, decimals and percents using hundredths grids
• Determine the unknown part given the percent and the whole
• Order fractions, decimals and percents
• How to use ratios and proportions to convert between different units of measure
NJSLS
Please link all standards that apply in this section within the curriculum of the unit being written. Please include all New Jersey Student Learning Standards. MA.6.RP Ratios and Proportional Relationships
MA.6.RP.A Understand ratio concepts and use ratio reasoning to solve problems.
MA.6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
MA.6.RP.A.2 Understand the concept of a unit rate 𝑎 /𝑏 associated with a ratio 𝑎 :𝑏 with 𝑏 ≠0, and use rate language in the context of a ratio relationship.
MA.6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
MA.6.RP.A.3a Make tables of equivalent ratios relating quantities with whole number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
MA.6.RP.A.3b Solve unit rate problems including those involving unit pricing and constant speed.
MA.6.RP.A.3c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
MA.6.RP.A.3d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
Exit Skills
By the end of Unit 2 students will be able to:
Connect ratio and rate to whole number multiplication and division and use concepts of ratio and rate to solve problems.
Students use reasoning about multiplication and division to solve ratio and rate problems about quatities. By viewing equivalent ratios and rates as deriving from, and extending, pairs of rows (or colums) in the multiplication table and by analyzing simple drawings that indicate the relative size of quatities, students connect their understanding of multiplication and division with ratios and rates. Thus students expand the scope of problems for which they can use multiplication and division to solve problems, and they connect ratios and fractions. Students solve a wide variety of problems involving rations and rates.
Enduring Understanding
Students will be able to understand that:
• Proportional relationships express how quantities change in relationship to each other
• Rates compare two values
• Rates can be part-to-part or part-to-whole
• Ratios can be expressed in colon notation, written notation or fractional notation
• A proportion is a relationship between two ratios
• An equivalent ratio is also an equivalent fraction
• Define rate, ratio and percent
• Fractions may represent a ratio • A unit rate is a ratio relationship
• A percent is always a fraction based on 100
• Percents always have a denominator of 100
• Proportions can be used to convert between different units of measure
Essential Questions
• Why does one need to compare numbers?
• When does one need ratios to compare quantities?
• How can one compare and contrast quantities?
• How can one use ratios to compare two values in everyday life?
• What is a rate and how is it related to proportional reasoning?
• How are ratios related to fractions and division?
• Why is it important to know how to solve for unit rates?
• How can one model and represent rates and ratios?
• What are similarities and differences between fractions and ratios?
• How can one compare unit rates to determine cost, unit pricing and constant rate?
• How can one use rates, ratios and percents to solve real world situations?
• What type(s) of problems can be solves using rates, ratios and percents?
• How are ratios and proportions used to convert between different units of measure?
Learning Objectives
• Distinguish between rates and ratios
• Define percent, ratio proportion, rates, and ratios
• Express and recognise ratios written in different forms
• Use ratio reasoning to solve problems
• Identify and create equivalent decimals and percents using hundredth grids
• Create and analyze models of rates and ratios
• Compare and order percents
• Convert unit rates of measurement
• Represent ratios and proportions using models
• Relate percents to fractions and decimals
• Analyze and solve percent problems
• Translate real world problems into ratios and percents
Action Verbs
Below are examples of action verbs associated with each level of the Revised Bloom’s Taxonomy. These are useful in writing learning objectives, assignment objectives and exam questions. If you are utilizing the objectives, but want to address rigor, use the chart below.
Remember Understand Apply Analyze Evaluate Create
Choose Describe Define Label List Locate Match Memorize Name Omit Recite Select State Count Draw Outline Point Quote Recall Recognize Repeat Reproduce
Classify Defend Demonstrate Distinguish Explain Express Extend Give Examples Illustrate Indicate Interrelate Interpret Infer Match Paraphrase Represent Restate Rewrite Select Show Summarize Tell Translate Associate Compute Convert Discuss Estimate Extrapolate Generalize Predict
Choose Dramatize Explain Generalize Judge Organize Paint Prepare Produce Select Show Sketch Solve Use Add Calculate Change Classify Complete Compute Discover Divide Examine Graph Interpolate Manipulate Modify Operate Subtract
Categorize Classify Compare Differentiate Distinguish Identify Infer Point out Select Subdivide Survey Arrange Breakdown Combine Detect Diagram Discriminate Illustrate Outline Point out Separate
Appraise Judge Criticize Defend Compare Assess Conclude Contrast Critique Determine Grade Justify Measure Rank Rate Support Test
Combine Compose Construct Design Develop Formulate Hypothesize Invent Make Originate Organize Plan Produce Role Play Drive Devise Generate Integrate Prescribe Propose Reconstruct Revise Rewrite Transform
Interdisciplinary Connections
Please list all and any cross-curricular content standards that link to this Unit.
Science
Social Studies
Health & Nutrition
Music
Alignment to 21st Century Skills & Technology
Key SUBJECTS AND 21st CENTURY THEMES
Mastery of key subjects and 21st century themes is essential for all students in the 21stcentury.
Key subjects include:
• English, reading or language arts
• World languages
• Arts
• Mathematics
• Economics
• Science
• Geography
• History
• Government and Civics
21st Century/Interdisciplinary Themes
• Civic Literacy .
• Environmental Literacy .
• Financial, Economic, Business and Entrepreneurial Literacy .
• Global Awareness .
• Health Literacy .
21st Century Skills
• Communication and Collaboration .
• Creativity and Innovation .
• Critical thinking and Problem Solving .
• ICT (Information, Communications and Technology) Literacy .
• Information Literacy .
• Life and Career Skills .
• Media Literacy .
Technology Infusion
What technology can be used in this unit to enhance learning? Differentiation
As a Reminder: The basis of good differentiation in a lesson lies in differentiating by content, process, and/or product.
Resources:
• NJDOE: Instructional Supports and Scaffolds for Success in Implementing the Common Core State Standards
• Pre-teach vocabulary and meaning of symbols
• Connect new vocabulary and symbols to background knowledge
• Breal down nterms to familiar parts, suffixes and prefixes
• Make dictionaries available to learners
• Increase experience to academic vocabulary and language
• Provide flash cards
• Incorporate as many learners senses as possible to enhance the learning experience
• Brainstorm examples of new use of new terms or symbols making real world applications
• Engage student in relevant discussion about conceptual process
• Post and refer to math guides and anchor charts when applicable
• Clarify the relationship between operations
• Develop graphic representations of math procesess
• Make connections to formulas, concepts or structures previously learned
• Utiliza manipulatives to display structures
• Offer various ways to solve math problems
• Provide opportunites to integrate math, technology and art
• Provide graphic organizers and anchor charts for all symbols and formulas
• Create math journals for terms, formulas and symbols
• Develop interactive games and activities to promote retention
• Integrate videos
• Utilize grahics, diagrams, charts
Special Education
•printed copy of board work/notes provided .
•additional time for skill mastery .
•assistive technology .
•behavior management plan .
•Center-Based Instruction .
•check work frequently for understanding .
•computer or electronic device utilizes .
•extended time on tests/ quizzes .
•have student repeat directions to check for understanding .
• highlighted text visual presentation .
• modified assignment format .
• multi-sensory presentation .
• multiple test sessions .
• preferential seating .
• preview of content, concepts, and vocabulary .
• reduced/shortened reading assignments .
• secure attention before giving instruction/directions .
• shortened assignments .
• student working with an assigned partner .
• teacher initiated weekly assignment sheet .
• Use open book, study guides, test prototypes .
ELL
• using videos, illustrations, pictures, and drawings to explain or clarif .
• allowing products (projects, timelines, demonstrations, models, drawings, dioramas, poster boards, charts, graphs, slide shows, videos, etc.) to demonstrate student’s learning;
.
• allowing students to correct errors (looking for understanding) .
• allowing the use of note cards or open-book during testing .
• decreasing the amount of workpresented or required .
• having peers take notes or providing a copy of the teacher’s notes .
• providing study guides .
• tutoring by peers .
• using true/false, matching, or fill in the blank tests in lieu of essay tests .
Intervention Strategies
• allowing students to correct errors (looking for understanding) .
• teaching key aspects of a topic. Eliminate nonessential information .
• allowing products (projects, timelines, demonstrations, models, drawings, dioramas, poster boards, charts, graphs, slide shows, videos, etc.) to demonstrate student’s learning
.
• allowing students to select from given choices .
• allowing the use of note cards or open-book during testing .
• collaborating (general education teacher and specialist) to modify vocabulary, omit or modify items to reflect objectives for the student, eliminate sections of the test, and determine how the grade will be determined prior to giving the test.
.
• decreasing the amount of workpresented or required .
• having peers take notes or providing a copy of the teacher’s notes .
• marking students’ correct and acceptable work, not the mistakes .• providing study guides .
• reducing the number of answer choices on a multiple choice test .
• tutoring by peers .
• using authentic assessments with real-life problem-solving .
• using true/false, matching, or fill in the blank tests in lieu of essay tests .
• using videos, illustrations, pictures, and drawings to explain or clarify .
Evidence of Student Learning-CFU's
Please list ways educators may effectively check for understanding in this secion.
• Admit Tickets .
• Anticipation Guide .
• Compare & Contrast .
• Create a Multimedia Poster .
• Define .
• Describe .
• Evaluate .
• Evaluation rubrics .
• Exit Tickets .
• Explaining .
• Fist- to-Five or Thumb-Ometer .
• Illustration .
• Journals .
• KWL Chart .
• Outline .
• Question Stems .
• Quickwrite .
• Quizzes .
• Self- assessments .
• Socratic Seminar .
• Study Guide .
• Teacher Observation Checklist .
• Think, Pair, Share .
• Unit tests .
Primary Resources
Carnegie Learning Textbook Course 1
Standards Solution NJ Model Curriculum
PARCC/Pearson
Ancillary Resources
Dinah Zike's Foldables-Interactive Study Guides, Macmillan/MacGraw Hill
Glencoe/MacGraw Hill Workbooks
Dan Meyer's Three-Act Lessons wTEE#gid=0
Sample Lesson
Unit Name:Ratios and Proportions CCSS/NJCCCS:6RPA3c Interdisciplinary Connection: Art Statement of Objective:Students will be able to demonstrate their ability to determine, analyze and convert equivalent fractions, decimals, and percents using a 100's grid and student supplied data in a hands-on project. Anticipatory Set/Do Now: Using 1 large 100's grid students create a design in three colors. Students determine the fraction out of 100 for each color and checking that their amount equals 1 whole, i.e; 100/100. Students convert these to decimals and percents. Learning Activity: Students use one 100's grid for each letter of their name, a with minimum of 6 letters. Advanced or accellerated students may create more than 6. Students will create a block style letter using units of the 100's grid. Students then determine (for each letter) a portion out of 100, turn that fraction into an equivalent decimal and percent, also simplifying the original fraction. Fractions are randomly created by the students as they use the letters in their names to color the grids. Students will have previously learned how to convert fractions with denominators of 100 into decimals and percents.Letters are mounted to construction paper. This cal also be done on a computer in excel.
