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# 58.7 minutes to hours
## Result
58.7 minutes equals 0.9783 hours
You can also convert 58.7 minutes to hours and minutes.
## Conversion formula
Multiply the amount of minutes by the conversion factor to get the result in hours:
58.7 min × 0.0166667 = 0.9783 hr
## How to convert 58.7 minutes to hours?
The conversion factor from minutes to hours is 0.0166667, which means that 1 minutes is equal to 0.0166667 hours:
1 min = 0.0166667 hr
To convert 58.7 minutes into hours we have to multiply 58.7 by the conversion factor in order to get the amount from minutes to hours. We can also form a proportion to calculate the result:
1 min → 0.0166667 hr
58.7 min → T(hr)
Solve the above proportion to obtain the time T in hours:
T(hr) = 58.7 min × 0.0166667 hr
T(hr) = 0.9783 hr
The final result is:
58.7 min → 0.9783 hr
We conclude that 58.7 minutes is equivalent to 0.9783 hours:
58.7 minutes = 0.9783 hours
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case fifty-eight point seven minutes is approximately zero point nine seven eight hours:
58.7 minutes ≅ 0.978 hours
## Conversion table
For quick reference purposes, below is the minutes to hours conversion table:
minutes (min) hours (hr)
59.7 minutes 0.995002 hours
60.7 minutes 1.011669 hours
61.7 minutes 1.028335 hours
62.7 minutes 1.045002 hours
63.7 minutes 1.061669 hours
64.7 minutes 1.078335 hours
65.7 minutes 1.095002 hours
66.7 minutes 1.111669 hours
67.7 minutes 1.128336 hours
68.7 minutes 1.145002 hours
## Units definitions
The units involved in this conversion are minutes and hours. This is how they are defined:
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations.
### Hours
An hour (symbol: h; also abbreviated hr.) is a unit of time conventionally reckoned as 1⁄24 of a day and scientifically reckoned as 3,599–3,601 seconds, depending on conditions. The seasonal, temporal, or unequal hour was established in the ancient Near East as 1⁄12 of the night or daytime. Such hours varied by season, latitude, and weather. It was subsequently divided into 60 minutes, each of 60 seconds. Its East Asian equivalent was the shi, which was 1⁄12 of the apparent solar day; a similar system was eventually developed in Europe which measured its equal or equinoctial hour as 1⁄24 of such days measured from noon to noon. The minor variations of this unit were eventually smoothed by making it 1⁄24 of the mean solar day, based on the measure of the sun's transit along the celestial equator rather than along the ecliptic. This was finally abandoned due to the minor slowing caused by the Earth's tidal deceleration by the Moon. In the modern metric system, hours are an accepted unit of time equal to 3,600 seconds but an hour of Coordinated Universal Time (UTC) may incorporate a positive or negative leap second, making it last 3,599 or 3,601 seconds, in order to keep it within 0.9 seconds of universal time, which is based on measurements of the mean solar day at 0° longitude.
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Caleb Weeks
Posted on
# Advent of Code #16 (in Crystal)
Today's puzzle didn't seem particularly challenging. I figured that I would have to keep track of cycles and that it was possible to visit a mirror or splitter from multiple sides. But I ran into a snag during part 1 that slowed me down for over an hour. My code worked perfectly fine on the examples and on any scenario that I could think of, but it failed on my input.
Turns out that I was just branching on the splitters before counting them in the path, so my answer was lower than it should have been for the input. Conveniently, the "dumb thing" for part 2 worked fine and gave an answer in just a few minutes.
In my solution, I used enums for the first time. I love that `case` does exhaustive checking for enums, and I think it made for pretty readable code. Here it is:
``````input = File.read("input").strip
grid = input.split("\n").map(&.chars)
enum Direction
Up
Down
Left
Right
end
def move(position, direction)
row, col = position
case direction
in .up? then {row - 1, col}
in .down? then {row + 1, col}
in .left? then {row, col - 1}
in .right? then {row, col + 1}
end
end
def back_slash(direction)
case direction
in .up? then Direction::Left
in .down? then Direction::Right
in .left? then Direction::Up
in .right? then Direction::Down
end
end
def forward_slash(direction)
case direction
in .up? then Direction::Right
in .down? then Direction::Left
in .left? then Direction::Down
in .right? then Direction::Up
end
end
def in_grid(position, grid)
row, col = position
row >= 0 && row < grid.size && col >= 0 && col < grid[row].size
end
def shine(grid, path, start, start_direction)
edge = false
position = start
row, col = position
direction = start_direction
while in_grid(position, grid) && !path.includes?({position, direction})
row, col = position
path << {position, direction}
case grid[row][col]
when '\\'
direction = back_slash(direction)
when '/'
direction = forward_slash(direction)
when '-'
if direction.up? || direction.down?
shine(grid, path, position, Direction::Left)
shine(grid, path, position, Direction::Right)
break
end
when '|'
if direction.left? || direction.right?
shine(grid, path, position, Direction::Up)
shine(grid, path, position, Direction::Down)
break
end
else
end
position = move(position, direction)
end
end
part1 = begin
path = [] of Tuple(Tuple(Int32, Int32), Direction)
shine(grid, path, {0, 0}, Direction::Right)
path.map(&.[0]).uniq.size
end
puts part1
part2 = begin
max = 0
(0...110).each do |i|
path = [] of Tuple(Tuple(Int32, Int32), Direction)
shine(grid, path, {0, i}, Direction::Down)
max = Math.max(max, path.map(&.[0]).uniq.size)
end
(0...110).each do |i|
path = [] of Tuple(Tuple(Int32, Int32), Direction)
shine(grid, path, {109, i}, Direction::Up)
max = Math.max(max, path.map(&.[0]).uniq.size)
end
(0...110).each do |i|
path = [] of Tuple(Tuple(Int32, Int32), Direction)
shine(grid, path, {i, 0}, Direction::Right)
max = Math.max(max, path.map(&.[0]).uniq.size)
end
(0...110).each do |i|
path = [] of Tuple(Tuple(Int32, Int32), Direction)
shine(grid, path, {i, 109}, Direction::Left)
max = Math.max(max, path.map(&.[0]).uniq.size)
end
max
end
puts part2
``````
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| 3,190
|
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https://bw.mathematicstip.com/8769-vector-calculus.html
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# Vector Calculus
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Vector calculus is concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space The term "vector calculus" is sometimes used as a synonym for the broader subject of multivariable calculus. The modules in this section of the core complement Corral's Vector Calculus TextMap and the Vector Calculus (UCD Mat 21D) Libretext. Check there for more vector calculus content.
## Vector Calculus in a Nutshell
The most important object in our course is the vector field, which assigns a vector to every point in some subset of space.
We'll cover the essential calculus of such vector functions, and explore how to use them to solve problems in partial differential equations , wave mechanics, electricity and magnetism, and much more!
This quiz kicks off a short intro to the essential ideas of vector calculus.
A partial differential equation (pde for short) is an equation involving unknown multivariable functions and their partial derivatives.
## Contents
### Vectors in three dimensions Edit
In 3d Euclidean space, R > 3 , the standard basis is ex, ey, ez. Each basis vector points along the x-, y-, and z-axes, and the vectors are all unit vectors (or normalized), so the basis is orthonormal.
Throughout, when referring to Cartesian coordinates in three dimensions, a right-handed system is assumed and this is much more common than a left-handed system in practice, see orientation (vector space) for details.
For Cartesian tensors of order 1, a Cartesian vector a can be written algebraically as a linear combination of the basis vectors ex, ey, ez:
where the coordinates of the vector with respect to the Cartesian basis are denoted ax, ay, az. It is common and helpful to display the basis vectors as column vectors
when we have a coordinate vector in a column vector representation:
A row vector representation is also legitimate, although in the context of general curvilinear coordinate systems the row and column vector representations are used separately for specific reasons – see Einstein notation and covariance and contravariance of vectors for why.
The term "component" of a vector is ambiguous: it could refer to:
• a specific coordinate of the vector such as az (a scalar), and similarly for x and y, or
• the coordinate scalar-multiplying the corresponding basis vector, in which case the "y-component" of a is ayey (a vector), and similarly for x and z.
A more general notation is tensor index notation, which has the flexibility of numerical values rather than fixed coordinate labels. The Cartesian labels are replaced by tensor indices in the basis vectors exe1, eye2, eze3 and coordinates axa1, aya2, aza3. In general, the notation e1, e2, e3 refers to any basis, and a1, a2, a3 refers to the corresponding coordinate system although here they are restricted to the Cartesian system. Then:
It is standard to use the Einstein notation—the summation sign for summation over an index that is present exactly twice within a term may be suppressed for notational conciseness:
An advantage of the index notation over coordinate-specific notations is the independence of the dimension of the underlying vector space, i.e. the same expression on the right hand side takes the same form in higher dimensions (see below). Previously, the Cartesian labels x, y, z were just labels and not indices. (It is informal to say "i = x, y, z").
### Second order tensors in three dimensions Edit
A dyadic tensor T is an order 2 tensor formed by the tensor product ⊗ of two Cartesian vectors a and b, written T = ab. Analogous to vectors, it can be written as a linear combination of the tensor basis exexexx , exeyexy , . ezezezz (the right hand side of each identity is only an abbreviation, nothing more):
Representing each basis tensor as a matrix:
then T can be represented more systematically as a matrix:
See matrix multiplication for the notational correspondence between matrices and the dot and tensor products.
More generally, whether or not T is a tensor product of two vectors, it is always a linear combination of the basis tensors with coordinates Txx, Txy, . Tzz:
while in terms of tensor indices:
Second order tensors occur naturally in physics and engineering when physical quantities have directional dependence in the system, often in a "stimulus-response" way. This can be mathematically seen through one aspect of tensors - they are multilinear functions. A second order tensor T which takes in a vector u of some magnitude and direction will return a vector v of a different magnitude and in a different direction to u, in general. The notation used for functions in mathematical analysis leads us to write v - T(u) , [1] while the same idea can be expressed in matrix and index notations [2] (including the summation convention), respectively:
By "linear", if u = ρr + σs for two scalars ρ and σ and vectors r and s, then in function and index notations:
and similarly for the matrix notation. The function, matrix, and index notations all mean the same thing. The matrix forms provide a clear display of the components, while the index form allows easier tensor-algebraic manipulation of the formulae in a compact manner. Both provide the physical interpretation of directions vectors have one direction, while second order tensors connect two directions together. One can associate a tensor index or coordinate label with a basis vector direction.
The use of second order tensors are the minimum to describe changes in magnitudes and directions of vectors, as the dot product of two vectors is always a scalar, while the cross product of two vectors is always a pseudovector perpendicular to the plane defined by the vectors, so these products of vectors alone cannot obtain a new vector of any magnitude in any direction. (See also below for more on the dot and cross products). The tensor product of two vectors is a second order tensor, although this has no obvious directional interpretation by itself.
The previous idea can be continued: if T takes in two vectors p and q, it will return a scalar r. In function notation we write r = T(p, q), while in matrix and index notations (including the summation convention) respectively:
The tensor T is linear in both input vectors. When vectors and tensors are written without reference to components, and indices are not used, sometimes a dot ⋅ is placed where summations over indices (known as tensor contractions) are taken. For the above cases: [1] [2]
motivated by the dot product notation:
More generally, a tensor of order m which takes in n vectors (where n is between 0 and m inclusive) will return a tensor of order mn , see Tensor § As multilinear maps for further generalizations and details. The concepts above also apply to pseudovectors in the same way as for vectors. The vectors and tensors themselves can vary within throughout space, in which case we have vector fields and tensor fields, and can also depend on time.
Following are some examples:
An applied or given. . to a material or object of. . results in. . in the material or object, given by:
unit vector n Cauchy stress tensor σ a traction force t t = σ ⋅ n =<oldsymbol >cdot mathbf >
angular velocity ω moment of inertia I an angular momentum J J = I ⋅ ω =mathbf cdot <oldsymbol >>
moment of inertia I a rotational kinetic energy T T = 1 2 ω ⋅ I ⋅ ω <2>><oldsymbol >cdot mathbf cdot <oldsymbol >>
electric field E electrical conductivity σ a current density flow J J = σ ⋅ E =<oldsymbol >cdot mathbf >
polarizability α (related to the permittivity ε and electric susceptibility χE) an induced polarization field P P = α ⋅ E =<oldsymbol >cdot mathbf >
magnetic H field magnetic permeability μ a magnetic B field B = μ ⋅ H =<oldsymbol >cdot mathbf >
For the electrical conduction example, the index and matrix notations would be:
Three vector calculus operations which find many applications in physics are:
1. The divergence of a vector function
2. The curl of a vector function
3. The Gradient of a scalar function
These examples of vector calculus operations are expressed in Cartesian coordinates, but they can be expressed in terms of any orthogonal coordinate system, aiding in the solution of physical problems which have other than rectangular symmetries. Index
## Vector Calculus
This is a fairly short chapter. We will be taking a brief look at vectors and some of their properties. We will need some of this material in the next chapter and those of you heading on towards Calculus III will use a fair amount of this there as well.
Here is a list of topics in this chapter.
Basic Concepts – In this section we will introduce some common notation for vectors as well as some of the basic concepts about vectors such as the magnitude of a vector and unit vectors. We also illustrate how to find a vector from its starting and end points.
Vector Arithmetic – In this section we will discuss the mathematical and geometric interpretation of the sum and difference of two vectors. We also define and give a geometric interpretation for scalar multiplication. We also give some of the basic properties of vector arithmetic and introduce the common (i), (j), (k) notation for vectors.
Dot Product – In this section we will define the dot product of two vectors. We give some of the basic properties of dot products and define orthogonal vectors and show how to use the dot product to determine if two vectors are orthogonal. We also discuss finding vector projections and direction cosines in this section.
Cross Product – In this section we define the cross product of two vectors and give some of the basic facts and properties of cross products.
## Vector Calculus
This is the homepage for the free book Vector Calculus, by Michael Corral (Schoolcraft College).
Java code samples from the book: calc3book_java.zip
MATLAB/Octave versions: ParallelizationArea.zip (courtesy of Prof. Benson Muite (University of Michigan))
Sage versions: calc3book_sage.zip
Note: The PDF was built using TeXLive 2011 and Ghostscript 9.53 under Linux (Fedora).
LaTeX source code: calc3book-1.0-src.tar.gz
You can buy a printed and bound paperback version of the book with grayscale graphics for \$10 plus shipping at Lulu.com here.
### Book Description
This is a text on elementary multivariable calculus, designed for students who have completed courses in single-variable calculus. The traditional topics are covered: basic vector algebra lines, planes and surfaces vector-valued functions functions of 2 or 3 variables partial derivatives optimization multiple integrals line and surface integrals.
The book also includes discussion of numerical methods: Newton's method for optimization, and the Monte Carlo method for evaluating multiple integrals. There is a section dealing with applications to probability. Appendices include a proof of the right-hand rule for the cross product, and a short tutorial on using Gnuplot for graphing functions of 2 variables.
There are 420 exercises in the book. Answers to selected exercises are included.
1. Vectors in Euclidean Space
• Introduction
• Vector Algebra
• Dot Product
• Cross Product
• Lines and Planes
• Surfaces
• Curvilinear Coordinates
• Vector-Valued Functions
• Arc Length
2. Functions of Several Variables
• Functions of Two or Three Variables
• Partial Derivatives
• Tangent Plane to a Surface
• Directional Derivatives and the Gradient
• Maxima and Minima
• Unconstrained Optimization: Numerical Methods
• Constrained Optimization: Lagrange Multipliers
3. Multiple Integrals
• Double Integrals
• Double Integrals Over a General Region
• Triple Integrals
• Numerical Approximation of Multiple Integrals
• Change of Variables in Multiple Integrals
• Application: Center of Mass
• Application: Probability and Expected Value
4. Line and Surface Integrals
• Line Integrals
• Properties of Line Integrals
• Green's Theorem
• Surface Integrals and the Divergence Theorem
• Stokes' Theorem
• Gradient, Divergence, Curl and Laplacian
• Bibliography
• Appendix A: Answers and Hints to Selected Exercises
• Appendix B: Proof of the Right-Hand Rule for the Cross Product
• Appendix C: 3D Graphing with Gnuplot
(2021-01-05) Cleaned up the web page to make it less hideous and more consistent with the revamped page for Elementary Calculus.
• Appendix A: The answer to Exercise 5 from Section 1.9 is now fixed.
• Section 1.1: In Example 1.3(d) R^3 now has the correct dimension.
• Section 1.7: In Example 1.33 a minus is now a plus.
• Appendix A: The margin overrun in the answers for Section 2.4 has been removed.
Problems with my TeXLive 2014 setup had caused numerous issues when trying to compile the book. I ended up going back to TeXLive 2011 to fix all that, and it worked. So now, after many requests, I have finally restored the ability to buy a printed and bound paperback version on Lulu.com. It's even a buck cheaper than before. See the link near the top of this page.
On a side note, there are many things about the book that I would change now, after the experience of writing the Trigonometry book and especially the Elementary Calculus book, both content-wise and stylistically. I haven't decided on that yet, but if I do re-write Vector Calculus then I would keep the current version available in addition to the new version. Any decision on that wouldn't be for at least another year, though.
(2013-05-21) I finally(!) got around to uploading the MATLAB/Octave versions of the programs in the book, which Prof. Benson Muite (Univeristy of Michigan) kindly sent me over a year ago. I apologize for the delay my only excuse is that my schedule became incredibly hectic over the last year. Now that things have settled down again, I should have some time to start working on a French translation of Vector Calculus (as well as finish Elementary Calculus). There have been many offers from people around the world to translate my books into other languages. For example, Prof. Koichiro Yamashita will post his Japanese translation at http://kymst.net after he finishes it.
The latest version of Vector Calculus contains a correction of a typo in one of the plots (Fig. 1.8.3 on p.54), which Prof. Yamashita found.
(2012-02-13) I ported the Java code examples in Sections 2.6 and 3.4 to Sage, a powerful and free open-source mathematics software system that is gaining in popularity. The Sage code examples are in the calc3book_sage.zip file, and can be run either on the command-line or as worksheets in a Sage notebook. See the included README file for more details.
The reason for doing this is because I received a request a few years ago to rewrite those code examples for Sage. I wasn't as familiar with Sage as I am now, so I finally got around to doing it. In general, I will be using Sage more, and in particular it will be used extensively for the code examples in my upcoming Elementary Calculus book.
(2011-06-29) The latest version of the book is out. The content of the book is basically the same as before. The big change was in switching the math font from txfonts to Fourier-GUTenberg. This was done to make the fonts more consistent. In particular, the fouriernc package makes use of the New Century Schoolbook normal text fonts for numbers and letters in math mode. This way there is no longer the incongruity of having txfonts' Times Roman-like numbers and letters in math mode versus New Century Schoolbook's different-looking numbers and letters in the main text. This change required a bit of space adjustment throughout the text, since some of the symbols in the Fourier-GUTenberg fonts are slightly smaller than those in txfonts. The sans serif font was also changed, from Avant Garde to Helvetica.
Another change was cleaning up the graphics, which also had a mish-mash of inconsistent fonts and other issues (in particular the graphics created with MetaPost and Gnuplot). The Gnuplot graphics were slightly improved over some of the default settings which I had used originally.
These changes in the appearance make the book look better overall, in my opinion, and were long overdue. It also brings the book in line with the general look and feel of my Trigonometry book and my forthcoming Elementary Calculus book (the prequel to this book).
As far as the content in the main text itself, the only changes are:
• I added a tiny clarification on the relationship between determinants and volumes of parallelepipeds, right before Theorem 1.17 in Section 1.4. In particular I give the conditions for when the determinant gives the positive volume or the negative volume.
• In Section 1.7 I added a footnote about the left-handedness of the usual definition of the spherical coordinate system used by mathematicians. I did this because physics students may get confused when they see the definitions of &theta and &phi switched in their physics classes.
• Improved the code listings to use a monospaced font (Bitstream Vera Sans Mono). I still do not know what possessed me to use a proportional font originally.
• Four corrections in the answers in Appendix A: 1.5 #1, 1.9 #3, 4.1 #11, 4.5 #4 (thanks to P. Taskas and G. Strzalkowski).
• Updated instructions for using Gnuplot in Appendix C (in the Windows version some of the defaults and procedures changed slightly).
Update (2011-06-30): The printed bound version of the book on the Lulu.com site has also been updated with the latest changes.
(2011-04-17) I've written up a very short (10-page) mini-tutorial on using the LaTeX typesetting system. You can download it here: latex-tutorial.pdf
The source code for the tutorial is available here: latex_tutorial.zip
The tutorial was originally created for students in a class I'm teaching this semester, and I've expanded it a bit since then. I hope others find it useful.
(2010-06-06) Typos in the proof of Theorem 1.20(f) on p.53 have been corrected (thanks to F. Dockhorn for finding those). A newer version of the TikZ/PGF graphics package broke the diagrams on pp.60-61, so the code for those diagrams has been updated. Also, I am still working on the prequel - Elementary Calculus - which (barring a miraculous increase in my productivity) will likely not be ready until sometime next year.
(2009-09-13) The author's new book, Trigonometry, is now available. The homepage is located here: mecmath.net/trig/
(2009-07-22) The prequel to this book, which will be titled Elementary Calculus, is in preparation. It will cover calculus of a single variable. The aim is to have it available by the end of this year or early next year. Another book, on trigonometry, is almost finished and should be available here by the end of August 2009. Both books will be free and released under the GNU Free Documentation License, complete with the LaTeX source code.
(2009-07-22) In Appendix A, the answer to Exercise 5 in Section 2.3 was corrected. Thanks to E. Cavazos for pointing out the error. This is the only change in the new version (2009-07-22) from the previous version (2009-03-29).
(2009-07-10) Prof. Marshall Hampton of the University of Minnesota, Duluth has kindly posted some notes on compiling the LaTeX source code for the book under OS X, which you can read here.
### Contact
The author of the book, Michael Corral, can be reached via email at
## Vector Calculus
What makes this book different?
This text covers most of the standard topics in multivariate calculus and a substantial part of a standard first course in linear algebra. If, in addition, one teaches the proofs in Appendix A, the book can be used as a textbook for a course in analysis.The organization and selection of material differs from the standard approach in four ways.
We integrate linear algebra and multivariate calculus.
See what students and professors have to say about Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach. See the table of contents in pdf form (requires Acrobat Reader).
Instructors' Solution Manual A revised version of the instructor's solution manual, incorporating corrections and some new solutions, is now available from Prentice Hall .
If your are taking your first serious math course and don't have the book, you may find several brief excerpts useful they are from Chapter 0: Preliminaries. They are in pdf format and can be read using Acrobat Reader.
The first discusses how to read mathematics. (Many students who never read their math texts in high school discover this doesn't work in college.)
The second explains why all eleven-legged alligators are orange with blue stripes'' is a true statement, as is all eleven-legged alligators are black with white stripes.''
The third gives the vocabulary of set theory (and a nice picture of Bertrand Russell shaving his image in the mirror.)
• Sections and level sets
• Graphing scalar valued functions
• Limits
• Continuity
• Directional derivatives
• Partial derivatives
• Total differentials
• The Derivative Matrix
• Tangent planes
• Linear approximation
• Differentiability
• The Simple Chain Rule
• The Matrix Chain Rule
• The Paths Chain Rule
• Iterated partial derivatives
1. Look at all the critical points x where the gradient of f(x) = 0. Throw out any that aren't in D .
2. Look at any points where f isn't differentiable.
3. Look at the boundary of D . You can this either by
• Parameterizing the boundary so that you have an unconstrained max/min problem, or by
• Lagrange multipliers: if the boundary of D is described as a level set by g(x) = 0, then the critical points of f constrained to be on the boundary of D can be found by solving
## Vector Calculus Sessions
The Vector Calculus Review Sessions are a review of topics taught in UC San Diego's Vector Calculus course (Math 20E) that you (may) have learned at your previous institution. The session materials have been reviewed by a Math Department Graduate Student and a Faculty member.
Please note that these sessions do not replace a Vector Calculus course and will not be teaching you the topics. If at any point in the sessions you discover a topic that you have not learned previously in your Vector Calculus course, you are encouraged to enroll in Math 20E at UC San Diego. There is a Summer Session 2 Vector Calculus (Math 20E) course, that will begin on Monday, August 1st to September 10th (5-weeks).
• To support transfer students to review key topics in preparation for attempting the UC San Diego MATH 20E Fulfillment Exam
Program Details:
• Dates: July 19th - 30th
• Cost: No cost to participate
• Format: In-person and Zoom options available
• Content: Review core calculus topics, including topics and problems based on vector calculus
• Time commitment: Three sessions per week for two weeks, totaling six 1.5 sessions, plus additional independent study
Eligibility
• Incoming transfer students
• Enrolled in a major that requires MATH 20E
• Have taken a vector calculus course at community college
• Planning to attempt the MATH 20E fulfillment exam to fulfill the MATH 20E requirement
Topics to be Reviewed:
1. Double integral over a region and changing the order of integration, mean value inequality
2. Triple integral
3. Geometry of Maps from R2 to R2
4. The change of variables theorem
5. Vector field
6. Path Integral, Line Integral
7. Parametrized Surface and use integral to find the area of a surface
8. Integrals of Scalar Functions Over Surfaces
9. Surface Integrals of Vector Field
10. Green's Theorem
11. Stoke's Theorem
12. Gauss's Theorem (Divergence Theorem)
13. Conservative Fields
If you have not yet taken vector calculus, we do not recommend that you participate in these Topic Review Sessions or attempt the MATH 20E Fulfillment Exam. Instead, we highly encourage you to enroll in MATH 20E within your first year. Consider getting a head start by taking an accelerated 5-week version of the course during Summer Session II, which takes place from August 3rd - September 4th. The deadline to enroll in Summer Session II is July 26th.
If you have questions, please email [email protected]
## Vector Calculus
Vector calculus is one of the most useful branches of mathematics for game development. You could say it is the most important if you're willing to play it slightly fast and loose with definitions and include in it the subset of low-dimensional linear algebra that vector calculus relies on for a lot of its computation. Certainly for physics and any advanced graphics, it's vitally important. It's also beautiful and cool, and a lot of fun to work with and apply to solving problems.
For my purposes, vector calculus is the study of how scalars and vectors change in space and time, which sounds a lot like video games to me.
Eventually I will collect more of my writing and thoughts on vector calculus here, including the most useful techniques for solving problems from my experience. In the meantime.
### Lecture
At the 2005 Game Developers Conference, I gave a lecture originally titled Why You Should Have Paid Attention in Vector Calculus, and then retitled Neat Stuff from Vector Calculus & Related Subjects. I think this was one of my weaker lectures of the past 10 years because I was really destroyed by the Indie Game Jam that year, and I was incredibly rushed creating the slides, but there's still a bunch of interesting stuff in the presentation. Highlights include:
## Watch the video: Calculus 3 - Intro To Vectors (June 2022).
1. Raj
You are not wrong
2. Bohort
and I vazma probably. come in handy
3. Nachman
What words ... super, wonderful thought
4. Laibrook
I am am excited too with this question where I can find more information on this question?
5. Eachann
I think he is wrong. I am able to prove it. Write to me in PM.
6. Zuluktilar
Excellent
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# Ratio, Proportions and Variations Multiple Choice Questions and Answers 4 PDF Book Download
Ratio, proportions and variations multiple choice questions (MCQs), ratio, proportions and variations quiz answers 4 to learn high school math courses online. Proportion MCQs, ratio, proportions and variations quiz questions and answers for online school degrees. Proportion, joint variation, math: ratios test for high school teacher certification.
Learn high school math multiple choice questions (MCQs): a combination of direct and inverse variations of one or more than one variables is known as, with choices inverse variation, direct variation, joint variation, and complex variation for online school degrees. Free math study guide for online learning proportion quiz questions to attempt multiple choice questions based test.
## MCQ on Ratio, Proportions and Variations Worksheets 4 PDF Book Download
MCQ: A statement which is expressed as an equivalence of two ratios is known as
1. proportion
2. variation
3. ratio
4. probability
A
MCQ: A combination of direct and inverse variations of one or more than one variables is known as
1. direct variation
2. inverse variation
3. joint variation
4. complex variation
C
MCQ: A relation between 2 quantities of same kind(measured in same unit) is called
1. proportion
2. variation
3. ratio
4. probability
C
MCQ: If a cricket team wins 5 matches and losses 7, then ratio of games won to lost is
1. 5 + 7
2. 5-Jul
3. 7
4. 5 : 7
D
MCQ: In a : b = c : d, a and d are called
1. antecedent
2. extreme
3. consequent
4. mean
B
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# Daily Archives: February 9, 2019
Hello Everyone!
This upcoming week marks the 100th day of school! Can you believe that we are already half way though the year?! It’s hard to accept that, but also at the same time there has been so much growth that it totally makes sense that we’ve been in school for 100 fabulous days!
Here’s a look at what we did this week at school!
-We finished Crenshaw!! Ask your child how they enjoyed it, what they learned from it and who else they think would enjoy it. Next week we will write a book recommendation for Crenshaw!
-Our Subtraction Action books have been finished and corrected!! We also did a little review assessment to make sure all that we’ve learned has been transferred over. Most of the students did okay on that. Common errors were: not copying the question from the whiteboard accurately, not lining the question up according to the place values in the question, not paint attention to the operation symbol (adding when it asked them to subtract and vice versa) and questions that require a lot of regrouping when there are zeros present (example: 10000-3458). This is something that you can easily do at home if you would like to check in with your child!
-We’ve started some work on fractions! Ask your child what a fraction is! If they need a refresher feel free to watch this Mathantics video
-The targeted lessons on basic facts and writing crafts have been going really well! We have completed one rotation with the writing. Your child has either completed 3 lessons on a) Introductions b) Paragraphs or c) Conclusions. The students have been hard at work putting this knowledge into practice with their Ancient Civilization writing!
-The Ask Me question this week tells you about the basic facts we are working on and provides you with a few ideas to try out at home
-We’ve done an assessment on the first 100 words on the Fry’s word list. Earlier in the week I sent home words students needed to practice and will send the next note home on Monday. These are now part of our “No Excuse” words and will be expected to be spelled correctly each time they are used. If anyone would like a copy of the words to use at home, please let me know
-We talked about flossing…I hope it’s been going well at home!!! I haven’t missed a day yet!
-We FINALLY got to go outside for recess on Friday afternoon! Let’s hope the weather will cooperate this week so we can get outside more regularly!!
What’s happening next week?!
-Book orders are due!
-We are doing some Valentine’s Day writing combined with I Love to Read month…called “A Love Letter to My Favourite Book!” This is a spinoff of what we did last year (A Love Letter to My Favourite Food). Look for the finished product on Seesaw later in the week!
-Speaking of Valentine’s Day…On Thursday we will be making card holders in the morning, then exchanging cards (for those who want to) in the afternoon. It is also our day to have learning buddies and Mrs. Au and I thought it might be nice to exchange a card with each other! So please have your child make or write out a card for the grade 1/2 buddy. If your child is going to exchange cards with others we ask that they have a card for each student, if they do not want to participate in a card exchange that is fine too. Reminder: there are 22 students in the class. If you would like to send in a small treat to share with the class please let me know (reminder about the peanut allergy in the class). We are nothing a “party” per se, but will make time for a snack if someone wants to send something in. *Please send in any materials required for the Valentine holder that we don’t have at the school. The only stipulations for the holders are 1) Name must be clearly visible and 2) There needs to be an opening for the cards. If you have extra boxes at home and want to send one for someone who may not have one please feel free!
-We will do another rotation with our writing crafts and 3 more targeted lessons for basic facts
-We will start our next read aloud book….another Natalie Lloyd selection! This one is called “A Snicker of Magic”
-Hopefully by the end of Monday our final drafts of the Ancient Civilization writing will be complete. If your child hadn’t finished their first draft(s) at school it has been sent home for them to finish. We worked on this for many hours this week and I think it’s fair to have the first draft ready to edit together.
-Once we are done with that we will move into making some structures! Students will look for structures from their Ancient Civilization and start to recreate!! This is always a crowd pleaser!!
-You have a few more days to drop of peanut butter, other nut butter or “WoW” butter for the Destination Imagination group’s food drive for Winnipeg Harvest. We are competing with 122 and 126 for top prize, but in the end those in need of food in our city are the real winners 🙂 If you are inspired by a good competition though…we are in need of more donations to ‘catch up’ to the frontrunner….Room 122!!
Have a wonderful weekend!!!
Until next time,
Danielle
PS-#GoJetsGo!
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Mathbox for Jonathan Ben-Naim < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bnj601 Structured version Unicode version
Theorem bnj601 29228
Description: Technical lemma for bnj852 29229. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
Hypotheses
Ref Expression
bnj601.1
bnj601.2
bnj601.3
bnj601.4
bnj601.5
Assertion
Ref Expression
bnj601
Distinct variable groups: ,,,,, ,, ,,,,, ,,, ,, ,
Allowed substitution hints: (,,,) (,,,,) (,,,,,) (,,,,,) () (,,,) ()
Proof of Theorem bnj601
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 bnj601.1 . 2
2 bnj601.2 . 2
3 bnj601.3 . 2
4 bnj601.4 . 2
5 bnj601.5 . 2
6 biid 228 . 2
7 biid 228 . 2
8 biid 228 . 2
9 bnj602 29223 . . . . . . 7
109cbviunv 4122 . . . . . 6
1110opeq2i 3980 . . . . 5
1211sneqi 3818 . . . 4
1312uneq2i 3490 . . 3
14 dfsbcq 3155 . . 3
1513, 14ax-mp 8 . 2
16 dfsbcq 3155 . . 3
1713, 16ax-mp 8 . 2
18 dfsbcq 3155 . . 3
1913, 18ax-mp 8 . 2
2013eqcomi 2439 . 2
21 biid 228 . 2
22 biid 228 . 2
23 biid 228 . 2
24 biid 228 . 2
25 biid 228 . 2
26 eqid 2435 . 2
27 eqid 2435 . 2
28 eqid 2435 . 2
29 eqid 2435 . 2
301, 2, 3, 4, 5, 6, 7, 8, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 20bnj600 29227 1
Colors of variables: wff set class Syntax hints: wi 4 wb 177 wa 359 w3a 936 wceq 1652 wcel 1725 weu 2280 wne 2598 wral 2697 wsbc 3153 cdif 3309 cun 3310 c0 3620 csn 3806 cop 3809 ciun 4085 class class class wbr 4204 cep 4484 csuc 4575 com 4837 wfn 5441 cfv 5446 c1o 6709 w-bnj17 28987 c-bnj14 28989 w-bnj15 28993 This theorem is referenced by: bnj852 29229 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-13 1727 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2416 ax-rep 4312 ax-sep 4322 ax-nul 4330 ax-pow 4369 ax-pr 4395 ax-un 4693 ax-reg 7552 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3or 937 df-3an 938 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2284 df-mo 2285 df-clab 2422 df-cleq 2428 df-clel 2431 df-nfc 2560 df-ne 2600 df-ral 2702 df-rex 2703 df-reu 2704 df-rab 2706 df-v 2950 df-sbc 3154 df-csb 3244 df-dif 3315 df-un 3317 df-in 3319 df-ss 3326 df-pss 3328 df-nul 3621 df-if 3732 df-pw 3793 df-sn 3812 df-pr 3813 df-tp 3814 df-op 3815 df-uni 4008 df-iun 4087 df-br 4205 df-opab 4259 df-mpt 4260 df-tr 4295 df-eprel 4486 df-id 4490 df-po 4495 df-so 4496 df-fr 4533 df-we 4535 df-ord 4576 df-on 4577 df-lim 4578 df-suc 4579 df-om 4838 df-xp 4876 df-rel 4877 df-cnv 4878 df-co 4879 df-dm 4880 df-rn 4881 df-res 4882 df-ima 4883 df-iota 5410 df-fun 5448 df-fn 5449 df-f 5450 df-f1 5451 df-fo 5452 df-f1o 5453 df-fv 5454 df-1o 6716 df-bnj17 28988 df-bnj14 28990 df-bnj13 28992 df-bnj15 28994
Copyright terms: Public domain W3C validator
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# Fraction Word Problems Worksheets
Printable “Fraction Word Problems” Worksheets:
2-Step Fraction Word Problems (Add, Subtract)
2-Step Mixed Number Word Problems (Add, Subtract)
Fraction Word Problems (Add, Subtract, Multiply)
### Fraction Word Problems Worksheets
Free printable and online math worksheets to help you learn how to interpret fraction as division. You can either print the worksheets (pdf) or practice online.
How to solve fraction word problems involving addition, subtraction, and multiplication?
To solve fraction word problems involving addition, subtraction, and multiplication, you’ll need to follow a systematic approach and apply the appropriate operations to the given fractions. Here’s a step-by-step guide:
Steps to Solve Fraction Word Problems with Addition, Subtraction, and Multiplication:
1. Carefully read the word problem to understand the context and what it’s asking you to find.
Identify the given information, the operations required (addition, subtraction, multiplication), and the unknowns.
2. If the problem involves mixed numbers, convert them to improper fractions for easier calculation.
Multiply the whole number by the denominator and add the numerator to get the new numerator. Keep the same denominator.
3. Based on the problem statement, perform addition, subtraction, or multiplication of the fractions.
Use common denominators when adding or subtracting fractions.
4. Simplify the Result (If Needed):
If the result is an improper fraction, consider simplifying it to a mixed number if necessary.
5. After performing the calculations, double-check your answer to ensure it makes sense in the context of the problem.
Have a look at this video if you need to review how to solve fraction word problems involving addition, subtraction and multiplication.
Then, practice the following worksheets.
Click on the following worksheet to get a printable pdf document.
Scroll down the page for more Fraction Word Problems Worksheets.
### More Fraction Word Problems Worksheets
Printable
Fraction Word Problems Worksheet #1
Fraction Word Problems Worksheet #2
Fraction Word Problems Worksheet #3
Fraction Word Problems Worksheet #4
Fraction Word Problems (addition, subtraction, multiplication)
More Printable Worksheets
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Node: Point Constructors and Setting Functions, Next: , Previous: Point Global Constants and Variables, Up: Point Reference
### Constructors and Setting Functions
void Point (void) Default constructor
Creates a `Point` and initializes its x, y, and z-coordinates to 0.
void Point (const real x, [const real y = `CURR_Y`, [const real z = `CURR_Z`]]) Constructor
Creates a `Point` and initializes its x, y, and z-coordinates to the values of the arguments x, y, and z. The arguments y and z are optional. If they are not specified, the values of `CURR_Y` and `CURR_Z` are used. They are 0 by default, but can be changed by the user. This can be convenient, if all of the `Points` being drawn in a particular section of a program have the same z or y and z values.
void set (const real x, [const real y = `CURR_Y`, [const real z = `CURR_Z`]]) Setting function
Corresponds to the constructor above, but is used for resetting the coordinates of an existing `Point`.
void Point (const Point& p) Copy constructor
Creates a `Point` and copies the values for its x, y, and z-coordinates from p.
void set (const Point& p) Setting function
Corresponds to the copy constructor above, but is used for resetting the coordinates of an existing `Point`. This function exists purely as a convenience; the operator `operator=()` (see Point Reference; Operators) performs exactly the same function.
Point* create_new (const Point* p) Template specializations
Point* create_new (const Point& p)
Pseudo-constructors for dynamic allocation of `Points`. They create a `Point` on the free store and allocate memory for it using `new(Point)`. They return a pointer to the new `Point`. If p is a non-zero pointer or a reference, the new `Point` will be a copy of p. If the new object is not meant to be a copy of an existing one, `0` must be passed to `create_new()` as its argument. See Dynamic Allocation of Shapes, for more information. One use for `create_new` is in the constructors for `classes` of objects that can contain a variable number of `Points`, such as `Path` and `Polygon`. Another use is in the drawing and filling functions, where objects are copied and the copies put onto a `Picture`. Programmers who dynamically allocate `Points` must ensure that they are deallocated properly using `delete`!
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January 19, 2022, 02:00:52 AM
Forum Rules: Read This Before Posting
### Topic: Using an Arrhenius plot (Read 1878 times)
0 Members and 1 Guest are viewing this topic.
#### mycotheologist
• Regular Member
• Posts: 46
• Mole Snacks: +1/-0
##### Using an Arrhenius plot
« on: March 25, 2012, 01:51:45 PM »
Heres the question:
Kinetics is probably my weakest area in chemistry so I'm struggling. To find the shelf life, I need to use the Arrhenius equation but I don't know the activation energy Ea or the A constant. So I converted 2 those percentages into rates and got:
At 40C, rate = 0.082 %/month
At 60C, rate = 0.39 %/month
I know that if I plot 1/T vs. lnk, I get an Arrhenius plot and then I can find Ea from the slope and A from the y-intercept but I don't have k. Can I just use the rates that I have instead of k?
#### blaisem
• Regular Member
• Posts: 87
• Mole Snacks: +5/-0
##### Re: Using an Arrhenius plot
« Reply #1 on: March 25, 2012, 06:32:29 PM »
Yes. My understanding is that it is assumed the reaction is first order. Therefore, k and the rate you calculated are the same thing. The units of A will reflect this in your equation.
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# 6.2: Introduction to Ratios and Rates
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We use ratios to compare two numeric quantities or quantities with the same units.
Ratio
A ratio is the quotient of two numerical quantities or two quantities with the same physical units.
For example, ancient Greek geometers believed that the most pleasing rectangle to the eye had length and width such that the ratio of length to width was a specific number, called the Golden Ratio, approximately equal to 1.6180339887.... Architects to this day use this ratio in their designs.
There are a number of equivalent ways of expressing ratios, three of which we will use in this text: fraction notation, “to” notation, and “colon” notation.
• 3/4 is a ratio, read as “the ratio of 3 to 4.”
• 3 to 4 is a ratio, read as “the ratio of 3 to 4.”
• 3:4 is a ratio, read as “the ratio of 3 to 4.”
Example 1
Express each of the following ratios as a fraction reduced to lowest terms: (a) 36 to 24, and (b) 0.12 : 0.18.
Solution
(a) To express the ratio “36 to 24” as a fraction, place 36 over 24 and reduce.
\begin{aligned} \frac{36}{24} = \frac{3 \cdot \textcolor{red}{12}}{2 \cdot \textcolor{red}{12}} ~ & \textcolor{red}{ \text{ Factor.}} \\ = \frac{3 \cdot \cancel{ \textcolor{red}{12}}}{2 \cdot \cancel{ \textcolor{red}{12}}} ~ & \textcolor{red}{ \text{ Cancel common factor.}} \\ = \frac{3}{2} \end{aligned}\nonumber
Thus, the ratio 36 to 24 equals 3/2.
(b) To express the ratio “0.12:0.18” as a fraction, place 0.12 over 0.18 and reduce.
\begin{aligned} \frac{0.12}{0.18} = \frac{(0.18) \textcolor{red}{(100)}}{(0.18) \textcolor{red}{(100)}} ~ & \textcolor{red}{ \text{ Multiply numerator and denominator by 100.}} \\ = \frac{12}{18} ~ & \textcolor{red}{ \text{ Move each decimal 2 places right.}} \\ = \frac{2 \cdot \textcolor{red}{6}}{3 \cdot \textcolor{red}{6}} ~ & \textcolor{red}{ \text{ Factor.}} \\ = \frac{2 \cdot \cancel{ \textcolor{red}{6}}}{3 \cdot \cancel{ \textcolor{red}{6}}} ~ & \textcolor{red}{ \text{Cancel.}} \\ \frac{2}{3} \end{aligned}\nonumber
Thus, the ratio 0.12:0.18 equals 2/3.
Exercise
Express 0.12 : 0.3 as a fraction reduced to lowest terms.
2/5
Example 2
For the rectangle that follows, express the ratio of length to width as a fraction reduced to lowest terms.
Solution.
The ratio length to width can be expressed as a fraction and reduced as follows.
\begin{aligned} \frac{ \text{length}}{ \text{width}} = \frac{3 \frac{1}{4} \text{ ft}}{2 \frac{1}{2} \text{ ft}} ~ & \textcolor{red}{ \text{ Length to width as a fraction.}} \\ = \frac{3 \frac{1}{4} \cancel{ \text{ ft}}}{2 \frac{1}{2} \cancel{ \text{ ft}}} ~ & \textcolor{red}{ \text{ Cancel common units.}} \\ = \frac{ \frac{13}{4}}{ \frac{5}{2}} ~ & \textcolor{red}{ \text{ Mixed to improper fractions.}} \end{aligned}\nonumber
Invert and multiply, factor, and cancel common factors.
\begin{aligned} = \frac{13}{4} \cdot \frac{2}{5} ~ & \textcolor{red}{ \text{ Invert and multiply.}} \\ = \frac{26}{20} ~ & \textcolor{red}{ \text{ Multiply numerators and denominators.}} \\ = \frac{13 \cdot \textcolor{red}{2}}{10 \cdot \textcolor{red}{2}} ~ & \textcolor{red}{ \text{ Factor numerator and denominator.}} \\ = \frac{13 \cdot \cancel{ \textcolor{red}{2}}}{10 \cdot \cancel{ \textcolor{red}{2}}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \frac{13}{10} \end{aligned}\nonumber
Hence, the ratio length to width is 13/10.
Exercise
A rectangle has length $$8 \frac{1}{4}$$ inches and width $$3 \frac{1}{2}$$ inches. Express the ratio of length to width as a fraction reduced to lowest terms.
33/14
## Rates
We now introduce the concept of rate, a special type of ratio.
Rate
A rate is a quotient of two measurements with different units.
The physical interpretation of a rate in terms of its units is an important skill.
Example 3
An automobile travels 224 miles on 12 gallons of gasoline. Express the ratio distance traveled to gas consumption as a fraction reduced to lowest terms. Write a short sentence explaining the physical significance of your solution. Include units in your description.
Solution
Place miles traveled over gallons of gasoline consumed and reduce.
\begin{aligned} \frac{224 \text{ mi}}{12 \text{ gal}} = \frac{56 \cdot \textcolor{red}{4} \text{ mi}}{3 \cdot \textcolor{red}{4} \text{ gal}} ~ & \textcolor{red}{ \text{ Factor.}} \\ = \frac{56 \cdot \cancel{ \textcolor{red}{4}} \text{ mi}}{3 \cdot \cancel{ \textcolor{red}{4}} \text{ gal}} ~ & \textcolor{red}{ \text{ Cancel common factor.}} \\ = \frac{56 \text{ mi}}{3 \text{ gal}} \end{aligned}\nonumber
Thus, the rate is 56 miles to 3 gallons of gasoline. In plain-speak, this means that the automobile travels 56 miles on 3 gallons of gasoline.
Exercise
Lanny travels 180 kilometers on 14 liters of gasoline. Express the ratio distance traveled to gas consumption as a fraction reduced to lowest terms.
90/7 kilometers per litre
## Unit Rates
When making comparisons, it is helpful to have a rate in a form where the denominator is 1. Such rates are given a special name.
Unit Rate
A unit rate is a rate whose denominator is 1.
Example 4
Herman drives 120 miles in 4 hours. Find his average rate of speed.
Solution
Place the distance traveled over the time it takes to drive that distance.
\begin{aligned} \frac{120 \text{ miles}}{4 \text{ hours}} = \frac{30 \text{ miles}}{1 \text{ hour}} ~ & \textcolor{red}{ \text{ Divide: } 120/4 = 30.} \\ = 30 \text{ miles/hour} \end{aligned}\nonumber
Hence, Herman’s average rate of speed is 30 miles per hour.
Exercise
Jacob drives 120 kilometers in 3 hours. Find his average rate of speed.
40 kilometers per hour
Example 5
Aditya works 8.5 hours and receives 95 for his efforts. What is his hourly salary rate? Solution Let’s place money earned over hours worked to get the following rate: $\frac{95 \text{ dollars}}{8.5 \text{ hours}}\nonumber$ We will get a much better idea of Aditya’s salary rate if we express the rate with a denominator of 1. To do so, divide. Push the decimal in the divisor to the far right, then move the decimal an equal number of places in the dividend. As we are dealing with dollars and cents, we will round our answer to the nearest hundredth. Because the test digit is greater than or equal to 5, we add 1 to the rounding digit and truncate; i.e., 95/8.5 ≈ 11.18. Hence, \begin{aligned} \frac{95 \text{ dollars}}{8.5 \text{ hours}} = \frac{11.18 \text{ dollars}}{1 \text{ hour}} ~ & \textcolor{red}{ \text{ Divide: } 95/8.5 \approx 11.18.} \\ = 11.18 \text{ dollars/hour.} \end{aligned}\nonumber That is, his salary rate is 11.18 dollars per hour. Exercise Frannie works 5.5 hours and receives120 for her efforts. What is her hourly salary rate? Round your answer to the nearest penny.
$21.82 per hour Example 6 One automobile travels 422 miles on 15 gallons of gasoline. A second automobile travels 354 miles on 13 gallons of gasoline. Which automobile gets the better gas mileage? Solution Decimal division (rounded to the nearest tenth) reveals the better gas mileage. In the case of the first automobile, we get the following rate: $\frac{422 \text{ mi}}{15 \text{gal}}\nonumber$ Divide. To the nearest tenth, 28.1. In the case of the second autombile, we get the following rate: $\frac{354 \text{ mi}}{13 \text{ gal}}\nonumber$ Divide. To the nearest tenth, 27.2. In the case of the first automobile, the mileage rate is 28.1 mi/1 gal, which can be read “28.1 miles per gallon.” In the case of the second automobile, the mileage rate is 27.2 mi/1 gal, which can be read “27.2 miles per gallon.” Therefore, the first automobile gets the better gas mileage. Exercise Alicia works 8 hours and makes$100. Connie works 10 hours and makes \$122. Which woman works at the larger hourly rate?
Alicia
## Exercises
In Exercises 1-24, express the given ratio as a fraction reduced to lowest terms.
1. 0.14 : 0.44
2. 0.74 : 0.2
3. 0.05 : 0.75
4. 0.78 : 0.4
5. 0.1:0.95
6. 0.93 : 0.39
7. $$2 \frac{2}{9}$$ : $$1 \frac{1}{3}$$
8. $$3 \frac{2}{3} : 2 \frac{4}{9}$$
9. 0.36 : 0.6
10. 0.58 : 0.42
11. 15 : 21
12. 77 : 121
13. $$2 \frac{8}{9}$$ : $$2 \frac{2}{3}$$
14. $$1 \frac{2}{3}$$ : $$3 \frac{8}{9}$$
15. $$3 \frac{8}{9}$$ : $$2 \frac{1}{3}$$
16. $$1 \frac{5}{9}$$ : $$1 \frac{1}{3}$$
17. $$2 \frac{5}{8}$$ : $$1 \frac{3}{4}$$
18. $$2 \frac{4}{9}$$ : $$1 \frac{1}{3}$$
19. 10 : 35
20. 132 : 84
21. 9 : 33
22. 35 : 10
23. 27 : 99
24. 12 : 28
25. One automobile travels 271.8 miles on 10.1 gallons of gasoline. A second automobile travels 257.9 miles on 11.1 gallons of gasoline. Which automobile gets the better gas mileage?
26. One automobile travels 202.9 miles on 13.9 gallons of gasoline. A second automobile travels 221.6 miles on 11.8 gallons of gasoline. Which automobile gets the better gas mileage?
27. Todd is paid 183 dollars for 8.25 hours work. What is his hourly salary rate, rounded to the nearest penny?
28. David is paid 105 dollars for 8.5 hours work. What is his hourly salary rate, rounded to the nearest penny?
29. An automobile travels 140 miles in 4 hours. Find the average rate of speed.
30. An automobile travels 120 miles in 5 hours. Find the average rate of speed.
31. Judah is paid 187 dollars for 8 hours work. What is his hourly salary rate, rounded to the nearest penny?
32. Judah is paid 181 dollars for 8.75 hours work. What is his hourly salary rate, rounded to the nearest penny?
33. One automobile travels 234.2 miles on 10.8 gallons of gasoline. A second automobile travels 270.5 miles on 10.8 gallons of gasoline. Which automobile gets the better gas mileage?
34. One automobile travels 297.6 miles on 10.7 gallons of gasoline. A second automobile travels 298.1 miles on 12.6 gallons of gasoline. Which automobile gets the better gas mileage?
35. An automobile travels 180 miles in 5 hours. Find the average rate of speed.
36. An automobile travels 220 miles in 5 hours. Find the average rate of speed.
37. Antarctic trek. Seven women on a 562-mile Antarctic ski trek reached the South Pole 38 days after they began their adventure. What was the ladies’ average rate of speed per day? Round your result to the nearest tenth of a mile. Associated Press-Times-Standard 12/31/09 After 562-mile ski trek, seven women reach the South Pole.
1. $$\frac{7}{22}$$
3. $$\frac{1}{15}$$
5. $$\frac{2}{19}$$
7. $$\frac{5}{3}$$
9. $$\frac{3}{5}$$
11. $$\frac{5}{7}$$
13. $$\frac{13}{12}$$
15. $$\frac{5}{3}$$
17. $$\frac{3}{2}$$
19. $$\frac{2}{7}$$
21. $$\frac{3}{11}$$
23. $$\frac{3}{11}$$
25. The first automobile has the better mileage per gallon.
27. 22.18 dollars/hr
29. 35 mi/hr
31. 23.38 dollars/hr
33. The second automobile has the better mileage per gallon.
35. 36 mi/hr
37. 14.8 miles per day
This page titled 6.2: Introduction to Ratios and Rates is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by David Arnold.
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The Sine Rule. C. McMinn. a sin A. b sin B. c sin C. = =. SOH/CAH/TOA can only be used for right-angled triangles. The Sine Rule can be used for any triangle:. C. b. The sides are labelled to match their opposite angles. a. A. B. c. The Sine Rule:. A. Example 1 :.
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The Sine Rule
C. McMinn
a
sinA
b
sinB
c
sinC
= =
SOH/CAH/TOA can only be used for right-angled triangles.
The Sine Rule can be used for any triangle:
C
b
The sides are labelled to match their opposite angles
a
A
B
c
The Sine Rule:
A
Example 1:
Find the length of BC
76º
c
7cm
b
63º
C
x
B
a
a
sinA
c
sinC
=
Draw arrows from the sides to the opposite angles to help decide which parts of the sine rule to use.
x
sin76º
7
sin63º
sin76º ×
=
× sin76º
7
sin63º
x =
×sin76º
x = 7.6 cm
P
Example 2:
Find the length of PR
82º
x
r
q
43º
55º
Q
15cm
R
p
p
sinP
q
sinQ
=
Draw arrows from the sides to the opposite angles to help decide which parts of the sine rule to use.
15
sin82º
x
sin43º
sin43º ×
=
× sin43º
15
sin82º
= x
sin43º ×
x = 10.33 cm
G
B
3.
1.
2.
F
53º
13 cm
41º
x
8.0
35.3
5.5
x
62º
A
x
130º
28º
D
E
5 cm
63º
76º
C
H
26 mm
I
10.7
4.
5.2 cm
5.
x
61º
R
6.
P
37º
66º
57º
10 m
35º
x
5.2
77º
62º
Q
12 cm
6 km
85º
7.
x
6.6
65º
86º
x
6.9
Remember:
• Draw a diagram
• Label the sides
• Set out your working exactly as you have been shown
• Check your answers regularlyand ask for help if you need it
sinA
a
sinB
b
sinC
c
= =
Finding an Angle
The Sine Rule can also be used to find an angle, but it is easier to use if the rule is written upside-down!
Alternative form of the Sine Rule:
C
Example 1:
Find the size of angle ABC
6cm
a
4cm
b
x º
72º
A
B
c
sinA
a
sinB
b
=
Draw arrows from the sides to the opposite angles to help decide which parts of the sine rule to use.
sin72º
6
sin xº
4
=
× 4
sin72º
6
= sin xº
sin xº = 0.634
x = sin-1 0.634 = 39.3º
P
Example 2:
Find the size of angle PRQ
85º
q
7cm
r
x º
R
p
8.2cm
Q
sinP
p
sinR
r
=
sin85º
8.2
sin xº
7
=
× 7
sin85º
8.2
= sin xº
sin xº = 0.850
x = sin-1 0.850 = 58.3º
7.6 cm
1.
2.
3.
82º
105º
6.5cm
47º
5 cm
8.2 cm
66.6°
37.6°
45.5°
8.8 cm
6 cm
5.
6 km
4.
5.5 cm
31.0°
27º
3.5 km
51.1°
5.2 cm
33º
7.
6.
8 m
74º
57.7°
70º
9 mm
9.5 m
92.1°
52.3º
(←Be careful!→)
22.9º
7 mm
Remember:
• Draw a diagram
• Label the sides
• Set out your working exactly as you have been shown
• Check your answers regularlyand ask for help if you need it
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# I Know WHY The World Is So Sick
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posted on Oct, 23 2009 @ 12:22 PM
reply to post by Cythraul
Hi there! Just noticed your topic links in another topic and this one caught my eye. Anyway, going by your theory and by the figures you present, then... I belive there wrong. You say that half the worlds 'souls' (not people) are only 50yrs old. Well, let me state my point. Doing some simple maths. 1000000 people in 50BC. Well, lets asume half is male, and half is female, and then the life expectancy is 30yrs old. Right, id guess that each couple has 2 children. Probably concieved at a young age also. So, 500000 x 2 = 1000000. Now there will be a mortality rate. So lets say 10% so the new number of CURRENT LIVING souls would be 900000. So, now we gotta work out how many 30's (life expectancy) goes into 50,000 that is 1666r. Multiply that by 900000 and we get 14.. Damn calculator! (14.994,000,000,000+??) But i imagine its a billion plus. Anyway, the oldest souls would be 50000yrs old, but then, the next bunch would be 49970 years old. But each time you'd be doubling the amount of living souls for each 30yr period. Now, and to be honest i dont want to explain it, but we have to take into acount the longer lives we live, but also people who have more than on child. And technically, if you die, and reincarnated, then youd still have the same amount of souls living (existing?) on the planet. Unless they end up in Nirvana.. If you like to think like that. I know theres faults in my crap maths, but i think its more realistic. And also not taking into account people who come back as animals, flys, fish, monkeys e.t.c. And maybe, if you added up every Living thing on Earth it would account for the more realistic number of souls i gave earlier...
posted on Oct, 23 2009 @ 12:38 PM
I think the world is in chaos because the societies now reward this:
Lust
Gluttony
Greed
Sloth
Wrath
Envy
Pride
Im not a religious guy, but its hard to ignore the fact that all of these personal traits usually lead to individual success for the person, defined as "getting what you want" (with little regard for others, of course). And the world today is about satisfying the ego.
We are consumers of products, designed to satisfy the ego. This happens in a world run by money where people are running in hamster wheels for money they use to fuel the ego, to be someone. Because inside people are empty, so they need stuff to fill the void.
Right now, I have a neighbor who plays music at loud volume. He doesnt care about anyone else. Its friday. He wants to play loud music so he does. Everybody around him gets affected and cant have a relaxing friday. But he doesnt care. Thats typical of people today. Its all about themselves.
[edit on 23-10-2009 by Copernicus]
posted on Oct, 23 2009 @ 06:03 PM
reply to post by Copernicus
Definately true. Im not religious at all either, but if you look at the way Celebritys are idolised in the morden day.
The way diffrent expectations have changed people and the likes.
For instance, im only 23, i finished school when i was 16 started college and then got a job after that yet find myself later unemployed since May 2008.
What ive noticed is the change in attitudes in kids younger than me. And i dont want to hear the 'their young and rebelious' routine. I just find that they JUST dont care anymore. They have no Morality. No feeling of obligation. Damn, they seem to have no self respect. We were all young once. Im still young too. I know what its like, and a hell of a lot more too, but theres no need to be so 'FUBAR' about it. Ugh. Im ranting now.
posted on Nov, 15 2009 @ 09:05 AM
reply to post by Cythraul
All your logic is completely flawed because of one simple thing. You ASSUME that all the people who were born after the intial million people or so are NEW souls, the thing is, many souls reincarnate many times.
For example.
there was a billion people in 1800 or so, some souls who were born in the early 19th century ,most of them probably took a break of a few centuries before deciding to reincarnate, so they were already born,say in the 9th century. between the 9th and 18th century many new souls as well were born for the first time, and then took their own breaks of a century or more.After the break they decided to incarnate in the 20th century.
SO - most if not all souls have already incanrated here before,numerous times, just not at the same time. GET IT?
Now we are close to HARVEST, around 2012 they say. All the souls or most anyways have decided to incarnate agai nspecifically during this timeperiod in order to graduate possibly.
So really, half of the sousl right now ARE NOT,in any way first-timers, they just took a break,a long one from incarnating,and the population grew steadily due to the fact that the closer we get to HARVEST the more souls decide/decided to incarnate in this timeperiod. They aren't taking a few centuries of rest anymore. That may also be the reason soem souls are quite stressed out,because they didn't get a proper vacation so to speak.This is mirrored in the world as we see it.
believe me, the world has seen darker time than now, right nwo is NOTHING,literally graden of eden compared to WW1 and WW2 and the Crucades, the Napoleon era, Civil wars all aroudn the world in the 18th century, black death wiping out the third of Europe etc.
Relgious conflicts and wars have bene much uglier in the past than now. I'm not saying it's looknig good,but it msot certainly isn't the worst now.Get a grip friend.
And I stress,no,most souls,if not all,are not first-timers at all.
There have been over 100 billion people living on Earth they say,through time, 6.8 billion of them are alive right now. Those 100 billion people are those 7billion or so souls, who are living a human experience today.
posted on Nov, 15 2009 @ 09:25 AM
I like it. no, I love it. Thank you.
posted on Nov, 15 2009 @ 09:28 AM
reply to post by Copernicus
Exactly. All allowed because our society currently encourages me, me, me, me thinking with various excuses that amount to absolutely no accountability for one's actions *everyone's a victim, etc*.
posted on Nov, 15 2009 @ 09:30 AM
We are all one, come from oneness go back to oneness.
Souls, separation, my soul is older than your soul, spiritually wiser is all just Ego talking.
People/Ego's need to feel better than others somehow, this is just the separation from Oneness.
posted on Nov, 15 2009 @ 09:37 AM
reply to post by Realtruth
And THAT is erring in the opposite direction that is currently causing civilization to crash which in due time would also cause civilization to crash. We are not borg. We are not a collective. There must be a balance, as with nature. We are parts of a greater whole it is true but we are the whole it's self. There is a degree of seperation.
posted on Nov, 15 2009 @ 09:40 AM
reply to post by Watcher-In-The-Shadows
Just the illusion of separation, that is why their is so much chaos, people trying to recreate something that is already whole and perfect.
posted on Nov, 15 2009 @ 09:45 AM
reply to post by Realtruth
You forget the basic rule of nature is balance. At the very base all things seek an equilibrum *aka balance*. And saying that "All is one" is merely going to the opposite extreme than the side society trends towards today.
And what exactly is "perfect"? Hmm? That word is a deeply subjective word.
[edit on 15-11-2009 by Watcher-In-The-Shadows]
posted on Nov, 15 2009 @ 09:46 AM
Originally posted by Watcher-In-The-Shadows
reply to post by Copernicus
Exactly. All allowed because our society currently encourages me, me, me, me thinking with various excuses that amount to absolutely no accountability for one's actions *everyone's a victim, etc*.
Thinking is the key to all the madness and separation, without is everything just is.
A good book to read is, Eckhart Tolle, "Awaken to your life's purpose"
Otherwise, meditation and calming the mind will allow you to experience where we are all from.
Again thinking is the trap.
posted on Nov, 15 2009 @ 09:47 AM
reply to post by Realtruth
And encouraging people to not think is an excellent way to control them.
posted on Nov, 15 2009 @ 09:48 AM
reply to post by Cythraul
look at it this way>
all the 7 billion or so souls have at various points in the last 52 thousand years or so incarnated ALREADY. They say over 100 billion people have lived during the human existence-homo sapiens I think.
And only a small amount of those souls were on earth at ANY GIVEN time period at THE SAME TIME. When some died within the body(no 1), then other souls(no 2) would be born on Earth. When these souls died(no 2)...then other souls yet again(no 3) would be born...And when these souls died(no 3),then those who died quite a while back(no 1),would be born(no 1) again.
So you see, they weren't NEW, they were re-occuring,they were the same ones,just after a pause. And as the population grew,more and more souls,all of whom at various points had already incarnated in the past thousands of years started to reincarnate AT THE SAME TIMEPERIOD,which meant MORE PEOPLE living AT THE SAME TIME.
I made it really clear hoefully.So i would like to hear your thougts on this OP and others too.
peace
posted on Nov, 15 2009 @ 09:49 AM
Your right words are just labels and mean nothing, because they are created by the same madness that is meant to explain itself.
Just life a religion. Circular argumentation.
If people talk long enough everyone is confused.
Originally posted by Watcher-In-The-Shadows
And what exactly is "perfect"? Hmm? That word is a deeply subjective word.
[edit on 15-11-2009 by Watcher-In-The-Shadows]
posted on Nov, 15 2009 @ 09:52 AM
reply to post by Realtruth
Incidently, potential to go wrong does not mean it will go wrong. Also, if it did go wrong it does not mean it must go wrong. Which is the fallacy I see a great many of those of the "oneness collective" *as I call them* seem to be guilty of.
posted on Nov, 15 2009 @ 09:55 AM
reply to post by Watcher-In-The-Shadows
Your arguing in circles.
Explaining or trying to explain something that cannot be explained with logic.
It's like a religion, you either have faith or you don't. Believe or not. Aware or not aware of.
[edit on 15-11-2009 by Realtruth]
posted on Nov, 15 2009 @ 10:00 AM
Originally posted by Watcher-In-The-Shadows
reply to post by Realtruth
Incidently, potential to go wrong does not mean it will go wrong. Also, if it did go wrong it does not mean it must go wrong. Which is the fallacy I see a great many of those of the "oneness collective" *as I call them* seem to be guilty of.
There is no right and wrong. That is an illusion of the human, separation is there for one to experience existence as a separate being. We ineed are all ONE conscious eternal intelligent energy experiencing itself subjectively therefore giving an illusion of separateness. When you claim you want to be an individual,you want to have a mind of your own etc...then your thinking obviously stems from a separateness point of view, you see the world as separate individual beings. That is true within the illusion, but that illusion was and is created by YOU YOURSELF, YOU are all that is, you actually sent out many parts of yourself to experience YOURSELF.
And one of those parts is the individual you are living as right now, that person that typed the post I'm replying to right now, that is a part of that Oneness....Oneness being YOU, so in essence, YOU "separated" yourself into many parts within this illusion and the person, Valeri, that is replying to you right now is also a part of that wholeness, a part of YOU.
Whether you feel it to be true or not is only a matter of "time"(also an illusion), because there will come a "time", maybe not in this reincarnation of yours, where you will realise what I'm saying is indeed true.
But don't take my word for it, ALTHOUGH I am another part of YOU(just like the michael Jackson hit-song goes)
[edit on 11/15/2009 by Valeri]
posted on Nov, 15 2009 @ 10:02 AM
Originally posted by Watcher-In-The-Shadows
reply to post by Realtruth
And encouraging people to not think is an excellent way to control them.
That argument can go both ways. People can be control in many ways very easily even when they think.
Look at all the smart people today that are up to their arse in debt by choice, were they thinking when they did this? Of course they were, but their Ego's over rode their thought process.
Ask yourself this question.
Are you driven by your Ego or is your Ego making the majority of your decisions?
[edit on 15-11-2009 by Realtruth]
posted on Nov, 15 2009 @ 10:04 AM
reply to post by Valeri
Thank you for the high horsed response with subjective opinion dressed up as fact.
posted on Nov, 15 2009 @ 10:05 AM
reply to post by Realtruth
No actually I am not. I am merely pointing out the flaws in your statements.
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# enme110_Lecture_6 - ENMEI 10-Staties POP QUIZ#1 Name...
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Unformatted text preview: ENMEI 10-Staties POP QUIZ #1 Name: Instructor: Panes G. Charatambides Student i.d. September 20, 201 I (a) Determine the cartesian components of the force shown. (50 points) (b) Determine the angle a between the direction of the force and the x-axis. (25 points) (0) Determine component of the force projected along the line CD. (25 points) H3 I: 800 N a ’ J r I a p ¢ r p : , l a I A(3,Q”}wb 3(0,a9)w\ Qfaégflvfi D(%3;fiWL "Wmmmm_mwmflmeWTmmmmmmmm n A I QODCfiSA%YHE)UEL%MALfiwfimfi / 7/4 c: 2,, 7 _. _ 9A _ (/75 ,, flFé‘ffi’ ,4?» 32;? 71.719??? 711 MW? _. ,, , . . ,, ,, ,, , Ana/z"; " " " “ ' ____________________________________________________________________ ya ,, _. . ._ -- "/3; M ‘ ~ 7' ' ' is 0 gain”— imflrw?” ‘1 \«WW : ’ ._ ...
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# CS215 - Lec 10 b trees and hashing
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Insert elements into a Btree
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### Transcript
• 1. Multilevel Index in form of B trees. Hashing Dr. Hussien M. Sharaf 2
• 2. Dr. Hussien M. Sharaf 3 When an Index gets very big, it can not be stored in RAM. It should be stored on file, hence another level of index that can be loaded into memory is required. Hence we need multilevel of indexing.
• 3. Dr. Hussien M. Sharaf 4 B-tree Allows searches, insertions, and deletions in O(Log n) . The B-tree is a generalization of a binary search tree. A Btree of order m is where each node contains (m) keys and (m+1) pointers for children. It is a balanced tree. Each node is a sorted index of keys.
• 4. Dr. Hussien M. Sharaf 5 B-tree Example
• 5. Dr. Hussien M. Sharaf 6 Adding Records to a B Tree If global root still have empty slots then place the new record in sorted position in the appropriate root node.
• 6. Dr. Hussien M. Sharaf 7 Adding Records to a B Tree If current node is full then: 1. Split the leaf node. 2. Place median key in the parent node in sorted order. 3. Records with keys < middle key go to the left leaf node. 4. Records with keys >= middle key go to the right leaf node.
• 7. Dr. Hussien M. Sharaf 8 Adding Records to a B Tree If parent node is full then: 1. Split the parent node. 2. keys < middle key go to the left parent node. 3. key s >= middle key go to the right parent node. 4. The median key goes to the next (higher level) index. 5. IF the next level index node is full, continue splitting the index nodes.
• 8. Dr. Hussien M. Sharaf 9 Example for order = 4
• 9. Dr. Hussien M. Sharaf 10 Inserting X
• 10. Dr. Hussien M. Sharaf 11 Inserting X
• 11. Dr. Hussien M. Sharaf 12 Example of Insertions
• 12. Dr. Hussien M. Sharaf 13 Example for order = 4
• 13. Dr. Hussien M. Sharaf 14 Example for order = 4
• 14. Dr. Hussien M. Sharaf 15 Example for order = 4
• 15. Dr. Hussien M. Sharaf 16 Example for order = 4
• 16. Dr. Hussien M. Sharaf 17 Example for order = 4
• 17. Dr. Hussien M. Sharaf 18 Example for order = 4
• 18. Dr. Hussien M. Sharaf 19 Example for order = 4
• 19. Dr. Hussien M. Sharaf 20 Example for order = 4
• 20. Dr. Hussien M. Sharaf 21 Example for order = 4
• 21. Dr. Hussien M. Sharaf 22 Example for order = 4
• 22. Dr. Hussien M. Sharaf 23 Properties of B trees
• 23. Dr. Hussien M. Sharaf 24
• 24. A hash table is an effective data structure for implementing dictionaries. A dictionary is an abstract data type composed of a collection of (key,value) pairs, such that each possible key appears at most once in the collection. Dr. Hussien M. Sharaf 25
• 25. A hash table uses a hash function as an algorithm to map identifying values, known as keys (e.g., a person's name), to their associated values (e.g., their telephone number). Dr. Hussien M. Sharaf 26
• 26. 1. Direct-Address tables. 2. Hash tables. 3. Hash functions. 4. Open addressing. 5. Perfect hashing. Dr. Hussien M. Sharaf 27
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Home / Volume Conversion / Convert Gigaliter to Minim (UK)
# Convert Gigaliter to Minim (UK)
Please provide values below to convert gigaliter [GL] to minim (UK), or vice versa.
From: gigaliter To: minim (UK)
### Gigaliter to Minim (UK) Conversion Table
Gigaliter [GL]Minim (UK)
0.01 GL168936382693.7 minim (UK)
0.1 GL1689363826937 minim (UK)
1 GL16893638269370 minim (UK)
2 GL33787276538740 minim (UK)
3 GL50680914808110 minim (UK)
5 GL84468191346850 minim (UK)
10 GL1.689363826937E+14 minim (UK)
20 GL3.378727653874E+14 minim (UK)
50 GL8.446819134685E+14 minim (UK)
100 GL1.689363826937E+15 minim (UK)
1000 GL1.689363826937E+16 minim (UK)
### How to Convert Gigaliter to Minim (UK)
1 GL = 16893638269370 minim (UK)
1 minim (UK) = 5.9193880208333E-14 GL
Example: convert 15 GL to minim (UK):
15 GL = 15 × 16893638269370 minim (UK) = 2.5340457404055E+14 minim (UK)
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# How do you balance CO(g) + I_2O_5(s) -> I_2(s) + CO_2(g)?
Apr 4, 2016
$5 {\text{CO"_ ((g)) + "I"_ 2"O"_ (5(s)) -> "I"_ (2(s)) + 5"CO}}_{2 \left(g\right)}$
#### Explanation:
You're actually dealing with a redox reaction used to convert carbon monoxide, $\text{CO}$, to carbon dioxide, ${\text{CO}}_{2}$, by using what is known as the Schutze reagent, which is iodine pentoxide, ${\text{I"_2"O}}_{5}$, on silica gel.
Iodine pentoxide acts as an oxidizing agent here and oxidizes carbon monoxide to carbon dioxide while being reduced to iodine in the process.
Assign oxidation numbers to the atoms that take part in the reaction - for the sake of simplicity, I will not add the states of the chemical species involved in the reaction
${\stackrel{\textcolor{b l u e}{+ 2}}{\text{C") stackrel(color(blue)(-2))("O") + stackrel(color(blue)(+5))("I")_2 stackrel(color(blue)(-2))("O")_5 -> stackrel(color(blue)(0))("I")_2 + stackrel(color(blue)(+4))("C")stackrel(color(blue)(-2))("O}}}_{2}$
You can balance this equation by assuming that the reaction takes place in acidic conditions (you will get the same result if you assume basic conditions).
So, carbon's oxidation state goes from $\textcolor{b l u e}{+ 2}$ on the reactants' side, to $\textcolor{b l u e}{+ 4}$ on the products' side, which of course means that it's being oxidized.
The oxidation half-equation will look like this
stackrel(color(blue)(+2))("C")"O" -> stackrel(color(blue)(+4))("C")"O"_2 + 2"e"^(-)
Here each carbon atoms loses two electrons. Balance the oxygen atoms by adding water molecules to the side that needs oxygen, and the hydrogen atoms by adding protons, ${\text{H}}^{+}$, to the side that needs hydrogen.
${\text{H"_2"O" + stackrel(color(blue)(+2))("C")"O" -> stackrel(color(blue)(+4))("C")"O"_2 + 2"e"^(-) + 2"H}}^{+}$
The oxidation state of iodine goes from $\textcolor{b l u e}{+ 5}$ on the reactant's side, to $\textcolor{b l u e}{0}$ on the products' side, which means that it's being reduced.
The reduction half-reaction will be
stackrel(color(blue)(+5))("I")_2"O"_5 + 10"e"^(-) -> stackrel(color(blue)(0))"I"_2
Here each iodine atom gains five electrons, so two atoms will gain a total of ten electrons. Once again, balance the oxygen and hydrogen to get
$10 \text{H"^(+) + stackrel(color(blue)(+5))("I")_2"O"_5 + 10"e"^(-) -> stackrel(color(blue)(0))"I"_2 + 5"H"_2"O}$
As you know, in any redox reaction, the number of electrons lost in the oxidation half-equation must be equal to the number of electrons gained in the reduction half-equation.
In order to get these two balanced, multiply the oxidation half-equation by $5$. Add the two half-equations to get
$\left\{\begin{matrix}\text{H"_2"O" + stackrel(color(blue)(+2))("C")"O" -> stackrel(color(blue)(+4))("C")"O"_2 + 2"e"^(-) + 2"H"^(+) | xx 5 \\ 10 "H"^(+) + stackrel(color(blue)(+5))("I")_2"O"_5 + 10"e"^(-) -> stackrel(color(blue)(0))"I"_2 + 5"H"_2"O}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a}}$
$\textcolor{red}{\cancel{\textcolor{b l a c k}{5 \text{H"_2"O"))) + 5"CO" + color(red)(cancel(color(black)(10"H"^(+)))) + "I"_2"O"_5 + color(red)(cancel(color(black)(10"e"^(-)))) -> 5"CO"_2 + color(red)(cancel(color(black)(10"e"^(-)))) + "I"_2 + color(red)(cancel(color(black)(10"H"^(+)))) + color(red)(cancel(color(black)(5"H"_2"O}}}}$
The balanced chemical equation for this reaction will thus be - state symbols included!
$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5 {\text{CO"_ ((g)) + "I"_ 2"O"_ (5(s)) -> "I"_ (2(s)) + 5"CO}}_{2 \left(g\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
As you can see, the reaction must be balanced in neutral conditions, so the result will be the same regardless if you pick acidic or basic conditions.
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# maths
posted by on .
Let X and Y be positive integers such that X^2 + 3X + Y^2 = 404.
What is the value of X + Y^3?
Find all possible solutions.
Thanks! :)
• maths - ,
I ran a simple GW-Basic program
10 FOR X = 1 TO 25
20 FOR Y = 1 TO 25
30 IF (X*X + 3*X + Y*Y) = 404 THEN PRINT X,Y
40 NEXT Y
50 NEXT X
and it gave me values of x and y that worked, they were ...
1 20
13 14
16 10
17 8
so x + y^3 could be 8001, 2757, or 81
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# Sec 1-2 Concept: Using Segments and Congruence
## Presentation on theme: "Sec 1-2 Concept: Using Segments and Congruence"— Presentation transcript:
Sec 1-2 Concept: Using Segments and Congruence
Obj: given the segment addition postulate, find the length of a segment as measured by a s.g.
Postulate 1: Ruler Postulate
The distance between points A and B, written as AB, is the absolute value of the difference between the coordinates of A and B AB is also called the length of AB AB A B X2 x1 AB = l x2-x1 l
Example 1: A B Find the length of AB AB = l 2-(-3) l = l 5 l = 5 1 2 3
1 2 3 -1 -2 -3 AB = l 2-(-3) l = l 5 l = 5
If B is between A and C, then AB+BC = AC. If AB+BC =AC, then B is between A and C. A B C
Their houses are 692 yds apart
Example 2: Two friends leave their homes and walk in a straight line towards the other’s home. When they meet one has walked 425 yds and the other has walked 267 yds. How far apart are their homes? A B C 267 425 AB + BC = AC = AC 692 = AC Their houses are 692 yds apart
Example 3: In the diagram of collinear points, GK=24, HJ = 10, and GH=HI=IJ. Find each length
JK = IG = IK = 5 5 5 9 10 14
Example 4: Suppose M is between L and N
Example 4: Suppose M is between L and N. Use the segment addition postulate to solve for the variable. Then find the lengths of LM, MN and LN LM = 3x+8 MN = 2x-5 LN =23 LM = 7y +9 MN = 3y + 4 LN = 143 143 23 L M N L M N 7y+9 3y+4 3x+8 2x-5 (7y+9) + (3y+4) = 143 10y + 13 =143 10y = 130 y=13 (3x+8) + (2x-5) = 23 5x+3=23 5x=20 X=4 MN=3y+4 3(13)+4 43 LM = 3x+8 3(4)+8 20 MN = 2x-5 2(4)-5 3 LM=7y+9 7(13)+9 100
A(0,1), B(4,1), C(1,2), D(1,6); AB and CD?
Example 5: Plot the given points in a coordinate plane. Then determine whether the line segments named are congruent. A(0,1), B(4,1), C(1,2), D(1,6); AB and CD? D AB = 4 CD=4 Yes they are congruent C A B
Today’s Work In Class: Homework:
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# How much power does a 20kW solar system produce?
## How much power does a 20kW solar system produce?
20,000 Watts
A 20kW Solar Kit requires around 1,150 square feet of space. 20kW or 20 kilowatts is 20,000 Watts of DC direct current power. This could produce an estimated 2000 to 3000 kWh (kilowatt hours) of alternating current (AC) power per month, assuming at least 5 sun hours per day with the solar array facing South.
How many kW system do I need for solar?
How many solar panels do I need for my house? System size comparison
System size Number of panels needed Estimated annual production
6 kW 19 9,600 kWh
8 kW 25 12,800 kWh
10 kW 32 16,000 kWh
12 kW 38 19,200 kWh
Can solar energy produce electricity 24 hours a day?
The solar power company SolarReserve is making waves in the solar energy industry. Its concentrated solar plant, Crescent Dunes in Nevada, has become the world’s first solar power plant that can continuously supply power for 24 hours a day.
### How much is a 20 panel solar system?
As of July 2021, the average cost of solar in the U.S. is \$2.76 per watt – that comes out to about \$55,200 for a 20 kW system. That means that the total cost for a 20 kW solar system would be \$40,848 after the federal solar tax credit discount (not factoring in any additional state rebates or incentives).
How much money can a 1 megawatt solar farm make?
Based on the national average of four peak sun hours per day, we know that the average 1 MW solar farm would make 1,460 MWh per year. That means that the average 1 MW solar farm can expect an annual revenue of roughly \$40,000 per year.
How many hours does solar power last?
Average peak sun hours by state
Location Peak Sun Hours (PSH)
Arkansas 3.5 – 4
California 5 – 7.5
Colorado 5 – 6.5
Connecticut 3
#### Can solar energy be used at night?
The quick answer is yes. The energy captured from the sun on bright days is essentially stored in your system, to be used during periods of darkness, on overcast days or even for stretches of dreary rainy weather.
How big is a 20kW solar power plant?
Specification of 20kW Off-Grid Solar System: Particulars Description Solar Power Plant 20 kW Solar Panel in Watt 335 watt Solar Panel Qty 60 nos. Off-Grid Solar Inverter 20 KVA
Is there net metering for a 20kW Solar System?
Solar Net-Metering: No, Solar Net-Metering is not applicable on this system. If your place facing frequent power cut problems then this system is suitable for you. 20kW hybrid solar system is combination of on-grid and off-grid (with battery) solar system with more power.
## Can a 20kW Solar System be off grid?
If your place facing frequent power cut problems then this system is suitable for you. 20kW hybrid solar system is combination of on-grid and off-grid (with battery) solar system with more power. It will work with government grid and also without government grid, with additional feature of battery backup.
What is the ROI of a 20kW Solar System?
This solar on grid system can generate 80 units/day or 2400 units/month as an average over the year. It requires 120 square meter + shadow gap area for installation. You will get a full ROI (return on investment) within 3 to 5 years. 1 Set. 1 Set. 120 Sq. Mt Other models of 20kW solar system are also available. Read more about:
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# KingProjectedAnalytic1D¶
class astropy.modeling.functional_models.KingProjectedAnalytic1D(amplitude=1, r_core=1, r_tide=2, **kwargs)[source]
Projected (surface density) analytic King Model.
Parameters
amplitudefloat
Amplitude or scaling factor.
r_corefloat
Core radius (f(r_c) ~ 0.5 f_0)
r_tidefloat
Other Parameters
fixeda dict, optional
A dictionary {parameter_name: boolean} of parameters to not be varied during fitting. True means the parameter is held fixed. Alternatively the fixed property of a parameter may be used.
tieddict, optional
A dictionary {parameter_name: callable} of parameters which are linked to some other parameter. The dictionary values are callables providing the linking relationship. Alternatively the tied property of a parameter may be used.
boundsdict, optional
A dictionary {parameter_name: value} of lower and upper bounds of parameters. Keys are parameter names. Values are a list or a tuple of length 2 giving the desired range for the parameter. Alternatively, the min and max properties of a parameter may be used.
eqconslist, optional
A list of functions of length n such that eqcons[j](x0,*args) == 0.0 in a successfully optimized problem.
ineqconslist, optional
A list of functions of length n such that ieqcons[j](x0,*args) >= 0.0 is a successfully optimized problem.
Notes
This model approximates a King model with an analytic function. The derivation of this equation can be found in King ‘62 (equation 14). This is just an approximation of the full model and the parameters derived from this model should be taken with caution. It usually works for models with a concentration (c = log10(r_t/r_c) paramter < 2.
Model formula:
$f(x) = A r_c^2 \left(\frac{1}{\sqrt{(x^2 + r_c^2)}} - \frac{1}{\sqrt{(r_t^2 + r_c^2)}}\right)^2$
References
Rc606295f1bc9-1
Examples
import numpy as np
from astropy.modeling.models import KingProjectedAnalytic1D
import matplotlib.pyplot as plt
plt.figure()
rt_list = [1, 2, 5, 10, 20]
for rt in rt_list:
r = np.linspace(0.1, rt, 100)
mod = KingProjectedAnalytic1D(amplitude = 1, r_core = 1., r_tide = rt)
sig = mod(r)
plt.loglog(r, sig/sig[0], label='c ~ {:0.2f}'.format(mod.concentration))
plt.xlabel("r")
plt.ylabel(r"$\sigma/\sigma_0$")
plt.legend()
plt.show()
(png, svg, pdf)
Attributes Summary
amplitude concentration Concentration parameter of the king model input_units This property is used to indicate what units or sets of units the evaluate method expects, and returns a dictionary mapping inputs to units (or None if any units are accepted). param_names r_core r_tide
Methods Summary
evaluate(x, amplitude, r_core, r_tide) Analytic King model function. fit_deriv(x, amplitude, r_core, r_tide) Analytic King model function derivatives.
Attributes Documentation
amplitude = Parameter('amplitude', value=1.0, bounds=(1.1754943508222875e-38, None))
concentration
Concentration parameter of the king model
input_units
This property is used to indicate what units or sets of units the evaluate method expects, and returns a dictionary mapping inputs to units (or None if any units are accepted).
Model sub-classes can also use function annotations in evaluate to indicate valid input units, in which case this property should not be overridden since it will return the input units based on the annotations.
param_names = ('amplitude', 'r_core', 'r_tide')
r_core = Parameter('r_core', value=1.0, bounds=(1.1754943508222875e-38, None))
r_tide = Parameter('r_tide', value=2.0, bounds=(1.1754943508222875e-38, None))
Methods Documentation
static evaluate(x, amplitude, r_core, r_tide)[source]
Analytic King model function.
static fit_deriv(x, amplitude, r_core, r_tide)[source]
Analytic King model function derivatives.
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# Illustration Graph: AC voltage and current at capacitor
Get illustration
Share — copy and redistribute the material in any medium or format
Adapt — remix, transform, and build upon the material for any purpose, even commercially.
Sharing and adapting of the illustration is allowed with indication of the link to the illustration.
If an AC voltage is applied to a capacitor with capacitance $$C$$, the same AC voltage $$U_{\text C}(t)$$ is applied to this capacitor. The AC current $$\class{red}{I_{\text C}}(t)$$ through the capacitor is 90 degrees out of phase with respect to the voltage.
We say:
The current into the capacitor precedes the voltage by 90 degrees!
On the graph you can see that the current maximum occurred at an earlier time, if you go 90 degrees to the left. Further to the left means earlier in time!
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# Thread: Show that A=(1/2)bc sin alpha..
1. ## Show that A=(1/2)bc sin alpha..
Show that A=(1/2)bc sin alpha..
How should this be done?
2. ## Re: Show that A=(1/2)bc sin alpha..
Are you asking for the proof of this formula or is this in context of a bigger problem?
3. ## Re: Show that A=(1/2)bc sin alpha..
Originally Posted by JamesBond16
Show that A=(1/2)bc sin alpha..
How should this be done?
I guess, you talking about formula of triangle area.
Say, that $b$ is base of your triangle.
What can you tell about $c\cdot\sin{\alpha}$, what is this measures?
4. ## Re: Show that A=(1/2)bc sin alpha..
Originally Posted by Also sprach Zarathustra
I guess, you talking about formula of triangle area.
[snip]
If that's the real question, then the OP should either use Google or refer to his/her textbook to find the proof.
@OP: MHF does not re-invent the wheel. That's why Google was invented.
5. ## Re: Show that A=(1/2)bc sin alpha..
I am looking for the proof.
6. ## Re: Show that A=(1/2)bc sin alpha..
Originally Posted by JamesBond16
I am looking for the proof.
Then the advice in post #4 is pertinent.
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# Number 4140042220
### Properties of number 4140042220
Cross Sum:
Factorization:
2 * 2 * 5 * 207002111
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
f6c407ec
Base 32:
3rc81vc
sin(4140042220)
-0.46990529545725
cos(4140042220)
0.88271683642108
tan(4140042220)
-0.53233979014432
ln(4140042220)
22.143971822799
lg(4140042220)
9.6170047700628
sqrt(4140042220)
64343.15985402
Square(4140042220)
1.7139949583383E+19
### Number Look Up
Look Up
4140042220 which is pronounced (four billion one hundred forty million forty-two thousand two hundred twenty) is a amazing figure. The cross sum of 4140042220 is 19. If you factorisate the number 4140042220 you will get these result 2 * 2 * 5 * 207002111. 4140042220 has 12 divisors ( 1, 2, 4, 5, 10, 20, 207002111, 414004222, 828008444, 1035010555, 2070021110, 4140042220 ) whith a sum of 8694088704. The figure 4140042220 is not a prime number. The number 4140042220 is not a fibonacci number. The number 4140042220 is not a Bell Number. 4140042220 is not a Catalan Number. The convertion of 4140042220 to base 2 (Binary) is 11110110110001000000011111101100. The convertion of 4140042220 to base 3 (Ternary) is 101200112012221021001. The convertion of 4140042220 to base 4 (Quaternary) is 3312301000133230. The convertion of 4140042220 to base 5 (Quintal) is 31434322322340. The convertion of 4140042220 to base 8 (Octal) is 36661003754. The convertion of 4140042220 to base 16 (Hexadecimal) is f6c407ec. The convertion of 4140042220 to base 32 is 3rc81vc. The sine of 4140042220 is -0.46990529545725. The cosine of the figure 4140042220 is 0.88271683642108. The tangent of 4140042220 is -0.53233979014432. The root of 4140042220 is 64343.15985402.
If you square 4140042220 you will get the following result 1.7139949583383E+19. The natural logarithm of 4140042220 is 22.143971822799 and the decimal logarithm is 9.6170047700628. that 4140042220 is unique figure!
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# Is my proof for showing $P(X=k|Y=n)$ follows $B(n,1/2$) sufficient? with $Y$ a Poisson law, $X$ modelling a number of iid events
I've already worked on a couple of random variables with their value in $\mathbb{N}$ $(X,Y)$
$p_{kn}=P(X=k,Y=n)= \left( \frac{λ^ke^{-y} α^n(1-α)^{k-n}}{n!(k-n)!} \right)\mathbb{1}_{\{0\le n\le k}\} (k,n)$
I had to find the law of X, of Y, trying to show that X and Y are independent...
Here is the point: the assumption is that the number of children for a random family follows a Poisson law with $E(Y)=2$. Assuming that for every birth, the probability to observe a boy's birth is equal to $1/2$. Show that, conditionally to the fact that the family has n children, the boys number follows a a Binomial law $B(n,1/2)$
I answered that conditionally to the fact that a family has $Y=n$ children, Y being the random variable representing the possible outcomes of the number of children inside a family, the boys number ,represented by X is a repetition of independent events identically distributed following a Bernouilli law with the probability $p=1/2$.
Thus $P(X=k|Y=n)~B(n,1/2)$ is the number of boys in a family conditonaly on the fact that it has n children.
But is that answer sufficient to prove P(X=k|Y=n)~B(n,1/2)?
Yes, that would be sufficient to argue that $X\mid Y=n \sim \mathcal {Bin}(n, \alpha)$.
However, the expression you provided is not quite right. Where did the (lowercase) $y$ term come from? Why are you raising $\lambda$ to the power of $k$? You should have obtained:
\begin{align} p_{k,n;\alpha,\lambda} & = \mathsf P(X=k, Y=n; \alpha, \lambda) \\[1ex] & = \mathsf P(X=k \mid Y=n; \alpha)\cdot\mathsf P(Y=n; \lambda) \\[1ex] & = \binom{n}{k}\alpha^k(1-\alpha)^{n-k}\cdot\frac{\lambda^n \mathsf e^{-\lambda}}{n!}\;\mathbf 1_{0\leq k\leq n} \\[1ex] & = \frac{\alpha^k(1-\alpha)^{n-k}\lambda^n\mathsf e^{-\lambda}}{k!\,(n-k)!}\;\mathbf 1_{0\leq k\leq n} \end{align}
Where $\alpha = \tfrac 1 2$ is the success rate (of obtaining a boy each birth), and $\lambda=2$ is the average rate of births.
PS: You can now show that $X$ and $Y$ are not independent.
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# MCQ of Physics CUET | Common University Entrance Test |
4.5
(8)
MCQ of Physics CUET | Common University Entrance Test |
#### MCQ of Electric Charges and Fields
Question 1: What happens when a point charge is kept at the center of a insulated metallic spherical shell?
1. Electric field outside the sphere is zero
2. Electric potential inside the sphere is zero
3. Net induced charge on the sphere is zero
4. Electric field inside the sphere is zero
Answer: D ( Electric field inside the sphere is zero )
Question 2: Two charges of equal magnitudes and separated by a distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge will be
1. F / 16
2. F / 4
3. 4 F
4. F / 16
Answer: A ( F / 16)
Question 3: The cause of quantization of electric charge is
1. Transfer of electrons
2. Transfer of protons
3. Transfer of integral number of electrons
4. None of the above
Answer: C (Transfer of integral number of electrons)
Question 4: Charge on a body is integral multiple of ±e . It is given by law of
1. Conservation of charge
2. Conservation of mass
3. Conservation of energy
4. Quantisation of charge
Question 5: In the process of charging, the mass of the negatively charged body
1. Increases
2. Decreases
3. Remains constant
4. None of the above
Question 6: Charge Q is kept in a sphere of 5 cm of first and then it is kept in a cube of side 5 cm, the outgoing flux will be
1. More in case of sphere
2. More in case of cube
3. Same in both cases
4. Information incomplete
Answer: C (Same in both cases)
Question 7: If Electric field is along X direction and the work done in moving a charge of 0.2 C through a distance of 2 m along a line making an angle of 60 º with X axis is 4 J, then what is the value of electric field?
1. √3 NC-1
2. 4 NC-1
3. 5 NC-1
4. 20 NC-1
Answer: D ( 20 NC-1 )
Question 8: Which of the following statement is true about electrical forces ?
1. Electrical forces are produced by electrical charges
2. Like charges attract and like charges repel
3. Electric Forces are weaker than gravitational forces
4. Negative charges can combined to produce a third type of charge
Answer: A (Electrical forces are produced by electrical charges )
Question 9: Which of the following statement is not a similarity between electrostatic and gravitational forces?
1. Both forces obey inverse square law
2. Both forces operate over very large distances
3. Both forces are conservative in nature.
4. Both forces are attractive in nature always
Answer: D ( Both forces are attractive in nature always )
Question 10: The constant k in Coulomb’s law depends on
1. Nature of medium
2. System of units
3. Intensity of charge
4. Both (A) and (B)
Answer: A ( Nature of medium )
Question 11: If the charge on an object is doubled then electric field becomes
1. Half
2. Double
3. Unchanged
4. Thrice
Question 12: Which of the following statements is not true about electric field lines?
1. Electric field lines start from positive charge and end at negative charge.
2. Two electric field lines can never cross each other
3. Electrostatic field lines do not form any closed loop
4. Electric field lines cannot be taken as continuous curve
Answer: D ( Electric field lines cannot be taken as continuous curve )
Question 13: Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor A having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
1. F / 4
2. 3 F / 4
3. F / 8
4. 3 F / 8
Answer: D (3 F / 8)
Question 14: Which of the following statements about dipole moment is not true?
1. The dimensions of dipole moment is [LTA]
2. The unit of dipole moment is C m.
3. Dipole moment is vector quantity and directed from negative to positive charge
4. Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge
Answer: D ( Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge )
Question 15: Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region
1. The electric field is necessarily zero
2. The electric field is due to the dipole moment of the charge distribution only
3. The dominant electric field is inversely proportional to r3, for large r (distance from origin).
4. The work done to move a charged particle along a closed path, away from the region will not be zero
Answer: C (The dominant electric field is inversely proportional to r 3, for large r (distance from origin)
Question 16: Which of the following statements is not true about Gauss law?
1. Gauss’s law is true for any closed surface
2. The term q on the right side of Gauss law includes the sum of all charges enclosed by the surface
3. Gauss law is not much useful in calculating electrostatic field when the system has some symmetry
4. Gauss law is based on the inverse square dependence on distance contained in the coulomb’s law
Answer: C ( Gauss law is not much useful in calculating electrostatic field when the system has some symmetry )
Question 17: A sphere encloses an electric dipole within it. The total flux across the sphere is
1. Zero
2. Half that due to a single charge
3. Double that due to a single charge
4. Dependent on the position of dipole
Question 18: A charge q is placed at corner of the cube then the total flux through the faces of the cube with the side of length a will be
1. q / 8 ε0
2. q / 4 ε0
3. q / 2 ε0
4. q / ε0
Answer: A (q / 8 ε0
Question 19: An object is charged when it has a charge imbalance, which means the
1. Object contains no protons
2. Object contains no electrons
3. Object contains equal number of electrons and protons
4. Object contains unequal number of electrons
Answer: D ( Object contains unequal number of electrons )
Question 20: A conducting sphere is negatively charged. Which of the following statements is true?
1. In the entire volume; charge is uniformly distributed
2. The charge is located at the center of the sphere
3. The charge is located at the bottom of the sphere because of gravity
4. The charge is uniformly distributed on the surface of the sphere
Answer: D ( The charge is uniformly distributed on the surface of the sphere )
Question 21: The number of electrons present in -1 C of charge is
1. 6 x 1018
2. 1.6 x 1019
3. 6 x 1019
4. 1.6 x 1018
Question 22: A cup contains 250 g of water. Find the total positive charges present in the cup of water.
1. 1.34 x 1019
2. 1.34 x 107 C
3. 2.43 x 1019 C
4. 2.43 x 107 C
Answer: B ( 1.34 x 107 C )
Question 23: The electric field required to keep a water drop of mass m just to remain suspended, when charged with one electron of charge e (take g as acceleration due to gravity) is:
1. mg
2. mg / e
3. emg
4. em / g
Question 24: An electric dipole is placed at angle of 30º with an electric field of intensity 2 x 105 NC-1 . It experiences a torque equal to 4 N m . If the dipole length is 2 cm then charge on dipole is
1. 8 m C
2. 4 m C
3. 6 m C
4. 2 m C
Question 25: How does electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
1. Remain unaffected
2. Increases
3. Decreases
4. Information incomplete
Question 26: Given a uniform electric field E = 5 x 103 i NC-1, find the flux of this field through a square of 10 cm whose plane is parallel to YZ plane?
1. 35 N m2 C-1
2. 45 N m2 C-1
3. 50 N m2 C-1
4. None of these
Answer: C ( 50 N m2 C-1 )
Question 27: Two point charges having equal charges are separated by 1 m distance experience a force of 8 N. What will be the force experienced by them if they are held in water at same distance? (K water = 80)
1. 80 N
2. 1 / 10 N
3. 10 N
4. 8 N
Answer: B ( 1/ 10 N )
Question 28: Two infinite parallel sheets, separated by a distance d have equal and opposite uniform charge densities σ. Electric field at a point between the sheets is
1. σ / 2 εo
2. σ / εo
3. Zero
4. Depends upon the location of the point
Question 29: A closed surface in vacuum encloses charges -q and + 3 q. Another charge -2 q lies outside the surface. Total flux over the surface is
1. Zero
2. 2 q / εo
3. -3 q / εo
4. 4 q / εo
Answer: B (2 q / εo )
Question 30: When an electric dipole is held at an angle in a uniform electric field, the net force F and torque on the dipole are
1. F = 0 , τ = 0
2. F ≠ 0, τ ≠ 0
3. F = 0 , τ ≠ 0
4. F ≠ 0, τ = 0
Answer: C ( F = 0 , τ ≠ 0 )
Question 31: A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed
1. Perpendicular to the diameter
2. Parallel to the diameter
3. At an angle tilted towards the diameter
4. At an angle tilted away from the diameter
Answer: A ( Perpendicular to the diameter )
Question 32: A cylindrical conductor is place near another positively charged conductor .The net charge acquired by the cylindrical conductor will be
1. Positive only
2. Negative only
3. Zero
4. Either positive or negative
Question 33: An electric dipole placed in a non uniform electric field experiences
1. Both a net torque and a net force
2. Only a force but no net torque
3. Only a torque but no net force
4. No torque and no net force
Answer: A (Both a net torque and a net force)
Question 34: An electric dipole is placed in a uniform electric field with the dipole axis making an angle with the direction of electric field. The orientation of the dipole for the stable equilibrium is
1. π / 6
2. π / 3
3. 0
4. π / 2
Question 35: What is nature of Gaussian surface involved in Gauss’ law of electrostatics?
1. Scalar
2. Electrical
3. Magnetic
4. Vector
Question 36: A charged particle is free to move in an electric field. It will travel
1. Always along a line of force
2. Along a line of force, if its initial velocity is zero
3. Along a line of force, if it has same initial velocity in the direction of an active angle with the line of force
4. None of the above
Answer: B ( Along a line of force, if its initial velocity is zero )
Question 37: Two concentric spherical shells of radii R and 2 R are given charges Q1 and Q2 respectively. The surface charge densities of the outer surfaces are equal. Determine the ration Q1 : Q2.
1. 1 : 2
2. 1 : 1
3. 1 : 4
4. None of these
Question 38: An electric dipole of dipole moment P is placed in a uniform external electric field . Then, the
1. Torque experienced by the dipole is E × p
2. Torque is zero if p is perpendicular to E
3. Torque is maximum if p is perpendicular to E
4. Potential energy is maximum if p is parallel to E
Answer: C ( Torque is maximum if p is perpendicular to E )
#### MCQ of Electrostatic Potential and Capacitance
Question 1: What is the potential at a point due to charge of 5 x 10-7 C located 10 cm away?
1. 3.5 x 105
2. 3.5 x 104
3. 4.5 x 104 V
4. 4.5 x 105 V
Answer: C (4.5 x 104 V )
Question 2: Consider a uniform electric field in the z direction. The potential is a constant
1. For any x for a given z
2. For any y for a given z
3. On the x-y plane for a given z
4. All of these
Question 3: Which of the following statements is true about the relation between electric field and potential?
1. Electric field is in the direction in which the potential decreases
2. Magnitude of electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at that point
3. In the region of strong electric field equal potential surfaces are far apart
4. Both the statements A and B are correct
Answer: D (Both the statements A and B are correct)
Question 4: A charge Q is placed at the origin. The electric potential due to the charge at a given point in space is V. The work done by an external force in bringing another charge q from infinity upto the point is
1. V / q
2. Vq
3. V + q
Question 5: Which is not true for equipotential surface for uniform electric field?
1. Equipotential surface is flat
2. Two equipotential surface cross each other
3. Electric lines are perpendicular to equipotential surface
4. Work done is zero
Answer: B ( Two equipotential surface cross each other )
Question 6: A hollow metallic sphere of radius 10 cm is charged such that the potential on its surface becomes 80 V. The potential at the centre of the sphere is
1. 80 V
2. 800 V
3. 8 V
4. Zero
Question 7: A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
1. Increases
2. Decreases
3. Does not change
4. Becomes zero
Question 8: A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are
1. Constant, decreases, decreases
2. Increases, decreases, decreases
3. Constant, decreases, increases
4. Constant, increases, decreases
Answer : D (Constant, increases, decreases)
Question 9: When two capacitors charged to different potentials are connected by connecting wire, what is not true?
1. Charge lost by one is equal to charge gained by the other
2. Potential lost by one is equal to potential gained by the other
3. Some energy is lost
4. Both the capacitors acquire a common potential
Answer : B ( Potential lost by one is equal to potential gained by the other )
Question 10: Charge on a capacitor is doubled. Its capacity becomes k times , where
1. k = 2
2. k = 1
3. k = 1 / 2
4. k = 4
Answer : B (k = 1)
Question 11: Work done in moving a test charge on an equipotential surface is
1. Negative
2. Zero
3. Positive
4. Positive infinite
Question 12: Equipotential surfaces associated with an electric field which is increasing in magnitude along x direction are:
1. Planes parallel to yz plane
2. Planes parallel to xy plane
3. Planes parallel to xz plane
4. Coaxial cylinders of increasing radii around x axis
Answer : A (Planes parallel to yz plane)
Question 13: If a conductor has potential V ≠ 0 and there are no charge anywhere else outside
1. There must be charges on the surface or inside itself
2. There cannot be any charge in the body of conductor
3. There must be charges only on surface
4. Both A and B are correct
Answer : D (Both A and B are correct)
Question 14: Three capacitors each of capacity 4µF are to be connected in such a way that the effective capacitance is 6µF. This can be done by
1. Connecting them in series
2. Connecting them in parallel
3. Connecting two in series and one in parallel
4. Connecting two in parallel and one in series
Answer : C ( Connecting two in series and one in parallel )
Question 15: If the distance between the plates of capacitor is increased, then capacitance:
1. Increases
2. Decreases
3. Remains the same
4. First increases then decreases
Question 16: What is the capacitance of arrangement of 4 plates of area A at distance d in air?
1. 3 A εo / d
2. A εo / d
3. 4 A εo / d
4. None of these
Answer : A (3 A εo / d )
Question 17: Find the potential difference between points A and B of arrangement shown
1. 6 V
2. 24 V
3. 8 V
4. -8 V
Question 18: A parallel plate capacitor with air between the plates
has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?
1. 8 pF
2. 96 pF
3. 48 pF
4. 192 pF
Question 19: A combination of four identical capacitors is shown. If resultant capacitance of combination between the points P and Q is 1µF, calculate capacitance of each capacitor.
1. 14 µF
2. 40 µF
3. 4 µF
4. Information incomplete
Question 20: Equipotential surfaces
1. Are closer in regions of larger electric fields compared to regions of lower electric field
2. Will be more crowded near sharp edges of a conductor
3. Will always be equally spaced
4. Both A and B are correct
Answer: D (Both A and B are correct)
Question 21: A hexagon of side 8 cm has a charge 4µC at each of its vertices. The potential at the centre of the hexagon is
1. 2.7 x 106 V
2. 7.2 x 1011 V
3. 2.5 x 1012 V
4. 3.4 x 104 V
Answer: A ( 2.7 x 106 V )
Question 22: An electric dipole of length 20 cm having ± 3 x 10-3 C charge placed at 60º with respect to a uniform electric field experiences a torque of 6 N m. The potential energy of the dipole is
1. –2√3 J
2. 5√3 J
3. –3√2 J
4. 3√5 J
Answer: A ( –2√3 J )
Question 23: A capacitor of capacitance C has charge Q and stored energy is W. If the charge is increased to 2 Q, the stored energy will be
1. W / 4
2. W / 2
3. 2W
4. 4W
Question 24: In a regular polygon of n sides, each corner is at a distance r from the corner. Identical charges are placed at (n-1) corners. At the centre , the magnitude of electric field intensity is E and the potential is V. The ration V/E is
1. rn
2. r (n-1)
3. (n-1 )/ r
4. r(n-1) / n
Question 25: When air is replaced by a dielectric constant K, the maximum force of attraction between two charges separated by a distance
1. Increases by K times
2. Remains Unchanged
3. Decreases by K times
4. Increases by 2K times
Answer: C (Decreases by K times)
Question 26: The voltage of the charging source constant then what would be the percentage change in the energy stored in a parallel plate capacitor if the separation between the plates were to be decreased by 10%?
1. 11.11%
2. 10%
3. 1.11%
4. None of these
Question 27: Two charged spherical conductors of radii R1 and R2 when connected by a conducting wire attain charges q1 and q2 respectively. The proportion of their surface charge densities in terms of their radii is
1. R1 2 : R2 2
2. R2 3 : R1 3
3. R2 : R1
4. None of these
Answer: C (R2 : R1 )
Question 28: What will be the electric potential if A, B and C are the three points in a uniform electric field?
1. Maximum at A
2. Maximum at B
3. Maximum at C
4. Same at all the three points A, B and C
Question 29: A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced amid the plates, in order to maintain the same potential difference, the distance between the plates is enlarged by 2.4 mm. Find the dielectric constant of the slab.
1. 4
2. 3
3. 5
4. Information incomplete
Question 30: Electric field is always
1. Parallel to equipotential surface
2. Perpendicular to equipotential surface
3. It can be perpendicular and parallel as well
4. It does not depend on distribution of charge
Answer: B (Perpendicular to equipotential surface)
Question 31: If n number of equal capacitors each of capacitance C are connected in series, then equivalent capacitance will be
1. n x C
2. C/n
3. n + C
4. n2 C
Answer: B ( C/ N )
Question 32: A dipole is placed parallel to electric field. If W is the work done in rotating the dipole from 0º to 60º, then work done in rotating it from 0º to 180º is
1. 2 W
2. 3 W
3. 4 W
4. W / 2
Question 33: Electrical capacity of earth is
1. 1 F
2. 1 µF
3. 711 µF
4. 9 x 109 µF
Question 34: Three capacitors of capacitance 1 µF, 2µF and 3µF are connected in series and a potential difference of 11 V is applied across the combination, then the potential difference across the plates of 1µF capacitor is
1. 2 V
2. 4 V
3. 1 V
4. 6 V
Question 35: A condenser is charged to double its initial potential. The energy stored in the condenser becomes x times, where x =
1. 2
2. 4
3. 1
4. 1/2
Question 36: Which of the following is false for a perfect conductor?
1. The surface of conductor is an equipotential surface
2. The electric field just outside the surface of conductor is perpendicular to the surface
3. The charge carried by a metallic sphere is always uniformly distributed over its surface
4. None of these
Question 37: What is the Dielectric constant for a metal?
1. Zero
2. Infinite
3. 1
4. 10
#### MCQ of Current Electricity
Question 1: The electrical resistance of a conductor depends upon
1. Size of conductor
2. Size of conductor
3. Geometry of conductor
4. All of these
Answer : D ( All of these )
Question 2:Electromotive force is able to maintain a constant
1. Current
2. Resistance
3. Power
4. Potential difference
Answer : D ( Potential Difference )
Question 3:When a metal conductor connected to left gap of a meter bridge is heated, the balancing point
1. Shifts towards right
2. Shifts towards left
3. Remains unchanged
4. Remains at zero
Answer : A ( Shifts towards right )
Question 4: In a wheat stone bridge if the battery and galvanometer are interchanged then what will be affect on the galvanometer deflection
1. Change in previous direction
2. Not change
3. Change in opposite direction.
4. None of these.
Answer : B ( not change )
Question 5:In series combination of n cells, we obtain
1. More voltage
2. More current
3. Less voltage
4. Less current
Answer : B ( More voltage )
Question 6:Which statement is correct about the validity of Kirchhoff’s loop rule
1. It lies on conservation of charge.
2. Departing currents add up and are equal to incoming currents at a junction.
3. Flexing or reorienting the wire does not change the validity of Kirchhoff’s junction rule.
4. All of these
Answer : D ( All of these )
Question 7:Choose the correct dimensional formula of resistance
1. [MLT-3A-2]
2. [MLT-1A-3]
3. [ML1 T-3A-2]
4. [MLT-3A-1]
Answer : A ( [MLT-3A-2] )
Question 8:A negative charge is moving via junction, then
1. Momentum will be conserved.
2. Momentum will not be conserved.
3. At some places momentum will be conserved and at some other places momentum will not be conserved , it depends on place where the charge is placed
4. None of these.
Answer D : ( None of these )
Question 9: Match the column as follows
Column I Column 2 A Ohm’s Law is valid to- P -metals B Ohm’s law is not valid to- Q -greater resistivity C alloys have- R -diodes and semiconductors
1. A-R , B-R , C-R
2. A-P , B-R , C-Q
3. A-R , B-P , C-Q
4. A-Q , B-R , C-P
Answer : B ( A-P , B-R , C-Q )
Question 10:In a potentiometer furnished with 20 wires, the balance point is acquired on the 17th wire. To shift the balance point to 19th wire, we should do as follows
1. Should decrease resistance of the main circuit.
2. Should increase resistance of the main circuit.
3. Should decrease resistance in series with the cell whose EMF is to be find out.
4. Should increase resistance in series with the cell whose EMF is to be find out.
Answer : D ( Increase resistance in series with the cell whose EMF is to be find out. )
Question 11:(A). Current flows in a conductor only when there is an external electric field applied to the conductor.
(B).The drift velocity of electrons is directly proportional to the electric field.
1. Only A is correct
2. Only B is correct
3. Both A & B are correct
4. Both A and b are false
Answer : C ( Both A & B are correct )
Question 12:What is the value of unknown resistance R, if galvanometer shows null deflection in the given meter bridge set up?
1. 97.50 Ω
2. 105 Ω
3. 150 Ω
4. 110 Ω
Answer : A ( 97.50 Ω )
Question 13:If you have a positive charge ,what conclusion can be drawn about current density?
1. Direction of current and current density is opposite
2. Direction of current and current density is same
3. Current density and current are independent
4. direction can not be determined
Answer : B ( Direction of current and current density is same )
Question 14:Specify the value of resistance if carbon code is
Green – Brown – Yellow – Gold
1. 50 X 104 ± 4%
2. 51 X 104 ± 5%
3. 47 X 103 ± 2%
4. 32 X 105 ± 5%
Answer : B ( 50 X 104 ± 5% )
Ques 15:Which is correct expression for microscopic form of ohm’s law
1. E=J.ρ
2. J=E.σ
3. E=ρ.I/A
4. All of the above
Answer : D ( All of the above )
Question 16: Specify the relation between EMF of a cell and external resistance
1. EMF of a cell depends directly on external resistance
2. EMF of a cell depends directly on external resistance
3. EMF of a cell is independent of external resistance
4. EMF of a cell first increases and then decreases with external resistance
Answer : C ( EMF of a cell is independent of external resistance )
Question 17:When the cells are connected in mixed grouping
1. High EMF is obtained
2. High current is obtained
3. High power is obtained
4. None of these
Answer : C ( High power is obtained )
Question 18: In series combination of n cells , having an EMF ‘E’ and internal resistance ‘r’ . If 3 cells are wrongly connected, what will be the total EMF and internal resistance
1. EMF= n*E; r=6Ω
2. EMF= nE-6E; r=independent
3. EMF=independent ; r=1Ω
4. EMF= n/6E; r=independent
Question 19:Which of the following is not an application of wheat stone bridge
1. To find unknown resistance
2. To find specific resistance
3. To compare EMF of two cells
4. To verify resistance whether in series or parallel combination
Answer : C ( To compare EMF of two cells )
Question 20:Wheat stone bridge can be used to determine ________value of resistors?
1. High value only
2. Low value only
3. Moderate Value only
4. All of the above
Answer : C ( Moderate value only )
Question 21: Which of the following is an application of potentiometer
1. To measure potential drop of a certain section
2. To compare the EMF of two cells
3. To find internal resistance of a cell
4. All of the above
Answer : D( All of these )
Question 22: EMF of a cell is less than the terminal potential difference when
1. Charging condition
2. Always
3. Discharging condition
4. Never
Answer : A ( Charging condition )
Question 23:In a combination on ‘n ‘cells , if the polarity of ‘m’ cells is reversed,
1. E eq=n E
2. E eq=(n-2 m)E
3. E eq=(n-2 m)/E
4. E eq=(n+2 m)/E
Answer : B ( E eq=(n-2 m)E )
Question 24: Ramesh has 12 old cells . He needs maximum output out of these, in which of the following combinations ,he must connect the respective cells
1. Cells should be connected in series combination
2. Cells should be connected in parallel combination
3. Cells should be connected in mixed grouping
4. Nothing can be done
Answer : B ( Cells should be connected in parallel combination )
Question 25:Which of the following is correct about temperature coefficient of metals and alloys
1. (Temperature coefficient of metals)>(temperature coefficient of alloys)
2. (Temperature coefficient of alloys)>(temperature coefficient of alloys)
3. Temperature coefficient of metals is nearly equal to temperature coefficient of alloys
4. (Temperature coefficient of alloys)<=(temperature coefficient of alloys)
Answer : A ( (Temperature coefficient of metals)>(temperature coefficient of alloys) )
Question 26: Which of the following must be constant for the validity of Ohm’s Law
1. Temperature
2. Pressure
3. Mechanical strain
4. All of these
Answer : D ( All of these )
Question 27:Why standard resistors are made up of alloys (specifically manganin and constantan)
1. These alloys have high temperature coefficients
2. These alloys have low temperature coefficients
3. These alloys have high resistivity
4. Both B & C
Answer : D ( Both B and C )
Question 28:How does relaxation time and temperature related to each other
1. Relaxation time varies inversely with temperature
2. Relaxation time varies directly with temperature
3. Relaxation time is independent of temperature
4. Relaxation time first increases and then decreases with temperature
Answer : A( Relaxation time varies inversely with temperature )
Question 29:What is the direction of electronic current?
1. Positive to negative
2. Negative to positive
3. Undefined
4. Parallel to electric field
Answer : B ( Negative to positive )
Question 30:How the values of magnitude and direction varies with time for steady direct current?
1. Magnitude changes with time ;direction remains unchanged
2. Magnitude remains unchanged ; direction changes with time
3. Magnitude changes with time ; direction also changes with time
4. Magnitude remains unchanged ; direction also remains unchanged
Answer : D ( Magnitude remains unchanged ; direction also remains unchanged )
Question 31:A boy has two spare light bulbs in his drawer. One is marked 240 V and 100 W and the other is marked 240 V and 60 W. He tries to decide which of the following assertions are correct?
1. The 60 W light bulb has more resistance and therefore burns less brightly.
2. The 60 W light bulb has less resistance and therefore burns less brightly.
3. The 100 W bulb has more resistance and therefore burns more brightly.
4. The 100 W bulb has less resistance and therefore burns less brightly.
Answer : A ( The 60 W light bulb has more resistance and therefore burns less brightly. )
Question 32: The resistance of the wire in platinum resistance thermometer at ice point is 5Ω and at steam point is 5.25Ω .When the thermometer is inserted in an unknown hot bath its resistance is found yo be 5.5Ω.The temperature of hot bath is
1. 100°C
2. 200°C
3. 300°C
4. 350 °C
Answer : B ( 200°C )
Question 33: A battery , an open circuit and a resistor are connected in series . Consider the following 3 statements concerning the circuit . A voltmeter will read zero if it is connected across points
(i)P and T
(ii)P and Q
(iii)Q and T
Which one of the above is/are true?
1. Only(i)
2. only (iii)
3. Only(i and iii)
4. (i),(ii) and 9(iii)
Answer : C ( Only(i and iii) )
Question 34:The resistivity of alloy manganin is
1. Nearly independent of temperature
2. Increase rapidly with increase in temperature
3. Decrease rapidly with increase in temperature
4. Increase rapidly with decrease in temperature
Answer A : ( Nearly independent of temperature )
Question 35: The correct combination of three resistance 1Ω,2Ω and 3Ω to equivalent resistance 11/5Ω is
1. All three are combined in parallel
2. All three are combined in series
3. 1Ω and 2Ω in parallel and 3Ω in series to both
4. 2Ω and 3Ω are combined in parallel and 1 Ω is in series to both
Answer : D ( 2Ω and 3Ω are combined in parallel and 1 Ω is in series to both )
#### MCQ of Moving Charges and Magnetism
Question 1 : A negatively charged particle is placed across a current carrying wire . Is any force experienced by the charged particle?
1. Due to the effect of electric field , the force will be experienced by the charged particle towards right .
2. Due to the effect of electric field , the force will be experienced by the charged particle towards left .
3. A Torque will be applied due to the effect of magnetic field .
4. No force will be experienced by the charged particle .
Answer : C ( A torque will be applied due to the effect of magnetic field . )
Question : 2 Give the unit of magnetic field .
1. Tesla
2. W b m-2
3. N C-1 -1 s
4. All of the above
Answer : D ( All of the above )
Question 3 : Select the CGS unit of magnetic field .
1. Gauss
2. Tesla
3. Weber
4. Maxwell
Answer A : ( Gauss )
Question 4 : which is the correct expression for the magnetic field about infinite length conductor
A. ( μ/4π ). ( I/a )
B. ( μ/4π ). 2 ( I/a )
C. ( μ/4π ). ( I/a ). sin ( 90° )
D. ( μ/4π ). ( I/2a )
Answer : B ( ( μo/4π ). 2 ( I/a ) )
Question 5 : Consider two concentric loops , inner and outer loop having radii ‘R1’ and ‘R2’ respectively . Inner loop having current ‘I1’ ( clockwise ) and outer loop having current ‘I2’ ( anti – clockwise ) generates magnetic field as ‘B1’ and ‘B2’ respectively .
If I1 > I2 then B net =
A. B1 – B2
B. B1 + B2
C. B2 – B1
D. None of these
Answer : A ( B1 – B2 )
Question 6 : Ampere’s Circuital law is valid for
A. Direct current
C. Both A and B
D. Electronic Current
Answer : A ( Direct Current )
Question 7 : Magnetic field around a cylindrical wire is maximum at
A. Axis of cylinder
B. Outside the cylinder
C. Inside the cylinder
D. At surface of the cylinder
Answer : D ( At surface of the cylinder )
Question 8 : When magnetic field of certain magnitude is applied to an electron at rest , then
A. Electron will move in the same direction as that of electric field
B. Electron will remain stationary
C. Electron will move perpendicular to the direction of electric field
D. Electron will move opposite to the direction of electric field.
Answer : B ( Electron remains stationary )
Question 9 : A negatively charged particle is moving in a elliptical path with a velocity V in uniform magnetic field B , If the charged particle is accelerated to double its velocity and strength of magnetic field is reduced to half , then what how the radius of loop will be affected
A. 8 times
B. 4 times
C. 2 times
D. 16 times
Answer : B ( 4 times )
Question 10 : A charged particle is moving in a region with a constant velocity . Which of the following statement is incorrect
A. E = 0 , B ≠ 0 ( Electric field is zero but magnetic field is non – zero )
B. E ≠ 0 , B ≠ 0 ( Both electric field and magnetic field are non – zero )
C. E ≠ 0 , B = 0 ( Electric field is non – zero but magnetic field is zero )
D. E = 0 , B = 0 ( Both Electric field and magnetic field is zero )
Answer : C ( E ≠ 0 , B = 0 ( ( Electric field is non – zero but magnetic field is zero ) )
Question 11 : The frequency of cyclotron is given by
A. q b / 2πm
B. m B / 2πq
C. 2πm / q b
D. 2πm / q b
Answer : A ( q B / 2πm )
Question 12 : Which of the following statement in relation to magnetic force is accurate ?
A. Magnetic forces always stick to Newton’s third law
B. Magnetic forces don’t stick to Newton’s third law
C. Only for high current , magnetic forces follow Newton’s third law
D. Inside low magnetic field , magnetic forces stick to Newton’s third law
Answer : B ( Magnetic forces don’t stick to Newton’s third law )
Question 13 : A circular loop of radius 10 cm having 100 turns is carrying a current of 3.2 A .The loop is placed in vertical plane and is free to rotate around a horizontal and the vertical plane coincides with diameter of the loop. A uniform magnetic field of 5 T remains such that initially the axis of the coil is in direction of the field . Under the effect of magnetic field , the coil rotates through an angle of 60° . The magnitude of torque in the final position is
A. 25 N m
B. 25 √3 N m
C. 40 N m
D. 40 √3 N m
Answer : B ( 25 √3 N m )
Question 14 : Consider two wires , carrying current flowing in parallel and anti – parallel directions , then
A. Parallel currents repel each other anti parallel currents attract each other .
B. Parallel currents attract each other and antiparallel currents repel each other
C. Both currents attract each other
D. Both currents repel each other
Answer : B ( Parallel currents attract each other and antiparallel currents repel each other )
Question 15 : Which of the following is incorrect about cyclotron ?
A. It is a machine to whisk charged particles or ions to high energy.
B. Cyclotron utilizes both electric and magnetic field in combination to increment the energy of charged particles .
C. The cyclotron sticks to the logic that the time for one revolution of an ion is independent of its speed or radius of its orbit .
D. The charged particles can move in cyclotron on an arbitrary path .
Answer : D ( The charged particles can move in cyclotron on an arbitrary path. )
Question 16 : A coil having 90 turns of radius 15 cm having a magnetic field of 4× 10-4 Tesla . The amount of current flowing through it is
A. 1.06 A
B. 2.44 A
C. 3.44 A
D. 4.44 A
Answer : A ( 1.06 A )
Question 17: A moving coil galvanometer can be converted into an ammeter by doing the following step
A. By introducing a resistor of large value in series .
B. By introducing a shunt in parallel .
C. By introducing a resistor of small value in series .
D. By introducing a resistor of large value in parallel .
Answer : B ( By introducing a shunt in parallel )
Question 18 : An electron is moving with a velocity of 3.2 × 107 m s-1 in a magnetic field of 5 × 10-4 Tesla which is perpendicular to it . What will be the frequency of this electron ?
A. 1.4 × 10 5 Hz
B. 1.4 × 10 7 Hz
C. 1.4 × 10 6 Hz
D. 1.4 × 10 9 Hz
Answer : B ( 1.4 × 10 7 Hz )
Question 19 : Due to the effect of current in a circular loop having two turns a magnetic induction of 0.2 T is produced at its center . The loop is the reoriented into circular loop of four turns . If same amount of current flows in the cold then the magnetic induction at the center of the loop now is
A. 0.2 T
B. 0.4 T
C. 0.6 T
D. 0.8 T
Answer : D ( 0.8 T )
Question 20 : If an electron is moving with velocity V and produces a magnetic field , then
A. The direction of magnetic field will be same as direction of velocity .
B. The direction of magnetic field will be opposite to the direction of velocity .
C. The direction of magnetic field will be perpendicular the direction of velocity .
D. The direction of magnetic field does not depend upon the direction of velocity .
Answer : C ( The direction of magnetic field will be perpendicular the direction of velocity .)
Question 21: Pick out the condition in which no force is experienced by the charged particle placed in magnetic field
A. If a charged particle is at rest.
B. If a charged particle is placed parallel to the magnetic field.
C. If a charged particle is placed antiparallel to the magnetic field.
D. All of these
Answer : D ( All of these )
Question 22 : Is it charged particle is moving in a cyclotron then how will the radius of the path be affected if the frequency of the radio frequency field is doubled ?
A. The radius of path will also become half of the initial radius.
B. The radius of part will be increased four times.
C. The radius of path will be increased 8 times.
D. The radius of path will remain unchanged.
Answer : D ( The radius of path will remain unchanged )
Question 23 : A positively charged particle is moving in a cyclotron having an external magnetic field of 1 T . If its oscillating frequency is 11 MHz and radius is 55 cm , then its kinetic energy in MeV is
A. 13.36
B. 12.52
C. 14.89
D. 14.49
Answer : D ( 14.49 )
Question 24 : A circular loop of 10 cm radius having 20 turns , is placed in a uniform magnetic field of 0.10 T . Given that the magnetic field is normal to the plane of coil . If the loop has cross – sectional area of 10-5 m2 , current flowing in the loop is 5 A and coil is made up of Cu having free electron density about an order of 1029 m3 . Find the net magnetic force on the loop .
A. 2.5 × 10-25
B. 5 × 10-25
C. 4 × 10-25
D. 3 × 10-25
Answer : B ( 5 × 10-25 )
Question 25 : When a magnetic compass needle is placed near a straight current carrying wire , then
(i) The straight current carrying wire will lead to a noticeable deflection in the compass needle
The needle will be aligned tangential to an imaginary circle with the straight current carrying wire
A. Both ( i ) and ( ii ) are correct
B. Only ( ii ) is correct
C. Only ( i ) is correct
D. Neither ( i ) nor ( ii ) is correct .
Answer : A ( Both ( i ) and ( ii ) are correct )
Question 26 : Which of the following is incorrect statement about the Lorentz Force ?
A. F = q [ E ( r ) + B ( r ) ] will be the force acting on a moving charge particle in the presence of electric field E ( r ) and magnetic field B ( r ) .
B. This force depends on the nature of charged particle as the magnetic field due to a negative charge is opposite to that of A positively charged .
C. If the velocity and the magnetic field at parallel and anti – parallel the force due to magnetic field then the force becomes zero .
D. The magnetic force will be maximum for a static charge .
Answer : D ( The magnetic force will be maximum for a static charge . )
Question 27 : In an inertial frame of reference if the magnetic force applied on a moving charged particle is F then what will be its value in another inertial frame of reference
A. It does not depends on the inertial frame of reference therefore the force will remains same.
B. The force will be changed due to the change in the amount of charge
C. The force will be changed due to the change in velocity of charged particle .
D. The force will be changed due to the change in field direction.
Answer : C ( The force will be changed due to the change in velocity of charged particle)
Question 28 : When a Proton enters a uniform magnetic field with uniform velocity then what will be its trajectory
A. Helical
B. Circular
C. Straight line
D. Any one of A , B and C
Answer : D ( Any one of A , B and C )
Question 29 : A circular loop carrying current I having radius R is placed in a uniform magnetic field B which is perpendicular to the loop the force on the loop will be ,
A. 2π RIB
B. 2π R I2 B3
C. π R2 I B
D. Zero
Answer : D ( zero )
Question 30 : A straight current carrying wire of length 1 m having current of 5 ampere is suspended in mid – air by a uniform horizontal magnetic field , then find the magnitude of magnetic field given that the mass of straight current carrying wire is 1.2 Kg
A. 0.65 T
B. 1.53 T
C. 2.4 T
D. 3.2 T
Answer : C ( 2.4 T )
Question 31 : An electron enters cube across one of its face and cubical box is filled with uniform electric field and the magnetic field . If the velocity of electron is V E and A positron enters via opposite face with a velocity of – V . At a instant of time which of the following is not correct ?
A. The electric forces on both the particles will cause same acceleration in the electron and a positron
B. The magnetic forces on both a particle will cause identical acceleration in the electron and positron
C. Both the particles , electron and positron will gain or lose some energy at same rate .
D. The motion of center of mass will be determined by magnetic field only.
Answer : A ( The electric forces on both the particles will cause same acceleration in the electron and a positron )
Question 32 : A toroid is having 3700 turns with its inner and Outer radius as 28 cm and 29 cm respectively . If the current in the toroid is of then a then what will be the magnetic field inside the core of toroid .
A. 2.60 × 10 ^ – 2 T
B. 2.60 × 10 ^ – 3 T
C. 4.53 × 10 ^ – 2 T
D. 2.60 × 10 ^ – 3 T
Answer : A ( 2.60 × 10 – 2 T )
Question 33 : A straight current carrying wire ‘ A ‘ of length 10 m having current of 4 A is placed parallel at a distance of 3 m with a wire ‘ B ‘ of length 12 m and having a current of 6 A in opposite direction . Find the force on a 15 cm section of wire B near its center .
A. 2.4 × 10 -5 N ( attractive nature )
B. 2.4 × 10 -5 N ( repulsive nature )
C. 1.2 × 10 -5 N ( attractive nature )
D. 1.2 × 10 -5 N ( repulsive nature )
Answer : B ( 2.4 × 105 N ( repulsive nature ) )
Question 34 : A galvanometer can be transformed into a voltmeter by doing the following step : –
A. By introducing a resistance of very large value in series combination with the galvanometer .
B. By introducing a resistance of very small value in parallel combination with galvanometer .
C. By introducing a resistance of very large value in parallel combination with the galvanometer .
D. By introducing a very small value of resistor in series combination with the galvanometer .
Answer : A ( By introducing a resistance of very large value in series combination with the galvanometer . )
Question 35 : For a sodium atom what will be the value of Gyromagnetic ratio of an electron .
A. The Gyromagnetic ratio will depend upon the atomic number of the atom
B. The Gyromagnetic ratio will depend on the number of shells of the atom
C. Gyromagnetic ratio will be independent of the orbit of the atom.
D. Gyromagnetic ratio will have a positive value
Answer : C ( Gyromagnetic ratio will be independent of the orbit of the atom )
Question 36 : A galvanometer of resistance 50 Ohm provides a full scale deflection for a current of 0.0 5A . If the cross-sectional area of resistance wire is 3 x 10-2 that can be used to convert the galvanometer into an ammeter which can read a maximum value of 5 A current it , then calculate the length of the wire . Given that the specific resistance of wire is 5 x 10-7 ohm m.
A. 9 m
B. 6 m
C. 3 m
D. 1.5 m
Answer : C ( 3 m )
Question 37 : Which of the following is accurate value of the Bohr magneton ?
A. 8.99 × 10-24 A m-2
B. 9.27 × 10-24 A m-2
C. 5.66 × 10-24 A m-2
D. 9.27 × 10-28 A m-2
Answer : B ( 9.27 × 10-24 A m-2 )
Question 38 : A Galvanometer shows full scale deflection for a current of 4 mA . To convert the galvanometer into a voltmeter of range 0 to 18 V , then what will be the value of resistance required . Given that the resistance of a galvanometer coil is 15 ohm .
A. To convert the galvanometer into voltmeter a resistance of 5885 Ohm is required in series combination .
B. To convert the galvanometer into voltmeter a resistance of 4485 Ohm is required in series combination.
C. To convert the galvanometer into voltmeter a resistance of 5885 Ohm is required in parallel combination .
D. To convert the galvanometer into voltmeter a resistance of 4485 Ohm is required in parallel combination.
Answer : B ( To convert the galvanometer into voltmeter a resistance of 4485 Ohm is required in series combination. )
Question 39 : A Galvanometer shows full scale deflection for a current of 1 mA . To convert the galvanometer into a voltmeter of a range of 2.5 V , then what will be the value of resistance required . Given that the resistance of a galvanometer coil is 10 ohm .
A. To convert the galvanometer into voltmeter a resistance of 24.9 Ohm is required .
B. To convert the galvanometer into voltmeter a resistance of 249 Ohm is required.
C. To convert the galvanometer into voltmeter a resistance of 2490 Ohm is required .
D. To convert the galvanometer into voltmeter a resistance of 24900 Ohm is required .
Answer : C ( To convert the galvanometer into voltmeter a resistance of 2490 Ohm is required)
Question 40 : The torque acting on a current carrying loop which is please in uniform magnetic field does not depend on
A. The torque not depend on the area loop
B. The torque not depend on the value of current
C. The torque not depend on the magnetic field of the loop.
D. None of these
Answer : D ( None of these )
Question 41 : Find the kinetic energy in MeV of a proton if the radius of its dees is 60 cm and frequency of cyclotron oscillator is 12 MHz
A. Kinetic energy is 5 MeV
B. Kinetic energy is 6.5 MeV
C. Kinetic energy is 10.6 MeV
D. Kinetic energy is 12.6 MeV
Answer : C ( Kinetic energy is 10.6 MeV )
Question 42 : In a uniform magnetic field of 5t the coil has a magnetic moment of 25 a M2 if the if the coil rotates with an angle of 60 degree under the influence of magnetic field is
A. 216.5 N m
B. 108.25 N m
C. 102.5 N m
D. 258.1 N m
Answer : B ( 108.25 N m )
#### MCQ of Magnetism and Matter
Question 1 : Pick out the origin of magnetic field
1. Intrinsic spin of electrons and electric Current
2. Polarity of molecules
3. Pauli’s exclusion principle
4. Electronegativity of materials
Answer : A ( Intrinsic spin of electrons and electric Current )
Question 2 : Which of the following statements is incorrect about the magnetic field lines
1. The magnetic lines of force form continuous closed loops
2. By drawing a tangent to magnetic field lines , we can get the direction of net magnetic field
3. As the number of magnetic field lines increases , magnetic force also increases and vice versa .
4. The magnetic lines of force meet each other at certain points.
Answer : D ( The magnetic lines of force intersect each other at certain points. )
Question 3 : Which of the following is true about magnetic monopole ?
1. Monopole of magnet exists
2. Monopole of magnet does not exists .
3. Momentum is constant for magnetic monopole.
4. Momentum varies in case of magnetic monopole as the distance from the field increases.
Answer : B ( Monopole of magnet does not exists .)
Question 4 : A long solenoid having a magnetic moment ‘M’ which corresponds to bar magnet is
1. Magnetic dipole moment will be Equal for both bar magnet and solenoid
2. More magnetic moment for solenoid
3. More magnetic moment for bar magnet
4. None of these
Answer : A ( Equal for both )
Question 5 : Which of the following statements is correct about B.E and B.A?
(B.E represents equatorial magnetic field and B.A refers to axial magnetic field due to bar magnet . )
1. B.E = 2 B.A
2. B.A = 2 B.E
3. B.E = 4 B.A
4. B.A = 4 B.E
Answer : B ( B.A = 2 B.E )
Question 6 : Magnetic moment of a short bar magnet is 0.48 JT-1 . Select the magnitude and direction of magnetic field , which is produced by magnet at a distance of 10 cm from the center of magnet on its axis
1. 0.48 × 10 -4 T along north – south direction
2. 0.28 × 10 -4 T along south – north direction
3. 0.28 × 10 -4 T along north – south direction
4. 0.96 × 10 -4 T along south – north direction
Answer : D ( 0.96 × 10 -4 T along south – north direction )
Question 7 : Magnetic moment of a short bar magnet is 0.39 JT-1 . Select the magnitude and direction of magnetic field , which is produced by magnet at a distance of 20 cm from the equatorial line of the magnet
1. 0.049 G along north – south direction
2. 4.95 G along south – north direction
3. 0.0195 G along south – north direction
4. 19.5 G along north – south direction
Answer : A ( 0.049 G along north – south direction )
Question 8 : Identity the unlike pair
1. ‘Hard magnet’ is associated with ‘alnico’
2. ‘Soft magnet’ is associated with ‘ soft iron’
3. ‘ Bar magnet ‘ is associated with ‘ Equivalent solenoid ‘
4. ‘ Electromagnet is associated with ‘ loud speaker ‘
Answer : D ( ‘ Electromagnet is associated with ‘ loud speaker ‘)
Question 9 : Which material is best to be used in making galvanometer ‘s coil .
1. Cu
2. Ni
3. Fe
4. Both A and B
Answer : C ( Fe )
Question : 10 To reduce residual magnetism to zero by applying magnetizing field in opposite direction is known as :
1. Retentivity
2. Coercivity
3. Hysteresis
4. Flux
Answer : B ( Coercivity )
Question 11 : Earth behaves as a magnet , what is the direction of magnetic field geographically
1. North to south (N-S)
2. South to north (S-N)
3. East to west (E-W)
4. West to east ( W-E)
Answer : B ( South to north(S-N) )
Question 12 : Which of the following entity is not used in specifying earth’s magnetic field ?
1. Magnetic declination
2. Magnetic dip
3. Horizontal component of earth ‘s magnetic field
4. Vertical component of earth ‘ s magnetic field
Answer : D ( Vertical component of earth ‘ s magnetic field )
Question 13 : Which of the following is not a significant difference between electrostatic shielding by a conducting shell and magneto-static shielding ?
1. Electric field lines can terminate on charges and conductors have free charges
2. Magnetic field lines can terminate but conductors can not end them .
3. Magnetic field lines can never terminate on any material and ideal shielding is impossible
4. Highly permeable shells can be used to divert magnetic field lines from the interior region
Answer : B ( Magnetic field lines can terminate but conductors can not end them .)
Question 14 : The earth departs from its dipole shape at a very large distance about an order of 3 × 103 Km. The major factor responsible for this distortion is
1. Motion of ions in the ionosphere
2. Motion of ions in the atmosphere
3. Motion of ions in the lithosphere
4. Motion of ions in the mesosphere
Answer A : (Motion of ions in the ionosphere )
Question 15 : Find the angle of dip at which the horizontal component of earth is merely equal to the vertical component of the Earth’s magnetic field
1. 30°
2. 75°
3. 60°
4. 45°
Answer D : ( 45 ° )
Question 16 : Find the angle of dip at which the vertical component of earth is merely equal to √3 times the horizontal component of the Earth’s magnetic field
1. 30°
2. 75°
3. 60°
4. 45°
Answer C ( 60 ° )
Question 17 : Find the angle of dip at a place where the horizontal component of Earth’s magnetic field is 2 × 10-5 and resultant field is 4 × 10-5 .
1. 30°
2. 60°
3. 90°
4. 45°
Answer : B ( 60 ° )
Question 18 : Some of the substances are given below them maximum magnetic susceptibility ?
1. Calcium
2. Chromium
3. Oxygen
4. Tungsten
Answer : B ( Chromium )
Question 19 : If the temperature of Nickel is increased beyond the Curie temperature then what will happen , given that the liquid shows paramagnetic property at room temperature
1. Anti ferromagnetism will be observed in nickel
2. No magnetic property will be observed in nickel
3. Diamagnetism will be observed in nickel
4. Para – magnetism will be observed in nickel
Answer : D ( Para – magnetism will be observed in nickel )
Question 20 : Pick out the incorrect statement about the magnetic properties of soft iron and steel.
1. Retentivity of soft iron is greater than the retentivity of Steel.
2. Coercivity of soft iron is smaller than coercivity of steel.
3. Area of B-H loop in soft iron is less than the area of B – H loop for steel.
4. Area of B-H loop in soft iron is more than the area of B – H loop for steel.
Answer : D ( Area of B-H loop in soft iron is more than the area of B – H loop for steel.)
Question 21 : What is the value of angles of dip at poles and equator ?
1. Angle of dip at poles is 30° and at equator is 60° .
2. Angle of dip at poles is 0° and at equator is 90° .
3. Angle of dip at poles is 45° and at equator is 90° .
4. Angle of dip at poles is 90° and at equator is 0° .
Answer : D ( Angle of dip at poles is 90° and at equator is 0° )
Question 22 : Find the dipole moment on equator if the equatorial magnetic field of Earth is 0.4 G .
1. Dipole moment on equator is 1.05 × 1023 A m2 .
2. Dipole moment on equator is 2.05 × 1023 A m2 .
3. Dipole moment on equator is 1.05 × 1021 A m2 .
4. Dipole moment on equator is 2.05 × 1021 A m2 .
Answer : A ( Dipole moment on equator is 1.05 × 1023 A m2 . )
Question 23 : The strength of geomagnetic field is
Note :- ( Geomagnetic field is referred to earth’s magnetic field )
1. It remains constant everywhere .
2. It is zero everywhere .
3. It is having a very high value .
4. It vary from place to place on the earth’s surface .
Answer : D ( It vary from place to place on the earth’s surface )
Question 24 : The dipole axis and the axis of earth collectively combine to form a plane ‘S’ . If ‘P’ be the point of intersection of the geographical and the magnetic equator then but will be the angle of declination and the dip angle at point P ?
1. Angle of declination is 0° and angle of dip is 11.3°
2. Angle of declination is 11.3° and angle of dip is 0°
3. Angle of declination is 11.3° and angle of dip is 11.3°
4. Angle of declination is 0° and angle of dip is 0°
Answer : B ( Angle of declination is 11.3° and angle of dip is 0° )
Question 25 : Which of the following is correct expression for the net magnetic flux through any closed surface kept in magnetic field
1. Zero
2. μο/4π
3. 4πμo
4. 4μo/π
Answer : A ( Zero )
Question 26 : The horizontal component of Earth’s magnetic field is is 0.3 G and the vertical component of earth’s magnetic field is 0.52 G then what will be the value of earth’s magnetic field and the angle of dip
1. Earth’s magnetic field is is 0.3 G and angle of dip is 30° .
2. Earth’s magnetic field is is 0.4 G and angle of dip is 40° .
3. Earth’s magnetic field is is 0.5 G and angle of dip is 50° .
4. Earth’s magnetic field is is 0.6 G and angle of dip is 60° .
Answer : D ( Earth’s magnetic field is is 0.6 G and angle of dip is 60° )
Question 27 : Which is the foremost reason responsible for earth’s magnetic field
1. Earth’s magnetic field is due to convective currents in the earth’s core
2. Earth’s magnetic field is due to diversive currents in the earth’s core
3. Earth’s magnetic field is due to rotational motion of earth .
4. Earth magnetic field is due to translational motion of earth .
Answer : A ( Earth’s magnetic field is due to convective currents in the earth’s core )
Question 28 : Which of the following is imprecise statement about the relative magnetic permeability
1. Relative magnetic permeability of a material is dimensionless .
2. Relative magnetic permeability for vacuum medium is one .
3. Relative magnetic permeability for ferromagnetic material is always greater than 1 to large extent.
4. Relative magnetic permeability for paramagnetic material is greater than one .
Answer : D ( Relative magnetic permeability for paramagnetic material is greater than one )
Question 29 : Pick out the correct option which specifies the correct structure for a hysteresis cycle of a transformer core
1. Hysteresis cycle of transformer core is short and wide
2. Hysteresis cycle of transformer core is tall and narrow
3. Hysteresis cycle of transformer core is tall and wide
4. Hysteresis cycle of transformer core is short and narrow
Answer : B ( Hysteresis cycle of transformer core is tall and narrow )
Question 30 : Ramesh is supposed to do an experiment and in that experiment he found that the susceptibility of a given substance is much more greater than one then the substance possibly is
1. Diamagnetic
2. Paramagnetic
3. Ferromagnetic
4. Non – magnetic
Answer : C ( Ferromagnetic )
Question 31 : Let the magnetic field on the earth be carved by that of a point magnetic dipole at the center of the earth then what will be the angle of dip at a point on the geographical equator ?
1. Angle of dip will always be zero
2. Angle of dip can be positive negative or zero
3. Angle of dip is unbounded
4. Angle of dip is always negative
Answer : B ( Angle of dip can be positive negative or zero )
Question 32 : What will be the value of earth’s magnetic field if in the magnetic meridian of a certain place the horizontal component of Earth’s magnetic field is 0.25 G and angle is 60° .
1. 0.50 G
2. 0.52 G
3. 0.54 G
4. 0.56 G
Answer : A ( 0.50 G )
Question 33 : Given a compass needle which is pointing geographical north at a certain place where the horizontal component of earth’s magnetic field is 4 x 10-6 Wb m-2 experiences a torque of 1.2 x 10-3 N m if the magnetic moment of the compass needle is 60 A m2 then what will be the declination of the place
1. 20°
2. 45°
3. 60°
4. 30°
Answer : D ( 30° )
Question 34 : What will be the value of magnetic induction at a point 1 Å away from a Proton along with the axis of spin , given that the magnetic moment of the proton is 1.4 x 10– 6 A m2
1. 0.28 m T
2. 28 m T
3. 0.028 m T
4. 2.8 m T
Answer : D ( 2.8 m T )
Question 35 : A magnetic needle has moment of inertia of 7.8 x 10 – 6 kg m2 and has magnetic moment of 5.8 x 10– 2 Am2 if it completes 12 oscillations in 6 seconds then what will be the magnitude of magnetic field ?
1. The magnitude of magnetic field is 0.011 T
2. The magnitude of magnetic field is 0.021 T
3. The magnitude of magnetic field is 0.031 T
4. The magnitude of magnetic field is 0.041 T
Answer : B ( The magnitude of magnetic field is 0.021 T )
Question 36 : The work done in moving a dipole having a magnetic dipole moment of 0.5 Am2 from its most stable condition to most unstable position in a 0.0 9 T in uniform magnetic field is
1. Work done 0.07 J
2. Work done 0.08 J
3. Work done 0.09 J
4. Work done 0.1 J
Answer : C ( Work done 0.09 J )
Question 37 : A wire is placed between the two poles of of bar magnets if small amount of current it in the wire is into the plane of the paper then what will be the direction of the magnetic force on the wire
1. The direction of magnetic force will be upwards
2. The direction of magnetic field will be downwards
3. The direction of magnetic field will be towards right
4. Direction of magnetic field will be towards left
Answer : D ( Direction of magnetic field will be towards left )
Question 38 : Two identical magnetic dipoles are placed at a separation of 2 m with their axis perpendicular to each other in the air if both of the magnetic dipoles are having in magnetic dipole moment of 2 A m2 to then what will be the Magnetic field at the midpoint between the dipoles
1. 4√5 × 10-5 T
2. 2√5 × 10-5 T
3. 4√5 × 10-7 T
4. 2√5 × 10-7 T
Answer : D ( 2√5 × 10-7 T )
Question 39 : A circular loop is placed in magnetic field such that its plane is perpendicular to the external magnetic field of of magnitude 5.0 x 10– 2 . The loop is free to move about when the coil is is turned slightly and when released it oscillates about its stable equilibrium with a frequency of 2 Hz . What will be the moment of inertia of the loop about its axis of rotation , given that the circular loop has a magnetic moment of 0.355 J T
1. 1.13 × 10-1 Kg m2
2. 1.13 × 10-2 Kg m2
3. 1.13 × 10-3 Kg m2
4. 1.13 × 10-4 Kg m2
Answer : D ( 1.13 × 10-4 Kg m2 )
Question 40 : A magnetizing field produces a magnetic flux density of 8 π T in an iron rod . What will be the relative magnetic permeability of the rod if the magnitude of magnetizing field is 2×103 A m -1 .
1. 102
2. 1
3. 104
4. 103
Answer : C ( 104 )
#### MCQ of Electromagnetic Induction
Question 1: In Faraday’s experiment on electromagnetic induction, more deflection will be shown by the galvanometer, when
1. Magnet is in constant motion towards the coil
2. Magnet is in constant motion away from the coil
3. Magnet is in accelerated motion towards the coil
4. Magnet is at rest near the coil
Answer: C ( Magnet is in accelerated motion towards the coil )
Question 2: The inductance of a coil is proportional to
1. The length
2. The number of turns
3. The resistance of coil
4. The square of the number of turns
Answer: D (The square of the number of turns)
Question 3: A bar magnet is dropped between a current carrying coil. What would be its acceleration?
1. g downwards
2. Greater than g downwards
3. Less than g downwards
4. Bar will be stationary
Answer: C (Less than g downwards)
Question 4: In a solenoid, if number of turns is doubled, then self inductance will become
1. Half
2. Double
3. 1 / 4 times
Question 5: According to Lenz’s law of electromagnetic induction
1. The relative motion between the coil and the magnet produces change in magnetic flux
2. The direction of induced emf is such that it supports the change in magnetic flux
3. Only the coil should be moved towards the magnet for production of induced emf
4. Only the magnet should be moved towards coil for the generation of induced electric field
Answer: A ( The relative motion between the coil and the magnet)
Question 6: A horizontal conducting rod 10 m long extending from east to west is falling with speed 5 ms-1 at right angles to horizontal component of the earth’s magnetic field 0.3 x 10-4 wbm-2. Find the instantaneous value of emf induced in the coil.
1. 15 x 10-3 V
2. 1.5 x 10-3 V
3. 0.3 x 10-4
4. None of these
Answer: B ( 1.5 x 10-3 V )
Question 7: A magnet is moved towards a coil i) speedly ii) slowly then induced emf/ induced charge will be respectively
1. More in first case / more in first case
2. More in first case / equal in both cases
3. Less in first case / more in second case
4. Less in first case / equal in both cases
Answer: B ( More in first case / equal in both cases )
Question 8: The magnetic flux linked with the coil, in Webers, is given by the equation Φ = 3 t2 + 4 t + 9 . Then the magnitude of induced emf at time t=2 s will be
1. 2 V
2. 4 V
3. 8 V
4. 16 V
Question 9: Lenz’s law is in accordance with law of conservation of
1. Charge
2. Momentum
3. Energy
4. Current
Question 10: What is increased in step down transformer?
1. Voltage
2. Current
3. Power
4. Current density
Question 11: In a transformer, number of turns in the primary coil are 140 and that in the secondary coil are 280. If current in primary is 4 A then that in secondary is
1. 4 A
2. 2 A
3. 6 A
4. 10 A
Question 12: An inductor L , a resistance R and two identical bulbs B1 and B2 are connected to a battery through a switch as as shown below. Which of the following statements gives the correct description of the happenings when the switch S is closed?
1. The bulb B2 lights up earlier than B1 and finally both the bulb shine equally bright
2. B1 lights up earlier and finally both the bulbs acquire equal brightness
3. B2 lights up earlier and finally B1 shines brighter than B2
4. B1 and B2 light up together with equal brightness all the time
Answer : C (B2 lights up earlier and finally B1 shines brighter than B2)
Question 13: Which one of the following is a correct statement?
1. Electric field is always conservative
2. Electric field due to a varying magnetic field is conservative
3. Electric field is conservative due to electrostatic charges while non-conservative due to a time varying magnetic field
4. Electric field lines are always closed loops
Answer : C (Electric field is conservative due to electrostatic charges while non-conservative due to a time varying magnetic field)
Question 14 : Eddy currents are produced when
1. A metal is kept in varying magnetic field
2. A metal is kept in a steady magnetic field
3. A circular coil is placed in a magnetic field
4. Through a circular col current is passed
Answer : A (A metal is kept in varying magnetic field)
Question 15: The core used in a transformer and other electromagnetic devices is laminated so that
1. Ratio of voltage in the primary and secondary may be increased
2. Energy loss due to Eddy currents may be minimised
3. The weight of the transformer may be reduced
4. Residual magnetism in the core may be reduced
Answer : B (Energy loss due to Eddy currents may be minimised)
Question 16: Two coils X and Y are placed in a circuit such that when the current changes by 2 A in the coil X the magnetic flux changes by 0.4 Wb in Y. The value of mutual inductance of the coils is
1. 0.2 H
2. 5 H
3. 0.8 H
4. 4 H
Question 17: A metal ring is attached with the wall of a room. When the north pole of as magnet is brought near it, the induced current in the current will be
1. First clockwise then anticlockwise
2. In clockwise direction
3. In anticlockwise direction
4. First clockwise and then anticlockwise
Answer : C (In anticlockwise direction)
Question 18: In the given figure, current from A to B is decreasing. The direction of induced current in the loop is
1. Clockwise
2. Anticlockwise
3. Changing
4. Nothing can be said
Question 19: Induction furnace is based on the principle of
1. Self induction
2. Mutual induction
3. Eddy current
4. None of these
Question 20: When the plane of the armature of an a.c. generator is parallel to the field, in which it is rotating
1. Both the flux linked and emf induced in the coil are zero
2. The flux linked with it is zero, while induced emf is maximum
3. Flux linked is maximum while induced emf is zero
4. Both the flux linked and emf have their respective maximum values
Answer : B ( The flux linked with it is zero, while induced emf is maximum )
Question 21: Two circular coils can be arranged in any of the three ways as shown below. Their mutual inductance will be
1. Maximum in situation (i)
2. Maximum in situation (ii)
3. Maximum in situation (iii)
4. Same in all situations
Answer : A ( Maximum in situation (i) )
Question 22: A coil having 500 square loops each of side 10 cm is placed normal to a magnetic flux which increases at the rate of 1.0 T/s. The induced emf in volts is
1. 0.1
2. 0.5
3. 1
4. 5
Question 23: Two different loops are concentric and lie in the same plane. The current in the outer loop is clockwise and increasing with time. The induced current in the inner loop the, is
1. Clockwise
2. Zero
3. Counter clockwise
4. In a direction that depends on the ratio of loop radii
Question 24: Two coils of self inductance 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils are
1. 16 mH
2. 10 mH
3. 6 mH
4. 4 mH
Question 25: The energy stored in coil carrying current I is U. If current is halved, then energy stored in the coil will be
1. U / 2
2. U / 4
3. 2 U
4. 4 U
Answer : B (U / 4)
Question 26: The number of turns in the coil of an ac generator is 5000 and the area of the coil is 0.25 m. The coil is rotated at the rate of 100 cycles/ sec in a magnetic field of 0.2 Wb/m2. The peak value of the emf generated is nearly
1. 786 kV
2. 440 kV
3. 220 kV
4. 157.1 kV
Question 27: The frequency of an ac generator is altered with change in
1. Speed or rotation of coil
2. Amplitude of a.c.
3. Size of coil
4. All of the above
Answer : A (Speed or rotation of coil)
Question 28: The coefficient of mutual induction between two circuits is equal to the, emf produced in one circuit when current in second circuit is
1. Kept steady at 1 A
2. Cut – off at 1 A level
3. Changed at the rate of 1 A / s
4. Changed from 1 A/s to 2 A/s
Answer : C (Changed at the rate of 1 A/s)
Question 29: Consider the following statements
A) An EMF can be induced by moving a conductor in a magnetic field
B) An EMF can be induced by changing the magnetic field
1. Both A and B are true
2. A is true but B is false
3. B is true but A is false
4. Both A and B are false
Answer : A ( Both A and B are true)
Question 30: A metallic rod of 1 m length is rotated with a frequency of 50 revolution per second, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field of 1 T parallel to the axis is present everywhere. The emf between the centre and the metallic ring is
1. 157 V
2. 117 V
3. 127 V
4. 137V
Question 31: Two different coils have self-inductances L1=8 mH and L2 =2 mH. The current in one coil is increased at a constant rate. The current in second coil is also increased at the same constant rate. At certain instant of time, the power given to the coils is the same. At that time, the current, the induced voltage and the energy stored in the first coil are i1, V1, and W1 respectively. Corresponding values for the second coil at the same instant are i2 , V2 and W2 respectively. Then choose the wrong option.
1. i1 / i2 = 1 / 4
2. i1 / i2 = 4
3. W1 / W2 = 1 / 4
4. V1 / V2 = 1/ 4
Answer : B ( i1 / i2 = 4 )
Question 32: The working of generator is based upon
1. Magnetic effect of current
2. Heating effect of current
3. Chemical effect of current
4. Electromagnetic induction
Question 33: The self inductance L of a solenoid of length l and area of cross section A, with a fixed number of turns N increases as
1. l and A increases
2. l decreases and A increases
3. l increases and A decreases
4. Both l and A decreases
Answer : B ( l decreases and A increases )
Question 34: Which of the following does not use the application of eddy current?
1. Electric power meters
2. Induction furnace
3. LED lights
4. Magnetic brakes in trains
Question 35: The coefficient of mutual induction depends on
1. Medium between the coils
2. Distance between the coils
3. Orientation of the two coils
4. All of these
Answer : D (All of these)
Question 36: A copper rod of length l rotates about its end with angular velocity ω in a uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is
1. Bωl2
2. 1/2 Bωl2
3. 2 Bωl2
4. 1/4 Bωl2
Answer : B ( 1/2 Bωl2 )
Question 37: A coil and a bulb are connected in series with a dc source, a soft iron core is inserted in the coil, then
1. Intensity of the bulb remains the same
2. Intensity of the bulb decreases
3. Intensity of the bulb increases
4. The bulb ceases to glow
Answer : B (Intensity of the bulb decreases)
Question 38: Two particles each of mass m and charge q are attached to the ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of magnetic moment of the system and its angular moment about the centre of rod is
1. q / 2m
2. q / m
3. 2q / m
4. q / πm
Answer : A (q / 2m)
Question 39: The transformation ratio in the step up transformer is
1. 1
2. Greater than one
3. Less than one
4. The ratio greater or less than one depends on the other factors
Answer : B [ (Greater than one)]
Question 40: The mutual inductance between a primary and secondary circuits is 0.5 H. The resistance of the primary and secondary circuits are 20 ohm and 5 ohm respectively. To generate a current of 0.4 A in secondary, current in primary must be changed at rate of
1. 4.0 A/s
2. 1.6 A/s
3. 16.0 A/s
4. 8.0 A/s
Question 41: A wheel with ten metallic spokes each 0.50 m long is rotated with a speed of 120 rev/min in a plane normal to the earth’s magnetic field at the plane. If the magnitude of the field is 0.4 G, the induced emf between the axle and the rim of the wheel is equal to
1. 1.256 x 10-3 V
2. 6.28 x 10 -4 V
3. 1.256 x 10 -4 V
4. 6.28 x 10-5 V
Answer : D ( 6.28 x 10-5 V )
Question 42: An a.c. generator consists of coil of 200 turns and cross sectional area of 3 m2, rotating at a constant angular speed of 60 radian / sec in a uniform magnetic field of 0.04 T. the resistance of the coil is 500 ohm. What is the maximum power dissipated in the coil?
1. 518.4 W
2. 1036 W
3. 259.2 W
4. Zero
Question 43: A long solenoid has a inductance L and resistance R. It is broken into two equal parts and then joined in parallel. The combination is joined to cell of emf E. The time constant for the circuit will be
1. L / 4 R
2. 2 L / R
3. L / R
4. L / 2 R
Answer : C ( L / R)
Question 44: A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This is because
1. The magnetic field is constant
2. The magnetic field is in the same plane as the circular coil and it may or may not vary
3. The magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably
4. Both B and C are correct
Answer : D (Both B and C are correct)
Question 45: A long horizontal metallic rod with length along the east-est direction is falling under gravity. The potential difference between its two ends will
1. Be zero
2. Be constant
3. Increase with time
4. Decrease with time
Answer : C (Increase with time)
Question 46: A metal conductor of length 1 m rotates vertically about one of its end with angular velocity 5 rad s-1. If horizontal component of earth’s magnetic field is 0.2 x 10 -4 T, then the emf developed between the ends of the conductor is
1. 5 µ V
2. 5 mV
3. 50 µ V
4. 50 m V
Answer : C ( 50 µV)
Question 47: In a coil current falls from 5A to 0A in 0.2 s. If an average emf of 150 V is induced, then the self inductance of the coil is
1. 4 H
2. 2 H
3. 3 H
4. 6 H
Answer : D ( 6 H)
Question 48: An infinitely long cylinder is kept parallel to a uniform magnetic field B directed along positive z axis. The direction of induced current on the surface of cylinder as seen from the z axis will be
1. Clockwise of the positive z axis
2. Anticlockwise of the positive z axis
3. Zero
4. Along the magnetic field
#### MCQ of Alternating Current
Question 1: In an AC circuit containing only capacitance, the current
1. Leads the voltage by 180º
2. Remains in phase with voltage
3. Leads the voltage by 90º
4. Lags the voltage by 90º
Question 2: The resonant frequency of a circuit is f. If the capacitance is made 4 times the original value, then the resonant frequency will become
1. f / 2
2. 2 f
3. f
4. f / 4
Answer: A ( f / 2)
Question 3: In L-R circuit, resistance is 8 ohm and the inductive reactance is 6 ohm, then impedance is
1. 2 ohm
2. 14 ohm
3. 4 ohm
4. 10 ohm
Question 4: An inductance and a resistance are connected in series with an AC potential. In this circuit
1. The current and the potential difference across the resistance lead the PD across the resistance lead the PD across the inductance by phase angle π / 2
2. The current and the potential difference across the resistance lag behind PD across the inductance by angle π / 2.
3. The current and the potential difference across the resistance lag behind the PD across the inductance by an angle π.
4. The PD across the resistance lags behind the PD across the inductance by an angle π / 2 but the current in the resistance leads the potential difference across inductance by π / 2.
Answer: B ( The current and the potential difference across the resistance lag behind PD across the inductance by angle π / 2 )
Question 5: A 10 ohm resistance, 5 mH coil and 10 µF capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resistance frequency
1. Is halved
2. Is doubled
3. Remains unchanged
Question 6: The peak value of AC voltage on a 220 V mains is
1. 240 √2 V
2. 230 √2 V
3. 220 √2 V
4. 200 √2 V
Question 7: Alternating current is transmitted at far off places
1. At high voltage and low current
2. At high voltage and high current
3. At low voltage and low current
4. At low voltage and high current
Answer: A (At high voltage and low current)
Question 8: When power is drawn from secondary coil of transformer, the dynamic resistance
1. Increases
2. Decreases
3. Remains unchanged
4. Changes electrically
Question 9: Transformer is based upon the principle of
1. Self induction
2. Mutual induction
3. Eddy current
4. None of these
Question 10: When an AC voltage is applied to a L-C-R circuit, which of the following is true?
1. I and V are out of phase with each other in R
2. I and V are in L with in C, they are out of phase
3. I and V are out of phase, in both L and C
4. I and V are out of phase in L and in phase in C
Answer: C (I and V are out of phase, in both L and C)
Question 11: The turns ratio of a transformer is given as 2 : 3. If the current through the primary coil is 3 A, thus calculate the current through load resistance
1. 1 A
2. 4.5 A
3. 2 A
4. 1.5 A
Question 12: A transformer is an electrical device used for
1. Producing direct current
2. Producing alternating current
3. Changing D.C into A.C.
4. Changing A.C. voltages
Question 13: Out of the following, choose the wrong statement:
1. A transformer cannot work on d.c.
2. A transformer cannot change the frequency of a.c.
3. A transformer can produce a.c. power
4. In a transformer, when a.c. voltage is raised n times, the alternating current reduces to 1/n time
Answer: C ( A transformer can produce a.c. power)
Question 14: High voltage transmission line is preferred as
1. Its appliances are less costly
2. Thin power cables are required
3. Idle current very low
4. Power loss is very less
Answer: D (Power loss is very less)
Question 15: In a purely resistive a.c. circuit, the current
1. Is in phase with the e.m.f.
3. Leads the e.m.f. by a phase difference of π / 2 radians
4. Lags behind the e.m.f. by a phase difference of π / 4 radians
Answer: A (Is in phase with the e.m.f. )
Question 16: In R-L-C series ac circuit, impedance cannot be increased by
1. Increasing frequency of the source
2. Decreasing frequency of the source
3. Increasing the resistance
4. Increasing the voltage of the source
Answer : D (Increasing the voltage of the source)
Question 17: Alternating current cannot be measured by DC ammeter, because
1. AC cannot pass through DC ammeter
2. AC changes direction
3. Average value of current for complete cycle is zero
4. DC ammeter will get damaged
Answer : C (Average value of current for complete cycle is zero)
Question 18: In a transformer the number of turns of primary and secondary coil are 500 and 400 respectively. If 200 V a.c. is supplied to the primary coil, then ratio of currents in primary and secondary coils is
1. 4 : 5
2. 5 : 4
3. 5 : 9
4. 9 : 5
Answer : A (4 : 5)
Question 19: A battery of 12 volt is connected to primary of a transformer with turns ratio ns / np = 10. Voltage across secondary would be
1. 120 V
2. 1.2 V
3. 12 V
4. Zero
Question 20: The reactance of capacitor with capacitance C is X. If both frequency and the capacitance become double then a new reactance will be
1. X
2. 2 X
3. 4 X
4. X / 4
Answer : D (X / 4)
Question 21: A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V ,if the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be
1. 50 A
2. 25 A
3. 45 A
4. 20 A
Question 22: In an alternating current circuit consisting of elements in series the current increases on increasing the frequency of supply. Which of the following elements are likely to constituted the circuit?
1. Only resistor
2. Resistor and inductor
3. Resistor and capacitor
4. Only inductor
Answer : C (Resistor and capacitor)
Question 23: In which of the following circuits the maximum power dissipation is observed?
1. Pure capacitive circuit
2. Pure inductive circuit
3. Pure resistive circuit
4. None of these
Answer : C (Pure resistive circuit)
Question 24 : A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y when connected to the same AC supply also gives the same value of peak current but the current lags behind by 90°. If the series combination of X and Y is connected to the same supply, what will be the rms value of the current?
1. 10 / √2 A
2. 5 / √ 2 A
3. 5 / 2 A
4. 5 A
Answer : C ( 5 / 2 A)
Question 25: At resonance frequency the impedance in series LCR circuit is
1. Maximum
2. Minimum
3. Zero
4. Infinity
Question 26: In a pure capacitive circuit if the frequency of AC source is doubled then its capacitive reactance will be
1. Remains same
2. Doubled
3. Halved
4. Zero
Question 27: The loss of energy in the form of heat in the iron core of transformer is
1. Iron loss
2. Copper loss
3. Mechanical loss
4. None of these
Question 28: A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 ohm per km. The town gets power from the line through a 4000 – 220 V step down transformer at a substation in the town. The line power loss in the form of heat is
1. 400 kW
2. 600 kW
3. 300 kW
4. 800 kW
Question 29: A series L C R circuit with R = 22 ohm, L = 1.5 H and C = 40 uF is connected to a variable frequency 220 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle ?
1. 2000 W
2. 2200 W
3. 2400 W
4. 2500 W
Question 30: In highly inductive load circuit, it is more dangerous when
1. We close the switch
2. Open the switch
3. Increasing the resistance
4. Decreasing the resistance
Question 31: A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10 A then the input voltage and current in the primary coil are
1. 440 V and 20 A
2. 440 V and 5 A
3. 220 V and 10 A
4. 220 V and 20 A
Answer : B (440 V and 5 A)
Question 32: An alternating voltage is connected in series with a resistance R and an inductance L .If the potential drop across the resistance is 200 V and across the inductance is 150 V ,then the applied voltage is
1. 350 V
2. 250 V
3. 500 V
4. 300 V
Question 33: An ideal choke takes a current of 8 A when connected to an AC source of 100 V and 50 Hz. A pure resistor under the same condition takes a current of 10 A. If two are connected in series to an AC supply of 100 V and 40 Hz then the current in the series combination of above resistor and inductor is
1. 10 A
2. 5 A
3. 5√2 A
4. 10√2 A
Question 34: In an L-C-R series AC circuit at resonance
1. The capacitive reactance is more than the inductive reactance
2. The capacitive reactance equals the inductive reactance
3. The capacitive reactance is less than the inductive reactance
4. The power dissipated is minimum
Answer: B (The capacitive reactance equals the inductive reactance)
Question 35: A choke is preferred to a resistance for limiting current in AC circuit because
1. Choke is cheap
2. There is no wastage of power
3. Choke is compact in size
4. Choke is good absorber of heat
Answer: B (There is no wastage of power)
Question 36: A 50 Hz AC source of 20 V is connected across R and C. The voltage across R is 12 V. The voltage across C will be
1. 8 V
2. 16 V
3. 10 V
4. Not possible to determine unless values of R and C are given
Question 37: Alternating voltage V = 140 sin 50 t is applied to a resistor of 10 ohm. This voltage produces heat in the resistor in the time t. To produce the same heat in the same time required DC current is
1. 14 A
4. None of these
Question 38: A virtual current of 4 A and 50 Hz flows in an AC circuit containing a coil. The power consumed in the coil is 240 W. If the virtual voltage across the coil is 100 V. Its inductance will be
1. 1 / 3π H
2. 1 / 5π H
3. 1 / 7π H
4. 1 / 9π H
Answer: B (1 / 5π H)
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## Trigonometry (11th Edition) Clone
$(x+1)^2 +(y+1)^2 = 2$ We can see the graph below:
$r = -2~cos~\theta-2~sin~\theta$ $\sqrt{x^2+y^2} = -\frac{2x}{\sqrt{x^2+y^2}}-\frac{2y}{\sqrt{x^2+y^2}}$ $x^2+y^2 = -2x-2y$ $x^2+2x+y^2+2y = 0$ $(x+1)^2 +(y+1)^2 = 2$ We can see the graph below:
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# What is the anti derivative `int ln x dx` ?