Student Assessment/CFU's: Thumb-o-meter/five finger, grading rubric Materials:Copies of 10 x 10 grids, pencils, erasers, glue, scissors, construction paper, coloring materials, rulers, calculators 21st Century Themes and Skills: Computer (if that version is used.) Differentiation/Modifications: Students may use larger grids, or do fewer letters. Students with motor skills challenges may try this on excel. Integration of Technology: calculator and/or computer skills |
13686 | https://kenurotor.weebly.com/uploads/1/4/2/1/142118309/gojogija.pdf | Cipolletti weir pdf Over the years, there are have been a number of developments to the thin-plate weir – expanding weirs beyond the V-notch and Rectangular styles commonly used. Each special weir type is an attempt to overcome very specific limitations of the thin-plate weirs in general. Cipolletti The Cipolletti weir is a fully contracted weir with a trapezoidal notch section. The sides slope outwards at a 1 (horizontal) to 4 (vertical) slope. While the main notch section has a horizontally flat edge. The weir was an attempt to reduce produce a simple discharge relationship by reducing the side contraction effects of other weir shapes. The accuracy of the Cipolletti weir has been found to be less than that of V-notch or contracted rectangular weirs, with an accuracy of the discharge coefficient of +/- 5%. Cipolletti weirs should not be used for heads less than 0.2 feet or for heads greater than 1/3 of the crest length unless field rated. Circular Circular weirs are designed to the flow of water in partially full pipes. The weirs are intended to either bolt onto the end of the pipe or inserted into it. The weirs are independent of the size of the pipe that they are installed on as the height of the crest can be sized to fit the anticipated flow rates. In using Circular weirs, keep in mind that the flow measurement takes place in the half-circle crest of the weir. It is not a compound weir where flow can top the half-circle and still be measured accurately. Compound For widely varying flows, where the main flows can be handled by one weir type and maximum flows by another weir type, the two types of weirs can be joined into one hydraulic structure. The combination of weir types usually consists of a V-notch weir for the lower / normal flow rates and a Rectangular weir for the higher flow rates. Combining two weir types does have a disadvantage: when the flow begins to exceed the capacity of the smaller (lower) weir, thin sheets of water will begin to pass over the larger (upper) weir. This overflow causes a discontinuity in the flow curve. As a result, transition from the smaller weir to the larger weir should be selected where it is of minimum importance. Proportional Notch The Proportional Notch weir is designed to give a linear relationship between the discharge flow rate and the head. Above the base and the crest of the notch tapers in a parabola on each side. As the weir gets taller, the space between the two parabolas gets smaller. Two types of Proportional Notches have been developed: the Sutro and the Rettger. The two primarily differ in the that parabolas of the upper section of the weir join the base section at its outside width, while the parabola of the Rettger joins the rectangular base section closer to the centerline of the weir. Of the two Proportional Notch weirs, the Sutro is more commonly used. Image: Andre Simoes This calculates the flow rate over a Cipolletti weir, a commonly used weir in many irrigation districts. The weir opening has a flat, level bottom and the sides that have a particular slope. The water before the weir should be held in a relatively calm and smooth pool. There should be air (not trapped) underneath the water leaving the weir. The Length is the bottom width of the weir. The height is measured from the bottom of the weir opening to the top of the water level ponded behind the weir (not the water level right as it leaves the weir). Learn more about the units used on this page. Note: 1 point = 1/100 ft. The equation to determine the flow rate for a Cipolletti (Trapezoidal) Weir is: Where: = Flow Rate in cfs. = Width of the weir crest in feet. = Height of the upstream water above the weir crest in feet. Print Save Cite Email this content Copy this link, or click below to email it to a friend Email this content or copy the link directly: Show Summary Details Page of PRINTED FROM OXFORD REFERENCE (www.oxfordreference.com). (c) Copyright Oxford University Press, 2021. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single entry from a reference work in OR for personal use (for details see Privacy Policy and Legal Notice). date: 23 September 2022A trapezoidal-shaped weir with 1:4 side slopes. Used for measuring the flow in open channels such as streams and rivers.... ... Access to the complete content on Oxford Reference requires a subscription or purchase. Public users are able to search the site and view the abstracts and keywords for each book and chapter without a subscription.Please subscribe or login to access full text content.If you have purchased a print title that contains an access token, please see the token for information about how to register your code.For questions on access or troubleshooting, please check our FAQs, and if you can''t find the answer there, please contact us. Weirs are structures consisting of an obstruction such as a dam or bulkhead placed across the open channel with a specially shaped opening or notch. The flow rate over a weir is a function of the head on the weir. Common weir constructions are the rectangular weir, the triangular or v-notch weir, and the broad-crested weir. Weirs are called sharp-crested if their crests are constructed of thin metal plates, and broad-crested if they are made of wide timber or concrete. If the notch plate is mounted on the supporting bulkhead such that the water does not contact or cling to the downstream weir plate or supporting bulkhead, but springs clear, the weir is a sharp-crested or thin-plate weir. Water level-discharge relationships can be applied and meet accuracy requirements for sharp-crested weirs if the installation is designed and installed consistent with established ASTM and ISO standards. Common Standards and Specifications for Weir Flow Measurements Rectangular weirs and triangular or v-notch weirs are often used in water supply, wastewater and sewage systems. They consist of a sharp edged plate with a rectangular, triangular or v-notch profile for the water flow. Broad-crested weirs can be observed in dam spillways where the broad edge is beneath the water surface across the entire stream. Flow measurement installations with broad-crested weirs will meet accuracy requirements only if they are calibrated. Other available weirs are the trapezoidal (Cipolletti) weir, the Sutro (proportional) weir and the compound weirs (combination of the previously mentioned weir shapes). By combining V-notch weirs with broad chested weirs - larger range of flow can be measured with accuracy. Rectangular Weir The flow rate measurement in a rectangular weir is based on the Bernoulli Equation principles and can be expressed as: q = 2/3 cd b (2 g)1/2 h3/2 (1) where q = flow rate (m3/s) h = elevation head on the weir (m) b = width of the weir (m) g = 9.81 (m/s2) - gravity cd = discharge constant for the weir - must be determined cd must be determined by analysis and calibration tests. For standard weirs - cd - is well defined or constant for measuring within specified head ranges. The lowest elevation (h = 0) of the overflow opening of the sharp-crested weirs or the control channel of broad-crested weirs is the head measurement zero reference elevation. Rectangular Weir Flow Rate Measurement Calculator cd - discharge constant b - width of weir (m) h -height of weir (m) The Francis Formula - Imperial Units Flow through a rectangular weir can be expressed in imperial units with the Francis formula q = 3.33 (b - 0.2 h) h3/2 (1b) where q = flow rate (ft3/s) h = head on the weir (ft) b = width of the weir (ft) Alternative with height in inches and flow in gpm: Triangular or V-Notch Weir The triangular or V-notch, thin-plate weir is an accurate flow measuring device particularly suited for small flows. For a triangular or v-notch weir the flow rate can be expressed as: q = 8/15 cd (2 g)1/2 tan(θ/2) h5/2 (2) where θ = v-notch angle Broad-Crested Weir For the broad-crested weir the flow rate can be expressed as: q = cd h2 b ( 2 g (h1 - h2) )1/2 (3) Measuring the Levels For measuring the flow rate it's obviously necessary to measure the flow levels, then use the equations above for calculating. It's common to measure the levels with: ultrasonic level transmitters, or pressure transmitters Ultrasonic level transmitters are positioned above the flow without any direct contact with the flow. Ultrasonic level transmitters can be used for all measurements. Some of the transmitters can even calculate a linear flow signal - like a digital pulse signal or an analog 4 - 20 mA signal - before transmitting it to the control system. Pressure transmitters can be used for the sharp-crested weirs and for the first measure point in broad-crested weir. The pressure transmitter outputs a linear level signal - typical 4-20 mA - and the flow must be calculated in the transmitter or the control system.
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13687 | https://www.geeksforgeeks.org/dsa/partition-an-array-of-non-negative-integers-into-two-subsets-such-that-average-of-both-the-subsets-is-equal/ | Partition an array of non-negative integers into two subsets such that average of both the subsets is equal - GeeksforGeeks
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Partition an array of non-negative integers into two subsets such that average of both the subsets is equal
Last Updated : 12 Jul, 2025
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Given an array of size N. The task is to partition the given array into two subsets such that the average of all the elements in both subsets is equal. If no such partition exists print -1. Otherwise, print the partitions. If multiple solutions exist, print the solution where the length of the first subset is minimum. If there is still a tie then print the partitions where the first subset is lexicographically smallest.
Examples:
Input : vec[] = {1, 7, 15, 29, 11, 9}
Output : [9, 15] [1, 7, 11, 29]
Explanation : Average of the both the subsets is 12
Input : vec[] = {1, 2, 3, 4, 5, 6}
Output : [1, 6] [2, 3, 4, 5].
Explanation : Another possible solution is [3, 4] [1, 2, 5, 6],
but print the solution whose first subset is lexicographically
smallest.
Observation :
If we directly compute the average of a certain subset and compare it with another subset's average, due to precision issues with compilers, unexpected results will occur. For example, 5/3 = 1.66666.. and 166/100 = 1.66. Some compilers might treat them as same, whereas some others won't.
Let the sum of two subsets under consideration be sub1 and sub2, and let their sizes be s1 and s2. If their averages are equal, sub1/s1 = sub2/s2 . Which means sub1s2 = sub2s1.
Also total sum of the above two subsets = sub1+sub2, and s2= total size - s1.
On simplifying the above, we get
(sub1/s1) = (sub1+sub2)/ (s1+s2) = (total sum) / (total size).
Now this problem reduces to the fact that if we can select a particular size
of subset whose sum is equal to the current subset's sum, we are done.
Approach :
Let us define the function partition(ind, curr_sum, curr_size), which returns true if it is possible to construct subset using elements with index equals to ind and having size equals to curr_size and sum equals to curr_sum.
This recursive relation can be defined as:
partition(ind, curr_sum, curr_size) = partition(ind+1, curr_sum, curr_size) || partition(ind+1, curr_sum - val[ind], curr_size-1).
Two parts on the right side of the above equations represent whether we including the element at index ind or not.
This is a deviation from the classic subset sum problem, in which subproblems are being evaluated again and again. Therefore we memorize the subproblems and turn it into a Dynamic Programming solution.
C++14
```cpp14
// C++ program to Partition an array of
// non-negative integers into two subsets
// such that average of both the subsets are equal
include
using namespace std;
vector > > dp;
vector res;
vector original;
int total_size;
// Function that returns true if it is possible to
// use elements with index = ind to construct a set of s
// ize = curr_size whose sum is curr_sum.
bool possible(int index, int curr_sum, int curr_size)
{
// base cases
if (curr_size == 0)
return (curr_sum == 0);
if (index >= total_size)
return false;
// Which means curr_sum cant be found for curr_size
if (dp[index][curr_sum][curr_size] == false)
return false;
if (curr_sum >= original[index]) {
res.push_back(original[index]);
// Checks if taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum -
original[index],
curr_size - 1))
return true;
res.pop_back();
}
// Checks if not taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum, curr_size))
return true;
// If no solution has been found
return dp[index][curr_sum][curr_size] = false;
}
// Function to find two Partitions having equal average
vector > partition(vector& Vec)
{
// Sort the vector
sort(Vec.begin(), Vec.end());
original.clear();
original = Vec;
dp.clear();
res.clear();
int total_sum = 0;
total_size = Vec.size();
for (int i = 0; i < total_size; ++i)
total_sum += Vec[i];
// building the memoization table
dp.resize(original.size(), vector<vector<bool> >
(total_sum + 1, vector(total_size, true)));
for (int i = 1; i < total_size; i++) {
// Sum_of_Set1 has to be an integer
if ((total_sum i) % total_size != 0)
continue;
int Sum_of_Set1 = (total_sum i) / total_size;
// We build our solution vector if its possible
// to find subsets that match our criteria
// using a recursive function
if (possible(0, Sum_of_Set1, i)) {
// Find out the elements in Vec, not in
// res and return the result.
int ptr1 = 0, ptr2 = 0;
vector<int> res1 = res;
vector<int> res2;
while (ptr1 < Vec.size() || ptr2 < res.size())
{
if (ptr2 < res.size() &&
res[ptr2] == Vec[ptr1])
{
ptr1++;
ptr2++;
continue;
}
res2.push_back(Vec[ptr1]);
ptr1++;
}
vector<vector<int> > ans;
ans.push_back(res1);
ans.push_back(res2);
return ans;
}
}
// If we havent found any such subset.
vector<vector<int> > ans;
return ans;
}
// Function to print partitions
void Print_Partition(vector > sol)
{
// Print two partitions
for (int i = 0; i < sol.size(); i++) {
cout << "[";
for (int j = 0; j < sol[i].size(); j++) {
cout << sol[i][j];
if (j != sol[i].size() - 1)
cout << " ";
}
cout << "] ";
}
}
// Driver code
int main()
{
vector Vec = { 1, 7, 15, 29, 11, 9 };
vector<vector<int> > sol = partition(Vec);
// If partition possible
if (sol.size())
Print_Partition(sol);
else
cout << -1;
return 0;
}
```
// C++ program to Partition an array of
// non-negative integers into two subsets// such that average of both the subsets are equal#include using namespace std;vector > > dp;vector res;vector original;int total_size;// Function that returns true if it is possible to // use elements with index = ind to construct a set of s// ize = curr_size whose sum is curr_sum.bool possible(int index, int curr_sum, int curr_size){ // base cases if (curr_size == 0) return (curr_sum == 0); if (index >= total_size) return false; // Which means curr_sum cant be found for curr_size if (dp[index][curr_sum][curr_size] == false) return false; if (curr_sum >= original[index]) { res.push_back(original[index]); // Checks if taking this element at // index i leads to a solution if (possible(index + 1, curr_sum - original[index], curr_size - 1)) return true; res.pop_back(); } // Checks if not taking this element at // index i leads to a solution if (possible(index + 1, curr_sum, curr_size)) return true; // If no solution has been found return dp[index][curr_sum][curr_size] = false;}// Function to find two Partitions having equal average vector > partition(vector& Vec){ // Sort the vector sort(Vec.begin(), Vec.end()); original.clear(); original = Vec; dp.clear(); res.clear(); int total_sum = 0; total_size = Vec.size(); for (int i = 0; i < total_size; ++i) total_sum += Vec[i]; // building the memoization table dp.resize(original.size(), vector >(total_sum + 1, vector(total_size, true))); for (int i = 1; i < total_size; i++) { // Sum_of_Set1 has to be an integer if ((total_sum i) % total_size != 0) continue; int Sum_of_Set1 = (total_sum i) / total_size; // We build our solution vector if its possible // to find subsets that match our criteria // using a recursive function if (possible(0, Sum_of_Set1, i)) { // Find out the elements in Vec, not in // res and return the result. int ptr1 = 0, ptr2 = 0; vector res1 = res; vector res2; while (ptr1 < Vec.size() || ptr2 < res.size()) { if (ptr2 < res.size() && res[ptr2] == Vec[ptr1]) { ptr1++; ptr2++; continue; } res2.push_back(Vec[ptr1]); ptr1++; } vector > ans; ans.push_back(res1); ans.push_back(res2); return ans; } } // If we havent found any such subset. vector > ans; return ans;}// Function to print partitions void Print_Partition(vector > sol){ // Print two partitions for (int i = 0; i < sol.size(); i++) { cout << "["; for (int j = 0; j < sol[i].size(); j++) { cout << sol[i][j]; if (j != sol[i].size() - 1) cout << " "; } cout << "] "; }}// Driver code int main(){ vector Vec = { 1, 7, 15, 29, 11, 9 }; vector > sol = partition(Vec); // If partition possible if (sol.size()) Print_Partition(sol); else cout << -1; return 0;}
Java
```java
// Java program to Partition an array of
// non-negative integers into two subsets
// such that average of both the subsets are equal
import java.io.;
import java.util.;
class GFG
{
static boolean[][][] dp;
static Vector<Integer> res = new Vector<>();
static int[] original;
static int total_size;
// Function that returns true if it is possible to
// use elements with index = ind to construct a set of s
// ize = curr_size whose sum is curr_sum.