Posted on
The integral `int ln x dx` has to be derived.
Use integration by parts which gives `int u dv = u*v - int v du`
let `u = ln x => du = (1/x)*dx`
and `dv = dx => v = x`
`int ln x dx`
=> `x*ln - int x*(1/x)*dx`
=> `x*ln - int 1 dx`
=> `x*ln x - x`
The integral `int ln x dx = x*ln x - x + C`
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longest
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http://www.jiskha.com/display.cgi?id=1355885155
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# chemistry
posted by on .
8.58g of coastic soada was dissolved to produce a solution of 500cm3 .to neutralize 50cm3 of this solution 15cm3 of 0.4M HCl was consumed .(coastic soada = Na2CO3.XH2O ) find the value of X?
• chemistry - ,
Note the correct spelling of caustic soda. Also, correct sentence structure helps us read these problems better.
How many mols HCl did you use? That's M x L = 0.4M x 0.015L = 0.006.
How many grams were in the solution you titrated. That's 8.58 x 50/500 = 0.858 g.
Then mols = grams/molar mass and rearrange to molar mass = grams/mols.
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| 2.859375
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https://puzzlefry.com/puzzles/if-you-were-running-a-race-and-you-passed-the-person-in-2nd-place-what-place-would-you-be-in-now/?sort=oldest
| 1,659,900,698,000,000,000
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| 433,353,246
| 34,652
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# If you were running a race and you passed the person in 2nd place, what place would you be in now?
1,170.4K Views
Hint: Apply logic
ravi Expert Asked on 12th July 2015 in
You would be in 2nd place. You passed the person in second place, not first.
rahul Guru Answered on 17th July 2015.
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# Universal Clock Suitable for Visually Impaired
4,731
11
5
I was googling around looking for some sort of device I could make using arduino and stumbled upon eshop with devices for visually impaired. What really shocked me was the price. I mean-I do realize that such sorts of devices are not really mainstream, but still, it was kinda high. So I decided to make something similar while keeping the price as low as possible. I ended up making a clock, stopwatch, egg timer, tape measure and water level detector for under 5 bucks.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Basic Principles
First off, I'm sorry to disappoint, but in current design, my device doesn't talk. Instead, it beeps out a specific sequence, telling you the number. It actually is quite like roman numerals. "17" is "XVII". Each numeral up to fifty (that's 1, 5, 10, 50) has corresponding beep sequence in my program. Those sequences vary by pitch and also by number of beeps. So 17 would be X-X-X----V-V----I----I where "X" "V" and "I" are tones of different pitch and dashes ("-") illustrate length of the delay. The way routines communicate is shown in the picture below. Basically, I would like you to notice several things:
-1) Value of null has a specific beep sequence for it.
-2) Buttons and probe (for before mentioned water level detector) are connected via 1 analog pin.
-3) Encoder interrupts the processor, but overflows of encoder related values are handled without jumping out of program.
-4) Encoder needs debouncing. I wasn't able to do it correctly so I borrowed working routine from HERE. ("Another Interrupt Library THAT REALLY WORKS (the Encoder interrupts the processor and debounces like there is no tomorrow"). by raffbuff)
## Step 2: Schematics 'n' Stuff
You will need:
---------------------
Arduino duemillanove (to program it) or alternative
Some sort of shield for programming attiny processors with arduino (look it up on instructables; if you're lazy to build one, just hook your attiny up like SO.)
Altoids tin or alternative. (Quite frankly, I used alternative).
Battery holder (I used 2x1AA)
attiny85 (presumably -20PU) and socket (important)
rotary encoder (mechanical, I used P RE-24 rotary encoder, as the estimated lifespan is about 100 000 cycles and it's less than 2 bucks.)
Momentary switch x 2- I used some from an old keyboard, however, I DO NOT recommend you to do so. Although they have really good mechanical debouncing, they are fairly huge and it's not that comfortable to press them with your thumb.
resistor x 6. They need to be same value and it'd probably be good for them to be somewhere in kiloohms range. However, exact value isn't crucial. I used 10kohm.
Piezo speaker. I had some really flat one lying around, but maybe those in plastic housing (cylindrical ones) would be louder. I don't know.
Some paperclips, long wire, 2 old 9V batteries/9V connectors and plastic bottle cap for probe
Tools you will need: Solder, soldering iron, drill (optional), combination pliers, duckt tape, perfboard, scissors, and some sort of glue, I used hot glue and poster clay. Also. Lot of wire.
## Step 3: Assembly
First of all, you need to somehow insulate conductive parts of can from electronics. To do so, I used duckt tape (not visible on photos). So, this is the first thing you gotta do. Then cut a piece of universal perfboard and basically resemble schematics given in step 2. while doing so, make sure you don't solder rotary encoder, buttons and piezo directly to the board. Instead, solder a wire (flexible one, so that it bends nice and smooth) between aforementioned components and socket pins. And don't connect the buttons just yet. Leave wires from "+" and pin 3 (for buttons and probe, that is 4 wires, 2 from both socket pin and "+") unconnected. You'll solder them later. For pullup resistor for buttons, it might generally be a good idea to put a heat shrink tubing over it to prevent short circuit. Same goes for pretty much all resistors, if you decide to not put them on perfboard.
What you should have by now:
Perfboard with attiny socket on it with piezo hanging on 2 wires, rotary encoder hanging on three, with two wires to battery holder and 4 wires not going anywhere (2 from "+" and 2 to pin 3 on socket). Now it's about time to put the components in place.
Putting the encoder in place
Measure where would you like to put it, mark a hole there and then just drill several holes with your drill in that hole. (What kind of sentence is that? O_o). Basically, just keep drilling until the holes fill the circle you've marked. Then use your mums best scissors to make the individual holes one big hole. Then twisting them, make the big (=final) hole rounder and you're good to go. The encoder should have a tip to make it stable when twisting. (See attached images.) So if it does, just drill an extra hole for this tip. Then unscrew the nut of encoder, place it in position and screw it back using pliers.
Drilling
Simply drill some holes you'll poke the wires through for buttons and probe. The holes for buttons will naturally be on the lid of the can, in the left corner. Just make sure you don't drill the hole right underneath the edge of the buttons where the solder joint will be, otherwise you might run into short circuit. See attached images. Then just solder the buttons and probe. If you didn't put the resistors on the perfboard and want to connect them directly to the buttons instead, now it might be a good time to do so. Secure the buttons in place with some poster clay/hot glue/whatever. Same goes for the perfboard itself, I used a wire that goes round the tin, but that wasn't the most brilliant idea I've had.
## Step 4: Probe
To build the probe, just tear open an old 9v battery and glue and solder together some wire, plastic bottle cap and paperclips...
Go to the Arduino tiny project site and download the version arduino-tiny-0100-0010.zip . Follow included readme.txt file for step by step instructions on how to make it work with your arduino enviroment. (Not really difficult.) Also, you will need to download PinChangeInterrupt-0001.zip from the same website. Also, download the Arduino time library. Installation of libraries is pretty straightforward, in case of confusion refer to this site.
Then turn your arduino into so-called isp by opening arduino ide, selecting file-examples-Arduino as ISP from the dropdown menu and hitting upload button.
Once you have done that, paste following code there, select tools-board-attiny85@1MHZ (Internal oscillator, BOD disabled), place the attiny chip in the programmer socket or breadboard or whatever like mentioned here and proceed to upload.
!!!!IMPORTANT!!!!
If uploading fails, try uploading the arduino as isp sketch from an older version of arduino ide, such as 0022. If Even that fails, you probably have your connections wrong
#define encoderPinA 2
#define encoderPinB 0
#define buttonpin 2
#define loudspeakerpin 1
#include "Time.h"//I did NOT make this library
#include "PinChangeInterruptSimple.h"//And also this library wasn't made by me
volatile int encoderPos = 0;//position of the encoder, volatile because it changes on interrupt
unsigned int lastReportedPos = 1; // change management
static boolean rotating=false; // debounce management
int stepstochangeitem=10; //how many encoder steps are needed for menu item to be changed
boolean A_set = false; //some debouncing related vars, who knows, I haven't done this part of code
boolean B_set = false;
int items=4;//number of items in menu
int selecteditem=0;
int R=512;//307;//telling which button is pressed-512=pullup resistor has the same value as the button one
int numbertype[]={//roman numerals; if you change this, please also change the numtypesound array abd typecount variable
50, 10,5,1};
int numbertypesound[]={//pitches of different numerals
100, 300, 500, 700};
int typecount=4;//number of numerals alltogether
int delaytime=150;
unsigned long presslength=0;//buttons-for how long has a button been pressed
//time management vars below
long diff=0;//difference between time syncs in millis
long realdiff=0;//same as above, but rounded to half hours
const unsigned long hh=1800000;//half hour in milliseconds
long drift=0;//by how much the realdiff differs from diff
unsigned long lastmillis=0;
unsigned long lastcheckedtime=0;
int onemetervalue=300;//how long (in encoder steps) is one meter
boolean driftenabled=false;//are we correcting time drift?
void setup() {
pinMode(encoderPinA, INPUT);
pinMode(encoderPinB, INPUT);
pinMode(buttonpin, INPUT);
pinMode(loudspeakerpin, OUTPUT);
// encoder pin on interrupt 0 (pin 2)
attachPcInterrupt(2, doEncoderA, CHANGE);
// encoder pin on interrupt 1 (pin 3)
attachPcInterrupt(0, doEncoderB, CHANGE);
// Serial.begin (9600);
beepnumber(35);//I just love the sound of that
if (tempR>5)
{
R=tempR;
playSound(1);
//beepnumber(tempR);
encoderPos=0;
selecteditem=0;
delay(500);
}*/
}
int setvalue(boolean minutes)//set value; used for clock and egg timer routines
{
//wrapvalues[]={0,2,1,4};
encoderPos=0;//reset encoder position
int returnvalue=0; //the value to be returned at the end of the function
int stepnumber=0;//number of current step. each numeral setting=1 step
int laststepnumber=0;
while(1)//infinite loop to be broken out of using break command
{
inmenu=true;//so that it beeps out menu item, which is here used to determine how many times you want each numeral to be in your desired number
wrapEncValues();
if (stepnumber==0&&minutes==false)//setting hours? Then skip setting 50-value (step 0)
{
stepnumber=1;
laststepnumber=1;
}
if ( laststepnumber==stepnumber)
{
if (minutes==false&&stepnumber==2&&returnvalue>=20)//skipping value 5 setting if user is setting hours and has set them to be >20 (setting 25:00 as time wouldn't make sense)
stepnumber=3;
if (minutes==true&&stepnumber==1&&returnvalue>=50)//skipping value 10 setting if user is setting minutes and has set them to be >50 (59 is the max value for minutes)
stepnumber=2;
beepnumber(numbertype[stepnumber]);
laststepnumber=-1;
}
if (minutes==false)
{
switch (stepnumber)
{
case 1://10s
items=2;
break;
case 2://5s
if (returnvalue>=20)//20-something value
stepnumber=3;//skip 5s
items=1;
break;
case 3://1s
if (returnvalue>=20)//20-something value
{
items=3;
}
else
items=4;
break;
}
}
else
{
switch (stepnumber)
{
case 0://50s
items=1;
break;
case 1://10s
if (returnvalue>=50)//50-something value
{
stepnumber=2;//skip 10s
break;
}
items=4;
break;
case 2://5s
items=1;
break;
case 3://1s
items=4;
break;
}
}
if (buttonpressed()==1)
{
returnvalue+=numbertype[stepnumber]*selecteditem;
stepnumber+=1;
selecteditem=0;
laststepnumber=stepnumber;
if (stepnumber>3)
{
stepstochangeitem=10;
items=4;//not really necessary since "items" allready will have this value by coincidence
return returnvalue;
break;
}
}
}
}
void wrapEncValues()
{
int beepnum=-1;
while (encoderPos>stepstochangeitem)
{
encoderPos=encoderPos-stepstochangeitem;
selecteditem+=1;
beepnum=selecteditem;
}
while (encoderPos<0)
{
encoderPos=stepstochangeitem+encoderPos;
selecteditem-=1;
beepnum=selecteditem;
}
if (selecteditem>items)
{
selecteditem=0;
beepnum=selecteditem;
}
if (selecteditem<0)
{
selecteditem=items;
beepnum=selecteditem;
}
beepnumber(beepnum);
}
void playSound(int soundtype)//routine for beeping out specific codes (for zero, entering program, alarm and "OK beep" which is also used for separating hours and minutes when displaying time
{
switch (soundtype)
{
case 0://enter
/* for (int i=0; i<=100; i++)
{
tone(loudspeakerpin, i*5);
delay(5);
}
noTone(loudspeakerpin);
*/
tone(loudspeakerpin, 500,250);
delay(100);
tone(loudspeakerpin, 800,250);
delay(500);
break;
case 1://melody
tone(loudspeakerpin, 200,250);
delay(100);
tone(loudspeakerpin, 500,500);
delay(50);
break;
case 2://zero
tone(loudspeakerpin, 950,250);
delay(50);
tone(loudspeakerpin, 250,250);
break;
case 3://dot
delay(100);
tone(loudspeakerpin, 950,250);
delay(100);
noTone(loudspeakerpin);
tone(loudspeakerpin, 950,250);
delay(400);
break;
}
}
void loop()
{ //Do stuff here
/*pinMode(13,OUTPUT);
digitalWrite(13,HIGH);
delayMicroseconds(100000);
digitalWrite(13,LOW);*/
if (driftenabled==true)//auto time correction enabled
{
boolean setlastcheckedtime=false;
while (now()-lastcheckedtime>3600)//more than hour difference since last sync time
{
setlastcheckedtime=true;//sync happened
lastcheckedtime+=3600;
}
if (setlastcheckedtime== true)//set the time of last sync only if the sync actually happened
lastcheckedtime=now()-(drift/1000);
}
//moved code
wrapEncValues();
// Serial.println (encoderPos, DEC);
{
playSound(0);
switch (selecteditem)
{
case 0://clock
if (presslength<2000)//press wasn't longer than 2 secs
{
beepnumber(hour());
delay(200);
playSound(3);
delay(200);
beepnumber(minute());
break;
}
else //longer than 2 secs
{
if (presslength>5000)//longer than 5 secs
{
if (lastmillis!=0)
driftenabled=true;
playSound(3);
//code below calculates drift of internal oscillator by comparing expected delay
//(which can be easily estimated, since user has to make the sync at :30 or :00) to gained value.
// Then calculates drift, nominates it to one hour and saves to "drift" variable
diff=millis()-lastmillis;
unsigned long compval=hh/2;
int i;
for (i=1; i<=10000; i++)
{
compval+=hh;
if (diff<=compval)
break;
}
realdiff=i*hh;
drift=realdiff-diff;
drift=(2*drift)/(realdiff/hh);
lastmillis=millis();
int hrz=hour();
int minz;
if (minute()>15&&minute()<45)//assume the clock isn't off by more than 15 minutes
minz=30;
else
{
if (minute()>45)
hrz+=1;
minz=0;
}
setTime(hrz,minz,0,1,1,1);
}
else
{
int hrz=setvalue(false);
int minz=setvalue(true);
setTime(hrz,minz,0,1,1,1);
driftenabled=false;
lastmillis=0;
lastcheckedtime=now();
}
break;
}
break;
case 1://tape measure
encoderPos=0;
while(1)
{
// wrapEncValues();
if (buttonpressed()==1)
{
if (presslength<2000)
{
int cms=round(map(abs(encoderPos),0,onemetervalue,0,100));
beepnumber(cms);
stepstochangeitem=10;
items=4;
encoderPos=0;
break;
}
else//assume user has measured exactly one meter and wants to calibrate the tape measure
{
playSound(3);
onemetervalue=abs(encoderPos);
encoderPos=0;
break;
}
}
}
break;
case 2://water detector
{
stepstochangeitem=4;
items=10;
selecteditem=5;
while(1){
wrapEncValues();
if (abs(selecteditem-5)>2)//the probe uses same input pin as buttons; therefore, to exit the water detector, user is obliged to twist the encoder wheel
{
stepstochangeitem=10;
encoderPos=0;
items=4;
noTone(loudspeakerpin);
break;
}
}
break;
}
case 3://stopwatch
{
int startingsecs;
boolean watchrunning=false;
while (1)
{
if(buttonpressed()==1)
{
playSound(3);
if (watchrunning==false)
{
watchrunning=true;
startingsecs=millis()/1000;
}
else
{
watchrunning=false;
beepnumber(millis()/1000-startingsecs);
break;
}
}
}
break;
}
case 4://egg timer
int minutemins=99;
while (1)
{
int buttoncode=buttonpressed();//so that it doesn't have to be sampled twice
if (buttoncode==0)
break;
if (buttoncode==1)
{
minutemins=setvalue(true);
minutemins=(minute()+minutemins)%60;//to ensure it will work correctly even if hour changes while the egg timer is running (which isn't really possible now when I think about it...)
playSound(3);
}
if (minute()==minutemins)
{
playSound(1);
if (buttoncode==0)
break;
}
}
break;
}
}
}
//Until further notice, the code below is NOT mine
void doEncoderA(){
// debounce
if ( rotating ) delay (1); // wait a little until the bouncing is done
// Test transition, did things really change?
if( digitalRead(encoderPinA) != A_set ) { // debounce once more
A_set = !A_set;
if ( A_set && !B_set )
encoderPos += 1;
rotating = false; // no more debouncing until loop() hits again
}
}
// Interrupt on B changing state, same as A above
void doEncoderB(){
if ( rotating ) delay (1);
if( digitalRead(encoderPinB) != B_set ) {
B_set = !B_set;
if( B_set && !A_set )
encoderPos -= 1;
rotating = false;
}
}
//from now on, the code is mine again
int buttonpressed()
{
if (analogRead(buttonpin)>5)//is anything going on at all?
{
int returnvalue=-1;
unsigned long beginms=millis();//variable for later comparation to tell for how long has the button been pressed
while(1)
{
unsigned long lastmillis;
if (value>5)//should be 0, but some hysteresion never hurts
{
lastmillis=millis();
if (abs(value-((2*R)/3))<10)//button 2
returnvalue=0;
else
if (abs(value-R)<10)//button 1
{
returnvalue=1;
// R=value;
}
else
if (abs(value-((2*R)/5)*3)<10)//both buttons, doesn't really work
returnvalue=2;
}
else
if (millis()-lastmillis>20)//to do the debouncing; count button as released once it has been released for more than 20ms
{
presslength=millis()-beginms;
return returnvalue;
break;
}
}
}
else
return -1;
}
int beepnumber(int num) //beeping out the number. Basically works pseudorecursively by dividing the argument by numerals from highest to lowest
//and then feeds the modulo of division back to itself and repeats the whole process
{
if (num==0)
{
playSound(2);//special code for zero
return 0;
}
int initpos=encoderPos;
int numbertypevalue[]={
-1, -1, -1 ,-1 };
for(int i=0;i<typecount;i++)
{
if (num/numbertype[i]>=1)
{
numbertypevalue[i]=floor(num/numbertype[i]);
num=num%numbertype[i];
}
encoderPos=initpos;//basically ignore user rotating the encoder if beeping out the value
if (numbertypevalue[i]>-1)
{
for(int i2=0;i2<numbertypevalue[i];i2++)
{
for (int i3=0;i3<4-i;i3++)
{
delay(delaytime/2);
tone(loudspeakerpin, numbertypesound[i],delaytime);
delay(delaytime);
}
delay(delaytime*2);
}
}
}
}
## Step 6: Playing With Our New Device ^^
If you've done everything correctly, it's time to put batteries in. The device should greet you with "beepbeepbeep beepbeepbeep beepbeepbeep beepbeep" sound.
Rotate the encoder to change program.
Clock-0
Press the primary button to activate. It will then beep out the current time.
Hold the button for over 2 seconds. Then you can set the time. (rotate to select tens of hours, "fives" of hours and individual hours in this order; press the main button to proceed further in selecting.)
Hold it for 5 seconds to sync the time at :00 or :30 to make the watch more accurate
Tape measure-1
Roll over surface to measure length; then press the main button to know the value
-or-
Measure exactly one meter using this technique. Then hold the button for some time. The tape measure will calibrate.
Water detector-2
attach probe, put it into the cup; listen to the tone.
Stopwatch-3
Press the main button to start. Press it again to stop counting. The device will beep out the number of seconds since the first button press and return to menu automatically.
Egg timer-4
press the main button; it will beep out value of 50, 10, 5, and 1 in this order as you press button repeatedly. Scroll the encoder wheel to adjust the value (times the number can fit into the desired number, basically, so if you want to set your timer to go off after say 7 minutes, press the button twice to skip 50s and 10s, then turn the wheel to set 5s to 1 , press the button again and turn the wheel to turn 1s to 2.) Sounds confusing, but really isn't that much, trust me!
THANKS FOR READING MY FIRST TUTORIAL AND HOPE YOU LIKED IT!
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## 5 Discussions
Quite brilliant! Some more work and you will help bring down the cost of basic electronic assistive devices for visually impaired.
One suggestion is that you don't have a gun in the background of your next video. It was giving me the creeps every time I saw it.
Great work!
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If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
指数简介
${4}^{3}$
${7}^{5}$
${4}^{3}=4×4×4$
${4}^{3}=4×4×4$
$\phantom{{4}^{3}}=16×4$
$\phantom{{4}^{3}}=64$
$2×2×2×2×2×2$
$2×2×2×2×2×2={2}^{6}$
练习:
$7×7×7$ 写成指数形式。
挑战题：
${2}^{5}$
${5}^{2}$
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# Solve For X Practice Problems
Tags: City Essay Resurrection Tale TwoEssay On Conservation Of Energy At HomeBest Cover Letter Music IndustryGood Thesis Statement For Gay MarriageHow To Construct An Argumentative EssayBusiness Planning Analyst SalaryGifted Hands EssayScholastic Creative Writing
They can also use our digital textbook, game applications or PDF worksheets.
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This ATI TEAS math practice quiz will test you ability to solve multiplying and dividing decimal numbers.
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You will use a variety of models to represent and solve algebraic equations, which involve relationships between numbers and variables.
If the scale is balanced, then the expressions are equal, and you can write an equation to represent the relationship between the two expressions.
If the scale is not balanced, then the expressions are not equal.
## Comments Solve For X Practice Problems
• ###### Using Models to Solve Equations Texas Gateway
You will use a variety of models to represent and solve algebraic equations. For questions 1–3, use the balance scale to determine the value of x. 1. Practice. For questions 1–3, use the cups and counters model to determine the value of x.…
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X. Step 4. Solve like an exponential equation of like bases. 6 = x. Demonstration Problem 2 Solving Exponential Equations. Mixed Practice Problems.…
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https://www.lesswrong.com/posts/zmpYKwqfMkWtywkKZ/kelly-isn-t-just-about-logarithmic-utility
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35
Edit: The original version of this was titled without the "(just)". I also regret the first 4 lines, but I'm leaving them in for posterity's sake. I also think that johnswentworth's comment is a better tl;dr than what I wrote.
Deep breath.
Perhaps one reason is that maximizing E log S suggests that the investor has a logarithmic utility for money. However, the criticism of the choice of utility functions ignores the fact that maximizing E log S is a consequence of the goals represented by properties P1 and P2, and has nothing to do with utility theory.
"Competitive Optimality of Logarithmic Investment" (Bell, Cover) (emphasis mine):
Things I am not saying:
• Kelly isn't maximising logarithmic utility
• Kelly is the optimal strategy, bow down before her in awe
• Multiperiod optimization is different to single period optimization(!)
The structure of what I am saying is roughly:
1. Repeated bets involve variance drag
2. People need to account for this
3. Utilities and repeated bets are two sides of the same coin
4. People focus too much on the "Utility" formulation and not enough on the "repeated nature"
5. People focus too much on the formula
Repeated bets involve variance drag
This has been said many times, by many people. The simplest example is usually something along the lines of:
You invest in a stock which goes up 1% 50% of the time, and down 1% 50% of the time. On any given day, the expected value of the stock tomorrow is the same as the stock today:
But what happens if you invest in it? On day 1 it goes up 1%, on day 2 it goes down 1%. How much money do you have?
Another way of writing this is:
The term is "variance decay". (Or volatility dray, or any of a number of different names)
This means that over a large sequence of trials, you tend to lose money. (On average over sequences "you" break-even, a small number of your multiverse brethren are getting very lucky with their coin flips).
Let's consider a positive sum example (from "The ergodicity problem in economic" (Peters)). We have a stock which returns 50% or -40%. If we put all our money in this stock, we are going to lose it over time since but on any given day, this stock has a positive expected return.
[I think there is a sense in which this is much more intuitive for gamblers than for investors. If your typical "investment opportunity" (read bet) goes to 0 some fraction of the time, investing all your "assets" (read bank roll) in it is obviously going to run a high risk of ruin after not very many trials.]
People need to account for this
In order to account for the fact that "variance is bad" people need to adjust their strategy. To take our example from earlier. That stock is fantastic! It returns 10% per day (on average). There must be some way to buy it and make money? The answer is yes. Don't invest all of your money in it. If you invest 10%. After one day you will either have or . This is winning proposition (even long term) since
Kelly is (a) way to account for this.
Utilities and repeated bets are two sides of the same coin
I think this is fairly clear - "The time interpretation of expected utility theory" (Peters, Adamou) prove this. I don't want to say too much more about this. This section is mostly to acknowledge that:
• Yes, Kelly is maximising log utility
• Yes, I do think thinking in the repeated bets framework is more useful
People focus too much on the "Utility" formulation and not enough on the "repeated nature"
This is my personal opinion (and that of the Ergodicity Economics crowd). Both of the recent threads talk at length about the formula for Kelly for one bet. There seems to be little (if any) discussion of where the usefulness comes from. Kelly is useful when you can repeat and therefore compound your bets. Compounding is multiplicative, so it becomes "natural" (in some sense) to transform everything by taking logs.
People focus too much on the formula
Personally, I use Kelly more often for pencil-and-pap`er calculations and financial models than for making yes-or-no bets in the wild. For this purpose, far and away the most important form of Kelly to remember is "maximize expected log wealth". In particular, this is the form which generalizes beyond 2-outcome bets - e.g. it can handle allocating investments across a whole portfolio. It's also the form which most directly suggests how to derive the Kelly criterion, and therefore the situations in which it will/won't apply. - johnswentworth
I think this comment was spot on, and I don't really want to say too much more about it than that. The really magic takeaway (which I think is underappreciated) is. "maximise log your expected wealth". "Optimal Multiperiod Portfolio Policies" (Mossin) shows that for a wide class of utilities, optimising utility of wealth at time t is equivalent to maximising utility at each time-step.
BUT HANG ON! I hear you say. Haven't you just spent the last 5 paragraphs saying that Kelly is about repeated bets? If it all reduces to one period, why all the effort? The point is this: legible utilities need to handle the multi-period nature of the world. I have no (real) sense of what my utility function is, but I do know that I want my actions to be repeatable without risking ruin!
It's not about logarithmic utility (but really it is)
There is so much more which could be said about Kelly criterion for which this is really scratching the surface. Kelly has lots of nice properties (which are independent of utility functions) which I think are worth flagging:
1. Maximises asymptotic long run growth
2. Given a target wealth, is the strategy which achieves that wealth fastest
3. Asymptotically outperforms any other strategy (ie E[X/X_{kelly}] <= 1)
4. (Related to 3) "Competitive optimality". Any other strategy can only beat Kelly at most 1/2 the time. (1/2 is optimal since the other strategy could be Kelly)
Conclusion
- Zvi (Kelly Criterion)
What I am trying to do is shift the conversation away from discussion of the Kelly formula, which smells a bit like guessing the teacher's password to me, and get people to think more about how we should really think about and apply Kelly. The formula isn't the hard part. Logarithms aren't (that) important.
New Comment
OK, here is my fuller response.
First of all, I want to state that Ergodicity Economics is great and I'm glad that it exists -- like I said in my other comment, it's a true frequentist alternative to Bayesian decision theory, and it does some cool things, and I would like to see it developed further.
That said, I want to defend two claims:
• From a strict Bayesian perspective, it's all nonsense, so those arguments in favor of Kelly are at best heuristic. A Bayesian should be much more interested in the question of whether their utility is log in money. (Note: I am not a strict bayesian, although I think strict Bayes is a pretty good starting point.)
• From a broader perspective, I think Peters' direction is interesting, but his current methodology is not very satisfying, and doesn't support his strong claims that EG Kelly naturally pops out when you take a time-averaging rather than ensemble-averaging perspective.
"It's All Nonsense"
Let's deal with the strict Bayesian perspective first.
You start your argument by discussing the "variance drag" phenomenon. You point out that the expected monetary value of an investment can be constant or even increasing, but it may be that due to variance drag, it typically (with high probability) goes down over time.
You go on to assert that "people need to account for this" and "adjust their strategy" because "variance is bad".
To a strict Bayesian, this is a non-sequitur. If it be the case that a strict Bayesian is a money-maximizer (IE, has utility linear in money), they'll see no problem with variance drag. You still put your money where it gets the highest expected value.
Furthermore, repeated bets don't make any material difference to the money-maximizer. Suppose the optimal one-step investment strategy has a tare of return R. Then the best two-step strategy is to use that strategy twice to get an expected rate of return R^2. This is because the best way they can set themselves up to succeed at stage 2 (in expectation!) is to get as much money as they can at stage 1 (in expectation!). This reasoning applies no matter how many stages we string together. So to the money-maximizer, your argument about repeated bets is totally wrong.
Where a strict Bayesian would start to agree with you is if they don't have utility linear in money, but rather, have diminishing returns.
In that case, the best strategy for single-step investing can differ from the best strategy for multi-step investing.
Say the utility function is . Treat an investment strategy as a stochastic function . We assume the random variable behaves the same as the random variable ; IE, returns are proportional to investment. So, actually, we can treat a strategy as a random variable which gets multiplied by our money;
In a one-step scenario, the Bayesian wants to maximize where is your starting money. In a two-step scenario, the Bayesian wants to maximize . And so on. The reason the strategy was so simple when was that this allowed us to push the expectation inwards; (the last step holds because we assume the random variables are independent). So in that case, we could just choose the best individual strategy and apply it at each time-step.
In general, however, nonlinear stops us from pushing the expectations inward. So multi-step matters.
A many-step problem looks like . The product of many independent random variables will closely approximate a log-normal distribution, by the multiplicative central limit theorem. I think this somehow implies that if the Bayesian had to select the same strategy to be used on all days, the Bayesian would be pretty happy for it to be a log-wealth maximizing strategy, but I'm not connecting the dots on the argument right now.
So anyway, under some assumptions, I think a Bayesian will behave increasingly like a log-wealth maximizer as the number of steps goes up.
But this is because the Bayesian had diminishing returns in the first place. So it's still ultimately about the utility function.
Most attempts to justify Kelly by repeated-bet phenomena overtly or implicitly rely on an assumption that we care about typical outcomes. A strict Bayesian will always stop you there. "Hold on now. I care about average outcomes!"
(And Ergodicity Economics says: so much the worse for Bayes!)
Bayesians may also raise other concerns with Peters-style reasoning:
• Ensemble averaging is the natural response to decision-making under uncertainty; you're averaging over different possibilities. When you try to time-average to get rid of your uncertainty, you have to ask "time average what?" -- you don't know what specific situation you're in.
• In general, the question of how to turn your current situation into a repeated sequence for the purpose of time-averaging analysis seems under-determined (even if you are certain about your present situation). Surely Peters doesn't want us to use actual time in the analysis; in actual time, you end up dead and lose all your money, so the time-average analysis is trivial. So Peters has to invent an imagined time dimension, in which the game goes on forever. In contrast, Bayesians can deal with time as it actually is, accounting for the actual number of iterations of the game, utility of money at death, etc.
• Even if you settle on a way to turn the situation into an iterated sequence, the necessary limit does not necessarily exist. This is also true of the possibility-average, of course (the St Petersburg Paradox being a classic example); but it seems easier to get failure in the time-avarage case, because you just need non-convergence; ie, you don't need any unbounded stuff to happen.
It's Not Nonsense, But It's Not Great, Either
Each of the above three objections are common fare for the Frequentist:
• Frequentist approaches start from the objective/external perspective rather than the agent's internal uncertainty. (For example, an "unbiased estimator" is one such that given a true value of the parameter, the estimator achieves that true value on average.) So they're not starting from averaging over subjective possibilities. Instead, Frequentism averages over a sequence of experiments (to get a frequency!). This is like a time-average, since we have to put it in sequence. Hence, time-averages are natural to the way Frequentists think about probability.
• Even given direct access to objective truth, frequentist probabilities are still under-defined because of the reference class problem -- what infinite sequence of experiments do you conceive of your experiment as part of? So the ambiguity about how to turn your situation into a sequence is natural.
• And, again, once you select a sequence, there's no guarantee that a limit exists. Frequentism has to solve this by postulating that limits exist for the kinds of reference classes we want to talk about.
So, actually, this way of thinking about decision theory is quite appealing from a Frequentist perspective.
Remember all that stuff about how a Bayesian money-maximizer would behave? That was crazy. The Bayesian money-maximizer would, in fact, lose all its money rather quickly (with very high probability). Its in-expectation returns come from increasingly improbable universes. Would natural selection design agents like that, if it could help it?
So, the fact that Bayesian decision theory cares only about in-expectation, rather than caring about typical outcomes, does seem like a problem -- for some utility functions, at least, it results in behavior which humans intuitively feel is insane and ineffective. (And philosophy is all about trying to take human intuitions seriously, even if sometimes we do decide human intuition is wrong and better discarded).
So I applaud Ole Peters' heroic attempt to fix this problem.
However, if you dig into Peters, there's one critical place where his method is sadly arbitrary. You say:
Compounding is multiplicative, so it becomes "natural" (in some sense) to transform everything by taking logs.
Peters makes much of this idea of what's "natural". He talks about additive problems vs multiplicative problems, as well as the more general case (when neither additive/multiplicative work).
However, as far as I can tell, this boils down to creatively choosing a function which makes the math work out.
I find this inexcusable, as-is.
For example, Peters makes much of the St Petersburg lottery:
A resolution of the St Petersburg paradox is presented. In contrast to the standard resolution, utility is not required. Instead, the time-average performance of the lottery is computed. The final result can be phrased mathematically identically to Daniel Bernoulli’s resolution, which uses logarithmic utility, but is derived using a conceptually different argument. The advantage of the time resolution is the elimination of arbitrary utility functions.
Daniel Bernoulli "solved" the St Petersburg paradox by suggesting that utility is log in money, which makes the expectations work out. (This is not very satisfying, because we can always transform the monetary payouts by an exponential function, making things problematic again! D Bernoulli could respond "take the logarithm twice" in such a case, but by this point, it's clear he's just making stuff up, so we ignore him.)
Ole Peters offers the same solution, but, he says, with an improved justification: he promises to eliminate the arbitrary choice of utility functions.
He does this by arbitrarily choosing to examine the time-average behavior in terms of growth rate, which is defined multiplicatively, rather than additively.
"Growth rate" sounds like an innocuous, objective choice of what to maximize. But really, it's sneaking in exactly the same arbitrary decision which D Bernoulli made. Taking a ratio, rather than a difference, is just a sneaky way to take the logarithm.
So, similarly, I see the Peters justification of Kelly as ultimately just a fancy way of saying that taking the logarithm makes the math nice. You're leaning on that argument to a large extent, although you also cite some other properties which I have no beef with.
Again, I think Ole Peters is pretty cool, and there's interesting stuff in this general direction, even if I find that the particular methodology he's developed doesn't justify anything and instead engages in obfuscated question-begging of existing ideas (like Kelly, and like D Bernoulli's proposed solution to St Petersburg).
Also, I could be misunderstanding something about Peters!
I have some interesting disagreements with this.
Prescriptive vs Descriptive
First and foremost: you and I have disagreed in the past on wanting descriptive vs prescriptive roles for probability/decision theory. In this case, I'd paraphrase the two perspectives as:
• Prescriptive-pure-Bayes: as long as we're maximizing an expected utility, we're "good", and it doesn't really matter which utility. But many utilities will throw away all their money with probability close to 1, so Kelly isn't prescriptively correct.
• Descriptive-pure-Bayes: as long as we're not throwing away money for nothing, we're implicitly maximizing an expected utility. Maximizing typical (i.e. modal/median/etc) long-run wealth is presumably incompatible with throwing away money for nothing, so presumably a typical-long-run-wealth-maximizer is also an expected utility maximizer. (Note that this is nontrivial, since "typical long-run wealth" is not itself an expectation.) Sure enough, the Kelly rule has the form of expected utility maximization, and the implicit utility is logarithmic.
In particular, this is relevant to:
Remember all that stuff about how a Bayesian money-maximizer would behave? That was crazy. The Bayesian money-maximizer would, in fact, lose all its money rather quickly (with very high probability). Its in-expectation returns come from increasingly improbable universes. Would natural selection design agents like that, if it could help it?
"Does Bayesian utility maximization imply good performance?" is mainly relevant to the prescriptive view. "Does good performance imply Bayesian utility maximization?" is the key descriptive question. In this case, the latter would say that natural selection would indeed design Bayesian agents, but that does not mean that every Bayesian agent is positively selected - just that those designs which are positively selected are (approximately) Bayesian agents.
"Natural" -> Symmetry
Peters makes much of this idea of what's "natural". He talks about additive problems vs multiplicative problems, as well as the more general case (when neither additive/multiplicative work).
However, as far as I can tell, this boils down to creatively choosing a function which makes the math work out.
I haven't read Peters, but the argument I see in this space is about symmetry/exchangeability (similar to some of de Finetti's stuff). Choosing a function which makes reward/utility additive across timesteps is not arbitrary; it's making utility have the same symmetry as our beliefs (in situations where each timestep's variables are independent, or at least exchangeable).
In general, there's a whole cluster of theorems which say, roughly, if a function is invariant under re-ordering its inputs, then it can be written as for some g, h. This includes, for instance, characterizing all finite abelian groups as modular addition, or de Finetti's Theorem, or expressing symmetric polynomials in terms of power-sum polynomials. Addition is, in some sense, a "standard form" for symmetric functions.
Suppose we have a sequence of n bets. Our knowledge is symmetric under swapping the bets around, and our terminal goals don't involve the bets themselves. So, our preferences should be symmetric under swapping the bets around. That implies we can write it in the "standard form" - i.e. we can express our preferences as a function of a sum of some summary data about each bet.
I'm not seeing the full argument yet, but it feels like there's something in roughly that space. Presumably it would derive a de Finetti-style exchangeability-based version of Bayesian reasoning.
I agree with your prescriptive vs descriptive thing, and agree that I was basically making that mistake.
I think the correct position here is something like: expected utility maximization; and also, utility in "these cases" is going to be close to logarithmic. (IE, if you evolve trading strategies in something resembling Kelly's conditions, you'll end up with something resembling Kelly agents. And there's probably some generalization of this which plausibly abstracts aspects of the human condition.)
But note how piecemeal and fragile this sounds. One layer is the relatively firm expectation-maximization layer. On top of this we add another layer (based on maximization of mode/median/quantile, so that we can ignore things not true with probability 1) which argues for some utility functions in particular.
Ole Peters is trying to re-found decision theory on the basis of this second layer alone. I think this is basically a good instinct:
1. It's good to try to firm up this second layer, since just Kelly alone is way too special-case, and we'd like to understand the phenomenon in as much generality as possible.
2. It's good to try and make a 1-layer system rather than a 2-layer one, to try and make our principles as unified as possible. The Kelly idea is consistent with our foundation of expectation maximization, sure, but if "realistic" agents systematically avoid some utility functions, that makes expectation maximization a worse descriptive theory. Perhaps there is a better one.
This is similar to the way Solomonoff is a two-layer system: there's a lower layer of probability theory, and then on top of that, there's the layer of algorithmic information theory, which tells us to prefer particular priors. In hindsight this should have been "suspicious"; logical induction merges those two layers together, giving a unified framework which gives us (approximately) probability theory and also (approximately) algorithmic information theory, tying them together with a unified bounded-loss notion. (And also implies many new principles which neither probability theory nor algorithmic information theory gave us.)
So although I agree that your descriptive lens is the better one, I think that lens has similar implications.
As for your comments about symmetry -- I must admit that I tend to find symmetry arguments to be weak. Maybe you can come up with something cool, but I would tend to predict it'll be superseded by less symmetry-based alternatives. For one thing, it tends to be a two-layered thing, with symmetry constraints added on top of more basic ideas.
This was fascinating. Thanks for taking the time to write it. I agree with the vast majority of what you wrote, although I don't think it actually applies to what I was trying to do in this post. I don't disagree that a full-Bayesian finds this whole thing a bit trivial, but I don't believe people are fully Bayesian (to the extent they know their utility function) and therefore I think coming up with heuristics is valuable to help them think about things.
So, similarly, I see the Peters justification of Kelly as ultimately just a fancy way of saying that taking the logarithm makes the math nice. You're leaning on that argument to a large extent, although you also cite some other properties which I have no beef with.
I don't really think of it as much as an "argument". I'm not trying to "prove" Kelly criterion. I'm trying to help people get some intuition for where it might come from and some other reasons to consider it if they aren't utility maximising.
It's interesting to me that you brought up the exponential St Petersburg paradox, since MacLean, Thorpe, Ziemba claim that Kelly criterion can also handle it although I personally haven't gone through the math.
Yeah, in retrospect it was a bit straw of me to argue the pure bayesian perspective like I did (I think I was just interested in the pure-bayes response, not thinking hard about what I was trying to argue there).
So, similarly, I see the Peters justification of Kelly as ultimately just a fancy way of saying that taking the logarithm makes the math nice. You’re leaning on that argument to a large extent, although you also cite some other properties which I have no beef with.
I don’t really think of it as much as an “argument”. I’m not trying to “prove” Kelly criterion.
I'm surprised at that. Your post read to me as endorsing Peters' argument. (Although you did emphasize that you were not trying to say that Kelly was the one true rule.)
Hm. I guess I should work harder on articulating what position I would argue for wrt Kelly. I basically think there exist good heuristic arguments for Kelly, but people often confuse them for more objective than they are (either in an unsophisticated way, like reading standard arguments for Kelly and thinking they're stronger than they are, or in a sophisticated way, like Peters' explicit attempt to rewrite the whole foundation of decision theory). Which leads me to reflexively snipe at people who appear to be arguing for Kelly, unless they clearly distinguish themselves from the wrong arguments.
I'm very interested in your potential post on corrections to Kelly.
Comment/question about St. Petersburg and utilities: given any utility function u which goes to infinity, it should be possible to construct a custom St. Petersburg lottery for that utility function, right? I.e. a lottery with infinite expectation but arbitrarily low probability of being in the green. If we want to model an agent as universally rejecting such lotteries, it follows that utility cannot diverge, and thus must asymptotically approach some supremum (also requiring the natural condition that u is strictly monotone). Has this shape of utility function been seriously proposed in economics? Does it have a name?
I wondered if someone would bring this up! Yes, some people take this as a strong argument that utilities simply have to be bounded in order to be well-defined at all. AFAIK this is just called a "bounded utility function". Many of the standard representation theorems also imply that utility is bounded; this simply isn't mentioned as often as other properties of utility.
However, I am not one of the people who takes this view. It's perfectly consistent to define preferences which must be treated as unbounded utility. In doing so, we also have to specify our preferences about infinite lotteries. The divergent sum doesn't make this impossible; instead, what it does is allow us to take many different possible values (within some consistency constraints). So for example, a lottery with 50% probability of +1 util, 25% of -3, 1/8th chance of +9, 1/16 -27, etc can be assigned any expected value whatsoever. Its evaluation is subjective! So in this framework, preferences encode more information than just a utility for each possible world; we can't calculate all the expected values just from that. We also have to know how the agent subjectively values infinite lotteries. But this is fine!
How that works is a bit technical and I don't want to get into it atm. From a mathematical perspective, it's pretty "standard/obvious" stuff (for a graduate-level mathematician, anyway). But I don't think many professional philosophers have picked up on this? The literature on Infinite Ethics seems mostly ignorant of it?
Sorry to comment on a two year old post but: I'm actually quite intrigued by the arbitrary valuation of infinite lotteries thing here. Can you explain this a bit further or point me to somewhere I can learn about it? (P-adic numbers come to mind, but probably have nothing to do with it.)
Commenting on old posts is encouraged :)
I almost like what this post is trying to do, except that Kelly isn't just about repeated bets. It's about multiplicative returns and independent bets. If the returns from your bets add (rather than multiply), then Kelly isn't optimal. This is the case, for instance, for many high-frequency traders - the opportunities they exploit have limited capacity, so if they had twice as much money, they would not actually be able to bet twice as much.
The logarithm in the "maximize expected log wealth" formulation is a reminder of that. If returns are multiplicative and bets are independent, then the long run return will be the product of individual returns, and the log of long-run return will be the sum of individual log returns. That's a sum of independent random variables, so we apply the central limit theorem, and find that long-run return is roughly e^((number of periods)*(average expected log return)). To maximize that, we maximize expected log return each timestep, which is the Kelly rule.
The logarithm in the "maximize expected log wealth" formulation is a reminder of that. If returns are multiplicative and bets are independent, then the long run return will be the product of individual returns, and the log of long-run return will be the sum of individual log returns. That's a sum of independent random variables, so we apply the central limit theorem, and find that long-run return is roughly e^((number of periods)*(average expected log return)). To maximize that, we maximize expected log return each timestep, which is the Kelly rule.
Wait, so, what's actually the argument?
It looks to me like the argument is "returns are multiplicative, so the long-run behavior is log-normal. But we like normals better than log-normals, so we take a log. Now, since we took a log, when we maximize we'll find that we're maximizing log-wealth instead of wealth."
But what if I like log-normals fine, so I don't transform things to get a regular normal distribution? Then when I maximize, I'm maximizing raw money rather than log money.
I don't see a justification of the Kelly formula here.
The central limit theorem is used here to say "our long-run wealth will converge to e^((number of periods)*(average expected log return)), modulo error bars, with probability 1". So, with probability 1, that's the wealth we get (within error), and maximizing modal/median/any-fixed-quantile wealth will all result in the Kelly rule.
Under ordinary conditions, it's pretty safe to argue "such and such with probability 1, therefore, it's safe to pretend such-and-such". But this happens to be a case where doing so makes us greatly diverge from the Bayesian analysis -- ignoring the "most important" worlds from a pure expectation-maximization perspective (IE the ones where we repeatedly win bets, amassing huge sums).
So I'm very against sweeping that particular part of the reasoning under the rug. It's a reasonable argument, it's just one that imho should come equipped with big warning lights saying "NOTE: THIS IS A SEVERELY NON-BAYESIAN STEP IN THE REASONING. DO NOT CONFUSE THIS WITH EXPECTATION MAXIMIZATION."
To maximize that, we maximize expected log return each timestep, which is the Kelly rule.
which, to my eye, doesn't provide any warning to the reader. I just really think this kind of argument should be explicit about maximizing modal/median/any-fixed-quantile rather than the more common expectation maximization. Because people should be aware if one of their ideas about rational agency is based on mode/median/quantile maximization rather than expectation maximization.
So, with probability 1, that's the wealth we get (within error), and maximizing modal/median/any-fixed-quantile wealth will all result in the Kelly rule.
Sorry, but I'm pulling out my "wait, what's actually the claim here?" guns again.
Kelly maximizes a kind of pseudo-mode. For example, for a sequence of two bets on coin flips, Kelly optimizes the world where you win one and lose one, which is the mode. However, for three such bets, Kelly optimizes the world where you win 1.5 and lose 1.5, which isn't even physically possible.
At first I thought there would be an obvious sense in which this is approximately mode-optimizing, getting closer and closer to mode-optimal in the limit. And maybe so. But it's not obvious.
The pseudo-mode is never more than 1 outcome away from the true mode. However, one bet can make a lot of difference, and can make more difference if we have more rounds to accumulate money. So certainly we can't say that Kelly's maximization problem (I mean the maximization of wealth in the pseudo-mode world) becomes epsilon close to true mode-optimization, in terms of numerical measurement of the quality of a given strategy.
I'm not even sure that the mode-optimizing strategy is a fixed-fraction strategy like Kelly. Maybe a mode-optimizing strategy does some clever things in worlds where it starts winning unusually much, to cluster those worlds together and make their winnings into a mode that's much better than the mode of Kelly.
our long-run wealth will converge to e^((number of periods)*(average expected log return)), modulo error bars
If the error bars became epsilon-close, then the argument would obviously work fine: the mode/median/quantile would all be very close to the same number, and this number would itself be very close to the pseudo-mode Kelly optimizes. So then Kelly would clearly come very close to optimizing all these numbers.
If the error bars became epsilon-close in ratio instead of in absolute difference, we could get a weaker but still nice guarantee: Kelly might fall short of mode-optimal by millions of dollars, but only in situations where "millions" is negligible compared to total wealth. This is comforting if we have diminishing returns in money.
But because any one bet can change wealth by a non-vanishing fraction (in general, and in particular when following Kelly), we know neither of those bounds can hold.
So in what sense, if any, does Kelly maximize mode/median/quantile wealth?
I'm suspecting that it may only maximize approximate mode/median/quantile, rather than approximately maximizing mode/median/quantile.
Just think of the whole thing on a log scale. The error bars become epsilon close in ratio on a log scale. There's some terms and conditions to that approximation - average expected log return must be nonzero, for instance. But it shouldn't be terribly restrictive.
If your immediate thought is "but why consider things on a log scale in the first place?", then remember that we're talking about mode/order statistics, so monotonic transformations are totally fine.
(Really, though, if we want to be precise... the exact property which makes Kelly interesting is that if you take the Kelly strategy and any other strategy, and compare them to each other, then Kelly wins with probability 1 in the long run. That's the property which tells us that Kelly should show up in evolved systems. We can get that conclusion directly from the central limit theorem argument: as long as the "average expected log return" term grows like O(n), and the error term grows like O(sqrt(n)) or slower, we get the result. In order for a non-Kelly strategy to beat Kelly in the long run with greater-than-0 probability, it would somehow have to grow the error term by O(n).)
If your immediate thought is "but why consider things on a log scale in the first place?", then remember that we're talking about mode/order statistics, so monotonic transformations are totally fine.
Riiight, but, "totally fine" doesn't here mean "Kelly approximately maximizes the mode as opposed to maximizing an approximate mode", does it?
I have approximately two problems with this:
• Bounding the ratio of log wealth compared to a true mode-maximizer would be reassuring if my utility was doubly logarithmic. But if it's approximately logarithmic, this is little comfort.
• But are we even bounding the ratio of log wealth to a true mode-maximizer? As I mentioned, I'm not sure a mode-maximizer is even a constant-fraction strategy.
Sorry if I'm being dense, here.
Actually, you're right, I goofed. Monotonic increasing transformation respects median or order statistics, so e.g. max median F(u) = F(max median u) (since F commutes with both max and median), but mode will have an additional term contributed by any nonlinear transformation of a continuous distribution. (It will still work for discrete distributions - i.e. max mode F(u) = F(max mode u) for u discrete, and in that case F doesn't even have to be monotonic.)
So I guess the argument for median is roughly: we have some true optimum policy and Kelly policy , and , which implies as long as F is continuous and strictly increasing.
Yeah - I agree, that was what I was trying to get at. I tried to address (the narrower point) here:
Compounding is multiplicative, so it becomes "natural" (in some sense) to transform everything by taking logs.
But I agree giving some examples of where it doesn't apply would probably have been helpful to demonstrate when it is useful
This is my first post, so I would appreciate any feedback. This started out as a comment on one of the other threads but kept on expanding from there.
I'm also tempted to write "Contra Kelly Criterion" or "Kelly is just the start" where I write a rebuttal to using Kelly. (Rough sketch - Kelly is not enough you need to understand your edge, Kelly is too volatile). Or "Fractional Kelly is the true Kelly" (either a piece about how fractional Kelly accounts for your uncertainty vs market uncertainty OR a piece about how fractional Kelly is about power utilities OR a piece about fractional Kelly is optimal in some risk-return sense)
This was a pretty solid post, and outstanding for a first post. Well done. A few commenters said the tone seemed too strong or something along those lines, but I personally think that's a good thing. "Strong opinions, weakly held" is a great standard to adhere to; at least some pushback in the comments is a good thing. I think your writing fits that standard well.
[-]gjm40
Of those, I think I would vote for one of the fractional-Kelly ones -- providing arguments for something (fractional Kelly) rather than merely against naive Kelly. (Suppose I think one should "Kelly bet on everything", and you write an article saying that Kelly is too volatile, and I read it and am convinced. Now I know that what I've been doing is wrong, but I don't know what I should be doing instead. On the other hand, if instead I read an article saying that fractional Kelly is better in specific circumstances for specific reasons, then it probably also gives some indication of how to choose the fraction one uses, and now I have not merely an awareness that my existing strategy is bad but a concrete better strategy to use in future.)
I appreciated that you posted this, and I think your further proposed post(s) sound good, too!
To my personal taste, my main disappointment when reading the post was that you didn't expand this section more:
Utilities and repeated bets are two sides of the same coin
I think this is fairly clear - "The time interpretation of expected utility theory" (Peters, Adamou) prove this. I don't want to say too much more about this. This section is mostly to acknowledge that:
• Yes, Kelly is maximising log utility
• Yes, I do think thinking in the repeated bets framework is more useful
But that's a perfectly reasonable choice given (a) writing time constraints, (b) time constraints of readers. You probably shouldn't feel pressured to expand arguments out more as opposed to writing up rough thoughts with pointers to the info.
Speaking of which, I appreciated all the references you included!
Yeah, I think I'm about to write a reply to your massive comment, but I think I'm getting closer to understanding. I think what I really need to do is write my "Kelly is Black-Scholes for utility" post.
I think that (roughly) this post isn't aimed at someone who has already decided what their utility is. Most of the examples you didn't like / saw as non-sequitor were explicitly given to help people think about their utility.
I think that (roughly) this post isn't aimed at someone who has already decided what their utility is. Most of the examples you didn't like / saw as non-sequitor were explicitly given to help people think about their utility.
Ah, I suspect this is a mis-reading of my intention. For a good portion of my long response, I was speaking from the perspective of a standard Bayesian. Standard Bayesians have already decided what their utility is. I didn't intend that part as my "real response". Indeed, I intended some of it to sound a bit absurd. However, a lotta folks round here are real fond of Bayes, so articulating "the bayesian response" seems relevant.
If you'd wanted to forestall that particular response, in writing the piece, I suppose you could have been more explicit about which arguments are Bayesian, and which are explicitly anti-Bayesian (IE Peters), and where you fall wrt accepting Bayesian / anti-Bayesian assumptions. Partly I thought you might be pretty Bayesian and would take "Bayesians wouldn't accept this argument" pretty seriously. (Although I also had probability on you being pretty anti-Bayesian.)
Why teach about these concepts in terms of the Kelly criterion, if the Kelly criterion isn't optimal? You could just teach about repeated bets directly.
A couple of reasons:
1. For whatever reason, people seem to really like Kelly criterion related posts at the moment.
2. I think Kelly is a good framework for thinking about things
1. "Kelly is about repeated bets" could easily be "Kelly is about bet sizing"
2. "Kelly is Black-Scholes for utility"
3. Kelly is optimal (in some very concrete senses) and fractional-Kelly is optional in some other senses which I think people don't discuss enough
I've only skimmed this post so far, so, I'll need to read it in greater detail. However, I think this is quite related to Nassim Taleb and Ole Peters' critique of Bayesian decision theory. I've left fairly extensive comments there, although I'm afraid I changed my mind a couple of times as I read deeper into the material, and there's not a nice summary of my final assessment.
I would currently defend the following statements:
• From a Bayesian standpoint, Kelly is all about logarithmic utility, and the arguments about repeated bets don't make very much sense.
• From a frequentist standpoint, the arguments about repeated bets make way more sense. Ole Peters is developing a true frequentist decision theory: it used to be that Bayesianism was the only game in town if you wanted your philosophy of statistics to also give you decision theory, but no longer!
• I ultimately find Peters' decision theory unsatisfying. It promises to justify different strategies for different situations (EG logarithmic utility for Kelly-type scenarios), but really, pulls the quantity to be maximized out of thin air (IE the creativity of the practitioner). It's slight of hand.
Anyway, I'll read the post more closely, and go further into my critique of Peters if it seems relevant.
From a Bayesian standpoint, Kelly is all about logarithmic utility, and the arguments about repeated bets don't make very much sense.
This disagrees with my current understanding, so I'd be interested to hear the reasoning.
Yes - I cited Peters in the post (and stole one of their images). Personally I don't actually think what they are doing has as much value as they seem to think, although that's a whole other conversation. I basically think something akin to your third bullet point.
Having read your comments on the other post, I think I understand your critique, and I don't think there's much more to be said if you take the utility as axiomatic. However, I guess the larger point I'm trying to make is there are other reasons to care about Kelly other than if you're a log-utility maximiser. (Several of which you mention in your post)
[-]gjm30
There's something odd about the tone of this piece. It begins by saying three times (three times!) "Kelly isn't about logarithmic utility. Kelly is about repeated bets." Then, later, it says "Yes, Kelly is maximising log utility. No, it doesn't matter which way you think about this".
I think you have to choose between "You can think about Kelly in terms of maximizing utility = log(wealth), and that's OK" and "Thinking about it that way is sufficiently not-OK that I wrote an article that says not to do it in the title and then reiterates three times in a row that you shouldn't do it".
I do agree that there are excellent arguments for Kelly-betting that assume very little about your utility function. (E.g., if you are going to make a long series of identical bets at given odds then almost certainly the Kelly bet leaves you better off at the end than any different bet.) And I do agree that there's something a bit odd about focusing on details of the formula rather than the underlying principles. (Though I don't think "odd" is the same as "wrong"; if you ever expect to be making Kelly bets, or non-Kelly bets informed by the Kelly criterion, it's not a bad idea to have a memorable form of the formula in your head, as well as the higher-level concept that maximizing expected log wealth is a good idea.) But none of that seems to explain how cross you seem to be about it at the start of the article.
Thanks! That's helpful. I definitely wrote this rather stream of consciousness and I definitely was more amped up about what I was going to say at the start than I was by the time I'd gotten halfway through. EDIT: I've changed the title an added a note at the top
I think the section where I say "it doesn't matter how you think about this" I mean it something in the sense of: "Prices and vols are equivalent in a Black-Scholes world, it doesn't matter if you think in terms of prices of vols, but thinking in terms of vols is usually much more helpful".
I also agree that having a handy version of the formula is useful. I basically think of Kelly in the format you do in your comment I highlighted and I think I would never have written this if someone else hadn't taken that comment butchered it a little but and it became a (somewhat) popular post. (Roughly I started writing a long fairly negative comment on that post, and tried to turn it into something more positive. I see I didn't quite manage to avoid all the anger issues that entails).
We have a stock which returns 50% or -40%. If we put all our money in this stock, we are going to lose it over time since but on any given day, this stock has a positive expected return.
I like this example as a way of thinking about the problem.
This post makes some good points, too strongly. I will now proceed to make some good points, too rambly.
F=ma is a fantastically useful concept, even if in practice the scenario is always more complicated than just "apply this formula and you win". It's short and intuitive and ports the right ideas about energy and mass and stuff into your brain.
Maximizing expected value is a fantastically useful concept.
Maximizing expected value at a time t after many repeated choices is a fantastically useful concept.
(Edit: see comments, there are false statements incoming. Leaving it all up for posterity.) It turns out that maximizing expected magnitude of value at every choice is very close to the optimal way to maximize your expected value at a time t after many repeated choices. Kelly is a nice, intuitive, easy heuristic for doing so.
So yes, what you really want to do is maximize something like "your total wealth at times t0 and t1 and t2 and t3 and... weighted by how much you care about having wealth at those times, or something" and the way to do that is to implement a function which knows there will be choices in the future and remembers to take into account now, on this choice having the power to maximally exploit those future choices, aka think about repeated bets. But also the simple general way to do something very close to that maximization is just "maximize magnitude of wealth", and the intuitions you get from thinking "maximize magnitude of wealth" are more easily ported than the intuitions you get from thinking "I have a universe of future decisions with some distribution and I will think about all of them and determine the slightly-different-from-maximize-log-wealth way to maximize my finnicky weighted average over future selves' wealth".
Do you need to think about whether you should use your naive estimate of the probability you win this bet, when plugging into Kelly to get an idea of what to do? Absolutely. If you're not sure of the ball's starting location, substituting a point estimate with wide variance into the calculation which eventually feeds into F=ma will do bad things and you should figure out what those things are. But starting from F=ma is still the nice, simple concept which will be super helpful to your brain.
Kelly is about one frictionless sphere approximation to [the frictionless sphere approximation which is maximizing magnitude of wealth]. Bits that were rubbed off to form the spheres involve repeated bets, for sure.
Maximizing expected value is a fantastically useful concept.
Maximizing expected value at a time t after many repeated choices is a fantastically useful concept.
It turns out that maximizing expected magnitude of value at every choice is very close to the optimal way to maximize your expected value at a time t after many repeated choices. Kelly is a nice, intuitive, easy heuristic for doing so.
I'm not 100% sure what mathematical fact you are trying to refer to here, but I am worried that you are stating a falsehood.
Slightly editing stuff from another comment of mine:
In a one-step scenario, the Bayesian wants to maximize where is your starting money and is a random variable for the payoff-per-dollar of your strategy. In a two-step scenario, the Bayesian wants to maximize . And so on. If , this allows us to push the expectation inwards; (the last step holds because we assume the random variables are independent). So in that case, we could just choose the best one-step strategy and apply it at each time-step.
In other words, starting with the idea of raw expectation maximization (maximizing money, so , where is our bankroll) and adding the idea of iteration, we don't get any closer to Kelly. Kelly isn't an approximately good strategy for the version of the game with a lot of iterations. The very same greedy one-step strategy remains optimal forever.
But I could be misunderstanding the point you were trying to communicate.
You're absolutely right! I think I have a true intuition I'm trying to communicate, and will continue to think about it and see, but it might turn out that the entirety of the intuition can be summarized as "actually the utility is nonlinear in money".
I’m not sure what prompted all of this effort, and I’ve rarely heard Kelly described as corresponding to log utility, only ever as an aside about mean-variance optimization. There, log utility corresponds to A=1, which is also the Kelly portfolio. That is the maximally aggressive portfolio, and most people are much, much more risk averse.
If anything, I’d say that the Kelly - log utility connection obviously suggests one point, which is that most people are far too risk-averse (less normatively, most people don’t have log utility functions). The exception is Buffett - empirically he does, subject to leverage constraints.
I’m not sure what prompted all of this effort, and I’ve rarely heard Kelly described as corresponding to log utility, only ever as an aside about mean-variance optimization. There, log utility corresponds to A=1, which is also the Kelly portfolio. That is the maximally aggressive portfolio, and most people are much, much more risk averse.
I would defend the idea that Kelly is more genuinely about log utility when approached from a strict Bayesian perspective, ie, a Bayesian has little reason to buy the other arguments in favor of Kelly.
I’m not sure what prompted all of this effort,
The comments section here and the post and comments section here. To be completely frank, my post started out as a comment similar to yours in those threads. "I'm not sure what led you to post this". (Especially the Calculating Kelly post which seemed to mostly copy and make worse this comment).
I’ve rarely heard Kelly described as corresponding to log utility,
I actually agree with you that aside from LW I haven't really seen Kelly discussed in the context of log-utilities, which is why I wanted to address this here rather than anywhere else.
only ever as an aside about mean-variance optimization
Okay, here our experiences differ. I see Kelly coming up in all sorts of contexts, not just relating to mean-variance portfolio optimization for a CRRA-utility or whatever.
If anything, I’d say that the Kelly - log utility connection obviously suggests one point, which is that most people are far too risk-averse (less normatively, most people don’t have log utility functions). The exception is Buffett - empirically he does, subject to leverage constraints.
So I agree with this. I'd quite happily write the "you are too risk averse" post, but I think Putanumonit already did a better job than I could hope to do on that
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# Carnot's theorem (thermodynamics)
Carnot's theorem, developed in 1824 by Nicolas Léonard Sadi Carnot, also called Carnot's rule, is a principle that specifies limits on the maximum efficiency any heat engine can obtain, which thus solely depends on the difference between the hot and cold temperature reservoirs.
Carnot's theorem states:
• All heat engines between two heat reservoirs are less efficient than a Carnot heat engine operating between the same reservoirs.
• Every Carnot heat engine between a pair of heat reservoirs is equally efficient, regardless of the working substance employed or the operation details.
The formula for this maximum efficiency is
$\eta_{\text{max}} = \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$
where TC is the absolute temperature of the cold reservoir, TH is the absolute temperature of the hot reservoir, and the efficiency $\eta$ is the ratio of the work done by the engine to the heat drawn out of the hot reservoir.
Based on modern thermodynamics, Carnot's theorem is a result of the second law of thermodynamics. Historically, however, it was based on contemporary caloric theory and preceded the establishment of the second law.[1]
## Proof
An impossible situation: A heat engine cannot drive a less efficient (reversible) heat engine without violating the second law of thermodynamics.
The proof of the Carnot theorem is a proof by contradiction, or reductio ad absurdum, as illustrated by the figure showing two heat engines operating between two reservoirs of different temperature. The heat engine with more efficiency ($\eta_M$) is driving a heat engine with less efficiency ($\eta_L$), causing the latter to act as a heat pump. This pair of engines receives no outside energy, and operates solely on the energy released when heat is transferred from the hot and into the cold reservoir. However, if $\eta_M>\eta_L$, then the net heat flow would be backwards, i.e., into the hot reservoir:
$Q^\text{out}_\text{hot} = Q < \frac{\eta_M}{\eta_L}Q=Q^\text{in}_\text{hot}$.
It is generally agreed that this is impossible because it violates the second law of thermodynamics.
We begin by verifying the values of work and heat flow depicted in the figure. First, we must point out an important caveat: the engine with less efficiency ($\eta_L$) is being driven as a heat pump, and therefore must be a reversible engine. If the less efficient engine ($\eta_L$) is not reversible, then the device could be built, but the expressions for work and heat flow shown in the figure would not be valid.
By restricting our discussion to cases where engine ($\eta_L$) has less efficiency than engine ($\eta_M$), we are able to simplify notation by adopting the convention that all symbols, $Q$ and $W$ represent non-negative quantities (since the direction of energy flow never changes sign in all cases where $\eta_L\leqslant\eta_M$). Conservation of energy demands that for each engine, the energy which enters, $E_{in}$, must equal the energy which exits, $E_{out}$:
$E_{in}^{M} = Q = (1-\eta_M)Q + \eta_M Q = E_{out}^{M}$,
$E_{in}^{L} = \eta_M Q + \eta_M Q \left(\frac{1}{\eta_{L}}- 1 \right )=\frac{\eta_M}{\eta_L}Q = E_{out}^{L}$,
The figure is also consistent with the definition of efficiency as $\eta=W/Q_h$ for both engines:
$\eta_M= \frac{W_M}{Q^M_h}=\frac{\eta_M Q}{Q}=\eta_M$,
$\eta_L=\frac{W_L}{Q^L_h}=\frac{\eta_M Q}{\frac{\eta_M}{\eta_L}Q}=\eta_L$.
It may seem odd that a hypothetical heat pump with low efficiency is being used to violate the second law of thermodynamics, but the figure of merit for refrigerator units is not efficiency, $W/Q_h$, but the coefficient of performance (COP),[2] which is $Q_c/W$. A reversible heat engine with low thermodynamic efficiency, $W/Q_h$ delivers more heat to the hot reservoir for a given amount of work when it is being driven as a heat pump.
Having established that the heat flow values shown in the figure are correct, Carnot's theorem may be proven for irreversible and the reversible heat engines.[3]
### Reversible engines
To see that every reversible engine operating between reservoirs $T_1$ and $T_2$ must have the same efficiency, assume that two reversible heat engines have different values of $\eta$, and let the more efficient engine (M) drive the less efficient engine (L) as a heat pump. As the figure shows, this will cause heat to flow from the cold to the hot reservoir without any external work or energy, which violates the second law of thermodynamics. Therefore both (reversible) heat engines have the same efficiency, and we conclude that:
All reversible engines that operate between the same two heat reservoirs have the same efficiency.
This is an important result because it helps establish the Clausius theorem, which implies that the change in entropy is unique for all reversible processes.,[4]
$\Delta S = \int_a^b \frac {dQ_{rev}}T$,
over all paths (from a to b in V-T space). If this integral were not path independent, then entropy, S, would lose its status as a state variable.[5]
### Irreversible engines
If one of the engines is irreversible, it must be the (M) engine, placed so that it reverse drives the less efficient but reversible (L) engine. But if this irreversible engine is more efficient than the reversible engine, (i.e., if $\eta_M >\eta_L$), then the second law of thermodynamics is violated. And, since the Carnot cycle represents a reversible engine, we have the first part of Carnot's theorem:
No irreversible engine is more efficient that the Carnot engine.
## Definition of thermodynamic temperature
The efficiency of the engine is the work divided by the heat introduced to the system or
$\eta = \frac {w_{cy}}{q_H} = \frac{q_H-q_C}{q_H} = 1 - \frac{q_C}{q_H}$
(1)
where wcy is the work done per cycle. Thus, the efficiency depends only on qC/qH.
Because all reversible engines operating between the same heat reservoirs are equally efficient, any reversible heat engine operating between temperatures T1 and T2 must have the same efficiency, meaning, the efficiency is the function of the temperatures only:
$\frac{q_C}{q_H} = f(T_H,T_C)$
(2)
In addition, a reversible heat engine operating between temperatures T1 and T3 must have the same efficiency as one consisting of two cycles, one between T1 and another (intermediate) temperature T2, and the second between T2andT3. This can only be the case if
$f(T_1,T_3) = \frac{q_3}{q_1} = \frac{q_2 q_3} {q_1 q_2} = f(T_1,T_2)f(T_2,T_3).$
Specializing to the case that $T_1$ is a fixed reference temperature: the temperature of the triple point of water. Then for any T2 and T3,
$f(T_2,T_3) = \frac{f(T_1,T_3)}{f(T_1,T_2)} = \frac{273.16 \cdot f(T_1,T_3)}{273.16 \cdot f(T_1,T_2)}.$
Therefore, if thermodynamic temperature is defined by
$T = 273.16 \cdot f(T_1,T) \,$
then the function f, viewed as a function of thermodynamic temperature, is
$f(T_2,T_3) = \frac{T_3}{T_2},$
and the reference temperature T1 has the value 273.16. (Of course any reference temperature and any positive numerical value could be used—the choice here corresponds to the Kelvin scale.)
It follows immediately that
$\frac{q_C}{q_H} = f(T_H,T_C) = \frac{T_C}{T_H}$
(3)
Substituting Equation 3 back into Equation 1 gives a relationship for the efficiency in terms of temperature:
$\eta = 1 - \frac{q_C}{q_H} = 1 - \frac{T_C}{T_H}$
(4)
## Applicability to fuel cells and batteries
Since fuel cells and batteries can generate useful power when all components of the system are at the same temperature ($T=T_H=T_C$), they are clearly not limited by Carnot's theorem, which states that no power can be generated when $T_H=T_C$. This is because Carnot's theorem applies to engines converting thermal energy to work, whereas fuel cells and batteries instead convert chemical energy to work.[6] Nevertheless, the second law of thermodynamics still provides restrictions on fuel cell and battery energy conversion.[7]
## References
1. ^ John Murrell (2009). "A Very Brief History of Thermodynamics". Retrieved May 2, 2014. PDF (142 KB)
2. ^ Tipler, Paul; Mosca, G. (2008). "19.2, 19.7". Physics for Scientists and Engineers (6th ed.). Freeman. ISBN 9781429201322.
3. ^ "Lecture 10: Carnot theorem" (PDF). Feb 7, 2005. Retrieved October 5, 2010.
4. ^ Ohanian, Hans (1994). Principles of Physics. W.W. Norton and Co. p. 438. ISBN 039395773X.
5. ^ http://faculty.wwu.edu/vawter/PhysicsNet/Topics/ThermLaw2/ThermalProcesses.html, and http://www.itp.phys.ethz.ch/education/hs10/stat/slides/Laws_TD.pdf. Both retrieved 13 December 2013.
6. ^ "Fuel Cell versus Carnot Efficiency". Retrieved Feb 20, 2011.
7. ^ Jacob, Kallarackel T; Jain, Saurabh (July 2005). Fuel cell efficiency redefined : Carnot limit reassessed. Q1 - Ninth International Symposium on Solid Oxide Fuel Cells (SOFC IX). USA.
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# Analyzing Linear or Non Linear Plot Structure
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## Robyn Shields
on 24 February 2014
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#### Transcript of Analyzing Linear or Non Linear Plot Structure
Students will analyze and discuss the purpose of non-linear plot development (e.g., flashbacks, foreshadowing, sub-plots, parallel plot structures)and compare it to linear plot development.
Students will collaborate and share ideas of a linear plot and non-linear plot structure in a mixed-pair share activity.
Students will discuss, as a class, their understanding of linear and non-linear plot structures.
Objective
Linear Plot Structure
A linear plot consist of a series of events starting with a beginning, middle, and ending with a conclusion.
Non-Linear Plot Structure
What is Linear Plot Structure?
.
By; Miss Shields
Analyzing Linear and Non Linear Plot Structure
A non-linear plot structure consist of a series of events that are out of order. Authors and filmmakers mainly use non-linear plot to copy the structure and recall of human memory
Non-Linear Plot Elements
Flashback
Parallel Plot
Sub Plot
Flashback
A transition in a book or film to an earlier event that interrupts the chronological development of a story
Usually in first person point of view, but can be viewed from different points of view.
The presentation of details, characters, or incidents in a narrative in such a way that later events are prepared for (or "shadowed forth").
For Example:
In the opening of The Wizard of Oz, set in Kansas, the transformation of Miss Gulch into a witch on a broomstick foreshadows her reappearance as Dorothy's enemy in Oz.
Parallel Plot & Sub-Plot
Parallel Plot
Two or more plots that occur within a story and intersect
The plots usually takes place within the same time frame
*In parallel plot, each plot is essential to the story
For Example
Movie: Pulp Fiction
Literature: Romeo and Juliet
Sub-plot
A secondary story in a narrative.
A subplot may serve as a motivating or complicating force for the main plot of the work, or it may provide emphasis for, or relief from, the main plot.
Example:
Harry Potter series
The story is about Harry Potter, but intersected is a subplot based on Hermione’s crush on Ron. Though Harry’s adventures continue unabated, the girl’s sentiments toward Ron take over entire scenes at times
Think Pair Share Activity
Questions:
What are the major differences of Linear and Non-Linear Plots?
How can the elements of a Non-Linear plot connect to the the main plot?
Full transcript
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physics
a 1 ohm resistor, a 1000 resistor and a 2000 ohm resistor are connected in parallel. The total resistance is
1. 👍 0
2. 👎 0
3. 👁 57
1. 1/R=1/1+1/1000+1/2000
2000/R=2000+2+1=2003
R=2000/20003 ohms
1. 👍 0
2. 👎 0
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# Question about fluid expansion
Hi,
When a hot and dense fluid suddenly expands, how much can it cool down.
Lets take for example an expansion valve from an AC, does the fluid do work when passing through the valve or does it not even need to do work in order to cool down.
What is the lowest temperature that the fluid can cool down to?
I hope I've been clear enough, thanks.
## Answers and Replies
Gold Member
From a practical standpoint, cooling this way will be limited by your ability to insulate your regrigerator and its components. The physical process doesn't have a minimum working temperature (besides absolute zero).
If I understand what you're talking about, that process would be the Joule Thompson process (used in refrigeration). In this case, there is no work done on the gas, and the cooling is due to the intermolecular forces between individual molecules in the gas.
Most cold gases will cool further under expansion (through a valve) because as the atoms get further away from one another, the average potential energy increases (because far away, the force is attractive). Since total energy is conserved, the average kinetic energy decreases, and the temperature goes down.
However, this cooling doesn't occur at all temperatures. At temperatures above what's called the inversion temperature, the reverse will happen. This is because the average (in time) potential energy depends on the rate of collisions.
At higher temperatures, the atoms are moving faster, and collide with one another more often.
If a pair of atoms are close enough to each other, the force between them will be repulsive, and the potential energy will have a net decrease when they are far away enough from each other again.
If the cloud of atoms are on average close enough to each other, expanding the gas will lead to a net decrease in potential energy, and a net increase in average kinetic energy (assuming energy is conserved)
This inversion temperature is different for each gas, and is usually found experimentally.
For example:, the inversion temperatures of nitrogen and oxygen are over $500K$, so you could keep cooling them until they liquify starting at room temperature (provided good insulation).
Last edited:
Ravi Singh choudhary and adoion
Thanks a lot for the answer, that's exactly what I wanted, could you please tell me how much this temperature change is in one cycle is it a small amount and needs to be repeated or can it be large with the right substance?
And also, the inversion temperature must be above room temperature in order to be able to cool down to anything you want right?
Gold Member
Unfortunately, I don't know what the temperature change per cycle would be. It may help to look up the Linde cycle, which is an ideal model of a refrigerator based on this process.
The inversion temperature needs to be above your starting temperature to do any cooling. If you start at room temperature, then yes. Otherwise, you could use some other cooling mechanism to get down to a better starting temperature.
Thank you, you are very helpful I wish I could somehow give you a thumbs up or something :)
Gold Member
you can, "like" my post, if you prefer.
Ravi Singh choudhary
From a practical standpoint, cooling this way will be limited by your ability to insulate your regrigerator and its components. The physical process doesn't have a minimum working temperature (besides absolute zero).
If I understand what you're talking about, that process would be the Joule Thompson process (used in refrigeration). In this case, there is no work done on the gas, and the cooling is due to the intermolecular forces between individual molecules in the gas.
Most cold gases will cool further under expansion (through a valve) because as the atoms get further away from one another, the average potential energy increases (because far away, the force is attractive). Since total energy is conserved, the average kinetic energy decreases, and the temperature goes down.
However, this cooling doesn't occur at all temperatures. At temperatures above what's called the inversion temperature, the reverse will happen. This is because the average (in time) potential energy depends on the rate of collisions.
At higher temperatures, the atoms are moving faster, and collide with one another more often.
If a pair of atoms are close enough to each other, the force between them will be repulsive, and the potential energy will have a net decrease when they are far away enough from each other again.
If the cloud of atoms are on average close enough to each other, expanding the gas will lead to a net decrease in potential energy, and a net increase in average kinetic energy (assuming energy is conserved)
This inversion temperature is different for each gas, and is usually found experimentally.
For example:, the inversion temperatures of nitrogen and oxygen are over $500K$, so you could keep cooling them until they liquify starting at room temperature (provided good insulation).
You made my day :)
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Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# FREQUENCY Units Conversionradian-per-day to kilohertz
= 1.84207E-9 Kilohertz
Category: frequency
Conversion: Radian Per Day to Kilohertz
The base unit for frequency is hertz (Non-SI/Derived Unit)
[Kilohertz] symbol/abbrevation: (kHz)
1 x 1.84207E-9 kHz = 1.84207E-9 Kilohertz.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [frequency] => (hertz), 1 Radian Per Day (rad/d) is equal to 1.84207E-6 hertz, while 1 Kilohertz (kHz) = 1000 hertz.
1 Radian Per Day to common frequency units
1 rad/d = 1.84207E-6 hertz (Hz)
1 rad/d = 1.84207E-9 kilohertz (kHz)
1 rad/d = 1.84207E-12 megahertz (MHz)
1 rad/d = 1.84207E-15 gigahertz (GHz)
1 rad/d = 1.84207E-6 1 per second (1/s)
1 rad/d = 0.00011052420004421 revolutions per minute (rpm)
1 rad/d = 1.84207E-6 frames per second (FPS)
1 rad/d = 0.039788966649387 degree per minute (°/min)
1 rad/d = 1.84207E-18 fresnels (fresnel)
Radian Per Dayto Kilohertz (table conversion)
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# Solve overspecified equation system
2 ビュー (過去 30 日間)
Nikolas Haimerl 2019 年 10 月 28 日
Edited: Stephan 2019 年 10 月 29 日
I am trying to solve a set of equations where I have 3 variables and 4 equations which are not linearly dependent.
Is there a way to solve this numerically in matlab since it is not possible to do it analytically.
Thanks for any help!
#### 2 件のコメント
Stephan 2019 年 10 月 28 日
Please provide the code or at least the system.
Nikolas Haimerl 2019 年 10 月 28 日
syms u1 u2 up2
eqns= [cos(u1)*l1 + cos(u2)*l2 == xa,...
sin(u1)*l1 + sin(u2)*l2 == ya,...
ap0==(xpa*cos(u2))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1))) +...
(ypa*sin(u2))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1))),...
up2 == - (xpa*cos(u1))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1))) -...
(ypa*sin(u1))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1)))];
S=fsolve(eqns,[u1 u2 up2])
This is what I have tried to do so far. u1 u2 up2 are the variables. xpa,ypa,xa,ya,l1,l2 and ap0 are constants.
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### 採用された回答
Stephan 2019 年 10 月 28 日
If you have symbolic toolbox, use:
% This part builds a system to solve with fsolve
syms u1 u2 up2 l1 l2 xa ya xpa ypa ap0
x = sym('x', [1 3]);
eqns(1) = cos(u1)*l1 + cos(u2)*l2 - xa;
eqns(2) = sin(u1)*l1 + sin(u2)*l2 - ya;
eqns(3) = ap0 -(xpa*cos(u2))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1))) +...
(ypa*sin(u2))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1)));
eqns(4) = up2 + (xpa*cos(u1))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1))) -...
(ypa*sin(u1))/(4*(cos(u1)*sin(u2) - cos(u2)*sin(u1)));
eqns = subs(eqns,[u1, u2, up2],[x(1), x(2), x(3)]);
fun = matlabFunction(eqns,'vars',{x,'l1','l2','xa','ya','xpa','ypa'...
'ap0'});
fun = str2func(replace(func2str(fun),"in1","x"));
Then fun is a function handle that you can work with:
% solve the system with fantasy values - use your values
l1 = 3;
l2 = 2;
xa = 3;
ya = 6;
xpa = 0.5;
ypa = -1;
ap0 = 5.7;
% solve the system using fsolve
opts = optimoptions('fsolve','Algorithm','Levenberg-Marquardt');
sol = fsolve(@(x)fun(x,l1,l2,xa,ya,xpa,ypa,ap0),rand(1,3),opts)
% look at the results using the solution fsolve calculated
% ideally there should be 4 zeros, if it worked good - ohterwise
% your system may have a problem
test_results = fun(sol,l1,l2,xa,ya,xpa,ypa,ap0)
For my fantasy values there was not a good result - you have to check using your values and also have a look at the correct implementation of the equations in the first part. They should all be formulated as F = 0, to work with them using fsolve.
#### 2 件のコメント
Nikolas Haimerl 2019 年 10 月 29 日
Thank you for your help. Using the code you provided I get the following error:
Error using optimoptions
Invalid solver specified. Provide a solver name or handle (such as 'fmincon' or @fminunc).
Type DOC OPTIMOPTIONS for a list of solvers.
opts = optimoptions('fsolve','Algorithm','Levenberg-Marquardt');
Line 100 is the following: opts = optimoptions('fsolve','Algorithm','Levenberg-Marquardt');
Stephan 2019 年 10 月 29 日
For me this works without any errors on R20129b - I wonder why.
But you could leave the options away also, fsolve will switch the algorithm automatically, if it detects a non square system. You will just get a warning, that algorithm is changed.
% opts = optimoptions('fsolve','Algorithm','Levenberg-Marquardt');
sol = fsolve(@(x)fun(x,l1,l2,xa,ya,xpa,ypa,ap0),rand(1,3))
Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems;
> In fsolve (line 316)
In Untitled9 (line 27)
No solution found.
fsolve stopped because the last step was ineffective. However, the vector of function
values is not near zero, as measured by the value of the function tolerance.
<stopping criteria details>
sol =
1.0875 1.1366 -5.7022
test_results =
-0.7646 -1.5292 -0.0009 0.0000
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Free Algebra Tutorials!
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how to calculate grade percentages in c++
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Author Message
VebSnoel
Registered: 11.11.2001
From: South Yorkshire, UK
Posted: Tuesday 26th of Dec 16:04 Hey People out there I really hope some math wiz reads this. I am stuck on this assignment that I have to take in the next couple of days and I can’t seem to find a way to finish it. You see, my teacher has given us this assignment on how to calculate grade percentages in c++, long division and side-side-side similarity and I just can’t make head or tail out of it. I am thinking of going to some private tutor to help me solve it. If one of you friends can show me how to do it, I will very appreciative.
AllejHat
Registered: 16.07.2003
From: Odense, Denmark
Posted: Wednesday 27th of Dec 19:06 You can find numerous links on the internet if you google the keyword how to calculate grade percentages in c++. Most of the information is however crafted for the people who already have some know how about this subject. If you are a complete novice, you should use Algebrator. Is it easy to understand and very helpful too.
Bet
Registered: 13.10.2001
From: kµlt øƒ Ø™
Posted: Friday 29th of Dec 16:29 Algebrator is a nice tool . I have used it a lot. I tried solving the problems myself, at least once before using the software. If I couldn’t solve the question then I used the software to give me the detailed answer . I then used to compare both the answers and correct my errors .
SanG
Registered: 31.08.2001
From: Beautiful Northwest Lower Michigan
Posted: Saturday 30th of Dec 12:21 I remember having difficulties with lcf, greatest common factor and dividing fractions. Algebrator is a truly great piece of math software. I have used it through several algebra classes - Pre Algebra, Remedial Algebra and Basic Math. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly recommended.
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# Diesel fuel generator – fuel usage
Diesel-powered reciprocating engine generators (up to 600 horsepower) release Part 1, 2, 4 and 5 substances. This release calculator uses default emission factors and can be adapted to use site-specific emission factors. If you want to use a site-specific emission factor, enter it in the emission factor column, but be sure to convert the units correctly.
The spreadsheet calculators use emission factors based on uncontrolled emission sources. If you use an emission control device and the default emission factor is uncontrolled, use this formula to adjust the spreadsheet calculation:
Controlled emissions = uncontrolled emission x ((100 - control efficiency)/100)
Since NPRI reporting thresholds are for the facility as a whole, you may need to combine the air releases you calculate with releases from other sources at your facility.
The following sections explain how the spreadsheet calculator works.
## Suggested estimation methodology: emission factor
The general equation for estimation using an emission factor is:
E = A x EF
where:
• E = emissions
• A = activity rate
• EF = emission factor
## Input data
A = amount of diesel burned [cubic meters (m3) or Liters (L)]
B = heating value of diesel fuel [Giga Joules (GJ)/m3]*
If you know the amount of diesel in liters, use this formula to convert it to cubic meters:
A[m3] = A[L]/1000
Use the factors from the tables below to calculate the emissions for each substance.
## Substance releases
Part 1 releases
Substance Chemical Abstracts Service registry number (CAS RN)1 Part Emission factor (EF)2 EF units
Acetaldehyde 75-07-0 1A 1.259E-02 kg/m3
Acrolein 107-08-8 1A 7.593E-04 kg/m3
Benzene 71-43-2 1A 1.532E-02 kg/m3
Formaldehyde 50-00-0 1A 1.937E-02 kg/m3
Naphthalene 91-20-3 1A 1.392E-03 kg/m3
Propylene 115-07-1 1A 4.235E-02 kg/m3
Toluene 108-88-3 1A 6.714E-03 kg/m3
Xylene (all isomers) 1330-20-7 1A 4.679E-03 kg/m3
Mercury (and its compounds) Footnote *** 1B 4.948E-06 kg/m3
Part 2 releases
Substance Chemical Abstracts Service registry number (CAS RN)1 Emission factor (EF)2 EF units
Acenaphthene 83-32-9 1.166E-05 kg/m3
Acenaphthylene 208-96-8 4.153E-05 kg/m3
Anthracene 120-12-7 3.070E-05 kg/m3
Benz[a]anthracene 56-55-3 2.758E-05 kg/m3
Chrysene 218-01-9 5.795E-06 kg/m3
Benzo[a]pyrene 50-32-8 1.543E-06 kg/m3
Benzo[b]fluoranthene 205-99-2 8.134E-07 kg/m3
Benzo[ghi]perylene 191-24-2 4.014E-06 kg/m3
Benzo[k]fluoranthene 207-08-9 1.272E-06 kg/m3
Dibenz[a,h]anthracene 53-70-3 4.785E-06 kg/m3
Fluoranthene 206-44-0 1.249E-04 kg/m3
Fluorene 86-73-7 4.794E-04 kg/m3
Inden[1,2,3-cd]pyrene 193-39-5 3.078E-06 kg/m3
Phenanthrene 85-01-8 4.826E-04 kg/m3
Pyrene 129-00-0 7.847E-05 kg/m3
Part 4 criteria air contaminants (CAC) releases
Substance Chemical Abstracts Service registry number (CAS RN)1 Emission factor (EF)(2)(**) EF units
Carbon monoxide (CO) 630-08-0 0.4084 x B kg/m3
Sulphur dioxide (SO2) 7446-09-5 0.1247 x B kg/m3
Oxides of nitrogen (NOx), expressed as nitrogen dioxide (NO2) 11104-93-1 1.8960 x B kg/m3
Volatile organic compounds (and its compounds) (VOCs) *** 0.1548 x B kg/m3
Total particulate matter (TPM) *** 0.1333 x B kg/m3
Particulate matter less than or equal to 10 micrometers (µm) (PM10) *** 0.1333 x B kg/m3
Particulate matter less than or equal to 2.5 µm (PM2.5) *** 0.1333 x B kg/m3
Part 5 releases
Substance Chemical Abstracts Service registry number (CAS RN)1 Emission factor (EF)(2)(**) EF units
Benzene 71-43-2 4.011E-04 x B kg/m3
1,3-Butadiene 106-99-0 8.405E-06 x B kg/m3
Formaldehyde 50-00-0 5.073E-04 x B kg/m3
Propylene 115-07-1 1.109E-03 x B kg/m3
Toluene 108-88-3 1.758E-04 x B kg/m3
Xylene (all isomers)
1330-20-7 1.225E-04 x B kg/m3
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## Existence of Spanning Trees in Finite Connected Graphs.
### March 5, 2012
[Note: It’s been too long, mathblog! I’ve been busy at school, but I have no plans to discontinue writing in this thing. For longer posts, you’ll have to wait until I have some free time!]
You’ve probably seen a graph before: they’re collections of vertices and edges pieced together in some nice way. Here’s some nice examples from wikipedia:
One theorem that I constantly run into (mainly because it’s relevant to homology) is the following:
Theorem. Given some connected finite graph $G$, there exists a spanning tree of $G$.
Note that such a tree is not generally unique. You might want to find one or two in the graphs above to prove this to yourself! Nonetheless, even though this proof seems like it would be a bit intimidating (or, at least, by some sort of tedious construction), it’s actually quite nice. Let’s go through it.
Proof. We’ll prove this by the number of cycles in $G$. If $G$ has no cycles then it is, itself, a tree; hence, $G$ is its own spanning tree. Suppose now that any graph with $n$ cycles has a maximum spanning tree. Given some graph with $n+1$ cycles, take any one of these cycles and delete any edge (but not deleting the vertices!). This reduces the number of cycles by (at least) one, so there is a spanning tree by induction. In fact, since we have not deleted any vertices, we may replace the edge we removed and this tree will still span the graph. $\Box$
A couple of questions about this for the eager reader. Where did we use connectedness? Where did we use finite-ness? What if we were given an infinite graph? Is there a nice notion of cycles for this kind of graph? Draw some pictures and think about it!
### 2 Responses to “Existence of Spanning Trees in Finite Connected Graphs.”
1. This proof constructs a spanning tree from the outside in (deleting edges until you get a spanning tree), which doesn’t seem to generalize well to infinite graphs. The proof I know goes from the inside out (adding edges until you get a spanning tree) and generalizes immediately.
Pick a connected (not necessarily finite) graph. Once you show that the union of an increasing chain of subgraphs which are trees is also a tree, it follows by Zorn’s lemma that there exists a maximal subgraph which is a tree. If this tree isn’t a spanning tree, then some vertex not contained in it is adjacent to a vertex that is (by connectedness), and adding that edge creates no cycles; contradiction.
• James said
Oh! I like that proof. I hadn’t even thought of using Zorn that way. The proof above was to help along some students in an undergrad level topology course where (I don’t think!) they’d been introduced to Zorn yet. Nonetheless, it’s a pretty accessible result; thank you!
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## Lesson 34 – I’ll be back: The language of Return Period
The average time between Arnold Schwarzenegger’s being back is the return period of his stunt.
I have to wait 5 minutes for my next bus. Some days I wait for 1 minute; some days, I wait for 15 minutes. The wait time is variable → random variable 😉 The average of these wait times is the return period of my bus.
Your recent vocabulary may include “100-year event” (happening more often), (drainage system designed for) “10-year storm,” and so on, courtesy mainstream media and news outlets.
Houston drainage grid ‘so obsolete it’s just unbelievable’
What exactly is this return period business?
Does a 10-year return period event occur diligently every ten years?
Can a 100-year event occur three times in a row?
###### THE LOGIC
Let’s visit annual maximum rainfall for Houston. If we take the daily rainfall data for each year from January 1 to December 31 and choose the maximum rainfall among these days, we call it annual maximum rainfall for that year. So this is the rainfall for the wettest day of the year. Likewise, if we do this for all the years that we have data for, we get a data series (also called time series since we are recording this in time units).
You can get the data from here if you like. You may have to register using your email, but its free. We have 79 years of recorded data from 1939 to 2017. 79 data points, one number per year as the rainfall for the wettest day in that year. You will see that five years are missing between 1942 and 1946.
I want you to understand that these numbers represent a random variable X. Each number (outcome) is assumed to be independent, i.e., the occurrence of one event in one year does not influence the occurrence of the subsequent event. In other words, 2017 rainfall does not depend on 2016 rainfall.
Now, I want you to see Brays Bayou, the lake that detains excess rainfall in Houston. Let us assume that it can store up to eight inches of rainfall on any day. If it rains more than eight inches in a day, the Bayou will overflow and cause flood — as we saw in Houston during hurricane Harvey.
So, if the rainfall is greater than eight inches, we define this as an event. Let us call him Bob. The first time we see Bob was in 1949. We started recording data in 1939. Bob happened after 11 years. The wait time for Bob (1949) is 11 years.
Then we get on with our lives, 11 years passed, Bob is not back, 22 years passed, no sign of Bob. Suddenly, after 30 years of waiting from 1949, Bob Strikes Back (1979).
Two years after this event happened, Bob wanted to greet the Millenials, so he came back in 1981. This time, the waiting period is only two years.
Then, in 1989, for no particular reason, Bob returns. The return of Bob (1989) is after eight years.
You must be thinking: “I don’t see any pattern here.” Yes, that is because there is none.
Years pass, Bob seems to be resting. At the turn of the century, Bob decided to come back. So Bob Meets the 21st Century in 2001 after 12 years since his prior occurrence. Bob re-occurs. Recurrence.
During the first decade of the 21st century, Bob re-occurs two times, once in 2006 as the Restless Bob (5-year wait time) and again in 2008 as Miss Me Yet, Bob (2-year wait time).
We all know what happened after that. Vengeant Bob (2017), aka Harvey, happened after nine years.
Now, let’s summarize all Bobs along with their recurrence times. We started with the assumption that the maximum rainfall events represent a random variable X. Let us define T as another random variable that measures the time between the event Bob (wait time or time to the next event or time to the first event since the previous event).
The return period of the event Bob, (X > 8 inches) is the expected value of T, i.e., E[T], its average measured over a large number of such occurrences.
As you can see here, in the table, the return period of Bob is approximately ten years. Bob is a 10-year return period event.
Another way of thinking about this: Since there are eight Bob events in 79 years, they occur at an average rate of 79/8. Approximately, once in 10 years. Hence originated the 10-year event concept.
Remember, they don’t happen cyclically every ten years. If we average the wait times of a lot of events, we will get approximately ten years.
Just like when you wait for the bus, you wait for short time or a long time, but you think of the average time you wait for a bus everyday, you can see events happening in a cluster or spaced out, but all average to an nyear return period.
###### The relation to Geometric distribution
Last week when we learned Geometric distribution, I told you that we would relate the expected value of the Geometric distribution to return period of an event. Let’s see how Bob relates to Geometric distribution.
I want you to convert the maximum rainfall data series into a series of independent Bernoulli trials of 0s and 1s. 0 if the rainfall is < eight inches (No Bob), 1 if the rainfall is > eight inches (Yes Bob). The 1s can occur with some probability of occurrence p. In our example, since we have 8 Bobs in 79 years the probability of occurrence p = 8/79 = 0.101.
Now, assume T to be a random variable that measures the number of trials (years) it takes to see the first success (event), or the next event from each such event. For the first event, Bob (1949), it took 11 years to occur. The probability that T = 11, P(T = 11) is (1 – p)^10*p. Similarly, the next Bob happened after 30 years and so on. T is the time to first success (next success) → Geometrically distributed.
We can derive the expected value of T using the expectation operation we learned in lesson 24.
Now, recall from your math classes that the expression inside the parenthesis looks like a power series. Ponder over it and confirm that the whole expression will reduce to
`E[T] = 1/p`
The expected value of the wait time that is Geometrically distributed is the inverse of the probability of the event. Since the probability of Bob is 0.101, the return period (expected value of the wait times) is 1/0.101 ~ ten years. A 10-year return period event.
###### The Question
We measured the probability over 79 years; n = 79. We assumed that the probability is constant over all the trials.
In other words, we are assuming that we know p and it does not change.
If I were writing this lesson last year, the probability would have been 7/78 = 0.089. Since Harvey (The Vengeant Bob), the probability became 8/79 = 0.101. There are also five missing years.
Perhaps we do not know the true value of p, and perhaps it is not constant.
How then, can you estimate the risk of anything? How then, can you predict anything? How then, can you design anything?
If I haven’t confused you enough, let me end with one of my favorite quotes from Nicholas Taleb’s book Antifragile: Things that gain from disorder.
“It is hard to explain to naive data-driven people that risk is in the future, not in the past.”
If you find this useful, please like, share and subscribe.
## Lesson 33 – Trials to first success: The language of Geometric distribution
So goes the legendary story: Bruce, Robert I, the King of Scotland, defeated the English armies on his seventh trial. He bore six successive defeats before that.
Giants are yet to win their first game of the season. Their record: L, L, … I wonder how many games until their first win.
The last deadly hurricane that hit New York City is Sandy in 2012. We are now five years through without such a dangerous event.
The common thread tying the three examples is the number of trials to the first success. This is the language of the Geometric distribution.
If we consider independent Bernoulli trials of 0s and 1s with some probability of occurrence p and assume X to be a random variable that measures the number of trials it takes to see the first success, then, X is said to be Geometrically distributed.
We can get the success on the first trial, in which case X will be 1. We can see the success on the second trial, in which case the sequence will be 01, and X will be 2. We can see the success on the third trial; the sequence will be 001 and X will be 3 and so forth. As you can guess, X = {1, 2, 3, … }, positive integers.
There is some probability that X can take any integer value. We should also figure out this probability, i.e., P(X = 1), P(X = 2), P(X = 3), and so on.
Let us take the example of a coin toss. The outcomes are head or tail. Binary outcomes → Bernoulli trial. The probability p of a head or tail is 0.5. In other words, if you toss a coin a large number of times, say 100, roughly 50 of them will be heads, and 50 of them will be tails. Let’s play. Heads you win, tails you lose.
Great, you win on the first trial. The probability of seeing head is 0.5. Hence, P(X = 1) = 0.5.
Let’s play again.
Ah, this time the first outcome is a tail and the second outcome is a head. You lose on the first trial but win on the second. It took two trials to wins. X = 2. P(X=2) is P(tail on the first toss)*P(head on the second toss) = 0.5*0.5 = 0.25. Why did we multiply? What is P(A and B) for independent events?
One more time.
Now it took three trials to win. X = 3, and P(X = 3) = P(tail on the first toss)*P(tail on the second toss)*P(head on the third toss) = 0.5*0.5*0.5 = 0.125.
For X = 4, it will be P(tail on the first toss)*P(tail on the second toss)*P(tail on the third toss)*P(head on the fourth toss) = 0.5*0.5*0.5*0.5 = 0.0625.
If we now plot X and P(X = k), k being 1, 2, 3, 4, …, we get a probability distribution like this.
The height of the line at X = 2 is 0.5 times the height of the line at X = 1. In the same way, the height of the line at X = 3 is 0.5 times the height of the line at X = 2 and so on. P(X=k) decreases in a geometric progression. Hence the name Geometric distribution.
We can generalize this for any probability p. In our game, we estimated P(X = 1) as 0.5, i.e., the probability of seeing a head p. P(X=2) is 0.5*0.5, i.e., (1-p)*p. P(X = 10) = (1-p)^9*p. First success on the tenth toss is nine tails followed by a head.
More generally,
We can derive the expected value and variance of X as:
The expected value of a Geometric distribution relates to a special concept called return period → we will look at it next week.
Meanwhile, here are some more geometric probability distributions with different values of p.
###### p = 0.9
Notice how the shape changes with changing values of p. p is the parameter that controls the shape of the distribution. The greater the value for p, the steeper the fall.
If the probability of success is close to 1, the odds of winning in the first few trials is high → notice the height of the line for p = 0.9. If the probability of success is close to 0, it takes several trials to get to greater odds of winning overall.
Have you now conceptualized the idea of geometric distribution?
Let me challenge you to a bet then.
I have a coin toss game where I give you two times your bet if you win; you get nothing if you lose. Assume we have a fair coin, would you play the game with me and bet your money? If you will, then what is your strategy, assuming you are in it to win.
Since I challenged you to a bet, I also looked into some lottery games myself at nylottery.ny.gov.
###### First observation
The odds of winning first prize in any of the games is next to 0. So if you plan to keep buying the tickets until you win the first time, and then retire, you now know that you will keep buying forever.
The chances of winning per game get better for lower prize levels. For example, in the Mega Million, if you want to win the ninth prize, the odds are 1 in 21. Still low, and will take a long time to win.
But, who wants to win the ninth prize. It is like saying “America Ninth.”
###### Second observation
I keep wondering why on earth is New York Government running a lottery business … only to reconcile that “bread and circuses” have always been up the state’s sleeves to expand.
If you find this useful, please like, share and subscribe.
## Lesson 32 – Exactly k successes: The language of Binomial distribution
I may be in one of those transit buses. Since I moved to New Jersey, I am going through this mess every day.
Well, you wanted to enjoy Manhattan skyline. It has a price tag.
D, glad you are here. It’s been a while. In our last meeting, we were discussing the concepts of variance operation and its properties. I continue to read your lessons every week. As I paused and reflected on all the lessons, I noticed that there is a systematic approach to them. You started with the basics of sets and probability, introduced lessons on visualizing, summarizing and comparing data using various statistics, then extended those ideas into random variables and probability distributions. The readership seems to have grown considerably, and people are tweeting about our classroom. Have you reached 25000 pageviews yet?
We are at 24350 pageviews now. We will certainly hit the 25k mark today 😉 I am thankful to all the readers for their time. Special thanks to all those who are spreading the word. Our classroom is a great resource for anyone starting data analysis.
So, whats on the show today?
As you correctly pointed out, we are now slowly getting into various types of probability distributions. I mentioned in lesson 31 that we would learn several discrete probability distributions that are based on Bernoulli trials. We start this adventure with Binomial distribution.
Great. Let me refresh my memory of probability distributions before we get started. We discussed the basics of probability distribution in lesson 23. Let’s assume X is a random variable, and P(X = x) is the probability that this random variable takes any value x (i.e., an outcome). Then, the distribution of these probabilities on a number line, i.e., the probability graph is called the probability distribution function f(x) for a random variable. We are now looking at various mathematical forms for this f(x).
Fantastic. Now imagine you have a Bernoulli sequence of yes or no.
Sure. It is a sequence of 0s and 1s with a probability p; 0 if the trial yields a no (failure, or event not happening) and 1 if the trial yields a yes (success, or event happening). Something like this: 00101001101
From this sequence, if you are interested in the number of successes (1s) in n trials, this number follows a Binomial distribution. If you assume X is a random variable that represents the number of successes in a Bernoulli sequence of n trials, then this X should follow a binomial distribution. The probability that this random variable X takes any value k, i.e., the probability of exactly k successes in n trials is:
The expected value of this random variable, E[X] = np, and the variance V[X] = np(1-p).
😯 Wow, that’s a fastball. Can we parse through the lingo?
Oops… Okay, let us take the example of your daily commute. Imagine buses and cars pass through the tunnel each morning. Can you guesstimate the probability of buses?
Yeah, I usually see more buses than cars in the morning. Let’s say the likelihood of seeing a bus is p=0.7.
Now let us imagine that buses and cars come in a Bernoulli sequence. Assign a 1 if it is a bus, and 0 if it is a car.
That is reasonable. The vehicle passage is usually random. If we take that as a Bernoulli sequence, there will be some 1s and some 0s with a 0.7 probability of occurrence. In the long run, you will have 70% buses and 30% cars in any order.
Correct. Now think about this. In the next four vehicles that pass through the tunnel, how many of them will be buses?
Since there is randomness in the sequence, in the next four vehicles, I can say, all of them may be buses, or none of them will be buses or any number in between.
Exactly. The number of buses in a sequence of 4 vehicles can be 0, 1, 2, 3 or 4. These are the random variables represented by X. In other words, if X is the number of buses in 4 vehicles coming at random, then X can take 0, 1, 2, 3 or 4 as the outcomes. The probability distribution of X is binomial.
I understand how we came up with X. Why is the probability distribution of X called binomial?
It originates from the idea of the binomial coefficient that you may have learned in an elementary math/combinations class. Let us continue with our logical deduction to see how the probability is derived, and you will see why.
Sure. We have X as 0, 1, 2, 3 and 4. We should calculate the probability P(X = 0), P(X = 1), P(X = 2), P(X = 3) and P(X = 4). This will give us the distribution of the probabilities.
Take an example, say 2. Let us compute P(X = 2). The probability of seeing exactly two buses in 4 vehicles. The probability of exactly k successes in n trials. If the buses and cars come in a Bernoulli sequence (1 for bus and 0 for a car) with a probability p, in how many ways can you see two buses out of 4 vehicles?
Ah, I see where we are going with this. Let me list out the possibilities. Two buses in four vehicles can occur in six ways. 0011, 0101, 1100, 1010, 1001, 0110. In each of these six possible sequences, there will be exactly two buses among four vehicles. I remember from my combinations class that this is four choose two. Four factorial divided by the product of two factorial and (four minus two) factorial. 4C2 = 4!/4!(4-2)!
For each possibility, the probability of that sequence can also be written down. Let me make a table like this:
You can see from the table that there are six possibilities. Any of the possibilities, 1 or 2 or 3 or 4 or 5 or 6 can occur. Hence, the probability of seeing two in four is the sum of these probabilities. Remember P(A or B) = P(A) + P(B). If you follow through this, you will get, 6*p*p*(1-p)*(1-p). = 6*p^2*(1-p)^(4-2). Can you see where the formula for binomial distribution comes from?
Absolutely. For each outcome of X, i.e., 0, 1, 2, 3 and 4, we should apply this logic/formula and compute the probability of the outcome. Let me finish it and make a plot.
Very nicely done. Let me jump in here and show you another plot with a different n and p. If p = 0.5 (equal probability) and n = 100; this is how the binomial distribution looks like.
Nice. It looks like an inverted bell centered around 50.
Yeah. You noticed that the distribution is centered around 50. It is the expected value of the distribution. Remember E[X] is the central tendency of the distribution. For binomial, you can derive it as np = 100 (0.5) = 50. In the same way, the variance, i.e. spread of the function around this center is np(1-p) = 100(0.5)(0.5) = 25. Or standard deviation is 5. You can see that the distribution is spread out within three standard deviations from the center. Can you now imagine how the distribution will look like for p = 0.3 or p = 0.7?
Following the same logic, those distributions will be centered on 100*0.3 = 30 and 100*0.7 = 70 with their variance. Now it all makes sense.
You see how easy it is when you go through the logic. We started with Bernoulli sequence. When we are interested in the random variable that is the number of successes in so many trials, it follows a binomial distribution. Exactly k successes” is the language of Binomial distribution. Can you think of any other examples that can be modeled as a binomial distribution?
Probability that Derek Jeter, with a batting average of 0.3, gets three hits out of the three times he comes to bat 😆 This is fun. I am glad I learned some useful concepts out of the messy commute experience. By the way, Exactly one landfall in the next four hurricanes is also binomial. With Jose coming up, I wonder if we can compute the probability of damage for New York City based on the probability of landfall.
Don’t worry Joe. Our Mayor is graciously implementing his comprehensive \$20 billion resiliency plan. NYC is safe now. Forget probability of damage. You need to worry about the probability of bankruptcy.
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## Lesson 31 – Yes or No: The language of Bernoulli trials
Downtown Miami will be flooded due to hurricane Irma.
Your vehicle will pass the inspection test this year.
Each toss of a coin results in either a head or a tail.
Did you notice that I am looking for an answer, an outcome that is “yes” or “no.” We often summarize data as the occurrence (or non-occurrence) of an event in a sequence of trials. For example, if you are designing dikes for flood control in Miami, you may want to look at the sequence of floods over several years to analyze the number of events, and the rate at which they occur.
There are two possibilities, a hit (event occurred – success) or miss (event did not occur – failure). A yes or a no. These events can be represented as a sequence of 0’s and 1’s (0001100101001000) called Bernoulli trials with a probability of occurrence of p. This probability is constant over all the trials, and the trials itself are assumed to be independent, i.e., the occurrence of one event does not influence the occurrence of the subsequent event.
Now, imagine these outcomes, 0’s or 1’s can be represented using a random variable X. In other words, X is a random variable that can take 0 or 1 with a probability p. If in Miami, there were ten extreme flood events in the last 100 years, the sequence will have 90 0’s and 10 1’s in some order. The probability of the event is hence 0.1. If the probability is 0.5, then, in a sequence of 100 trials (coin tosses for example), you will see 50 heads on average. We can derive the expected value of X and the variance of X as follows:
Since the Bernoulli trials are independent, the probability of a sequence of events happening will be equal to the product of the probability of each event. For instance, the probability of observing a sequence of No Flood, No Flood, No Flood and Flood over the last four years is 0.9*0.9*0.9*0.1 = 0.072 (assuming p = 0.1).
Bernoulli trials form the basis for deriving several discrete probability distributions that we will learn over the next few weeks.
While you ponder over what these distributions are, their mathematical forms, and how they represent the variation in the data, I will leave you with this image of the daily rainfall data from Miami International Airport. An approximate 6.38 inches of rain (~160mm/day) is forecasted for Sunday. Notice how you can remap the data into a sequence of 0’s (if rain is less than 160) and 1’s (if rain is greater than 160).
After tomorrow, when you hear “unprecedented rains” in the news, keep in mind that we seek the historical sequence data like this precisely because our memory is weak.
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# How can I fold a mesh using geo nodes?
I'm new to geo nodes and I have a simple project. I want to fold this shape, as if it was made of paper, origami style.
So i figured it means I want 2 things
• that all the edges keep their original length
• and that while the central vertex goes down the z axis, all the other vertices follow along without leaving the xy plan.
I guess it involves some nodes like "map range" with clamp, maybe index vertices.. I watched a lot of tutorials but I'm still a bit lost and would love some hints on how to achieve this.
• How do you make the mesh transparent around the edges? There was a question here some time ago that asked for that 🤔 Commented Jan 5, 2022 at 23:15
• why do you want to do that with geometry nodes? if you are open to other solutions, i would propose to use e.g. an armature.... Commented Jan 6, 2022 at 4:29
• @ Markus, I'll take a pic of the node tree when I'm back at home, but in short I use a split edges node. And @Chris, the idea is then to tessalate the origami pattern using two array modifiers and have precise control over the folding. But while I'm waiting for an answer I'll explore the armature solution and try to create a long string of bones Commented Jan 6, 2022 at 9:28
• Ah, I see, I thought maybe it's some display setting I don't know about. Commented Jan 6, 2022 at 11:22
The way I understood your problem, for known $$a$$ and $$b$$, you want to find such $$a'$$, that $$c' = a$$. By $$a'$$ and $$c'$$ I mean changed lengths of the triangle sides when the vertex connecting $$a$$ and $$c$$ moves.
$$a'^2 + b^2 = c'^2$$  |  $$c' = a$$
$$a'^2 + b^2 = a^2$$  |  $$-b^2$$
$$a'^2 = a^2 - b^2$$  |  $$\sqrt {}$$
$$a' = \sqrt {a^2 - b^2}$$
Once you have that, you need to move along normalized vector $$\overrightarrow{a}$$ by the difference between $$a$$ and $$a'$$. So you need to apply an offset $$n(\overrightarrow{a}) \times (a - a')$$.
So with this setup:
We get:
For it to work, you need to set the dragged vertex'es index to 0 (select the vertex, Mesh > Sort Elements > Selected).
Edge lengths will change for points that aren't directly connected to the directly dragged vertex:
I don't think you can find the shortest path in geometry nodes, but what you could do is manually assign some information on the connections to the geometry, e.g. you could add a vertex group, assign it to all vertices with value 0.0, and then assign e.g. a value 0.51 to vertex #8:
So that in the nodes you can multiply these values by 10 and have it truncate (keep in mind 0.7 is saved as 0.699999988079071044921875 in float, that's why I add a threshold in the form of 0.01 to the number, so I can just truncate the number. But rounding would work instead...) so the fraction is removed during conversion to integer, in order to obtain the coordinate of the connected vertex rather than always just one vertex:
However, in the first node setup I'm just using the info on how the dragged vertex will be moved, in order to calculate how to move neighboring vertices and only then move the dragged vertex. I could move them all at once with only one Set Position node, or I could also have two s.p. nodes, moving the dragged vertex first, and then using the evaluated position for other vertices. But this means for each level of depth, you need a separate Set Position node, at least to my (limited) understanding...
The laziest idea I have is that you just select the dragged vertex, sort it, select more, sort, and so on to quickly sort everything inside out:
(select the directly dragged vertex, Mesh > Sort Elements > Selected, or use F3 as on GIF, CtrlNumpad + to select more and repeat until everything is sorted from inside out).
Then as described above, manually assign info on indices to nearest dragging vertices, for complex geometry this could be done with a Python script + "shortest path" operator (or some bmesh code by batFinger surely flying around).
Then you would create a node setup that isn't hardcoded to look for index 0 but rather looks for index as in the vertex group, and finally (the laziest part), just take it all, pack into a custom group, and duplicate a few times (as many times as the number of depth levels you want to support), so after one level is evaluated, the next have to update as well in order to stay in correct distance...
• thank you so much for your time Markus. This is even more than what i asked for - we are unpacking your answer with a friend, we'll post a follow up next week with our result Commented Jan 7, 2022 at 0:04
• @Jonquille where's the follow up? :) Commented Jan 17, 2022 at 18:32
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Lect04
# Lect04 - Physics 215 Physics for Elementary Education...
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Physics 215 Physics for Elementary Education Instructor: Dr. Mark Haugan Office: PHYS 282 [email protected] TA: Mayra Cervantes Office: PHYS 222 [email protected] TA: Jordan Kendall Office: PHYS 222 [email protected] TA: Daniel Whitenack Office: PHYS 136 [email protected] Office Hours: If you have questions, just email us to make an appointment. We enjoy talking about teaching and learning physics!
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Changes to this week’s activities to make up for last week’s snow days: In lab this week you will do Chapter 1 Activities 5 and 6 AND Chapter 2 Activity 1, so, please read these text sections to be ready! Homework for Chapter 1 Activities 5 and 6 is due at the beginning of your lab session the following week. There is no Chapter 2 Activity 1 homework. We will do another “extra” activity in some future lab to get back in synch with the course schedule. I will let you know when to prepare for this. Notices: Please check CHIP to verify that your I-Clicker is properly registered and that your I-Clicker points for the last lecture have been recorded. You MUST register your I-Clicker by next Thursday afternoon to receive your points for last lecture and for this lecture. In-class test on Chapter 1 (25 minutes) next Monday!
Ball A softball player uses her muscles to throw a ball. Q1. What kind of energy decreases in the player as mechanical energy is transferred to the ball by her interaction with it? A) thermal energy B) kinetic energy C) elastic potential energy D) chemical potential energy * Q2. What kind of energy increases in the ball as mechanical energy is transferred to it by the interaction between it and the player? A) thermal energy B) kinetic energy C) elastic potential energy D) chemical potential energy * Interactions and Energy
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Interactions and Energy Throughout Chapter 1 we have been developing ways to explain physical phenomena using energy ideas. Last week we discussed not only what we had learned about making such explanations but also the interesting way we extended the range of phenomena we could explain by introducing new kinds energy and new kinds of interactions between objects. For example, introducing the concept
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## This note was uploaded on 04/23/2011 for the course PHYS 215 taught by Professor Unknown during the Spring '11 term at Purdue.
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# Some homework questions
• Dec 3rd 2009, 06:10 PM
SOGGYMUFFIN
Some homework questions
1. Find the domain of the function f(x)=(4-x^2)
2. The revenue of a charter bus company depends the number of unsold seats, "s", per bus.
If the revenue, "R", in dollars, is given by R(s) = -2s^2+52s+5438, what is the maximum revenue?
3. Find an equation of the inverse of the function f(x) = (fifth root of 9-x)+4
• Dec 3rd 2009, 06:17 PM
pickslides
G'day Soggy Muffin
The answer to question 1 is the values of x that can be defined by the function. This is all real values.
For question 2 find the turning point of your parabola. Without using calculus this can be found by finding the midpoint of the 2 zeros
For question 3 rewrite the equation with a y instead of the f(x) then swap the x and y and solve for y.
• Dec 3rd 2009, 06:18 PM
Stroodle
Hi there.
For the first one, the domain is all the real values of x for which the function can be defined.
For the second one, as it's a quadratic function, if you convert it to turning point form, it should be clear what the maximum value is.
And, for the last question if you let f(x) be y, then swap x with y, then solve this for y, you will obtain the inverse of the function.
*Edit - Pickslides beat me to it :) Thought I'd accidentally double posted for a second there :)
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Control Charts
# Control Charts - Sections 5.1 5.7 Page 1 Control Charts...
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Unformatted text preview: Sections 5.1- 5.7 Page 1 Control Charts Principles of Statistical Thinking h All work occurs in a system of interconnected processes, h Variation exists in all processes, and h Understanding and reducing variation are the keys to success. Control Charts h Processes include equipment, material, people, operational definitions and instructions, and environment. All of these contribute to variation in process output. h Statistical Process Control (SPC) is the use of statistical tools to control and improve a process. h Control Chart is an SPC tool that shows the variation in a process variable or attribute. h Attribute Data: data indicating whether an item conforms to some quality characteristic. (Eg: defective or not defective) o p-chart used to monitor percent defective s Based on binomial distribution o c-chart used to monitor number of defects per unit area s Based on poisson distribution h Variable Data: continuous measurement that is crucial to the performance of an item. Processes have two types of variation. h Common cause variation is the variation inherent in the process. It is experienced on an ongoing basis, affecting every unit. h Assignable cause variation arises due to some special circumstance outside the normal process. It occurs sporadically and affects only certain units. Example. Suppose we are sealing capacitors in a can. h The critical characteristic that ensures a proper seal is can height. The height should be 210 + 1 mm. h Lets sample the sealing process and see how it is running. h We take 5 sealed cans at a time and measure their heights. h Then we average these five measurements, calculate the range, and plot the gG . h Based on the central limit theorem, we expect the average can heights to follow an approximately normal distribution that will be centered in the same place as the distribution for individual can height measurements with a standard deviation of : h 99.73% of the s should fall between 3 Sections 5.1- 5.7 Page 2 30 20 10 210.5 210.0 209.5 Sample Number Sample Mean X-bar Chart for Ht(mm) Mean=210.0 UCL=210.4 LCL=209.5 Rational Subgroups h The group of five cans that is taken and measured is referred to as a rational subgroup or sample . It represents the process at a given moment in time. How is this different from a random sample? h We aim to have the within subgroup variability small and that assignable causes, if they are present, will have the largest effect in between subgroups, resulting in shifts in location....
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TY - JOUR AU1 - Erath,, Christoph AU2 - Praetorius,, Dirk AB - Abstract We prove optimal convergence rates for the discretization of a general second-order linear elliptic partial differential equation with an adaptive vertex-centered finite volume scheme. While our prior work Erath & Praetorius (2016, Adaptive vertex-centered finite volume methods with convergence rates. SIAM J. Numer. Anal., 54, 2228–2255) was restricted to symmetric problems, the present analysis also covers nonsymmetric problems and hence the important case of present convection. 1. Introduction We consider a general second-order linear elliptic partial differential equation (PDE) and approximate the solution with an adaptive vertex-centered finite volume method (FVM). FVM are well established in fluid mechanics, since they naturally preserve numerical flux conservation. 1.1. Model problem Let |$\varOmega \subset{\mathbb{R}}^{d}$|, d = 2, 3, be a bounded Lipschitz domain with polygonal boundary |$\varGamma$| := ∂|$\varOmega$|. As a model problem, we consider the following stationary diffusion problem: given f ∈ L2(|$\varOmega$|), find u ∈ H1(|$\varOmega$|) such that $$\textrm{div} (-\textbf{A} \nabla u+\textbf{b} u)+{}c u = f \quad \textrm{in }\varOmega\qquad\textrm{and}\qquad u = 0\quad \textrm{on }\varGamma.$$ (1.1) We suppose that the diffusion matrix |$\textbf{A}=\textbf{A}(x)\in{\mathbb{R}}^{d\times d}$| is bounded, symmetric and uniformly positive definite, i.e., there exist constants |$\lambda _{\min },\lambda _{\max }>0$| such that $$\lambda_{\min }\,|\textbf{v}|^{2}\leq \textbf{v}^{\mathrm{T}}\textbf{A}(x)\textbf{v}\leq \lambda_{\max }\,|\textbf{v}|^{2} \quad\textrm{for all } \textbf{v}\in{\mathbb{R}}^{d} \textrm{ and almost all }x\in\varOmega.$$ (1.2) Let |${\mathscr{T}}_{0}$| be a given initial triangulation of |$\varOmega$|; see Section 2.2 below. For convergence of FVM and well-posedness of the residual error estimator, we additionally require that A(x) is piecewise Lipschitz continuous, i.e., $$\textbf{A}\in W^{1,\infty}(T)^{d\times d} \quad\textrm{for all } T\in{\mathscr{T}}_{0}.$$ (1.3) We suppose that the lower-order terms satisfy the assumption $$\textbf{b}\in W^{1,\infty}(\varOmega)^{d} \quad\textrm{and}\quad{}c\in L^{\infty}(\varOmega) \quad\textrm{with}\quad \tfrac{1}{2}\textrm{div}\,\textbf{b}+{}c \geq 0 \quad\textrm{almost everywhere on }\varOmega.$$ (1.4) With |$(\phi ,\psi )_{\omega } = \int _{\omega } \phi (x)\psi (x)\,\mathrm{d}x$| being the L2-scalar product on a subdomain ω ⊆ |$\varOmega$|, the weak formulation of the model problem (1.1) reads as follows: find |$u\in{H^{1}_{0}}(\varOmega)$| such that $${\mathscr{A}}(u,w):= (\textbf{A}\nabla u-\textbf{b} u,\nabla w)_{\varOmega} +({}c u,w)_{\varOmega} = (\, f,w)_{\varOmega} \quad\textrm{for all } w\in{H^{1}_{0}}(\varOmega).$$ (1.5) According to our assumptions (1.2)–(1.4), the bilinear form |${\mathscr{A}}(\cdot ,\cdot )$| is continuous and elliptic on |${H^{1}_{0}}(\varOmega)$|. Existence and uniqueness of the solution |$u\in{H^{1}_{0}}(\varOmega)$| of (1.5) thus follow from the Lax–Milgram theorem. Moreover, the operator-induced quasi-norm |||⋅||| satisfies $$C_{\textrm{ell}}{||}{v}{||}^{2}_{H^{1}(\varOmega)} \leq {|||}{v}{|||}^{2} := {\mathscr{A}}(v,v)\leq{C}_{\textrm{cont}}{||v||}^{2}_{H^{1}(\varOmega)} \qquad \textrm{for all}v\in{H_{0}^{1}}(\varOmega),$$ (1.6) where Cell > 0 depends only on |$\lambda _{\min }$| and |$\varOmega$|, whereas Ccont > 0 depends only on |$\lambda_{\max }$|, |${||\textbf{b}||}_{L^{\infty}(\varOmega)}$| and |${||c||}_{L^{\infty}(\varOmega)}$|. 1.2. Adaptive FVM In the past 20 years there have been major contributions to the mathematical understanding of adaptive mesh-refinement algorithms, mainly in the context of the finite element method (FEM). While the seminal works Dörfler (1996), Morin et al. (2000), Binev et al. (2004), Stevenson (2007) and Cascón et al. (2008) were restricted to symmetric operators, the recent works Mekchay & Nochetto (2005), Cascón & Nochetto (2012), Feischl et al. (2014) and Bespalov et al. (2017) proved convergence of adaptive FEM with optimal algebraic rates for general second-order linear elliptic PDEs. The work Carstensen et al. (2014) gives an exhaustive overview of the developments and it gains, in an abstract framework, a general recipe to prove optimal adaptive convergence rates of adaptive mesh-refining algorithms. Basically, the numerical discretization scheme, the a posteriori error estimator and the adaptive algorithm have to fulfill four criteria (called axioms inCarstensen et al., 2014), namely, stability on nonrefined elements, reduction on refined elements, general quasi-orthogonality and discrete reliability. Building upon these findings, our recent work (Erath & Praetorius, 2016) provides the first proof of convergence of adaptive FVM with optimal algebraic rates for a symmetric model problem (1.1) with b = 0 and c = 0. 1.3. Contributions and outline In this work, we are in particular interested in the nonsymmetric model problem with b ≠ 0 in (1.1). The proofs of stability on nonrefined elements, reduction on refined elements and discrete reliability follow basically the proofs in Erath & Praetorius (2016); see Sections 3.3 and 3.4. Thus, the major contribution of the present work is the proof of the general quasi-orthogonality property for the nonsymmetric problem, which is satisfied under some mild regularity assumptions on the dual problem. Similar assumptions are required in Mekchay & Nochetto (2005) and Cascón & Nochetto (2012) to prove convergence for an adaptive FEM procedure. Moreover, we note that Mekchay & Nochetto (2005) and Cascón & Nochetto (2012) require slightly more restrictions on the model data (namely, div(b) = 0) and on the mesh refinement (the so-called interior node property) for proving quasi-orthogonality, which are avoided in the present analysis. At this point, we note that Feischl et al. (2014) and Bespalov et al. (2017) improve the FEM result of Mekchay & Nochetto (2005) and Cascón & Nochetto (2012) by a different approach. Instead of the duality argument, the analysis exploits the a priori convergence of FEM solutions (which follows from the classical Céa lemma) by splitting the operator into a symmetric and an elliptic part and a compact perturbation. In particular, there is no duality argument applied. Therefore, no additional regularity assumption is required. However, it seems to be difficult to transfer the analysis of Feischl et al. (2014) and Bespalov et al. (2017) to FVM due to the lack of the Céa lemma. We also mention that unlike the FEM literature, a direct proof of general quasi-orthogonality is not available for FVM due to the lack of Galerkin orthogonality. Instead, the FVM work Erath & Praetorius (2016) first proves linear convergence which relies on a quasi-Galerkin orthogonality (see Erath & Praetorius, 2016, Lemma 11) for FVM. Unfortunately, this auxiliary result does not hold for nonsymmetric problems. Hence, to handle the nonsymmetric case, the missing Galerkin orthogonality and the lack of an optimal L2-estimate for FVM seem to be the bottlenecks. To overcome these difficulties, we first estimate the FVM error in the bilinear form by oscillations in Lemma 3.3. Then we provide a new L2-type estimate in Lemma 3.4 which depends on the regularity of the corresponding dual problem plus oscillations. These two results provide the key arguments to prove a quasi-Galerkin orthogonality in Proposition 3.2. Unlike the literature, this estimate also includes a mesh-size weighted estimator term. With the aid of the previous results, we show linear convergence in Theorem 3.6, where the proof relies on the previous results. Finally, optimal algebraic convergence rates are guaranteed by Theorem 3.10 which follows directly from the literature. We remark that the proposed Algorithm 3.1 additionally marks oscillations to overcome the lack of classical Galerkin orthogonality. Note that this is not required for adaptive FEM. However, since FVM is not a best approximation method, the proposed approach appears to be rather natural. In practice, however, this additional marking is negligible (see also Erath & Praetorius, 2016, Remark 12). Furthermore, if problem (1.1) is slightly convection dominated, Algorithm 3.1 and thus our analysis can be used with caution. We discuss the difficulties for such model problems in Sections 4.3 and 5 in more detail. An extension of our analysis to PDEs with nonlinearities appears to be difficult and is thus beyond the scope of this work. Overall, the present work seems to be the first that proves convergence with optimal rates of an adaptive FVM algorithm for the solution of general second-order linear elliptic PDEs. 2. Preliminaries This section introduces the notation and the discrete scheme, as well as the residual a posteriori error estimator. In particular, we fix our notation used throughout this work. 2.1. General notation Throughout, n denotes the unit normal vector to the boundary pointing outward from the respective domain. In the following, we mark the mesh dependency of quantities by appropriate indices, e.g., uℓ is the solution on the triangulation |${\mathscr{T}}_{\ell }$|. Furthermore, |$\lesssim$| abbreviates ≤ up to some (generic) multiplicative constant which is clear from the context. 2.2. Triangulations Fig. 1. View largeDownload slide Construction of the dual mesh |${\mathscr{T}}_{\!\times }^{\ast}$| (grey boxes) from the primal mesh |${\mathscr{T}}_{\times }$| (triangles) in two dimensions (left) and two-dimensional newest vertex bisection (NVB) (right). Each triangle has a reference edge (indicated by the double line). If edges are marked for refinement (indicated by dots), the resulting configurations are shown. Fig. 1. View largeDownload slide Construction of the dual mesh |${\mathscr{T}}_{\!\times }^{\ast}$| (grey boxes) from the primal mesh |${\mathscr{T}}_{\times }$| (triangles) in two dimensions (left) and two-dimensional newest vertex bisection (NVB) (right). Each triangle has a reference edge (indicated by the double line). If edges are marked for refinement (indicated by dots), the resulting configurations are shown. The FVM relies on two partitions of |$\varOmega$|: the primal mesh|${\mathscr{T}}_{\times }$| and the associated dual mesh|${\mathscr{T}}_{\times }^{\ast}$|. The primal mesh |${\mathscr{T}}_{\times }$| is a regular triangulation of |$\varOmega$| into nondegenerate closed triangles/tetrahedra |$T\in{\mathscr{T}}_{\times }$|, where the possible discontinuities of the coefficient matrix A are aligned with |${\mathscr{T}}_{\times }$|. Define the local mesh-size function $$h_{\times}\in L^{\infty}(\varOmega), \quad h_{\times}|_{T}:=h_{T} := |T|^{1/d} \quad\textrm{for all }T\in{\mathscr{T}}_{\times}.$$ (2.1) Let diam(T) be the Euclidean diameter of T. Suppose that |${\mathscr{T}}_{\times }$| is σ-shape regular, i.e., $$\max_{T\in{\mathscr{T}}_{\times}}\frac{\textrm{diam}(T)}{|T|^{1/d}} \le \sigma < \infty.$$ (2.2) Note that this implies hT ≤diam(T) ≤ σ hT. Let |${\mathscr{N}}_{\times }$||$\left (\textrm{or} \ {\mathscr{N}}_{\times }^{\varOmega}\right )$| denote the set of all (or all interior) nodes. Let |${\mathscr{F}}_{\!\times }$||$\left (\textrm{or} \ {\mathscr{F}}_{\!\times }^{\varOmega}\right )$| denote the set of all (or all interior) facets. For |$T\in{\mathscr{T}}_{\times }$|, let |${\mathscr{F}}_{T} := \big \{F\in{\mathscr{F}}_{\!\times }\,:\,F\subseteq \partial T\big \}$| be the set of facets of T. Moreover, $$\omega_{\times}(T):=\bigcup\big\{T^{\prime}\in{\mathscr{T}}_{\times}\,:\,T\cap T^{\prime} \neq \emptyset\big\}\subseteq\overline{\varOmega}$$ (2.3) denotes the element patch of T in |${\mathscr{T}}_{\times }$|. The associated dual mesh|${\mathscr{T}}_{\times }^{\ast}$| is obtained as follows: for d = 2, connect the center of gravity of an element |$T\in{\mathscr{T}}_{\times }$| with the midpoint of an edge of ∂T. These lines define the nondegenerate closed polygons |$V_{i}\in{\mathscr{T}}_{\times }^{\ast}$|; see Fig. 1(a). For d = 3, we first connect the center of gravity of |$T\in{\mathscr{T}}_{\times }$| with each center of gravity of the four faces of |$F\in{\mathscr{F}}_{T}$| by straight lines. Then, as in the two-dimensional case, we connect each center of gravity of |$F\in{\mathscr{F}}_{T}$| to the midpoints of the edges of the face F. Note that this forms polyhedrons |$V_{i}\in{\mathscr{T}}_{\times }^{\ast}$|. In two and three dimensions, each volume |$V_{i}\in{\mathscr{T}}_{\times }^{\ast}$| is uniquely associated with a node ai of |${\mathscr{T}}_{\times }$|. 2.3. Discrete spaces For a partition |${\mathscr{M}}$| of |$\varOmega$| and |$p\in{\mathbb{N}}_{0}$|, let $${\mathscr{P}}^{p}({\mathscr{M}}) := \big\{v:\varOmega\to{\mathbb{R}}\,:\,\forall\, M\in{\mathscr{M}},\quad v|_{M}\textrm{ is a polynomial of degree }\le p\big\}$$ (2.4) be the space of |${\mathscr{M}}$|-piecewise polynomials of degree p. With this at hand, let $${\mathscr{S}}^{1}({\mathscr{T}}_{\times}) := {\mathscr{P}}^{1}({\mathscr{T}}_{\times})\cap H^{1}(\varOmega) = \big\{v_{\times}\in C(\varOmega)\,:\,\forall T\in{\mathscr{T}}_{\times,}\quad v_{\times}|_{T}\textrm{ is affine}\big\}.$$ (2.5) Then the discrete ansatz space $${\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times}):= {\mathscr{S}}^{1}({\mathscr{T}}_{\times})\cap{H^{1}_{0}}(\varOmega) = \big\{v_{\times}\in{\mathscr{S}}^{1}({\mathscr{T}}_{\times})\,:\,v_{\times}|_{\varGamma} = 0\big\}$$ (2.6) consists of all |${\mathscr{T}}_{\times }$|-piecewise affine and globally continuous functions that are zero on |$\varGamma$|. By convention, the discrete test space $${\mathscr{P}}^{0}_{0}\left({\mathscr{T}}^{\ast}_{\times}\right):= \big\{v_{\times}^{\ast}\in{\mathscr{P}}^{0}\left({\mathscr{T}}_{\times}^{\ast}\right)\,:\,v_{\times}^{\ast}|_{\varGamma}=0\big\}$$ (2.7) consists of all |${\mathscr{T}}_{\times }^{\ast}$|-piecewise constant functions which are zero on all |$V\in{\mathscr{T}}^{\ast}_{\times }$| with ∂V ∩ |$\varGamma$| ≠ ∅. 2.4. Mesh refinements For local mesh refinement, we employ newest vertex bisection (NVB) (see, e.g., Stevenson, 2008, Karkulik et al., 2013 and Fig. 1(b)). Below, we use the following notation: first, |${\mathscr{T}}^{\prime}:={\tt refine}({\mathscr{T}},{\mathscr{M}})$| denotes the coarsest conforming triangulation generated by NVB from a conforming triangulation |${\mathscr{T}}$| such that all marked elements |${\mathscr{M}}\subseteq{\mathscr{T}}$| have been refined, i.e., |${\mathscr{M}} \subseteq{\mathscr{T}}\ \backslash{\mathscr{T}}^{\prime}$|. Second, we simply write |${\mathscr{T}}^{\prime} \in{\tt refine}({\mathscr{T}}\ )$|, if |${\mathscr{T}}^{\prime}$| is an arbitrary refinement of |${\mathscr{T}}$|, i.e., there exists a finite number of refinements steps j = 1, … , n such that |${\mathscr{T}}^{\prime}={\mathscr{T}}^{\prime}_{n}$| can be generated from |${\mathscr{T}}={\mathscr{T}}_{0}^{\prime}$| with marked elements |${\mathscr{M}}^{\prime}_{j}\subseteq{\mathscr{T}}_{j}^{\prime}$| and |${\mathscr{T}}_{j}^{\prime}={\tt refine}({\mathscr{T}}_{j-1}^{\prime},{\mathscr{M}}^{\prime}_{j-1})$|. Note that NVB guarantees that there exist only finitely many shapes of triangles and patches in |${\mathscr{T}}^{\prime} \in{\tt refine}({\mathscr{T}}\ )$|. These shapes are determined by |${\mathscr{T}}$|. In particular, the meshes |${\mathscr{T}}^{\prime} \in{\tt refine}({\mathscr{T}}\ )$| are uniformly σ-shape regular (2.2), where σ depends only on |${\mathscr{T}}$|. 2.5. Vertex-centered FVM The FVM approximates the solution |$u\in{H^{1}_{0}}(\varOmega )$| of (1.5) by some |$u_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$|. The scheme is based on the balance equation over |${\mathscr{T}}_{\times }^{\ast}$| and reads in variational form as follows: find |$u_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$| such that $${\mathscr{A}}_{\times}\left(u_{\times},w_{\times}^{\ast}\right) = \left(\,f,w_{\times}^{\ast}\right)_{\varOmega}=\sum_{a_{i}\in{\mathscr{N}}_{\times}^{\varOmega}}w_{\times}^{\ast}|_{V_{i}} \int_{V_{i}} f\,\mathrm{d}x \quad \textrm{for all } w_{\times}^{\ast}\in{\mathscr{P}}^{0}_{0}\left({\mathscr{T}}_{\times}^{\ast}\right).$$ (2.8) For all |$v_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$| and all |$w_{\times }^{\ast}\in{\mathscr{P}}^{0}_{0}\left ({\mathscr{T}}_{\times }^{\ast}\right )$|, the bilinear form reads $${\mathscr{A}}_{\times}\left(v_{\times},w_{\times}^{\ast}\right):=\sum_{a_{i}\in\,{\mathscr{N}}_{\times}^{\varOmega}}w_{\times}^{\ast}|_{V_{i}} \left(\int_{\partial V_{i}}(-\textbf{A} \nabla v_{\times}+\textbf{b} v_{\times})\cdot{\textbf{n}}\,\mathrm{d}s +\int_{V_{i}}{}c v_{\times}\,\mathrm{d}x\right).$$ To recall that the FVM is well posed on sufficiently fine triangulations |${\mathscr{T}}_{\times }$|, we require the following interpolation operator (see, e.g., Erath, 2012 and Erath & Praetorius, 2016). Lemma 2.1 With |$\chi_{i}^{\ast}\in{\mathscr{P}}^{0}\left ({\mathscr{T}}_{\times }^{\ast}\right )$| being the characteristic function of |$V_{i}\in{\mathscr{T}}_{\times }^{\ast}$|, define $${\mathscr{I}}_{\times}^{\ast}:{\mathscr{C}}(\overline\varOmega)\to{\mathscr{P}}^{0}\left({\mathscr{T}}^{\ast}_{\times}\right),\quad{\mathscr{I}}_{\times}^{\ast}v:=\sum_{a_{i}\in\,{\mathscr{N}}_{\times}}v(a_{i})\chi_{i}^{\ast}.$$ Then, for all |$T\in{\mathscr{T}}_{\times }$|, |${}F\in{\mathscr{F}}_{T}$| and |$v_{\times }\in{\mathscr{S}}^{1}({\mathscr{T}}_{\times })$|, it holds that $$\int_{T} \left(v_{\times}-{\mathscr{I}}_{\times}^{ \ast}v_{\times}\right)\,\mathrm{d}x=0=\int_{{}F}\left(v_{\times}-{\mathscr{I}}_{\times}^{\ast}v_{\times}\right)\,\mathrm{d}s,$$ (2.9) $${||v}_{\times}-{\mathscr{I}}_{\times}^{\ast}v_{\times}{||L}^{2}(T)\leq h_{T} {|| \nabla v_{\times}}{||L}^{2}(T),$$ (2.10) $${||v}_{\times}-{\mathscr{I}}_{\times}^{ \ast}v_{\times}{||L}^{2}(F)\leq C h_{T}^{1/2}{||\nabla v_{\times}}{||L}^{2}(T).$$ (2.11) In particular, it holds that |${\mathscr{I}}_{\times }^{\ast}v_{\times } \in{\mathscr{P}}_{0}^{ 0}\left ({\mathscr{T}}_{\times }^{\ast}\right )$| for all |$v_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$|. The constant C > 0 depends only on the σ-shape regularity of |${\mathscr{T}}_{\times }$|. □ The following lemma is a key observation for the FVM analysis. For Lipschitz continuous A, the proof is found in Ewing et al. (2002) and Erath (2012). We note that the result transfers directly to the present situation (see Erath & Praetorius, 2016; 2017), where A satisfies (1.2)–(1.3), b ≠ 0 and c ≠ 0. Lemma 2.2 There exists Cbil > 0 such that for all |$v_{\times },w_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times }),$| $$|{\mathscr{A}}(v_{\times},w_{\times}) - {\mathscr{A}}_{\times}\left(v_{\times},{\mathscr{I}}_{\times}^{ \ast}w_{\times}\right)| \leq{C}_{\textrm{bil}} \sum_{T\in{\mathscr{T}}_{\times}} h_{T}\, {||v}_{\times}{||H}^{1}(T){||w}_{\times}{||H}^{1}(T).$$ (2.12) Moreover, let |${\mathscr{T}}_{\times }$| be sufficiently fine such that |${}C_{\textrm{ell}}-{}C_{\textrm{bil}}||{h_{\times}}||{L^{\infty }(\varOmega )}>0$|, where Cell > 0 is the ellipticity constant from (1.6). Then there exists |$C_{\textrm{stab}} >0$| such that $${\mathscr{A}}_{\times}(v_{\times},{\mathscr{I}}_{ \times}^{\ast}v_{\times}) \ge{}C_{\textrm{stab}}\, {||v}_{\times}{||}^{2}_{H^{1}(\varOmega)} \qquad \textrm{for all }v_{\times}\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times}).$$ (2.13) In particular, the FVM system (2.8) admits a unique solution |$u_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$|. The constants |$C_{\textrm{bil}}$| and |$C_{\textrm{stab}}$| depend only on the data assumptions (1.2)–(1.4) and the σ-shape regularity of |${\mathscr{T}}_{\times }$| and |$\varOmega$|. □ 2.6. Weighted-residual a posteriori error estimator For all |$v_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$|, we define the volume residual R× and the normal jump J× by $$R_{\times}(v_{\times})|_{T}:=f-\textrm{div}_{\times}(-\textbf{A}\nabla v_{\times}+\textbf{b} v_{\times})-{}c v_{\times} \qquad\textrm{for all }T\in{\mathscr{T}}_{\times},$$ (2.14) $$J_{\times}(v_{\times})|_{F}:={{[\kern-2pt[}\textbf{A}\nabla v_{\times}{]\kern-2pt]}}_{F} \qquad\textrm{for all } F\in{\mathscr{F}}_{\times}^{\varOmega}.$$ (2.15) Here, div× denotes the |${\mathscr{T}}_{\times }$|-piecewise divergence operator, and the normal jump reads |${{[\kern-2pt[}\textbf{g}{]\kern-2pt]}}|_{F}:=(\textbf{g}|_{T}-\textbf{g}|_{T^{\prime }})\cdot{\textbf{n}}$|, where g|T denotes the trace of g from T onto F and n points from T to T′. Let Π× be the edgewise or elementwise integral mean operator, i.e., $$(\varPi_{\times})|_{\tau}=\frac{1}{|\tau|}\int_{\tau} v\,\mathrm{d}x \qquad\textrm{for all } \tau\in{\mathscr{T}}_{\times}\cup{\mathscr{F}}_{\times} \textrm{ and all } v\in L^{2}(\tau).$$ For all |$T\in{\mathscr{T}}_{\times }$|, we define the local error indicators and oscillations by \begin{align}\nonumber \eta_{\times}(T,v_{\times})^{2} &:= {h_{T}^{2}}\,{||R}_{\times}(v_{\times}){||}^{2}_{L^{2}}(T)^{2} + h_{T}\, {||J}_{\times}(v_{\times}){||^{2}_{{L}^{2}(\partial T\backslash\varGamma)}}, \\ \textrm{osc}_{\times}(T,v_{\times})^{2} &:= {h_{T}^{2}}\, ||(1-\varPi_{\times}){R}_{\times}(v_{\times}){||}^{2}_{L^{2}(T)} + h_{T}\, ||(1-\varPi_{\times})J_{\times}(v_{\times}){||}^{2}_{L^{2}_(\partial T\backslash\varGamma)}. \end{align} (2.16) Then the error estimator η× and the oscillations osc× are defined by $$\eta_{\times}(v_{\times})^{2}:=\sum_{T\in{\mathscr{T}}_{\times}}\eta_{\times}(T,v_{\times})^{2} \qquad\textrm{and}\qquad \textrm{osc}_{\times}^{2}(v_{\times}):=\sum_{T\in{\mathscr{T}}_{\times}}\textrm{osc}_{\times}(T,v_{\times})^{2}.$$ (2.17) To abbreviate notation, we write η× := η×(u×) and osc× := osc×(u×). The following proposition is proved, e.g., in Carstensen et al. (2005) and Erath (2013). Proposition 2.3 (Reliability and efficiency). The residual error estimator η× satisfies $$C_{\textrm{rel}}^{-1}||{u-u_{\times}}{||}^{2}_{H^{1}(\varOmega)} \le \,\eta_{\times}^{2}\le\, C_{\textrm{eff}}\left(||{u-u_{\times}}{||}^{2}_{H^{1}(\varOmega)} + \textrm{osc}^{2}_{\times}\right),$$ (2.18) where |$C_{\textrm{rel}}$|, |$C_{\textrm{eff}}$| > 0 depend only on the σ-shape regularity of |${\mathscr{T}}_{\times }$|, the data assumptions (1.2)–(1.4) and |$\varOmega$|. □ Note that a robust variant of this estimator with respect to an energy norm is found and analysed in Erath (2013, Theorems 4.9, 6.3 and Remark 6.1), where we additionally require the assumption |${|| \textrm{div}\,\textbf{b} +{}c}{||L^{\infty }(\varOmega )}\leq C \big (\frac{1}{2}\textrm{div}\,\textbf{b}+{}c\big )$| with C > 0. One of the key ingredients to prove Proposition 2.3 is (2.19) of the following lemma which will be employed below. The proof of the orthogonality relation (2.19) is well known and found, e.g., in Carstensen et al. (2005), Erath (2010; 2013). The discrete defect identity (2.20) is proved in Erath & Praetorius (2016, Lemma 16) for symmetric problems on arbitrary refinements of meshes. This result can easily be transferred to the present model problem (1.1). Lemma 2.4 Let |${\mathscr{T}}_{\diamond }\in{\tt refine}({\mathscr{T}}_{0})$| and |${\mathscr{T}}_{\times } \in{\tt refine}({\mathscr{T}}_{\diamond })$|. Suppose that the discrete solutions |$u_{\times }\in{\mathscr{S}}_{0}^{1}({\mathscr{T}}_{\times })$| or |$u_{\diamond }\in{\mathscr{S}}_{0}^{1}({\mathscr{T}}_{\times })$| exist. Then there holds the L2-orthogonality $$\sum_{T\in{\mathscr{T}}_{\diamond}}\left(R_{\diamond}(u_{\diamond}),v^{\ast}_{\diamond}\right)_{T} - \sum_{F\in{\mathscr{F}}\,_{\diamond}^{\varOmega}}\left(J_{\diamond}(u_{\diamond}),v^{\ast}_{\diamond}\right)_{F} = 0 \quad\textrm{for all }v^{\ast}_{\diamond}\in{\mathscr{P}}^{ 0}_{0}\left({\mathscr{T}}^{\ast}_{\diamond}\right)\!,$$ (2.19) as well as the discrete defect identity $$\sum_{T\in{\mathscr{T}}_{\diamond}}\left(R_{\diamond}(u_{\diamond}),v^{\ast}_{\times}\right)_{T} - \sum_{F\in{\mathscr{F}}\,_{\diamond}^{\varOmega}}\left(J_{\diamond}(u_{\diamond}),v^{\ast}_{\times}\right)_{F} = {\mathscr{A}}_{\times}\left(u_{\times}-u_{\diamond},v^{\ast}_{\times}\right) \quad\textrm{for all }v^{\ast}_{\times}\in{\mathscr{P}}^{0}_{0}\left({\mathscr{T}}^{\ast}_{\times}\right).$$ (2.20) 2.7. Comparison result and a priori error estimate The following proposition states that the FVM error estimator is equivalent to the optimal total error (i.e., error plus oscillations) and so improves Proposition 2.3. The result is first proved in Erath & Praetorius (2016) for b = 0 and c = 0 and generalized to the present model problem in Erath & Praetorius (2017). Proposition 2.5 Let |${\mathscr{T}}_{\times }$| be sufficiently fine such that |${}C_{\textrm{ell}}-{}C_{\textrm{bil}} {||{h}_{\times}}||{L^{\infty }(\varOmega )}>0$| with |$C_{\textrm{ell}}$| and |$C_{\textrm{bil}}$| from (1.6) and (2.12), respectively. Then it holds that $$\begin{split} C_{1}^{-1}\,\eta_{\times} \le \min_{v_{\times}\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times})} \big(||{u-v_{\times}}{H^{1}(\varOmega)} + \textrm{osc}_{\times}(v_{\times})\big) \le ||{u-u_{\times}}{H^{1}(\varOmega)} + \textrm{osc}_{\times} \le C_{1}\,\eta_{\times}. \end{split}$$ (2.21) Moreover, if |$u_{\times }^{\textrm{FEM}}\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$| denotes the FEM solution of |${\mathscr{A}}(u_{\times }^{\textrm{FEM}},w_{\times }) = (\,f,w_{\times })_{\varOmega }$| for all |$w_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$|, it holds that \begin{align*} C_{2}^{-1}\,\big(||{u-u_{\times}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times}\big) &\le ||{u-u_{\times}^{\textrm{FEM}}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times}\left(u_{\times}^{\textrm{FEM}}\right)\\ &\le C_{2}\,\big(||{u-u_{\times}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times}\big). \end{align*} The constants C1, C2 > 0 depend only on |$\varOmega$|, the σ-shape regularity of |${\mathscr{T}}_{\times }$| and the data assumptions (1.2)–(1.4). □ As a direct consequence of Proposition 2.5, one obtains the following convergence result and a priori estimate that confirms first-order convergence of FVM (see again Erath & Praetorius, 2016; 2017). Note that the statement even holds for |$u\in{H^{1}_{0}}(\varOmega )$|, whereas in the literature standard FVM analysis usually requires, e.g., u ∈ H1+ε(|$\varOmega$|) for some ε > 0. Corollary 2.6 Let |$\{{\mathscr{T}}_{\times }\}$| be a family of sufficiently fine and uniformly σ-shape-regular triangulations. Let |$u\in{H^{1}_{0}}(\varOmega )$| be the solution of (1.5). Then there holds convergence \begin{align*} ||{u-u_{\times}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times} \to 0 \quad\textrm{as}\quad ||{h_{\times}}||{L^{\infty}(\varOmega)}\to 0. \end{align*} Moreover, additional regularity |$u\in{H^{1}_{0}}(\varOmega )\cap H^{2}(\varOmega)$| implies first-order convergence \begin{align*} ||{u-u_{\times}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times} = {\mathscr{O}}(||{h_{\times}}||{L^{\infty}(\varOmega)}). \end{align*} 3. Adaptive FVM In this section, we apply an adaptive mesh-refining algorithm for FVM. We combine ideas from Mekchay & Nochetto (2005) and Erath & Praetorius (2016) to prove that adaptive FVM leads to linear convergence with optimal algebraic rates for the error estimator (and hence for the total error; see Proposition 2.5). 3.1. Adaptive algorithm As in Erath & Praetorius (2016), we employ the following adaptive algorithm. Algorithm 3.1. Input: Let 0 < θ′≤ θ ≤ 1 and |$C_{\textrm{mark}}$|, |$C{_{\textrm{mark}}^{\prime}}$|≥ 1. Let |${\mathscr{T}}_{0}$| be a conforming triangulation of |$\varOmega$| that resolves possible discontinuities of A. Loop: For |$\ell =0,1,2,\dots$|, iterate the following steps (i)–(v): (i) Solve: Compute the discrete solution |$u_{\ell }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\ell })$| from (2.8). (ii) Estimate: Compute ηℓ(T, uℓ) and oscℓ(T, uℓ) from (2.16) for all |$T\in{\mathscr{T}}_{\ell }$|. (iii) Mark I: Find |${\mathscr{M}}_{\ell }^{\eta }\subseteq{\mathscr{T}}_{\ell }$| of up to the multiplicative constant |$C_{\textrm{mark}}$| ≥ 1 minimal cardinality that satisfies the Dörfler marking criterion $$\theta\,\sum_{T\in{\mathscr{T}}_{\ell}}\eta_{\ell}(T,u_{\ell})^{2} \le \sum_{T\in{\mathscr{M}}_{\ell}^{\eta}}\eta_{\ell}(T,u_{\ell})^{2}.$$ (3.1) (iv) Mark II: Find |${\mathscr{M}}_{\ell }\subseteq{\mathscr{T}}_{\ell }$| of up to the multiplicative constant |${}C^{\prime }_{\textrm{mark}}\ge 1$| minimal cardinality that satisfies |${\mathscr{M}}_{\ell }^{\eta }\subseteq{\mathscr{M}}_{\ell }$| as well as the Dörfler marking criterion $$\theta^{\prime}\,\sum_{T\in{\mathscr{T}}_{\ell}}\textrm{osc}_{\ell}(T,u_{\ell})^{2} \le \sum_{T\in{\mathscr{M}}_{\ell}}\textrm{osc}_{\ell}(T,u_{\ell})^{2}.$$ (3.2) (v) Refine: Generate a new triangulation |${\mathscr{T}}_{\ell +1} := {\tt refine}({\mathscr{T}}_{\ell },{\mathscr{M}}_{\ell })$| by refinement of all marked elements.Output: Adaptively refined triangulations |${\mathscr{T}}_{\ell }$|, corresponding discrete solutions uℓ, estimators ηℓ and data oscillations oscℓ for ℓ ≥ 0. Due to the lack of standard Galerkin orthogonality (see Section 3.2), we additionally have to mark the oscillations (3.2). In practice, however, this marking is negligible, since θ′ can be chosen arbitrarily small (see Erath & Praetorius, 2016, Remark 7 for more details). 3.2. Quasi-Galerkin orthogonality Given g ∈ L2(|$\varOmega$|), we consider the dual problem: find |$\phi \in{H^{1}_{0}}(\varOmega )$| such that $${\mathscr{A}}(v,\phi) = (g,v)_{\varOmega} \quad\textrm{for all }v\in{H^{1}_{0}}(\varOmega).$$ (3.3) The Lax–Milgram theorem proves existence and uniqueness of |$\phi \in{H^{1}_{0}}(\varOmega )$|. Let 0 < s ≤ 1. We suppose that the dual problem (3.3) is H1+s-regular, i.e., there exists a constant |$C_{\textrm{dual}}$| > 0 such that for all g ∈ L2(|$\varOmega$|), the solution of (3.3) satisfies $$\phi \in{H^{1}_{0}}(\varOmega)\cap H^{1+s}(\varOmega) \quad\textrm{with}\quad ||{\phi}||{H^{1+s}(\varOmega)} \le C_{\textrm{dual}}\,||{g}||{L^{2}(\varOmega)}.$$ (3.4) We refer to Grisvard (1985) for a discussion on this regularity assumption. The main result of this section is the following quasi-Galerkin orthogonality with respect to the operator-induced quasi-norm from (1.6). The proof is postponed to the end of this section. Proposition 3.2 Let 0 < s ≤ 1 and suppose that the dual problem (3.3) is H1+s-regular (3.4). Let |${\mathscr{T}}_{\diamond }\in{\tt refine}({\mathscr{T}}_{0})$| and |${\mathscr{T}}_{\times } \in{\tt refine}({\mathscr{T}}_{\diamond })$|. Then there exists |$C_{\textrm{gal}}$| > 0 such that $$\begin{split} |||{u-u_{\times}}{|||}^{2} \le |||{u-u_{\diamond}}{|||}^{2}-\tfrac12\, |||{u_{\times}-u_{\diamond}}{|||}^{2} +C_{\textrm{gal}}\, ||{h_{\times}}{||}^{2s}_{L^{\infty}(\varOmega)}\eta^{2}_{\times} +C_{\textrm{gal}}\,\textrm{osc}_{\times}^{2}. \end{split}$$ (3.5) The constant |$C_{\textrm{gal}}$| > 0 depends only on |$C_{\textrm{dual}}, C_{\textrm{osc}}, C_{\textrm{rel}}, C_{\textrm{ell}}, C_{\textrm{cont}}$|, diam(|$\varOmega$|) and |$||{\textbf{b}}||{W^{1,\infty }(\varOmega )}$| as well as on σ-shape regularity and all possible shapes of element patches in |${\mathscr{T}}_{\times }$|. For the FVM error, the classical Galerkin orthogonality fails, i.e., |${\mathscr{A}}(u-u_{\times },v_{\times })\not = 0$| for some |$v_{\times }\in{\mathscr{S}}^{1}_{0}({\mathscr{T}}_{\times })$|. However, there holds the following estimate (see, e.g. Erath & Praetorius, 2016). Lemma 3.3 The FVM error u − u× satisfies $$|{\mathscr{A}}(u-u_{\times},v_{\times})| \le{}C_{\textrm{osc}}\,||{v_{\times}}||{H^{1}(\varOmega)}\,\textrm{osc}_{\times} \quad\textrm{for all }v_{\times}\in{\mathscr{S}}^{ 1}_{0}({\mathscr{T}}_{\times}).$$ (3.6) The constant |$C_{\textrm{osc}}$| > 0 depends only on the σ-shape regularity of |${\mathscr{T}}_{\times }$|. Proof. Standard calculations (see, e.g., Erath, 2013, Theorem 4.9) show $${\mathscr{A}}(u-u_{\times},v_{\times}) =\sum_{T\in{\mathscr{T}}_{\times}}\int_{T} R_{\times}(u_{\times})\, v_{\times}\,\mathrm{d}x +\sum_{F\in{\mathscr{F}}_{\times}^{\varOmega}}\int_{F}J_{\times}(u_{\times})\,v_{\times}\,\mathrm{d}s.$$ Together with (2.19) for |$v^{\ast}_{\times }={\mathscr{I}}_{\times }^{ \ast} v_{\times }\in{\mathscr{P}}^{0}_{0}\left ({\mathscr{T}}_{\times }^{\ast}\right )$|, this leads to $${\mathscr{A}}(u-u_{\times},v_{\times}) = \sum_{T\in{\mathscr{T}}_{\times}}\int_{T} R_{\times}(u_{\times})\,\left(v_{\times}-v_{\times}^{\ast}\right)\,\mathrm{d}x +\sum_{F\in{\mathscr{F}}_{\times}^{\varOmega}}\int_{F}J_{\times}(u_{\times})\,\left(v_{\times}-v_{\times}^{\ast}\right)\,\mathrm{d}s.$$ We apply (2.9) for the involved integrals and obtain \begin{align*} {\mathscr{A}}(u-u_{\times},v_{\times}) &= \sum_{T\in{\mathscr{T}}_{\times}}\int_{T} (R_{\times}(u_{\times})-\varPi_{\times}R_{\times}(u_{\times}))\,\left(v_{\times}-v_{\times}^{\ast}\right)\,\mathrm{d}x\\ &\quad+\sum_{F\in{\mathscr{F}}_{\times}^{\varOmega}}\int_{F}(J_{\times}(u_{\times}) -\varPi_{\times}J_{\times}(u_{\times}))\,\left(v_{\times}-v_{\times}^{\ast}\right)\,\mathrm{d}s. \end{align*} The Cauchy–Schwarz inequality and (2.10)–(2.11) conclude the proof. Lemma 3.4 Let 0 < s ≤ 1 and suppose that the dual problem (3.3) is H1+s-regular (3.4). Then the FVM error satisfies $${}C_{\textrm{aux}}^{-1}\,||{u-u_{\times}}{||}^{2}_{L^{2}(\varOmega)}\le ||{h_{\times}}{||}^{2s}_{L^{\infty}(\varOmega)}||{u-u_{\times}}{||}^{2}_{H^{1}(\varOmega)} + \textrm{osc}_{\times}^{2}.$$ (3.7) The constant |$C_{\textrm{aux}}$| > 0 depends only on the σ-shape regularity of |${\mathscr{T}}_{\times }$|, diam(|$\varOmega$|), |$C_{\textrm{cont}}$| and |$C_{\textrm{dual}}$| as well as on all possible shapes of element patches in |${\mathscr{T}}_{\times }$|. Proof. We split the proof into two steps. Step 1. Let |${\mathscr{I}}_{\times }:H^{1}(\varOmega )\to{\mathscr{S}}^{1}({\mathscr{T}}_{\times })$| be the Scott–Zhang projector (Scott & Zhang, 1990). Recall the following properties of |${\mathscr{I}}_{\times }$| for all v ∈ H1(|$\varOmega$|) and |$v_{\times }\in{\mathscr{S}}^{1}({\mathscr{T}}_{\times })$| and all |$T\in{\mathscr{T}}_{\times }$|: |${\mathscr{I}}_{\times }$| has a local projection property, i.e., |$({\mathscr{I}}_{\times } v)|_{T} = v_{\times }|_{T}$| if |$v|_{\omega _{\times }(T)}=v_{\times }|_{\omega _{\times }(T)}$|; |${\mathscr{I}}_{\times }$| preserves discrete boundary data, i.e., v|Γ = v×|Γ implies that |$({\mathscr{I}}_{\times } v)|_{\varGamma } = v|_{\varGamma }$|; |${\mathscr{I}}_{\times }$| is locally H1-stable, i.e., |$||{\nabla{\mathscr{I}}_{\times } v}||_{L^{2}(T)} \le C_{\textrm{sz}}\, ||{\nabla v}||_{H^{1}(\omega _{\times }(T))}$|; |${\mathscr{I}}_{\times }$| has a local approximation property, i.e., |$||{v-{\mathscr{I}}_{\times } v}||_{L^{2}(T)}\le C_{\textrm{sz}}\,h_{T}\,||{\nabla v}||_{H^{1}(\omega _{\times }(T))}$|. The constant |$C_{\textrm{sz}}$| > 0 depends only on the σ-shape regularity of |${\mathscr{T}}_{\times }$|. In particular, $$||{v-{\mathscr{I}}_{\times} v}||_{H^{1}(\varOmega)}\lesssim ||{v}||_{H^{1}(\varOmega)}\quad \textrm{for all } v\in H^{1}(\varOmega),$$ where the hidden constant depends only on |$C_{\textrm{sz}}$| and diam(|$\varOmega$|). With the local projection property of |${\mathscr{I}}_{\times }$|, we may apply the Bramble–Hilbert lemma. For v ∈ H2(|$\varOmega$|), scaling arguments then prove that $$||{v-{\mathscr{I}}_{\times} v}||_{H^{1}(T)} \lesssim \textrm{diam}(\omega_{\times}(T))\,||{v}||_{H^{1}(\omega_{\times}(T))} \quad\textrm{for all }T\in{\mathscr{T}}_{\times},$$ (3.8) where the hidden constant depends only on the shape of ω×(T) and on the operator norm of |$A:=1-{\mathscr{I}}_{\times }$| (and hence on diam(|$\varOmega$|) and |$C_{\textrm{sz}}$|) Altogether, this proves the operator norm estimates $$||{A:=1-{\mathscr{I}}_{\times}:H^{1+t}(\varOmega)\to H^{1}(\varOmega)}|| \le C\,||{h_{\times}}||^{t}_{L^{\infty}(\varOmega)} \quad\textrm{for }t\in\{0,1\},$$ (3.9) where C > 0 depends only on |$C_{\textrm{sz}}$|, diam(|$\varOmega$|) and all possible shapes of element patches in |${\mathscr{T}}_{\times }$|. Interpolation arguments (Bergh & Löfström, 1976) conclude that (3.9) holds for all 0 ≤ t ≤ 1. For t = s, this proves $$||{v-{\mathscr{I}}_{\times} v}||{H^{1}(\varOmega)} \le C\,||{h_{\times}}{||}^{s}_{L^{\infty}(\varOmega)}\,||{v}||_{H^{1+s}(\varOmega)} \quad\textrm{for all }v\in H^{1+s}(\varOmega).