static boolean possible(int index, int curr_sum,
int curr_size)
{
// base cases
if (curr_size == 0)
return (curr_sum == 0);
if (index >= total_size)
return false;
// Which means curr_sum cant be found for curr_size
if (dp[index][curr_sum][curr_size] == false)
return false;
if (curr_sum >= original[index])
{
res.add(original[index]);
// Checks if taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum - original[index],
curr_size - 1))
return true;
res.remove(res.size() - 1);
}
// Checks if not taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum, curr_size))
return true;
// If no solution has been found
return dp[index][curr_sum][curr_size] = false;
}
// Function to find two Partitions having equal average
static Vector<Vector<Integer>> partition(int[] Vec)
{
// Sort the vector
Arrays.sort(Vec);
original = Vec;
res.clear();
int total_sum = 0;
total_size = Vec.length;
for (int i = 0; i < total_size; ++i)
total_sum += Vec[i];
// building the memoization table
dp = new boolean[original.length][total_sum + 1][total_size];
for (int i = 0; i < original.length; i++)
for (int j = 0; j < total_sum + 1; j++)
for (int k = 0; k < total_size; k++)
dp[i][j][k] = true;
for (int i = 1; i < total_size; i++)
{
// Sum_of_Set1 has to be an integer
if ((total_sum i) % total_size != 0)
continue;
int Sum_of_Set1 = (total_sum i) / total_size;
// We build our solution vector if its possible
// to find subsets that match our criteria
// using a recursive function
if (possible(0, Sum_of_Set1, i))
{
// Find out the elements in Vec, not in
// res and return the result.
int ptr1 = 0, ptr2 = 0;
Vector<Integer> res1 = res;
Vector<Integer> res2 = new Vector<>();
while (ptr1 < Vec.length || ptr2 < res.size())
{
if (ptr2 < res.size() &&
res.elementAt(ptr2) == Vec[ptr1])
{
ptr1++;
ptr2++;
continue;
}
res2.add(Vec[ptr1]);
ptr1++;
}
Vector<Vector<Integer>> ans = new Vector<>();
ans.add(res1);
ans.add(res2);
return ans;
}
}
// If we havent found any such subset.
Vector<Vector<Integer>> ans = new Vector<>();
return ans;
}
// Function to print partitions
static void Print_Partition(Vector<Vector<Integer>> sol)
{
// Print two partitions
for (int i = 0; i < sol.size(); i++)
{
System.out.print("[");
for (int j = 0; j < sol.elementAt(i).size(); j++)
{
System.out.print(sol.elementAt(i).elementAt(j));
if (j != sol.elementAt(i).size() - 1)
System.out.print(" ");
}
System.out.print("]");
}
}
// Driver Code
public static void main(String[] args)
{
int[] Vec = { 1, 7, 15, 29, 11, 9 };
Vector<Vector<Integer>> sol = partition(Vec);
// If partition possible
if (sol.size() > 0)
Print_Partition(sol);
else
System.out.println("-1");
}
}
// This code is contributed by
// sanjeev2552
Pythonpython3
Python3 program to partition an array of
non-negative integers into two subsets
such that average of both the subsets are equal
dp = []
res = []
original = []
total_size = int(0)
Function that returns true if it is possible
to use elements with index = ind to construct
a set of s ize = curr_size whose sum is curr_sum.
def possible(index, curr_sum, curr_size):
index = int(index)
curr_sum = int(curr_sum)
curr_size = int(curr_size)
global dp, res
# Base cases
if curr_size == 0:
return (curr_sum == 0)
if index >= total_size:
return False
# Which means curr_sum cant be
# found for curr_size
if dp[index][curr_sum][curr_size] == False:
return False
if curr_sum >= original[index]:
res.append(original[index])
# Checks if taking this element
# at index i leads to a solution
if possible(index + 1,
curr_sum - original[index],
curr_size - 1):
return True
res.pop()
# Checks if not taking this element at
# index i leads to a solution
if possible(index + 1, curr_sum, curr_size):
return True
# If no solution has been found
dp[index][curr_sum][curr_size] = False
return False
Function to find two partitions
having equal average
def partition(Vec):
global dp, original, res, total_size
# Sort the vector
Vec.sort()
if len(original) > 0:
original.clear()
original = Vec
if len(dp) > 0:
dp.clear()
if len(res) > 0:
res.clear()
total_sum = 0
total_size = len(Vec)
for i in range(total_size):
total_sum += Vec[i]
# Building the memoization table
dp = [[[True for _ in range(total_size)]
for _ in range(total_sum + 1)]
for _ in range(len(original))]
for i in range(1, total_size):
# Sum_of_Set1 has to be an integer
if (total_sum i) % total_size != 0:
continue
Sum_of_Set1 = (total_sum i) / total_size
# We build our solution vector if its possible
# to find subsets that match our criteria
# using a recursive function
if possible(0, Sum_of_Set1, i):
# Find out the elements in Vec,
# not in res and return the result.
ptr1 = 0
ptr2 = 0
res1 = res
res2 = []
while ptr1 < len(Vec) or ptr2 < len(res):
if (ptr2 < len(res) and
res[ptr2] == Vec[ptr1]):
ptr1 += 1
ptr2 += 1
continue
res2.append(Vec[ptr1])
ptr1 += 1
ans = []
ans.append(res1)
ans.append(res2)
return ans
# If we havent found any such subset.
ans = []
return ans
Driver code
Vec = [ 1, 7, 15, 29, 11, 9 ]
sol = partition(Vec)
if len(sol) > 0:
print(sol)
else:
print("-1")
This code is contributed by saishashank1
C#csharp
// C# program to Partition an array of
// non-negative integers into two subsets
// such that average of both the subsets are equal
using System;
using System.Collections;
class GFG{
static bool[,,] dp;
static ArrayList res = new ArrayList();
static int[] original;
static int total_size;
// Function that returns true if it is possible to
// use elements with index = ind to construct a set of s
// ize = curr_size whose sum is curr_sum.
static bool possible(int index, int curr_sum,
int curr_size)
{
// base cases
if (curr_size == 0)
return (curr_sum == 0);
if (index >= total_size)
return false;
// Which means curr_sum cant be
// found for curr_size
if (dp[index, curr_sum, curr_size] == false)
return false;
if (curr_sum >= original[index])
{
res.Add(original[index]);
// Checks if taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum -
original[index], curr_size - 1))
return true;
res.Remove(res[res.Count - 1]);
}
// Checks if not taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum, curr_size))
return true;
dp[index, curr_sum, curr_size] = false;
// If no solution has been found
return dp[index, curr_sum, curr_size];
}
// Function to find two Partitions
// having equal average
static ArrayList partition(int[] Vec)
{
// Sort the vector
Array.Sort(Vec);
original = Vec;
res.Clear();
int total_sum = 0;
total_size = Vec.Length;
for(int i = 0; i < total_size; ++i)
total_sum += Vec[i];
// Building the memoization table
dp = new bool[original.Length,
total_sum + 1,
total_size];
for(int i = 0; i < original.Length; i++)
for(int j = 0; j < total_sum + 1; j++)
for(int k = 0; k < total_size; k++)
dp[i, j, k] = true;
for(int i = 1; i < total_size; i++)
{
// Sum_of_Set1 has to be an integer
if ((total_sum i) % total_size != 0)
continue;
int Sum_of_Set1 = (total_sum i) / total_size;
// We build our solution vector if its possible
// to find subsets that match our criteria
// using a recursive function
if (possible(0, Sum_of_Set1, i))
{
// Find out the elements in Vec, not in
// res and return the result.
int ptr1 = 0, ptr2 = 0;
ArrayList res1 = new ArrayList(res);
ArrayList res2 = new ArrayList();
while (ptr1 < Vec.Length || ptr2 < res.Count)
{
if (ptr2 < res.Count &&
(int)res[ptr2] == Vec[ptr1])
{
ptr1++;
ptr2++;
continue;
}
res2.Add(Vec[ptr1]);
ptr1++;
}
ArrayList ans = new ArrayList();
ans.Add(res1);
ans.Add(res2);
return ans;
}
}
// If we havent found any such subset.
ArrayList ans2 = new ArrayList();
return ans2;
}
// Function to print partitions
static void Print_Partition(ArrayList sol)
{
// Print two partitions
for(int i = 0; i < sol.Count; i++)
{
Console.Write("[");
for(int j = 0; j < ((ArrayList)sol[i]).Count; j++)
{
Console.Write((int)((ArrayList)sol[i])[j]);
if (j != ((ArrayList)sol[i]).Count - 1)
Console.Write(" ");
}
Console.Write("] ");
}
}
// Driver Code
public static void Main(string[] args)
{
int[] Vec = { 1, 7, 15, 29, 11, 9 };
ArrayList sol = partition(Vec);
// If partition possible
if (sol.Count > 0)
Print_Partition(sol);
else
Console.Write("-1");
}
}
// This code is contributed by rutvik_56
JavaScriptjavascript
// JS program to Partition an array of
// non-negative integers into two subsets
// such that average of both the subsets are equal
let dp = [];
let res = [];
let original = [];
let total_size;
// Function that returns true if it is possible to
// use elements with index = ind to construct a set of s
// ize = curr_size whose sum is curr_sum.
function possible(index, curr_sum, curr_size)
{
// base cases
if (curr_size == 0)
return (curr_sum == 0);
if (index >= total_size)
return false;
// Which means curr_sum cant be found for curr_size
if (dp[index][curr_sum][curr_size] == false)
return false;
if (curr_sum >= original[index]) {
res.push(original[index]);
// Checks if taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum -
original[index],
curr_size - 1))
return true;
res.pop();
}
// Checks if not taking this element at
// index i leads to a solution
if (possible(index + 1, curr_sum, curr_size))
return true;
// If no solution has been found
dp[index][curr_sum][curr_size] = false
return dp[index][curr_sum][curr_size];
}
// Function to find two Partitions having equal average
function partition(Vec)
{
// Sort the vector
Vec.sort();
original = [];
original = Vec;
dp = [];
res = [];
let total_sum = 0;
total_size = Vec.length;
for (var i = 0; i < total_size; ++i)
total_sum += Vec[i];
// building the memoization table
dp = new Array(original.length);
for (var i = 0; i < original.length; i++)
{
dp[i] = new Array(total_sum + 1);
for (var j = 0; j <= total_sum; j++)
{
dp[i][j] = new Array(total_size).fill(true);
}
}
for (var i = 1; i < total_size; i++) {
// Sum_of_Set1 has to be an integer
if ((total_sum i) % total_size != 0)
continue;
var Sum_of_Set1 = (total_sum i) / total_size;
// We build our solution vector if its possible
// to find subsets that match our criteria
// using a recursive function
if (possible(0, Sum_of_Set1, i)) {
// Find out the elements in Vec, not in
// res and return the result.
var ptr1 = 0, ptr2 = 0;
var res1 = res;
var res2 = [];
while (ptr1 < Vec.length || ptr2 < res.length)
{
if (ptr2 < res.length &&
res[ptr2] == Vec[ptr1])
{
ptr1++;
ptr2++;
continue;
}
res2.push(Vec[ptr1]);
ptr1++;
}
let ans = [];
ans.push(res1);
ans.push(res2);
return ans;
}
}
// If we havent found any such subset.