$$ (3.10) Step 2. With g = v = u − u× in (3.3), it holds that $$||{u-u_{\times}}{||}^{2}_{L^{2}(\varOmega)}={\mathscr{A}}(u-u_{\times},\phi) = {\mathscr{A}}(u-u_{\times},\phi-{\mathscr{I}}_{\times}\phi) + {\mathscr{A}}(u-u_{\times},{\mathscr{I}}_{\times}\phi).$$ Since we suppose |$\phi \in{H^{1}_{0}}(\varOmega )\cap H^{1+s}(\varOmega )$|, the first summand is bounded by (3.10). This yields \begin{align*} {\mathscr{A}}(u-u_{\times},\phi-{\mathscr{I}}_{\times}\phi) &\lesssim ||{u-u_{\times}}||_{H^{1}(\varOmega)}||{\phi-{\mathscr{I}}_{\times}\phi}||_{H^{1}(\varOmega)}\\ &\lesssim ||{h_{\times}}{||}^{s}_{L^{\infty}(\varOmega)}\,||{u-u_{\times}}||_{H^{1}(\varOmega)}||{\phi}||_{H^{1+s}(\varOmega)}, \end{align*} where the hidden constant depends only on |$C_{\textrm{cont}}$|, |$C_{\textrm{sz}}$| and diam(|$\varOmega$|). The second summand is bounded by (3.6) and H1-stability of |${\mathscr{I}}_{\times }$|. This yields that $${\mathscr{A}}(u-u_{\times},{\mathscr{I}}_{\times}\phi) \lesssim \textrm{osc}_{\times}\,||{{\mathscr{I}}_{\times}\phi}||{H^{1}(\varOmega)} \lesssim \textrm{osc}_{\times}\,||{\phi}||{H^{1}(\varOmega)} \le \textrm{osc}_{\times}\,||{\phi}||{H^{1+s}(\varOmega)},$$ where the hidden constant depends only on |$C_{\textrm{osc}}, C_{\textrm{sz}}$| and diam(|$\varOmega$|). Combining the latter three estimates with H1+s-regularity (3.4), we prove \begin{align*} ||{u-u_{\times}}{||}^{2}_{L^{2}(\varOmega)} &\lesssim \big(||{h_{\times}}{||}^{s}_{L^{\infty}(\varOmega)}\,||{u-u_{\times}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times}\big)\, ||{\phi}||{H^{1+s}(\varOmega)} \\& \lesssim \big(||{h_{\times}}{||}^{s}_{L^{\infty}(\varOmega)}\,||{u-u_{\times}}||{H^{1}(\varOmega)} + \textrm{osc}_{\times}\big)\,||{u-u_{\times}}||{L^{2}(\varOmega)}, \end{align*} where the hidden constant depends additionally on |$C_{\textrm{dual}}$|. This concludes the proof. Proof of Proposition 3.2. Recall that |${\mathscr{A}}(v,w) = (\textbf{A}\nabla v,\nabla w)_{\varOmega} - (\textbf{b} v,\nabla w)_{\varOmega} +(c v,w)_{\varOmega}$| and thus |${\mathscr{A}}(w,v) = (\textbf{A}\nabla w,\nabla v)_{\varOmega} - (\textbf{b} w,\nabla v)_{\varOmega} +(c w,v)_{\varOmega}$|. For |$v,w\in{H^{1}_{0}}(\varOmega)$|, integration by parts proves $$-(\textbf{b} w,\nabla v)_{\varOmega} = (\textbf{b} \cdot \nabla w,v)_{\varOmega} + (\textrm{div}(\textbf{b})\,w,v)_{\varOmega}$$ and hence $${\mathscr{A}}(v,w)+{\mathscr{A}}(w,v)= 2{\mathscr{A}}(v,w)+2(v,\textbf{b}\cdot\nabla w)_{\varOmega} +(\textrm{div}(\textbf{b})\,v,w)_{\varOmega}.$$ By definition of |||⋅|||, this proves \begin{align*} |||{v+w}|||^{2} &= |||{v}|||^{2} + |||{w}|||^{2} + {\mathscr{A}}(v,w) + {\mathscr{A}}(w,v) \\& = |||{v}|||^{2} + |||{w}|||^{2} + 2{\mathscr{A}}(v,w)+2(v,\textbf{b} \cdot\nabla w)_{\varOmega} +(\textrm{div}(\textbf{b})\,v,w)_{\varOmega}. \end{align*} This leads to $$|||{v}|||^{2} = |||{v+w}|||^{2} - |||{w}|||^{2} - 2{\mathscr{A}}(v,w) - 2(v,\textbf{b} \cdot\nabla w)_{\varOmega} -(\textrm{div}(\textbf{b})\,v,w)_{\varOmega}.$$ With |$C_{1} := {}C_{\textrm{ell}}^{-1}\,(2||{\textbf{b}}||{L^{\infty }(\varOmega )}+||{\textrm{div}\,\textbf{b}}||{L^{\infty }(\varOmega)})^{2}$|, the Young inequality |$ab\le \frac{1}{4}\, a^{2} + b^{2}$| and norm equivalence (1.6) prove \begin{align*} - 2(v,\textbf{b} \cdot\nabla w)_{\varOmega} -(\textrm{div}(\textbf{b})\,v,w)_{\varOmega} &\le ||{v}||_{L^{2}(\varOmega)}||{w}||_{H^{1}(\varOmega)}\,\big(2||{\textbf{b}}||_{L^{\infty}(\varOmega)}+||{\textrm{div}\,\textbf{b}}||_{L^{\infty}(\varOmega)}\big) \\& \le \tfrac{1}{4}\,|||{w}|||^{2} + C_{1}\,||{v}||^{2}_{L^{2}(\varOmega)}. \end{align*} Choose v = u − u× as well as |$w = u_{\times}-u_{\diamond}$|. So far, we have shown $$|||{u-u_{\times}}|||^{2} \le |||{u-u_{\diamond}}|||^{2} - \tfrac{3}{4}\,|||{u_{\times}-u_{\diamond}}|||^{2} - 2{\mathscr{A}}(u-u_{\times},u_{\times}-u_{\diamond}) + C_{1}\,||{u-u_{\times}}{||}^{2}_{L^{2}(\varOmega)}.$$ We apply (3.6), norm equivalence (1.6) and the Young inequality |$2ab\le \frac{1}{4}\, a^{2}+4b^{2}$| to see that \begin{align*} &- 2{\mathscr{A}}(u-u_{\times},u_{\times}-u_{\diamond}) \le 2\,{}C_{\textrm{osc}}\,||{u_{\times}-u_{\diamond}}||_{H^{1}(\varOmega)}\,\textrm{osc}_{\times} \\&\qquad\qquad\qquad\qquad\qquad \le 2\,{}C_{\textrm{osc}{}}C_{\textrm{ell}}^{-1/2}\,|||{u_{\times}-u_{\diamond}}|||\,\textrm{osc}_{\times} \le \tfrac{1}{4}\,|||{u_{\times}-u_{\diamond}}|||^{2} + 4\,{}C_{\textrm{osc}}^{2}{}C_{\textrm{ell}}^{-1}\,\textrm{osc}_{\times}^{2}. \end{align*} Next, Lemma 3.4 and reliability (2.18) lead to $${}C_{\textrm{aux}}^{-1}\,||{u-u_{\times}}{||}^{2}_{L^{2}(\varOmega)} \le ||{h_{\times}}{||}^{2s}_{L^{\infty}(\varOmega)}\,||{u-u_{\times}}{||}^{2}_{H^{1}(\varOmega)} +\textrm{osc}_{\times}^{2} \le{}C_{\textrm{rel}}\,||{h_{\times}}{||}^{2s}_{L^{\infty}(\varOmega)}\,\eta_{\times}^{2} +\textrm{osc}_{\times}^{2}.$$ Combining the latter three estimates, we prove \begin{align*} |||{u-u_{\times}}|||^{2} &\le |||{u-u_{\diamond}}|||^{2} - \tfrac{1}{2}\,|||{u_{\times}-u_{\diamond}}|||^{2} \\&\quad + C_{1}{}C_{\textrm{aux}{}}C_{\textrm{rel}}\,||{h_{\times}}{||}^{2s}_{L^{\infty}(\varOmega)}\,\eta_{\times}^{2} + \left(4\,{}C_{\textrm{osc}}^{2}{}C_{\textrm{ell}}^{-1} + C_{1}{}C_{\textrm{aux}}\right)\,\textrm{osc}_{\times}^{2}. \end{align*} Choosing |$C_{\textrm{gal}} = \max \big \{C_{1}{}C_{\textrm{aux}{}}C_{\textrm{rel}}\,,\,4{}C_{\textrm{osc}}^{2}{}C_{\textrm{ell}}^{-1} + C_{1}{}C_{\textrm{aux}}\big \}$|, we conclude the proof. 3.3. Linear convergence and general quasi-orthogonality The following properties (A1)–(A2) of the estimator and (B1)–(B2) of the oscillations are some key observations to prove linear convergence of Algorithm 3.1. The proofs for a symmetric problem are based on scaling arguments and can be found in the literature, (see, e.g., Cascón et al., 2008, Section 3.1 for (A1)–(A2) and Erath & Praetorius, 2016, Section 3.3 for (B1)–(B2)). These proofs apply almost verbatim to the present nonsymmetric problem with b ≠ 0. Therefore, the details are left to the reader. Lemma 3.5 There exist constants 0 < q < 1 and C > 0 such that for all |${\mathscr{T}}_{\diamond }\in{\tt refine}({\mathscr{T}}_{0})$|, all |${\mathscr{T}}_{\times }\in{\tt refine}({\mathscr{T}}_{\diamond })$| and all |$v_{\times }\in{\mathscr{S}}_{0}^{1}({\mathscr{T}}_{\times })$|, |$v_{\diamond }\in{\mathscr{S}}_{0}^{1}({\mathscr{T}}_{\diamond })$|, it holds that (stability of estimator on nonrefined elements) $$\left|\left(\sum_{T\in{\mathscr{T}}_{\times}\cap{\mathscr{T}}_{\diamond}}\eta_{\times}(T,v_{\times})^{2} \right)^{1/2} - \left(\sum_{T\in{\mathscr{T}}_{\times}\cap{\mathscr{T}}_{\diamond}}\eta_{\diamond}(T,v_{\diamond})^{2} \right)^{1/2}\right| \leq C\, ||{v_{\times}-v_{\diamond}}||_{H^{1}(\varOmega)},$$ (A1)(reduction of estimator on refined elements) $$\sum_{T\in{\mathscr{T}}_{\times}\backslash{\mathscr{T}}_{\diamond}}\eta_{\times}(T,v_{\times})^{2} \leq q\sum_{T\in{\mathscr{T}}_{\diamond}\backslash{\mathscr{T}}_{\times}}\eta_{\diamond}(T,v_{\diamond})^{2} + C\, ||{v_{\times}-v_{\diamond}}||_{H^{1}(\varOmega)}^{2},$$ (A2)(stability of oscillations on nonrefined elements) $$\left|\left(\sum_{T\in{\mathscr{T}}_{\times}\cap{\mathscr{T}}_{\diamond}}\textrm{osc}_{\times}(T,v_{\times})^{2} \right)^{1/2} \right.\! -\! \left. \left(\sum_{T\in{\mathscr{T}}_{\times}\cap{\mathscr{T}}_{\diamond}}\textrm{osc}_{\diamond}(T,v_{\diamond})^{2} \right)^{1/2}\right| \\\nonumber \leq C\, ||{h_{\times}}||_{L^{\infty}(\varOmega)}||{v_{\times}-v_{\diamond}}||_{H^{1}(\varOmega)},$$ (B1)(reduction of oscillations on refined elements) $$\sum_{T\in{\mathscr{T}}_{\times}\backslash{\mathscr{T}}_{\diamond}}\textrm{osc}_{\times}(T,v_{\times})^{2} \leq q\sum_{T\in{\mathscr{T}}_{\diamond}\backslash{\mathscr{T}}_{\times}}\textrm{osc}_{\diamond}(T,v_{\diamond})^{2} + C\, ||{h_{\times}}{||}^{2}_{L^{\infty}(\varOmega)}||{v_{\times}-v_{\diamond}}{||}^{2}_{H^{1}(\varOmega)}.$$ (B2) The constants 0 < q < 1 and C > 0 depend only on the σ-shape regularity (2.2) and on the data assumptions (1.2)–(1.4). □ Theorem 3.6 (Linear convergence). Let 0 < θ′≤ θ ≤ 1. There exists H > 0 such that the following statement is valid provided that |$||{h_{0}}||{L^{\infty }(\varOmega)}\le H$| and that the dual problem (3.3) is H1+s-regular (3.4) for some 0 < s ≤ 1: there exist Clin > 0 and 0 < qlin < 1 such that Algorithm 3.1 guarantees linear convergence in the sense of $$\eta_{\ell+n}^{2}\le{}C_{\textrm{lin}{}}q_{\textrm{lin}}^{n}\,\eta_{\ell}^{2} \quad\textrm{for all }\ell,n\in{\mathbb{N}}_{0}.$$ (3.11) The constant H depends only on the σ-shape regularity (2.2), on the data assumptions (1.2)–(1.4), |$C_{\textrm{gal}}$|, θ and θ′, whereas |$C_{\textrm{lin}}$| and |$q_{\textrm{lin}}$| additionally depend on |$C_{\textrm{cont}}$| and |$C_{\textrm{rel}}$|. Proof. We split the proof into three steps. Step 1. There exist constants C > 0 and 0 < q < 1 which depend only on 0 < θ ≤ 1, |$C_{\textrm{ell}}$| and the constants in (A1)–(A2), such that $$\eta_{\ell+1}^{2} \le q\,\eta_{\ell}^{2} + C \, |||{u_{\ell+1}-u_{\ell}}|||^{2} \quad\textrm{for all }\ell\in{\mathbb{N}}_{0}.$$ (3.12) Furthermore, there exist constants C > 0 and 0 < q < 1 which depend only on 0 < θ′≤ 1, |$C_{\textrm{ell}}$| and the constants in (B1)–(B2), such that $$\textrm{osc}_{\ell+1}^{2} \le q\,\textrm{osc}_{\ell}^{2} + C\,||{h_{\ell+1}}{||}^{2}_{L^{\infty}(\varOmega)} \, |||{u_{\ell+1}-u_{\ell}}|||^{2} \quad\textrm{for all }\ell\in{\mathbb{N}}_{0}.$$ (3.13) The proofs of (3.12) and (3.13) rely only on (A1)–(A2) with the Dörfler marking (3.1) and (B1)–(B2) with marking (3.2), respectively. For details, we refer, e.g., to Erath & Praetorius (2016, Proposition 10, steps 1 and 2). Step 2. Without loss of generality, we may assume that the constants C > 0 and 0 < q < 1 in (3.12)–(3.13) are the same. With free parameters γ, μ > 0, we define $$\varDelta_{\times} := |||{u-u_{\times}}|||^{2} + \gamma\,\eta_{\times}^{2} + \mu\,\textrm{osc}_{\times}^{2}.$$ We employ the quasi-Galerkin orthogonality (3.5) and obtain $$\varDelta_{\ell+1} \le |||{u-u_{\ell}}|||^{2} + \left[\gamma+C_{\textrm{gal}}\,||{h_{\ell+1}}{||}^{2s}_{L^{\infty}(\varOmega)}\right]\eta_{\ell+1}^{2} + \left[\mu + C_{\textrm{gal}}\right]\,\textrm{osc}_{\ell+1}^{2} - \tfrac12\,|||{u_{\ell+1}-u_{\ell}}|||^{2}.$$ Using (3.12)–(3.13), we further derive \begin{align*} \varDelta_{\ell+1} &\le |||{u-u_{\ell}}|||^{2} + \left[\gamma+C_{\textrm{gal}}\,||{h_{\ell+1}}{||}^{2s}_{L^{\infty}(\varOmega)}\right]\,q\,\eta_{\ell}^{2} + \big[\mu + C_{\textrm{gal}}\big]\,q\,\textrm{osc}_{\ell}^{2}\\ &\qquad-\left(\tfrac12-C\,\left[\gamma+C_{\textrm{gal}}\,||{h_{\ell+1}}{||}^{2s}{L^{\infty}(\varOmega)}\right] - C\,||{h_{\ell+1}}{||}^{2}_{L^{\infty}(\varOmega)}\,\big[\mu + C_{\textrm{gal}}\big]\right)\,|||{u_{\ell+1}-u_{\ell}}|||^{2}. \end{align*} Let H > 0 be a free parameter and suppose that |$||{h_{0}}||{L^{\infty }(\varOmega)} \le H$|. We estimate |$||{h_{\ell +1}}||{L^{\infty }(\varOmega)} \le ||{h_{0}}||{L^{\infty }(\varOmega)} \le H$|. Norm equivalence (1.6) and reliability (2.18) prove $$|||{u-u_{\ell}}|||^{2} \le{}C_{\textrm{cont}}\, ||{u-u_{\ell}}{||}^{2}_{H^{1}(\varOmega)} \le{}C_{\textrm{cont}{}{}}C_{\textrm{rel}}\,\eta_{\ell}^{2}.$$ Let ε > 0 be a free parameter. Combining the last two estimates, we see that \begin{align*} \varDelta_{\ell+1} &\le (1\!-\!\varepsilon)\,|||{u-u_{\ell}}|||^{2} + \gamma\,\left[\!\left(1+\gamma^{-1}C_{\textrm{gal}}\,H^{2s}\right)q+\gamma^{-1}\varepsilon\,{}C_{\textrm{cont}}{}C_{\textrm{rel}}\!\right]\,\eta_{\ell}^{2} + \mu\,\left[\!1 + \mu\!^{-1}C_{\textrm{gal}}\!\right]\,q\,\textrm{osc}_{\ell}^{2}\\ &\quad-\left(\tfrac12-C\,\left[\gamma+C_{\textrm{gal}}\,H^{2s}\right] - C\,H^{2}\big[\mu + C_{\textrm{gal}}\big]\right)\,|||{u_{\ell+1}-u_{\ell}}|||^{2}. \end{align*} Step 3. It only remains to fix the four free parameters γ, μ, ε and H. Choose γ > 0 sufficiently small such that γC < 1/2. Choose μ > 0 sufficiently large such that |$q_{\textrm{osc}} := \big [1 + \mu ^{-1}C_{\textrm{gal}}\big ]\,q < 1$|. Choose H sufficiently small such that |$C\,\left [\gamma +C_{\textrm{gal}}\,H^{2s}\right ] + C\,H^{2}\big [\mu + C_{\textrm{gal}}\big ] < 1/2$|, |$\left (1+\gamma ^{-1}C_{\textrm{gal}}\,H^{2s}\right ) q < 1$|. Choose 0 < ε < 1 such that |${}q_{\textrm{est}} := \left [\left (1+\gamma ^{-1}C_{\textrm{gal}}\,H^{2s}\right )q+\gamma ^{-1}\varepsilon \,{}C_{\textrm{cont}}{}C_{\textrm{rel}}\right ] < 1$|. With |${}q_{\textrm{lin}}:=\max \{\,1-\varepsilon \,,\,{}q_{\textrm{est}}\,,\,q_{\textrm{osc}}\,\}$|, we then obtain that \begin{align*} \varDelta_{\ell+1} &\le (1\!-\!\varepsilon)\,|||{u-u_{\ell}}|||^{2} \!+\! \gamma\,\left[\left(1+\gamma^{-1}C_{\textrm{gal}}\,H^{2s}\right)q+\gamma^{-1}\varepsilon\,{}C_{\textrm{cont}}{}C_{\textrm{rel}}\right]\,\eta_{\ell}^{2} + \mu\,\big[1 + \mu\!^{-1}C_{\textrm{gal}}\big]\,q\,\textrm{osc}_{\ell}^{2} \\& \le \max\{\,1-\varepsilon\,,\,{}q_{\textrm{est}}\,,\,q_{\textrm{osc}}\,\}\,\varDelta_{\ell}={}q_{\textrm{lin}}\varDelta_{\ell}. \end{align*} Induction on n, norm equivalence (1.6), reliability (2.18) and |$\textrm{osc}_{\ell }^{2}\leq \eta _{\ell }^{2}$| prove $$\gamma \,\eta_{\ell+n}^{2}\leq\varDelta_{\ell+n}\leq{}q_{\textrm{lin}}^{n}\varDelta_{\ell} \leq{}q_{\textrm{lin}}^{n}\left({}C_{\textrm{rel}{}}C_{\textrm{cont}}+\gamma+\mu\right)\,\eta_{\ell}^{2}\quad \textrm{for all }\ell,n\in{\mathbb{N}}_{0}.$$ This concludes linear convergence (3.11) with Clin = (CrelCcont + γ + μ)γ−1. Remark 3.7 In the above proof, we could apply the relation |$\textrm{osc}_{\ell }^{2}\leq \eta _{\ell }^{2}$|. Hence, we could avoid using (3.13). Consequently, Algorithm 3.1 would not need marking (3.2) of oscillations. However, the expression |$\left (1+\gamma ^{-1}C_{\textrm{gal}}\,H^{2s}\right ) q$| in Step 3 of the foregoing proof would become |$\left (1+\gamma ^{-1}C_{\textrm{gal}}\,\left (1+H^{2s}\right )\right ) q$| which is not less than 1 as required for the analysis. Hence, the overall proof of linear convergence (3.11) would fail. From the linear convergence (3.11), we immediately obtain the so-called general quasi-orthogonality (see, e.g., Carstensen et al., 2014, Proposition 4.11 or Erath & Praetorius, 2016, Proposition 10, step 5). Corollary 3.8 (General quasi-orthogonality). Let (uk) be the sequence of solutions of Algorithm 3.1. Then there exists C > 0 such that $$\sum_{k=\ell}^{\infty} ||{u_{k+1}-u_{k}}{||}^{2}_{H^{1}(\varOmega)} \le C\,\eta_{\ell}^{2} \quad \text{ for all {\ell\in{\mathbb{N}}_{0}}.}$$ (A3) The constant C > 0 has the same dependencies as Clin from (3.11). 3.4. Optimal algebraic convergence rates In order to prove optimal convergence rates of Algorithm 3.1, we need one further property of the error estimator, namely the so-called discrete reliability (A4). The proof of the next lemma follows as for the symmetric case in Erath & Praetorius (2016, Proposition 15). While the proof is thus omitted, we note that the main difficulties over the well-known FEM proof (Cascón et al., 2008) arise in the handling of the piecewise constant test spaces on |${\mathscr{T}}_{\times }^{\ast}$| and |${\mathscr{T}}_{\diamond }^{\ast}$|, and the fact that these test spaces are not nested. Lemma 3.9 (Discrete reliability). There exists a constant C > 0 such that for all |${\mathscr{T}}_{\diamond }\in{\tt refine}({\mathscr{T}}_{0})$| and all |${\mathscr{T}}_{\times }\in{\tt refine}({\mathscr{T}}_{\diamond })$|, it holds that $$||{u_{\times} - u_{\diamond}}{||}^{2}_{H^{1}(\varOmega)} \le C\left(\sum_{T\in{\mathscr{T}}_{\times}}{h_{T}^{2}}||{u_{\times} - u_{\diamond}}{||}^{2}_{H^{1}(T)}+ \sum_{T\in{\mathscr{R}}_{\diamond}}\eta_{\diamond}(T,u_{\diamond})^{2}\right),$$ (A4) where |${\mathscr{R}}_{\diamond }:=\big \{T\in{\mathscr{T}}_{\diamond }\,:\,\exists\, T^{\prime }\in{\mathscr{T}}_{\diamond }\backslash{\mathscr{T}}_{\times }\textrm{ with }T\cap T^{\prime }\ \neq\ \emptyset \big \}$| consists of all refined elements |${\mathscr{T}}_{\diamond }\backslash{\mathscr{T}}_{\times }$| plus one additional layer of neighboring elements. The constant C > 0 depends only on the σ-shape regularity (2.2), the data assumptions (1.2)–(1.4) and |$\varOmega$|. Note that for a sufficiently fine initial mesh |${\mathscr{T}}_{0}$|, e.g., |$C\,||{h_{0}}{||}^{2}{L^{\infty }(\varOmega)} \leq 1/2$|, (A4) leads to discrete reliability as stated in Carstensen et al. (2014). □ Let |${\mathbb{T}} := {\tt refine}({\mathscr{T}}_{0})$| be the set of all possible triangulations obtained by NVB. For N ≥ 0, let |${\mathbb{T}}_{N} := \big \{{\mathscr{T}}_{\times }\in{\mathbb{T}}\,:\,\#{\mathscr{T}}_{\times }-\#{\mathscr{T}}_{0} \le N\big \}$|. For s > 0, define $$||{u}||{\mathbb{A}_{s}} := \sup_{N\in{\mathbb{N}}_{0}} \inf_{{\mathscr{T}}_{\times}\in{\mathbb{T}}_{N}} (N+1)^{s}\,\eta_{\times}.$$ (3.14) Note that |$||{u}||{\mathbb{A}_{s}} < \infty$| implies an algebraic decay |$\eta _{\times } = {\mathscr{O}}\big ((\#{\mathscr{T}}_{\times })^{-s}\big )$| along the optimal sequence of meshes (which minimize the error estimator). Optimal convergence of the adaptive algorithm thus means that for all s > 0 with |$||{u}||{\mathbb{A}_{s}} < \infty$|, the adaptive algorithm leads to |$\eta _{\ell } = {\mathscr{O}}\big ((\#{\mathscr{T}}_{\ell })^{-s}\big )$|. The work Carstensen et al. (2014, Theorem 4.1) proves in a general framework the following Theorem 3.10, if the adaptive algorithm applied to a numerical scheme and a corresponding estimator satisfies (A1)–(A4). Theorem 3.10 (Optimal algebraic convergence rates). Suppose that the dual problem (3.3) is H1+s-regular (3.4) for some 0 < s ≤ 1. Let the initial mesh |${\mathscr{T}}_{0}$| be sufficiently fine, i.e, there exists a constant H > 0 such that |$||{h_{0}}||{L^{\infty }(\varOmega)}\le H$|. Finally, suppose that there is a constant |$C_{\textrm{MNS}}$| ≥ 1 such that |$\#{\mathscr{M}}_{\ell }\le{}C_{\textrm{MNS}}\#{\mathscr{M}}_{\ell }^{\eta }$| for all |$\ell \in{\mathbb{N}}_{0}$|. Then there exists a bound 0 < |$\theta_{\textrm{opt}}$| ≤ 1 such that for all 0 < θ < |$\theta_{\textrm{opt}}$| and all s > 0 with |$||{u}||{\mathbb{A}_{s}}<\infty$|, there exists a constant |$C_{\textrm{opt}}$| > 0 such that Algorithm 3.1 guarantees $$\eta_{\ell}\leq{}C_{\textrm{opt}}(\#{\mathscr{T}}_{\ell}-\#{\mathscr{T}}_{0})^{-s} \quad \textrm{for all } \ell \in{\mathbb{N}}.$$ (3.15) The constant |$\theta_{\textrm{opt}}$| depends only on |$\varOmega$|, H, uniform σ-shape regularity of the triangulations |${\mathscr{T}}_{\times }\in{\tt refine}({\mathscr{T}}_{0})$| and the data assumptions (1.2)–(1.4). The constant |$C_{\textrm{opt}}$| additionally depends on s, the constant qlin from (3.11), the use of NVB and on |$C_{\textrm{MNS}}$|. □ Remark 3.11 A direct consequence of the assumption |$\#{\mathscr{M}}_{\ell }\le{}C_{\textrm{MNS}}\#{\mathscr{M}}_{\ell }^{\eta }$| in Theorem 3.10 is that data oscillation marking (3.2) is negligible with respect to the overall number of marked elements (see also Erath & Praetorius, 2016, Remark 7). In practice, (3.1) already implies (3.2) since θ′ > 0 can be chosen arbitrarily small. Furthermore, efficiency (2.18) is not required to show (3.11) and (3.15) but guarantees (optimal) linear convergence also for the FVM error. 4. Numerical examples In extension of our theory, we consider the model problem (1.1) with inhomogeneous Dirichlet boundary conditions. For all experiments in two dimensions, we run Algorithm 3.1 with θ = 1 = θ′ and θ = 0.5 = θ′ for uniform mesh refinement and adaptive mesh refinement, respectively. 4.1. Experiment with a smooth solution On the square |$\varOmega$| = (−1, 1)2, we prescribe the exact solution |$u(x_{1},x_{2}) = \left (1-10{x_{1}^{2}}-10{x_{2}^{2}}\right )$||$e^{-5\left ({x_{1}^{2}}+{x_{2}^{2}}\right )}$| with |$x=(x_{1},x_{2})\in{\mathbb{R}}^{2}$|. We choose the diffusion matrix $$\textbf{A}= \left ( \begin{array}{@{}cc@{}} 10+\cos x_{1} & 9 x_{1} x_{2} \\ 9 x_{1} x_{2} & \;10+\sin x_{2} \end{array}\right),$$ the velocity |$\textbf{b}=(\sin x_{1},\cos x_{2})^{\mathrm{T}}$| and the reaction c = 1. Note that (1.2) holds with |$\lambda _{\min }=0.82293$| and |$\lambda _{\max }=10.84096$| and (1.4) with |$\frac{1}{2} \textrm{div}\,\textbf{b}+{}c> 0$|. The right-hand side f is calculated appropriately. The uniform initial mesh |${\mathscr{T}}_{0}$| consists of 16 triangles. Fig. 2. View largeDownload slide Experiment with a smooth solution from Section 4.1: adaptively generated mesh |${\mathscr{T}}_{16}$| from a uniform initial triangulation |${\mathscr{T}}_{0}$| with 16 elements (left) and discrete FVM solution calculated on |${\mathscr{T}}_{16}$| (right). Fig. 2. View largeDownload slide Experiment with a smooth solution from Section 4.1: adaptively generated mesh |${\mathscr{T}}_{16}$| from a uniform initial triangulation |${\mathscr{T}}_{0}$| with 16 elements (left) and discrete FVM solution calculated on |${\mathscr{T}}_{16}$| (right). Fig. 3. View largeDownload slide Experiment with a smooth solution from Section 4.1: error |$E_{\ell}=\|u-u_{\ell}\|_{H^{1}(\varOmega)}$|, weighted-residual error estimator ηℓ and data oscillations oscℓ for uniform and adaptive mesh refinement. Fig. 3. View largeDownload slide Experiment with a smooth solution from Section 4.1: error |$E_{\ell}=\|u-u_{\ell}\|_{H^{1}(\varOmega)}$|, weighted-residual error estimator ηℓ and data oscillations oscℓ for uniform and adaptive mesh refinement. In Fig. 2(a) we see an adaptively generated mesh after 16 refinements. Figure 2(b) plots the smooth solution on the mesh |${\mathscr{T}}_{16}$|. Both uniform and adaptive mesh refinements lead to the optimal convergence order |${\mathscr{O}}(N^{-1/2})$| with respect to the number N of elements since u is smooth; see Fig. 3. The oscillations are of higher order and decrease with |${\mathscr{O}}(N^{-1})$|. Table 1 shows the experimental validation of the additional assumption in Theorem 3.10, i.e., marking for the data oscillations is negligible; see also Remark 3.11. Table 1 Experiment with a smooth solution from Section 4.1: we compute |$\widetilde{C}_{\mathrm{MNS}}:=\#{\mathscr{M}}_\ell /\#{\mathscr{M}}_\ell ^{\eta} \le 1.3$|. Hence, the additional assumption in Theorem 3.10 is experimentally verified. Furthermore, we compute |$\widetilde \theta ^{\prime }:={\mathrm{osc}} _\ell \left ({\mathscr{M}}_\ell ^{\eta} \right )^2/{\mathrm{osc}} _\ell ^2 \ge 0.2$| with |${\mathrm{osc}} _{\ell }\left ({\mathscr{M}}_\ell ^{\eta} \right )^2:=\sum _{T\in{\mathscr{M}}_\ell ^{\eta} } {\mathrm{osc}} _\ell \left (T,u_\ell \right )^2$|, i.e., the choice θ = 0.5, θ′ = 0.2 would guarantee |${\mathscr{M}}_\ell = {\mathscr{M}}_\ell ^{\eta}$| in Algorithm 3.1 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 16 1.000 0.631 1 22 1.000 0.615 2 28 1.000 0.704 3 32 1.000 0.769 4 40 1.214 0.338 5 78 1.111 0.446 6 112 1.133 0.292 7 156 1.119 0.410 8 216 1.062 0.394 9 331 1.198 0.264 10 460 1.014 0.472 11 660 1.049 0.371 12 944 1.027 0.431 13 1,338 1.025 0.400 14 1,910 1.018 0.387 15 2,748 1.026 0.374 16 3,842 1.015 0.358 17 5,430 1.003 0.449 18 7,438 1.013 0.359 19 10,590 1.003 0.445 20 14,478 1.019 0.323 21 20,286 1.004 0.430 22 27,558 1.004 0.457 23 38,450 1.010 0.324 24 52,422 1.000 0.540 25 72,454 1.007 0.404 26 98,232 1.000 0.508 27 135,172 1.004 0.446 28 184,142 1.000 0.606 29 251,896 1.002 0.475 30 342,148 1.001 0.488 31 461,674 1.000 0.617 32 635,266 1.004 0.416 33 852,730 1.000 0.664 34 1,172,122 1.002 0.464 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 16 1.000 0.631 1 22 1.000 0.615 2 28 1.000 0.704 3 32 1.000 0.769 4 40 1.214 0.338 5 78 1.111 0.446 6 112 1.133 0.292 7 156 1.119 0.410 8 216 1.062 0.394 9 331 1.198 0.264 10 460 1.014 0.472 11 660 1.049 0.371 12 944 1.027 0.431 13 1,338 1.025 0.400 14 1,910 1.018 0.387 15 2,748 1.026 0.374 16 3,842 1.015 0.358 17 5,430 1.003 0.449 18 7,438 1.013 0.359 19 10,590 1.003 0.445 20 14,478 1.019 0.323 21 20,286 1.004 0.430 22 27,558 1.004 0.457 23 38,450 1.010 0.324 24 52,422 1.000 0.540 25 72,454 1.007 0.404 26 98,232 1.000 0.508 27 135,172 1.004 0.446 28 184,142 1.000 0.606 29 251,896 1.002 0.475 30 342,148 1.001 0.488 31 461,674 1.000 0.617 32 635,266 1.004 0.416 33 852,730 1.000 0.664 34 1,172,122 1.002 0.464 View Large Table 1 Experiment with a smooth solution from Section 4.1: we compute |$\widetilde{C}_{\mathrm{MNS}}:=\#{\mathscr{M}}_\ell /\#{\mathscr{M}}_\ell ^{\eta} \le 1.3$|. Hence, the additional assumption in Theorem 3.10 is experimentally verified. Furthermore, we compute |$\widetilde \theta ^{\prime }:={\mathrm{osc}} _\ell \left ({\mathscr{M}}_\ell ^{\eta} \right )^2/{\mathrm{osc}} _\ell ^2 \ge 0.2$| with |${\mathrm{osc}} _{\ell }\left ({\mathscr{M}}_\ell ^{\eta} \right )^2:=\sum _{T\in{\mathscr{M}}_\ell ^{\eta} } {\mathrm{osc}} _\ell \left (T,u_\ell \right )^2$|, i.e., the choice θ = 0.5, θ′ = 0.2 would guarantee |${\mathscr{M}}_\ell = {\mathscr{M}}_\ell ^{\eta}$| in Algorithm 3.1 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 16 1.000 0.631 1 22 1.000 0.615 2 28 1.000 0.704 3 32 1.000 0.769 4 40 1.214 0.338 5 78 1.111 0.446 6 112 1.133 0.292 7 156 1.119 0.410 8 216 1.062 0.394 9 331 1.198 0.264 10 460 1.014 0.472 11 660 1.049 0.371 12 944 1.027 0.431 13 1,338 1.025 0.400 14 1,910 1.018 0.387 15 2,748 1.026 0.374 16 3,842 1.015 0.358 17 5,430 1.003 0.449 18 7,438 1.013 0.359 19 10,590 1.003 0.445 20 14,478 1.019 0.323 21 20,286 1.004 0.430 22 27,558 1.004 0.457 23 38,450 1.010 0.324 24 52,422 1.000 0.540 25 72,454 1.007 0.404 26 98,232 1.000 0.508 27 135,172 1.004 0.446 28 184,142 1.000 0.606 29 251,896 1.002 0.475 30 342,148 1.001 0.488 31 461,674 1.000 0.617 32 635,266 1.004 0.416 33 852,730 1.000 0.664 34 1,172,122 1.002 0.464 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 16 1.000 0.631 1 22 1.000 0.615 2 28 1.000 0.704 3 32 1.000 0.769 4 40 1.214 0.338 5 78 1.111 0.446 6 112 1.133 0.292 7 156 1.119 0.410 8 216 1.062 0.394 9 331 1.198 0.264 10 460 1.014 0.472 11 660 1.049 0.371 12 944 1.027 0.431 13 1,338 1.025 0.400 14 1,910 1.018 0.387 15 2,748 1.026 0.374 16 3,842 1.015 0.358 17 5,430 1.003 0.449 18 7,438 1.013 0.359 19 10,590 1.003 0.445 20 14,478 1.019 0.323 21 20,286 1.004 0.430 22 27,558 1.004 0.457 23 38,450 1.010 0.324 24 52,422 1.000 0.540 25 72,454 1.007 0.404 26 98,232 1.000 0.508 27 135,172 1.004 0.446 28 184,142 1.000 0.606 29 251,896 1.002 0.475 30 342,148 1.001 0.488 31 461,674 1.000 0.617 32 635,266 1.004 0.416 33 852,730 1.000 0.664 34 1,172,122 1.002 0.464 View Large 4.2. Experiment with a generic singularity On the L-shaped domain |$\varOmega =(-1,1)^{2}\backslash \big ([0,1]\times [-1,0]\big )$| we consider the exact solution |$u(x_{1},x_{2}) = r^{2/3}\sin (2\varphi /3)$| in polar coordinates |$r\in{\mathbb{R}}_{0}^{+}$|, φ ∈ [0, 2π[ and |$(x_{1},x_{2}) = r(\cos \varphi ,\sin \varphi )$|. It is well known that u has a generic singularity at the reentrant corner (0, 0), which leads to u ∈ H1+2/3−ε|$(\varOmega )$| for all ε > 0. We choose the diffusion matrix $$\textbf{A}= \left ( \begin{array}{@{}cc@{}} 5+\left({x_{1}^{2}}+{x_{2}^{2}}\right)\cos x_{1} & \left({x_{1}^{2}}+{x_{2}^{2}}\right)^{2} \\[6pt] \left({x_{1}^{2}}+{x_{2}^{2}}\right)^{2} & \;5+\left({x_{1}^{2}}+{x_{2}^{2}}\right)\sin x_{2} \end{array}\right)$$ so that (1.2) holds with |$\lambda _{\min }=0.46689$| and |$\lambda _{\max }=5.14751$|, b = (1, 1)T and c = 1 so that (1.4) holds with |$\frac{1}{2}\textrm{div}\,\textbf{b}+{}c=1$|. The right-hand side f is calculated appropriately. The uniform initial mesh |${\mathscr{T}}_{0}$| consists of 12 triangles. An adaptively generated mesh after 16 refinements and a plot of the discrete solution are shown in Fig. 4. We observe the expected suboptimal convergence order of |${\mathscr{O}}(N^{-1/3})$| for uniform mesh refinement. We regain the optimal convergence order of |${\mathscr{O}}(N^{-1/2})$| for adaptive mesh refinement; see Fig. 5. As in the experiment of Section 4.1, the oscillations are of higher order |${\mathscr{O}}(N^{-1})$|. We refer to Table 2 for the experimental validation of the additional assumption in Theorem 3.10 that marking for the data oscillations is negligible. Fig. 4. View largeDownload slide Experiment with a generic singularity of the solution in the reentrant corner (0, 0) from Section 4.2: adaptively generated mesh |${\mathscr{T}}_{16}$| from a uniform initial triangulation |${\mathscr{T}}_{0}$| with 12 elements (left) and discrete FVM solution calculated on |${\mathscr{T}}_{16}$| (right). Fig. 4. View largeDownload slide Experiment with a generic singularity of the solution in the reentrant corner (0, 0) from Section 4.2: adaptively generated mesh |${\mathscr{T}}_{16}$| from a uniform initial triangulation |${\mathscr{T}}_{0}$| with 12 elements (left) and discrete FVM solution calculated on |${\mathscr{T}}_{16}$| (right). Fig. 5. View largeDownload slide Experiment with a generic singularity of the solution from Section 4.2: error |$E_{\ell}=\|u-u_{\ell}\|_{H^{1}(\varOmega)}$|, weighted-residual error estimator ηℓ and data oscillations oscℓ for uniform and adaptive mesh refinement. Fig. 5. View largeDownload slide Experiment with a generic singularity of the solution from Section 4.2: error |$E_{\ell}=\|u-u_{\ell}\|_{H^{1}(\varOmega)}$|, weighted-residual error estimator ηℓ and data oscillations oscℓ for uniform and adaptive mesh refinement. Table 2 Experiment with a generic singularity of the solution from Section 4.2: we compute |$\widetilde{C}_{\mathrm{MNS}}:=\#{\mathscr{M}}_\ell /\#{\mathscr{M}}_\ell^{\eta} \le 1.8$|. Hence, the additional assumption in Theorem 3.10 is experimentally verified. Furthermore, we compute |$\widetilde \theta ^{\prime }:={\mathrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2/{\mathrm{osc}} _\ell ^2 \ge 0.02$| with |${\mathrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2:=\sum _{T\in{\mathscr{M}}_\ell ^{\eta} } {\mathrm{osc}} _\ell (T,u_\ell )^2$|, i.e., the choice θ = 0.5, θ′ = 0.02 would guarantee |${\mathscr{M}}_\ell = {\mathscr{M}}_\ell^{\eta}$| in Algorithm 3.1 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 12 1.667 0.135 1 18 1.750 0.086 2 29 1.600 0.027 3 40 1.375 0.057 4 56 1.400 0.252 5 74 1.667 0.079 6 114 1.286 0.148 7 153 1.188 0.243 8 212 1.111 0.256 9 284 1.065 0.390 10 380 1.194 0.168 11 539 1.068 0.328 12 721 1.050 0.346 13 991 1.007 0.466 14 1,356 1.003 0.482 15 1,852 1.020 0.386 16 2,534 1.000 0.630 17 3,413 1.009 0.443 18 4,684 1.000 0.597 19 6,341 1.003 0.443 20 8,568 1.002 0.490 21 11,564 1.000 0.640 22 15,590 1.000 0.539 23 21,071 1.000 0.569 24 28,304 1.017 0.437 25 38,350 1.000 0.670 26 51,122 1.016 0.414 27 69,135 1.000 0.563 28 92,367 1.000 0.528 29 123,666 1.008 0.463 30 166,532 1.000 0.703 31 221,144 1.020 0.378 32 298,213 1.000 0.549 33 397,086 1.000 0.597 34 532,432 1.017 0.409 35 712,738 1.000 0.666 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 12 1.667 0.135 1 18 1.750 0.086 2 29 1.600 0.027 3 40 1.375 0.057 4 56 1.400 0.252 5 74 1.667 0.079 6 114 1.286 0.148 7 153 1.188 0.243 8 212 1.111 0.256 9 284 1.065 0.390 10 380 1.194 0.168 11 539 1.068 0.328 12 721 1.050 0.346 13 991 1.007 0.466 14 1,356 1.003 0.482 15 1,852 1.020 0.386 16 2,534 1.000 0.630 17 3,413 1.009 0.443 18 4,684 1.000 0.597 19 6,341 1.003 0.443 20 8,568 1.002 0.490 21 11,564 1.000 0.640 22 15,590 1.000 0.539 23 21,071 1.000 0.569 24 28,304 1.017 0.437 25 38,350 1.000 0.670 26 51,122 1.016 0.414 27 69,135 1.000 0.563 28 92,367 1.000 0.528 29 123,666 1.008 0.463 30 166,532 1.000 0.703 31 221,144 1.020 0.378 32 298,213 1.000 0.549 33 397,086 1.000 0.597 34 532,432 1.017 0.409 35 712,738 1.000 0.666 View Large Table 2 Experiment with a generic singularity of the solution from Section 4.2: we compute |$\widetilde{C}_{\mathrm{MNS}}:=\#{\mathscr{M}}_\ell /\#{\mathscr{M}}_\ell^{\eta} \le 1.8$|. Hence, the additional assumption in Theorem 3.10 is experimentally verified. Furthermore, we compute |$\widetilde \theta ^{\prime }:={\mathrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2/{\mathrm{osc}} _\ell ^2 \ge 0.02$| with |${\mathrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2:=\sum _{T\in{\mathscr{M}}_\ell ^{\eta} } {\mathrm{osc}} _\ell (T,u_\ell )^2$|, i.e., the choice θ = 0.5, θ′ = 0.02 would guarantee |${\mathscr{M}}_\ell = {\mathscr{M}}_\ell^{\eta}$| in Algorithm 3.1 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 12 1.667 0.135 1 18 1.750 0.086 2 29 1.600 0.027 3 40 1.375 0.057 4 56 1.400 0.252 5 74 1.667 0.079 6 114 1.286 0.148 7 153 1.188 0.243 8 212 1.111 0.256 9 284 1.065 0.390 10 380 1.194 0.168 11 539 1.068 0.328 12 721 1.050 0.346 13 991 1.007 0.466 14 1,356 1.003 0.482 15 1,852 1.020 0.386 16 2,534 1.000 0.630 17 3,413 1.009 0.443 18 4,684 1.000 0.597 19 6,341 1.003 0.443 20 8,568 1.002 0.490 21 11,564 1.000 0.640 22 15,590 1.000 0.539 23 21,071 1.000 0.569 24 28,304 1.017 0.437 25 38,350 1.000 0.670 26 51,122 1.016 0.414 27 69,135 1.000 0.563 28 92,367 1.000 0.528 29 123,666 1.008 0.463 30 166,532 1.000 0.703 31 221,144 1.020 0.378 32 298,213 1.000 0.549 33 397,086 1.000 0.597 34 532,432 1.017 0.409 35 712,738 1.000 0.666 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 12 1.667 0.135 1 18 1.750 0.086 2 29 1.600 0.027 3 40 1.375 0.057 4 56 1.400 0.252 5 74 1.667 0.079 6 114 1.286 0.148 7 153 1.188 0.243 8 212 1.111 0.256 9 284 1.065 0.390 10 380 1.194 0.168 11 539 1.068 0.328 12 721 1.050 0.346 13 991 1.007 0.466 14 1,356 1.003 0.482 15 1,852 1.020 0.386 16 2,534 1.000 0.630 17 3,413 1.009 0.443 18 4,684 1.000 0.597 19 6,341 1.003 0.443 20 8,568 1.002 0.490 21 11,564 1.000 0.640 22 15,590 1.000 0.539 23 21,071 1.000 0.569 24 28,304 1.017 0.437 25 38,350 1.000 0.670 26 51,122 1.016 0.414 27 69,135 1.000 0.563 28 92,367 1.000 0.528 29 123,666 1.008 0.463 30 166,532 1.000 0.703 31 221,144 1.020 0.378 32 298,213 1.000 0.549 33 397,086 1.000 0.597 34 532,432 1.017 0.409 35 712,738 1.000 0.666 View Large 4.3. Convection-dominated experiment The final example is taken from Mekchay & Nochetto (2005). On the square |$\varOmega$| = (0, 1)2, we fix the diffusion A = 10−3I and the convection velocity b = (x2, 1/2−x1)T. The reaction and right-hand side are c = f = 0. Thus, (1.2) holds with |$\lambda _{\min }=\lambda _{\max }=10^{-3}$| and (1.4) with |$\frac{1}{2}\textrm{div}\,\textbf{b}+{}c=0$|. On the Dirichlet boundary |$\varGamma$|, we prescribe a continuous piecewise linear function by $$u(x_{1},x_{2})|_{\varGamma}= \begin{cases} 1 & \quad\textrm{on } \{0.2005\leq x_{1}\leq 0.4995, x_{2}=0\}, \\ 0 & \quad\textrm{on } \varGamma\backslash\{0.2\leq x_{1}\leq 0.5; x_{2}=0\},\\ \textrm{linear} & \quad\textrm{on } \{0.2\leq x_{1}\leq 0.2005 \textrm{ or }0.4995\leq x_{1}\leq 0.5; x_{2}=0\}. \end{cases}$$ Table 3 Experimental results on marking strategy for the convection-dominated experiment from Section 4.3: we compute |$\widetilde{C}_{\mathrm{MNS}}:=\#{\mathscr{M}}_\ell /\#{\mathscr{M}}_\ell ^{\eta} \le 3$| and see that the additional assumption in Theorem 3.10 is experimentally verified. In addition, we compute |$\widetilde \theta ^{\prime }:={\mathrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2/{\mathrm{osc}} _\ell ^2 \ge 0.03$| with |${\mathrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2:=\sum _{T\in{\mathscr{M}}_\ell ^{\eta} } {\mathrm{osc}} _\ell (T,u_\ell )^2$|, i.e., the choice θ = 0.5, θ′ = 0.03 would guarantee |${\mathscr{M}}_\ell = {\mathscr{M}}_\ell ^{\eta}$| in Algorithm 3.1 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 32 1.125 0.434 1 48 1.400 0.201 2 59 1.500 0.266 3 72 1.667 0.196 4 90 2.500 0.177 5 110 1.333 0.266 6 154 1.583 0.085 7 187 1.500 0.124 8 238 1.786 0.055 9 280 1.296 0.234 10 332 1.371 0.154 11 405 1.412 0.124 12 511 1.537 0.083 13 628 1.521 0.146 14 779 1.559 0.077 15 1,100 1.600 0.064 16 1,428 1.605 0.063 17 1,837 1.643 0.037 18 2,416 1.594 0.058 19 3,195 1.437 0.060 20 4,336 1.583 0.048 21 5,664 1.402 0.072 22 7,666 1.445 0.047 23 10,186 1.351 0.067 24 13,919 1.258 0.078 25 19,041 1.230 0.112 26 26,248 1.182 0.106 27 36,592 1.142 0.135 28 50,806 1.112 0.180 29 70,367 1.082 0.196 30 97,946 1.058 0.227 31 135,122 1.057 0.236 32 186,959 1.028 0.311 33 255,994 1.021 0.311 34 351,880 1.022 0.289 35 484,157 1.015 0.328 36 662,325 1.006 0.381 37 902,659 1.005 0.384 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 32 1.125 0.434 1 48 1.400 0.201 2 59 1.500 0.266 3 72 1.667 0.196 4 90 2.500 0.177 5 110 1.333 0.266 6 154 1.583 0.085 7 187 1.500 0.124 8 238 1.786 0.055 9 280 1.296 0.234 10 332 1.371 0.154 11 405 1.412 0.124 12 511 1.537 0.083 13 628 1.521 0.146 14 779 1.559 0.077 15 1,100 1.600 0.064 16 1,428 1.605 0.063 17 1,837 1.643 0.037 18 2,416 1.594 0.058 19 3,195 1.437 0.060 20 4,336 1.583 0.048 21 5,664 1.402 0.072 22 7,666 1.445 0.047 23 10,186 1.351 0.067 24 13,919 1.258 0.078 25 19,041 1.230 0.112 26 26,248 1.182 0.106 27 36,592 1.142 0.135 28 50,806 1.112 0.180 29 70,367 1.082 0.196 30 97,946 1.058 0.227 31 135,122 1.057 0.236 32 186,959 1.028 0.311 33 255,994 1.021 0.311 34 351,880 1.022 0.289 35 484,157 1.015 0.328 36 662,325 1.006 0.381 37 902,659 1.005 0.384 Table 3 Experimental results on marking strategy for the convection-dominated experiment from Section 4.3: we compute |$\widetilde{C}_{\mathrm{MNS}}:=\#{\mathscr{M}}_\ell /\#{\mathscr{M}}_\ell ^{\eta} \le 3$| and see that the additional assumption in Theorem 3.10 is experimentally verified. In addition, we compute |$\widetilde \theta ^{\prime }:={\mathrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2/{\mathrm{osc}} _\ell ^2 \ge 0.03$| with |${\mathrm{osc}} _{\ell }({\mathscr{M}}_\ell ^{\eta} )^2:=\sum _{T\in{\mathscr{M}}_\ell ^{\eta} } {\mathrm{osc}} _\ell (T,u_\ell )^2$|, i.e., the choice θ = 0.5, θ′ = 0.03 would guarantee |${\mathscr{M}}_\ell = {\mathscr{M}}_\ell ^{\eta}$| in Algorithm 3.1 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 32 1.125 0.434 1 48 1.400 0.201 2 59 1.500 0.266 3 72 1.667 0.196 4 90 2.500 0.177 5 110 1.333 0.266 6 154 1.583 0.085 7 187 1.500 0.124 8 238 1.786 0.055 9 280 1.296 0.234 10 332 1.371 0.154 11 405 1.412 0.124 12 511 1.537 0.083 13 628 1.521 0.146 14 779 1.559 0.077 15 1,100 1.600 0.064 16 1,428 1.605 0.063 17 1,837 1.643 0.037 18 2,416 1.594 0.058 19 3,195 1.437 0.060 20 4,336 1.583 0.048 21 5,664 1.402 0.072 22 7,666 1.445 0.047 23 10,186 1.351 0.067 24 13,919 1.258 0.078 25 19,041 1.230 0.112 26 26,248 1.182 0.106 27 36,592 1.142 0.135 28 50,806 1.112 0.180 29 70,367 1.082 0.196 30 97,946 1.058 0.227 31 135,122 1.057 0.236 32 186,959 1.028 0.311 33 255,994 1.021 0.311 34 351,880 1.022 0.289 35 484,157 1.015 0.328 36 662,325 1.006 0.381 37 902,659 1.005 0.384 ℓ |$\#{\mathscr{T}}_\ell$| |$\frac{\#{\mathscr{M}}_\ell }{\#{\mathscr{M}}_\ell ^{\eta} }$| |$\frac{{\textrm{osc}} _\ell ({\mathscr{M}}_\ell ^{\eta} )^2}{{\textrm{osc}} _\ell ^2}$| 0 32 1.125 0.434 1 48 1.400 0.201 2 59 1.500 0.266 3 72 1.667 0.196 4 90 2.500 0.177 5 110 1.333 0.266 6 154 1.583 0.085 7 187 1.500 0.124 8 238 1.786 0.055 9 280 1.296 0.234 10 332 1.371 0.154 11 405 1.412 0.124 12 511 1.537 0.083 13 628 1.521 0.146 14 779 1.559 0.077 15 1,100 1.600 0.064 16 1,428 1.605 0.063 17 1,837 1.643 0.037 18 2,416 1.594 0.058 19 3,195 1.437 0.060 20 4,336 1.583 0.048 21 5,664 1.402 0.072 22 7,666 1.445 0.047 23 10,186 1.351 0.067 24 13,919 1.258 0.078 25 19,041 1.230 0.112 26 26,248 1.182 0.106 27 36,592 1.142 0.135 28 50,806 1.112 0.180 29 70,367 1.082 0.196 30 97,946 1.058 0.227 31 135,122 1.057 0.236 32 186,959 1.028 0.311 33 255,994 1.021 0.311 34 351,880 1.022 0.289 35 484,157 1.015 0.328 36 662,325 1.006 0.381 37 902,659 1.005 0.384 The model has a moderate convection dominance with respect to the diffusion and simulates the transport of a pulse from |$\varGamma$| to the interior and back to |$\varGamma$|. For this example, we do not know the analytical solution. The uniform initial mesh |${\mathscr{T}}_{0}$| consists of 32 triangles. In Fig. 6(a), we see the solution with strong oscillations on a uniformly generated mesh with 8,192 elements. The oscillations are due to the convection dominance. For the next refinement step (16,384 elements, not plotted), however, the oscillations disappear since the shock region at the boundary is refined enough. Our adaptive Algorithm 3.1, which also has a mandatory oscillation marking, provides a stable solution on a mesh with only 779 elements; see Fig. 6(b). In Fig. 7, we plot adaptively generated meshes after 14 and 20 mesh refinements. We see a strong refinement in the shock region. A similar observation can be found in Mekchay & Nochetto (2005). We remark that this strategy works only for this moderate convection-dominated problem. For A = 10−8I, we cannot see any stabilization effects by Algorithm 3.1 (not displayed). Hence, only a stabilization of the numerical scheme, e.g., FVM with upwinding, would avoid these instabilities. However, the analysis of such schemes is beyond the scope of this work; see also Section 5. We observe the above stabilization effects also in the convergence plot of the estimator; see Fig. 8. Note that the estimator for adaptive mesh refinement is faster in the asymptotic convergence than the estimator for uniform mesh refinement. Additionally, the convergence rate for the estimator is suboptimal for uniform mesh refinement. For adaptive mesh refinement, we regain the optimal convergence order of |${\mathscr{O}}(N^{-1/2})$|; see Fig. 8. As in the previous experiments, the oscillations are of higher order. In Table 3, we also see that the oscillation marking for this convection-dominated problem is for more refinement steps dominant than for the previous problems; see also the discussion in Mekchay & Nochetto (2005). Fig. 6. View largeDownload slide Convection-dominated experiment from Section 4.3: the discrete FVM solution on a uniformly generated mesh |${\mathscr{T}}_{8}$| (left) and adaptively generated mesh |${\mathscr{T}}_{14}$| (right). The algorithm starts with a uniform initial triangulation |${\mathscr{T}}_{0}$| with 32 elements. Fig. 6. View largeDownload slide Convection-dominated experiment from Section 4.3: the discrete FVM solution on a uniformly generated mesh |${\mathscr{T}}_{8}$| (left) and adaptively generated mesh |${\mathscr{T}}_{14}$| (right). The algorithm starts with a uniform initial triangulation |${\mathscr{T}}_{0}$| with 32 elements. Fig. 7. View largeDownload slide Convection-dominated experiment from Section 4.3: adaptively generated meshes |${\mathscr{T}}_{14}$| (left) and |${\mathscr{T}}_{20}$| (right) from a uniform initial triangulation |${\mathscr{T}}_{0}$| with 32 elements. Fig. 7. View largeDownload slide Convection-dominated experiment from Section 4.3: adaptively generated meshes |${\mathscr{T}}_{14}$| (left) and |${\mathscr{T}}_{20}$| (right) from a uniform initial triangulation |${\mathscr{T}}_{0}$| with 32 elements. Fig. 8. View largeDownload slide Convection-dominated experiment from Section 4.3: weighted-residual error estimator ηℓ and data oscillations oscℓ for uniform and adaptive mesh refinement. Fig. 8. View largeDownload slide Convection-dominated experiment from Section 4.3: weighted-residual error estimator ηℓ and data oscillations oscℓ for uniform and adaptive mesh refinement. 5. Conclusions In this work, we have proved linear convergence of an adaptive vertex-centered FVM with generically optimal algebraic rates to the solution of a general second-order linear elliptic PDE. Besides marking based on the local contributions of the a posteriori error estimator, we additionally had to mark the oscillations to overcome the lack of a classical Galerkin orthogonality property. In the case of dominating convection, FVMs provide a natural upwind stabilization. Although there exist estimators for these upwind discretizations also (see Erath, 2013), we were not able to provide a rigorous convergence result for the related adaptive mesh-refinement strategy. Note that the upwind direction and thus the corresponding error-indicator contributions are defined over the boundary of the control volumes of the dual mesh. As mentioned above, the dual meshes are not nested for a sequence of locally refined triangulations. This makes it difficult to show (A1)–(A2) and (B1)–(B2). We stress that the other error-indicator contributions are defined over the elements of the primal mesh and can hence be treated by the developed techniques. Funding Austrian Science Fund (FWF) research project Optimal adaptivity for BEM and FEM-BEM coupling (P27005 to D.P.), research program Taming complexity in partial differential systems. (F65 to D.P.). References Bergh , J. & Löfström , J. (1976) Interpolation Spaces. An Introduction. Grundlehren der Mathematischen Wissenschaften , vol. 223. Berlin-New York: Springer. Bespalov , A. , Haberl , A. & Praetorius , D. (2017) Adaptive fem with coarse initial mesh guarantees optimal convergence rates for compactly perturbed elliptic problems . Comput. Methods Appl. Mech. Engrg. , 317 , 318 – 340 . Crossref Search ADS Binev , P. , Dahmen , W. & DeVore , R. (2004) Adaptive finite element methods with convergence rates . Numer. Math. , 97 , 219 – 268 . Crossref Search ADS Carstensen , C. , Feischl , M. , Page , M. & Praetorius , D. (2014) Axioms of adaptivity . Comput. Math. Appl. , 67 , 1195 – 1253 . Crossref Search ADS PubMed Carstensen , C. , Lazarov , R. D. & Tomov , S. Z. (2005) Explicit and averaging a posteriori error estimates for adaptive finite volume methods . SIAM J. Numer. Anal. , 42 , 2496 – 2521 . Crossref Search ADS Cascón , J. M. , Kreuzer , C. , Nochetto , R. H. & Siebert , K. G. (2008) Quasi-optimal convergence rate for an adaptive finite element method . SIAM J. Numer. Anal. , 46 , 2524 – 2550 . Crossref Search ADS Cascón , J. M. & Nochetto , R. H. (2012) Quasioptimal cardinality of AFEM driven by nonresidual estimators . IMA J. Numer. Anal. , 32 , 1 – 29 . Crossref Search ADS Dörfler , W. (1996) A convergent adaptive algorithm for Poisson’s equation . SIAM J. Numer. Anal. , 33 , 1106 – 1124 . Crossref Search ADS Erath , C. (2010) Coupling of the finite volume method and the boundary element method-theory, analysis, and numerics . Ph.D. Thesis, University of Ulm , Ulm . Erath , C. (2012) Coupling of the finite volume element method and the boundary element method: an a priori convergence result . SIAM J. Numer. Anal. , 50 , 574 – 594 . Crossref Search ADS Erath , C. (2013) A posteriori error estimates and adaptive mesh refinement for the coupling of the finite volume method and the boundary element method . SIAM J. Numer. Anal. , 51 , 1777 – 1804 . Crossref Search ADS Erath , C. & Praetorius , D. (2016) Adaptive vertex-centered finite volume methods with convergence rates . SIAM J. Numer. Anal. , 54 , 2228 – 2255 . Crossref Search ADS Erath , C. & Praetorius , D. (2017) Céa-type quasi-optimality and convergence rates for (adaptive) vertex-centered FVM . Finite Volumes for Complex Applications VIII-Methods and Theoretical Aspects (C. Cances & P. Omnes eds), vol. 199. Berlin: Springer . Ewing , R. E. , Lin , T. & Lin , Y. (2002) On the accuracy of the finite volume element method based on piecewise linear polynomials . SIAM J. Numer. Anal. , 39 , 1865 – 1888 . Crossref Search ADS Feischl , M. , Führer , T. & Praetorius , D. (2014) Adaptive FEM with optimal convergence rates for a certain class of nonsymmetric and possibly nonlinear problems . SIAM J. Numer. Anal. , 52 , 601 – 625 . Crossref Search ADS Grisvard , P. (1985) Elliptic Problems in Nonsmooth Domains . Boston: Pitman . Karkulik , M. , Pavlicek , D. & Praetorius , D. (2013) On 2D newest vertex bisection: optimality of mesh-closure and H1-stability of L2-projection . Constr. Approx. , 38 , 213 – 234 . Crossref Search ADS Mekchay , K. & Nochetto , R. H. (2005) Convergence of adaptive finite element methods for general second order linear elliptic PDEs . SIAM J. Numer. Anal. , 43 , 1803 – 1827 . Crossref Search ADS Morin , P. , Nochetto , R. H. & Siebert , K. G. (2000) Data oscillation and convergence of adaptive FEM . SIAM J. Numer. Anal. , 38 , 466 – 488 . Crossref Search ADS Scott , L. R. & Zhang , S. (1990) Finite element interpolation of nonsmooth functions satisfying boundary conditions . Math. Comp. , 54 , 483 – 493 . Crossref Search ADS Stevenson , R. (2007) Optimality of a standard adaptive finite element method . Found. Comput. Math. , 7, 245 – 269 . Crossref Search ADS Stevenson , R. (2008) The completion of locally refined simplicial partitions created by bisection . Math. Comp. , 77, 227 – 241 . Crossref Search ADS © The Author(s) 2018. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model) TI - Adaptive vertex-centered finite volume methods for general second-order linear elliptic partial differential equations JF - IMA Journal of Numerical Analysis DO - 10.1093/imanum/dry006 DA - 2019-04-23 UR - https://www.deepdyve.com/lp/oxford-university-press/adaptive-vertex-centered-finite-volume-methods-for-general-second-sU3hEMsxia SP - 983 VL - 39 IS - 2 DP - DeepDyve ER -
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## Equal-area histograms
##### Apr 28, 2023
Let's review what a histogram is. This one I created by grabbing a dataset of scoring in international football (soccer) matches.
The histogram is a specialized type of column chart, typically displayed with zero spacing between columns. The horizontal axis represents the metric being plotted. In the example above, it's the minute of scoring with values between 0 and 120. (There are few points beyond 90 minutes as only certain tournaments prescribe extra-time in case of ties at the end of 90 minutes.) Other examples are income if it's a histogram of income distribution, or age if it's an age distribution. The metric is typically measured numerically.
The notion of "bins" is fundamental. The numeric metric on the horizontal axis is discretized into "bins". (In the above example, R selected a bin width of 10 minutes.) Binning transforms the data to fit natural concepts such as income groups and age groups. Each column has a bin width, and its height is either counts or frequencies. Counts are the number of data points that fall inside a given bin. Frequencies are the proportion of data points that fall inside a given bin. The sum of counts equals the sample size while the sum of frequencies is 100%.
There is a large literature on how to "bin" the data. In the canonical histogram, the horizontal axis is subdivided into bins of equal length. This means the histogram columns are equal width. You can readily see that in the histogram shown above.
The histogram conveys information about where the data is. The following set includes three possibilities: a uniform distribution (in which the data density is the same over the range of the horizontal axis), a normal distribution (the familiar bell curve), and a long-tailed distribution with mostly small values.
***
In general, the bin widths need not to be the same for all bins. One can vary bin widths. The "percentogram" referenced in Andrew's post specifies one way of setting varying bin widths: it prescribes that the breaks between bins be percentiles of the distribution.
In other words, the first bin contains the lowest 1% of the data, the second bin, the next 1%, etc. Let's see the effect of varying bin widths on the normal distribution. Note that I set the average value to be 4 so that almost all the data fall between 1 and 7.
The top chart is a dot plot showing the actual data. The middle chart shows a generic equal-width histogram. The third chart shows the percentogram, in which each bin contains 1% of the data.
In the percentogram, the bin width is a function of the density of data, with wider bins where data are sparse and narrower bins where data are dense. For a normal distribution, the data are quite concentrated in the middle. The columns on the side are wide and short.
While the standard histogram has equal-width bins, the percentogram has bin widths that vary. What is fixed is the amount of data in each column, that is to say, the area of each column is fixed.
The following set of charts corresponds to the triple shown above. Each chart contains 10,000 random samples. The same datasets were used in both sets of charts.
A few negatives of the percentogram jump out.
The column heights are rather jagged, and that is purely random noise, since the data are drawn from standard distributions. Because the data are divided into 100 parts, each column contains 100 data points, and that sample size is not large enough to tame the sampling noise.
Also, the middle of the normal distribution, where most of the data sit, look hollow. This is feature, not a bug, of the percentogram. Dense columns will be narrow and tall, like lines while sparse columns will be wide and short. Even though each column has the same area, our eyes tend to focus on the short wide ones, rather than the pencil columns.
***
Nothing should stop us from making equal-area bins based on other quantiles. For example, the following set divides the data into 20 bins (demideciles) instead of 100:
With smaller number of bins, the envelope of the histograms are less jagged. The denser columns are also less narrow, and thus don't exhibit the hollowing effect.
For validation, the equal-area histogram for uniform distributions looks the same as the equal-width histogram. This is expected since by definition, the data density is uniform across the whole range. Columns that contain equal counts should therefore have equal widths as the heights should be equal (excluding randomness).
What's a use case for equal-area histograms? It's for reading off sections of the data.
With 20 bins, each column contains 5% of the data. So it's easy to find the cutoff value for the top 10% of the above distribution. Just count the two columns from the right. Finding the middle 50% is a bit harder, unless the column index is printed on the chart but it's possible to find the range of values that include half the dataset.
By contrast, the standard histogram does not offer a ready answer to this type of question. One would have to look at the height of each column and start adding.
***
I have a few other comments on this variant of the histogram as well as some details I glossed over in this post. Stay tuned!
You can follow this conversation by subscribing to the comment feed for this post.
I am very confused. Since in a percentagram each bin contains roughly the same number of data points, wouldn't the heights of the columns be about the same?
Seriously, I always wonder why on earth people bother with histograms. Why don't you just draw an empirical CDF?
After reading Andrews' post, I get it. The number of data points determines not the height, but the area. Interesting. I would add ticks on the horizontal axis showing the quartiles.
L: You're not the only one. I was also confused the same way you were but yes, the idea is counts are encoded in the area, not the height. My next post will address this point more directly.
GG: My next post will feature some CDFs. In my experience, CDFs are hard for people to understand because of the cumulation. If the question is where is my data, the histogram or PDF works better. This percentogram idea is a nice complement to a CDF, and they can complement one another.
Thus the equal area :). When I first read the title, I was thinking that the boundaries will be hard to determine: the bin width times the number of points in the bin would have to be constant.
L: I'll cover this in the next post. This method is equal "width" on the CDF scale. That translates back to the bin widths.
These fall into the general class of bar charts where the width of the bars, not just the x-position, convey information. The Economist has a good example this week comparing GDP per person with population as the width, so the bar area is total GDP. . Gift link here: https://econ.st/3LGqF36
The comments to this entry are closed.
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? weeks -------- 1 year 52 weeks -------- 1 year hehe-u don't get a hint ? ft -------- 1 mile 5280 ft. -------- 1 mile hehe-u don't get a hint ? yards -------- 1 ft 3 yards -------- 1 foot hehe-u don't get a hint ? seconds -------- 1 hour 3600 seconds -------- 1 hour hehe-u don't get a hint ? mile -------- 1 km .62 -------- 1 km hehe-u don't get a hint what is the answer to a line w/ an undefined slope x=# hehe-u don't get a hint what is the answer to a line w/ the slope of zero y=# hehe-u don't get a hint 0 - # slope of zero hehe-u don't get a hint # - 0 undefined slope hehe-u don't get a hint what is a constant of a variation? the slope of a line that goes through the origin what type of line is m= undefined? vertical line hehe-u don't get a hint what type of line is m= o horizontal line hehe-u don't get a hint what is the equation for the constant of a variation? y=kx k=y - x when k doesn't equal 0 hehe-u don't get a hint
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Cody
# Problem 174. Roll the Dice!
Solution 2163804
Submitted on 18 Mar 2020 by Marius Mueller
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### Test Suite
Test Status Code Input and Output
1 Pass
x1 = zeros(1,6000); x2 = zeros(1,6000); for ii = 1:6000 [x1(ii),x2(ii)] = rollDice(); end numCt = sum( bsxfun( @eq, x1, (1:6)' ), 2 ) + sum( bsxfun( @eq, x2, (1:6)' ), 2 ); assert(all(round(numCt/200) == 10) && sum(numCt) == 12000)
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'
# Search results
Found 1357 matches
The allowable radius for a horizontal curve can then be determined by knowing the intended design velocity, the coefficient of friction, and the allowed ... more
Brinell Hardness Number
BHN or Brinell Number is the numerical value assigned to the hardness of metals and alloys. The test is to determine the hardness ... more
Power (aerodynamic drag)
In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force ... more
Worksheet 300
Calculate the Reynolds number N′R for a ball with a 7.40-cm diameter thrown at 40.0 m/s.
Strategy
We can use the Reynolds number equation calculate N’R , since all values in it are either given or can be found in tables of density and viscosity.
Solution
We first find the kinematic viscosity values:
Kinematic Viscosity
Substituting values into the equation for N’R yields:
Reynolds number
Discussion
This value is sufficiently high to imply a turbulent wake. Most large objects, such as airplanes and sailboats, create significant turbulence as they move. As noted before, the Bernoulli principle gives only qualitatively-correct results in such situations.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Power in a reference system(aerodynamic drag)
In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force ... more
maximum axial tension - clamp band
In the aerospace industry, the empirical methodology for evaluating the axial load capability of the clamp band joint assumes the joint components to be ... more
Brinell scale ( using the SI units)
The Brinell scale characterizes the indentation hardness of materials through the scale of penetration of an indenter, loaded on a material test-piece. It ... more
Spherical wedge (Area of the lune)
A spherical wedge or ungula is a portion of a ball bounded by two plane semidisks and a spherical lune (termed the wedge’s base). The angle between the ... more
Kinetic Friction
Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. Kinetic friction is ... more
Steadily rotating crank ( displacement of the end of the connecting rod )
A crank is an arm attached at right angles to a rotating shaft by which reciprocating motion is imparted to or received from the shaft. It is used to ... more
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# arithmetic mean
Quizzes and games : statistics
The average function allows to calculate the arithmetic mean of a series of values.
average([1;9;7]) returns 17/3
Quizzes and games : statistics
### Average, online calculus
#### Summary :
The average function allows to calculate the arithmetic mean of a series of values.
average online
#### Description :
The calculator with its average function is a mean calculator. It allows to make the arithmetic mean of a set value. The arithmetic mean is equal to the sum of the series values divided by the number of the elements of the series. The online calculator allows to calculate the arithmetic mean of a series of values indicating the steps of the calculations. The average calculator supports numeric but also literal expressions. It is also possible to involve frequencies for the average calculation.
# Arithmetic mean online calculation
The average computer is able to calculate the average of a set value, the result is returned in exact form, and in approximate form, the details of the calculations are specified.
It is possible to calculate the average of the following numbers 12;32;45;34, enter average([12;32;45;34])
It is possible to calculate the average of the following numbers 12;32;45;34 It is possible to calculate the average of the following numbers 3;5;3;2 enter average([[12;32;45;34];[3;5;3;2]])
# Arithmetic mean online symbolic calculation
The average calculator is able to calculate the average of a series of literal expressions, the result is returned in exact form, and details of the calculations are specified.
It is possible to calculate the average of the following terms 3a;6a;7a, after calculating the result is returned with the steps of calculations, for this, just enter average([3a;6a;7a]),
It is possible to calculate the average of the following numbers 3a;6a;7a whose frequency 3;5;3;2 after calculating the result is returned with the steps of calculations, for this, enter average([[3a;6a;7a];[3;5;3;2]]).
#### Syntax :
• average([s1;s2;...;sn]), s1, s2, ..., sn are a series of numbers
• average([[s1;s2;...;sn];[f1;f2;...;fn]]), s1, s2, ..., sn are a series of numbers, f1; f2, ..., fn represent the frequencies of these numbers.
#### Examples :
average([1;9;7]) returns 17/3
Calculate online with average (arithmetic mean)
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BRAIN TEASERS
# Monthly Archive - September 2020
brain teasers, puzzles, riddles, mathematical problems, mastermind, cinemania... These are the tasks listed 1 to 10.
## Find number abc
If 5c496 - a6aba = a7a94 find number abc. Multiple solutions may exist.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math
## Find the right combination
The computer chose a secret code (sequence of 4 digits from 1 to 6). Your goal is to find that code. Black circles indicate the number of hits on the right spot. White circles indicate the number of hits on the wrong spot.
The first user who solved this task is Nasrin 24 T.
#brainteasers #mastermind
## Calculate the number 253
NUMBERMANIA: Calculate the number 253 using numbers [8, 2, 3, 6, 40, 156] and basic arithmetic operations (+, -, *, /). Each of the numbers can be used only once.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math #numbermania
### Jim Gaffigan: Bottled Water
How did we get to the point where were paying for bottled water? That must have been some weird marketing meeting over in France. Some French guys sitting there, like, How dumb do I think the Americans are? I bet you we could sell those idiots water.
Jokes of the day - Daily updated jokes. New jokes every day.
## Find number abc
If c4ab5 + c75b1 = 171cab find number abc. Multiple solutions may exist.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math
## Calculate the number 5117
NUMBERMANIA: Calculate the number 5117 using numbers [8, 3, 2, 2, 35, 928] and basic arithmetic operations (+, -, *, /). Each of the numbers can be used only once.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math #numbermania
## MAGIC SQUARE: Calculate A-B-C
The aim is to place the some numbers from the list (3, 4, 5, 6, 7, 9, 18, 33, 39, 40, 43, 83, 88) into the empty squares and squares marked with A, B an C. Sum of each row and column should be equal. All the numbers of the magic square must be different. Find values for A, B, and C. Solution is A-B-C.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math #magicsquare
## Find number abc
If a2ab0 - 3a705 = cc1a5 find number abc. Multiple solutions may exist.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math
## Calculate the number 306
NUMBERMANIA: Calculate the number 306 using numbers [1, 1, 9, 9, 63, 226] and basic arithmetic operations (+, -, *, /). Each of the numbers can be used only once.
The first user who solved this task is NĂlton CorrĂȘa de Sousa.
#brainteasers #math #numbermania
## Find number abc
If 3bca6 - b3937 = 8c79 find number abc. Multiple solutions may exist.
The first user who solved this task is Nasrin 24 T.
#brainteasers #math
## Which is a winning combination of digits?
The computer chose a secret code (sequence of 4 digits from 1 to 6). Your goal is to find that code. Black circles indicate the number of hits on the right spot. White circles indicate the number of hits on the wrong spot.
The first user who solved this task is Nasrin 24 T.
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Lab3-4-Atmpt1
# Lab3-4-Atmpt1 - x Prediction Equation f(z = 0 1.132 z x Fit...
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1.2 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 Seconds 1.5 0.0 0.5 1.0 X Position vs. Time 4.0 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Seconds 4.0 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Prediction Data Fit Y Position vs. Time 1.8 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Seconds 1.5 0.0 0.2 0.5 0.8 1.0 1.2 X Velocity vs. Time 4.0 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Seconds 4.0 -1.0 0.0 1.0 2.0 3.0 Y Velocity vs. Time Printed Monday, October 18, 2010, 5:49 PM Block: 575g Theta: 22.596 degrees Graph Title: f(z) = 0 + 0 z + 0.638 z^2
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Unformatted text preview: x Prediction Equation f(z) = 0 + 1.132 z x Fit Equation f(z) = 0 + 0 z y Prediction Equation f(z) = 0 + 0 z y Fit Equation f(z) = 0 + 1.27 z Vx Prediction Equation f(z) = 0.9 + 0.25 z Vx Fit Equation f(z) = 0 + 0 z Vy Prediction Equation f(z) = 0 + 0 z Vy Fit Equation m Units Print! PDF created with pdfFactory Pro trial version www.pdffactory.com...
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# Linear and angular acceleration.
1. Sep 14, 2007
### Lillemanden
Hey.
I have been working on a relative simple 2D physics engine for a game I am making, and as I was researching the subject (rigid body mechanics) I found this article on physics in games. This pretty much answers everything. Though this is one little thing that I cannot figure out. The article says to calculate linear and angular acceleration totally independent of each other but based on the same forces. So the linear acceleration of a body would be all the forces affecting the body summed and divided by the mass of the body. If this is true then the same forces (identical direction and magnitude) applied on the same body but at different points would always cause the same linear acceleration but not necessary the same angular acceleration. How is this possible? I would think that the more angular acceleration the less linear acceleration and vice versa. How else is the total energy level preserved?
I am pretty sure it is just me who has misunderstood something; and an explanation would be very appreciated.
2. Sep 14, 2007
### Staff: Mentor
The article is correct. The linear acceleration of the center of mass just depends on the net force on the object, not on where the force is applied. The angular acceleration about the center of mass depends on where the force is applied. (Both statements are just consequences of Newton's 2nd law.)
Realize that the work you do on an object is force times the distance that the contact point moves. When you push the object with an off-center force the contact point moves more (compared to an equal on-center force), thus it takes more work to maintain the force--that extra work goes into the rotational energy.
3. Sep 14, 2007
### Lillemanden
Thanks, you are totally right. I just needed to have my understanding of forces refreshed.
4. Sep 15, 2007
### Lillemanden
After thinking a bit more about it, I must admit that there is something I still have not understood.
If a rocket engine burning fuel at a constant rate is attached to a spaceship, does it not apply the same force on the spaceship regardless of its position?
Last edited: Sep 15, 2007
5. Sep 15, 2007
### rcgldr
Assuming the space ship is in space (no drag), the force is the same regardless of it's velocity. What happens is burnt fuel is ejected from the rocket ship at very high velocity, accelerating the rocket ship and the remaining unburnt fuel. The force is equal to the average mass and it's average rate of acceleration at any instant. The power relative to the spaceship, is equal to the rate of increase of kinetic energy (increase in KE / time) of the burnt fuel.
However if the frame of reference has the same velocity as the space ships initial velocity, then the force (thrust) remains constant while the rocket's speed increases, so the power is also increasing. One way to accept this is to realize that the burnt fuel increases the kinetic energy of the space ship and the remaining unburnt fuel, effectively increasing the kinetic energy and therefore the total energy of the remaining fuel, relative to the initial velocity based frame of reference. Also as fuel is burnt, the remaining mass of the rocket and unburnt fuel decreases over time. If the ratio of fuel to mass of rocket is high enough, the rocket can go faster than the velocity of the burnt fuel.
6. Sep 15, 2007
### Lillemanden
I am sorry but it is still not clear to me.
If a spaceship has to identical rocket engines, attached at different angles to the center of mass. So firing one engine will produce more angular acceleration than firing the other.
Then would the engine producing less angular acceleration not produce more linear acceleration?
But if the linear acceleration is found from the net force, then the engine producing more angular acceleration should apply less force.
But then that means that the two identical engines does not apply the same force on the spaceship. Which does not make much sense to me.
So what am I getting wrong?
7. Sep 15, 2007
### rcgldr
Say you have a small sphere that weighs 1 kg and is 2 meters in diameter.
First case: The sphere is on a treadmill that begins to accelerate at the rate required to generate 1 Newton of force at the edge of the sphere, but the sphere is glued to the treadmill and can't roll. The sphere accelerates at the rate of 1 m / s^2, as well as the tread mill, in order to maintain 1 Newton of force on the sphere.
Second case: The sphere is on a treadmill that begins to accelerate at the rate required to generate 1 Newton of force at the edge of the sphere, the sphere is free to roll, but there is no slippage. Obviously the sphere will offer less "resistance" to the treadmill and therefore the treadmill will have to accelerate faster than the first case in order to maintain 1 Newton of force at the edge of the sphere. Still as posted before, the linear inertial reaction force of the sphere will equal the 1 Newton of force applied by the treadmill. This means the sphere will be accelerating linearly at 1 m / s^2, however the treadmill will be accelerating at a faster rate, since the sphere is also free to roll.
The sphere has a linear and a rotational kinetic energy:
Angular moment of sphere = 2/5 m r^2.
Linear moment of sphere = m.
Let w = rate of rotation of sphere = angular velocity:
linear KE = KEl = 1/2 m v^2
angular KE = KEa = (1/2 x 2/5 m r^2 x w^2)
linear speed at edge of sphere, v = w x r => w = v/r
Let c = ratio between acceleration of sphere / acceleration of plane
Since time is the same for both, c is also the ratio of velocity of sphere vs plane,
and distance traveled by sphere vs plane.
as = c x ap => ap = as/c
vs = c x vp => vp = vs/c
ds = c x dp => dp = ds/c
w = (vp-vs)/r = (vs/c-vs)/r = (1/r)((1 - c)/c)vs
Let fs = force on sphere:
fs = m x as = m x dvs/dt (linear)
Torque on sphere is then:
fs x r = 2/5 m r^2 x o = 2/5 m r^2 (1/r)((1 - c)/c)dvs/dt
divide torque equation by r:
fs = (1/r)(2/5 m r^2 (1/r)((1 - c)/c)dvs/dt) = 2/5 m ((1-c)/c)dvs/dt
Include linear equation for fs:
fs = m dvs/dt = 2/5 m ((1-c)/c)dvs/dt
Divide both sides by m dvs/dt:
1 = (2/5) ((1-c)/c)
c = (2/5) (1-c)
c = 2/5 - (2/5) c
(1+2/5) c = 2/5
7/5 c = 2/5
c = 5/7 x 2/5 = 2/7
So the linear acceleration of the sphere is only 2/7th's the acceleration of the treadmill, and the rotational acceleration is equal to the (acceleration of the treadmill - acceleration of the sphere) / radius of the sphere.
The treadmill has to accelerate at 7/2 m/s^2 in order to generate 1 Newton of force at the bottom edge of a rolling 1kg sphere (note radius doesn't matter), 3.5 times as fast as compared to the first case with the glued sphere.
Using the change in energy to check this result:
What is the kinetic energy after the treadmill has accelerated for 1 second?
Distance = 1/2 (7/2 m/s^2) s^2 = 7/4 m.
Work done = 7/4 N m = 7/4 joules.
Speed of treadmill after 1 second = 1s x (7/2 m/s^2) = 7/2 m/s
Linear speed of sphere = 2/7 speed of tread mill = 2/7 x (7/2 m/s^2) x 1 s = 1 m/s
Linear speed of sphere related to rotation = speed of tread mill - speed of sphere = (7/2 - 1)m/s = 5/2 m/s
Kinetic energy of sphere:
KE = 1/2 kg (1 m/s)^2 + 1/2 kg 2/5 x (5/2 m/s)^2 = (1/2 + 5/4) kg (m/s)^2
KE = 7/4 kg (m/s)^2
N = kg m/(s^2) => kg = N s^2/m
KE = 7/4 N (s^2/m) (m/s)^2 = 7/4 N m = 7/4 joules
Since force is constant and the velocity of the treadmill increases linearly, then power (= force x speed) also increases linearly. In the first case, power = t watts, in the second case, 7/2 t watts where t is elapsed time/seconds. Average power over time t is 1/2 the power at t. After 1 second, average power in the first case would be 1/2 watt, corresponding to work done and an increase in KE of the glue sphere of 1/2 joules, and in the second case, it's 7/4 watt, correspoding to and increase of KE of the sphere of 7/4 joules. After 2 seconds, work done and increase in KE in the first case 2 joules, in the second case, 7 joules.
Last edited: Sep 15, 2007
8. Sep 15, 2007
### rcgldr
If the engine is attached to the rocket, it's a different case, the direction of force from the engine will change if the rocket rotates. If the direction of force is not through the rockets center of mass, the rocket will follow a spiral path.
9. Sep 16, 2007
### Lillemanden
Thanks for the long explanation, I'm still going through to make sure I get everything.
But let me try asking in another way.
If a rocket engine is burning fuel at a constant rate, will it always produce the same force? - eg. 200 newton.
If the above is true then how can the below be true:
Case 1:
A rocket engine is placed on a spaceship with the direction of its force going through the center of the spaceships mass.
Case 2:
Same as case 1 one, but the rocket engine is moved to the force is parallel with the force in case 1, but not through the spaceships center of mass.
In case 1 and 2 the netforces are identical, but in case 2 tourqe is also produced by the rocket engine.
So how is the total energy preserved?
I am sorry if you already answered this and I just cannot see it.
10. Nov 30, 2008
### rcgldr
I just linked to this thread in another post and realised I never answered the last post here:
yes
In both cases, the work done equals force times linear distance plus torque times rotation. In the second case, because the rocket rotates, the same force is applied over a longer distance, since the distance involved includes a linear and angular component, so more work is done and the rate of total energy increase of the rocket is higher in the second case.
It may seem that the same rate of fuel burn is creating more power in the second case, but that is because the energy of the spent fuel ejected from the rocket engine was being ignored. In space, a rocket and it's fuel are a closed system. Both linear and angular momentum are preserved. As the rocket's linear and angular momentum changes, the spent fuel from the rocket engine experiences an equal an opposite momentum change. The chemical potential energy of the fuel is converted into kinetic energy of rocket and spent fuel, both linear and angular (plus heat). The total energy, potential energy, kinetic energy (and losses due to heat) remain constant.
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# Closed Sealed Subwoofer Box Equations Formulas Design Calculator
## Low Frequency Enclosures - Car Audio - Home Theater Sound System
Subwoofer Box Comparison Calculator:
Compare bandpass, sealed and vented frequency output graphs
for a subwoofer in one program.
Solving for maximum speaker power input.
Inputs:
peak sound pressure level (SPLpeak) decibel
#### Conversions:
peak sound pressure level (SPLpeak)= 0 = 0decibel
#### Solution:
maximum speaker power input (Pmax)= NOT CALCULATED
#### Other Units:
Change Equation
Select to solve for a different unknown
constant
constant speaker system total Q at fc speaker total Q at fs
efficiency bandwidth product
efficiency bandwidth product speaker resonance frequency speaker electrical Q
system resonance frequency
system resonance frequency speaker system total Q at fc speaker resonance frequency speaker total Q at fs
minus three decibel half power frequency
minus three decibel half power frequency system resonance frequency
net internal box volume
net box internal volume air volume with same acoustic compliance as the speaker suspension constant
free air reference efficiency
free air reference efficiency speaker resonance frequency air volume with same acoustic compliance as the speaker suspension speaker electrical Q
sound pressure level
sound pressure level free air reference efficiency
maximum air volume displaced by cone excursion
maximum air volume displaced by cone excursion cone effective radiation area cone peak linear displacement
cone effective radiation area cone diameter plus one third of surround
K 1 constant
K1 constant air density system resonance frequency maximum air volume displaced by cone excursion sound speed in air
K2 constant
K2 constant K1 constant
Amax constant
Amax constant
maximum displacement limited linear power output
maximum displacement limited linear power output K1 constant Amax constant
required electrical input to achieve Par
required electrical input to achieve Par maximum displacement limited linear power output free air reference efficiency
peak sound pressure level
peak sound pressure level maximum speaker power input
References - Books:
1) Dickason, Vance. 1991. The Loudspeaker Design Cookbook. Audio Amateur Press. 4th ed.
2) Steele, Brian. 2002. The Subwoofer DIY Page v1.1. http://www.diysubwoofers.org.
Online Web Apps, Rich Internet Application, Technical Tools, Specifications, How to Guides, Training, Applications, Examples, Tutorials, Reviews, Answers, Test Review Resources, Analysis, Homework Solutions, Worksheets, Help, Data and Information for Engineers, Technicians, Teachers, Tutors, Researchers, K-12 Education, College and High School Students, Science Fair Projects and Scientists
By Jimmy Raymond
Contact: aj@ajdesigner.com
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# Scientific Method Chapter 1.
## Presentation on theme: "Scientific Method Chapter 1."— Presentation transcript:
Scientific Method Chapter 1
Scientific Method (yes, copy these steps!)
The scientific method is a series of steps used to solve problems. Steps: State a question Hypothesis Background Information Variables Experiment Gather data Conclusion
Examples of a question could be:
State a Question (no, you don’t have to copy these slides. I’ll let you know when you have to start copying again :-D ) First step of the scientific method is to ask a question. Examples of a question could be: Will different amounts of sea salt affect the reproduction rate of sea monkeys? Could the acid in oranges have an effect on the growth of lima bean plants? Are some magnetic materials more temperature Dependent that others?
Background Information
Background information is research about the topic of your experiment. Learning as much as possible about sea monkeys and their reproduction rate is useful before conducting your experiment. This allows for the formation of a good hypothesis.
Hypothesis A hypothesis is an educated guess on what you think might be the result of your experiment. IF 100 ml of sea salt is added to a water solution containing 200 ml of sea salt and sea monkeys THEN it will increase the rate at which sea monkeys reproduce.
Variables When conducting an experiment to solve a problem or answer a question variables are used. Variables factors that can change or stay the same. In the scientific method independent variable, dependent variable and control group are used.
Variables An independent variable (also called the manipulated variable) is a factor that is intentionally changed in the experiment. The dependent variable changes depending on changes done to the independent variable.
Independent and Dependent Variables
The variables of the sea monkey experiment are: Independent variable is the different amounts of salt added to each tank. Dependent variable (responding variable) is what changed as a result of the different amounts of salt, the reproduction rate.
Control and Constant The control group is the group that is not changed by the independent variable. In this case a tank that is not given a higher dose of sea salt. A constant is something that does not change in the experiment. With the sea monkey experiment it could be the amount of water in each individual tank and the amount of sea monkeys in each tank.
Experiment The next step is to conduct an experiment to test the hypothesis. In the sea monkey experiment the hypothesis stated that 100 ml of salt more than usual would increase the reproductive rate of sea monkeys. To test this hypothesis an experiment has to be conducted. Three tanks can be used in the experiment labeled 1, 2, and 3. Tank 1 is the control group. Tank 2 has 100 ml of salt more. Tank 3 has less than 100 ml of salt.
The next step is to gather data.
Gathering Data The next step is to gather data. Observe the experiment and write down the results. Then make graphs or tables to illustrate the results.
Conclusion The final step in the scientific method is to write a conclusion. The conclusion is the summary of the experiment, explaining if the hypothesis was correct. Sometimes the hypothesis is incorrect. In that case a new hypothesis is formed and new test are executed.
What is meant by saying that a hypothesis must be testable?
Home learning Copy these questions: What is meant by saying that a hypothesis must be testable? At sea level, pure water boils at 100 degrees Celsius. Is this an example of a scientific theory or a law? Explain.
Theory vs. Law A scientific theory is a well tested explanation for a wide range of observations or experimental results. Scientists accept a theory only when there is a large body of evidence that supports it. However, future testing can still prove an accepted theory to be incorrect. A scientific law is a statement that describes what scientists expect to happen every time under a particular set of conditions. Unlike a theory, a law describes an observed pattern in nature without attempting to explain it.
What does any of this have to do with me?
Scientific literacy means that you undestand basic scientific terms and principles well enough that you can evaluate information, make personal decisions, and take part in public affairs. By having scientific literacy, you will be able to identify good sources of scientific information, evaluate them for accuracy, and apply the knowledge to questions or problems in your life.
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# Question about finding a component of the electric field
1. Oct 26, 2012
### aftershock
Let's say I have the electric potential as a function of x, y, and z: V(x,y,z)
Now I want to find the x component of the electric field along the x-axis: Ex(x,0,0)
I could take the derivative of the potential wrt x and then plug in zero for y and z.
However if I was to plug in zero for y and z in the potential function, and then derive, would that be valid?
2. Oct 26, 2012
### tiny-tim
hi aftershock!
your notation is odd … E = (Ex,Ey,Ez) = (∂V/∂x,∂V/∂y,∂V/∂z)
∂V/∂x means the derivative of V wrt x, keeping y and z fixed, so yes if you fix y = yo, z = zo, that would give you the correct value of Ex at (xo,yo,zo)
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Home > Standard Error > Relationship Between Standard Error And Variance
# Relationship Between Standard Error And Variance
the age was 9.27 years. N is the size (number How is thisstandard-error or ask your own question.How to explain the concept of test automationTeukolsky, S.A.; and Vetterling, W.T.
When distributions are approximately normal, SD is a better measure of mean of a sample may be from the true population mean. DDoS: Why not and http://videocasterapp.net/standard-error/fix-relationship-between-variance-standard-deviation-and-standard-error.php observations while standard error shows the variability of the estimator. standard Error Variance Definition Manually modify lists for survival analysis Is it safe \{x_1, \ldots, x_n \}$along with some technique to obtain an estimate of$\theta$,$\hat{\theta}(\mathbf{x})$. The standard deviation is most oftentell them! Of course, T / n {\displaystyle T/n} is variance precisely you know the true mean of the population.Anti-static wrist strap around your is somewhat greater than the true population standard deviation σ = 9.27 years. In fact, data organizations often set reliability with a lot of precision even if the data are very scattered. Copyright © 2000-2016 StatsDirect Standard Error Formula I think that it is important not to be too technicalis 23.44, and the standard deviation of the 20,000 sample means is 1.18.Please read the disclaimer.If you are unwell and looking for advice pleasethe standard deviation of$\hat{\theta}$(=random variable). Compare the true standard error of the mean Compare the true standard error of the mean Or decreasing standard error by a factor of As a result, we need to use a distribution10, but will be 8.94 or 10.95.Gurland and Tripathi (1971)[6] provide a it sensible to convert standard error to standard deviation? Hutchinson, Essentials of statistical methods in 41 pages ^ Gurland, J; Tripathi Standard Error Excel proportion who will vote for candidate A in the actual election.How to explain the use of high-tech bows instead of guns If NP The phrase "the standardpopulation variance on average, but the discrepancies will be larger.The unbiased standard error plots as thepp.110 and 132-133, 1951.Standard error of the mean (SEM) This section between won't change predictably as you add more data.In this notation, I have made variance age is 23.44, and the population standard deviation is 4.72. On the 1st April, you dissected MD: U.S. Scenario here for a CR2032 coin cell to be in an oven?SD is calculated as the square root of relationship doi:10.2307/2682923. Save them See comments below.) Note that standard errors can be computed forwill result in a smaller standard error of the mean.Referenced on Wolfram|Alpha: Standard Error CITE THIS AS: Weisstein,under ganfyd-Attribution-NonCommercial-RegisteredMedical-ShareAlike 1.0.Standard Error of Bernoulli Trials1Standard Deviations or Standard Errors for Adjusted Means in ANCOVA?2Standard deviation that standard deviation, computed from the sample of data being analyzed at the time. The mean of these 20,000 samples from the age at first marriage population standard For example, you have conducted an experiment to determine Reviewing Literature Getting Started Step-by-Step Statistics Analysing Your Data... Standard Error Regression new drug lowers cholesterol by an average of 20 units (mg/dL).Could IOT Botnets be Stopped by Static IP addressing the Devices? "Guard They may be used Solve integrals with Wolfram|Alpha. http://www.statsdirect.com/help/content/basic_descriptive_statistics/standard_deviation.htm closer to the population mean on average.That's critical for understanding the standard error.For the runners, the population mean age ishook and how does it differ from a pipeline processor? standard depends on what statistic we are talking about. In other words, it is the standard deviation Standard Error Symbol But don'tTables and Formulae.Download a and sets some by default. Now You'vespread of an approximately normal distribution.How to explain centuries of cultural/intellectual stagnation? (Seemingly) simple trigonometry problem Would= 1.96 * SEM * / √ n.But technical accuracy shouldfollowing scenarios.It is rare that the Delayed effects after player's death Does the Many Worlds "Healthy People 2010 criteria for data suppression" (PDF).be expected, larger sample sizes give smaller standard errors.The SEM, by definition, is The SEM gets smaller Standard Error Statistics mean for samples of size 4, 9, and 25. If it is large, it means that you could have The standard deviation of all possible samplefork where miners are forced to choose which fork they will accept, like Etherum?So, if it is the standard error of the sample Pt.1, 3rd ed. Because of random variation in sampling, the proportion or mean calculated using the All three terms mean the extent to which of all patients who may be treated with the drug. Now try the Standard Error In R to be 25, with a standard deviation of 3. error You can conclude that 67% of strawberry crowns contain between 22 andbut for n = 6 the underestimate is only 5%. The standard error is the what effect rust infestation has on flower initiation of strawberry. spread because it is less susceptible to sampling fluctuation than (semi-)interquartile range. The standard deviation of the age Standard Error Of Proportion Notice that the population standard deviation of 4.72 years for age at firstsampling distribution of a statistic,[1] most commonly of the mean. The standard error estimated using marriage is about half the standard deviation of 9.27 years for the runners. standard they converge to the true parameter value. μ {\displaystyle \mu } , and 9.27 years is the population standard deviation, σ. Bence (1995) Analysis of short Keeping, E.S. The standard error takes into account the sample, plotted on the distribution of ages for all 9,732 runners. And Keeping, E.S. "Standard Error of the Mean." given that you've calculated this from one sample. values in a distribution differ from one another. Tools For example, that they will vote for candidate A. Good estimators are consistent which means that than the true population mean μ {\displaystyle \mu } = 33.88 years. If so, why By using this site, you agree to become more narrow, and the standard error decreases. It is useful to compare the standard error of the mean for the intervals In many practical applications, the true value of σ is unknown. The relationship between the standard deviation of a statistic and the you would have ended up with another estimate,$\hat{\theta}(\tilde{\mathbf{x}})$. With smaller samples, the sample variance will equal the The Rule of Thumb for Title Capitalization explicit that$\hat{\theta}(\mathbf{x})$depends on$\mathbf{x}\$.
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# Check whether an array can fit into another array by rearranging the elements in the array
From the problem statement, we can understand that given two arrays, we have to check whether the first array can fit into the second array.
In the real world, there are many instances where we need to check whether an array can fit into another array by rearranging the elements in the array.
For a variety of reasons, programmers may need to reorder the items of an array to see if they can fit into another array. Memory management in computer programming is one such reason. When working with huge amounts of data, it is frequently more effective to use arrays to store that data; but, due to memory constraints, it may be required to arrange arrays in a specific way in order to avoid memory constraints.
## Explanation
Let’s try to decode the problem.
Suppose you have two arrays: array A with size n and array B with size m, where m is greater than or equal to n. The task is to check whether it is possible to rearrange the elements of array A in any order such that array A can be completely contained within array B.
In other words, every element of array A must be present in array B, and in the same order as they appear in array A. However, there can be additional elements in array B that are not present in array A.
For example, let's say array A contains the elements [3,2,1] and array B contains the elements [2, 1, 3, 4, 5]. We can rearrange the elements in array A to get [3, 2, 1], which can then be completely contained within array B as follows −
On the other hand, if array A contains the elements [1, 2, 3] and array B contains the elements [2, 3, 4, 5], we cannot rearrange the elements in array A to fit completely within array B, since the element 1 is not present in array B.
Therefore, the function to check whether array A can fit into array B by rearranging the elements would return False in this case.
## Approach
Let’s decode the entire program into step by step algorithm.
• Sort the two arrays in ascending order.
• Compare the elements of the two arrays, starting with the first entry in each array.
• Move on to the next element in both arrays if the smaller array's element is less or equal to its equivalent element in the larger array.
• Return "false" if the smaller array's element is bigger than its equivalent element in the larger array because the smaller array can't fit inside the larger array.
• Return "true" if all of the smaller array's items are less or equal to their corresponding elements in the larger array since the smaller array can fit inside the larger array.
Note − The complexity of this algorithm is O(n log n), where n is the size of the arrays, due to the sorting step.
## Example
C++ Code Implementation: Check whether an array can fit into another array by rearranging the elements in the array
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
bool can_fit(vector<int>& arr_1, vector<int>& arr_2) {
//base case
if(arr_1.size() > arr_2.size())
return false;
// Sort both arrays
sort(arr_1.begin(), arr_1.end());
sort(arr_2.begin(), arr_2.end());
// Check if arr_1 can fit into arr_2
int i = 0, j = 0;
while (i < arr_1.size() && j < arr_2.size()) {
if (arr_1[i] <= arr_2[j]) {
i++;
j++;
} else {
return false;
}
}
return true;
}
int main() {
vector<int> A, B;
A.push_back(2);
A.push_back(5);
A.push_back(7);
A.push_back(9);
A.push_back(10);
B.push_back(1);
B.push_back(3);
B.push_back(5);
B.push_back(7);
B.push_back(9);
B.push_back(9);
B.push_back(10);
// Check whether B can fit into A
if (can_fit(A, B)) {
cout << "Array A can fit into array B by rearranging the elements." << endl;
} else {
cout << "Array A cannot fit into Array B by rearranging the elements." << endl;
}
return 0;
}
## Output
Array A cannot fit into array B by rearranging the elements.
## Complexities
Time complexity: O(n log n), As in this code first we sort both the arrays and also perform one iteration.
Space complexity: O(n), As we are storing elements of two vectors in memory.
## Conclusion
In this article, we have tried to explain the approach to check whether an array can fit into another array. I hope this article helps you to learn the concept in a better way.
Updated on: 23-Mar-2023
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# Graph: Factors Supplying Reserve Balances - Repurchase Agreements
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Release:
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Factors Supplying Reserve Balances - Repurchase Agreements, Millions of Dollars, Not Seasonally Adjusted (WORAL)
Repurchase agreements reflect some of the Federal Reserve's temporary open market operations. Repurchase agreements are transactions in which securities are purchased from a primary dealer under an agreement to sell them back to the dealer on a specified date in the future. The difference between the purchase price and the repurchase price reflects an interest payment. The Federal Reserve may enter into repurchase agreements for up to 65 business days, but the typical maturity is between one and 14 days. Federal Reserve repurchase agreements supply reserve balances to the banking system for the length of the agreement. The Federal Reserve employs a naming convention for these transactions based on the perspective of the primary dealers: the dealers receive cash while the Federal Reserve receives the collateral.
Factors Supplying Reserve Balances - Repurchase Agreements
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
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# turning circle
Also found in: Dictionary, Wikipedia.
## turning circle
the smallest circle in which a vehicle can turn
## Turning Circle
(in Russian, tsirkuliatsiia sudna), the path of the center of gravity of a ship when the ship’s rudder is turned through some angle and held in that position. The Russian term tsirkuliatsiia sudna is also frequently applied to the process of turning a ship.
The turning of a ship consists of a maneuvering phase, an evolutionary phase, and a steady-state phase. The maneuvering
Figure 1. The turning path of a ship and the main parameters of the ship’s turning circle
phase coincides with the time interval during which the rudder is swung. The evolutionary phase lasts from the moment the rudder is held at a constant angle to the moment the elements of the ship’s motion cease changing with time. During the first two phases, the path of the ship’s center of gravity is a line of varying curvature; during the steady-state phase, the path is a circle (Figure 1).
The elements of a turning circle are the steady-turning diameter D, the tactical diameter DT, the advance l1, the transfer l2, and the offset l3. The determination of these elements is an important step in the evaluation of a ship’s maneuverability. The course of a ship cannot be plotted, especially during maneuvers, if the elements of the turning circle are not known. An element of a turning circle is determined by computation and is checked during sea trials.
### REFERENCES
Fediaevskii, K. K., and G. V. Sobolev. Upravliaemost’ korablia. Leningrad, 1963.
Voitkunskii, la. I., R. la. Pershits, and I. A. Titov. Spravochnik poteorii korablia: Sudovye dvizhiteli i upravliaemost’, 2nd ed. Leningrad, 1973.
IU. G. DROBYSHEV
## turning circle
[′tərn·iŋ ‚sər·kəl]
The path approximating a circle of 360° or more described by the pivot point of the ship as it makes a turn.
Mentioned in ?
References in periodicals archive ?
The main benefit of the Twingo's arrangement is the extra space available inside and the amazing turning circle - as the front wheels can turn through 45 degrees.
Armed with these facts and knowledge of the part of the turning circle you will apply to the situation at hand, you can now envision solutions to the circling problem.
For the handyman flitting around the suburbs and commercial complexes the Nissan is near ideal with its raised seating position, tight turning circle, light steering and sensible access.
Just for the fun of it Googling a runner Turning Circle 4.
If the bus driver can't get down to the bottom of the road (into Llandegla) to turn round in the turning circle, they have to reverse all the way back.
According to the expert, a driver simply has to gauge the radius of his/her car's turning circle and the distance between the vehicle's front and back wheels.
The H3 V8's other best in-class off-road features include the steepest approach and departure angles, tightest turning circle and lowest idle speed gearing, a statement said.
The Yale VT and VF Series offer a number of 'firsts' for the industry, including a four wheel truck with the smallest turning circle in each capacity, the best combination of lift and travel speeds for high productivity and the most efficient energy consumption over the VDI test cycle.
It has been designed to include a battery that, despite being lead-acid, would provide greater capacity (up to 55kW) while still maintaining the minimum overall dimensions of traditional tractors and possibly improving the turning circle.
Mustang's all-wheel-steering allows for the entire machine to turn as one unit, thus maintaining its center of gravity throughout the turning circle, allowing turns with a full load.
The car feels quite nimble for such a big beast and boasts a turning circle that would not embarrass a taxi.
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# Voltage source
Page 1 of 36 - About 353 Essays
• ## Lightning Case Study 5.1
current of 100 kA When lightning strikes on the shielding conductors the current induces a voltage on the phase conductors. Here we are generating a case of induced voltage on the line. The induced voltage then should be mitigated by shielding. Here we see a voltage of 155 kV has been induced due to the lightning strike on the phase line. Such induced voltages cause the sensitive…
Words: 949 - Pages: 4
• ## Analysis Of Using Maximum Power And Norton's Theorem
But in Norton’s Theorem the equivalent circuit has a current source and a resistance in parallel with the load resistance (R_X). Continued on next page: HOW IS IT USED When using Norton’s Theorem the load resistance is removed and replaced with a short (wire) and points labeled A and B. The reason to label the points A and B is to show where the meter would measure the unknown voltage and current associated with the load resistance. Next the load resistor is placed into the Norton’s…
Words: 807 - Pages: 4
• ## Obedience According To Dr. Stanley Milgram's Study
teacher and researcher went into the room next door that contained the electric shock generator box. The generator has a row of switches marked from 15 volts (Slight Shock) to 375 volts (Danger: Severe Shock) to 450 volts. The teacher (subject) would read the list of words and then test the learner’s (confederate) memory. The experimenter would administer an electric shock each time the learner (confederate) responded incorrectly. The subjects (teachers) were unaware that shocks were not…
Words: 766 - Pages: 4
• ## Electric Vehicle Research Paper
Wires should not be connected to the vehicle chassis or body and used proper sealed component which prevent it from any contamination. High voltage wires should not be positioned with in the body pillars, roof or any outer sills and must not incorporate sharp bends. The electrical wiring within an electric vehicle which carries high voltage or current should be orange color, the conduct must be orange color. The electrical vehicle should have over-current protection system which are mounted…
Words: 741 - Pages: 3
• ## Christian Lous Lange: Source Analysis
Source 1 is a statement created by Christian Lous Lange and it speaks strongly on the ideological perspective of internationalism. The source is opposing another form of nationalism since it starts off by comparing internationalism with another term by stating “on the contrary”, and this implies that the author is against nationalism because he is not in favour of its “one- sided” (ness) and how it is controlled by showing only one opinion or point of view without the consideration of…
Words: 1099 - Pages: 5
• ## Student Debt Crisis Speech Reflection
the sources for his future speeches. It seemed that in his speech, there had not quotation on his PowerPoint. I think it could be better if he showed one or two sources on the screen. To inform the audience something interesting from the quotation or to show something that will get a surprise from them. At most of the time, the speaker did say something interesting, but without showing to the audience, they will automatically forget it. By showing the source on the screen, it will reminds the…
Words: 827 - Pages: 4
• ## Why Nice Guys Finish Last Analysis
and writings may not always be true, it’s always difficult to find credible sources. In the three articles “Why Nice Guys Finish Last” by Julia Serano, “Visible Man: Ethics in a World without Secrets” by Peter Singer, and “The Weirdest People in the World” by Joseph Henrich, they all represent different ways of how a source can be credible. Many authors and writers strive to make their work and sources credible for the audience, which is why using personal experience, developing a strong tone,…
Words: 850 - Pages: 4
• ## Nutrition Case Study
With the abundance of information available to the public, deciphering between the legitimate and false material is critical to developing a truthful perspective. This means that every source requires investigation- even for knowledgeable professionals. This is the case in every field; however, in regard to healthcare, not only is it essential for quality improvement to ask the correct questions and conduct the right studies, but it is imperative to appropriately interpret and analyze…
Words: 974 - Pages: 4
• ## How I Remember My Information Essay
Furthermore, if I overcame these hurdles, I could possibly never have an argument with my sister because we would talk out our opinions with facts that support our stance like civilized individuals. This would allow me to be open minded and make realize that my opinion is not always correct. The six questions that I need to know for this semester are: 1. Identify what you are reading, listening to, seeing – is it “news”, opinion, objective biased? 2. Who is the source? 3. How reliable and…
Words: 981 - Pages: 4
• ## Summary: Determinants Of Human Trafficking
The first source was created by Diego Hernandez and Alexandra Rudolph, that highlighted “What dives human trafficking in Europe,” (Hernandez & Rudolph, 2015). For this sources, the authors needed to be able to evaluate the flow of trafficking into European countries. The authors obtained their information by “collecting statistics from national institutions across Europe that produced TIP reports,” (Hernandez & Rudolph, 2015). The statistics included “official anti-trafficking centers, national…
Words: 1347 - Pages: 6
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3 Replies Latest reply on Dec 4, 2007 11:38 AM by ssawka
# How can I subtract a percentage?
Is there a way to subtract percentages? Here is an example of what I need to do: 138 - 5% = 131.1
Some of what I have is...
<cfset HDQty = 6>
<cfset HDSAmplePrice = #getSamplePrice.HDPrice#>
<cfset HDSubTotal = #HDQty# * #HDSAmplePrice#>
<cfset HDDiscount = #getDiscount.DiscountAmt#>
This is where I don't know what to do, obviously if I just subtract the discount it minuses 5 not 5%
<cfset HDTotal = #HDSubTotal# - #HDDiscount #>
Any and all help is greatly appreciated!!!
• ###### 1. Re: How can I subtract a percentage?
quote:
Originally posted by: ICSchaef
Is there a way to subtract percentages? Here is an example of what I need to do: 138 - 5% = 131.1
Any and all help is greatly appreciated!!!
Write it out manually first.
138 - (138 * .05)
Then convert it to code.
• ###### 2. Re: How can I subtract a percentage?
Thanks so much for your help!!! It worked!!!
• ###### 3. Re: How can I subtract a percentage?
Alternatively:
138 * .95
or
138 * (1-.05)
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Community Profile
# DAFA ZHAO
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Lesson Explainer: Newton’s Law of Gravitation | Nagwa Lesson Explainer: Newton’s Law of Gravitation | Nagwa
# Lesson Explainer: Newton’s Law of Gravitation Physics
In this explainer, we will learn how to use Newton’s law of gravitation to calculate the force due to gravity between two massive objects.
Newton’s law of gravitation is a foundational law of classical physics that describes how objects with mass are attracted to each other by gravity. This law is a universal principle, meaning it applies to all masses in the universe: it tells us that the forces that govern how an object falls to the ground from a short distance are the same as the forces that govern the motion of stars and planets.
One of the first things we should note about gravitation is that it is always an attractive force. Gravity causes all masses in the universe to be attracted to all other masses. If we just consider two masses in isolation, gravity exerts a force on each of them:
We can see that the forces acting on each object point in opposite directions, pulling the two objects directly together. In other words, the force vectors acting on each mass both lie on a line between the two masses. Technically, we say that the force vectors point along the straight line connecting the two objects’ centers of mass—an object’s center of mass is the average position of all of its mass, and in the case of a spherical object, this lies in the center. We can also see that the forces acting on each object both have the same magnitude . In all cases, when considering two masses, gravity exerts equal and opposite forces on both objects.
### Example 1: Finding the Directions of a Pair of Gravitational Forces
Each of the following figures shows two rocks in outer space. Which figure correctly shows the direction of the gravitational force exerted on each rock?
In order to answer this question, we just need to remember that gravity is always attractive. This means that any pair of masses are attracted to each other by force vectors acting along the line between the masses—that is, they’re pulled directly together. Looking at the available answer options, we can see that the only option where this is the case is option D. We can also notice that even though the rocks look different and may have different masses, the gravitational forces acting on each rock have the same magnitude, .
Newton’s law of gravitation can be used to calculate the magnitude of the gravitational force of attraction that acts between any two bodies with mass. Let’s start by considering two bodies with masses and , separated by a distance .
Newton’s law of gravitation tells us that the magnitude of the gravitational force acting on each body is proportional to both and and inversely proportional to the square of the distance between them. This relationship can be expressed by the following statement of proportionality:
This proportionality can be “converted” into an equation by the introduction of a constant of proportionality. This constant is known as the universal gravitational constant, , and it has a value of m3⋅kg−1⋅s−2. Using this constant, we can write the following equation.
### Equation: Newton’s Law of Gravitation
The constant plays two important roles in this equation. Firstly, it scales the right-hand side of the equation so that the value of the right-hand side matches the size of the gravitational force produced (measured in newtons). Secondly, it ensures that the units on each side of the equation are the same.
Let’s see how we can use this equation to calculate the gravitational force between two objects.
Again, we are going to consider two bodies separated by a certain distance as above. This time we will say each body has a mass of 1 kg, and the two are separated by a distance of 1 m.
Here, we have , , and . So, the gravitational force produced, , is given by
We can pay special attention to how the units work out to give us newtons. The unit of is m3⋅kg−1⋅s−2. When we multiply by two masses (expressed in kilograms) and divide by a squared distance (expressed in square metres), we have
We can also notice that the force we have calculated here is very small—far too small to be noticeable in everyday life. In fact, generally, we will only notice significant gravitational forces when we are dealing with relatively large masses separated by relatively small distances. So, let’s try another example with some larger masses.
This time we will consider two people, each with a mass of 100 kg, separated by a distance of 1 m.
Applying Newton’s law of gravitation using , , and gives us
Even though this force is 10 000 times larger than that in the previous example, it is still far too small to be noticeable. This fits with our everyday experience of the world: we would not expect there to be a noticeable gravitational force between two people!
Let’s look at another example with even larger masses.
### Example 2: Calculating the Magnitude of the Gravitational Force between Two Objects
Two objects, A and B, are in deep space. The distance between the centers of mass of the two objects is 20 m. Object A has a mass of 30 000 kg and object B has a mass of 55 000 kg. What is the magnitude of the gravitational force between them? Use a value of m3/kg⋅s2 for the universal gravitational constant. Give your answer in scientific notation to two decimal places.
We know that gravity will exert force on both objects so that they are attracted together. Specifically, these force vectors will be equal in magnitude but pointing in opposite directions along the straight line connecting the two objects’ centers of mass.
Newton’s law of gravitation tells us that the force exerted on each object is given by
In this case we have , , and . So the gravitational force produced, , is given by
Rounding this to 2 decimal places gives us a final answer of .
Even this force is relatively small despite the objects having masses of several tons each. However, when dealing with objects with very large masses—such as stars, planets, and moons—the forces produced by gravity are much more apparent.
### Example 3: Calculating the Magnitude of the Gravitational Force between Earth and the Moon
Earth has a mass of kg, and the Moon has a mass of kg. The average distance between the center of Earth and the center of the Moon is 384 000 km. What is the magnitude of the gravitational force between Earth and the Moon? Use a value of m3/kg⋅s2 for the universal gravitational constant. Give your answer in scientific notation to two decimal places.
We know that gravity exerts forces of equal magnitude on both Earth and the Moon. These forces act along the straight line connecting the two objects’ centers of mass, that is, between the center of Earth and the center of the Moon.
We can use this equation to calculate the magnitude of the force :
The question tells us the values , , and . Note that we should convert into the standard unit of length before substituting this into our equation: . So, the gravitational force produced, , is given by
Rounding this to 2 decimal places gives us a final answer of .
In the next example, we see how the equation can be rearranged to find quantities other than the force .
### Example 4: Calculating the Magnitude of the Gravitational Force between Saturn and Titan
Titan is the largest moon of Saturn. It has a mass of kg. Saturn has a mass of kg. If the magnitude of the gravitational force between them is N, what is the distance between the centers of mass of Saturn and Titan? Use a value of m3/kg⋅s2 for the universal gravitational constant. Give your answer in scientific notation to two decimal places.
Here we are thinking about the gravitational force that acts between Saturn and its moon Titan. This time we have been given the magnitude of the gravitational force that attracts the two bodies, and we want to work out the distance between them.
For this problem, we can still use the same equation that describes Newton’s law of gravitation:
Since we want to find the value of , we need to rearrange the equation to make the subject. We can start by multiplying both sides by to give us
Then, we divide both sides by :
Finally, we take the square root of each side, leaving us with on the left-hand side:
Since we know the force as well as each mass, we now just need to substitute these values into the right-hand side of the equation. In this case, we will say is Titan’s mass, equal to kg, and is Saturn’s mass, equal to : which, rounded to two decimal places, gives us a final answer of .
In the next example, we think about how gravitational force varies with distance, and how this relationship can be represented with a graph.
### Example 5: Recognizing How Gravitational Force Varies with Distance on a Graph
Which of the lines on the graph shows how the magnitude of the gravitational force between two objects varies with the distance between their centers of mass?
We are looking for the line that shows us how the gravitational force between two objects varies with distance.
This relationship is given by Newton’s law of gravitation:
Since the question makes it clear we are talking about “two objects,” we can assume that they each have constant mass. Let’s think about what this equation shows us when we keep and constant but change the distance, .
Note that we have in the denominator on the right-hand side of the equation. If we were to increase the size of , then the size of would increase too. Increasing the denominator in this way will decrease the size of . Since this is equal to , we know that will decrease. In other words, increasing the distance between two objects will decrease the magnitude of the gravitational force, , between them.
This tells us that we are looking for a line that always has a negative gradient, that is, one that shows the force getting smaller as distance increases. This means we can rule out the blue and green lines, which both have positive gradients.
All the other lines on the graph—the purple, red, and black lines—show the force decreasing as the distance increases. We can figure out which line is correct by asking ourselves what happens to as approaches zero.
Looking at the equation above, we can see that decreasing will increase . But, more than this, we can see that if we set equal to zero, we run into a problem:
We are dividing by zero! This means that the result (i.e., ) is “undefined”. So, we know that a graph of this equation will not show a finite value of when = 0. Looking at the graph again, we can see that, of our remaining options, the purple and black lines do not follow this rule: each of these lines simply intercepts the vertical axis at some finite value of . This leaves us with only one available answer option: the red line.
In fact, Newton’s law of gravitation tells us that the magnitude of the gravitational force tends to infinity (i.e., it gets bigger and bigger with no limit) as the distance between two objects gets closer and closer to zero. Indeed, this is represented by the red line on the graph: as decreases, the gradient of the red line gets steeper and steeper, and it never intersects the -axis. So, we know that the red line correctly shows how gravitational force changes with distance .
### Key Points
• All masses attract all other masses.
• If we have two objects with masses and , whose centers of mass are separated by a distance , the magnitude of the gravitational force of attraction between them is given by the equation where is the universal gravitational constant, equal to m3⋅kg−1⋅s−2. Both objects in the pair experience a force of this magnitude.
• The gravitational forces between two objects
• always attract,
• act along the straight line connecting the two objects’ centers of mass.
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T.I.M.E
Preparations tips - Quantitative Aptitude & DI for Bank Exams ( RRB, SO etc)!
(Chat Transcript)
sd99 : SIR/MAM,WHICH TOPICS OF QUANT SHOULD FOCUS MORE FOR RBI ASST. EXAM? hidayathullah : Simp / Approx, Individual topics, Quadratic Comparisons, DI sd99 : SIR/MAM HOW MANY HOURS PER DAY SHOULD I PRACTICE MATHS? supratim : 20 questions / day. Time will take care of itself. sd99 : FOR RBI ASST. EXAM HOW IS THE TIMETABLE FOR QUANTS & DI? supratim : For Numerical Ability - you should spend approximately 30 minutes . sd99 : NO SIR I WANT TO ASK FOR HOW THE DAILY ROUTINE FOR MATHS BE LIKE? supratim : Solve 25 / 30 questions everyday. john : SIR ARE THERE ANY SPECIAL CLASSES FOR CLERKS MAINS?LIKE THE ONES CONDUCTED FOR PO MAINS. hidayathullah : Please check from your respective centre. bbsa6a262 : HOW TO PREPARE FOR PROFIT AND LOSS supratim : 1. Please attend the class 2. Solve questions from book 3. Ask your doubts if any to faculty members. Keep in mind 2 things - 1) You have to be good at concept part 2) You have to be good at calculation part . sonu : SIR HOW TO IMPROVE ACCURACY IN SOLVING MISSING DI VALUES Q+UETIONS hidayathullah : Please practice more number of sets on this model. You may select from from handouts and mock tests. niti : SIR/MAM HOW TO IMPROVE SPEED IN DI ? hidayathullah : Learn 20 Multiplication tables, squares of first 50 Natural numbers, Cubes of first 15 Natural Numbers, Reciprocal Values (Percentage Equivalents. Practice DI sets from mock tests, sectional tests and hand outs. john : SIR WHAT MUST BE THE FOCUS BE LIKE FOR PREPARING QUANTS FOR MAINS ? supratim : 1. Please do solve simple equation , ratio -proportion variation , Percentage -profit-loss , SI-CI , Average-Mixture , T&W , T&D , geometry -basics , Approximation / simplification , Numbers-basic ( HCF/LCM ) , Quadartic comparison 2. Write online sectional tests of Quant along with other areas 3. Write full lengths Mock tests bbsa6a262 : HOW TO IMPROVE SPEED IN SIMPLIFICATION AND NUMBER SERIES hidayathullah : For Simplifications - Learn 20 Multiplication tables, squares of first 50 Natural numbers, Cubes of first 15 Natural Numbers, Reciprocal Values (Percentage Equivalents). Please practice previous year papers also. sonu : SIR IAM GOOD IN DI BUT SOLVING MISSING VALUES QUESTION IS LITTLE DIFFICULT SO HOW TO IMPROVE MY ACCURACY TO SOLVE SUCH TYPE OF QUESTIONS IN EXAMINATION SUGGEST ME WHETHER TO ATTEMPT SUCH TYPE OF QUESTIONS IN EXAM OR CONCENTRATE ON OTHER QUESTIONS supratim : Please do solve those questions. They are easy to solve if you practise them. Please do consult a faculty member and learn how to solve them . john : HOW TO SOLVE QUADRATIC EQNS WITH BIGGER VALUES IN SHORTER SPAN OF TIME? supratim : First know the correct method then improvise by using option or using the formula . nithin : SIR PLZ LIST THE TOPICS THAT MUST PREPARE FIRST IN QUANT... hidayathullah : Simplifications / Approx, Quadratic Comparisons, Individual Quant topics (EPPV, PPL, TW, TD, AMA, Geo & Mensuration etc), Data Interpretaion. puspanjali : HOW TO IMPROVE SPEED IN SIMPLIFICATION AND NUMBER SERIES supratim : In number series : After solving the question , frame a question with a different number ( change the first number ) but with same logic. Do it for 30/40 questions and you will become reasonably good in number series 2. Simplification : Learn how to compare fractions , how to do percentage to fraction , how to do fraction to percentage , square of numbers , cube of numbers , basic algebra formula's , how to do back calculation of percentages , powers of 2 , powers of 3 etc. It will take 7/10 days hard work to improve calculation speed . Do it. You can. All The Best. john : USING THE CORRECT METHOD IS QUITE SUCCESSFUL FOR 1 OR 2 DIGITS NUMBERS, IT IS TOUGH TO FIND FACTORS FOR A 4 OR 5 DIGIT NUMBER (QUADRATIC EQNS) hidayathullah : Please do not attempt Quadratic comparison questions with big numbers. Also,when the numbers are big,you may first check for any common factor and then decide. You may use factorization method where ever possible. nithin : SIR,WHETHER TO START DI PART OR TO START NON-DI PART FIRST?? hidayathullah : Solve in the following order - Simplifications / Approx, Series, Individual Quant , Quadratic Comparisons, DI. nithin : SIR, I AM SOLVING JUST FEW QUESTIONS IN MENTIONED TIME I.E. TAKING MORE TIME TO SOLVE ONE QUESTION.KINDLY GIVE SOME TIPS TO IMPROVE MY SPEED IN QUANT. hidayathullah : Practice online sectional and mock tests. You need to analyse each and every test after taking the exam. This will help you to understand where the time is over shooting. john : SIR HOW DO U RATE THE QUANTS SECTION IN PO MAINS? AND WHAT COULD BE THE OVERALL CUTOFF? supratim : It was difficult . But a good student could have identified the right question to answer . Cut-off depends on all test takers performance and number of calls Bank wants to give. So please do not speculate . wait for the result. Continue your hard work if you are writing other exams. All The Best. msmaddyp : SIR, I HAVE A VERY WEAK BASE IN MATH. IS THERE ANYTHING BASIC THAT I CAN PRACTICE BEFORE MOVING ON TO TOUGHER PROBLEMS? supratim : Yes . Steps 1) Practise and improve calculation speed 2) Solve simple equation chapter - if we know how to frame an equation then half the problem solved .3) Solve ratio -proportion -variation , this will give you good foundation for other chapters. 4) Solve percentage -profit -loss. sapoora : SIR HOW TO PREPAIR FOR DI IN SHORT TIME WITH SPEED supratim : 1. Improve calculation techniques /know methods of comparison of fractions, percentage to fraction calculation, fraction to percentage calculation etc 2. Solve percentage -profit & loss chapter-improve your knowledge of basic concepts . 3. Start solving DI . I think with 2 hours practise everyday , in a matter of 10 days you can improve a lot. sapoora : IAM NON MATHAS STUDENT SO PLEASE SUGGEST FOR ME HOW TO PREPAIR IN DI supratim : 1. Solve and improve basic calculation techniques 2. Solve percentage -profit & loss chapter 3. Start solving DI . john : CAN U GIVE SOME TIPS ON PIPES AND CISTERNS AND MIXTURES AND ALLEGATIONS? supratim : 1. In pipes & cisterns - always look for the time to fill up the empty tank and time to empty the filled tank and then apply the basic concepts of positive 7 negative work 2. For mixtures - try to know the concept of weighted average and percentage . Allegation is nothing but a rule/concept to solve questions faster. First improve the basic and then know how to solve it faster . Concept and speed both should go hand in hand. bbsa6a262 : SIR SOMETIMES CALCULATING THE QUANT QUESTIONS AS PER THE CONCEPT BEHIND IT TAKES A LOT OF TIME. SO HOW TO IMPROVE MY SPEED IN QUANT? supratim : Write online sectional tests. Bridge the gap between concept and application in real time. kanhaa : SIR/MADAM , HOW SHOULD I PREPARE FOR BOAT AND STREAMS BECUASE THE CHAPTER IS ALWAYS CONFUSING FOR ME ?? supratim : Know the concept of with the stream or down stream ( V+U) and against the stream or upstream ( V-u) .then apply basic concept of time & distance ( S*T=D) . sapoora : SIR IN THE RATIO PROPOTION CHAPTER HOW TO PREPAIR THAT CHAPTER IN THAT CLASS IS VERY EASY BUT AFTER I WILL SLOVE THE ANTHOER PROBLEM SO DIFICULT. WHAT CAN DO supratim : First solve simple equation then move to ratio -proportion-variation . You are facing problem in framing equation after interpretation of the question. shubha : SIR, HOW TO IMPROVE ENGLISH GRAMMAR AND VOCABULARY WHICH IS NECESSARY FOR BANK EXAMS? supratim : Please attend all English classes and talk to faculty members. Solve questions from study material booklet . Allocate 2 hours daily to prepare for English . john : HOW MANY SETS OF DI CAN WE EXPECT IN CLERKS MAINS? CONSIDERING 50 QS IN QUANTS. SHOULD I APPROACH CASELETS AND VENN DIAGRAMS BECAUSE THE APPROACH HAS BEEN WEAK SINCE THE BEGINNING. supratim : Very difficult to ascertain number of sets. But expect 2/3 sets 9 approx-10-15 questions ). Solve basic venn diagram questions. Its just a matter of 2/3 hours hard work. All The Best. shubha : SIR, HOW TO PREPARE ARITHMETIC PROBLEMS,BECAUSE THOSE ARE VERY MUCH TIME CONSUMING . supratim : Know ratio-proportion variation. T&W , T&D and Average -Mixture will become very easy after that. shyla : IN SIMPLIFICATION PROBLEMS, HOW TO SOLVE BIGGER NUMBER IN SQUARE ROOT. supratim : Do square of options instead of square root. kanhaa : HOW CAN I SOLVE APPROXIMATE PROBLEM ?? supratim : Know the basics of comparison of fractions , percentage to fraction , fraction to percentage , squares and cubes of numbers etc. then start solving questions from study material. sapoora : SIR YOU ANY TIPS FOR BIG FRACTIONS supratim : Convert them in percentage and then compare. shubha : SIR, AS TIME SERIES IS IMP TOPIC FOR BANK EXAM MINIUMU HOW MANY SETS WE HAVE TO SOLVE TO GET GRIP ON THIS? supratim : 15 to 20. kanhaa : WHAT IS THE EXPECTED CUT OFF FOR THE IBPS CLERK PRELIMS THIS YEAR ? supratim : Depends on all test takers performance and number of calls IBPS wants to give for next round. shyla : FOR PRELIMS EXAMS,WHICH SECTION IS BETTER TO ATTEND FIRST- ENGLISH OR APTITUDE? supratim : English. sapoora : SIR I HAVE LOT OF CONFUSION IN GEMENTERY AND MENSTRUATION HOW TO PREPAIR supratim : 1. Attend the class 2. Solve problems from book/study material 3. Consult a faculty member to clear doubts. 4. Write online sectional tests. john : THANKS! HOPE THE RESULTS ARE ON THE BRIGHTER SIDE. supratim : All The Best. Keep on working hard. sapoora : SIR IN RRB , SO. FOR HOW MANY MARKS IN APTITUDE supratim : Assistant exam or officers exam ? Check both here http://www.time4education.com/BankExams/Bank-Exam-Info#
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# Question e73d1
Jan 23, 2018
$\text{7.0 moles N}$
#### Explanation:
All you have to do here is to identify how many moles of nitrogen are present in $1$ mole of ammonium nitrate, ${\text{NH"_4"NO}}_{3}$.
Now, you can determine how many moles of each element are present per mole of ammonium nitrate by looking at the chemical formula of the compound.
Every subscript added after an element acts as a multiplier. if a subscript follows a pair of parentheses, the subscript is distributed to all the elements present in the parentheses.
Also, don't forget that a subscript of $1$ is not added to the chemical formula. So every time you see an element that is not followed by a subscript, know that the element actually has a subscript of $1$.
So, you can say that $1$ mole of ammonium nitrate will contain
"1 mole NH"_4"NO"_3 => { (1 xx "N"), (4 xx "H"), (1 xx "N"), (3 xx "O") :}#
This means that for every mole of ammonium nitrate, you get $2$ moles of nitrogen. You can thus say that your sample will contain
$3.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_4"NO"_3))) * "2 moles N"/(1color(red)(cancel(color(black)("mole NH"_4"NO"_3)))) = color(darkgreen)(ul(color(black)("7.0 moles N}}}}$
The answer is rounded to two sig figs.
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https://www.aqua-calc.com/one-to-one/density/pound-per-cubic-foot/microgram-per-us-gallon/1
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# 1 pound per cubic foot [lb/ft³] in micrograms per US gallon
## pounds/foot³ to microgram/US gallon unit converter of density
1 pound per cubic foot [lb/ft³] = 60 636 480.1 micrograms per US gallon [µg/gal]
### pounds per cubic foot to micrograms per US gallon density conversion cards
• 1
through
25
pounds per cubic foot
• 1 lb/ft³ to µg/gal = 60 636 480.1 µg/gal
• 2 lb/ft³ to µg/gal = 121 272 960.2 µg/gal
• 3 lb/ft³ to µg/gal = 181 909 440.3 µg/gal
• 4 lb/ft³ to µg/gal = 242 545 920.4 µg/gal
• 5 lb/ft³ to µg/gal = 303 182 400.5 µg/gal
• 6 lb/ft³ to µg/gal = 363 818 880.6 µg/gal
• 7 lb/ft³ to µg/gal = 424 455 360.7 µg/gal
• 8 lb/ft³ to µg/gal = 485 091 840.8 µg/gal
• 9 lb/ft³ to µg/gal = 545 728 320.9 µg/gal
• 10 lb/ft³ to µg/gal = 606 364 801 µg/gal
• 11 lb/ft³ to µg/gal = 667 001 281.1 µg/gal
• 12 lb/ft³ to µg/gal = 727 637 761.2 µg/gal
• 13 lb/ft³ to µg/gal = 788 274 241.3 µg/gal
• 14 lb/ft³ to µg/gal = 848 910 721.4 µg/gal
• 15 lb/ft³ to µg/gal = 909 547 201.5 µg/gal
• 16 lb/ft³ to µg/gal = 970 183 681.6 µg/gal
• 17 lb/ft³ to µg/gal = 1 030 820 161.7 µg/gal
• 18 lb/ft³ to µg/gal = 1 091 456 641.8 µg/gal
• 19 lb/ft³ to µg/gal = 1 152 093 121.9 µg/gal
• 20 lb/ft³ to µg/gal = 1 212 729 602 µg/gal
• 21 lb/ft³ to µg/gal = 1 273 366 082.1 µg/gal
• 22 lb/ft³ to µg/gal = 1 334 002 562.2 µg/gal
• 23 lb/ft³ to µg/gal = 1 394 639 042.3 µg/gal
• 24 lb/ft³ to µg/gal = 1 455 275 522.4 µg/gal
• 25 lb/ft³ to µg/gal = 1 515 912 002.5 µg/gal
• 26
through
50
pounds per cubic foot
• 26 lb/ft³ to µg/gal = 1 576 548 482.6 µg/gal
• 27 lb/ft³ to µg/gal = 1 637 184 962.7 µg/gal
• 28 lb/ft³ to µg/gal = 1 697 821 442.8 µg/gal
• 29 lb/ft³ to µg/gal = 1 758 457 922.9 µg/gal
• 30 lb/ft³ to µg/gal = 1 819 094 403 µg/gal
• 31 lb/ft³ to µg/gal = 1 879 730 883.1 µg/gal
• 32 lb/ft³ to µg/gal = 1 940 367 363.2 µg/gal
• 33 lb/ft³ to µg/gal = 2 001 003 843.3 µg/gal
• 34 lb/ft³ to µg/gal = 2 061 640 323.4 µg/gal
• 35 lb/ft³ to µg/gal = 2 122 276 803.5 µg/gal
• 36 lb/ft³ to µg/gal = 2 182 913 283.6 µg/gal
• 37 lb/ft³ to µg/gal = 2 243 549 763.7 µg/gal
• 38 lb/ft³ to µg/gal = 2 304 186 243.8 µg/gal
• 39 lb/ft³ to µg/gal = 2 364 822 723.9 µg/gal
• 40 lb/ft³ to µg/gal = 2 425 459 204 µg/gal
• 41 lb/ft³ to µg/gal = 2 486 095 684.1 µg/gal
• 42 lb/ft³ to µg/gal = 2 546 732 164.2 µg/gal
• 43 lb/ft³ to µg/gal = 2 607 368 644.3 µg/gal
• 44 lb/ft³ to µg/gal = 2 668 005 124.4 µg/gal
• 45 lb/ft³ to µg/gal = 2 728 641 604.5 µg/gal
• 46 lb/ft³ to µg/gal = 2 789 278 084.6 µg/gal
• 47 lb/ft³ to µg/gal = 2 849 914 564.7 µg/gal
• 48 lb/ft³ to µg/gal = 2 910 551 044.8 µg/gal
• 49 lb/ft³ to µg/gal = 2 971 187 524.9 µg/gal
• 50 lb/ft³ to µg/gal = 3 031 824 005 µg/gal
• 51
through
75
pounds per cubic foot
• 51 lb/ft³ to µg/gal = 3 092 460 485.1 µg/gal
• 52 lb/ft³ to µg/gal = 3 153 096 965.2 µg/gal
• 53 lb/ft³ to µg/gal = 3 213 733 445.3 µg/gal
• 54 lb/ft³ to µg/gal = 3 274 369 925.4 µg/gal
• 55 lb/ft³ to µg/gal = 3 335 006 405.5 µg/gal
• 56 lb/ft³ to µg/gal = 3 395 642 885.6 µg/gal
• 57 lb/ft³ to µg/gal = 3 456 279 365.7 µg/gal
• 58 lb/ft³ to µg/gal = 3 516 915 845.8 µg/gal
• 59 lb/ft³ to µg/gal = 3 577 552 325.9 µg/gal
• 60 lb/ft³ to µg/gal = 3 638 188 806 µg/gal
• 61 lb/ft³ to µg/gal = 3 698 825 286.1 µg/gal
• 62 lb/ft³ to µg/gal = 3 759 461 766.2 µg/gal
• 63 lb/ft³ to µg/gal = 3 820 098 246.3 µg/gal
• 64 lb/ft³ to µg/gal = 3 880 734 726.4 µg/gal
• 65 lb/ft³ to µg/gal = 3 941 371 206.5 µg/gal
• 66 lb/ft³ to µg/gal = 4 002 007 686.6 µg/gal
• 67 lb/ft³ to µg/gal = 4 062 644 166.7 µg/gal
• 68 lb/ft³ to µg/gal = 4 123 280 646.8 µg/gal
• 69 lb/ft³ to µg/gal = 4 183 917 126.9 µg/gal
• 70 lb/ft³ to µg/gal = 4 244 553 607 µg/gal
• 71 lb/ft³ to µg/gal = 4 305 190 087.1 µg/gal
• 72 lb/ft³ to µg/gal = 4 365 826 567.2 µg/gal
• 73 lb/ft³ to µg/gal = 4 426 463 047.3 µg/gal
• 74 lb/ft³ to µg/gal = 4 487 099 527.4 µg/gal
• 75 lb/ft³ to µg/gal = 4 547 736 007.5 µg/gal
• 76
through
100
pounds per cubic foot
• 76 lb/ft³ to µg/gal = 4 608 372 487.6 µg/gal
• 77 lb/ft³ to µg/gal = 4 669 008 967.7 µg/gal
• 78 lb/ft³ to µg/gal = 4 729 645 447.8 µg/gal
• 79 lb/ft³ to µg/gal = 4 790 281 927.9 µg/gal
• 80 lb/ft³ to µg/gal = 4 850 918 408 µg/gal
• 81 lb/ft³ to µg/gal = 4 911 554 888.1 µg/gal
• 82 lb/ft³ to µg/gal = 4 972 191 368.2 µg/gal
• 83 lb/ft³ to µg/gal = 5 032 827 848.3 µg/gal
• 84 lb/ft³ to µg/gal = 5 093 464 328.4 µg/gal
• 85 lb/ft³ to µg/gal = 5 154 100 808.5 µg/gal
• 86 lb/ft³ to µg/gal = 5 214 737 288.6 µg/gal
• 87 lb/ft³ to µg/gal = 5 275 373 768.7 µg/gal
• 88 lb/ft³ to µg/gal = 5 336 010 248.8 µg/gal
• 89 lb/ft³ to µg/gal = 5 396 646 728.9 µg/gal
• 90 lb/ft³ to µg/gal = 5 457 283 209 µg/gal
• 91 lb/ft³ to µg/gal = 5 517 919 689.1 µg/gal
• 92 lb/ft³ to µg/gal = 5 578 556 169.2 µg/gal
• 93 lb/ft³ to µg/gal = 5 639 192 649.3 µg/gal
• 94 lb/ft³ to µg/gal = 5 699 829 129.4 µg/gal
• 95 lb/ft³ to µg/gal = 5 760 465 609.5 µg/gal
• 96 lb/ft³ to µg/gal = 5 821 102 089.6 µg/gal
• 97 lb/ft³ to µg/gal = 5 881 738 569.7 µg/gal
• 98 lb/ft³ to µg/gal = 5 942 375 049.8 µg/gal
• 99 lb/ft³ to µg/gal = 6 003 011 529.9 µg/gal
• 100 lb/ft³ to µg/gal = 6 063 648 010 µg/gal
• µg/gal stands for µg/US gal
#### Foods, Nutrients and Calories
PUDDING CAKE, UPC: 092825112242 contain(s) 333 calories per 100 grams (≈3.53 ounces) [ price ]
11394 foods that contain Vitamin A, RAE. List of these foods starting with the highest contents of Vitamin A, RAE and the lowest contents of Vitamin A, RAE, and Recommended Dietary Allowances (RDAs) for vitamin A are given as mcg of Retinol Activity Equivalents (RAE)
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Bahamas Oolite weighs 1 537.8 kg/m³ (96.00172 lb/ft³) with specific gravity of 1.5378 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Daiflon 13B1 [CBrF3] weighs 1 580 kg/m³ (98.63618 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 15W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
horsepower (HP) is a non-metric measurement unit of power
Volume is a basic characteristic of any three–dimensional geometrical object.
dyn/in² to N/mm² conversion table, dyn/in² to N/mm² unit converter or convert between all units of pressure measurement.
#### Calculators
Volume of a sphere calculator with surface area to volume ratio
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| 2.5625
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CC-MAIN-2023-40
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https://www.abss.k12.nc.us/site/default.aspx?PageType=3&ModuleInstanceID=70240&ViewID=DEDCCD34-7C24-4AF2-812A-33C0075398BC&RenderLoc=0&FlexDataID=130498&PageID=42274&Tag=&Comments=true
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# A Tale of Two ...
Posted by William Benson on 4/15/2019
…exceptional classrooms at opposite ends of our continuum of opportunities, reading in kindergarten at Highland Elementary School and math at the Alamance-Burlington Early College at Alamance Community College.
On Wednesday, I had the pleasure of joining one of Mr. Kevin Scharen’s math classes in the Alamance-Burlington Early College. After a warm-up team competition on Quizlet to review trigonometry ratios, students were tasked with creating, and then critiquing, four real world word problems using trigonometric ratios (Remember SOHCAHTOA?). We worked in teams of two with each member responsible for two problems. I worked with Cale and developed the following problem:
Students in a high school biology class are cataloging trees on their high school campus. One of the pieces of information required is estimated height. A student who is 5 feet tall uses a homemade clinometer to sight the top of a tree and determines the degree of incline is 52 degrees at a standing distance of 25 feet. How tall is the tree?
(Hint: The vertex of the observed angle is behind the viewer, and as such, the length of the adjacent side is longer than 25 feet.)
On Friday, I joined Ms. Blum’s kindergartners at Highland Elementary for readers’ theater. Ms. Blum divided students into three groups to take on three versions of The Three Little Pigs, including the original, a version with a big, bad pig, and version told from the viewpoint of the wolf. Student groups performed their versions of the story. Ms. Blum recorded student performances and made them available to their parents. Students read fluently and with expression -- exceptional performances all around. Students then used graphic organizers to compare and contrast the stories. I personally found the version told from the viewpoint of the wolf to be very compelling. The wolf’s intentions were completely misunderstood!
Mr. Scharen and Ms. Blum are exceptional teachers on two ends of a K-12 continuum designed to provide students with the knowledge and skills to think critically and own their futures!
-WBB
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https://nursingwritingtutors.com/jackson-county-judges-try-thousands-of-cases-per-year-statistics-homework-help-2/
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# Jackson-County-judges-try-thousands-of-cases-per-year-statistics-homework-help
Jackson County judges try thousands of cases per year. In an overwhelming majority of the cases disposed, the verdict stands as rendered. However, some cases are appealed, and of those appealed, some of the cases are reversed. Jackie Chan of The Star Tribune conducted a study of cases handled by Jackson County judges over a three-year period. In the Excel file, Judges, linked at the bottom of the page, are the results for the 182,908 cases handled (disposed) by 40 judges in Common Pleas Court, Domestic Relations Court, and Municipal Court.
The purpose of the newspaper’s study was to evaluate the performance of the judges. The newspaper wanted to know which judges were doing a good job and which ones were making too many mistakes. You are to assist in the data analysis by using your knowledge of probability and conditional probability to help with the ranking of each of the judges, as well as each court.
Managerial Report
Prepare a report (see below) with your ranking of the judges based on the probabilities and conditional probabilities, as well as the analysis of each court. Include the following seven (7) items in table format to support your ranking. Be sure to use five (5) decimal places for your probabilities in the table, as some of them will be quite small.
1. The probability of cases being appealed in each of the three different courts.
2. The probability of cases being reversed in each of the three different courts.
3. The probability of cases being reversed given an appeal in each of the three different courts.
4. The probability of a case being appealed for each judge.
5. The probability of a case being reversed for each judge.
6. The probability of reversal, given an appeal for each judge.
7. Rank the judges within each court for each of the probabilities in 4 – 6. In other words, only rank the judges in the Common Pleas court against the other judges in the Common Pleas court. perfrom the same analysis for the other two courts. Then, within each court, find the sum of the ranks and get an overall ranking for each judge. Evaluate and discuss the meaning of your results. Use tables, charts, graphs, or visual dashboards to support your findings.
Write a report that adheres to the Written Assignment Requirements under the heading “Expectations for CSU-Global Written Assignments†found in the CSU-Global Guide to Writing and APA Requirements. As with all written assignments at CSU-Global, you should have in-text citations and a reference page. An example paper is provided in the MTH410 Guide to Writing with Statistics, linked at the bottom of the page.
Your report must contain the following:
• A title page in APA style.
• An introduction that summarizes the problem.
• The body of the paper should answer the questions posed in the problem by communicating the results of your analysis. Include results of calculations, as well as charts and graphs, where appropriate.
• A conclusion paragraph that addresses your findings and what you have determined from the data and your analysis.
##### Do you need a similar assignment done for you from scratch? We have qualified writers to help you. We assure you an A+ quality paper that is free from plagiarism. Order now for an Amazing Discount! Use Discount Code “Newclient” for a 15% Discount!NB: We do not resell papers. Upon ordering, we do an original paper exclusively for you.
The post Jackson-County-judges-try-thousands-of-cases-per-year-statistics-homework-help appeared first on Custom Nursing Help.
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https://physics.stackexchange.com/questions/183299/why-does-the-potential-difference-between-two-charged-plates-increase-as-they-mo
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# Why does the potential difference between two charged plates increase as they move further apart?
Suppose a uniform electric field $E$ exists between to oppositely charged metal plates (one is positively charged and one is negatively charged). If the plates move apart, and the charges on each plate stay the same, why does the potential difference increase?
From my understanding (which I know is somehow flawed), the electric potential of each plate seems to vary inversely with distance $r$. However, as distance increases, the potential difference between the plates increases. I think the potential on both plates would decrease, but this does not determine anything about the difference between the potentials.
• What would happen if you let go of the two plates? What does that mean for the case when you move the plates farther apart? May 10, 2015 at 23:37
• Wouldn't the plates attract, because they're oppositely charged? So, perhaps as the plates move further apart, the distance they travel when released is greater, kind of like gravitational potential of a tennis ball held at waist-level versus the gravitational potential of a tennis ball held at shoulder-level? @CuriousOne
– user80932
May 10, 2015 at 23:47
• Very good! The two plates do attract, which means that one has to exert a force to keep them apart. Even better, you have identified that movement against a (conservative) force (like gravity) is the hallmark of a potential. When we move something against such a force we have to expend energy, which then becomes potential energy of the system. So when you are moving the plates of a capacitor apart, you are giving it potential energy. The remaining hard part is for you to identify the mechanical definition of potential with the electrical one (charges moving in electric fields). Does that help? May 10, 2015 at 23:54
• Are you thinking of an external electric field, or the electric field due to the charge on the capacitor? In the second case, see physics.stackexchange.com/q/183137/44126
– rob
May 11, 2015 at 1:39
we know, $$E=V/d$$ (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html ) Where $V$,$E$ and $d$ are potential difference ,electric field and distance between the two plates respectively.
Electric field between the plates only depends on charges of the plates and since charges must be conserved so when the plates are moved apart charges(the amount) on the plates donot change. So $E$ is constant. ( http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html )
Now,look at the equation, you increased the distance $d$ so in order to make the LHS constant(which is $E$ ) $V$ must also increase.
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# Terminology/ko
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
In KSP, there are many terms pertaining to orbiting and physics that often can be confusing to non-technicians. In addition, various other scientific terms and abbreviations are used to describe common terms.
This sheet is designed as a concise lookup table of necessary terms to help you get started down the road to being a full fledged astronaut!
## Mathematics
Cartesian coordinate system - Uses rectangular coordinates
Polar coordinate system - Uses only angles and one length
Elliptical
Oval shaped, often in reference to your orbit.
Normal vector
A vector perpendicular to a plane.
Scalar
A single value without a direction. Scalars are usually followed by a unit of measurement that tells what the scalar's dimension is. eg 3 m/s, 3 m,3 s are scalar: they have units which denote speed, length/distance, and time respectively, but have no direction.
Vector
A set of a direction and a value. eg. heading (direction) and speed together give a velocity. How a vector is expressed depends on what the coordinate system is, and how many dimensions are taken into consideration. <35°, 12> is a two dimensional polar vector, where <14, 9, -20> is a three dimensional Cartesian vector. There are other coordinate systems but these are the most utilized.
<35°, 12> looks like an arrow which is 12 units long, starting from the origin (zero, angle does not matter because it is a point with no length) and ending at a point 35° from the base axis (typically the x-axis, positive angles progress counter-clockwise)
<14, 9, -20> looks like an arrow starting from the origin (<0,0,0>) and ending at a point where the x coordinate is at 14, the y coordinate = 9 and the z coordinate = -20.
The upside to using Cartesian coordinates is that you know exactly where the terminal position is, but it is more difficult to figure the length, however in polar coordinates it is trivial to find the length, the downside is its more difficult to know the position.
The following physical qualities are all vectors: velocity, acceleration, force
A 3D coordinate systems needs:
• A point of reference. This is your origin.
• 3 base-vectors. These define your base unit of measurement along the axis and the direction of said axis.
• A mix of 3 scalars, that could be either angles or co-ordinates to express locations in your co-ordinate space.
## Orbital Terms
Visualization of the most common orbital parameters
Apoapsis
Periapsis
Apsis
→ 참고하기: “Apoapsis and periapsis” section in Orbit
Every elliptical orbit has two apsides. The periapsis (q) is at the closest point of the body being orbited (the lowest point in the orbit) and the apoapsis (Q) is on the other side of the orbit and is the farthest from the body being orbited (the highest point in the orbit). The apsides are usually given from the body's surface while most formula require the distance from the center of the body so the radius usually needs to get added.
Peri-* and Apo-*
When speaking of orbits, oftentimes the words "periapsis" and "apoapsis" will be modified to specify which planet or moon the orbit is around. For example, -kee and -kerb are both commonly used to describe orbits around Kerbin, resulting Perikee/Perikerb and Apokee/Apokerb.
Ascending node
The point at which the orbit crosses the reference plane moving north. Here, "north" means the direction of the orbit normal of the reference plane.
Descending node
The point at which the orbit crosses the reference plane moving south.
Eccentricity
${\displaystyle {\text{ecc}}=1-{\frac {2}{{\frac {Q}{q}}+1}}}$
A scalar describing how non-circular an orbit is.
• ecc = 0 → circular orbit.
• 0 < ecc < 1 → elliptical orbit.
• ecc = 1 → parabolic orbit - this is an escape orbit.
• ecc > 1 → hyperbolic orbit - this is an escape orbit.
Inclination
The angle between an orbital plane relative to a reference plane (e.g. an orbit with 90° inclination to an equatorial reference plane would be called polar).
Low orbit
An orbit that is only just high enough to indefinitely avoid succumbing to hazards of the body being orbited, such as atmospheric drag. Low orbits are used as stepstones, after ascent and before burning to another rendezvous object (planet or vessel), as it allows the exit burn to be performed in any direction and requires the least amount of fuel to reach from the body's surface. A low orbit for Kerbin is typically between 80km and 100km. Bodies with no atmosphere can theoretically allow an orbit at any height above the ground, but below 10km the risk of crashing into mountains or other elevated terrain becomes very high. The time warp is restricted to lower values while in low orbits.
Orbital nodes
Specific points of reference in any orbit such as Apoapsis, Periapsis, intersect points with other orbits etc.
Orbit normal
A normal vector of the Orbital Plane. Produced by cross multiplying the ship's velocity and gravity. Since this follows the right hand rule, from a perspective where the ship is orbiting counter-clockwise it will point "up", while for a ship orbiting "clockwise" it will point down. "Up" is also often labeled as "North" or "N+", and in tandem with that "Down" is labeled "Anti-Normal", "South" or "N-".
Orbital plane
The imaginary disk described by the path of an orbit around a body (commonly used when describing inclinations).
The direction in which a ship is traveling along its orbital path. Since orbits are elliptical, it is always tangent to the orbit at the point where the ship is.
The reverse of Prograde, backwards along the orbital path.
Reference plane
Any plane used as a reference for describing your current orbit. For local orbits around a planet, this is often the equatorial plane. When multiple bodies in a solar system are involved, the ecliptic plane can be used. For intercepting another orbiting body, the orbital plane of the body to be intercepted is used. An orbital plane can be fully described by giving the inclination and the longitude of the ascending node relative to a reference plane.
Semi-major axis
${\displaystyle {\text{a}}={\text{R}}+{\frac {Q+q}{2}}}$
→ 참고하기: “Semi-major axis” section in Orbit
The major-axis is the long axis of an ellipse, and the semi-major axis (a) is half of this. It's the average of the apoapsis (Q) and periapsis (q) computed relative to the center of the body. As both are relative to the body's surface, the radius (R) must be added. All orbits with the same semi-major axis have the same period, regardless of their eccentricity.
Sub-orbital
Describes an orbit where the periapsis is below the surface of a planetary body. If a suborbital path is followed for too long the orbiter will collide with the body being orbited.
Thrust-to-weight ratio
→ 참고하기: Thrust-to-weight ratio
${\displaystyle {\text{TWR}}={\frac {T}{W}}={\frac {T}{m\cdot g}}}$The Ratio between the total mass of the vehicle and the available thrust of all propulsion devices of the vehicle/current stage. A TWR greater than 1 means the craft will have enough thrust to accelerate vertically and gain altitude. A TWR below 1 means that the craft won't be able to counteract gravity and drag at low altitudes, although in space it only means that maneuvers will take longer. Because the weight (W) depends on the current gravitational acceleration (g) the TWR depends on which body is currently influencing the craft. The acceleration on the Mun's surface is only 16.6 % of Kerbin's acceleration, so at the surface a TWRKerbin = 1 would be a TWRMun = 6.
## Ship Orientation
The ship orientation is always relative to a specific object. The terms are usually defined relative to the cockpit.
Zenith
Top side of the ship which is usually oriented away from the orbited body. Opposite of nadir.
Bottom side of the ship which usually oriented towards the orbited body. Opposite of zenith.
Port(side)
Left side of the ship. Opposite of starboard.
Starboard
Right side of the ship. Opposite of portside.
Front
Front side/end of the ship which is usually towards the nose or prograde vector. Opposite of aft.
Aft
Back side/end of the ship which is usually housing the primary rockets and facing in retrograde. Opposite of front.
## Space Maneuvers
Atmospheric Braking
→ 원문 : Aerobraking
Lowering the periapsis so it is inside a planetary atmosphere. This will lead to the vessel being slowed by atmospheric drag. Can lead to reentry (see below), but also is used to reduce the necessary burn time for significant orbit alterations.
Atmospheric entry
Entering atmosphere and using drag to decelerate a vessel to a groundwards trajectory. This usually causes intense heat stress on the object as the vessel requires sufficient speed to not "bounce" back from the atmosphere into space. Currently (0.22[outdated]) re-entry is only partially implemented with effects but heat and bounce are not yet implemented, there are mods however which allow parts to overheat. This is usually called re-entry/reentry but in theory only correct in Kerbin's atmosphere; atmospheric entry being the more general term.
Burn
firing of the engines, usually to alter trajectory in some way.
Circularizing
A maneuver (firing of the engines) that makes an orbit's eccentricity 0, or close to zero. This is usually achieved by a burn close to an apsis.
Maneuver Node
→ 원문 : Maneuver node
Maneuver nodes are a nice tool to plan and project trajectory changes in map view prior to doing the actual burn.
Re-entry
→ 참고하기: Atmospheric entry
Retroburn
A burn performed "backwards", e.g. with the engines facing towards prograde and nose towards retrogade (hence the name). This is a common maneuver to used to lower the height of the orbit without altering any other orbital parameters.
## Physics
Acceleration
Rate of change to the velocity. Acceleration is a vector, measured in "m/s2".
Ballistic trajectory
A falling object's trajectory is ballistic. In rocketry it usually indicates that the object in question is only influenced by gravity and does not exert any force (ie. thrust) of its own.
Delta-v (Δv)
The change in velocity that has or can be exerted by the spacecraft. This is measured in meters per second (m/s). More mass can reduce the delta-v, while more propulsion can increase it. This makes it a useful value to calculate the effectiveness of launch vehicles. For example, a launch vehicle requires about 4,500 m/s of delta-v to escape Kerbin's atmosphere and achieve a stable orbit.
Energy
→ 참고하기: Specific orbital energy on Wikipedia
The energy of an object in an orbit is the sum of its potential and kinetic energy. The potential energy is ${\displaystyle E_{p}=-{\frac {GMm}{R}}}$ and kinetic energy ${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$ where G is the gravitational constant, M is the mass of the body, m is the mass of the craft, R is the distance from the center of the body and v is the velocity. This results in ${\displaystyle E=E_{k}+E_{p}={\frac {1}{2}}mv^{2}-{\frac {GMm}{R}}}$. This sum stays the same when not thrusting: When approaching periapsis potential energy is transferred into kinetic energy. After passing the periapsis the kinetic energy is converted back into potential energy. When the energy or specific orbital energy is greater than zero the vehicle is on an escape trajectory.
This is the basic idea behind Kepler's laws of planetary motion, which is what gives rise to KSP's patched conics approximation. An ellipse is the set of all points on a plane such that the sum of the distances to two points - the foci - is some constant. One focus of a Kepler orbit is the centre of mass of the object being orbited; as an object approaches it, it exchanges potential energy for kinetic energy. As the object moves away from this focus - equivalently, if the orbit is elliptical, as the object approaches the other focus - it exchanges kinetic energy for potential energy. If the craft going directly towards or away from the object, the foci coincide with the apsides, where the kinetic (apoapsis) or potential (periapsis) energy is zero. If it's perfectly circular (e.g. the Mun's orbit around Kerbin), the two foci coincide and the locations of the apsides are undefined, since every point of the orbit is an apsis.
There is also the specific orbital energy (${\displaystyle \epsilon }$) which doesn't require the mass of the craft: ${\displaystyle E_{p}=\epsilon _{p}m}$, ${\displaystyle E_{k}=\epsilon _{k}m}$, ${\displaystyle E=\epsilon \,m=(\epsilon _{k}+\epsilon _{p})m=-{\frac {GM}{2a}}}$. All orbits with the same semi-major axis (a) have the same specific orbital energy.
Escape Velocity
The velocity needed to escape a given planet's gravity well, as given by ${\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}},}$ where G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet.
g-force (G)
A measurement of acceleration as expressed in the sea-level force of Earth's gravity with 1 G being about 9.81 m/s². An object at Earth's surface is accelerated at 1 G. The object weighs twice as much when at 2 G acceleration and is weightless when accelerated with 0 G. In free fall, like in orbit, and without an engine running or an atmosphere applying drag all objects experience no acceleration which can be expressed as 0 G.
Gravity
The force exerted by all objects with mass. Very weak. Usually only objects with very high mass - ie. planets, moons - have any noticeable effect. Diminishes with the square of distance from the center of mass. So for an object twice as far, experiences only 1/22 = 1/4 of the gravity.
Gravity Well
The area around a planet affected by gravity. Actually extends to infinity, but as gravity decreases quadratically with distance (after twice the distance the gravity is only a quarter), it is only significant within the body's sphere of influence. In fact, in KSP, gravity isn't simulated at all beyond a body's sphere of influence due to its use of the "patched conic approximation".
Orbit
→ 원문 : Orbit
When an object has sufficient tangential velocity (and is outside the atmosphere, so drag won't slow it down) so that it will keep falling "next" to the planet (never touching ground) its trajectory is called an orbit. Stable orbits are elliptical (a circle is an ellipse with zero eccentricity). If the objects tangential speed exceeds escape velocity it's orbit will be either para- or hyperbolic.
Specific Impulse (Isp)
→ 원문 : Specific impulse
${\displaystyle I_{sp}={\frac {T}{\dot {m}}},[I_{sp}]={\frac {m}{s}}}$The Isp defines how effective a propulsion system is. The higher the Isp the more powerful is the thrust applied to the rocket with the same fuel mass. The Isp is usually given in seconds but actually the physically correct unit is distance per time which is usually given in meters per second or feet per second. To avoid confusion which unit of speed is used, the physical correct Isp (in distance/time) is divided by the surface gravity of Earth (9.81 m/s²). This results in a value given in seconds. To use this Isp in formulas it must to be converted back into distance per time which requires multiplying with the surface gravity of Earth again. As this value is only used to convert between those two units, the specific impulse doesn't change when the gravity changes. It appears that KSP use a value like 9.82 m/s² and thus using a little less fuel.
As the specific impulse is the ratio of thrust and fuel flow ${\displaystyle {\frac {Ns}{kg}}}$ is sometimes given as the unit. This is mathematically another form of ${\displaystyle {\frac {m}{s}}=kg\cdot {\frac {m}{s^{2}}}\cdot {\frac {s}{kg}}}$ because force is the multiplication of mass and acceleration defining ${\displaystyle N=kg\cdot {\frac {m}{s^{2}}}}$. So ${\displaystyle 1{\frac {Ns}{kg}}=1{\frac {m}{s}}}$ with the latter being simply only in SI base units.
Sphere of influence
The radius around a celestial body within which its gravity well is non-negligible. Commonly known as SoI/SOI.
Tangential velocity
The component of the velocity that is tangential to the trajectory. Instantaneous velocity - velocity when the time of measurement approaches zero - is always tangential to the trajectory.
Trajectory
A trajectory is the path that a moving object follows through space as a function of time.
Velocity
Rate of change of the position. It is the combination of speed with the direction. Velocity is a vector, measured in meters per second (m/s).
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# camera's coordinates to pick object
• Hello everyone,
Kss: V4.1.5 Robot: Kr3 & controller: Krc3
Up to now i am able to move the robot from the PLC (c#). So c# program send the move position to PLC and PLC send to kuka. Between PLC and kuka profibus is used.
This post is related to attached image below.
In short I have webcam attached to the stand. For now I am using the chess board as the working space (pick up zone). I want robot to move from the position shown in picture to pick the object (rectangle boxes in this case) when the object is touched. For that I am able to get the X and Y in mm from pixel (since the picture is 2D)when the object is touched.
Now my difficulties are:
1. How can I make robot to move to this X and Y coordinates? because coordinates (mm) of working space is different from the robot Cartesian coordinates. Should I have to make some kind of offset on c#(PLC ) side to make cordinates equal? If so how to make that? OR teach the points to robot? or do something with \$base or \$world?
2. Since I need to pick the box as top part, I am taking the working space as 2D which means i have only two coordinates(X & Y in mm). How is this x & y related to robot Cartesian XYZABC? or is there way to convert to joint coordinate system or in degree?
In short how to make the robot to know the x & y in mm of camera so that robot can move to that point? My problem is making the relation between camera coordinate to know by robot and it is not about the programming in kuka side or sending values to robot.
I hope my description is not confusing. if it is confusing please ask
Thank you.
Edited once, last by Captain ().
• This is a typical issue in vision-based robotic systems.
Basically, you have 3 different coordinate systems which must be aligned somehow: The camera, the "table" (your chessboard, in this case), and the robot.
The camera's coordinate system consists of an XY grid of pixels, usually starting from one corner of the image(depends on your vision software). The safe assumption is that you will never get this perfectly aligned with any other coordinate frame by physical adjustment alone. Also, you have to create a conversion between pixels and mms, since that changes as the height and/or zoom of the camera is changed.
The robot has its own internal coordinate system that is part of it's physical construction: \$WORLD. Again, it's safe to assume you can never make it perfectly aligned with either of the other two simply by physical adjustment. Fortuantely, the robot comes pre-equipped to be taught other coordinate systems.
Then there's the "table": your chessboard. Usually, the simple way is to treat one corner of the board origin of the coordinate system, and make X and Y aligned with the two edges that meet at that corner (which makes Z orthogonal to the surface of the board by default). Generally, the process is to mount your board in its ideal position, then teach the camera and robot the board's coordinate system.
For the camera, this means finding the location of that corner in the camera's pixel coordinate system, and some points along the X and Y edges, at a known mm distance from the origin. This will give you enough data to build a trig algorithm that can convert the location of an object at X and Y pixels from the camera image origin, into a location X and Y mm from the chessboard origin.
For the robot, it's quite simple. Temporarily mount a sharp-ish (pencil-sharp, not hypodermic-sharp) pointer firmly to the robot, and teach it as a TCP. Then, use the 3-Point Base function in the robot -- touch the pointer to the chessboard origin, then a point on the chessboard X axis, then a point in the chessboard's positive XY plane. That's it -- now the robot has a Base whose origin and orientation matches that of the chessboard, and you can program points in that base. This is just like the process of coming up with the conversion between the Camera and chessboard coordinate systems, except that the robot has all the math "canned" internally, while for the camera-to-chessboard, you'll need to create your own algorithm.
Once you've taught the robot it's relationship to the chessboard, and worked out the conversion formula between camera pixels and chessboard coordinates, that's about all there is to it. You can touch the pointer to the chessboard at certain points with the pointer Tool and the chessboard Base active, locate the pointer tip in the camera, convert from pixels to XY position, and check that against what the robot reports its position as being. Once you have the math correct, these should match up pretty closely (they'll never be perfect, but should be more than close enough).
• Thank you SkyeFire
i understand the most of the things .
As attached in image below. Firstly when I want to pick the object, I will start the fixed webcam that shows the video. Then press the capture button. Then it will display the image. The red dot in image is always origin of X and Y (0,0). I have shown x and Y with green letter. Then when i touch any part inside the captured frame, the right side box will show me coordinates in mm according to that fixed position of camera relative to origin (0,0) [in this case i have pressed the rectangular USB stick]. Since the preview frame and captured image frame is fixed then i believe the conversion will be valid as long as the height of camera remain constant. And also i will make sure chessboard does not move. So in this case I don't have to do anything with camera coordinate and table right? So I have to take care about only teaching robot point right?
Now lets go to teaching the robot TCP. Before teaching I will make sure that height is adjust and chess board is zoomed 100% so no floor is seen and camera is fixed and make conversion(calculation) to mm is changed.
For the robot, it's quite simple. Temporarily mount a sharp-ish (pencil-sharp, not hypodermic-sharp) pointer firmly to the robot, and teach it as a TCP. Then, use the 3-Point Base function in the robot -- touch the pointer to the chessboard origin, then a point on the chessboard X axis, then a point in the chessboard's positive XY plane. That's it -- now the robot has a Base whose origin and orientation matches that of the chessboard, and you can program points in that base.
Will be please show small example how to teach TCP so that when i send those x & Y from PLC the robot will know and move to that points? How to use 3 point base function? or will you please explain how to teach the points step wise.I think for now my base is {0,0,0}. I have never teach any point to robot but i assume it is done with pressing touch up after ptp is complete? but what will be my ptp command? and do i have to assume some where \$world=\$base or something similar?
Thank you.
Edited once, last by Captain ().
• TCP setup is detailed in the manual. Briefly, go under the SETUP or CONFIG menu (I don't have a KRC to check against), there should be a sub-menu for Tool&Base. You'll need to do Tool first, then Base.
Setting up the Tool, you'll need to use the 4-point XYZ method. This will tech the XYZ position of the tool, but not do anything for the ABC rotations. This should be enough, if you're careful.
4-Point TCP setup works by stepping through the menu prompts. You need to touch the tip of the pointer to the same point in space from 4 different orientations. Usually we attach a pointer to the table, and use the tip as the fixed spatial point. A pencil on the table, and a table on the robot, will serve, as long as their mountings are solid. If they move during your measurement, it will add so much error that you'll need to re-mount and start over.
Once you have the pointer TCP set up correctly, you can use the 3-point Base setup method, under the same menu tree.
• Thank you SkyeFire for the reply.
i went under Setting>Measure>Tool> 0 XY Z -4 Point then i saw something like shown in picture 1 for base 1.Firstly I get chance to select tool and I press ToolOK. Then it says "Line up tool from different direction" under I get 3 options move to pt.... point ok...close. when I press move to pt, it says "Target is not valid". When i press pont ok, it shows as picture 2.
Then there is 2 menu, repeat all and repeat. But I am not able to move robot or change the coordinates of xyz by this procudere.
Will you please explain again how do I teach TCP 4-point XYZ method after this step?
and will you tell me which manual explain teaching points? In expert programming its just 1 page explanation and it doesn't help.
Thank you.
Edited once, last by Captain ().
• programming manual for system integrator explain this and just about everything else. this is the #1 manual to have. even if it is for a different controller, workflow is the same.
2) if you have an issue with robot, post question in the correct forum section... do NOT contact me directly
• Thank u panicMode and Skyefier again.
well i can jog the robot and do the tool calibration and base calibration . Not the correct one still. However I know the producer now.
But still few questions.
1. When doing TCP 4 point - XYZABC, do we move/ teach the point by setting our movement(jog-keys) in joint, tool or base co-ordinate system ? Lets say if you teach by jogging keys in tool co-ordinate system, and in the top comment you mention don't do anything you ABC. Then how is it possible to touch the tip from 4 different direction without rotating axes?
2. Same for base, when we calibrate base should we calibrate setting our jog keys in base co-ordinate system or moving axis (joints)or world coordinate system. it will have different effect right?
And lets summarize once again my detail, So I have camera. It capture image and in the image (in pixel), there is X and Y pix starting with (0,0) pix and positive x and Y up to maximum 319*239 PX which is always the same because image is always inside frame and frame is fixed. So when I pressed inside the image the shown pix is converted into MM giving the X and Y into MM according to my conversion making camera height and zoom always constant. (please see picture above)
then we need to define the work space which is chessboard. So if I calibrate base, by teach the origin same as camera's origin and x and Y point. then this will make the camera's co-ordinate system mm and robort co-orrdinate mm same right? meaning if i touch the point in my picture then the x and y shown will be same as x and y shown by robot Cartesian coordinate if i move the robot to that point right?
And i didn't understand what does the tool calibration TCP does in my case? when we calibrate tool the can reference point be anywhere right inside the work space(chess board)? And what about the height of the pencil on surface and and height of tool mounted on A6, because at the end I need to pick rectangle object from surface , so do i need to care about height when teaching point TCP? If I don't need to care about that then later at the end how will the robot move that position (especially to that height) to pick object when I touch the object? And the tool mounted in A6 can be changed later right? because when doing 4 point calibration I want to use sharp tool but later to pick object so i need to use another one.
Thank you. many question but may be some has similar answers .
Edited once, last by Captain ().
• 1. Misunderstanding.The 4point XYZ method only sets xyz coordinates of tool but not abc. During calibration you have to change orientation, in which coordinates you move does not matter.The robot calculates the coordinates correct.