return -1;
}
// Driver code
let Vec = [1, 7, 15, 29, 11, 9 ];
let sol = partition(Vec);
console.log(sol)
```
Output[9 15] [1 7 11 29]
Time Complexity:O(n 3)
Auxiliary Space: O(n 3)
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13688 | https://www.youtube.com/watch?v=KDKoBGY6u2w | Derivative of sinxcosx with Product Rule | Calculus 1 Exercises
Wrath of Math
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Posted: 1 Jun 2023
We differentiate sinxcosx using the product rule. After simplifying and using a trig identity, we find the derivative of sinxcosx is cos2x. #calculus #apcalculus
Product Rule for Derivatives Explained: (coming soon)
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Transcript:
let's find the derivative of sine times cosine notice this is a product of functions so rather than just our basic derivative rules we're going to need the product rule I've written the product rule here in case you've forgotten it and I'll leave a link in the description to my lesson introducing it if you need a more thorough review the derivative of a product of functions f of x times G of X is f Prime Times G Plus G Prime Times F since multiplication is commutative it doesn't actually matter which one of these functions we call F and which we call G but let's follow convention and call the function on the left F and we'll call the function on the right G then what is f Prime well F Prime is the derivative of sine which is cosine and don't forget it's cosine of x we then need to multiply that by G of x g of X in this case is cosine so it's cosine of x times cosine of x then we need to add ad and don't forget its addition some people screw it up and sometimes do multiplication or any number of other things then we need to add G Prime again G is cosine so what is g Prime well that's the derivative of cosine so negative sine X and then this needs to get multiplied by F that's the last step F we already said is sine X so we just multiply this by sine X we can of course do a little bit of simplification here cosine x times cosine X is cosine squared x and then sine X times sine X is sine squared x it's also negative so let's just write minus sine squared x and if you know your trig identities the Pythagorean identity doesn't apply here because they're not being added but there is another identity that does cosine squared of x minus sine squared of X is actually equal to cosine of 2x so that's probably the best way we could write our answer thanks for watching let me know in the comments if you have any questions Link in the description to some more calculus exercises [Music] on lunch |
13689 | https://blog.csdn.net/lzm12278828/article/details/145233361 | 极坐标与直角坐标之间变换的原理和应用示例_极坐标系与直角坐标系转换-CSDN博客
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极坐标与直角坐标之间变换的原理和应用示例
最新推荐文章于 2025-07-26 10:52:47 发布
搏博于 2025-01-18 21:58:53 发布
阅读量3.3k收藏 15
点赞数 30
CC 4.0 BY-SA版权
分类专栏：算法文章标签：算法人工智能python机器学习
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
本文链接：
极坐标变换的原理是将平面上的点从直角坐标系转换为极坐标系，或者从极坐标系转换为直角坐标系。以下是关于极坐标变换原理的详细解释：
一、极坐标系的基本概念
在极坐标系中，一个点的位置由两个参数确定：径向距离（ρ）和极角（θ）。
（1）径向距离（ρ）：点到原点的距离。
（2）极角（θ）：点与正x轴之间的夹角，通常以弧度为单位，逆时针方向为正。
图1 极坐标表示
二、直角坐标系与极坐标系之间的转换
1.直角坐标转极坐标
径向距离可以通过勾股定理计算得到：
极角可以通过反正切函数计算得到：
但需要注意的是，arctan函数返回的角度值在之间，因此需要根据和的符号来确定所在的象限。
图2 直角坐标转为极坐标
2.极坐标转直角坐标
直角坐标和可以通过极坐标
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图像 极坐标 变换 及其逆 变换 （ 笛卡尔 --＞ 极坐标 极坐标 --＞ 笛卡尔）
thequitesunshine007的博客
03-06 5574
声明： 摘自 并在此基础上加了自己的理解！！！ 极坐标 变换 就是将图像在 直角坐标系 与 极坐标系 中互相 变换，形式如图3-26所示，它可以将一圆形图像 变换 成一个矩形图像，常用于处理钟表、圆盘等图像。圆形图案边缘上的文字经过及坐标 变换 后可以垂直的排列在新图像的边缘，便于对文字的识别 和 检测。 注意opencv的warpPolar这里顺时针为旋转正方向！ OpenCV 4中提供了warpPolar()函数用于实现图像的 极坐标 变换，该函
把 极坐标 化为 直角坐标 c语言,极坐标 与 直角坐标 的互化
weixin_39768444的博客
05-16 1774
在二维平面中，要确定一个点的位置，需要两个独立的参数，比如在某班级可以用第几行第几列来确定某个同学的位置，这种确定方法其实就是建立了一个 直角坐标系.我们也可以选择长度及角度这两个独立的参数来确定平面中点的位置，长度就是极径，角度就是极角.当然，还可以选择其他的两个独立参数用以刻画平面中点的位置.但这两种是最经常使用的方法.因为 直角坐标 描述平移变动时很方便，在初中时我们就很熟悉在 直角坐标系 中函数图象...
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极坐标系 与 直角坐标系 转换
9-18
对于平面内任何一点M,用ρ表示线段OM的长度,θ表示从Ox到OM的角度,ρ叫做点M的极径,θ叫做点M的极角,有序数对 (ρ,θ)就叫点M的 极坐标,这样建立的坐标系叫做 极坐标系。 极坐标系 中的两个坐标 r 和 θ 可以由下面的公式 转换 为 直角坐标系 下的坐标值 x = rcos(θ), y = rsin(θ), 由上述二公式...
...测绘】极坐标 与 直角坐标 转换 代码_写一个程序把 极坐标 转换 为...
9-13
直角坐标系 图像 转换 为 极坐标系 格式:m 资源大小:602.0B 直角坐标 与 极坐标 相互 转换 浏览:76 c++的 直角坐标 与 极坐标 相互 转换 程序,很直观有例图,适合初学者学习 格式:rar 资源大小:413.9KB java编程实现 极坐标 的 转换 浏览:114 5星 · 资源好评率100% 将笛卡尔坐标系上的点定义为一个服务类Point,Point类提供求...
极坐标 下交换积分次序的方法 热门推荐
Art1st_D的博客
07-21 3万+
1、转换 直角坐标系 法 将ρ、θ换做 直角坐标系，画出原积分的草图(即θ对应x坐标，ρ对应y坐标)，再按照 直角坐标系 下交换积分次序的方法交换即可； 2、极坐标 常数穿越法 根据特定点划分两个积分域（ρ发生变化的角度），将D分为多个子积分域，确定每个子积分域ρ 和 θ的边界，累加即可； 3、分析法 参考资料：
读书笔记-opencv-极坐标 变换
aaron1996123456的博客
09-17 2778
读书笔记-opencv-极坐标 变换 原理 解析 极坐标 变换 用来矫正图像中的圆形物体，或者包含在圆形物体中。 笛卡尔坐标系xoy平面上任意一点(x,y),以(x1,y1)为中心通过以下计算公式对应到 极坐标系 上的 极坐标（θ，r） 极坐标 变换 后的角度范围[0, 360] 举例：（11, 13）以（3, 5）为中心进行 极坐标 变换 import math r = math.sqrt(...
MATLAB实现 极坐标 向 直角坐标 转换 资源
9-27
图象 直角坐标 转为 极坐标 软件 浏览:2 图象 直角坐标 转为 极坐标 软件,给人一种崭新观察世界的角度 直角坐标 转为 极坐标 浏览:0 将指教图像变为 极坐标 图像的vc代码,并且显示出处理前后的图片 直角坐标系 图像 转换 为 极坐标系 浏览:75 直角坐标系 图像 转换 为 极坐标系 雷达探测 极坐标系(AER)与 地球等经纬度坐标系(GEO)...
直角坐标系 与 极坐标系 了解 与 转换 _ 极坐标 与 直角坐标
9-18
直角坐标系 与 极坐标系 了解 与 转换 直角坐标系(Rectangularcoordinates) 在平面内画两条互相垂直,并且有公共原点的数轴。其中横轴为X轴,纵轴为Y轴。这样我们就说在平面上建立了平面 直角坐标系,简称 直角坐标系。 坐标系所在平面叫做坐标平面,两
实现图像 极坐标 转换 的完整指南 最新发布
weixin_34374684的博客
07-26 920
OpenCV，即开源计算机视觉库（Open Source Computer Vision Library），是一个跨平台的计算机视觉 和 机器学习 软件库。它提供了超过2500种优化的 算法，这些 算法 涵盖了图像处理、计算机视觉、图形处理 和 机器学习 等多个领域。OpenCV的高效性能得益于其底层用C++编写，并提供了 Python、Java等多种语言接口，方便不同背景的开发者使用。OpenCV的关键优势包括：性能高效：针对各种硬件架构优化，具有高度的优化性能，特别适合实时图像处理 和 分析。功能全面。
极坐标 转换
IAXin的博客
10-17 1181
/// /// 极坐标 转换 /// /// private Vector3 GetSpeedWithAngle1(int angle, float speed) { float x = speed Ma...
极坐标 转换 与 排序
9-17
当 极坐标系 中的极点 O 与 直角坐标系 中的原点 O 重合,极轴 OX 与 直角坐标系 中的 X 轴的正半轴重合,并且两种坐标系的单位长度相同,那么平面内任意一点 P 的 直角坐标 与 极坐标 可以互相 转换。 例如: 点p 直角坐标 为:(1,1),则对应的 极坐标 为:(1.4142,π/4)。
雷达探测 极坐标系(AER)与 地球等经纬度坐标系(GEO)转换 公式推导
9-24
雷达探测 极坐标系(AER)与 地球等经纬度坐标系(GEO) 图1中,O为地心,OD为地球半径R, A为雷达架设点,AD为雷达架设高度h,雷达探测水平面为AF,B为雷达探测的任意一点。 雷达探测 极坐标系 的三个参数为r,θ,δ。 雷达探测拟 直角坐标系 的三个参数为X,Y,H,其中H为距地面的高度。记雷达探测任意一点B,在雷达探测...
直角坐标 和 极坐标
吃葡萄不吐葡萄皮
06-26 7163
通过将图像从 直角坐标系 转换 为 极坐标系，可以将圆形图案边缘上的文字排列在新图像的边缘，使得文字的识别 和 检测更加方便。总而言之，极坐标 和 直角坐标 的互相 转换 提供了在不同坐标系 之间 进行方便的 变换 和 处理的能力，可以在图像处理、几何计算 和 其他领域中发挥重要作用。在二维 极坐标系 中，一个点的位置由两个坐标值（r，θ）表示，其中r表示点到原点的距离，θ表示点 与 参考线的夹角。在二维 直角坐标系 中，一个点的位置由两个坐标值（x，y）表示，其中x表示点在水平方向上的位置，y表示点在垂直方向上的位置。
C# OpenCvSharp 环形文字处理 直角坐标 与 极坐标 转换.rar
11-09
直角坐标 与 极坐标 的 转换 在计算机图形学 和 图像处理中是基础概念。直角坐标系 统是我们常见的x-y平面，而 极坐标 则用距离 和 角度表示位置。在C#和 OpenCvSharp中，这种 转换 可以通过数学公式实现。例如，将 直角坐标(x, y)...
gps.zip_GPS坐标 转换 _ 极坐标 笛卡尔坐标系_从 _直角坐标系(笛卡尔...
9-15
笛卡尔坐标系,也称为 直角坐标系,由两条互相垂直的数轴构成,通常标记为X轴 和 Y轴。在笛卡尔坐标系中,一个点的位置可以通过一对有序实数(x, y)来表示。在GIS中,笛卡尔坐标通常用于平面坐标系统,例如UTM(统一世界坐标系统)。 极坐标系 则以一个原点为中心,通过距离(半径r)和 角度(方位角θ)来定义点的位置。在GIS...
...算法 入门_cordic 算法 实现 直角坐标 转 极坐标 全过程示意图
9-20
图1 直角坐标系 到 极坐标系 的 转换 为了突出重点,这里我们只讨论X 和 Y都为正数的情况。这时θ=atan(y/x)。求θ的过程也就是求atan 函数的过程。Cordic 算法 采用的想法很直接,将(X,Y)旋转一定的度数,如果旋转完纵坐标变为了0,那么旋转的度数就是θ。坐标旋转的公式可能大家都忘了,这里把公式列出了。设(x,y)是...
Python 在OpenCV里实现 极坐标 变换 功能
09-18
通过本文，我们不仅学会了如何用代码实现 极坐标 变换 和 极坐标 绘图，也更深刻地理解了 极坐标系 在数据分析 和 可视化方面的 应用。此外，通过OpenCV的名称我们可以推测，文章的标题中可能出现了错误，因为OpenCV（开源...
直角坐标 与 极坐标 相互 转换
01-11
在C++中实现 直角坐标 与 极坐标 的 转换，通常需要定义两个函数：一个用于将 直角坐标 转换 为 极坐标，另一个用于相反的 转换。直角坐标 到 极坐标 的 转换 公式如下： 1. 半径r = √(x² + y²) 2. 角度θ = arctan(y / x) 或者...
极坐标 与 参数方程学习：直角坐标系 应用 解析
2. 示例 2涉及的是三角形边的关系，通过建立 直角坐标系，可以研究边中线BE 和 CF 之间 的位置关系，可能需要用到相似三角形或向量知识。 三、当堂检测 这部分提供了不同级别的练习题，从简单的轨迹方程求解到空间直角...
直角坐标系 图像 转换 为 极坐标系
08-16
直角坐标系 图像 转换 为 极坐标系
【Halcon视觉】极坐标 变换
文布斯的博客
07-19 1326
极坐标 变换，是将图像在 直角坐标系 与 极坐标系 中，相互 转换。常用于圆形图像的处理，如：圆形图案边缘上的文字，经过 极坐标 变换 后，可以垂直的排列在新图像的边缘，便于对文字的识别 和 检测。 polar_trans_image_ext() 极坐标 变换 polar_trans_image_inv() 极坐标 逆 变换 ...
直角坐标系 和 极坐标系
Monster
10-07 1万+
在二维坐标系下，某点坐标表示U(x，y),这种表示则是称之为在平面坐标系下，或者叫做x,y坐标系中。 极坐标 表示某点利用某点到原点距离 和 夹角表示，U(r,α)； 极坐标 和 平面坐标系 转换： x=rcos(α) y=rsin(α) 极坐标 使用弧度制， 平面坐标上的任意点都能在 极坐标 中 表示出来，而且不止一种表示方法。 ...
【C语言】极坐标 转换 为 直角坐标
点滴记忆
11-18 8929
写一个程序把 极坐标(r,θ) (θ之单位为度)转换 为 直角坐标( X,Y)。 转换 公式是x=r.cosθy=r.sinθ 程序输出；输出 转换 后的坐标。 弧度 和 角度的换算关系如下： 1弧度=180/π度 1度=π/180弧度 也就是说，180度=π 弧度 角度转弧度的计算，角度angle乘圆周率PI除以180.0得弧度。 代码： #include <stdio.h&...
音频处理七:(极坐标 转换)
taw19960426的博客
03-28 610
程序设计七：极坐标 转换 一：需求分析 在数学中，极坐标系 是一个二维坐标系统。该坐标系统中的点由一个夹角 和 一段相对中心点——极点（相当于我们较为熟知的 直角坐标系 中的原点）的距离来表示。极坐标系 的 应用 领域十分广泛，包括数学、物理、工程、航海以及机器人领域。在两点间的关系用夹角 和 距离很容易表示时，极坐标系 便显得尤为有用；而在平面 直角坐标系 中，这样的关系就只能使用三角函数来表示。对于很...