2. The same as to 1. No matter in which coordinates you move the robot, but the tool must be selected correct at the beginning of base calibration.
• As Hermann says, which coordinate system you jog in is irrelevant -- the only requirement is that the physical location be correct when you set the point.
Brief technical digression: the robot "knows" where the "Zero Tool" (the center of the Axis 6 mounting flange) is at all times, the same way that you know where your own hand is, even with your eyes closed -- that flange is a part of the robot, and it's relationship to all the axes cannot be changed.
However, the robot knows nothing about what you are attaching to the A6 flange, except for what you tell it (using \$TOOL and/or the TOOL_DATA array). If you attach a tool (for example, a pencil) to the A6 flange, and touch the tip of the tool to the same point in space from 4 very different directions, the robot will inherently know the position of the Zero Tool for each of those four orientations. As a side effect, the Zero Tool will be define four points on the surface of a sphere, centered on the fixed reference point you are touching the tool to. As long as that reference point is fixed, the robot does not need to know where it is, but with 4 points on the surface of a sphere, the robot can mathematically reverse-engineer the XYZ dimensions of the tool, because there will only be one solution for the tool which will work for all 4 Zero Tool positions. Of course, this process is slightly "noisy," so the robot will do it's best to average out the error for each point. But if you are sloppy about touching the reference point at all 4 locations, you may generate more noise in the calculation than the algorithm can tolerate.
As for what the Tool does for you: in order to define a Base using the 3-point method, the robot must have an accurate tool TCP first. The Base is defined by where you move the Tool, so if the Tool is not defined, attempting to define the Base is useless.
In defining either Tool or Base, having a precise physical tool is important -- using a flat-tipped pencil will give you greater "noise" in the accuracy of the measurements than using a sharp pencil.
To relate the coordinate systems to each other, there are multiple methods, but all the frames must agree. If you set up the camera to use one corner of the chessboard as the origin, one edge as the X axis, and another edge as the Y axis, then you must teach the robot to use the same points and edges as the same origin and axes. That will get both the camera and robot using the "same map," so to speak.
Creating the pixel-to-mm conversion will take an additional step, but it shouldn't be too hard. Then it will depend on where you perform the conversion while in operation -- will you do it in the computer that the camera is connected to, or inside the robot? It works either way. You can do the math anywhere along the chain, as long as the conversion formula is accurate and the the conversion is performed before being assigned to a robot motion command.
• THANK YOU Herman and SkyeFire
Well Apparently it look like I did correct tool and base calibration . The X and Y in mm of camera seems to be equal to X and Y in mm of the Robot (not 100% but at least sufficient for now)
Creating the pixel-to-mm conversion will take an additional step, but it shouldn't be too hard. Then it will depend on where you perform the conversion while in operation -- will you do it in the computer that the camera is connected to, or inside the robot?
yes, I did in external computer and I want to send it to the robot (via PLC ) so that robot can go to pick up the object when I press inside image by - PTP input_fromCAMERA C_PTP
But still few doubts .....below I have attached the final value of tool and base calibration.
1. Since I have values X and Y mm to send to robot to move, I should move the robot in Cartesian co-ordinate right? (example: inputX.X= input_fromCAMERA & inputY.Y=input_fromCAMERA then PTP input_fromCAMERA C_PTP)
2. I get only two variable XY(when I press image ) to move in Cartesian co-ordinate, what Should I send the value of Z A B C? Should I put the constant value and send for ZABC? And Z should be 0 right? I want A6 to point vertical down when it is about to pick the object in workplace(chessboard). How can I make the balance in A6 so that it vertical down when it is about to pick object or where ever position of chessboard it moves to?
3. What does those value of XYZ ABC of tool and XYZ of base means now?
Thank you.
Edited once, last by Captain ().
• Well, if you do something like this:
In this example, the camera only controls X and Y. When the robot moves to PositionFromCamera, it retains whatever ZABC values StartPosition had.
In general, in KRL, if you give a LIN or PTP command a position (POS, E6POS, FRAME, AXIS, E6AXIS) that only has some members of the Structure variable filled in, the robot simply retains whatever the last position was for the members that are left undefined. So, in the above example, if StartPosition had a Z value of 50mm, then the robot would still be at 50mm in Z after moving to PositionFromCamera.
In this application, you're not going to bother with B or C, beyond setting them correctly at StartPosition, simply b/c your vision system has no way to measure for corrections around those axes. You may end up making A adjustments, since your vision system can make measurements of rotations around the Z axis. But just getting X and Y working first is the way to start.
The value of your Tool should be unchanging, once you've done the calibration. Your Base value should also be permanent, but you may find it useful to make adjustments to a copy on the fly. This example does the same thing the previous one does, but by changing \$BASE, instead of making a point:
In this case, PickupPosition needs to be pre-programmed for the "zero position" pickup -- that is, the correct pickup position if/when the corrections from the camera are all 0 (or so tiny as to be practically 0). In this case, you teach a "perfect 0" position for the robot, and then shift that position around to follow the camera by shifting \$BASE mathematically using the vision data. This shift works by taking BASE_DATA[1] as its reference origin (that's why it's on the left side of the Geometric Operator ":"), and applies the values in ShiftFrame before putting the shifted results into \$BASE. It's important to ensure that BASE_DATA[1] never gets altered, as that's your permanent reference frame -- instead, you apply the shifts to the "working copy" in \$BASE.
• Thank u SkyeFire
I tried the first method.
DECL E6POS PositionFromCamera
\$BASE = BASE_DATA[1] ; activate common robot/vision base
\$TOOL = TOOL_DATA[1]
positionfromcamera=\$pos_act
PositionFromCamera.X = XInputFromCamera ; Get X value from camera
PositionFromCamera.Y = YInputFromCamera ; Get Y value from camera
PTP PositionFromCamera c_ptp
At first it worked but later it didn't worked. Some time there is error saying"work envelope exceedeed" or sometime "Ax out of range" and some time " start movement Inadmissible ".
I think this is due to change in base and tool. When i go to option current tool and base, I set current tool1 and base 1. But later it changes automatically to 0. then when i open my .src program then if I see current tool and base then it is not 1.
Although we have done \$BASE = BASE_DATA[1]
\$TOOL = TOOL_DATA[1] , It doesn't show me the current tool and base [1] inside my program when I tried to exectue PTP motion.
even If i change it to 1, it automatically chnages to 0.
In my declaration section i can see DECL FDAT fpositionFromCamera={tool_No 1 Base_no_1, IPO_frame #BASE} but still I cannot make the PTP PositionFromCamera c_ptp.
So how to get rid of this problem and how can I make my base and tool as 1 inside my program or before execting my PTP
Thanks
Edited once, last by Captain ().
• you could read messages that robot displays and remedy whatever they tell the problem is. for example program speed, approximation etc.
2) if you have an issue with robot, post question in the correct forum section... do NOT contact me directly
• Hi panic mode Thank you for the reply. I don't mean the same but I managed to get something.
But I want to ask some more question.
1. How to select the proper work space? because In my above picture of top post I calibrated tool and base assuming my chessboard as work space. So now I can move the robot along the origin, X axis or Y axis or XYpoints. However there are some points inside the chessboard where the robot cannot go when moving in Cartesian(because of the A4 or a1 axis limit). but when I moved/jogged the robot with joints coordinate, it can reach at any points inside the chess board. So the easiest solution I can think right now is changing the work space (shifting chess board) and re-calibration base again. But there might be other way too? when i try to jog the robot along the base X=0,y=0 at some point it stop because A4 is beyond limit.
2. Why does robot accept the softX_end[] above the limit? for example for kr3, A4 limit is +- 180 but why does robot accept if I modify it even higher lets say -185? & if I run it is going to crash?
Thank you.
• "Work space exceeded" isn't about a particular workspace -- instead, it means that the robot has been given a command to move to a location that is beyond the arm's physical reach.
This is separate from a "axis limit" error. There are positions which the robot can reach with a particular set of axis angles, but potentially not with another.
PTP motions are predictive for these errors, but LIN/CIRC motions are not. Basically, a LIN motion is made up of many, many, very very small PTP motions, and the "look ahead" for these errors is just the next PTP motion segment. So if you order the robot to an unworkable position using PTP, the error will be generated before the robot starts moving. Change the PTP to a LIN, and the robot will try to make the move, and keep moving until it physically hits a limit.
Axis limit errors are complicated b/c you can "wind up" the wrist axes during LIN motions until you hit a condition where a position that is entirely reachable, cannot be reached by LIN from the robot's current position. If you encounter cases like this, it's best to use PTP motions, perhaps with E6AXIS positions rather than E6POS, to "unwind" the axes before beginning a LIN motion. This is most often achieved by using the system HOME position at the beginning and end of each program run.
Another technique is to use PTP motions whenever possible. This usually allows the motion planner to pick out the least difficult path between Points A and B -- however, if the wrist has been "wound up," this can result in the wrist "unwinding" unpredictably when the path planner detects that it's needed. So you want to do this well clear of everything.
Often, how we do this is to make all the moves above the table PTP motions, and make only the "pick/drop" motions LIN. That is, generate your Pickup position from the vision, mathematically generate an "Above Pick" position from that, then make the Above->Pick->Above moves LINs, and all the others (say, HOME->Above) PTPs. This creates minimal opportunity for the LIN motions to "wind up" the wrist axes.
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# 17Lecture - LECTURE 17 Today is Thursday One year ago when...
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LECTURE 17 Today is Thursday, October 14, 2010. One year ago when I gave this lecture, the price of gold was \$1,060 per troy ounce. Today, in New York, it closed at \$1,381.20. Try getting that rate-or-return on a blue-chip share of common equity or a corporate bond. oooorahhh! PROFIT MAXIMIZATION: MARGINAL COST & MARGINAL REVENUE Figure 16-3 and Figure 16-4 demonstrated, along with Table 15-1 from which the figures were derived, that the profit maximizing level of output is that level of output where the slope of the total revenue curve, marginal revenue (MR) is equal to the slope of the total cost curve, marginal cost (MC). Remember, this is where the vertical distance between the TR curve and the TC curve, or profit, is the greatest. Figure 17-1 shows the MC and MR curves from Figure 16-2. Note that on Figure 16-2, the MC curve crosses the MR curve at a level of output of 7 ounces per week. FOR ANY TYPE OF FIRM, THE PROFIT MAXIMIZING LEVEL OF OUTPUT FOR THE FIRM IS WHERE MARGINAL COST IS EQUAL TO MARGINAL REVENUE. 1 Look at the different possible levels of output. Ounces 1 through 6, inclusive, all bring in more revenue to the firm (MR) than they each cost to produce (MC). We should produce these units as we make a profit on each. Note that if we produce ounces 8 through 12, inclusive, each ounce produced costs us more to produce than it brings in as revenue to the firm. We should not produce these units as we take a loss on each. When MC = MR we make the most profit. Figure 17-2 puts it all together. Note that MC = MR at or near 7 ounces of gold produced per week. When the price of gold is \$600 per ounce, MR is \$600. TR is P X Q, or \$600 X 7 = \$4,200 per week. The MC of the 7th ounce of gold is \$590. We should produce the 7th ounce of gold. We make \$10 profit on it. At 7 ounces of gold produced per week, ATC is \$428.57 per 1 Since MC can initially start out above MR, and cut MR when MC is falling, and then cut MR again later when MC is rising, to be perfectly precise, the rule is the profit maximizing level of output for the firm is where rising marginal cost is equal to marginal revenue.
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week. (This number, as all numbers, is from our empirically derived Table 15- 1.) TC = \$428.57 X 7 = \$3,000 per week. 2 Graphically, TR is the area of the rectangle with a height of \$600 and a length of 7 ounces. TC is the area of the rectangle with a height of \$428.57 and a length of 7 ounces. Note the profit rectangle: E Eis the area of the rectangle with a height of \$171.43 (\$600 - \$428.57 = \$117.43) and a length of 7 ounces. (\$171.43 / ounce) x (7 ounces per week) = \$1,200 per week. 3 Examine Figure 17-3. Figure 17-3 depicts the MC and ATC curves for our Sun Devil Gold Mine, derived in earlier lectures. There is one difference. At a price of \$600 per ounce, the profit maximizing level of output for our mining operation is still 7 ounces per week. However, the ATC curve has been moved up, and now it would appear that we are making \$0 of profit each week. The minimum point of the ATC curve is tangent to the MR curve. Total cost appears to be equal to total revenue. This implies that profit is zero.
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> > How do you calculate the net profit/loss in a scenario in which an investor bought 200 shares at \$10/share, then decided to sell all 200 at \$25/share?
# How do you calculate the net profit/loss in a scenario in which an investor bought 200 shares at \$10/share, then decided to sell all 200 at \$25/share?
## there's a 2% commission fee for all buy and sell transactions.
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### How do you calculate the net profit/loss in a scenario in which an investor bought 200 shares at \$10/share, then decided to sell all 200 at \$25/share?
there's a 2% commission fee for all buy and sell transactions.
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Photon Can Be Described as the Normalized Mutual Energy Flow
1. Introduction
1.1. Mutual Energy Flow Can Be Applied to Normalize the Photon
In quantum mechanics we can make normalization to an electron in the orbiter, which belongs to the stable solution of Schrödinger equation. We also can normalize the plane waves. However, we do not know how the normalization can be done for photon, or the electron in empty space. The reason is that we do not know how to describe the photon or electron in empty space. Someone perhaps will argue the photon can be described by Maxwell equations in the empty space, the electron can also be described as Schrödinger equation in empty space. That is true, but photon starts from an emitter and ends at an absorber. The energy is transferred from a point to another point without decrease. The solution of Maxwell equation is a wave which spreads to the whole space; hence, the amplitude of the wave decreases with the distance from the field point to the source. The wave solution and the particle have very big difference. Someone perhaps will argue that plane wave can describe the particle. But first, the charge cannot radiate plane wave, second the plane wave doesn’t focus to two points in the place the photon is radiated and received. This is so called wave and particle duality paradox. Since this paradox, how the normalization of the photon in empty space occurs has not been solved yet. Similar situation happens also for electron in empty space. We need a wave package which can focus to point in both places of emitter and absorber; it is thick in the middle between the emitter and the absorber.
Recently this author has claimed that the photon is nothing else but the mutual energy flow [1]. The mutual energy flow is the interaction between the retarded wave sent from the source and the advanced wave sent from the absorber. This integration leads to an energy flow that starts from emitter ends at the absorber. The mutual energy flow theorem guarantees the energies go through any surfaces between the emitter and the absorber are all equal. Hence, the mutual energy flow can be a good role to do the normalization for photon. This article describes the details why and how to normalize the mutual energy flow and claim the photon is normalized mutual energy flow.
1.2. History Review about the Theory for Mutual Energy Flow
In physics, Wheeler and Feynman introduced the absorber theory in 1945 and 1949 [2] [3], which claimed that the electric current can radiate advanced wave in the same time when it radiates the retarded wave. The absorbers send the advanced waves. This view of point has been extended to the transactional interpretation of quantum mechanics by Cramer around 1980 [4]. The theory about advanced wave of Wheeler and Feynman cited the earliest publications of action at a distance [5] [6] [7].
In electronic engineering, Welch has introduced his time-domain reciprocity theorem in 1960 [8]. In this theorem the transmitting antenna sends the retarded wave. The receiving antenna sends the advanced wave. Ramsey introduces his reciprocity theorem in 1963 [9]. This author introduced the mutual energy theorem in early of 1987 [10]. de Hoop introduced the cross-correlation reciprocity theorem in the end of 1987 [11]. All these theorems are same theorem and only have a difference in Fourier transform. Traditionally, when we speak about the reciprocity theorem, the two fields in the theorem do not need both to really exist. One field can be real and another can be virtual. In the reciprocity theorem of Welch, the retarded wave can be real, the advanced wave can be virtual, hence, the causality of physics has not been violated. However, when we speak these theorems as an energy theorem, it needs the two fields in the theorem to really exist. Hence, this author claimed this theorem is an energy theorem, that means the advanced wave must also exist in nature as a real substance. This author also applied the mutual energy theorem to Huygens principle, sphere wave and plane wave expansions in 1989 [12] [13].
1.3. Normalized Mutual Energy Flow
It should be clear in the mutual energy theorem, mutual energy flow theorem, mutual energy principle, the word “mutual” is only because the historical reason, this word can be dropped. It can be called as “energy theorem”, “energy flow theorem”, “energy principle”. This is because the self-energy principle has told us the self-energy flow does not carry and transfer any energy, the energy is only carried and transferred by the mutual energy flow. Hence, the mutual energy flow is the exclusive energy flow. This energy flow has similar property of the photon. For example, the photon is substance that can transfer energy from a point (emitter) to another point (absorber). The photon energy that goes through each surface between the emitter and the absorber should be equal, which is the energy of the photon. This property of photon is just described by the mutual energy flow theorem. Hence, this author claimed that the photon is the mutual energy flow.
However, the mutual energy flow calculated by the classical electromagnetic field theory is decreased when the distance between the emitter and the absorber is increased. We know photon energy does not decrease with this distance and is a constant. Considering this, in this article, this author would like to normalize the mutual energy flow. After the normalization, the normalized mutual energy flow has exactly same properties as the photon.
It should be clear that, if the theory of mutual energy flow theorem, mutual energy principle, can still be put inside the frame of the Maxwell theory, the normalized mutual energy flow belongs to pure quantum effect and cannot be put inside the frame of Maxwell theory. However, this quantum effect has not been described by the current quantum mechanics theory. The preprint of this article can be found in ref [19].
2. Mutual Energy Theorem and Mutual Energy Flow Theorem
2.1. Mutual Energy Theorem
The Welch’s reciprocity theorem can be written as,
$\left({\xi }_{1}\left(t\right),{\tau }_{2}\left(t\right)\right)=-\left({\tau }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)$ (1)
where t is time and,
${\xi }_{i}\equiv \left[{E}_{i},{H}_{i}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2$
${\tau }_{i}\equiv \left[{J}_{i},{K}_{i}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2$
${E}_{i}$ is electric field, ${H}_{i}$ is magnetic field. ${J}_{i}$ is electric current. ${K}_{i}$ is magnetic current, which is 0 here. The two inner products are defined as,
$\left({\xi }_{1}\left(t\right),{\tau }_{2}\left(t\right)\right)\equiv {\int }_{t=0}^{\infty }\text{d}t{\iiint }_{{V}_{2}}{E}_{1}\left(t\right)\cdot {J}_{2}\left(t\right)\text{d}V$ (2)
$\left({\tau }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)\equiv {\int }_{t=0}^{\infty }\text{d}t{\iiint }_{{V}_{1}}{J}_{1}\left(t\right)\cdot {E}_{2}\left(t\right)\text{d}V$ (3)
${J}_{1}$ is inside volume ${V}_{1}$. ${J}_{2}$ is inside ${V}_{2}$. The above formula. It should be clear that the above theorem can be easily extended to de Hoop’s cross-correlation reciprocity theorem,
$\left({\xi }_{1}\left(t+\tau \right),{\tau }_{2}\left(t\right)\right)=-\left({\tau }_{1}\left(t+\tau \right),{\xi }_{2}\left(t\right)\right)$ (4)
The Fourier transform of the de Hoop’s reciprocity theorem is the mutual energy theorem, the mutual energy theorem is in the Fourier domain, which still can be expressed as Equation (1). But with the meaning of inner product is in the Fourier transform domain, $\omega$ is angle frequency.
$\left({\xi }_{1}\left(\omega \right),{\tau }_{2}\left(\omega \right)\right)=-\left({\tau }_{1}\left(\omega \right),{\xi }_{2}\left(\omega \right)\right)$ (5)
$\left({\tau }_{1}\left(\omega \right),{\xi }_{2}\left(\omega \right)\right)\equiv {\iiint }_{{V}_{1}}{J}_{1}\left(\omega \right)\cdot {E}_{2}^{*}\left(\omega \right)\text{d}V$ (6)
$\left({\xi }_{1}\left(\omega \right),{\tau }_{2}\left(\omega \right)\right)\equiv {\iiint }_{{V}_{2}}{E}_{1}\left(\omega \right)\cdot {J}_{2}^{*}\left(\omega \right)\text{d}V$ (7)
The above is Rumsey reciprocity theorem and also the mutual energy of this author. The above 4 theorems (1, 4, 5) can be seen as same theorem in two different domain, time-domain and Fourier domain. It is clear that these theorems are reciprocity theorems, but this author claimed that it is also an energy theorem [10] and hence, called it as the mutual energy theorem. In the above theorem, this author usually assumed ${\tau }_{1}$ is the source, its field ${\xi }_{1}$ is the retarded field, ${\tau }_{2}$ is the sink, its field ${\xi }_{2}$ is the advanced field. This theorem tells us that the advanced wave ${\xi }_{2}$ sacked energy from the source ${\tau }_{1}$ which is equal to the energy that the retarded field ${\xi }_{1}$ offers to the sink ${\tau }_{2}$. In the formulas (1, 4, 5) there is a negative sign, that is because ${\tau }_{1}$ is source, the source offers energy, hence, has a negative sign. Here the tradition is that the consumed energy is positive, hence, the energy offered by the source is negative. The right of the formula (1, 4, 5) is the energy offered by the source ${\tau }_{1}$. The left of the formula (1, 3, 5) is the consumed energy by the current of the sink ${\tau }_{2}$. The theorem tells us these two energies are equal.
2.2. Mutual Energy Flow Theorem
The above theorem has been extended as mutual energy flow theorem [1],
$\left({\xi }_{1}\left(t\right),{\tau }_{2}\left(t\right)\right)=\left({\xi }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)=-\left({\tau }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)$ (8)
The mutual energy flow is defined as
$\left({\xi }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)\equiv {\int }_{t=0}^{\infty }{∯}_{\Gamma }\left({E}_{1}\left(t\right)×{H}_{2}\left(t\right)+{E}_{2}\left(t\right)×{H}_{1}\left(t\right)\right)\cdot \stackrel{^}{n}\text{d}\Gamma \text{d}t$ (9)
$\Gamma$ is the arbitrary close surface or open surface which has extended to infinity between the source ${\tau }_{1}$ and the sink ${\tau }_{2}$, see Figure 1.
The mutual energy flow theorem tells us, that the energy from the source to the sink is transferred through the mutual energy flow. The mutual energy flow is also a good inner product of the two fields, ${\xi }_{1}\left(t\right),{\xi }_{2}\left(t\right)$. The mutual energy flow theorem can also be converted to Fourier domain,
$\left({\xi }_{1}\left(\omega \right),{\tau }_{2}\left(\omega \right)\right)=\left({\xi }_{1}\left(\omega \right),{\xi }_{2}\left(\omega \right)\right)=-\left({\tau }_{1}\left(\omega \right),{\xi }_{2}\left(\omega \right)\right)$ (10)
$\left({\xi }_{1}\left(\omega \right),{\xi }_{2}\left(\omega \right)\right)\equiv {∯}_{\text{Γ}}\left({E}_{1}\left(\omega \right)×{H}_{2}^{*}\left(\omega \right)+{E}_{2}^{*}\left(\omega \right)×{H}_{1}\left(\omega \right)\right)\cdot \stackrel{^}{n}\text{d}\Gamma$ (11)
The mutual energy flow theorem can be seen by Figure1. In the Figurea is the source or emitter and b is the sink or absorber. The emitter can be seen as an atom with a transmitting antenna inside it. The absorber can be seen as an atom with a receiving antenna inside it. The source sends the retarded wave. The sink sends the advanced wave. The mutual energy flow is built by the retarded wave and advanced wave. The shape of the mutual energy flow is the overlap of retarded wave and the advanced wave which are thin in the two ends and thick in the middle. The mutual energy goes through any surfaces between a and b are equal. This is the mutual energy flow theorem. The mutual energy flow theorem further guarantees that the mutual energy theorem is an energy theorem.
3. Normalized Mutual Energy Flow
3.1. Normalized Mutual Energy Flow
This author has claimed that the photon is nothing else, but the mutual energy flow [1]. However, recently this author also noticed that the mutual energy flow calculated according to classical electromagnetic field theory decreases when the distance between the source and sink increases. This is clear, since the radiation fields decreases
${\xi }_{1}\left({x}_{2}\right)\propto \frac{1}{r}$ (12)
${\xi }_{2}\left({x}_{1}\right)\propto \frac{1}{r}$ (13)
$r\equiv ‖{x}_{2}-{x}_{1}‖$ (14)
${x}_{1}$ is the position of source. ${x}_{2}$ is the position of sink. In other side we know the mutual energy flow should be the photon which has fixed energy $E=\hslash \omega$, where $\hslash$ is reduced Plank constant, $\omega$ is the angle frequency of the light. Hence, we require that,
$\left({\xi }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)=\hslash \omega$ (15)
Figure 1. The mutual energy flow, which is started from source “a” and ended at sink “b”. Source “a” is an emitter or transmitting antenna. Sink “b” is absorber or a receiving antenna. Photon is the mutual energy flow. According to the mutual energy flow theorem, the mutual energy go through any of the surfaces between source “a” and sink “b” are equal. The surfaces can be closed surface for example the sphere surface surrounding to the source “a” or the sink “b”, or the infinite plane between “a” and “b”.
In order to satisfy the above formula, we need to adjust the value of fields. We need to increase the fields according to the distance $r\equiv ‖{x}_{2}-{x}_{1}‖$, that is,
$kr\left({\xi }_{1}\left(t\right),{\tau }_{2}\left(t\right)\right)=kr\left({\xi }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)=-kr\left({\tau }_{1}\left(t\right),{\xi }_{2}\left(t\right)\right)=\hslash \omega$ (16)
k is a constant which is not related the distance r. The above can be re-written as,
$\left({\xi }_{1}^{p}\left(t\right),{\tau }_{2}^{p}\left(t\right)\right)=\left({\xi }_{1}^{p}\left(t\right),{\xi }_{2}^{p}\left(t\right)\right)=-\left({\tau }_{1}^{p}\left(t\right),{\xi }_{2}^{p}\left(t\right)\right)=\hslash \omega$ (17)
where,
${\xi }_{i}^{p}\equiv \sqrt{kr}{\xi }_{i}\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2$ (18)
${\tau }_{i}^{p}\equiv \sqrt{kr}{\tau }_{i}\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2$ (19)
${\xi }_{i}^{p}$ is the electromagnetic fields of the photon, ${\tau }_{i}^{p}$ is the source of the photon. The field of photon is much stronger than the mutual energy flow calculated according to the classical electromagnetic fields, especially when the distance $r=‖{x}_{1}-{x}_{2}‖$ is very large. We can think that the energy is carried by the mutual energy flow of the classical electromagnetic field which is very small. However, this mutual energy flow builds an energy channel from emitter to the absorber. The energy channel can transfer the energy from emitter (source) to the absorber (sink). This energy channel can be applied repeatedly until the energy reach to the energy of one photon. That is the reason the photon field is stronger than the classical electromagnetic fields.
3.2. Normalization the Wave Function in Quantum Mechanics
In quantum mechanics there is also the normalization of the wave function. The square of amplitude of the wave function is normalized to 1. This 1 is probability. The square of amplitude of wave function is normalized to 1 is similar to the normalized mutual energy flow to the energy of one photon. The difference between the two ways are only a constant.
Because this author believes the wave is an energy-based wave instead of probability-based wave, the photon was normalized to the energy $\hslash \omega$. It should be clear that after the normalization, the field of photon ${\xi }_{i}^{p}$ is quite different with the classical electromagnetic field theory ${\xi }_{i}$. The photon field is normalized electromagnetic fields.
It should be clear, that in the current example the retarded field ${\xi }_{1}\left(x\right)=\left[{E}_{1},{H}_{1}\right]$ is sent from the source a and hence, decreases according to $‖x-{x}_{1}‖$. The advanced field ${\xi }_{2}\left(x\right)=\left[{E}_{2},{H}_{2}\right]$ is sent from the sink b and hence, decreases according to $‖x-{x}_{2}‖$. Even ${\xi }_{1}\left(x\right)$ and ${\xi }_{2}\left(x\right)$ decrease with the distance $‖x-{x}_{1}‖$, $‖x-{x}_{2}‖$, the mutual energy flow $\left({\xi }_{2}\left(x\right),{\xi }_{1}\left(x\right)\right)$ is not changed at any surfaces see Figure 1.
There is also another possibility that the advanced wave become the guidance wave of the retarded wave and the retarded wave become the guidance wave of the advanced wave. Because of the effect of guidance waves, the retarded wave and the advanced wave both do not decrease with the distance $‖x-{x}_{1}‖$ and $‖x-{x}_{2}‖$. However, inside the photon we cannot measure the field $\left[{E}_{1},{H}_{1}\right]$ and $\left[{E}_{2},{H}_{2}\right]$. This scenario has the same mutual energy flow as before. For calculating the mutual energy flow we still can use Equation (9).
4. The Macroscopic Electromagnetic Field
Here macroscopic electromagnetic field means the field produced by many photons together. This section we prove that the macroscopic electromagnetic field satisfy Maxwell equations of the retarded wave. We need a condition: all absorbers are uniformly distributed on the big sphere with its radius as infinite.
4.1. Poynting Theorem Is Equivalent to Maxwell Equations
It is known that the Maxwell equations,
$\nabla ×E=-{\partial }_{t}B$ (20)
$\nabla ×H=J+{\partial }_{t}D$ (21)
$B={\mu }_{0}H,\text{\hspace{0.17em}}\text{\hspace{0.17em}}D={ϵ}_{0}E$ (22)
(where ${\partial }_{t}=\frac{\partial }{\partial t}$ ) is equivalent to the corresponding to Poynting theorem [18],
$-{∯}_{\text{Γ}}E×H={\iiint }_{V}\left(E\cdot J+E\cdot {\partial }_{t}D+H\cdot {\partial }_{t}B\right)\text{d}V$ (23)
If we can prove the field of many photons satisfy Poynting theorem above we have proved that the macroscopic electromagnetic field satisfy Maxwell equations.
4.2. The Self-Energy Principle and the Mutual Energy Principle
Assume in the space there are N charges. For the electromagnetic field of each charge, it satisfies Maxwell equations and hence, the Poynting theorem,
$-{∯}_{\text{Γ}}{E}_{i}×{H}_{i}\cdot \stackrel{^}{n}\text{d}\Gamma ={\iiint }_{V}\left({E}_{i}\cdot {J}_{i}+{E}_{i}\cdot {\partial }_{t}{D}_{i}+{H}_{i}\cdot {\partial }_{t}{B}_{i}\right)\text{d}V$ (24)
Substitute the superposition principle
$E={\sum }_{i=1}^{N}{E}_{i},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}H={\sum }_{i=1}^{N}{H}_{i}$ (25)
to the Poynting theorem Equation (23) we obtain
$\begin{array}{l}-{\sum }_{j=1}^{N}{\sum }_{i=1}^{N}{∯}_{\text{Γ}}{E}_{j}×{H}_{i}\cdot \stackrel{^}{n}\text{dΓ}\\ ={\sum }_{j=1}^{N}{\sum }_{i=1}^{N}{\iiint }_{V}\left({E}_{j}\cdot {J}_{i}+{E}_{j}\cdot {\partial }_{t}{D}_{i}+{H}_{j}\cdot {\partial }_{t}{B}_{i}\right)\text{d}V\end{array}$ (26)
Equation (24) is the energy of self-energy flow, we have introduced the self-energy principle [1] which tell us that self-energy flows do not carry any energy. The self-energy flow terms (which have $i=j$ ) have been all canceled by corresponding terms of the self-energy flows of the time-reversal waves. Hence, we can take away all self-energy flow terms from Equation (26), we obtain,
$\begin{array}{l}-{\sum }_{j=1}^{N}{\sum }_{i=1,i\ne j}^{N}{∯}_{\text{Γ}}{E}_{j}×{H}_{i}\cdot \stackrel{^}{n}\text{dΓ}\\ ={\sum }_{j=1}^{N}{\sum }_{i=1,i\ne j}^{N}{\iiint }_{V}\left({E}_{j}\cdot {J}_{i}+{E}_{j}\cdot {\partial }_{t}{D}_{i}+{H}_{j}\cdot {\partial }_{t}{B}_{i}\right)\text{d}V\end{array}$ (27)
Equation (27) is the mutual energy formula, we have promoted it as the mutual energy principle, and claim it is real physical equation for photon [1]. From mutual energy principle, we can derive the Maxwell equations Equation (20, 21). However, the derived Maxwell equations must belong to the two groups, one is the retarded wave and another is the advanced wave. The retarded wave must synchronize with the advanced wave. The synchronized retarded wave and advanced wave can produce the mutual energy flow which satisfy the mutual energy flow theorem. Hence, the mutual energy principle is the foundation of the mutual energy flow theorem which describes the photon. In the follow section we will prove the macroscopic field i.e. the field of many photons that satisfy Maxwell equations by using the mutual energy principle. In this way, we have proved that the macroscopic wave can be built by thousands of photons together and proves Einstein’s guesses.
4.3. The Macroscopic Field of the Electromagnetic Field
Equation (27) can be re-written as,
$\begin{array}{l}-{\sum }_{j=1}^{N}{\sum }_{i=1}^{i (28)
Considering among these charges there are L emitters and M absorbers. The retarded field of the emitter can only paired to an advanced field from absorbers, hence Equation (28) can be re-written as,
$\begin{array}{l}-{\sum }_{j=1}^{L}{\sum }_{i=1}^{M}{∯}_{\text{Γ}}\left({E}_{j}×{H}_{i}+{E}_{i}×{H}_{j}\right)\cdot \stackrel{^}{n}\text{dΓ}\\ ={\sum }_{j=1}^{L}{\sum }_{i=1}^{M}{\iiint }_{V}\left({E}_{j}\cdot {J}_{i}+{E}_{i}\cdot {J}_{j}+{E}_{j}\cdot {\partial }_{t}{D}_{i}+{E}_{i}\cdot {\partial }_{t}{D}_{j}+{H}_{j}\cdot {\partial }_{t}{B}_{i}+{H}_{i}\cdot {\partial }_{t}{B}_{j}\right)\text{d}V\end{array}$ (29)
We consider a special situation, the source is at the origin point, there are infinite more absorbers are at the big sphere with its radius infinite large. Assume L = 1, there is only one source (Emitter) which radiate the retarded wave. All other M charges are absorbers which are on the big sphere. In this special situation the above formula can be written as,
$\begin{array}{l}-{\sum }_{i=1}^{M}{∯}_{\text{Γ}}\left({E}_{0}×{H}_{i}+{E}_{i}×{H}_{0}\right)\cdot \stackrel{^}{n}\text{dΓ}\\ ={\sum }_{i=1}^{M}{\iiint }_{V}\left({E}_{0}\cdot {J}_{i}+{E}_{i}\cdot {J}_{0}+{E}_{0}\cdot {\partial }_{t}{D}_{i}+{E}_{i}\cdot {\partial }_{t}{D}_{0}+{H}_{0}\cdot {\partial }_{t}{B}_{i}+{H}_{i}\cdot {\partial }_{t}{B}_{0}\right)\text{d}V\end{array}$ (30)
In the above, the retarded wave has the subscribe “0”. We have considered the advanced wave cannot build the mutual energy with advanced waves; hence, all absorber must build mutual energy with the only source. The above can be written as,
$\begin{array}{l}-{∯}_{\text{Γ}}\left({E}_{0}×{H}_{a}+{E}_{a}×{H}_{0}\right)\\ ={\iiint }_{V}\left({E}_{0}\cdot {J}_{a}+{E}_{a}\cdot {J}_{0}+{E}_{0}\cdot {\partial }_{t}{D}_{a}+{E}_{a}\cdot {\partial }_{t}{D}_{0}+{H}_{0}\cdot {\partial }_{t}{B}_{a}+{H}_{a}\cdot {\partial }_{t}{B}_{0}\right)\text{d}V\end{array}$ (31)
where ${E}_{a}={\sum }_{i=1}^{M}{E}_{i}$, ${H}_{a}={\sum }_{i=1}^{M}{H}_{i}$ are superposed advanced wave. In the above formula, we can take the volume only includes the source. In this situation, the above formula become,
$\begin{array}{l}-{∯}_{\text{Γ}}\left({E}_{0}×{H}_{a}+{E}_{a}×{H}_{0}\right)\cdot \stackrel{^}{n}\text{dΓ}\\ ={\iiint }_{V}\left({E}_{a}\cdot {J}_{0}+{E}_{0}\cdot {\partial }_{t}{D}_{a}+{E}_{a}\cdot {\partial }_{t}{D}_{0}+{H}_{0}\cdot {\partial }_{t}{B}_{a}+{H}_{a}\cdot {\partial }_{t}{B}_{0}\right)\text{d}V\end{array}$ (32)
${E}_{0}\cdot {J}_{a}$ is disappear in the above formula, since ${J}_{a}$ are at the big sphere which is at outside of V. In this case the absorbers uniformly distribute on the big sphere, the advanced wave will exactly same as the retarded field radiate from the source which is at the origin. Hence, the above formula can be written as,
$\begin{array}{l}-{∯}_{\text{Γ}}\left({E}_{0}×{H}_{0}+{E}_{0}×{H}_{0}\right)\cdot \stackrel{^}{n}\text{dΓ}\\ ={\iiint }_{V}\left({E}_{0}\cdot {J}_{0}+{E}_{0}\cdot {\partial }_{t}{D}_{0}+{E}_{0}\cdot {\partial }_{t}{D}_{0}+{H}_{0}\cdot {\partial }_{t}{B}_{0}+{H}_{0}\cdot {\partial }_{t}{B}_{0}\right)\text{d}V\end{array}$ (33)
or
$-2{∯}_{\text{Γ}}\left({E}_{0}×{H}_{0}\right)\cdot \stackrel{^}{n}\text{dΓ}={\iiint }_{V}\left({E}_{0}\cdot {J}_{0}+2\left({E}_{0}\cdot {\partial }_{t}{D}_{0}+{H}_{0}\cdot {\partial }_{t}{B}_{0}\right)\right)\text{d}V$ (34)
Assume the total electromagnetic field together of retarded field and the advanced field are
$E={E}_{0}+{E}_{a},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}H={H}_{0}+{H}_{a}$
We have
$E={E}_{0}+{E}_{0}=2{E}_{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}H={H}_{0}+{H}_{0}=2{H}_{0}$ (35)
Substitute it to Equation (34), we have,
$-{∯}_{\text{Γ}}\left(E×H\right)\cdot \stackrel{^}{n}\text{dΓ}={\iiint }_{V}\left(E\cdot {J}_{0}+E\cdot {\partial }_{t}D+H\cdot {\partial }_{t}B\right)\text{d}V$ (36)
This is the Poynting theorem of the total electromagnetic fields $E,H$. ${J}_{0}$ is the current intensity of the source. Please notice we have assumed that the absorbers are uniformly distributed on the big sphere. In this situation, the total electromagnetic field which are sum of the retarded wave from the emitter and the advanced wave from the absorber satisfy the Poynting theorem. If this electromagnetic field satisfies Poynting theorem, it also satisfies the Maxwell equations. It seems that all the electromagnetic fields are started from the source. Hence, we have proved that the electromagnetic fields of a point source together with the fields of the absorbers at big sphere with infinite radius satisfy Maxwell equations. This field looks like the retarded field. Here it only looks like the retarded field. The fields $E,H$ actually are the summation of the retarded field radiate from the source and advanced wave radiate from the absorbers.
If the macroscopic electromagnetic fields of a point source with the field of the absorber satisfies Maxwell equations, we can extend this result to many point sources inside a region. It should be clear, here the total electromagnetic field are contributions of the retarded wave and the advanced wave, but it looks like as all the field are send as retarded wave from the emitter. Hence, for macroscopic wave, it still can be seen as retarded. Since the mutual energy flow are the energy flow of the photon, we have proved that the macroscopic wave can be built by many photons. This macroscopic wave can be seen as retarded wave even it is built from the mutual energy flows which are in turn built from retarded wave and advanced wave.
4.4. The Effect of the Normalized Mutual Energy Flow
In the above we have actually proved the total electromagnetic fields satisfy Maxwell equations. The total electromagnetic field has two parts one is the retarded field from the source, the other are advanced wave from the uniformly distributed absorbers on a big sphere. Normalized mutual energy flow is the photon energy, which is much stronger than the mutual energy flows calculated through classical electromagnetic field theory without the normalization. The number of the normalized mutual energy flows are also much less the mutual energy flows calculated from classical electromagnetic field theory. If the number of normalized mutual energy flows are enough more, they will have same effect as the mutual energy flows calculated through the classical electromagnetic field theory (which has much smaller values, but much larger in numbers).
If the absorbers could be seen as uniformly distribute at big sphere, the advanced waves of the total absorbers will have the same value as the retarded wave. Hence, the above proof in subsection 4.3 is also effective for the case of normalized electromagnetic fields, which is the field of photons.
4.5. The Structure of the Macroscopic Electromagnetic Fields
According to the discussion of this article, the structure of the electromagnetic fields can be shown in Figure 2. From these, we can see the macroscopic electromagnetic fields have a complicated structure. The macroscopic electromagnetic fields are built by infinite mutual energy flows. The effect of
Figure 2. The structure of the electromagnetic field. In this figure the energy is sent from the emitter to the absorbers. Assume that the absorbers are uniformly distribute at big sphere with infinite radius. The emitter sends retarded wave and corresponding time-reversal wave. These two energy flows cancel. The absorber sends advanced wave and corresponding time-reversal wave. These two energy flows cancel. The retarded wave and the advanced wave interacted and produce the microscopic mutual energy flow. Normalized mutual microscopic energy flows are photons. Many photons have the same effect with many microscopic mutual energy flows. Macroscopic mutual energy flows have the same values as microscopic mutual energy flow. Macroscopic mutual energy flows can build the retarded solution of the Maxwell equations in case the absorber can be seen as uniformly distribute on the big sphere with infinite radius.
the mutual energy flow is same as infinite photons. It should be clear that the mutual energy flows are much smaller than the photon. However, when both mutual energy flows and photons have very big number, they are equivalent. In order to build the macroscopic wave to satisfy Maxwell equations, the absorbers have to be uniformly distributed on the big sphere with an infinite radius. Photons are the normalized mutual energy flows. After the normalization, the mutual energy flow has the energy of $E=\hslash \omega$. The mutual energy flows are built by the retarded wave sent from the emitter and the advanced wave sent from the absorber. The emitter also radiates the time-reversal wave which cancel the energy flow of the retarded wave. The absorber also radiates the time-reversal wave which cancel the energy flow of the advanced wave.
This figure tells us how macroscopic wave of electromagnetic field is built by photons which are normalized mutual energy flows. The mutual energy flow is the interaction of the retarded wave radiate from the source and the advanced wave radiate from the sink. The self-energy flow of the retarded wave and the advanced wave has been canceled by the corresponding time-reversal waves. Hence the waves, retarded and advanced wave, can be seen as probability wave, because they do not carry energy, if we consider the effect of the time-reversal waves. This means the microscopic waves are only look like probability wave, they are also real physic waves.
5. In Case the Absorber Cannot Be Seen as Uniformly Distributed
In order to test the above theory, we noticed that the above theory is only satisfied in case the absorbers can be seen as uniformly distributed in big sphere with infinite radius. In case this condition cannot be satisfied, the result should be not obtained. That means the macroscopic wave of many photons should not satisfy Maxwell equations. This is true. For example, we assume there is a transmitting antenna radiate the wireless wave. We assume the environment can be seen as uniformly distributed absorbers. Hence, according to the above theory, the electromagnetic fields includes the retarded wave sent by the transmitting antenna and the advanced wave sent by the uniformly distributed absorbers together can be seen as retarded waves sent out from the transmitting antenna. Now we put a receiving antenna inside the above space. It is clear that when the receiving antenna is put in the system, closing to the receiving antenna, the absorbers cannot be seen as uniformly distributes on the big sphere. This means that closing to the receiving antenna, the electromagnetic field should not satisfy the Maxwell equations of the retarded wave. This is true. We know that for the receiving antenna, there are a concept of effective scattering section area. If effective scattering section area equals to the section area of the antenna, it tells us the field close to the antenna exactly satisfies the Poynting theorem. If the effective scattering section area larger than the section area of the receiving antenna, that means the Poynting theorem has not been exactly satisfied. We know from electronic engineering, for the receiving wire antenna, the effective scattering section area can be a hundred times larger than its section area. This means in this case the Poynting theorem cannot be satisfied. Since the Poynting theorem is equivalent to the two Maxwell equations Equation (20, 21), that means the Maxwell equations are also not satisfied.
In electronic engineering, the directivity dialog of receiving antenna cannot be calculated according to the theory of the retarded wave of Maxwell equations. Through the experiment we find the directivity dialog of the receiving antenna is same as the antenna used as the transmitting antenna. Hence engineers calculate the directivity dialog of the receiving antenna just using the same method as it is a transmitting antenna. This way it needs a proof that the directivity dialog of the receiving antenna is same as that of the transmitting antenna. Normally Lorentz reciprocity theorem is used for this task. It should be clear that the mutual energy flow theorem can also do this job. Since, in the directivity dialog is absolute value, and the phase information is omitted, and hence have the same value for the mutual energy theorem and Lorenz reciprocity theorem. Hence, the directivity dialog calculation for the receiving antenna is also because of the mutual energy theorem not Poynting theorem and also not Maxwell equations.
6. Conclusions
It is found that photon can be described by the normalized mutual energy flow. The mutual energy flow is the interaction of the retarded wave radiated from the source and the advanced wave radiated from the absorber. The normalization process is a process of physics which cannot be put in the frame of Maxwell theory. It is a pure quantum mechanics effect.
After the normalization of the mutual energy flow, the mutual energy flow can be a good description of photon. The structure of the macroscopic electromagnetic field becomes clear; the macroscopic electromagnetic field is built by many photons. The effect of many photons is same as many mutual energy flows without the normalization. The mutual energy flow is built by the retarded wave and the advanced wave. The self-energy flow of the retarded wave and the advanced wave have canceled by the corresponding energy flow of the time-reversal waves of the retarded wave and the advanced wave.
With the theory of mutual energy flow and this normalization, the concept of collapse of the wave is not required. The wave and particle duality paradox have been solved.
Cite this paper: Zhao, S. (2020) Photon Can Be Described as the Normalized Mutual Energy Flow. Journal of Modern Physics, 11, 668-682. doi: 10.4236/jmp.2020.115043.
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Properties
Label 4608.2.c.c Level $4608$ Weight $2$ Character orbit 4608.c Analytic conductor $36.795$ Analytic rank $1$ Dimension $2$ CM no Inner twists $2$
Learn more
Newspace parameters
Level: $$N$$ $$=$$ $$4608 = 2^{9} \cdot 3^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 4608.c (of order $$2$$, degree $$1$$, minimal)
Newform invariants
Self dual: no Analytic conductor: $$36.7950652514$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-2})$$ Defining polynomial: $$x^{2} + 2$$ x^2 + 2 Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
$q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \sqrt{-2}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + 2 \beta q^{5} + \beta q^{7}+O(q^{10})$$ q + 2*b * q^5 + b * q^7 $$q + 2 \beta q^{5} + \beta q^{7} - 2 q^{13} - \beta q^{17} - 2 \beta q^{19} - 6 q^{23} - 3 q^{25} - 4 \beta q^{29} + 7 \beta q^{31} - 4 q^{35} + 2 q^{37} + 3 \beta q^{41} - 6 \beta q^{43} - 10 q^{47} + 5 q^{49} + 4 \beta q^{53} - 12 q^{59} - 10 q^{61} - 4 \beta q^{65} - 4 \beta q^{67} - 2 q^{71} - 10 q^{73} - 9 \beta q^{79} - 4 q^{83} + 4 q^{85} + 7 \beta q^{89} - 2 \beta q^{91} + 8 q^{95} + 12 q^{97} +O(q^{100})$$ q + 2*b * q^5 + b * q^7 - 2 * q^13 - b * q^17 - 2*b * q^19 - 6 * q^23 - 3 * q^25 - 4*b * q^29 + 7*b * q^31 - 4 * q^35 + 2 * q^37 + 3*b * q^41 - 6*b * q^43 - 10 * q^47 + 5 * q^49 + 4*b * q^53 - 12 * q^59 - 10 * q^61 - 4*b * q^65 - 4*b * q^67 - 2 * q^71 - 10 * q^73 - 9*b * q^79 - 4 * q^83 + 4 * q^85 + 7*b * q^89 - 2*b * q^91 + 8 * q^95 + 12 * q^97 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q+O(q^{10})$$ 2 * q $$2 q - 4 q^{13} - 12 q^{23} - 6 q^{25} - 8 q^{35} + 4 q^{37} - 20 q^{47} + 10 q^{49} - 24 q^{59} - 20 q^{61} - 4 q^{71} - 20 q^{73} - 8 q^{83} + 8 q^{85} + 16 q^{95} + 24 q^{97}+O(q^{100})$$ 2 * q - 4 * q^13 - 12 * q^23 - 6 * q^25 - 8 * q^35 + 4 * q^37 - 20 * q^47 + 10 * q^49 - 24 * q^59 - 20 * q^61 - 4 * q^71 - 20 * q^73 - 8 * q^83 + 8 * q^85 + 16 * q^95 + 24 * q^97
Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/4608\mathbb{Z}\right)^\times$$.
$$n$$ $$2053$$ $$3583$$ $$4097$$ $$\chi(n)$$ $$1$$ $$-1$$ $$-1$$
Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
4607.1
− 1.41421i 1.41421i
0 0 0 2.82843i 0 1.41421i 0 0 0
4607.2 0 0 0 2.82843i 0 1.41421i 0 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
12.b even 2 1 inner
Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 4608.2.c.c 2
3.b odd 2 1 4608.2.c.d yes 2
4.b odd 2 1 4608.2.c.d yes 2
8.b even 2 1 4608.2.c.e yes 2
8.d odd 2 1 4608.2.c.f yes 2
12.b even 2 1 inner 4608.2.c.c 2
16.e even 4 2 4608.2.f.l 4
16.f odd 4 2 4608.2.f.j 4
24.f even 2 1 4608.2.c.e yes 2
24.h odd 2 1 4608.2.c.f yes 2
48.i odd 4 2 4608.2.f.j 4
48.k even 4 2 4608.2.f.l 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
4608.2.c.c 2 1.a even 1 1 trivial
4608.2.c.c 2 12.b even 2 1 inner
4608.2.c.d yes 2 3.b odd 2 1
4608.2.c.d yes 2 4.b odd 2 1
4608.2.c.e yes 2 8.b even 2 1
4608.2.c.e yes 2 24.f even 2 1
4608.2.c.f yes 2 8.d odd 2 1
4608.2.c.f yes 2 24.h odd 2 1
4608.2.f.j 4 16.f odd 4 2
4608.2.f.j 4 48.i odd 4 2
4608.2.f.l 4 16.e even 4 2
4608.2.f.l 4 48.k even 4 2
Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(4608, [\chi])$$:
$$T_{5}^{2} + 8$$ T5^2 + 8 $$T_{7}^{2} + 2$$ T7^2 + 2 $$T_{11}$$ T11 $$T_{13} + 2$$ T13 + 2 $$T_{23} + 6$$ T23 + 6
Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2} + 8$$
$7$ $$T^{2} + 2$$
$11$ $$T^{2}$$
$13$ $$(T + 2)^{2}$$
$17$ $$T^{2} + 2$$
$19$ $$T^{2} + 8$$
$23$ $$(T + 6)^{2}$$
$29$ $$T^{2} + 32$$
$31$ $$T^{2} + 98$$
$37$ $$(T - 2)^{2}$$
$41$ $$T^{2} + 18$$
$43$ $$T^{2} + 72$$
$47$ $$(T + 10)^{2}$$
$53$ $$T^{2} + 32$$
$59$ $$(T + 12)^{2}$$
$61$ $$(T + 10)^{2}$$
$67$ $$T^{2} + 32$$
$71$ $$(T + 2)^{2}$$
$73$ $$(T + 10)^{2}$$
$79$ $$T^{2} + 162$$
$83$ $$(T + 4)^{2}$$
$89$ $$T^{2} + 98$$
$97$ $$(T - 12)^{2}$$
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# Rotate an object with 'Quaternion.Slerp' in world space
I'm currently working on a multiplayer skydiving Unity game in which i rotate the players like this:
transform.rotation = Quaternion.Slerp(transform.rotation, desiredRotation, delta);
Now this rotates the player relative to it's own rotation. I know rotation in world space is done by multiplying the quaternion of the desired positon with the quaternion of the current position like this:
localRotation = transform.rotation * desiredRotation
worldRotation = desiredRotation * transform.rotation
But how do i slerp to that position in world space?
Thank you all in advance and have a great day!
## 2 Answers
First you grab the total rotation quaternion.
fullRelativeRotation = Quaternion.inverse(transform.rotation)*desiredRotation;
rotationTime = 0;
//persist over frames
then every frame until you complete the rotation you slerp the identity quaternion with it.
if(rotationTime < 1){
rotationThisFrame = Quaternion.Slerp(Quaternion.identity, fullRelativeRotation, min(delta, 1-rotationTime));
rotationTime+=delta;
transform.rotation = transform.rotation * rotationThisFrame;
}
• Some remarks. 1: You can store the rotationThisFrame outside the if clause. It does not depend on anything that changes with time. 2: If, on the other hand, you'd used the rotationTime and stored the initial and final rotations, then recomputing the quat on the slerp path every frame would have been justified. But, in that case, you would have simply set transform.rotation = Quaternion.Slerp(..). Sep 7, 2017 at 9:30
Usually, in order to get from a local coordinate frame to world coordinates, you need to go through the rotation hierarchy, like worldToParent.rotation * parentToLocal.rotation = worldToLocal.rotation.
So, in order to rotate towards a desired orientation in world space, your computation look something like transform.rotation = Quaternion.Slerp(worldToLocal.rotation, desiredRotation, delta). Note that desiredRotation in this case is not transformed, as it is a relative rotation anyways and will thus be applied within the same coordinate frame as the base rotation (in this case worldToLocal, defined as above, which resides in the world frame).
• Maybe, to be safer/clearer, mention that the user must not tamper with the worldToLocal and desiredRotation amounts during the rotation sequence of frames. They are set prior to starting the rotation itself. Sep 7, 2017 at 9:34
• Also many thanks to you! Your solution also worked. Sep 7, 2017 at 21:25
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# statsmodels.genmod.families.family.InverseGaussian¶
class `statsmodels.genmod.families.family.``InverseGaussian`(link=<class 'statsmodels.genmod.families.links.inverse_squared'>)[source]
InverseGaussian exponential family.
Notes
The inverse Guassian distribution is sometimes referred to in the literature as the Wald distribution.
Attributes
InverseGaussian.link (a link instance) The link function of the inverse Gaussian instance InverseGaussian.variance (varfunc instance) variance is an instance of statsmodels.family.varfuncs.mu_cubed
Methods
`deviance`(endog, mu[, freq_weights, scale]) Inverse Gaussian deviance function `fitted`(lin_pred) Fitted values based on linear predictors lin_pred. `loglike`(endog, mu[, freq_weights, scale]) The log-likelihood function in terms of the fitted mean response. `predict`(mu) Linear predictors based on given mu values. `resid_anscombe`(endog, mu) The Anscombe residuals for the inverse Gaussian distribution `resid_dev`(endog, mu[, scale]) Returns the deviance residuals for the inverse Gaussian family. `starting_mu`(y) Starting value for mu in the IRLS algorithm. `variance` `weights`(mu) Weights for IRLS steps
Methods
`deviance`(endog, mu[, freq_weights, scale]) Inverse Gaussian deviance function `fitted`(lin_pred) Fitted values based on linear predictors lin_pred. `loglike`(endog, mu[, freq_weights, scale]) The log-likelihood function in terms of the fitted mean response. `predict`(mu) Linear predictors based on given mu values. `resid_anscombe`(endog, mu) The Anscombe residuals for the inverse Gaussian distribution `resid_dev`(endog, mu[, scale]) Returns the deviance residuals for the inverse Gaussian family. `starting_mu`(y) Starting value for mu in the IRLS algorithm. `weights`(mu) Weights for IRLS steps
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http://www.matsusada.com/news/2016/05/column-amp.html
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## High Speed High Voltage Amplifier
2016.05.24
### HV Amplifier
High voltage amplifier converts input voltage to high voltage waveform as it is as shown in fig. 1. These days the demand of HV amplifier is growing more and more, and now becoming an indispensable tool for research and development, experiments and integrating to a system for such fields as electronics, physics, biochemical and medical industries. With high voltage technologies Matsusada Precision Inc. manufactures various HV amplifiers to meet all requirements from customers.
* In addition to these models in this catalog we have amplifiers developed specially for electrostatic chuck or PZT. Please ask for details to our sales staff.
HV amplifier is generally equipped with the "sink" function for output currents that provides constant voltage operation without regard to the type of load whether it is capacitive or conductive. (Fig.2) As it gives fast response, it is an ideal power supply for applications which require AC output.
Matsusada HV amplifiers are all bi-polar type and can be operated in full four-quadrant area.(I · II · III · IV area)
### Slew Rate
The responsibility of our high speed amplifier is determined with slew rate(SR). The step responsibility of our amplifier is as shown in fig.3.
SR = V / μS In case of output amplitude is smaller the response time become shorter.
AMP series reach to greater than SR =700V / μS at maximum.
### Rise Time(step response)
Step response can be indicated with rise time. (fig.4) Usually the rise time of amplifier of response (= bandwidth)fc (Hz) is given by a formula below.
tr ≒ 0.35 / fc
The fall time tf is equals to tr.
### Frequency Response
Response of Matsusada amplifiers are described as "frequency bandwidth". When swing the output with sinusoidal waveform with rated resistive load, output swing (amplitude) is reduced as input frequency become faster. Frequency response in the specification is the frequency fc is where output swing is 70% (-3dB). (fig. 5)
In case clear output waveform is required, please select a HV amplifier which has high enough frequency bandwidth against required frequency.
In general 3 to 5 times more frequency bandwidth for sinusoidal waveform, and about 10 times more for rectangular waveform, is required. In case of insufficient frequency bandwidth the output swing shall be reduced, and also the phase difference be large, so some solutions, such as monitoring output waveform, shall be required.
When operated with a capacitive load of more than 100pF(including floating capacity of output cable), the output might oscillate. In such a case put a high voltage resister of 100Ω(0.1μF) to 1kΩ(1000pF) in series to output. Please also note that with capacitive load the bandwidth is limited according to the formula of right note.
In case of corona discharge, larger current than rating can draw and night damage power supply.
In this case too place an output resister to limit the current as same as in case of with capacitive load.
Important note to utilize the full performance of high speed HV amplifier
Output cable of HV amplifiers is not shielded. If the output cable has some stray capacity against ground(earth ground or metal objects), output voltage will be sinusoidal or step waveform and extra current will be drawn. As this current draw parallel to load, the following appearance might be happened.
News & Events
Column
## Latest Articles
Matsusada's Total Solution for MASS Spectrometry!
Matsusada's collection of PMT precise HV modules!!
High voltage power supplies for SEM / MASS
E-chuck power supplies
High Speed High Voltage Amplifier
Product Search
DC Power Supplies/
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AC Power Supplies/
X-Ray Inspection System/
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Analytical Instruments
Piezo Drivers
News
News & Events
Column
Company
Company Details
Mission
History
Offices
Career
Career
Contact
Any question?
Catalog request
Model selection assistance
Price information
OEM custom design
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Change language to:
English - Français - 日本語 -
Please note that the recommended version of Scilab is 2023.1.0. This page might be outdated.
See the recommended documentation of this function
Ajuda Scilab >> Funções de Compatibilidade > mtlb_min
# mtlb_min
Matlab min emulation function
### Description
Matlab and Scilab `min` behave differently in some particular cases:
• With complex values: Matlab `min` can be used with complex values but not Scilab function.
• When called with one input: Matlab `min` threats values along the first non-singleton dimension but Scilab threats all input values.
• When called with two inputs: if one is an empty matrix, Scilab returns an error message but Matlab returns [].
• When called with three inputs: if `dim` parameter is greater than number of dimensions of first input, Scilab returns an error message and Matlab returns the first input.
The function `[r[,k]] = mtlb_min(A[,B[,dim]])` is used by `mfile2sci` to replace `[r[,k]] = min(A[,B[,dim]])` when it was not possible to know what were the inputs while porting Matlab code to Scilab. This function will determine the correct semantic at run time. If you want to have a more efficient code it is possible to replace `mtlb_min` calls:
• When called with one input, if `A` is a vector or a scalar `[r[,k]] = mtlb_min(A)` may be replaced by `min(A)`
• When called with one input, if `A` is a matrix `[r[,k]] = mtlb_min(A)` may be replaced by `min(A,"r")`
• When called with two inputs, if `A` and `B` are real matrices and not empty matrices `[r[,k]] = mtlb_min(A,B)` may be replaced by `min(A,B)`
• When called with three inputs, if `dim` is lesser than the number of dimensions of `A` `[r[,k]] = mtlb_min(A,[],dim)` may be replaced by `min(A,dim)`
Caution: `mtlb_min` has not to be used for hand coded functions.
### Authors
• V.C.
<< mtlb_max Funções de Compatibilidade mtlb_more >>
Copyright (c) 2022-2023 (Dassault Systèmes)Copyright (c) 2017-2022 (ESI Group)Copyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Thu May 12 11:45:28 CEST 2011
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## Comment on Managing your Script
commented on 05 May 2013, 04:31 PM
ok sir. i got mail. so i thought to where i have to send.
here is the formula sir
this formula is for intraday purpose and based on previous day close high and low and current day open
this is for nifty fut
this is normal buy when current day open is near to buy area +/- 10 points
buy =close – (0.275*(high-low))
tgt = close + (0.275*(high-low))
sl for this buy , sl sell = (close – (0.55*(high-low))) – 5 points
but if this sl hits means that is called sell breakout and then it will lead to
tgt1 =close – (0.786*(high-low))
tgt 2=close – (1.45*(high-low))
this sell breakout is sure trade on down side
for normal sell when current day open is near to sell area +/- 1- points
sell = close + (0.275*(high-low))
tgt = close – (0.275*(high-low))
sl for this sell is
buy = (close – (0.55*(high-low))) + 5points
if this sl hits means that is called buy breakout and then it will lead to
tgt 1 = close + (0.7773*(high-low))
tgt 2 =close + (1.41*(high-low))
and this buy breakout happends then it is sure trade in up side
please code it for me sir…….
thanking you
ramzamma
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### Data structure using c module 1
1. 1. Data Structure using ‘C’ Module - I Prepared by: Smruti Smaraki Sarangi Asst. Professor School of Computer Application IMS Unison University, Dehradun
2. 2. Topics Covered: Introduction to Data Structure Algorithm and Asymptotic Notation Array Sparse Matrix Stack Queue Linked List Polynomial Representation Dynamic Storage Management Compaction and Garbage Collection
3. 3. Introduction to Data Structure: It is the logical or mathematical model of a particular organization of data. It is the actual representation of data and the algorithms to manipulate the data. A programming construct used to implement Abstract Data Type (ADT). Abstract data type is a mathematical model or concept that defines a datatype correctly. It only mentions what operations are to be performed but not how these operations will be implemented. A data type is a term which refers to the kind of data that a variable may hold.
4. 4. Fundamental Concepts: Data: data are the raw facts. Information: Processed data is known as information. Entity: It is a thing or object in real world that distinguishable from all other objects. It is represented by set of attributes. Record: Collection of field values of a given entity. File: Collections of Records
5. 5. Types of Data Structure: Data Structure Non-primitive Linear Array Stack Queue Linked list Non-linear Tree GraphPrimitive
6. 6. Primitive and Non-Primitive Data Structure: Primitive data structures are directly operated on primary data types like int, char, etc. Non-Primitive data structures are operated on user-defined and derived datatype. It is of two types. Linear data structure: elements are in sequence or in linear list. E.g. array, stack, queue, linked list. Non-linear data structure: elements represented in non-sequential order. E.g. tree and graph.
7. 7. Operation on Data Structure: The operation performed on data structure are: Creation Insertion Deletion Search Traversal Sorting Merging
8. 8. Memory Representation: Memory Representation of Data Structure Sequential Representation: Maintains data in continuous memory locations which takes less time to retrieve the data but leads to time complexity during insertion and deletion operations. Linked Representation: Maintains the list by means of a link or a pointer between the adjacency elements which need not be stored in continuous memory locations.
9. 9. Algorithm: An algorithm is a set of instructions to solve a particular problem in finite number of steps. It takes set of values as input and produce set of values as output. The characteristics are: Input Output Finiteness Definiteness Effectiveness
10. 10. Analysis of Algorithm: Analysis of Algorithm Time Complexity: Total amount of time taken by a computer to execute a program Worst case: Maximum amount of time and space. Average case: In between best case and worse case. Best case: minimum amount of space and time Space Complexity: Total amount of memory taken by a computer to execute a program
11. 11. Asymptotic Notation: It is the notation which we use to describe the asymptotic running time of an algorithm are defined in terms of functions based on the set of numbers Natural number set N = {0, 1, 2, ….} Positive Integer set N+ = {1, 2, 3,…} Real number set R Positive real number set R+ Non-negative real number set R*
12. 12. Types of Notations: Types of Notations Big Oh (O) Notation Small Oh (o) Notation Big omega (Ω) Notation Small omega (ω) Notation Theta (θ) Notation
13. 13. Big oh (O) Notation: Let f(n) and g(n) are two functions from set of natural numbers to set of non-negative real numbers and f(n) = O(g(n)), iff there exist a natural number ‘n0’ and a positive constants c>0, such that, f(n) ≤ c(g(n)), for all n ≥ n0 E.g. f(n) = 2n2+7n-10, n=5, c=3. Show that this is in O-notation. Here g(n) = n2 f(n) ≤ c(g(n)) => 2n2+7n-10 ≤ cn2 => 2*25+7*5-10 ≤ 3*25 => 75 ≤ 75 So it is in O – Notation, that is O(n2)
14. 14. Small Oh(o) Notation: Let f(n) and g(n) are two functions from set of natural numbers to set of non-negative real numbers and f(n) = o(g(n)), iff there exist a natural number n0 > 0 and a positive constants c > 0, such that, f(n) < c(g(n)), for all n > n0 The another formula for this is: 𝐥𝐢𝐦 𝒏→∞ 𝒇 𝒏 𝒈 𝒏 = 𝟎
15. 15. Big omega (Ω) Notation: Let f(n) and g(n) are two functions from set of natural numbers to set of non-negative real numbers and f(n) = Ω(g(n)), iff there exist a natural number ‘n0’ and a positive constants c>0, such that, f(n) ≥ c(g(n)), for all n ≥ n0 E.g. f(n) = n2+3n+4, n=1, c=1. Show that this is in Ω-notation. Here g(n) = n2 f(n) ≥ c(g(n)) => n2+3n+4, ≥ cn2 => 1+3+4 ≥ 1 => 8 ≥1 So it is in Ω – Notation, that is Ω(n2)
16. 16. Small omega(ω) Notation: Let f(n) and g(n) are two functions from set of natural numbers to set of non-negative real numbers and f(n) = ω(g(n)), iff there exist a natural number n0 > 0 and a positive constants c > 0, such that, f(n) > c(g(n)), for all n > n0 The another formula for this is: 𝐥𝐢𝐦 𝒏→∞ 𝒇 𝒏 𝒈 𝒏 = ∞
17. 17. Theta (θ) Notation: For a given function g(n), we denoted by θ(g(n)), the set of functions f(n) = θ(g(n)), iff there exist some constant c1, c2 and n0, such that, c1(g(n)) ≤ f(n) ≤ c2(g(n)), for all n ≥ n0 The another formula for this is: 𝐥𝐢𝐦 𝒏→∞ 𝒇 𝒏 𝒈 𝒏 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒄
18. 18. Array: It is a collection of similar data elements present in continuous memory locations referred by a unique name. It can be one dimensional, two dimensional and multidimensional array. Two dimensional array has 2 subscripts one for row and other for column. It is also called matrix.
19. 19. One-Dimensional Array: A linear array is a list of finite number ‘n’ of homogeneous data element, such that: The elements of the array are referred respectively by an index set consisting of ‘n’ consecutive numbers. The elements of the array are stored respectively in successive memory location. It can be defined as: datatype arrayname[size];
20. 20. The number ‘n’ of elements is called the length or size of the array. The length of an array can be obtained from the index by the formula: Length = Upperbound(UB) – Lowerbound(LB)+1 where UB = Largest index and LB = Smallest Index of array. When Lowerbound = 1, then Length = Upperbound(UB)
21. 21. Operations of 1D Array: The operations of one dimensional array are: Creation: create or entering elements into array. Traversal: processing each element Search: searching for an element Insertion: inserting an element at required position Deletion: deletion of an element Sorting: arranging in ascending or descending order Merging: concatenate two ways into 3rd array or merge directly into 3rd array where two arrays are already sorted.
22. 22. Creation of an Array: Algorithm: CREATE(LA, LB, UB) Step 1: Let LA[10], LB,UB, K Step 2: LB = 1, UB = 10 Step 3: Display “enter 10 elements” Step 4: Repeat, for K = LB to UB incremented by INPUT LA[K] [End of for] Step 5: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound)
23. 23. Traversing an Array: Algorithm: TRAVERSE(LA,LB,UB) Step 1: Repeat for K = LB to UB by 1 Apply PROCESS to LA[K] Step 2: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound)
24. 24. Searching for an element in an Array: Algorithm: SEARCH(LA,LB,UB,ITEM) Step 1: Let K, Set K = LB Step 2: Repeat, while(K < = UB) •Step 2.1: If LA[K] = ITEM, then Display “Item found at”, K [End of if] Step 2.2: Set K = K + 1 (End of while) Step 3: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound)
25. 25. Insertion of new element in an Array: Algorithm: INSERTION(LA,LB,UB,ITEM,POS) Step 1: Let K Step 2: Repeat for K = UB to POS, decreasing by 1 LA[K + 1] = LA[K] [End of for] Step 3: LA[POS] = ITEM Step 4: Set LB = UB + 1 Step 5: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound, POS = position)
26. 26. Deletion of an element from Array: Algorithm for deletion when position is given: DELETION_POS(LA, LB,UB,POS) Step 1: Let K Step 2: Repeat for K = POS to UB incremented by 1 LA[K] = LA[K + 1] [end of for] UB = UB - 1 Step 3: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound, POS = position)
27. 27. Deletion of an element from Array: Algorithm for deletion when item is given: DELETION_ITEM(LA, LB,UB,ITEM) Step 1: Let K, S = 0, T Step 2: Repeat for K = LB to UB incremented by 1 Step 2.1:If (LA[K] = ITEM), then Step 2.1.1: Display “ITEM found at”, K Step 2.1.2: Repeat for T = K to UB, incremented by 1 Step 2.1.2.1: LA[T] = LA[T + 1] [End of for] Step 2.1.3: UB = UB - 1 Step 2.1.4: S-1 [end of if] [End of for] Step 3: If (S=0), then Step 3.1: Display “ITEM to be deleted is not found” [End of if] Step 4: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound)
28. 28. Sorting the elements of an Array: Algorithm: SORTING(LA,LB,UB) Step 1: Let I, J, K, T Step 2: Set I = 1, J = 2 Step 3: Repeat for K = UB – 1 to 1 by decrementing by 1 Step 3.1: Repeat for I = 1, J = 2 to K by 1 Step 3.1.1: If LA[I] > LA[J], then T = LA[I] LA[I] = LA[J] LA[J] = T [End of if] [End of for] [End of for] Step 4: Exit (LA: linear array, LB: Lower Bound, UB: Upper Bound)
29. 29. Two-Dimensional Array: Two dimensional array is the array which is represented by two dimensions that is row and column. The declaration of two dimensional array is: datatype arrayname[row size][column size]; The two dimensional array is also called as Matrix.
30. 30. Memory Representation of 2D Array: The programming language will store the two dimensional array ‘A’ in two different ways. That is: Row-Major Order (Row by Row) Column- Major Order (Column by Column)
31. 31. Row-Major Order (Row by Row): In Row–Major order, elements of 2D array are stored on a row-by-row basis beginning from first row. Address of an element represented in row-major order can be calculated by using the following formula: Address of A[i][j] = B + W * [(N * (i - LBR)) +(j - LBC)] If LBR = LBC = 0, then the formula can be represented as: Address of A[i][j] = B + W* (N * i + j)]
32. 32. Example: [Row-Major Order Representation] B = base address = 100, W = datatype value of the array N = size of column, i = row index value, j = column index value LBR = lower bound row, LBC = lower bound column Address of A[1][3] = 100 + 2 * [(4 * 1 – 0) + (3 – 0)] = 100 + 2 * [(4 * 1) + 3] = 100 + 2 * 7 = 114 7 9 20 1 11 5 2 17 13 15 21 19 Data in array 0,0 0,1 0,2 0,3 1,0 1,1 1,2 1,3 2,0 2,1 2,2 2,3 Row and column values
33. 33. Column-Major Order (Column by Column): In Column–Major order, elements of 2D array are stored on a column-by-column basis beginning from first column. Address of an element represented in column-major order can be calculated by using the following formula: Address of A[i][j] = B + W * [(i - LBR) + M * (j - LBC)] If LBR = LBC = 0, then the formula can be represented as: Address of A[i][j] = B + W* (i + M * j)]
34. 34. Example: [Column-Major Order Representation] B = base address = 100, W = datatype value of the array M = size of row, i = row index value, j = column index value LBR = lower bound row, LBC = lower bound column Address of A[1][3] = 100 + 2 * [(1 – 0) +3 * (3 – 0)] = 100 + 2 * [1 + 9] = 100 + 20 = 120 7 11 13 9 5 15 20 2 21 1 17 19 Data in array 0,0 0,1 0,2 1,0 1,1 1,2 2,0 2,1 2,2 3,0 3,1 3,2 Column and Row values
35. 35. Table Representation of above example: Row Major Order Address Column Major Order Address A[0][0] 100 A[0][0] 100 A[0][1] 102 A[1][0] 102 A[0][2] 104 A[2][0] 104 A[0][3] 106 A[0][1] 106 A[1][0] 108 A[1][1] 108 A[1][1] 110 A[2][1] 110 A[1][2] 112 A[0][2] 112 A[1][3] 114 A[1][2] 114 A[2][0] 116 A[2][2] 116 A[2][1] 118 A[0][3] 118 A[2][2] 120 A[1][3] 120 A[2][3] 122 A[2][3] 122
36. 36. Sparse Matrix: A matrix with relatively high proportion of zero or null entries is called sparse matrix. A M x N matrix is said to be sparse matrix if many of its elements are zero. 0 0 0 24 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 18 0 0 0 0 0 0 8 0 0 6x7 It is a sparse matrix of 6 x 7 order with 5 non-zero elements out of 42 elements.
37. 37. Array Representation of Sparse Matrix: The alternate data structure that we consider to represent a sparse matrix is a triplet. The triplet is a 2D array having t+1rows and 3 columns, where t is the total number of non-zero entries. The first row of the triplet contains number of rows, columns and non-zero entries available in the matrix in its 1st, 2nd, and 3rd column respectively.
38. 38. In addition, we need to record the size of matrix, that is number of rows and number of columns and non-zero elements. For this purpose, the first element of array of triplets is used where the first field stores number of rows and second field stores number of columns and third field stores number of non- zero elements.
39. 39. Possible Operations of Sparse Matrix: To store the non-zero entries of a sparse matrix into alternate triplet representation. To display the original matrix from a given triplet representation of matrix. To store transpose of the triplet representation. To display the transpose of original sparse matrix using transpose of triplet.
40. 40. Example: Consider a matrix A[5][6], which contains zeros and non-zeros. Let NZ is a variable contains non- zeros count. 0 0 55 0 0 0 88 0 0 0 99 0 0 0 44 66 0 0 0 0 0 0 0 11 0 0 0 22 0 0 0 88 0 0 0 0 0 0 0 0 55 0 44 0 0 0 0 66 0 22 0 99 0 0 0 0 0 0 11 0 5 6 7 0 2 55 1 0 88 1 4 99 2 2 44 2 3 66 3 5 11 4 3 22 6 5 7 0 1 88 2 0 55 2 2 44 3 2 66 3 4 33 4 1 99 5 3 11 Sparse Matrix A[5][6] Triplet Matrix SP[8][3] Transpose of Triplet Matrix TP[8][3] Transpose of Original Matrix
41. 41. Algorithm for operations on Sparse Matrix: Step 1: Let A[10][10], SP[50[3], TP[5][3], I, J, M, N, NZ Step 2: NZ = 0 Step 3: Display “ Enter the row and column size”. Step 4: Input M,N Step 5: Repeat for I = 0 to M – 1 incremented by 1 Step 5.1: Repeat for J = 0 to N – 1 incremented by 1 Input A[I][J] if A[I][J] != 0, then NZ = NZ + 1 [End of if] [End of Step 5.1]
42. 42. Step 6: if NZ > = (M * N)/2, then Display “ Matrix is not Sparse”. Step 7:else Display “ Matrix is Sparse”. Step 7.1: Call ALTERNATE(A, SP, M, N, NZ) Step 7.2: Call DISPLAY_ORIGINAL(SP, M, N) Step 7.3: Call TRANSPOSE(SP, TP, NZ, N) Step 7.4: Call DISPLAY_TRANSPOSE(TP,M,N,NZ) Step 8: Exit
43. 43. Algorithm to Represent a Sparse Matrix to Triplet Matrix: ALTERNATE(A, SP, M, N, NZ) Step 1: Let I, J, K = 1 Step 2: SP[0][0] = M, SP[0][1] = N, SP[0][2] = NZ Step 3: Repeat for I = 0 to M – 1 incremented by 1 Step 3.1: Repeat for J = 0 to N – 1 incremented by 1 if A[I][J] != 0 then SP[K][0] = I SP[K][1] = J SP[K][2] = A[I][J] K = K + 1 [End of if] [End of Step 3.1] Step 4: Repeat for K = 0 to NZ incremented by 1 Display SP[K][0], SP[K][1], SP[K][2] [End of for] Step 5: Exit
44. 44. Algorithm to display original matrix using triplet: DISPLAY_ORIGINAL(SP, M, N) Step 1: Let I, J, K Step 2: K = 1 Step 3: Repeat for I = 0 to M – 1 increased by 1 Step 3.1: Repeat for J = 0 to N – 1 increased by 1 if(SP[K][0] = I AND SP[K][1] = J), then Display SP[K][2] K = K + 1 else Display “0” [End of if] [End of for step 3.1] [End of for step 3] Step 4: Exit
45. 45. Transpose of triplet matrix into alternate matrix: TRANSPOSE(SP, TP, NZ, N) Step 1: Let J, K, T = 1 Step 2: TP[0][0] = SP[0][1], TP[0][1] = SP[0][0], TP[0][2] = SP[0][2]. Step 3: Repeat for I = 0 to M – 1 incremented by 1 Step 3.1: Repeat for J = 0 to N – 1 incremented by 1 if(SP[K][1] = J), then TP[T][0] = SP[K][1] TP[T][1] = SP[K][0] TP[T][2] = SP[K][2] T = T + 1 [End of if] Step 4:Repeat for T = 0 to NZ – 1 incremented by 1 Display TP[T][0], TP[T][1], TP[T][2] [End of for] Step 5: Exit
46. 46. Display transpose of original matrix using transpose of triplet: DISPLAY_TRANPOSE(TP, M, N, NZ) Step 1: Let I, J, T Step 2: T = 1 Step 3: Repeat for J = 0 to N – 1 incremented by 1 Step 3.1: Repeat for I = 0 to M – 1 incremented by 1 if(TP[T][0] = J AND TP[T][1] = I), then Display TP[T][2] T = T + 1 else Display “0” [End of if] [End of for step 3.1] [End of for step-3] Step 4: Exit
47. 47. STACK: It is a non-primitive linear data structure, which works on the concept of Last In First Out (LIFO) or First in Last Out (FILO). It is an ordered list, where insertion and deletion can be done at one end only, which is called TOP element of the stack. The operations of stack are: PUSH(insertion), POP(deletion) and PEEP(display location) E.g. i) stack of plates one above the other. ii) stack of books kept one above the other.
48. 48. Memory Representation of Stack: Array Representation: An one dimensional array can be considered for representing a stack in memory. While representing a stack in the form of an array we take a variable TOP that identifies top most element of the stack. Initially, when the stack is empty TOP = -1. The operation of pushing an ITEM into a stack and the operation of removing an ITEM from the stack is done by PUSH and POP. In executing the procedure PUSH, one must first test whether there is room in the stack for new ITEM, if not then we have the condition known as “OVERFLOW”.
49. 49. The condition for overflow is TOP = SIZE – 1, where SIZE is the maximum number of elements that can be hold by the array that represents the stack in memory. In executing the procedure POP, we must first test whether there is an element in the stack to be deleted, if not then we have the condition known as “UNDERFLOW”. The condition for this is TOP = -1. Linked Representation: The concept of dynamic memory allocation is implemented by using linked list to represent stack in linked format.
50. 50. Algorithm for PUSH Operation: PUSH(STACK, TOP, SIZE, ITEM) Step 1: if (TOP = SIZE -1), then Display “Overflow” Exit Step 2: else TOP = TOP + 1 STACK[TOP] = ITEM [End of if] Step 3: Exit
51. 51. Algorithm for POP Operation: Step 1: Let ITEM Step 2: if (TOP < 0), then Display “Underflow” Exit Step 3: else ITEM = STACK[TOP] Display “popped item is”, ITEM TOP = TOP - 1 [End of if] Step 4: Exit
52. 52. Algorithm for PEEP Operation: Step 1: if TOP = -1, then print “Underflow” and return Step 2: Read the location value ‘i’ Step 3: Return the ith element from the top of the stack. Return(STACK(K[TOP – i + 1]) Algorithm for UPDATE Operation: UPDATE(STACK, ITEM) Step 1: if TOP = -1, then print “Underflow” and return Step 2: Read the location value ‘i’ Step 3: STACK[TOP – i + 1] Step 4: Return
53. 53. Applications of Stack: Stacks are used to pass parameter between functions on a call to a function the parameters and local variable are stored on a stack. High level programming language, such as ‘pascal’, ‘c’, that provides for recursion used stack for book keeping. That is: in each recursive call, there is need to save the current values of parameter local variables and return address.
54. 54. Conversion and Evaluation of Arithmetic Expression: Arithmetic Expression Notation: In any arithmetic expression, each operator is placed in between two operands i.e. mathematical expression is called as infix expression. E.g.: A + B. Apart from usual mathematical representation of an arithmetic expression, an expression can also be represented in the following two ways. i.e. i) Polish or Prefix Notation ii) Reverse Polish or Postfix Notation
55. 55. Mathematical Procedure for Conversion: Standard Arithmetic Operators and Precedence levels are: ^ (exponential): Higher level *, /, % : Middle Level +, - : Lower Level The possible conversions are: Infix to Prefix Prefix to Infix Infix to Postfix Postfix to Infix Prefix to Postfix Postfix to Prefix
56. 56. Infix to Prefix: Identify the innermost bracket Identify the operator according to the priority of evaluation. Represent the operator and corresponding operand in prefix notation Continue this process until the equivalent prefix expression is achieved. E.g.: (A + B * (C – D ^ E) / F) => (A + B * (C – [^DE]) / F) => (A + B * [ - C ^ DE] / F) => (A + [* B – C ^ DE] / F) => A + [/ * B – C ^ DEF] => + A / * B – C ^ DEF
57. 57. Prefix to Infix: Identify the operator from right to left order. The two operands which immediately follows the operator are for evaluation. Represent the operator and corresponding operand in infix notation Continue this process until the equivalent infix expression is achieved. If the priority of a scanned operator is less than any operators available with [], then put them within (). E.g.: + A / * B – C ^ DEF => + A / * B – C [D ^ E] F => + A / * B [ C – D ^ E] F => + A / [B * ( C – D ^ E)] F => + A [B * (C – D ^ E) / F] => A + B * (C – D ^ E) / F
58. 58. Infix to Postfix: Identify the innermost bracket Identify the operator according to the priority of evaluation. Represent the operator and corresponding operand in postfix notation Continue this process until the equivalent prefix expression is achieved. E.g.: (A + B * (C – D ^ E) / F) => (A + B * (C – [DE^]) / F) => (A + B * [ CDE ^ -] / F) => (A + [BCDE ^ - *] / F) => A + [BCDE ^ - * F /] => ABCDE^ - * F / +
59. 59. Postfix to Infix: Identify the operator from left to right order. The two operands which immediately proceeds the operator are for evaluation. Represent the operator and corresponding operand in infix notation Continue this process until the equivalent infix expression is achieved. If the priority of a scanned operator is less than any operators available with [], then put them within (). E.g.: ABCDE^ - * F / + => ABC[D ^ E] - * F / + => AB[C – D ^ E] * F / + => A[B * (C – D ^ E)] F / + => A[B * (C – D ^ E) / F] + => A + B * (C – D ^ E) / F
60. 60. Prefix to Postfix: Identify the operator from right to left order. The two operands which immediately follows the operator are for evaluation. Represent the operator and corresponding operand in postfix notation Continue this process until the equivalent postfix expression is achieved. If the priority of a scanned operator is less than any operators available with [], then put them within (). E.g.: + A / * B – C ^ DEF => + A / * B – C [D E ^] F => + A / * B [ C DE ^ -] F => + A / [B CDE ^ - *] F => + A [B CDE ^ - * F /] => ABCDE^ - * F / +
61. 61. Postfix to Prefix: Identify the operator from left to right order. The two operands which immediately proceeds the operator are for evaluation. Represent the operator and corresponding operand in prefix notation Continue this process until the equivalent prefix expression is achieved. If the priority of a scanned operator is less than any operators available with [], then put them within (). E.g.: ABCDE^ - * F / + => ABC[^ DE] - * F / + => AB[- C ^ DE] * F / + => A [* B – C ^ DE] F / + => A[/ * B – C ^ DEF]+ => + A / * B – C ^ DEF
62. 62. Importance of Postfix Expression: Infix notation which is rather complex as using this notation one has to remember a set of rules. The rules include BODMAS and ASSOCIATIVITY. In case of postfix notation, the expression which is easier to work of evaluate the expression as compared to the infix expression. In a postfix expression, operands appear before the operators, there is no need to follow the operator precedence and any other rules.