如何把 极坐标 化为 直角坐标 如何将 _极坐标 转化为 直角坐标
weixin_35735370的博客
12-24 9587
展开全部 极坐标 转换 为 直角坐标：32313133353236313431303231363533e58685e5aeb931333366306532转化方法及其步骤：第一步：把 极坐标 方程中的θ整理成cosθ 和 sinθ的形式 ；第二步：把cosθ化成x/ρ,把sinθ化成y/ρ；或者把ρcosθ化成x,把ρsinθ化成y；第三步：把ρ换成(根号下x2＋y2)；或将其平方变成ρ2,再变成x2＋y2 ；第...
极坐标 变换 和 对数 极坐标 变换
05-26 4186
极坐标 变换 在平面内选择一个定点O作为“极点”，从该点引出一条射线OX，叫做“极轴”，再选定一个长度单位 和 角度的正方向(通常取逆时针方向)。对于平面内任何一点M，可以用r来表示OM的长度，用表示OM到OX所转过的角度。那么，就被称为点M的 极坐标。用这种方法建立的坐标系叫做 极坐标系。显然，极坐标系 和 直角坐标系 之间 存在着对应关系，即：，也可以写为：。 下图
直角/极坐标系 之间 的 转换
weixin_43702663的博客
05-12 3702
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目录
一、极坐标系的基本概念
二、直角坐标系与极坐标系之间的转换
1.直角坐标转极坐标
2.极坐标转直角坐标
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目录
一、极坐标系的基本概念
二、直角坐标系与极坐标系之间的转换
1.直角坐标转极坐标
2.极坐标转直角坐标
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上一篇： 改进果蝇优化算法之二：基于极坐标变换的果蝇优化算法（PCT-FOA） 下一篇： 改进果蝇优化算法之三：基于分组搜索的果蝇优化算法（G-FOA）
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13690 | https://prep.math.lsa.umich.edu/cgi-bin/pmc/crtopic?sxn=1&top=1&crssxn=prep | Integer Exponents
| | | |
---
| a^0 = 1 | | (if a isn't zero) |
| a^1 = a | | |
| a^n = a...a | | (for n a positive integer) |
| a^{-1} = 1/a | | |
| a^{-n} = 1/a^n | | |
| | | |
---
| a^m a^n = a^{m+n} | | multiplying powers |
| (a^m/a^n) = a^{m-n} | | dividing powers |
| (a^m)^n = a^{mn} | | raising a power to a power |
| (ab)^n = a^n b^n | | power of a product |
| (a/b)^n = a^n/b^n | | power of a quotient |
Notice how the rules for exponents follow from our definition: for
example, when we multiply powers we have
Go through the remaining rules to see how you can derive them from the
definition. Yes, now. Get out some paper and write them down.
Note that while these definitions and rules are largely intuitive,
there are a couple of places where this may not be the case:
| | | |
---
| ab^n = a(b^n) | | (Notice that ab^n ne (ab)^n ) |
| -b^n = -(b^n) | | (Notice that -b^n ne (-b)^n ) |
| -a b^n | | |
| | | |
---
| (a+b)^n ne a^n + b^n | | (The power of a sum is not equal to the sum of the powers) |
Before doing these practice problems, make sure that you have explained
each step in the examples above. Then work the practice problems
until you are sure you understand them. You can get more practice in
the section test at the end of this section.
Note that you can get new practice problems by clicking the "Refresh"
button at the bottom of the practice set.
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--- --- --- --- --- --- --- --- --- |
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| a | x | ^b | ( | x | ^c | y | ^d | ) | ^o | | = | | | x | | y | |
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--- --- --- |
| c | | x | ^p | y | ^q |
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--- --- --- --- --- --- --- --- --- --- --- --- |
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| ( | g | x | ^h | y | ^j | ) | ^k | | ( | x | ^m | y | ^n | ) | ^p | | = | | | x | | y | |
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--- --- --- |
| c | | x | ^p | y | ^q |
| | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- ---
| | | | | | | | | | | | --- --- --- --- --- | | y | ^q | | ( | x | ^r | y | ^s | ) | ^t | | | = | | | x | | y | |
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| |
| |
| | | | | | | | | | --- --- --- --- | | ( | u | x | ^v | y | ^w | ) | ^z | |
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--- --- --- --- --- |
| y | ^q | | ( | x | ^r | y | ^s | ) | ^t |
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13691 | https://www.math.uci.edu/~ndonalds/math180a/5primes.pdf | 5 Primes 5.1 The Distribution of the Set of Primes Given the usefulness of primes as the ‘building blocks’ of the integers, we naturally want to investi-gate how they are distributed: we’d like answers to questions such as the following.
1. How many primes are there?
2. How many primes are there with a certain property? (e.g. congruent to 3 modulo 4) 3. If we have discovered the first n primes, how much larger is the next?
4. Can we write every even integer ≥4 as a sum of two primes?
5. Are there infinitely many primes p such that p + 2 is also prime?
6. Does there exist at least one prime between any consecutive squares?
7. Are there infinitely many primes of the form N2 + 1?
The first three questions can, more or less, be answered, whereas the remaining four are famous conjectures (the Goldbach, Twin Prime, Legendre’s and N2 + 1 conjectures respectively) that have remained unsolved for over a century.1 The first question has the oldest answer: we earlier saw Euclid’s Theorem stating that there are infinitely many primes. We can extend his approach to other situations. For example, it is clear that any prime p ≥3 cannot be even and must therefore be congruent to 1 or 3 modulo 4. Consider the following table of the primes p such that 3 ≤p ≤120, arranged by remainder modulo 4: p ≡1 (mod 4) 5 13 17 29 37 41 53 61 73 89 97 101 109 113 · · · p ≡3 (mod 4) 3 7 11 19 23 31 43 47 59 67 71 79 83 103 107 · · · It appears that the primes are fairly evenly distributed between the two classes, and we might rea-sonably conjecture that there are infinitely many primes of each type. This is indeed the case.
Theorem 5.1.
Infinitely many primes are congruent each to 1 and 3 modulo 4.
Proof of half the Theorem. We modify Euclid’s proof. Suppose that there are finitely many primes con-gruent to 3 modulo 4: list them as 3, p1, . . . , pn and define Π := 4p1p2p3 · · · pn + 3 Certainly Π ≡3 (mod 4) and therefore odd, so all primes dividing it are odd. Note that x, y ≡1 (mod 4) = ⇒xy ≡1 (mod 4) (∗) hence, if all primes dividing Π were congruent to 1, so also would be Π. Plainly Π is divisible by some prime p ≡3 (mod 4). By assumption we have all of these, and there are two possibilities: 1. p = 3 from which 3 | 4p1p2p3 · · · pn = ⇒3 | pi = ⇒pi = 3 for some i; a contradiction.
2. p = pi for some i, in which case p | 3 = ⇒p = 3; again a contradiction.
1Several results which are very close to these have been proved recently, for example the weak Goldbach conjecture states that every odd integer ≥9 is the sum of three odd primes was proved in 2013.
1 Before moving on, consider why the proof cannot be modified to show that infinitely many primes are congruent to 1 modulo 4. One issue is that the corresponding proposition to (∗) is false: in fact x, y ≡3 (mod 4) = ⇒xy ≡1 (mod 4)!
and we cannot therefore claim that any Π ≡1 (or ≡3) is divisible by a prime congruent to 1. Indeed: • Π := 21 = 3 · 7 ≡1 (mod 4) is not divisible by any primes congruent to 1.
• Π := 3 · 7 · 11 = 231 ≡3 (mod 4) is not divisible by any primes congruent to 1.
A simple proof of the ≡1 part of the Theorem will be given later using quadratic residues.
In fact a much harder and more general result is available.
Theorem 5.2 (Dirichlet).
If gcd(a, m) = 1, then infinitely many primes p satisfy p ≡a (mod m).
Counting Primes Now we turn to the third in our list of questions. To think about this, we intro-duce the concept of a counting function: a function f : N →N0 for which f (x) is the number of positive integers less or equal to x satisfying some property. Euler’s totient function φ is an example: φ(x) = |{n ∈N≤x : gcd(x, n) = 1}| Here is another.
Example 5.3.
Consider the counting function f (x) = |{n ∈N≤x : n ≡4 (mod 7)}| To get a feel for f, compute the first few values: x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 f (x) 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 It seems reasonable to claim that, for large x, f (x) is approximately a seventh of x. More precisely, x −3 7 ≤f (x) < x + 4 7 = ⇒x −3 7x ≤f (x) x < x + 4 7x Squeeze = ⇒ Thm lim x→∞ f (x) x = 1 7 There is terminology for this: ‘f (x) is asymptotic to 1 7x,’ and we write f (x) ∼1 7x Intuitively, f (x) grows like 1 7x. This is one way of giving precision to the statement, ‘one seventh of the integers are congruent to 4 modulo 7.’ Armed with our new notation, we consider the asymptotic behavior of the primes.
Definition 5.4.
π(x) := |{p : p ≤x}| is the number of primes less than x.
Theorem 5.5 (Prime number theorem).
π(x) ∼ x ln x. Otherwise said, lim x→∞ π(x) x/ ln x = 1.
2 A proof is too involved for this course; interpreting the result is tough enough! One approach in-volves probability: the chance of a random integer in the interval [1, x] being prime is P y ∈[1, x] prime = π(x) x ≈ 1 ln x While the expression x ln x estimates the function π(x), it is, in fact, always an under-estimate. A more accurate estimate involves an integral, albeit one that needs its own estimation!
π(x) ∼ Z x 2 1 ln tdt Example 5.6.
To check the veracity of these claims: consider the 1000th prime p1000 = 7919: π(7919) = 1000, 7919 ln 7919 ≈882, Z 7919 2 1 ln tdt ≈1016 A little extra algebra tells us that the nth prime should be located around pn ≈n ln n.
Indeed 1000 ln 1000 ≈6908, which is a 13% under-estimate.
Exercises 5.1 1. (a) Verify that every even number between 70 and 80 is a sum of two primes.
(b) How many different ways can 70 be written as a sum of two primes 70 = p + q with p ≤q?
Repeat the question for 80.
2.
(a) Show that if p ≥5 is prime, then p ≡±1 (mod 6).
(b) Mimic the half-proof of Theorem 5.1 to show that there are infinitely many primes congru-ent to 5 modulo 6.
(Hint: let Π := 6p1p2 · · · pn + 5 where p1, . . . , pn ≡5 (mod 6)) 3.
(a) Explain the statement “one-fifth of all numbers are congruent to 2 modulo 5” by using the counting function F(x) = |{positive numbers n ≤x satisfying n ≡2 (mod 5)}| (b) Explain the statement “most numbers are not squares” by using the counting function S(x) = |{square numbers less than x}| 4. Let n be large. By computing x ln x when x = n ln n, argue that pn ≈n ln n is a reasonable estimate for the value of the nth prime. Use this expression to argue that, for large n, pn+1 −pn ≈1 + ln(n + 1) Comment on the values of p1000 and p1001.
5. (Hard) Let p be an odd prime and consider the quantity Ap Bp := 1 + 1 2 + 1 3 + 1 4 + · · · + 1 p −1 where gcd(Ap, Bp) = 1 (a) Find the value of Ap (mod p) and prove that your answer is correct.
(b) (Even harder - also proves part (a)) Make a conjecture for Ap (mod p2) and prove it.
(Hint: try adding 1 k + 1 p−k in pairs) 3 5.2 Mersenne Primes and Perfect Numbers While Euclid assures us that the set of primes is infinite, this hasn’t prevented a semi-formal compe-tition to find the largest known prime. Prior to the advent of computers and mechanical calculators, the largest verified prime had 39 digits. As of early 2022, the largest known prime is 282,589,933 −1 with 24,862,048 digits! Such primes have a special name.
Definition 5.7.
A Mersenne prime is a prime of the form Mp = 2p −1 where p is itself prime.
These are named for Marin Mersenne, a 17th century French music theorist, mathematician and priest.
Examples 5.8.
M2 = 22 −1 = 3, M3 = 23 −1 = 7, M5 = 25 −1 = 31, M7 = 27 −1 = 127. Not all Mersenne numbers are prime, for instance M11 = 211 −1 = 2047 = 23 · 89 In fact most Mersenne numbers are not prime; the current largest known prime is only the 51st Mersenne prime to be discovered! It is merely conjectured that there are infinitely many of them.
Whenever the ‘world’s largest prime’ is announced, it is usually a Mersenne prime.2 There are several reasons for this: a simple motivator is the fact that exponentiation quickly provides large candidates.
A related reason is that similar-looking numbers with other bases are never prime: Theorem 5.9.
If P = an −1 is prime for some a, n ≥2, then a = 2 and n is prime: that is, P is a Mersenne prime.
Proof. If a ≥3, then an −1 = (a −1)(an−1 + an−2 + · · · + a + 1) is composite. By a similar factorization, if n = mk is composite, so also is 2n −1: 2n −1 = (2m)k −1 = (2m −1)((2m)k−1 + (2m)k−2 + · · · + 1) There are many known results about Mersenne primes; look them up if you are interested. We now turn our attention to an old problem which turns out to be related to Mersenne primes, using it partly as an excuse to introduce another commonly-used function.
Definition 5.10.
Let n ∈N. Define σ(n) = ∑ d|n d to be the sum of the (positive) divisors of n.
We say that n is perfect if it equals the sum of its proper (positive) divisors: that is σ(n) = 2n (= proper divisors + n) Examples 5.11.
6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14 are both perfect.
2The Great Internet Mersenne Prime Search is an ongoing collaborative project hunting for such: anyone with a com-puter can sign up. If you’re the first to find a prime with 100 million digits, $100,000 could be yours!
4 We can compute σ(n) similarly to how we evaluated Euler’s function. First observe a simple fact following from unique prime factorization: If gcd(m, n) = 1 and d|mn, then d = d1d2 is uniquely a product of divisors d1 |m and d2 |n (prime factorization!). When m, n are coprime, it is now immediate that σ(mn) = ∑ d|mn d = ∑ d1|m, d2|n d1d2 = ∑ d1|m d1 · ∑ d2|n d2 = σ(m)σ(n) Moreover, the geometric series formula allows us to easily compute σ applied to a prime power: σ(pµ) = µ ∑ j=0 pj = pµ+1 −1 p −1 and we’ve now proved the main result: Theorem 5.12.
σ is multiplicative. Moreover, if n = pµ1 1 · · · pµk k is the prime decomposition of n, then σ(n) = k ∏ j=1 p µj+1 j −1 pj −1 Examples 5.13.
The sum of the positive divisors of 260 = 22 · 5 · 13 is σ(260) = 23 −1 1 · 52 −1 4 · 132 −1 12 = 588 This can tediously be checked since 260 has divisors 1, 2, 4, 5, 10, 13, 20, 26, 52, 65, 130, 260.
Repeating with n = 1000 = 23 · 53, we see that σ(1000) = 24 −1 2 −1 · 54 −1 5 −1 = 2340 There is an intimate relation between perfect numbers and Mersenne primes: half of it indeed ap-pears in Euclid’s Elements.
Theorem 5.14.
If 2p −1 is a Mersenne prime, then 2p−1(2p −1) is perfect.
Proof. Suppose that Mp = 2p −1 is a Mersenne prime. Since 2p −1 is prime, σ(Mp) = σ(2p−1)σ(2p −1) = 2p −1 2 −1 · (2p −1 + 1) = 2 · 2p−1(2p −1) = 2Mp 5 For small values of p we have the following table: the numbers increase very quickly!
p 2p −1 n = 2p−1(2p −1) 2 3 6 = 1 + 2 + 3 3 7 28 = 1 + 2 + 4 + 7 + 14 5 31 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 7 127 8128 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064 13 8191 33,550,336 17 131,072 8,589,869,056 It was conjectured in the middle ages and proved in the 1700’s that all even perfect numbers have this form.
Theorem 5.15 (Euler).
Every even perfect number has the form 2p−1(2p −1) for some Mersenne prime Mp = 2p −1.
Proof. Suppose that n = 2km is an even perfect number, where k ≥1 and m is odd. Our goal is to prove that m is prime; we will do this by showing that σ(m) = m + 1.
Since n is perfect and gcd(2k, m) = 1, we have two expressions for σ(n): σ(n) = ( 2n = 2k+1m σ(2k)σ(m) = (2k+1 −1)σ(m) = ⇒(2k+1 −1)σ(m) = 2k+1m Since 2k+1 −1 is odd, we see that 2k+1 | σ(m) so that σ(m) = 2k+1α for some α ∈N. We now have (2k+1 −1)α = m If we can show that α = 1 then we are finished: in such a case σ(m) = 2k+1 = (2k+1 −1) + 1 = m + 1 whence m is prime.
To obtain a contradiction, assume that α > 1. Then m is divisible by the distinct divisors 1, α, m. But then 2k+1α = σ(m) ≥1 + α + m = 1 + α + (2k+1 −1)α = 1 + 2k+1α Contradiction!
We conclude that m = 2k+1 −1 is prime. By Theorem 5.9 we see that k + 1 = p must also be prime, whence m = Mp is a Mersenne prime.
Since only fifty-one Mersenne primes have thus far been discovered, only fifty-one perfect numbers are known to exist, with the currently known largest having 49,724,095 digits! Of course the con-jectured infinity of Mersenne primes would also imply the existence of infinitely many even perfect numbers. It remains unknown whether there are any odd perfect numbers.
6 Exercises 5.2 1. Prove that p is prime if and only if σ(p) = p + 1.
2. Suppose that Mp = 2p −1 is a Mersenne prime. List all the divisors of 2p−1(2p −1) and use the geometric sequence formula to explicitly sum them. Hence provide a more explicit proof of Theorem 5.14.
3. Define τ(n) to be the number of positive divisors of n. Prove that τ is multiplicative and find a formula for τ(n) in terms of the prime decomposition of n = pµ1 1 · · · pµk k . Hence or otherwise, find the number of positive divisors of 1,000,000.
4. If an + 1 is prime for some integers a ≥2 and n ≥1, show that n must be a power of 2.
(Hints: if n is odd, show that (a + 1) | (an + 1) similarly to the proof of Theorem 5.14. Then write n = 2m, a2 = b and repeat. . . ) 5. Primes of the form Fk = 22k + 1 are called Fermat primes.3 For instance F1 = 5, F2 = 17, F3 = 257, F4 = 65537 (a) If k ≥2, prove that the final digit of Fk is 7.
(Hint: Think modulo 2 and 5. What is the period of 2m modulo 5?) (b) Show that if k ̸= m, then Fk and Fm are coprime.
(Hint: if k > m, show that Fm divides Fk −2) 6. Suppose n| Mp where p is an odd prime. Prove that n = 2kp + 1 for some integer k.
(Hint: if q is a prime divisor of 2p −1, think about why p should divide q −1) The remaining questions consider the potential impossibility of odd perfect numbers.
7.
(a) Show that a power of 3 can never be a perfect number.
(b) More generally, if p is an odd prime, show that pk is not perfect.
8.
(a) Show that a number of the form 3i5j can never be perfect.
(b) More generally, if p ≥5 is an odd prime, show that the product 3ipj can never be perfect.
(c) Even more generally, show that if p and q are distinct odd primes, then a number of the form qipj can never be perfect.
9. (Hard) Show that 3i5j7k is never perfect.
(Hint: consider σ(n) and sums 1 + 3 + 32 + · · · , etc. modulo 4, then think modulo 5) 3Fermat thought that all the Fk might be prime, however Euler (1732) and Clausen/Landry (1855/1880) successively showed that F5 and F6 are composite with prime factorizations: F5 = 4,294,967,297 = 641 · 6700417, F6 = 18,446,744,073,709,551,617 = 274,177 · 67,280,421,310,721 These were incredible achievements for the time. As of 2022, no other Fermat primes have been discovered, and only up to F11 has been completely factored! A distributed computing project similar to GIMPS continues the search. . .
7 |
13692 | https://artofproblemsolving.com/community/c728438h2787257?srsltid=AfmBOoo_E_DrMihbka3RQcmrz_M_t-K7ZioD-yZ5cLbGXjGDLdHmcup5 | Notable algebra methods : GCD of 2^n-1 and 2^m+1
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Notable algebra methods =======================
GCD of 2^n-1 and 2^m+1
by fungarwai, Feb 23, 2022, 1:46 PM
Proof
Proof
If ,
Proof
Proof
Example
Showing gcd(2^m−1,2^n+1)=1 when m is odd
When m is odd, . Therefore
有序數對的個數
Count the number of ordered sets for when
That implies counting the number of ordered sets for
The number of when is
Proof
By LTE,
Proof
By LTE,
This post has been edited 3 times. Last edited by fungarwai, Feb 25, 2022, 12:32 PM
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Video: Transition Metals | Definition, Properties & Examples
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Contributors: Keta Bhakta, Elizabeth (Nikki) Wyman
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Video Summary for Transition Metals
Transition metals are defined as metals with inner d or f orbitals being filled, located in columns 3-12 on the periodic table along with the lanthanide and actinide series.
These elements typically share properties like being lustrous, silvery, hard, and good conductors of heat and electricity, though individual characteristics vary greatly.
Examples include common metals like iron, gold, silver, copper, and platinum, with nearly 70 transition metals in total.
Transition metals can form multiple types of cations, unlike other elements that only form one charge.
Their chemical reactivity ranges widely, with some elements like iron readily forming compounds while others like gold remain unreactive.
Used extensively in building materials (iron, titanium)
Critical for manufacturing processes (tungsten, cobalt)
Provide vibrant colors in pigments and dyes
Essential components in modern technology and infrastructure
The video provides a comprehensive overview of transition metals' definition, properties, and practical applications.
Read Transition Metals | Definition, Properties & Examples Lesson
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13694 | https://www.isbe.net/CTEDocuments/HST-L630124.pdf | Understand the Anatomy and Physiology of the Epidermis Unit: Human Structure and Function: The Integumentary System Problem Area: Identify the Structures of the Integumentary System Lesson: Understand the Anatomy and Physiology of the Epidermis Student Learning Objectives. Instruction in this lesson should result in students achieving the following objectives: 1 Identify the components of the epidermis.
2 Relate the structures of the epidermis to their functions.
3 Describe the growth, development, and desquamation of the epidermal cells.
Resources. The following resources may be useful in teaching this lesson: Brannon, Heather. “Epidermis Anatomy,” About.com: Dermatology. Accessed Nov. 2, 2009. epidermis.htm.
Marieb, Elaine N. Anatomy & Physiology Coloring Workbook, 9th ed. Pearson Benjamin Cummings, 2008.
Marieb, Elaine N., and Susan J. Mitchell. Human Anatomy and Physiology Laboratory Manual, 9th ed. Pearson Benjamin Cummings, 2008.
McCann, Stephanie, and Eric Wise. Anatomy Coloring Book, 3rd ed. Kaplan, 2008.
Miller-Keane, and Marie T. O’Toole. Miller-Keane Encyclopedia and Dictionary of Medicine, Nursing, and Allied Health, 7th ed. W.B. Saunders, 2005.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 1 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 “Skin (Integumentary System),” mercksource.com. Accessed Nov. 10, 2009.
reftext/html/skin_sys_fin.html.
Smeltzer, Suzanne C., and Brenda G. Bare. Textbook of Medical-Surgical Nursing, 10th ed. Lippincott, 2004.
Sorrentino, Sheila A., and Mosby. Mosby’s Textbook for Nursing Assistants, 7th ed. Mosby, 2009.
Thibodeau, Gary A., and Kevin T. Patton. Anthony’s Textbook of Anatomy and Physiology, 18th ed. Mosby, 2006.
Equipment, Tools, Supplies, and Facilities Overhead or PowerPoint projector Visual(s) from accompanying master(s) Copies of sample test, lab sheet(s), and/or other items designed for duplication Materials listed on duplicated items Computers with printers and Internet access Classroom resource and reference materials Key Terms. The following terms are presented in this lesson (shown in bold italics): desmosomes desquamation keratin keratinization keratinocytes Langerhans’ cells melanin melanocytes Merkel cells stratum basale stratum corneum stratum granulosum stratum lucidum stratum spinosum Interest Approach. Use an interest approach that will prepare the students for the lesson. Teachers often develop approaches for their unique class and student situations. A possible approach is included here.
Explain that the epidermis (the outer layer of the skin, with a surface area of about 18 square feet, but thinner than plastic wrap over most of the body) acts like a noble warrior, protecting us against a hostile outer world. It keeps Lesson: Understand the Anatomy and Physiology of the Epidermis Page 2 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 moisture in and germs out and protects against harmful radiation. As it does so, it dies, but not before replacing itself.
Tell the students that the average person sheds about 40 pounds of epidermis over an average lifetime. Show a box of plastic wrap. Ask the students: “How many boxes of plastic wrap do you think would equal 40 pounds?” Then ask: “Why do people cover leftovers in plastic wrap before storing them?” CONTENT SUMMARY AND TEACHING STRATEGIES Objective 1: Identify the components of the epidermis.
Anticipated Problem: What are the components of the epidermis?
I. Epidermal layers or strata A. Stratum corneum (horny layer) is the outermost layer of the epidermis.
1. It consists of rows of dead cells containing keratin, which is a fibrous protein.
a. Keratin keeps the skin elastic.
b. Keratin protects the underlying tissues from drying out.
2. Desquamation is a process by which the keratin-filled dead cells are sloughed off the top.
B. Stratum lucidum (clear layer) is a thick layer found only on frequently used areas of the body, such as the palms of the hands and soles of the feet.
C. Stratum granulosum (granular layer) is a thin middle layer that initiates the production of keratin. Keratinization is the production of keratin and the death of the epidermal cells.
1. Keratinocytes are epidermal cells that produce keratin; they comprise more than 90 percent of the epidermal cells.
2. Keratinocytes fill up with keratin and flatten as they are pushed toward the sur-face.
3. Cells at the upper border of this layer are beginning to die.
D. Stratum spinosum (spiny layer), located above the basal layer, is a layer that consists of irregularly shaped cells that interlock to support the skin.
E. Stratum basale (base layer) is a single layer of columnar cells resting on a basement membrane at the bottom of the epidermis.
1. The majority of cells in this layer are constantly dividing as older ones are being pushed toward the skin’s surface.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 3 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 2. Melanocytes are cells included in the base layer that produce the protective pigment melanin.
3. Langerhans’ cells are cells that play a role in immunity.
4. Merkel cells are spiky hemispherical cells that help form sensory receptors.
F. Only cells in the basal and spiny layers receive nourishment by diffusion from the blood vessels located in the dermis below.
Teaching Strategy: Use VM–A and VM–B to prompt a discussion about the epidermis. Then assign LS–A to further student comprehension of this objective.
Objective 2: Relate the structures of the epidermis to their functions.
Anticipated Problem: What are the functions of the epidermal structures?
II. Functions of the epidermis A. The dead, flat, keratin-filled outermost cells protect underlying cells from drying out and act as a physical barrier against harmful substances.
B. The stratum lucidum protects against physical abrasion in areas subject to friction.
Absent in thin skin, this layer is quite apparent in thick skin from the soles of the feet or palms of the hands.
C. The stratum granulosum, filled with granules involved in keratin production, begin the process of keratinization.
D. The interlocking cells of the stratum spinosum, with prominent intercellular bridges called desmosomes, give support to the skin.
E. The cells of the basal layer, the stratum basale (stratum germinativum), divide continually and push already formed cells into higher layers. They include specialized cells, such as the following: 1. Melanocytes are spidery black cells that make melanin—a brownish pigment that shields the nuclei of the basal cells from the sun’s harmful effects.
2. Langerhans’ cells are phagocytic cells that play a role in immunity.
3. Merkel cells are occasional cells that form touch receptors.
Teaching Strategy: Have students draw the different layers of cells. Assign LS–B to assist students with comprehension of the material in this objective.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 4 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 Objective 3: Describe the growth, development, and desquamation of the epidermal cells.
Anticipated Problem: How can the growth, development, and desquamation of the epidermal cells be described?
III. Epidermal cells A. The cells of the stratum basale divide by mitosis and push older cells closer to the surface.
1. Friction and abrasion accelerate the rate of cell division.
2. Prolonged abrasion may result in thickened stratum corneum called “callus.” B. As the cells move toward the surface, they are flattened and fill with the protein keratin.
C. The flat, keratinized cells—now dead—are shed from the surface about every two to four weeks in the process called desquamation.
Teaching Strategy: Have students work in small groups to draw a picture of the growth, development, and desquamation of the epidermal cells. Then assign LS–C.
Review/Summary. Use the student learning objectives to summarize the lesson.
Have students explain the content associated with each objective. Student responses can be used in determining which objectives need to be reviewed or taught from a different angle.
Application. Use the included visual master(s) and lab sheet(s) to apply the information presented in the lesson.
Evaluation. Evaluation should focus on student achievement of the objectives for the lesson. Various techniques can be used, such as student performance on the application activities. A sample written test is provided.
Answers to Sample Test: Part One: Matching 1. b 2. d 3. a 4. e 5. c 6. f Lesson: Understand the Anatomy and Physiology of the Epidermis Page 5 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 Part Two: Completion 1. epidermis 2. desquamation 3. melanin 4. keratin 5. desmosomes 6. Merkel cells Part Three: Short Answer 1. a. stratum corneum b. stratum lucidum c. stratum granulosum d. stratum spinosum e. stratum basale 2. Any two of the following would be acceptable: from drying out, from germs, from harmful substances, or from harmful radiation.
3. They occur in the following order: cell division of the basal cells, flattening and keratinization, and desquamation.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 6 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 Sample Test Name ______ Understand the Anatomy and Physiology of the Epidermis Part One: Matching Instructions: Match the term with the correct definition.
a. stratum basale d. keratinocytes b. stratum corneum e. melanocytes c. stratum granulosum f. Langerhans’ cells 1. The layer of dead, flat epidermal cells that are shed about every two weeks 2. The most abundant cells in the epidermis _3. The lowest layer of the epidermis consisting of a row of cells that divide continuously 4. The cells that produce the dark-colored pigment of the skin __5. The thin, middle layer of the epidermis that begins keratin production _6. The cells that act as a frontline defense of the immune system in the skin Part Two: Completion Instructions: Provide the word or words to complete the following statements.
1. The entire outer layer of the skin that is thinner than plastic wrap over much of the body is called the ___.
2. The shedding of the top layer of the epidermis is called ____.
3. The dark pigment produced in the epidermis is called ___.
4. The fibrous protein that keeps the skin elastic and protects the underlying tissues from drying out is called ____.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 7 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 5. Intercellular bridges are ___.
6. Spiky hemispherical cells that help form sensory receptors are ____.
Part Three: Short Answer Instructions: Answer the following.
1. List the following layers (strata) of the skin in the proper order, top to bottom: stratum basale, stratum corneum, stratum granulosum, stratum spinosum, and stratum lucidum 2. List two things from which the epidermis protects the body.
3. Arrange the following events in the life of an epidermal cell in the order in which they occur: desquamation, cell division of the basal cells, and flattening and keratinization.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 8 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 VM–A OVERVIEW OF SKIN ANATOMY Lesson: Understand the Anatomy and Physiology of the Epidermis Page 9 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 Epidermis Dermis Hypodermis VM–B LAYERS OF THE EPIDERMIS Lesson: Understand the Anatomy and Physiology of the Epidermis Page 10 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 Stratum corneum Stratum lucidum Stratum granulosum Stratum spinosum Stratum basal LS–A Name _______ Studying the Layers of the Epidermis Purpose The purpose of this activity is to expand comprehension of the epidermis layers.
Objectives 1. Identify the layers of the epidermis shown on the diagram.
2. Associate the layers with their functions.
Materials lab sheet VM–B writing utensil Procedure 1. Look at the diagram of the layers of the epidermis and listen to your teacher read the description of each layer.
2. Finish the phrases that briefly describe at least one function. Include another fact about each layer, either structure or function.
3. Write the description in the box to the right of the name of each layer.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 11 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 Stratum (Layer) Description of Structure or Function Corneum (horny) Flat, dead cells, composed mainly of: Lucidum (clear) Found only in: Granulosum (grainy) Begins the production of the protein called: Spinosum (spiny) Cells with shapes that: Basale (basal) also called germinativum Column-shaped cells that are always: Lesson: Understand the Anatomy and Physiology of the Epidermis Page 12 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 LS–B Name ______ Cell Types in the Epidermis Purpose The purpose of this activity is to identify epidermis cells and functions.
Objective Relate the types of epidermal cells to their functions.
Materials lab sheet writing utensil textbook or notes from class Procedure 1. Refer to your textbook or your notes from class, as needed.
2. Fill in the missing name, function, or description for each cell type.
Cell Types in the Epidermis Name of Cell Description and/or Function Make up most of the epidermis, fill with a fibrous protein Melanocytes Langerhans’cells Merkel cells Lesson: Understand the Anatomy and Physiology of the Epidermis Page 13 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 LS–C Name ________ The Stages in the Growth and Repair of Epidermal Cells Purpose The purpose of this activity is to become more familiar with epidermal cell stages.
Objective Describe the sequence of changes involved in epidermal growth and repair.
Materials lab sheet paper writing utensil Procedure 1. Use the following terms to indicate the order in which the epidermal cells form and migrate to the skin’s surface: desquamation, cell division, flattening, keratinization, and death.
2. Write the terms in a vertical column on the right side of your paper.
3. Turn in your work to your instructor.
Lesson: Understand the Anatomy and Physiology of the Epidermis Page 14 www.MyCAERT.com Copyright © by CAERT, Inc. | Reproduction by subscription only. | L630124 |
13695 | https://www.khanacademy.org/math/cc-fifth-grade-math/divide-fractions/relate-fraction-division-to-fraction-multiplication/v/multiplication-and-division-relationship-for-fractions | Multiplication and division relationship for fractions (video) | Khan Academy
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5th grade math
Course: 5th grade math>Unit 7
Lesson 2: Relate fraction division to fraction multiplication
Multiplication and division relationship for fractions
Relate fraction division to fraction multiplication
Math>
5th grade math>
Divide fractions>
Relate fraction division to fraction multiplication
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Multiplication and division relationship for fractions
CCSS.Math: 5.NF.B.7
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About About this video Transcript
We learn how multiplication and division are related, even when we're dealing with fractions. Watch how dividing by a number is the same as multiplying by its reciprocal. Then watch Sal practice expressing these relationships using both division and multiplication.
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Evangeline 5 years ago Posted 5 years ago. Direct link to Evangeline's post “So couldn't _42 = 7 / 1/...” more So couldn't 42 = 7 / 1/6 be written as 42 x 7 = 1/6 instead of 42 x 1/6 = 7?
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Danforth, David 3 years ago Posted 3 years ago. Direct link to Danforth, David's post “There is this to consider...” more There is this to consider.
"Example one":
2x3 = 6 and 3x2 = 6
This is not true for the order in division.
15/3 = 5 however 3/15 = 1/5 or 0.2 they both yield a smaller number however the one with a larger number for its denominator yields a significantly smaller number when compared to the one with a larger numerator.
When dividing fractions this has to be taken into consideration.
"Example two":
4 x 1/16 = 1/4 or 0.25
1/16 x 4 = 1/4 or 0.25
4 divided by 1/16 = 64
1/16 divided by 4 = 1/64 or 0.015625
"Example three":
A whole number divided by a fraction yields a larger "whole" number.
A fraction divided by a whole number yields an even "smaller fraction."
Example three simplified:
number/fraction = number or "n/f=n"
fraction/number = fraction or "f/n=f"
There needs to be a video called "The differences between multiplying and dividing fractions".
After wracking my brain on this for six hours I had to break this apart and figure out the underlying system so that it's possible to get the right answer every time.
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Micheal ross 5 years ago Posted 5 years ago. Direct link to Micheal ross's post “What's the reason of doin...” more What's the reason of doing 421/6=7? Can't I write as 427=1/6? Unfortunately, This is very common in Khan Academy videos. There isn't enough explanation in details.
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Stefan 5 years ago Posted 5 years ago. Direct link to Stefan's post “So is your question 421/...” more So is your question 421/6=7 = 421/6=7?
because 421/6=42/6 and fraction is basically division so 42÷6=7 that makes sense and 427=294 so did I answer your question?
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31pedro.esilva 2 years ago Posted 2 years ago. Direct link to 31pedro.esilva's post “Bro why show the hard way...” more Bro why show the hard way then the easier way?
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SaanviD a year ago Posted a year ago. Direct link to SaanviD's post “all u have to do is take ...” more all u have to do is take the first number multiplied by the last one in a equation and BAM-u have the answer
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Aileen Freckleton 5 years ago Posted 5 years ago. Direct link to Aileen Freckleton's post “I need more explanation ....” more I need more explanation . I don't understand which number I should use to divide. 5.NF.B7
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SirBLUE 7 months ago Posted 7 months ago. Direct link to SirBLUE's post “Imagine if you actually s...” more Imagine if you actually saw Sal, then he said ''hey, you, 42 x 1/6 is 7'''
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Mary DeTray a month ago Posted a month ago. Direct link to Mary DeTray's post “If your lost as to why yo...” more If your lost as to why you can only flip it one way just remember this: you want to multiply the answer of the division probem by the number you "DIVIDED BY" (that's the divisor, the last number in the division problem) not the amount you're dividing (the dividend, the first number in the division problem).
So, remember:
If. A ÷ B = C
Then A = B x C
Or. C x B =A
You can flip a multiplication problem and get the same answer but not a division problem. Do examples for yourself if you're still confused, it helps.
3÷2 is not going to be the same answer as 2÷3 in the same way that 2x3 is the same as 3x2. So, when doing the inverse operation of a division problem you need to make sure your multiplying the answer by the divisor, not the dividend.
Hope this helps anyone struggling with this.
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[Instructor] You are likely already familiar with the relationship between multiplication and division. For example, we know that three times six is equal to 18. But another way to express that same relationship is to say, all right, if three times six is 18, then if I were to start with 18 and divide it by three, that would be equal to six. Or you could say something like this, that 18 divided by, divided by six is equal to three. Now we're just going to extend this same relationship between multiplication and division to expressions that deal with fractions. So for example, if I were to tell you that 1/4 divided by, and I'm going to color-code it, divided by two is equal to 1/8, is equal to 1/8, how could we express this relationship, but using multiplication? Well, if 1/4 divided by two is equal to 1/8, that means that 1/8 times two is equal to 1/4. Let me write this down, or I could write it like this. I could write that 1/4 is going to be equal to, is going to be equal to 1/8 times two, times two. And we could do another example. Let's say that I were to walk up to you on the street and I were to tell you that, hey, you, 42 is equal to seven, seven divided by 1/6. In the future, we will learn to compute things like this. But just based on what you see here, how could we express this same relationship between 42, seven, and 1/6, but express it with multiplication? Pause this video, and think about that. Well, if 42 is equal to seven divided by 1/6, that means that 42 times 1/6 is equal to seven. Let me write that down. This is the same relationship as saying that 42 times 1/6, 1/6 is equal to seven. Now let's say I walk up to you on the street and I were to say, all right, you, I'm telling you that 1/4 divided by, divided by six is equal to some number that we will express as t. So can we rewrite this relationship between 1/4, six, and t, but instead of using division, use multiplication? Pause this video, and try to think about it. So if 1/4 divided by six is equal to t, based on all of the examples we've just seen, that means that if we were to take t times six, we would get 1/4. So we could write it this way, t times six, times six is going to be equal to 1/4. If this isn't making sense, I really want you to think about how this relationship is really just the same relationship we saw up here. The only new thing here is instead of always having whole numbers, we're having fractions and representing some of the numbers with letters.
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13696 | https://gandhiviswanathan.wordpress.com/2020/05/06/perfect-matchings-of-a-complete-graph-and-the-double-factorial/ | Skip to primary content
Gandhi Viswanathan's Blog
Physics, Math and Life as a Scientist
Perfect matchings of a complete graph and the double factorial
Posted on by Gandhi Viswanathan
I have a longstanding interest in graph theory applied to statistical mechanics. For example, a few years ago I published a paper in Physical Review E that explores the correspondence between spanning trees and the Ising model on the square lattice (see PDF here). More recently, I became fascinated by perfect matchings.
There is a deep relationship between perfect matchings in graph theory on the one hand and the theory of equilibrium statistical mechanics on the other hand. Perfect matchings are relevant to dimer models and to certain lattice spin models, for instance.
As entertainment and also as a birthday present to myself for my 50th anniversary this week, I decided to figure out for myself exactly why it is that the double factorial appears in the expression for the number of perfect matchings of complete graphs of vertices. The number counts the number of different ways of pairing all vertices with exactly edges, without any vertex being left out or having more than one connecting edge. It turns out that
where
It is important not to confuse the double factorial with the factorial applied twice,
rather the double factorial is defined as
where the ceiling function gives the smallest integer not smaller than . In other words,
Let us first calculate the number of perfect matchings. Given vertices, we can pair the 1st vertex with others with a single edge. Once the first edge is assigned, there are vertices left that need to be paired. So
We can iterate this recursion relation:
So this is why the double factorial appears in (1).
To arrive at (2), we proceed to express the double factorial in terms of the standard factorial. First, notice that can be written as
so that
For the double factorial with odd arguments we proceed as follows. Note that the double factorial contains either only odd or only even factors, which can can combine to obtain the standard factorial thus:
This expression relates the double factorial of odd and even arguments. Let . Then the above gives us
from which follows (2).
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13697 | https://testbook.com/maths/concave-function | Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/AMS/Regular/Main.js
Test Series
A concave function is a mathematical function that has a downward curve, meaning that any line segment drawn between any two points on the graph of the function will lie below or on the graph. Concave functions are important in mathematics, economics, optimization, and other fields as they describe situations where increasing returns to scale or decreasing marginal utility are present.
The concept of concavity is important in calculus as it helps in analyzing the behavior of functions, such as determining the maximum or minimum points. Understanding the properties and applications of concave functions is essential in various areas of mathematics and its applications.
In this mathematics article, we will study what is concave functions, their graphs, methods for determining convexity or concavity, techniques for identifying concavity, and properties of concave functions through worked-out examples.
Concave Function
A concave function is a mathematical function that has a downward curve, meaning that any line segment drawn between any two points on the graph of the function will lie below or on the graph. In other words, the function is “curving inward.”
Mathematically, a function f(x)f(x) is concave if its second derivative, f″(x), is negative for all values of x within a certain domain. Intuitively, a concave function is one that becomes less steep as you move from left to right on the graph.
A common example of a concave function is the square root function, f(x)=√x. Another example is the natural logarithm function, f(x)=ln(x). Concave functions have important applications in economics, optimization, and other fields, as they describe situations where increasing returns to scale or decreasing marginal utility are present.
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Concave Function Graph
The graph of a concave function has a shape that is similar to the interior of a bowl or a cup. The curve starts out steep and becomes less steep as it moves towards the right.
The concavity of a function determines the direction in which it opens and refers to its state or quality. If a function opens upwards, it is considered concave up, and if it opens downwards, it is considered concave down.
The image below displays two functions, one concave upwards and the other concave downwards.
To learn about types of functions based on set elements with solved examples.
How to Check if a Function is Convex or Concave?
To check if a function is convex or concave, you need to examine its second derivative. Here are the steps:
Step 1: Find the first derivative of the function f(x) to obtain f′(x).
Step 2: Find the second derivative of f(x) by differentiating f′(x) with respect to x.
Step 3: Determine the sign of the second derivative at each point in the function's domain.
Step 4: If the second derivative is positive for all x in the domain, i.e., if f″(x)>0 for all x in the domain, then f(x) is convex. This means that the graph of the function is curved upward.
Step 5: If the second derivative is negative for all x in the domain, i.e., if f″(x)<0 for all x in the domain, then f(x) is concave. This means that the graph of the function is curved downward.
Step 6: If the second derivative is zero for some x, i.e., if f″(x)=0 for some x in the domain, further investigation is needed to determine the concavity or convexity of the function at that point.
Note that if the second derivative changes sign at some point, the function is neither convex nor concave at that point. In such cases, the function is said to have an inflection point.
In practice, you can use the second derivative test to quickly determine the concavity or convexity of a function at a critical point (where the first derivative is zero). The second derivative test states that if f′(c)=0 and f″(c)>0, then f(x) has a local minimum at x=c, and if f′(c)=0 and f″(c)<0, then f(x) has a local maximum at x=c.
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How to Find Concavity of a Function?
The concavity of the function means the direction in which the function opens, concavity describes the state or the quality of a Concave function. For example, if the function opens upwards it is called concave up and if it opens downwards it is called concave down.
The concavity of a function can also be identified by drawing tangents at points on the graph. For example, when a tangent drawn at a point lies below the graph in the vicinity of that point, the graph is said to be concave up. Conversely, when a tangent drawn at a point lies above the graph in the vicinity of that point, the graph is said to be concave downward.
An illustration of this concept is shown in the figure below.
It's important to observe that there exists a point on a graph where the tangent line doesn't lie completely above or below it. Instead, the tangent intersects the graph at that point, indicating a change in concavity from upwards to downwards or vice versa. This specific point is commonly known as the point of inflection.
Now, we will examine the precise definition for each of these points.
f(x) is said to be concave up in the interval I if all the possible tangents drawn to the curve at different points in the interval I, lie below the graph.
f(x) is said to be concave downwards in the interval I if all the possible tangents drawn to the curve at different points in the interval I, lie above the graph.
A point x=a on the curve f(x) is called point of inflection if the function is continuous and the concavity of the graph changes at that point.
To learn about the concept of interval notation and their types with solved examples.
Concavity of a Function Using Derivatives
The method described above involves analyzing the concavity of a function graphically. However, when the function graph is not available, derivatives can be used as a solution. For instance, given a function f(x), the figure below illustrates different scenarios, where the slope of the tangent line provides the derivative value of the function at that particular point of intersection.
Observe the above figure, where a function exhibits a concave downward shape. The slope of the tangent line in this case is seen to decrease, signifying a decrease in derivative values. On the other hand, when the function is concave upward, the derivative values tend to increase. Based on these observations, we can draw the following conclusions:
If the function is concave up, its derivative f′(x) is increasing.
If the function is concave down, its derivative f′(x) is decreasing.
When the function f(x) has an inflection point at point x=a. f′(x) either goes from increasing to decreasing or vice-versa. That means the graph of the function f′(x) has a minimum/maximum at x=a.
We can draw mathematical conclusion from the observations that are given above. Consider a function f(x), we have:
For the interval I, if f″(x)>0 then the function f(x) is concave up in the interval I.
For the interval I, if f″(x)<0 then the function f(x) is concave down in the interval I.
If x=a is a point of inflection, then at x=a, f″(a)=0.
To learn about the derivative rules and differentiation rules in detail with examples.
Concave Function Properties
A concave function is a function where a line segment between any two points on the graph of the function lies entirely below the graph. Some properties of concave functions are listed below:
First Derivative: The first derivative of a concave function is decreasing.
Second Derivative: The second derivative of a concave function is negative everywhere.
Tangent lines: The tangent line to a concave function at any point lies below the graph of the function.
Jensen's Inequality: For a concave function, the expected value of the function is less than or equal to the function of the expected value. This property is known as Jensen's inequality and has important applications in probability theory and economics.
Maximum: A concave function has a unique maximum, which is attained at the point where the first derivative is equal to zero.
Local Maxima: A concave function has no local maxima other than the global maximum.
These properties make concave functions useful in optimization problems, where one seeks to find the maximum or minimum of a function subject to certain constraints.
To learn about the absolute maxima and minima values with solved examples.
Concave Function Summary
A concave function is a function where a line segment between any two points on the graph of the function lies entirely below the graph.
Concave functions are always continuous and differentiable over their domain, have a decreasing slope, and a negative second derivative over their entire domain.
The tangent line to a concave function at any point lies below the graph of the function, and a concave function has a global maximum at its leftmost point.
Concave functions are always convex down, meaning that the curve “opens downward” and may have multiple inflection points where the curvature changes.
Concave Function Solved Examples
1.What should be the value of “a” for the function f(x)=ax3+4x2+1 to be concave downward at x=1.
Solution:
Analyze the given function through the second derivative test explained above,
f(x)=ax3+4x2+1
Differentiating the function,
⇒ f′(x)=3ax2+8x
Differentiating it again to find the second derivative,
⇒ f″(x)=6ax+8
Now at x=1,
f″(1)=6a+8
For the function to be concave downward, f″(x)<0
⇒ 6a+8<0
⇒ a<−43
2.Tell whether the graph of the function f(x)=ex+cos(x) is concave up or concave downward at x=0.
Solution:
Analyze the given function through the second derivative test explained above.
Given function is f(x)=ex+cos(x)
Differentiating the function,
⇒ f′(x)=ex−sin(x)
Differentiating it again to find the second derivative,
⇒ f″(x)=ex−cos(x)
Now at x=0,
f″(0)=e0−cos(0)
⇒ f″(0)=1−1
⇒ f″(0)=0
Since, f″(0)=0. From the definition above we can say that the function is neither concave up or concave downward at x=0. So, x=0 is the point of inflection for the function f(x).
3.What is the shape of the graph for the function f(x)=x3+4x2+1 at x=0.
Solution:
Analyze the given function through the second derivative test explained above.
Given function is f(x)=x3+4x2+1
Differentiating the function,
⇒ f′(x)=3x2+8x
Differentiating it again to find the second derivative,
⇒ f″(x)=6x+8
Now at x=0,
f″(0)=6(0)+8=8
So, f″(0)>0. Thus, from the definition above we can say that, the shape of the function is concave upward at x=0.
We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
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Concave Function FAQs
What is concave function definition?
A concave function is a function where a line segment connecting any two points on the graph of the function always lies below the function itself. In other words, the function curves downward.
What are the properties of concave function?
Properties of concave functions are listed below:
A concave function is always continuous and differentiable over its domain.
The second derivative of a concave function is negative over its entire domain.
The tangent line to a concave function at any point lies below the graph of the function.
A concave function has a global maximum at its leftmost point.
How do you know if a function is concave?
A function is concave if its second derivative is negative over its entire domain or if every line segment connecting any two points on the function lies above or on the function's graph.
How to check concavity of function?
The concavity of a function can be identified by drawing tangents at points on the graph:
When a tangent drawn at a point lies below the graph in the vicinity of that point, the graph is said to be concave up.
When a tangent drawn at a point lies above the graph in the vicinity of that point, the graph is said to be concave downward
What are concave function examples?
Examples of concave functions include the logarithmic function, the square root function, and the exponential function.
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13698 | https://www.youtube.com/watch?v=vd4bXmTA70M | How to find the Oxidation Number for in BaH2 (Barium hydride)
Wayne Breslyn (Dr. B.)
890000 subscribers
18 likes
Description
2765 views
Posted: 24 Oct 2020
To find the correct oxidation state of in BaH2 (Barium hydride), and each element in the compound, we use a few rules and some simple math.
First, since the BaH2 doesn’t have an overall charge (like NO3- or H3O+) we could say that the total of the oxidation numbers for BaH2 will be zero since it is a neutral compound.
We write the oxidation number (O.N.) for elements that we know and use these to figure out oxidation number for .
RESOURCES
How to Find Oxidation Numbers (rules and examples):
Oxidation Numbers Practice:
GENERAL RULES
Free elements have an oxidation state of zero (e.g. Na, Fe, H2, O2, S8).
In an ion the all Oxidation numbers must add up to the charge on the ion.
In a neutral compound all Oxidation Numbers must add up to zero.
Group 1 = +1
Group 2 = +2
Hydrogen with Non-Metals = +1
Hydrogen with Metals (or Boron) = -1
Fluorine = -1
Oxygen = -2 (except in H2O2 or with Fluorine)
Group 17(7A) = -1 except with Oxygen and other halogens lower in the group
5 comments
Transcript:
Intro in this video we'll find the oxidation numbers for each element here in bah2 this is called barium hydride when we look at barium hydride there's no negative or positive after it so it's going to be a neutral compound all the oxidation numbers they'll add up to zero barium that's in group two on the periodic table elements in group two they have a plus two oxidation number hydrogen when it's bonded to metals and barium is a metal it's in group two on the the periodic table when it's bonded Periodic Table to metals or Boron it's going to be NE -1 so each one of the hydrogens here is going to be1 and that makes sense 2 - 1 that's -2 and a -2 and a positive2 Summary that adds up to zero this is Dr B with the oxidation numbers for each element here in bah2 barium hydride thanks for watching |
13699 | https://ahdictionary.com/word/search.html?q=bold | American Heritage Dictionary Entry: bold
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bold (b ō ld)
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adj.bold·er, bold·est
1.
a.Fearless and daring; courageous:a bold leader.
b.Requiring or exhibiting courage or daring:a bold voyage to unknown lands. See Synonyms at brave.
2.Unduly forward and brazen; impudent:a bold, sassy child.
3.Strikingly different or unconventional; arresting or provocative:"[He] laid out a bold, new vision for America's leading universities" (Jerome Karabel).
4.
a.Clear and distinct to the eye; conspicuous:bold colors; a bold pattern.
b.Strong or pronounced; prominent:the bold flavor of ginger.
5.Steep or abrupt in grade or terrain:"The two walk along the high, bold, rocky shore" (Harriet Beecher Stowe).
6.Printing Boldface.
[Middle English, from Old English bald; see bhel-2in the Appendix of Indo-European roots.]
boldlyadv.
boldnessn.
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