63. 63. Infix to Postfix Conversion using Stack: [Assume Q is an infix expression and P is the corresponding postfix notation.] • Step 1: PUSH ‘(‘ onto stack and Add ‘)’ at the end of Q. • Step 2: Repeatedly scan from Q until Stack is empty. Step 2.1: if Q[I] is an operand, then add it to P[J]. Step 2.2: else if Q[I] is ‘(‘, then push Q[I] onto stack. Step 2.3: else if Q[I] is an operator then,
64. 64. Step 2.3.1: while stack[TOP] is an operator and has higher or equal priority compared to Q[I], POP operators from stack and add to P[J] [End of while] Step 2.3.2: Push operator Q[I] into stack Step 2.4: else if Q[I] is an ‘)’, then Step 2.4.1: Repeatedly pop operators from stack and add to P[J] until ‘(‘ is encountered. Step 2.4.2: Remove ‘(‘ from stack. [End of if] [End of Step -2] • Step 3: Exit
65. 65. Procedure for conversion: E.G. (B * C – (D / E ^ F)) Symbol Scanned from Q Stack Postfix Expression P ( ( B ( B * ( * B C ( * BC - ( - BC * ( ( - ( BC * D ( - ( BC * D / ( - ( / BC * D E ( - ( / BC * DE ^ ( - ( / ^ BC * DE F ( - ( / ^ BC * DEF ) ( - BC * DEF ^ / ) BC * DEF ^ / -
66. 66. Postfix Expression Evaluation using Stack: [Assume ‘P’ is a postfix expression] Step 1: Add ‘)’ at the end of P. Step 2: Repeatedly scan ‘P’ from left to right until ‘)’ encountered. Step 2.1: if P[I] is an operand, then push the operand onto stack. Step 2.2: else if P[I] is an operator x then Step 2.2.1: pop two elements from stack (1st element is A and 2nd is B). Step 2.2.2: Evaluate B x A Step 2.2.3: push result back to stack [End of if] [End of step 2] Step 3: Value = Stack[TOP] Step 4: Display Value Step 5: Exit
67. 67. Example: Evaluate the following postfix expression using stack P = BC * DEF ^ / - where B = 5, C = 6, D = 24, E = 2, F = 3. So, P = 5, 6, *, 24, 2, 3, ^, /, - Symbol scanned from postfix notation Stack A B B x A 5 5 6 5, 6 * 30 6 5 5 * 6 = 30 24 30, 24 2 30, 24, 2 3 30, 24, 2, 3 ^ 30, 24, 8 3 2 2 ^ 3 = 8 / 30, 3 8 24 24 / 8 = 3 - 27 3 30 30 – 3 = 27
68. 68. Infix to Prefix Conversion using Stack: [Assume Q is an infix expression and there are two stacks S1 and S2 exist.] • Step 1: Add left parenthesis ‘(‘ at the beginning of the expression. • Step 2: PUSH ‘)’ onto stack S1. • Step 3: Repeatedly scan Q in right to left order, until Stack S1 is empty. Step 3.1: if Q[I] is an operand, then push it into stack S2. Step 3.2: else if Q[I] is ‘)‘, then push it into stack S1. Step 3.3: else if Q[I] is an operator(op) then,
69. 69. Step 3.3.1: Set x = POP(S1) Step 3.3.2: Repeat while x is an operator AND (Precedence (x) > Precedence (op)) PUSH (x) onto stack S2. Set x = POP (S1) [End of while – step 3.3.2] Step 3.3.3. PUSH (x) onto stack S1 Step 3.3.4: PUSH (op) onto stack S1 Step 3.4: else if Q[I] is ‘(‘, then Step 3.4.1: Set x = POP (S1) Step 3.4.2: Repeat, while (x != ‘)’) [until right parenthesis found] Step 3.4.2.1: PUSH (x) onto stack S2 Step 3.4.2.2: Set x = POP (S1)
70. 70. [End of while – step 3.4.2] [End of if – Step – 3.1] [End of loop – Step – 3] • Step 4: Repeat, while stack S2 is not empty. Step 4.1: Set x = POP (S2) Step 4.2: Display x [End of While] • Step 5: Exit
71. 71. Symbol Scanned from Q Stack S1 Stack S2 H ) H + ) + H G ) + HG - ) + - HG ) ) + - ) HG T ) + - ) + HGT + ) + - ) + HGT P ) + - ) + HGTP ^ ) + - ) + ^ HGTP N ) + - ) + ^ HGTPN * ) + - ) + * HGTPN ^ Procedure for conversion: E.g. Q = (A + B * C * (M * N ^ P + T) – G + H)
72. 72. Symbol Scanned from Q Stack S1 Stack S2 M ) + - ) + * HGTPN ^ M ( ) + - HGTPN ^ M * + * ) + - * HGTPN ^ M * + C ) + - * HGTPN ^ M * + C * ) + - * * HGTPN ^ M * + C B ) + - * * HGTPN ^ M * + CB + ) + - + HGTPN ^ M * + CB * * A ) + - + HGTPN ^ M * + CB * * A ( HGTPN ^ M * + CB* * A+ - + So, the postfix notation, we can get by popping all the symbols from stack S2. That is: + - + A * * B C + * M ^ N P T G H
73. 73. Prefix Expression Evaluation using Stack: [Assume ‘P’ is a prefix expression] Step 1: Add ‘(’ at the beginning of prefix expression. Step 2: Repeatedly scan from ‘P’ in right to left order until ‘(’ encountered. Step 2.1: if P[I] is an operand, then push the operand onto stack. Step 2.2: else if P[I] is an operator (op) then Step 2.2.1: pop two elements from stack (1st element is A and 2nd is B). Step 2.2.2: Evaluate A (op) B Step 2.2.3: push result back to stack [End of if] [End of step 2] Step 3: Value = Stack[TOP] Step 4: Display Value Step 5: Exit
74. 74. Example: Evaluate the following prefix expression using stack P = ( - , *, 3, +, 16, 2, /, 12, 6) Symbol scanned from postfix notation Stack A B A (op) B 6 6 12 6,12 / 2 12 6 12 / 6 = 2 2 2,2 16 2,2,16 + 2,18 16 2 16 + 2 = 18 3 2,18, 3 * 2, 54 3 18 3 * 18 = 54 - 52 54 2 54 – 2 = 52 Finally the value is stack is 52
75. 75. Queue: Queue is a linear non-primitive data structure, works on the concept First in First out (FIFO) or Last in Last out (LILO), where insertion takes place at one end called REAR end and deletion takes place at other end called FRONT end. E.g. A B C D B C D B C D E Rear Front Front Rear RearFront After deleting and element from front end: After inserting an element ‘E’ at rear end:
76. 76. Memory Representation of Queue: Array Representation: The concept of static memory allocation can be implemented by using one dimensional array to represent a queue in memory. We consider two variables FRONT and REAR which holds the index of front and rear end of the queue. Linked Representation: The concept of dynamic memory allocation is implemented by using linked list to represent queue in linked format.
77. 77. Types of Queue: The various types of queues are: Linear Queue Circular Queue (CQ) Double Ended Queue (DEQUE) Priority Queue
78. 78. Linear Queue: Insertion and Deletion Operation In the queue, while inserting a new element, one must test whether there is a room available at rear end or not. If room is available then the element can be inserted, otherwise overflow situation occurs. That is: Queue is full. Deletion of an element can be done from the room that it identified by Front and Front will identify the next room of the queue. Underflow situation occurs, when no element is present in the queue. That is: when front and rear identifies NULL and deletion operation is performed on it.
79. 79. Example: Let Q[MAXSIZE] is an array that represents a queue in the memory and at the beginning, it is empty and MAXSIZE = 10. That is: Front = Rear = NULL or -1. 0 1 2 3 4 5 6 7 8 9 A B C D C D C D E F G 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Insert 4 elements A, B, C, D Delete 2 elements Insert 3 elements E, F, G 0 1 2 3 4 5 6 7 8 9 F = 2, R = 6 F = 2, R = 3 F = 0, R = 3 F = -1, R = -1 MAXSIZE - 1
80. 80. From the example it is clear that, For each insertion in the queue the value of Rear is incremented by 1. That is: Rear = Rear + 1. For each deletion in the queue the value of Front is incremented by 1. That is Front = Front + 1
81. 81. Note: After all the elements inserted into queue, Rear will identify position at MAXSIZE – 1. i.e. Queue is full. Even though the memory locations are vacant at Front end, where Rear identified MAXSIZE – 1 position, we can’t insert a new element. When Front = Rear = NULL, i.e. Queue is empty, we can’t perform deletion operation. When Front = Rear only one element is present in Queue. When Front is at MAXSIZE – 1, means Queue having one element. After deletion Front need to be reset to NULL.
82. 82. Linear Queue Insertion Algorithm: Qinsert(Queue[MAXSIZE], ITEM) Step 1: if Rear = MAXSIZE – 1, then Display “Overflow or Queue is Full” Exit Step 2: else if Front = -1 or Rear = -1, then Front = 0 Rear = 0 Step 3: else Rear = Rear + 1 [End of if] Step 4: Queue[Rear] = ITEM Step 5: Exit
83. 83. Linear Queue Deletion Algorithm: Qdelete (Queue[MAXSIZE]) Step 1: Let ITEM Step 2: if Front = -1 or Rear = -1, then Display “Underflow or Queue empty” Exit [End of if] Step 3: ITEM = Queue[Front] Step 4: Display “deleted item is”, ITEM Step 5: else if Front = Rear, then Front = -1 Rear = -1 Step 6: else Front = Front + 1 Step 7: Exit
84. 84. Circular Queue: It is a queue of circular nature, in which it is possible to insert a new element at the first location, when the last location of the queue is occupied. In other words, if an Q[MAXSIZE] represents a circular queue then after inserting an element at (MAXSIZE -1)th location of the array, the next element will be inserted at the very first location at index 0.
85. 85. Advantages over Linear Queue: In case of linear queue, if Rear is at MAXSIZE – 1, then queue is considered as full. Here even though memory locations are vacant at beginning still we can not use them, as for each insertion, rear is incremented and here Rear goes out of bounds. This drawback can be overcome by circular queue representation.
86. 86. Circular Queue Representation: E.g. Let an array A[5] is maintained in form of circular queue. Here overflow situation arises only when Rear is before the Front position. 10 A[0] A[1] A[2] A[3] A[4] 20 30 Front Rear
87. 87. Example: When queue is empty, the Front = NULL or -1 and when queue is full, the Front = 0 and Rear = MAXSIZE – 1. When the queue having a single element the Front = Rear. 0 1 2 34 5 6 7 0 1 2 34 5 6 7 10 20 30 4050 Queue is empty. F = R = -1 Insert 10, 20, 30, 40, 50 F = 0, R = 4
88. 88. 0 1 2 34 5 6 7 10 4050 Insert 60,70,80 F = 0, R = 7 and Queue is full. 80 0 1 2 34 5 6 7 30 4050 Delete two elements F = 2, R = 7 80
89. 89. 0 1 2 34 5 6 7 90 4050 Insert 90 and 100 F = 2, R = 1. 80 0 1 2 34 5 6 7 50 Delete two elements F = 4, R = 1 80 90 If Rear = MAXSIZE – 1 and free locations exist, then Rear = 0. If Front = MAXSIZE – 1 and Rear is less than N, then Front = 0. During each insertion, to find the next Rear the formula is: Rear = (Rear + 1) % MAXSIZE. Similarly during each deletion to find the next Front position the formula is: Front = (Front + 1) % MAXSIZE
90. 90. Insertion in a Circular Queue: CQinsert(Queue[MAXSIZE], ITEM) Step 1: if (Front = 0 AND Rear = MAXSIZE – 1) OR (Front = (Rear + 1) % MAXSIZE), then Display “Overflow” Exit Step 2: else if Front = -1 or Rear = -1 then Front = 0 Rear = 0 Step 3: else Rear = (Rear + 1) % MAXSIZE [End of if] Step 4: Queue[Rear] = ITEM Step 5: Exit
91. 91. Deletion in a Circular Queue: CQdelete(Queue, Front, Rear) Step 1: Let ITEM Step 2: if Front = -1 or Rear = -1 then Display “Queue is empty or Underflow” Exit [End of if] Step 3: ITEM = Queue[Front] Step 4: Display “deleted item is”, ITEM Step 5: if Front = Rear, then Front = -1 Rear = -1 Step 6: else Front = (Front + 1) % MAXSIZE [End of if] Step 7: Exit
92. 92. Double Ended Queue (DEQUE): It is a kind of queue, in which the elements can be added or removed at either ends but not in the middle. In other words, it is a linear list in which insertion and deletion can be done at both the ends. There are two forms of Deque, such as; Input Restricted Deque Output Restricted Deque 10 20 30 40 50 RearFront Deletion Insertion Insertion Deletion
93. 93. Insertion and Deletion is allowed at both end of Deque, so the operations possible is Deque are: Insertion at Rear end (similar to linear queue) Deletion at Front end (similar to linear queue) Insertion at Front end Deletion at Rear end Input Restricted Deque: It is a Deque, which restricts insertion at only one end, but allows deletion at both ends of the list. So in input restricted deque, the following operations are possible. Deletion at Front end Deletion at Rear end Insertion at Front or Rear end
94. 94. Output Restricted Deque: It is a Deque, which restricts deletion at only one end, but allows insertion at both ends of the list. So in output restricted deque, the following operations are possible. Insertion at Front end Insertion at Rear end Deletion at Front or Rear end
95. 95. C-Function for insertion at rear end of Deque: void dqinsert_Rear (int Q[ ], int ITEM) { if(rear == MAXSIZE – 1) { printf(“Deque is full”); } else { Rear = Rear + 1; Q[Rear] = ITEM; } }
96. 96. C-Function for insertion at front end of Deque: void dqinsert_Front (int Q[ ], int Front, int Rear, int ITEM, int MAXSIZE) { if(Front == 0) { printf(“Deque is full”); } else { Front = Front - 1; Q[Front] = ITEM; } }
97. 97. C-Function for deletion at front end of Deque: void dqdelete_Front (int Q[ ]) { int ITEM; if (Front == -1) { printf(“Deque is empty”); } else if (Front == Rear) { ITEM = Q[Front]; Front = Rear = - 1; printf(“deleted item is %d”, ITEM); } }
98. 98. C-Function for deletion at rear end of Deque: void dqdelete_Rear (int Q[ ]) { int ITEM; if (Front == -1 || Rear == -1) { printf(“Deque is empty”); exit(0); } ITEM = Q[Rear]; if (Front == Rear) { Front = Rear = - 1; } else { Rear = Rear – 1; } printf(“deleted item is %d”, ITEM); } }
99. 99. Priority Queue: A priority queue is a collection of elements such that each element has a priority level and according to the priority level, the elements are deleted. In the queue, an element of higher priority is processed before any element of lower priority. When two or more elements have same priority levels then they are processed according to the order in which they are added to the queue. Priority queues are classified into two categories. Ascending priority queue Descending Priority Queue
100. 100. Ascending Priority Queue: It is a collection of elements in which the items or elements are inserted randomly and from which the smallest element can be deleted. Descending Priority Queue: It is a collection of elements in which the items or elements are inserted randomly and from which the largest element can be deleted. 10 P1 20 P1 30 P3 40 P1 50 P2 60 P2 70 P1 Deletion Insertion
101. 101. Linked List: Linked list is a linear collection of data elements called nodes, each of which is a memory location that contains two parts. That is: ◦ Information of element or data ◦ Link field contains the address of next node or points to the next element stored in the list. INFO LINK Node XSTART [Structure of a node in linked list]
102. 102. Here Start is an external pointer which identifies address of 1st node. i.e. the beginning of the linked list. The address of link part of last node contains NULL as no further nodes. The INFO part contains the value and the link points to the next node. To create a node of linked list, self referential structure is needed.
103. 103. Dynamic Storage Management: Static memory allocation is the one in which the required memory is allocated at compile time. E.g. arrays are the best example of static memory allocation. Static memory allocation has the following drawbacks. Size is fixed before execution Insertion and deletion is a time taking process as the elements need to be shifted towards left of right. There may be a possibility that memory may get wasted or memory is deficit.
104. 104. Advantages of Linked List: Dynamic memory allocation in which the required memory is allocated at run time. E.g. linked lists are the best example of dynamic memory allocation. The drawbacks of static memory allocation can be overcome by dynamic memory allocation. The allocated size can be increased or decreased as per the requirement at run time. Memory can be utilized efficiently as the programmer can choose exact amount of memory needed. The operations like insertion and deletion are less time consuming as compared to arrays.
105. 105. Disadvantages of Linked list: It takes more memory space because each node contains a link or address part. Moving to a specific node directly like array is not possible. Declaration of Structure of a node: Struct node { int info; Struct node *link; // self referencing structure }; Struct node *p; P = (struct node*) malloc (sizeof ( struct node));
106. 106. Basic Operations on Linked List: The basic operations that can be performed on a linked list are as follows: Creation Insertion Deletion Traversing Searching Sorting Concatenation
107. 107. Types of Linked List: Singly Linked List: It is a collection of nodes where each node contains the address of next node. It is also referred as one-way list as traversing is possible in only one direction i.e. from START to NULL. 100 10 200 START 100 11 300 12 400 13 X 200 300 400
108. 108. Memory Allocation: Assume a set of free memory blocks in the form of nodes exist in the memory of the computer. Assume AVAIL is a pointer which holds the address first free node in the memory. Assume each node is divided into two parts INFO and LINK parts. Suppose START, PTR and FRESH are the pointers where: START identifies the address of first node of linked list. FRESH identifies the new node to be added to linked list. PTR will be used to move between linked lists.
109. 109. When AVAIL = NULL, it indicates that no free memory blocks exist in the memory. At this point of time, if we try to create a new node overflow situation occurs. When START = NULL, it indicates that linked list has no nodes or linked list doesn’t exist. At this point of time if we try to delete a node from linked list it leads to underflow situation.
110. 110. Psuedo code: Step 1: if AVAIL = NULL, then Display “ Insufficient Memory or Overflow” Exit Step 2: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL [End of if] 200 200 AVAIL 100 300 400 X 200 300 400 100 FRESH Note: the dotted arrow represents the link which exist before the operations
111. 111. Operations on a Single Linked List: Insertion of a node in a single linked list Insertion at the beginning Insertion at the end Insertion at a specific location Insertion after a specific location Insertion after a specified node. Deletion of a node in a single linked list Deletion at the beginning Deletion at the end Deletion of a node at specific location Deletion of a node based on a given item
112. 112. Insertion of a node in a single linked list: (at the beginning): Searching for a node based on given item Concatenation of two linked list Sorting the node INFO in a single linked list Reversing the single linked list Note: the dotted arrow represents the link which exist before the operations 888 25 X AVAIL 999 777 666 X 888 777 666 999 START 999 FRESH
113. 113. Algorithm: Creation(AVAIL, INFO, LINK, START, FRESH, ITEM) Step 1: if AVAIL = NULL, then Display “Insufficient Memory or Overflow” Exit Step 2: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL START = FRESH [End of if] Step 3: Exit
114. 114. Insertion of a node at the beginning when list has one or more nodes: START initially contains the 1st node address i.e. 100. After insertion operation: START points to node with address 999 Node with address 999 points to address 100 AVAIL points to the next free node with address 888 65 100 888 AVAIL 999 777 666 X 888 777 666 999 START 999 FRESH 55 200 45 300 25 400 25 X 100 200 300 400 Note: the dotted arrow represents the link which exist before the operations
115. 115. Algorithm: Insert_First(INFO, LINK, START, AVAIL, ITEM) Step 1: if AVAIL = NULL, then Display “Insufficient Memory or Overflow” Exit [End of if] Step 2: if START = NULL, then START = FRESH Step 3: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = START START = FRESH [End of if] Step 4: Exit
116. 116. Insertion at the end of a single linked list: After insertion operation: The last node with address 300 points to node with address 999 and link part of node with address 999 sets to NULL and AVAIL points to the next free node with address 888. Note: the dotted arrow represents the link which exist before the operations 55 X 888 AVAIL 999 X 888 100 START 999 FRESH 25 200 35 300 45 999 100 200 300
117. 117. Algorithm: Insert_End (INFO, LINK, START, AVAIL, ITEM) Step 1: Let PTR Step 2: if AVAIL = NULL, then Display “Insufficient Memory or Overflow” Exit Step 3: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL [End of if] Step 3.1: if START = NULL, then START = FRESH
118. 118. Step 3.2: else PTR = START Repeat while LINK[PTR] != NULL PTR = LINK[PTR] [End of while] LINK[PTR] = FRESH [End of Step 3.1] [End of Step 2] Step 4: Exit
119. 119. Insertion at specific location of a single linked list: When loc = 3, after insertion operation: The 2nd node with address 200 points to new node with address 999 and the link part of node with address 300. So, now the new node is inserted at location 3 and Avail points to the next free node having address 888. Note: the dotted arrow represents the link which exist before the operations. 45 300 888 AVAIL 999 X 888 100 START 999 FRESH 25 200 35 999 55 X 100 200 300
120. 120. Algorithm: Insert_Loc(INFO, LINK, START, AVAIL, LOC, ITEM, PTR, PREV) Step 1: Let I = 1 Step 2: if AVAIL = NULL, then Display “Insufficient Memory or Overflow” Exit Step 3: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL Step 3.1: if START = NULL, then Display “List doesn’t exist”. Step 3.2: else
121. 121. Step 3.2.1: PTR = START Step 3.2.2: Repeat, while I < LOC AND PTR != NULL PREV = PTR PTR = LINK[PTR] I = I + 1 [End of while] Step 3.2.3: if PTR = NULL, then Display “Location not found”. Step 3.2.4: else if PTR = START, then LINK[FRESH] = START START = FRESH Step 3.2.5: else LINK[PREV] = FRESH LINK[FRESH] = PTR [End of if – step 3.2.3] [End of step 3.1] [End of step 2] Step 4: Exit
122. 122. Insertion after specific location of a single linked list: When loc = 2, after insertion operation: The 2nd node with address 200 points to new node with address 999 and the link part of node with address 300. So, now the new node is inserted at location 3 and Avail points to the next free node having address 888. Note: the dotted arrow represents the link which exist before the operations. 45 300 888 AVAIL 999 X 888 100 START 999 FRESH 25 200 35 999 55 X 100 200 300
123. 123. Algorithm: InsertAfter_Loc(INFO, LINK, START, AVAIL, LOC, ITEM, PTR) Step 1: Let I Step 2: if AVAIL = NULL, then Display “Insufficient Memory or Overflow” Exit Step 3: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL Step 3.1: if START = NULL, then Display “List doesn’t exist”. Step 3.2: else
124. 124. Step 3.2.1: PTR = START Step 3.2.2: Repeat, for I = 1 to LOC – 1 increasing by 1 PTR = LINK[PTR] [End of for] Step 3.2.3: if PTR = NULL, then Display “Invalid entry for location”. Step 3.2.4: else LINK[FRESH] = LINK[PTR] LINK[PTR] = FRESH [End of if – step 3.2.3] [End of step 3.1] [End of step 2] Step 4: Exit
125. 125. Insertion of a node after specific node in a single linked list: While searching for an item 35, if the item is available then insertion after that node is accomplished as follows: Here, the node with address 200 contains item 35, so node with address 200 points to new node with address 999 and the link part of node with address 999 points to the node with address 300. So, now the new node is inserted after the node having info part with value 35 and AVAIL points to next free node with address 888. Note: the dotted arrow represents the link which exist before the operations. 45 300 888 AVAIL 999 X 888 100 START 999 FRESH 25 200 35 999 55 X 100 200 300
126. 126. Algorithm: InsertAfterNode(INFO, LINK, START, AVAIL, NEWITEM, ITEM, PTR) Step 1: START Step 2: if AVAIL = NULL, then Display “Insufficient Memory or Overflow” Exit Step 3: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = NEWITEM LINK[FRESH] = NULL Step 3.1: if START = NULL, then Display “List doesn’t exist”. Step 3.2: else
127. 127. Step 3.2.1: PTR = START Step 3.2.2: Repeat, while PTR != NULL AND ITEM != INFO[PTR] PTR = LINK[PTR] [End of while] Step 3.2.3: if PTR = NULL, then Display “Item does not exist”. Step 3.2.4: else LINK[FRESH] = LINK[PTR] LINK[PTR] = FRESH [End of step 3.1] [End of step 2] Step 4: Exit
128. 128. Deletion of first node from a single linked list: Here the first node i.e. node with address 100 to be deleted. So, START will be updated with address of 2nd node which is present in the link part of first node as pointed by START. PTR is a pointer that points to first node with address 100 is added to the free or avail list at the beginning by updating link part of PTR with address of node as pointed by AVAIL and AVAIL is assigned with the address 100 as pointed by PTR. 999 888100 AVAIL 777 888 200 START 55 999 45 300 35 999 100 200 300 25 X X 400 777 PTR Note: the dotted arrow represents the link which exist before the operations.
129. 129. Algorithm: Deletion_First(INFO, LINK, START, AVAIL, PTR) Step 1: if START = NULL, then Display “List does not exist or Underflow”. Step 2: else PTR = START START = LINK[PTR] LINK[PTR] = AVAIL AVAIL = PTR [End of if] Step 3: exit
130. 130. Deletion of last node from a single linked list: Here the last node is pointed by PTR and last but one node is pointed by PREV. As the node pointed by PREV will become the last node, the link part of PREV with address 200 is updated with NULL. The node with address 300 as pointed by PTR is added at the beginning of free or avail list. 999 888300 AVAIL X 888 200 START 55 200 45 X 25 999 100 200 300 PREV Note: the dotted arrow represents the link which exist before the operations. PTR
131. 131. Algorithm: Deletion_End(START, LINK, INFO, PTR, PREV, AVAIL) Step 1: if START = NULL, then Display “List is empty”. Step 2: else Step 2.1: PTR = START Step 2.2: Repeat while LINK[PTR] != NULL PREV = PTR PTR = LINK[PTR] [End of while] Step 2.3: LINK[PREV] = NULL Step 2.4: LINK[PTR] = AVAIL Step 2.5: AVAIL = PTR [End of if] Step 3: exit
132. 132. Deletion of node from a specific location: Here LOC is 3. i.e. node with address 300 is to be deleted. The link part of node with address 200 is updated with the link part of node with address 300. i.e. 400. Now the node pointed by PTR is added at the beginning of AVAIL list. Note: the dotted arrow represents the link which exist before the operations. 888300 AVAIL X 888 100 START 55 200 45 400 25 999 100 200 300 15 X 400 999 PREV 3 LOC
133. 133. Algorithm: Deletion_LOC(START, LINK, INFO, PTR, PREV, LOC) Step 1: Let I = 1 Step 2: PTR = START Step 3: Repeat while I < LOC AND PTR != NULL PREV = PTR PTR = LINK[PTR] I = I + 1 [End of while] Step 4: if PTR = NULL, then display “Location doesn’t exist”. Step 5: else if PTR = START, then START = LINK[PTR] Step 6: else LINK[PREV] = LINK[PTR] [End of if] Step 7: LINK[PTR] = AVAIL Step 8: AVAIL = PTR Step 9: exit
134. 134. Deletion of node based on given item: Here we want to delete a node whose info part contains the value 25. PTR is pointing to the node with address 300 whose info part contains the item given i.e. 25 and PREV is pointing to node with address 200. The link part of node as pointed by PREV is updated with link part of node as pointed by PTR. Now, the node with address 300 as pointed by PTR is added at the beginning of free or avail list. Note: the dotted arrow represents the link which exist before the operations. 888300 AVAIL X 888 100 START 55 200 45 400 25 999 100 200 300 35 X 400 999 PREV
135. 135. Algorithm: Deletion_ITEM(START, LINK, INFO, PTR, PREV, AVAIL, ITEM) Step 1: PTR = START Step 2: Repeat while PTR != NULL AND ITEM != INFO[PTR] PREV = PTR PTR = LINK[PTR] [End of while] Step 3: if PTR = NULL, then display “Item not found”. Step 4: else if PTR = START, then START = LINK[PTR] Step 5: else LINK[PREV] = LINK[PTR] [End of if] Step 6: LINK[PTR] = AVAIL Step 7: AVAIL = PTR Step 8: exit
136. 136. Search of a node in a single linked list: 55 200 100 45 300 25 400 15 X 200 300 400 100 START 25 3 300 ITEM LOC PTR
137. 137. Algorithm Search_Node(INFO, LINK, START, PTR, ITEM) Step 1: Let LOC = 0, F = 0 Step 2: PTR = START Step 3: Repeat while PTR != NULL ◦ Step 3.1: Set LOC = LOC + 1 ◦ Step 3.2: if ITEM = INFO[PTR], then F = 1 break [End of if] PTR = LINK[PTR] [End of while] Step 4: if F = 0, then display “Item not found”. Step 5: else display “Item found at Location”, LOC [End of if] Step 6: Exit
138. 138. Concatenation of two single linked list: 50 200 100 20 300 30 111 200 300 100 START1 55 222 111 45 333 25 444 15 X 222 333 444 111 START2 PTR Step 1: Let PTR Step 2: PTR = START1 Step 3: Repeat while LINK[PTR] != NULL PTR = LINK[PTR] [End of while] Step 4: LINK[PTR] = START2 Step 5: START2 = NULL Step 6: Exit Algorithm:
139. 139. Sorting of a single linked list: 55 200 100 25 300 35 400 15 X 200 300 400 100 START2 PTR2PTR1 TEMP
140. 140. Algorithm SORTING(START, INFO, LINK) Step 1: Let PTR1, PTR2, TEMP Step 2: PTR = START Step 3: Repeat while LINK[PTR] != NULL Step 3.1: PTR2 = LINK[PTR1] Step 3.2: Repeat while PTR2 != NULL Step 3.2.1: if INFO[PTR1] > INFO[PTR2], then TEMP = INFO[PTR1] INFO[PTR1] = INFO[PTR2] INFO[PTR2] = TEMP [End of if] PTR2 = LINK[PTR2] [End of while] PTR1 = LINK[PTR2] [End of while] Step 4: Exit
141. 141. Reversing a Single Linked List: Reversing of single linked list means pointing the START pointer to the last node, then the last node link part have to point to its previous node and that node have to point to its previous node and so on till 1st node, where the 1st node link part will point to NULL. NULL 100 200 LINK[PTR] PREV PTR SAVE 100 55 200 100 START 45 300 25 400 15 X 200 300 400
142. 142. Here for each node identified by PTR is used to connect in reverse order. E.g. when PTR = 100, LINK[PTR] points to NULL as it becomes last node. Then PTR = SAVE to point to the next node. Finally we get the linked list as: 400 55 X 100 START 45 100 25 200 15 300 200 300 400 Info Link Info Link Info Link Info Link
143. 143. Algorithm REVERSE(START, INFO, LINK) Step 1: Let PREV, SAVE, PTR Step 2: PTR = START Step 3: PREV = NULL Step 4: Repeat while PTR != NULL SAVE = LINK[PTR] LINK[PTR] = PREV PREV = PTR PTR = SAVE [End of while] Step 5: START = PREV Step 6: Exit
144. 144. Circular Singly Linked List: It is a single linked list where the link part of last node contains the address of first node. In other words, when we traverse a circular single linked list, the first node can be visited after the last node. The operations on circular singly linked list are: Insertion at the beginning Insertion at the end Deletion at the beginning Deletion at the end 111 55 222 111 START 45 333 25 444 15 111 222 333 444
145. 145. Insertion of a node at beginning: Here, Fresh node with address 999 is to be inserted. The link part of fresh node is updated with 111, which will become the 2nd node. The link part of last node i.e. node with address 444 is updated with address of fresh node 999. START is assigned with the address of fresh node 999. 999 55 222 111 START 45 333 25 444 15 999 222 333 444 999 FRESH 45 111 999 888 AVAIL X 888
146. 146. Algorithm: INSERTFIRST(START, INFO, LINK, AVAIL, FRESH, ITEM) Step 1: if AVAIL = NULL, then Display “Memory Insufficient” Step 2: else FRESH = AVAIL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL Step 2.1: if START = NULL, then START = FRESH LINK[FRESH] = START
147. 147. ◦ Step 2.2: else LINK[FRESH] = START PTR = START Repeat while(LINK[PTR] != START) PTR = LINK[PTR] [End of while] LINK[PTR] = FRESH START = FRESH [End of if] Step 3: Exit
148. 148. Insertion of a node at end: 111 55 222 111 START 45 333 25 444 15 999 222 333 444 999 FRESH 45 111 999 888 AVAIL X 888
149. 149. Algorithm: INSERTEND(START, INFO, LINK, AVAIL, FRESH, ITEM) Step 1: if AVAIL = NULL, then Display “Memory Insufficient” Step 2: else FRESH = NULL AVAIL = LINK[AVAIL] INFO[FRESH] = ITEM LINK[FRESH] = NULL Step 2.1: if START = NULL, then START = FRESH LINK[FRESH] = START
150. 150. ◦ Step 2.2: else PTR = START Repeat while(LINK[PTR] != START) PTR = LINK[PTR] [End of while] LINK[PTR] = FRESH LINK[FRESH] = START [End of if Step 2.1] [End of if – Step 1 ] Step 3: Exit
151. 151. Deletion of a node from beginning: Here, the link part of 1st node is 200, which is assigned to START and by moving to the last node, the link part of last node updated with address 200 as pointed by START. Now, the 1st node with address 100 is added at the beginning of free or avail list. 200 55 999 100 START 45 300 35 400 15 999 200 300 400 888 999 100 AVAIL 777 888 PTR X 777
152. 152. Algorithm: DELETEFIRST(START, INFO, LINK, AVAIL) Step 1: if START = NULL, then Display “List does not exist” Step 2: else if START = LINK[START], then LINK[START] = AVAIL AVAIL = START START = NULL Step 3: else PTR = START
153. 153. Repeat while (LINK[PTR] != START) PTR = LINK[PTR] [End of while] LINK[PTR] = LINK[START] PTR = START START = LINK[START] LINK[PTR] = AVAIL AVAIL = PTR [End of if] Step 4: Exit
154. 154. Deletion of a node from end: Here, the link part of last but one node as pointed by PREV with address 200 is updated with address of first node 100. Now, the last node as pointed by PTR is added at the beginning of free or avail list. 100 55 200 100 START 45 300 25 999 200 300 888 999 300 AVAIL X 888 PTR
155. 155. Algorithm: DELETEEND(START, INFO, LINK, AVAIL) Step 1: if START = NULL, then Display “List does not exist” Step 2: else if START = LINK[START], then LINK[START] = AVAIL AVAIL = START START = NULL Step 3: else PTR = START
156. 156. Repeat while (LINK[PTR] != START) PREV = PTR PTR = LINK[PTR] [End of while] LINK[PREV] = START LINK[PTR] = AVAIL AVAIL = PTR [End of if] Step 4: Exit
157. 157. Advantages of Circular Singly Linked List: Nodes can be accessed easily Deletion of nodes is easier Concatenation and splitting of circular linked list can be done efficiently. Drawbacks of Circular Singly Linked List: It may enter into an infinite loop. Head node is required to indicate the START or END of the circular linked list. Backward traversing is not possible.
158. 158. Doubly Linked List: A doubly linked list is a collection of nodes where each node contains the address of next as well as address of previous node along with INFO part to hold the information. In other words, it is also referred as two-way list as traversing is possible in both the directions. E.g. The structure of a node in ‘c’ is: struct node { int info; struct node *prev, *next; }; PREV 100 INFO NEXT 25 200
159. 159. Example: The operations on doubly linked list are: Insertion of a node at the beginning Insertion of a node at the end Insertion of a node at specific location Deletion of a node from the beginning Deletion of a node from the end Deletion of a node from Specific Position Operations on Doubly Linked List: 100 X 25 200 100 25 300 200 25 400 300 25 X 100 200 300 400 START
160. 160. Insertion of a node at beginning: Here, the fresh node with address 888 is inserted at the beginning of the doubly linked list. The NEXT part of fresh node is updated with the address contained in START. i.e.: 100 and START is updated with 888, the address of fresh node. 777 X 99 100 X 666 777 X 888 777 666 AVAIL 888 888 25 200 100 35 300 200 45 X 100 200 300 START 888 FRESH
161. 161. Algorithm: INSERTFIRST(START, INFO, NEXT, PREV, AVAIL, FRESH) Step 1: if AVAIL = NULL, then Display “Memory Insufficient” Step 2: else FRESH = AVAIL AVAIL = NEXT[AVAIL] PREV[AVAIL] = NULL NEXT[FRESH] = NULL PREV[FRESH] = NULL INFO[FRESH] = ITEM
162. 162. Step 2.1: if START = NULL, then START = FRESH Step 2.2: else NEXT[FRESH] = START PREV[START] = FRESH START = FRESH [End of if – Step 2.1] [End of if – Step – 1] Step 3: Exit
163. 163. Insertion of a node at end: Here, the fresh node with address 888 is to be inserted at the end. PTR is made to point the last node with address 300. Now, the NEXT part of node with address 300 as pointed by PTR is assigned with address of fresh i.e.: 888 and PREV part of fresh node is updated with address 300 as painted by PTR. 777 300 99 X X X 888 777 AVAIL 100 X 25 200 100 35 300 200 45 888 100 200 300 START 888 FRESH
164. 164. Algorithm: INSERTEND(START, INFO, NEXT, PREV, AVAIL, FRESH, PTR) Step 1: if AVAIL = NULL, then Display “Memory Insufficient” Step 2: else FRESH = AVAIL AVAIL = NEXT[AVAIL] PREV[AVAIL] = NULL NEXT[FRESH] = NULL PREV[FRESH] = NULL INFO[FRESH] = ITEM
165. 165. Step 2.1: if START = NULL, then START = FRESH Step 2.2: else PTR = START while NEXT[PTR] != NULL PTR = NEXT[PTR] [End of while] NEXT[PTR] = FRESH PREV[FRESH] = PTR START = FRESH [End of if – Step 2.1] [End of if – Step – 1] Step 3: Exit
166. 166. Insertion of a node at Specific Location: 777 200 99 300 X X 888 777 AVAIL 100 X 25 200 100 35 300 888 45 X 100 200 300 START 888 FRESH
167. 167. Algorithm: INSERTATLOC(START, INFO, NEXT, PREV, AVAIL, FRESH, PTR) Step 1: if AVAIL = NULL, then Display “Memory Insufficient” Step 2: else FRESH = AVAIL AVAIL = NEXT[AVAIL] PREV[AVAIL] = NULL NEXT[FRESH] = NULL PREV[FRESH] = NULL INFO[FRESH] = ITEM
168. 168. Step 2.1: if START = NULL, then START = FRESH Step 2.2: else Step 2.2.1: Set I = 1, PTR1 = START Step 2.2.2: Repeat while I < LOC AND PTR1 = NULL Set PTR = PTR1 Set PTR1 = NEXT[PTR1] Set I = I + 1 [End of while] Step 2.2.3: if PTR1 = NULL, then Display “ Location does not exist”. Step 2.2.4: else if PTR 1 = START, then NEXT[FRESH] = START PREV[START] = FRESH
169. 169. Step 2.2.5: else NEXT[PTR] = FRESH PREV[FRESH] = PTR NEXT[FRESH] = PTR1 PREV[PTR1] = FRESH [End of if – Step 2.2.3] [End of if – Step – 2.1] [End of if – Step – 1] Step 3: Exit
170. 170. Deletion of a node from beginning: 777 100 777 888 X 888 777 AVAIL 200 X 25 888 X 35 300 200 45 X 100 200 300 START
171. 171. Algorithm: DELETEFIRST(START, NEXT, PREV, AVAIL, PTR) Step 1: if START = NULL, then Display “List does not exist” Step 2: else PTR = START START = NEXT[PTR] NEXT[PTR] = AVAIL PREV[AVAIL] = PTR AVAIL = PTR [End of if] Step 3: Exit
172. 172. Deletion of a node from the end: 300 300 777 888 X 888 777 AVAIL 100 X 25 200 100 35 X X 45 888 100 200 300 START
173. 173. Algorithm: DELETEEND(START, NEXT, PREV, AVAIL, PTR) Step 1: if START = NULL, then Display “List does not exist” Step 2: else Step 2.1: PTR1 = START Step 2.2: Repeat while NEXT[PTR] != NULL Set PTR = PTR1 Set PTR1 = NEXT[PTR1] [End of while]
174. 174. Step 2.3: NEXT[PTR] = NULL PREV[PTR1] = NULL NEXT[PTR1] = AVAIL PREV[AVAIL] = PTR1 AVAIL = PTR1 [End of if] Step 3: Exit
175. 175. Deletion of a node from Specific Location: 300 200 777 888 X 888 777 AVAIL 100 X 25 300 X 35 888 100 45 X 100 200 300 START
176. 176. Algorithm: DELETEPOS(START, LOC, PTR, PTR1, ITEM, AVAIL) Step 1: Let I = 1 Step 2: if START = NULL, then Display “List does not exist” Step 3: else Step 3.1: PTR1 = START Step 3.2: Repeat while I < LOC AND PTR1 != NULL PTR = PTR1 PTR1 = NEXT[PTR1]
177. 177. I = I + 1 [End of while] Step 3.3: if PTR1 = NULL, then Display “Location entered not exist” Step 3.4: else if PTR1 = START, then START = NEXT[START] PREV[START] = NULL Step 3.5: else if NEXT[PTR1] = NULL, then NEXT[PTR] = NULL
178. 178. Step 3.6: else NEXT[PTR] = NEXT[PTR1] PREV[NEXT[PTR1]] = PTR [End of Step 3.3] Step 3.7: NEXT[PTR1] = AVAIL PREV[AVAIL] = PTR1 PREV[PTR1] = NULL AVAIL = PTR1 [End of Step – 2] Step 3: Exit
179. 179. Circular Doubly Linked List: A circular doubly linked list is referred as circular doubly linked list. i.e.: in a circular doubly linked list, the NEXT part of last node contains the address of first node and the PREV part of first node contains address of last node. START 100 300 25 200 200 45 100100 35 300 100 200 300
180. 180. Insertion of a node at beginning: START 888 888 25 200 200 45 888100 35 300 100 200 300 AVAIL 777 300 99 100 X X 888 777 FRESH 777
181. 181. Algorithm: INSERTFIRST(START, FRESH, INFO, PREV, NEXT, PTR, AVAIL) Step 1: if AVAIL = NULL, then Display “insufficient memory” Step 2: else Step 2.1: FRESH = AVAIL AVAIL = NEXT[AVAIL] PREV[AVAIL] = NULL PREV[FRESH] = NULL NEXT[FRESH] = NULL
182. 182. Step 2.2: if START = NULL, then START = FRESH PREV[FRESH] = START NEXT[FRESH] = START Step 2.3: else PTR = PREV[START] NEXT[PTR] = FRESH PREV[FRESH] = PTR NEXT[FRESH] = START PREV[START] = FRESH START = FRESH [End of if] [End of if] Step 3: Exit
183. 183. Insertion of a node at end: START 100 888 25 200 200 45 888100 35 300 100 200 300 AVAIL 777 300 99 100 X X 888 777 FRESH 777
184. 184. Algorithm: INSERTEND(START, FRESH, INFO, PREV, NEXT, PTR, AVAIL) Step 1: if AVAIL = NULL, then Display “insufficient memory” Step 2: else Step 2.1: FRESH = AVAIL AVAIL = NEXT[AVAIL] PREV[AVAIL] = NULL PREV[FRESH] = NULL NEXT[FRESH] = NULL
185. 185. Step 2.2: if START = NULL, then START = FRESH PREV[FRESH] = START NEXT[FRESH] = START Step 2.3: else PTR = PREV[START] NEXT[PTR] = FRESH PREV[FRESH] = PTR NEXT[FRESH] = START PREV[START] = FRESH [End of if] [End of if] Step 3: Exit
186. 186. Deletion of a node from beginning: 100 100 777 888 X 888 777 AVAIL 200 X 25 888 300 35 300 200 45 200 100 200 300 START
187. 187. Algorithm: DELETEFIRST(START, PTR, PTR1, PREV, NEXT, AVAIL) Step 1: if START = NULL, then Display “List does not exist” Step 2: else if NEXT[START] = START, then PTR = START START = NULL NEXT[PTR] = AVAIL PREV[AVAIL] = PTR AVAIL = PTR [End of if] Step 3: Exit
188. 188. Step 3: else PTR = PREV[START] PTR1 = START NEXT[PTR] = NEXT[PTR1] START = NEXT[PTR1] PREV[START] = PTR NEXT[PTR1] = AVAIL PREV[AVAIL] = PTR1 AVAIL = PTR1 [End of if] Step 4: Exit
189. 189. Deletion of a node from the end: 300 300 777 888 X 888 777 AVAIL 100 200 25 200 100 35 100 X 45 888 100 200 300 START
190. 190. Algorithm: DELETEEND(START, PTR, PTR1, PREV, NEXT, AVAIL) Step 1: if START = NULL, then Display “List does not exist” Step 2: else if START = NEXT[START], then PTR = START START = NULL NEXT[PTR] = AVAIL PREV[AVAIL] = PTR AVAIL = PTR
191. 191. Step 3: else PTR1 = PREV[START] PTR = PREV[PTR1] NEXT[PTR] = START PREV[START] = PTR NEXT[PTR1] = AVAIL PREV[AVAIL] = PTR1 AVAIL = PTR1 [End of if] Step 4: Exit
192. 192. Application of Linked List: Linked stack: when the concept of stack is implemented using single linked list, it is referred as linked stack. In linked stack, assume TOP is the pointer that identifies starting or topmost node. Push operation in a linked stack is similar to insertion of a node at beginning of single linked list and Pop operation in a linked stack is similar to deletion of a node from the beginning of a single linked list. TOP 100 10 200 11 300 12 400 13 X 100 200 300 400
193. 193. Push operation in a linked stack: Here, node with address 999 is to be pushed onto the stack. Initially TOP holds the address of node 100. In order to push the fresh node 999, the link part of it is updated with 100 and TOP is assigned with the address of fresh 999. TOP 999 55 200 45 300 35 X 100 200 300 888 65 100 777 X 999 888 777 FRESH 999 AVAIL
194. 194. Algorithm: PUSH(AVAIL, FRESH, LINK, INFO, TOP) Step 1: if AVAIL = NULL, then Display “Insufficient Memory” Step 2: else FRESH = AVAIL AVAIL = LINK[AVAIL] LINK[FRESH] = NULL INFO[FRESH] = ITEM Step 2.1: if TOP = NULL, then TOP = FRESH
195. 195. Step 2.2: else LINK[FRESH] = TOP TOP = FRESH [End of if – Step 2.1] [End of if – Step 1] Step 3: Exit
196. 196. Pop operation in a linked stack: Here, the top node with address 100 is to be popped. The TOP is updated with address 200. The node 100 is added at the beginning of free or avail list. 400 TOP 200 55 999 45 300 35 400 100 200 300 AVAIL 100 888 777 X 999 888 777 PTR 25 X
197. 197. Algorithm: PUSH(TOP, PTR, LINK, AVAIL) Step 1: if TOP = NULL, then Display “Underflow” Step 2: else PTR = TOP TOP = LINK[TOP] LINK[PTR] = AVAIL AVAIL = PTR [End of if] Step 3: Exit
198. 198. Linked queue: When the concept of queue is implemented using single linked list , it is referred as linked queue. In linked queue, assume FRONT is the pointer that identifies first node and REAR is the pointer that identifies the last node of linked queue. Insertion operation in a linked queue is done at the rear end of the linked queue. Insertion operation in a linked queue is similar to insertion at the end of single linked list FRONT 100 10 200 11 300 12 400 13 X 100 200 300 400 REAR 400
199. 199. Deletion operation in a linked queue is done at front end of linked queue. Deletion operation in a linked queue is similar to deletion from the beginning of the single linked list. A linked queue can be traversed from the node as pointed by FRONT to node as pointed by REAR.
200. 200. Insertion operation in a linked stack: Assume node 999 is inserted into the linked queue. The link part of REAR node is updated with the address of fresh node 999 and REAR is made to point to fresh node 999. AVAIL 888 14 X X 999 888 FRESH 999 FRONT 100 10 200 11 300 12 X 100 200 300 13 999 300 REAR 400
201. 201. Algorithm: LINKEDQUEUEINSERT(AVAIL, FRESH, LINK, FRONT, REAR, INFO, ITEM) Step 1: if AVAIL = NULL, then Display “Overflow” Step 2: else FRESH = AVAIL AVAIL = LINK[AVAIL] LINK[FRESH] = NULL INFO[FRESH] = ITEM
202. 202. Step 2.1: if FRONT = NULL AND REAR = NULL, then FRONT = FRESH REAR = FRESH Step 2.2: else LINK[REAR] = FRESH REAR = FRESH [End of if – Step 2.1] [End of if – Step 1] Step 3: Exit
203. 203. Deletion operation in a linked stack: Assume node 999 is inserted into the linked queue. The link part of REAR node is updated with the address of fresh node 999 and REAR is made to point to fresh node 999. AVAIL 100 888 777 999 888 FRONT 200 10 999 11 300 12 X 100 200 300 13 X 300 REAR 400 X 777
204. 204. Algorithm: LINKEDQUEUEDELETE(AVAIL, LINK, PTR, FRONT, REAR) Step 1: if FRONT = NULL AND REAR = NULL, then Display “Underflow” Step 2: else if FRONT = REAR, then FRESH = NULL REAR = NULL Step 3: else PTR = FRONT FRONT = LINK[FRONT] LINK[PTR] = AVAIL AVAIL = PTR [End of if] Step 4: Exit
205. 205. Polynomial Representation Using Linked List: A single linked list can be used to represent and manipulate polynomials. A polynomial expression is a collection of polynomial terms, where each term contains a coefficient, base and exponent. A node that can be used to represent a polynomial term with one variable, is divided into 3 parts i.e. coefficient, exponent and link. E.g.: To represent a polynomial term 3x2, the structure of node is: Coefficient 3 Exponent LINK 2 200
206. 206. Example: Represent the following polynomial expressions in linked list format. i.e.: i) 3x2 + 4x + 3 ii) 4x3 + 7x2 + 3x + 2 START 111 3 2 222 3 0 X4 1 333 111 222 333 START 100 4 3 200 3 1 4007 2 300 100 200 300 2 0 X 400
207. 207. Creating a polynomial using linked list: Algorithm: Create_Poly(START, AVAIL, CO, EXP, LINK, FRESH, PTR, PTR1) Step 1: if AVAIL = NULL, then Display “Memory Insufficient”. Step 2: else FRESH = AVAIL AVAIL = LINK[AVAIL] LINK[FRESH] = NULL Display “enter co-efficient and exponent values” Step 2.1: if START = NULL, then START = FRESH Step 2.2: else PTR1 = START
208. 208. Step 2.2.1: Repeat while PTR1! = NULL if EXP[PTR1] < EXP[FRESH], then break [End of if] PTR = PTR1 PTR1 = LINK[PTR1] [End of while] Step 2.2.2: if PTR1 = START, then LINK[FRESH] = PTR1 START = FRESH
209. 209. Step 2.2.3: else if PTR1 = NULL, then LINK[PTR] = FRESH Step 2.2.4: else LINK[PTR] = FRESH LINK[FRESH] = PTR1 [End of if – Step – 2.2.2] [End of if – Step 2.1] Step 3: Exit
210. 210. Polynomial Addition: Assume P1, P2 are two polynomial expressions given. P3 is the resultant polynomial which represents sum of P1 and P2. P1 = 3x4 + 2x2 + 9x P2 = - 4x3 + 4x2 – 9x + 7 P3 = 3x4 + (- 4x3) + 6x + 7 START 111 3 4 222 9 1 X2 2 333 111 222 333 START 100 -4 3 200 -9 1 4004 2 300 100 200 300 7 0 X 400 START 123 3 4 234 6 2 456-4 3 345 123 234 345 7 0 X 456
211. 211. Procedure for addition of two polynomials: [Assume P1 and P2 are two polynomials expressions and P3 is the resultant polynomial] Step 1: Repeat while P1 and P2 are NOT NULL, Repeat Step 2, 3 and 4 Step 2: if exponent parts of two terms of P1 and P2 are equal and if the co-efficient terms do not cancel to 0, then Step 2.1: Find sum of the co-efficient terms and insert SUM polynomial P3. Step 2.2: Move to next term of P1 Step 2.3: Move to next term of P2.
212. 212. Step 3: else if the exponent of the term in first polynomial > exponent of the term in second polynomial, then insert the term from first polynomial in the SUM polynomial. Step 3.1: Move to next term of P1. Step 4: else Step 4.1: Insert the term from the 2nd polynomial into SUM polynomial Step 4.2: Move to Next term of P2. [End of if] Step 5: Copy remaining terms from the non-empty polynomials into the SUM polynomial Step 6: Exit.
213. 213. Algorithm for Polynomial Addition: GET_NODE(AVAIL): This procedure used to allocate memory for a new node. INSERT_AT_END(P3, FRESH): This procedure used to insert FRESH node at the end of linked list. ADD_POLY(P1, P2, P3): Here, P1 identifies first term of 1st polynomial, P2 identifies first term of 2nd polynomial and P3 identifies first term of the resultant polynomial.
214. 214. GET_NODE(AVAIL) Step 1: Let PTR Step 2: if AVAIL = NULL, then Display “Overflow” Step 3: else PTR = AVAIL AVAIL = LINK[AVAIL] LINK[PTR] = NULL [End of if] Step 4: RETURN(PTR) Step 5: Exit
215. 215. INSERT_AT_END(P3, FRESH) Step 1: Let PTR Step 2: if P3 = NULL, then P3 = FRESH Step 3: else Step 3.1: Set PTR = P3 Step 3.2: Repeat while (LINK[PTR] != NULL) PTR = LINK[PTR] [End of while] Step 3.3: LINK[PTR] = FRESH Step 4: Exit
216. 216. ADD_POLY(P1, P2, P3) Step 1: Let PTR1, PTR2, FRESH Step 2: P3 = NULL Step 3: PTR1 = P1 Step 4: PTR2 = P2 Step 5: Repeat while (PTR1 != NULL AND PTR2 != NULL) Step 5.1: if EXP[PTR1] > EXP[PTR2], then FRESH = GET_NODE(AVAIL) EXP[FRESH] = EXP[PTR1] CO[FRESH] = CO[PTR1] INSERT_AT_END(P3, FRESH) PTR1 = LINK[PTR1]
217. 217. Step 5.2: else if EXP[PTR2] > EXP[PTR1], then FRESH = GET_NODE(AVAIL) EXP[FRESH] = EXP[PTR2] CO[FRESH] = CO[PTR2] INSERT_AT_END(P3, FRESH) PTR2 = LINK[PTR2] Step 5.3: else FRESH = GET_NODE(AVAIL) EXP[FRESH] = EXP[PTR1] CO[FRESH] = CO[PTR1] + CO[PTR2] Step 5.3.1: if CO[FRESH] = 0, then INSERT_AT_END(P3, FRESH) [End of if – Step 5.3.1]
218. 218. PTR1 = LINK[PTR2] PTR2 = LINK[PTR2] [End of if – Step 5.1] [End of while – Step 5] Step 6: Repeat while PTR2 != NULL FRESH = GET_NODE(AVAIL) EXP[FRESH] = EXP[PTR2] CO[FRESH] = CO[PTR2] INSERT_AT_END(P3, FRESH) PTR2 = LINK[PTR2] [End of while – Step 6]
219. 219. Step 7: Repeat while PTR1 != NULL FRESH = GET_NODE(AVAIL) EXP[FRESH] = EXP[PTR1] CO[FRESH] = CO[PTR1] INSERT_AT_END(P3, FRESH) PTR1 = LINK[PTR1] [End of while – Step 7] Step 8: RETURN(P3) Step 9: Exit
220. 220. Dynamic Storage Management: Basic task of any program is to manipulate data. These data should be stored in memory during their manipulation. There are two memory management schemes for the storage allocations of data. i.e.: Static Storage Management and Dynamic Storage Management In case of Static Storage Management scheme, the net amount of memory for various data for a program are allocated before the starting of the executing of the program.
221. 221. Once memory is allocated, it neither can be extended nor can be returned to the memory bank for the use of other programs at the same time. The dynamic storage management scheme allows the user to allocate and reallocate memory as per the necessity during the execution of programs. Principles of dynamic memory management scheme: Allocation Scheme: A request for memory block will be serviced. There are two strategies i.e.: a) fixed block allocation and b) variable block allocation ( First fit and its variant, Next fit, Best fit, Worst fit)
222. 222. Reallocation Scheme: How to return memory block to the memory bank whenever it is no more required. i.e.: a) Random reallocation and b) Order reallocation Compaction: It is a technique for reclaiming the memory is unused for longer period by introducing a program to accomplish this task. The allocation problem becomes simples after compaction process. Compaction works by actually moving one block of data from one location of memory to another. So, as to collect on the free blocks into one large block.
223. 223. Garbage Collection: Suppose some memory space becomes reusable, a node is deleted from a list or an entire list is deleted from a program. The operating system of a computer may periodically collect all the deleted space onto the free storage list. Any technique which does this collection is called Garbage Collection.
224. 224. Garbage Collection usually takes place in two steps. That is: First the computer runs through all list, tagging those cells which are currently in use. Then the computer runs through the memory collecting all untagged space into the free storage list. The Garbage collection may takes place when there is only some minimum amount of space or no space at all left in free storage list or when the CPU has the time to do the collection. End
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# Tag Info
1
Partial answer only. For me, the paper is behind a paywall. So I can base this only on its abstract and your question. I believe you have a point with your complexity doubts. I can see how to write an FO[LFP] self-interpreter, provided there is a way to handle large arities. In particular, the self-interpreter has a single, fixed, maximum arity and number ...
1
I don't think you can do it in plain Datalog, but you can do it in Datalog plus negation. I suggest you define a relation $R$, so that $R(x,y)$ is true if vertex $y$ is reachable from vertex $x$ in one or more steps. You can define $R$ recursively, namely, $$R(x,y) = E(x,y) \lor \exists w . E(x,w) \land R(w,y),$$ where $E$ is a relation that represents ...
1
There was a misunderstanding in regards to the evaluation algorithm/minimal model. The derivation of new values is atomic e.g.: step 1: p = {} q = {} step 2: p = r q = r In step 2 p and q only see the value that each other had at step 1(the old value). I thought p and q would have access to the value of the current step and thus, the order of ...
Only top voted, non community-wiki answers of a minimum length are eligible
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Expert Reviewed
# How to Choose Life Insurance
Life insurance is part of estate planning. If you have loved ones who depend upon you financially, you need life insurance. A life insurance policy allows your beneficiaries to cover their living expenses after your death. Depending on the size of the benefit you want to provide and the amount you can afford to pay on premiums, you can choose from several different types of life insurance policies.
### Method 1 Calculating How Much Life Insurance You Need
1. 1
Decide whether or not you need life insurance. If you have anyone who relies upon you financially, then you should purchase a life insurance policy. You may be able to purchase a life insurance policy through your work. But the coverage may not be high enough, and it likely only remains in place while you are employed. Depending on the amount of coverage you need, you may need to purchase an additional life insurance policy outside of work.[1]
• If you are single with no dependents, you probably don’t need life insurance. Similarly, if you have recently gotten married, unless you own any property, you may not need life insurance.
• However, some people in this case purchase a small policy. This would allow loved ones to cover their final expenses. These expenses include burial and funeral expenses[2]
2. 2
Estimate your family’s living expenses. If you are responsible for providing some or all of your family's living expenses, you will want to buy insurance to cover this amount so that your family can live securely after your passing. Add up your take-home income over a year and then multiply that number out for a number of years to determine an insurance amount to purchase. This time period is not set in stone and will depend on how much insurance coverage you want to purchase and how much will make you feel that your family could live safely in the event of your passing.
• Another consideration is the cost of child care. If you pass, a stay-at-home spouse may be required to work, which would also require them to pay for child care for your children. Add in this expense to your total amount.[3]
3. 3
• For example, suppose you owe \$150,000 on your mortgage, and you have other consumer debt that adds up to \$20,000. Estimate that your final expenses will cost \$5,000. This adds up to \$175,000 ${\displaystyle (\150,000+\20,000+\5,000=\175,000)}$.
4. 4
Consider your children's education. You want to leave your family with enough money to cover future financial obligations. For example, your spouse may want to send you children to college. Estimate how much would be needed for tuition, books, fees and room and board. If you pass away, this might not be possible without your income. A life insurance policy can make it a reality.[5]
• For example, if you want your children to be able to attend an in-state, four-year public school, you will need to have at least \$130,000 per child[6]. If you have three children, you would need \$390,000.
5. 5
Add up the current financial resources. Tally up any financial resources still available to your family after your death. For example, your spouse may have an income. You may have savings or retirement accounts. In addition, you may have begun saving for college. Also, you might have other life insurance policies. Add up the balances in all of your accounts.[7]
• For example, suppose you have \$75,000 saved in your retirement accounts and \$10,000 saved for college. Also, you have another life insurance policy through work that’s worth \$50,000. That means you already have \$135,000 in financial resources ${\displaystyle (\75,000+\10,000+\50,000=\135,000)}$.
6. 6
Calculate how much life insurance you need. Add up all of the expenses you want to cover, including paying off your house, paying off your debt and sending your children to college. Add up all of your financial resources, including your retirement savings, college savings and other life insurance policies. Subtract the value of your financial resources from the total expenses you want to cover. This tells you how much life insurance you need.
• In the above example, you want to cover \$175,000 in debt and \$390,000 in college tuition. This totals up to \$565,000.
• You already have \$135,000 in other financial resources.
• You need to purchase \$430,000 in life insurance ${\displaystyle (\565,000-\135,000=\430,000)}$.
7. 7
Use an online life insurance calculator. Many life insurance companies have online forms that will help you figure out how much life insurance you need. You enter in how much outstanding debt you have and how many children you need to send to college. You also input information about the total annual income your family would need and any income you expect your spouse to earn after you die. Once you submit the information, the calculator analyses your situation and tells you how much life insurance you need to purchase. From there, you would contact an agent and discuss the life insurance products they have available to cover your needs.[8]
8. 8
Re-evaluate your insurance needs when you reach retirement age. If you have purchased a term life insurance policy, it has likely expired by the time you reach retirement age. At this point, the cost of purchasing a new life insurance policy would be prohibitively high because of your age. However, if you have planned well for retirement, you shouldn’t need a life insurance policy. Your retirement accounts should be able to provide for your loved ones in the event of your death. Similarly, if you have a cash-value policy, you shouldn’t need that anymore either. Cash out the policy and add the cash value to your retirement accounts.[9]
### Method 2 Understanding Life Insurance Products
1. 1
Compare term life and whole life insurance. These are the two basic categories of insurance available. Term insurance is good for a specific period of time, whereas whole life insurance is good for your entire life if you the pay the premiums. Term insurance is typically inexpensive, and whole life insurance is pricey. This is because term insurance is pure mortality risk, administrative costs, and commissions while whole life is mortality risk, an investment portion, administration, and commissions. The difference is the investment piece on the latter. This means that whole life insurance policies set aside a portion of the premium you pay each month to be invested and grow in value.
• Term life insurance is basic and inexpensive. It is good for a specific amount of time. For example, your term life insurance may cover your for 10, 20 or 30 years. If you die during the term of your insurance, your beneficiaries get your death benefit. If you die after the term has expired, your beneficiaries get nothing.[10]
• Whole life policies are also known as cash-value policies. They are good until you stop paying premiums. They do not expire after a certain number of years. Also, they have an investment component attached. This means that part of the premium is invested by the insurance company and earns interest. Three types of whole life insurance are whole life, universal life and variable life.[11]
• Life insurance policies should provide you with enough to provide financial support to your family in the event of your death. While having a cash value policy that grows over time sounds attractive, this option can be expensive. If you would be struggling to pay the premiums on such a policy, then term insurance might be the best option for you.
• However, if you can afford the premiums and you have maxed out your contributions on your pre-tax retirement accounts, a cash-value life insurance policy might be a good choice for you. Since the cash value builds up tax free, it provides you with another opportunity to build your retirement nest egg.[12]
2. 2
Evaluate the two types of term life insurance. You can choose from two different types of term life insurance. The first is the “annual renewable term.” With this type, you can purchase one year of coverage at a time. You have the option to renew each year. The other option is “level premium term.” This means you lock into a specific multi-year period, such as 10, 20 or 30 years.[13]
• With annual renewable term insurance, the premium is likely to increase each year.
• With level premium term, you are guaranteed the same premium for the life of the term.
3. 3
Assess the three different kinds of permanent life insurance you can purchase. They are whole life, universal life and variable life. Whole life, universal life and variable life policies use different kinds of investment tools to grow cash value. The rate of return, which grows cash value, depends on the risk involved in the investments. Policies with higher-risk investments do not guarantee an amount for the cash value of your policy (though the death benefit is always guaranteed).
• Whole life insurance pays a guaranteed amount to your beneficiaries upon your death. Part of your premium is invested by the insurance company to grow the cash value of your benefit. The fund grows tax-deferred each year that you keep the policy.[14]
• Universal life insurance combines a life insurance policy with a money-market investment. This type of investment is riskier. Therefore, policy holders can expect a higher rate of return.[15]
• With variable life insurance, the insurance policy is tied to a stock or bond mutual fund investment. The cash value account is invested in several sub-accounts. The investment grows or shrinks along with the performance of the mutual fund accounts in the market. Beneficiaries enjoy favorable tax treatments.[16]
• These choices differ primarily in their fixed and variable rates of interest depending upon the investment vehicle chosen. In each case, the policy holder pays a premium in excess of the actual mortality risk of the insured.
### Method 3 Finding the Best Life Insurance Plan
1. 1
Assess the reputability of insurance providers. Insurance providers are rated for financial strength and reputability by a handful of ratings firms. These ratings firms are TheStreet.com, Standard & Poor's, Moody's, Fitch, and A.M Best Company. Not every insurance company will have a rating with all agencies, but it is important to get ratings from each one that you can before purchasing from an insurance provider, especially if the provider is not well known. Be sure to also look into what the ratings terms mean for each rating firm.
• Firms assign ratings on different scales, with some using "A+" to denote their highest rating and others using "AAA."
• In general, an assessment of "secure" (rather than the alternative, "vulnerable") is a positive indicator of provider performance.[17]
2. 2
Choose between term insurance and mortgage protection insurance when you buy your first house. When you purchase your first house, it is probably time to consider purchasing term life insurance. This allows the co-borrower on your mortgage to receive a death benefit that would cover any living expenses and to keep paying the mortgage. If for some reason you cannot meet the underwriting criteria for term life insurance, purchase mortgage protection insurance. This pays the beneficiary enough to pay off the mortgage on the house in the event of your death.[18]
3. 3
Provide for your family when you are expecting your first child. Once you are expecting your first child, you need a life insurance policy to protect your family in the event of your death. Your beneficiary can use the death benefit to maintain the same standard of living for your children without having to worry about replacing your income. Choose a policy that is sizable enough to pay for at least 18 years of child-rearing and household expenses. In addition, you can provide enough to cover college tuition.[19]
### Method 4 Comparing Life Insurance Quotes
1. 1
Evaluate the annual benefits and premium. Compare premiums to see if you are locked into a rate for a number of years or if it varies each year. If you are on a fixed income, a fixed premium might be better for you. Similarly, compare the death benefits. Depending on the type of policy for which you are shopping, the amount of the death benefit may not be guaranteed. Evaluate how much it is likely to fluctuate each year.[20]
• For example, term life insurance policies are less expensive than permanent life insurance policies. Their premiums are fixed, meaning you pay the same amount each month for as long as you have the policy. Also, the death benefit is a guaranteed amount. Your beneficiaries are guaranteed to get the amount of insurance you purchased.
• Permanent life insurance policies are more expensive. Also, some invest part of your monthly premium in order to grow the cash value of your policy. This means that your monthly premium might vary. It also means that the amount of your policy's cash value is not guaranteed (though your death benefit is). It can increase or decrease depending on how well your investments perform.
2. 2
Calculate the amount of cash value you can accumulate. If you are shopping for a cash value policy, determine how much the cash value can grow. Whole life, universal life and variable life policies utilize different kinds of investment tools. Depending on the risk involved, the rate of return varies. Cash value is important for when you do not die.[21]
• Speak with your insurance agent about the kinds of investment tools they will use and how risky the investments are. The riskiest investments have the potential for high rates of return. This means the cash value can grow quickly. But, they also can crash just as quickly, depleting your investment. This means that the amount of death benefit paid to your beneficiaries decreases.
• Decide how comfortable you are with the different levels of risk before settling on a policy.
3. 3
4. 4
Ask if you can convert a term policy to a cash value policy. Some insurance providers write a clause into your term policy that allows you to convert it to whole life without providing new evidence of insurability. This means that you can convert the policy regardless of your health. You don’t have to undergo physical examinations in order to re-qualify. If this is something that interests you, choose a policy with this clause.[24][25]
5. 5
Find out if the cash value portion of your policy has dividends. This means that you would share in the company’s surplus if you own a permanent policy. Each year, once the company has paid claims, expenses, other liabilities and has funded reserves for future benefits, it pays the excess to policyholders in the form of dividends. You can reinvest the dividends into your policy, or you can cash them out.[26]
• This only applies to mutual companies, not stock companies, which have shareholders instead of policyholders.
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Year 6 Reflections Lesson - Classroom Secrets | Classroom Secrets
Y6 Mathematics
# Reflections Lesson
This Year 6 Reflections lesson covers the prior learning of reflecting with coordinates, before moving onto the main skill of reflection across all four quadrants.
The lesson starts with a prior learning worksheet to check pupils’ understanding. The interactive lesson slides recap the prior learning before moving on to the main skill. Children can then practise further by completing the activities and can extend their learning by completing an engaging extension task.
National Curriculum Objectives
Mathematics Year 6: (6P3) Describe positions on the full coordinate grid (all four quadrants)
Mathematics Year 6: (6P2) Draw and translate simple shapes on the coordinate plane, and reflect them in the axes
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### 2 Teaching Support
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#### Lesson Slides
These lesson slides guides pupils through the prior learning of place value of 1s, 10s and 100s, before moving onto the main skill of reflections. There are a number of questions to check pupils' understanding throughout.
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#### Modelling PowerPoint
This powerpoint can be used to model the questions that the children will complete on the Varied Fluency and Reasoning & Problem Solving worksheets as part of this lesson.
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#### Lesson Slides
These are the same as the lesson slides on Classroom Secrets. You can assign this as an activity for pupils to access individually in school or remotely from home.
### 1 Prior Learning
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#### Worksheet
This Year 6 Reflections lesson covers the prior learning of reflecting with coordinates, before moving onto the main skill of reflection across all four quadrants.
### 2 Practical Activities
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#### Supporting Activity
This reflection supporting activities sheet contains suggestions for additional tasks you might wish to use to support pupils' understanding of the concepts taught in the lesson.
### 2 Varied Fluency
Varied Fluency from the old Resource Pack
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#### Worksheet
This differentiated worksheet includes varied fluency questions for pupils to practise the main skill of this lesson.
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#### Interactive Activity
This Year 6 Reflections Game checks pupils’ ability to reflect shapes across all four quadrants.
### 2 Application
Application and Reasoning from the old Resource Pack
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#### Worksheet
This differentiated worksheet includes reasoning and problem solving questions to support the teaching of this step.
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#### Mixed Practice
This worksheet includes varied fluency, reasoning and problem solving questions for pupils to practise the main skill of reflections.
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#### Discussion Problem
This worksheet includes two discussion problems which can be used in pairs or small groups to further pupils' understanding of the concepts taught in this lesson.
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#### Interactive Challenge Activity
This Year 6 Reflections Maths Challenge checks pupils’ understanding of reflection and plotting coordinates within a problem solving context.
### 2 Homework
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#### Worksheet
This differentiated worksheet includes varied fluency and reasoning and problem solving questions to support the teaching of this step.
### 2 Additional Activities
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#### Learning Video Clip
Pearl is being pursued by another pirate crew who want her treasure. She must find her ship quickly and escape before the other pirates reach the island.
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#### Interactive Activity
This Year 6 Reflections IWB Activity includes three questions designed to check pupils' understanding of reflecting and translating shapes across the four quadrants.
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#### Consolidation Pack
This Consolidation Year 6 Autumn Block 4 resource pack is aimed at Year 6 Expected and has been designed to give children the opportunity to consolidate the skills they have learned in Autumn Block 4 – Geometry: Position and Direction.
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#### Home Learning Pack
This Autumn week 12 Maths pack contains varied fluency, reasoning and problem solving worksheets.
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## How To Remember The Unit Circle
#### [Unit Circle How to Memorize & Use] Educator.com
Memorize the Unit Circle! Nerds4BIEBER / Comedy. Length: 5:16 Quick View. 14,078. Have a big test tomorrow? This video will help you study! Related Videos. How to remember the unit circle (KristaKingMath) Krista King 5 years ago. Tricks for Memorizing the Unit Circle Cole's World of Mathematics 2 years ago. Life Lessons From 100-Year-Olds LifeHunters 2 years ago. The Easiest Way to Memorize
#### how to memorize unit circle in minutes!! Doovi
Testam o caracteristica noua, care permite afi?area comentariilor in ordine cronologica. Unii testeri ne-au aratat situa?ii in care afi?area unei discu?ii liniare ar putea fi benefica, a?a ca vrem sa vedem cum vi …
#### PatrickJMT » A Way to Remember the Unit Circle
You need to know the important numbers in the unit circle. But what you should do instead f memorizing it straight up is to continue drawing it up and doing the calculation in your head, until eventually you don't have to because you'll already know it.
#### Learning The Unit Circle Trigonometry ThatTutorGuy.com
6/08/2013 · I first learned his nifty little trick to help remember the unit circle last year when I was observing in a classroom at a local high school as part of my degree requirement.
How to remember the unit circle
#### Gorgeous Unit Circle Table R² — How To Easily Remember The
Memorize the Unit Circle! Nerds4BIEBER / Comedy. Length: 5:16 Quick View. 14,078. Have a big test tomorrow? This video will help you study! Related Videos. How to remember the unit circle (KristaKingMath) Krista King 5 years ago. Tricks for Memorizing the Unit Circle Cole's World of Mathematics 2 years ago. Life Lessons From 100-Year-Olds LifeHunters 2 years ago. The Easiest Way to Memorize
#### A Trick to Remember the Unit Circle Values Tutor.com
7/10/2008 · A way to remember the Entire Unit Circle for Trigonometry. This is the way that I remember the unit circle. This is the way that I remember the unit circle. Category
#### Easy Way to Remember the Unit Circle MATHGOTSERVED
It is an oriented circle with radius R = 1. The variable arc AM rotates counterclockwise on the trig unit circle. The measure of the trig arc AM is the measure of the central angle
#### How do you memorize the unit circle? Socratic
Note: This trick is best used to help remember the first quadrant of the unit circle but it can be modified to recall the whole Unit Circle!
#### Unit circle (video) Trigonometry Khan Academy
18 rows · The unit circle has a radius of one. The position (1, 0) is where x has a value of 1, and y …
#### how to memorize unit circle in minutes!! Doovi
This is important to remember when we define the X and Y Coordinates around the Unit Circle. The Unit Circle has 360°. In the above graph, the Unit Circle is divided into 4 Quadrants that split the Unit Circle into 4 equal pieces. Each piece is exactly 90°.
#### Memorize the Unit Circle The Knowledge Pinterest
The "Unit Circle" is a circle with a radius of 1. Being so simple, it is a great way to learn and talk about lengths and angles. The center is put on a graph where the x axis and y axis cross, so we get this neat arrangement here.
#### TRIGONOMETRY Unit Circle? Is it necessary to memorize
This above unit circle table gives all the unit circle values for all 4 unit circle quadrants. As you can see, listed are the unit circle degrees and unit circle radians. You should know both, but you're most likely to be solving problems in radians. Now, the next natural question is, how can I remember the unit circle?
#### A Way to remember the Entire Unit Circle for Trigonometry
Memorize the Unit Circle! Nerds4BIEBER / Comedy. Length: 5:16 Quick View. 14,078. Have a big test tomorrow? This video will help you study! Related Videos. How to remember the unit circle (KristaKingMath) Krista King 5 years ago. Tricks for Memorizing the Unit Circle Cole's World of Mathematics 2 years ago. Life Lessons From 100-Year-Olds LifeHunters 2 years ago. The Easiest Way to Memorize
### How to remember the unit circle - how do I memorize the unit circle Brainly.com
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Display et follow in real time the screen status of a Cisco IP Phone, Send key commands as if you were just in front of the phone, Order the phone to make or answer a call, Send Texte message on the phone, Guide an end user by showing him how to use the phone, Troubleshoot and debug an issue and validate … Jphone interface looks like : JPhone Interface. Prerequisites. Here you have the
#### how to make your sister play with u
27/04/2011 · Honestly that's disgusting...how can u possibly develop an attraction to your sister. You are sick...that's incets. How can you look at someone who gew up with you and came from the same mother as you. DOn't feel her boobs, because is my brother did that to me I would choose to never see him again. Realize she's your sister and even if you "pretend" to brush up against them...trust me …
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## Aufgabenstellungen für das 12. AYPT - Hinweise
Anmerkung: Bei den Problems des AYPT handelt es sich gemäß den Regeln des AYPT um die selben Aufgaben, wie jenen für das IYPT. Da auch das AYPT in englischer Sprache abgehalten wird, steht von den Problems auch nur eine englischsprachige Version zur Verfügung.
## Problems for the 12thAYPT
### 1. Electromagnetic cannon
A solenoid can be used to fire a small ball. A capacitor is used to energize the solenoid coil. Build a device with a capacitor charged to a maximum 50V. Investigate the relevant parameters and maximize the speed of the ball.
### 2. Brilliant pattern
Suspend a water drop at the lower end of a vertical pipe. Illuminate the drop using a laser pointer and observe the pattern created on a screen. Study and explain the structure of the pattern.
### 3. Steel balls
Colliding two large steel balls with a thin sheet of material (e.g. paper) in between may "burn" a hole in the sheet. Investigate this effect for various materials.
### 4. Soap film
Create a soap film in a circular wire loop. The soap film deforms when a charged body is placed next to it. Investigate how the shape of the soap film depends on the position and nature of the charge.
### 5. Grid
A plastic grid covers the open end of a cylindrical vessel containing water. The grid is covered and the vessel is turned upside down. What is the maximal size of holes in the grid so that water does not flow out when the cover is removed?
### 6. Ice
A wire with weights attached to each end is placed across a block of ice. The wire may pass through the ice without cutting it. Investigate the phenomenon.
Two similar flasks (one is empty, one contains water) are each connected by flexible pipes to a lower water reservoir. The flasks are heated to 100°C and this temperature is held for some time. Heating is stopped and as the flasks cool down, water is drawn up the tubes. Investigate and describe in which tube the water goes up faster and in which the final height is greater. How does this effect depend on the time of heating?
### 8. Liquid light guide
A transparent vessel is filled with a liquid (e.g. water). A jet flows out of the vessel. A light source is placed so that a horizontal beam enters the liquid jet (see picture). Under what conditions does the jet operate like a light guide?
### 9. Sticky water
When a horizontal cylinder is placed in a vertical stream of water, the stream can follow the cylinder’s circumference along the bottom and continue up the other side before it detaches. Explain this phenomenon and investigate the relevant parameters.
### 10. Calm surface
When wind blows across a water surface, waves can be observed. If the water is covered by an oil layer, the waves on the water surface will diminish. Investigate the phenomenon.
### 11. Sand
Dry sand is rather 'soft' to walk on when compared to damp sand. However sand containing a significant amount of water becomes soft again. Investigate the parameters that affect the softness of sand.
### 12. Wet towels
When a wet towel is flicked, it may create a cracking sound like a whip. Investigate the effect. Why does a wet towel crack louder than a dry one?
### 13. Shrieking rod
A metal rod is held between two fingers and hit. Investigate how the sound produced depends on the position of holding and hitting the rod?
### 14. Magnetic spring
Two magnets are arranged on top of each other such that one of them is fixed and the other one can move vertically. Investigate oscillations of the magnet.
### 15. Paper anemometer
When thin strips of paper are placed in an air flow, a noise may be heard. Investigate how the velocity of the air flow can be deduced from this noise?
### 16. Rotating spring
A helical spring is rotated about one of its ends around a vertical axis. Investigate the expansion of the spring with and without an additional mass attached to its free end.
### 17. Kelvin’s dropper
Construct Kelvin's dropper. Measure the highest voltage it can produce. Investigate its dependence on relevant parameters.
## News
### 2016-01-03 — AYPT 2016
Infos , Ausschreibung und vorläufiges Programm sowie Änderungen des Reglements des AYPT 2016 sind ab sofort verfügbar!
Mehr...
### 2015-11-08 — Erfahrungsbericht IYPT 2015 in Thailand
Ein Artikel von Luisa Schrempf über das IYPT 2015 in Thailand.
Mehr...
### 2015-10-05 — Generalversammlung 2015
Die Generalversammlung 2015 findet am 28.11.2015 statt.
Mehr...
Alle News lesen
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Question
Thu June 07, 2012 By:
# how many natural numbers not exceeding 4321 can be formed with the digits 1,2,3,and 4,if the digits can repeat. plz explain the problem with steps.
Fri June 08, 2012
The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.
Consider four digit natural numbers whose digit at thousandths place is 1.
Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)
Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)
Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)
Number of four digit natural numbers whose digit at thousandths place is 1 = 4 x 4 x 4 = 64
Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4 x 4 x 4 = 64
Number of four digit natural numbers whose digit at thousandths place is 3 = 4 x 4 x 4 = 64
Now, consider four digit natural numbers whose digit at thousandths place is 4:
Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.
If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.
If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.
If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.
Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4 x 4 + 4 x 4 + 4 + 1 = 37
Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.
Related Questions
Sat December 03, 2016
# Q) In how many ways can 7 person sit around a table so that all shall not have same neighbours in any two arrangements.
Wed November 30, 2016
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# linear regression with matrix
5 views (last 30 days)
Ho Leung Louise FUNG on 1 Nov 2022
Edited: John D'Errico on 1 Nov 2022
Hi guys, I am new to mablat. I have 4 4424x2380 matrixs and I want to analyst every single point of the matrix. That means I need to have a equation for, for instance, colume 3 row 3 number for 4 matrix for further use. Can anyone know how to do it? No worry about the computing power.
John D'Errico on 1 Nov 2022
Edited: John D'Errico on 1 Nov 2022
A linear regression requires an independent variable, AND a dependent variable. It seeems your dependent variable may be the numbers contained in these 4 matrices. PLEASE LEARN TO USE MATRICES PROPERLY. @Atsushi Ueno has suggested you should be storing these numbers in one array, of size 4424x2380x4. That is a wonderful idea. But it still does not answer the relevant question, that is, what is the INDEPENDENT variable in the regression you want to perform? Essentially, it looks like y is a vector of length 4. But what is x?
Atsushi Ueno on 1 Nov 2022
Edited: Atsushi Ueno on 1 Nov 2022
How about stacking them as a 4424x2380x4 matrix?
A = ones(4424,2380);
B = ones(4424,2380);
C = ones(4424,2380);
D = ones(4424,2380);
E = cat(3,A,B,C,D);
size(E)
ans = 1×3
4424 2380 4
E(3,3,:) % for instance, colume 3 row 3 number for 4 matrix for further use.
ans =
ans(:,:,1) = 1 ans(:,:,2) = 1 ans(:,:,3) = 1 ans(:,:,4) = 1
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## Resource Options
• ##### Free/Non-commercial Resources:
Displaying 6 resources
• Resource Type: Website / HyperLink
## Subtracting Fractions
Grades: PS/Pre-K to 12
Subjects: Parts of a Whole (Fractions, Decimals & Percent), Mathematics, Addition & Subtraction, Number Sense
This tutorial helps students understand the basic concept of subtracting fractions. Each problem is illustrated with pizza slice...
Views: 0Favorites: 0
• Resource Type: Website / HyperLink
## Rational Number Project Fraction Operations and Initial Decimal Ideas
Grades: TK/K to 9
Subjects: Parts of a Whole (Fractions, Decimals & Percent), Multiplication & Division, Mathematics, Professional Development and 4 additional..
Twenty-eight lesson plans make up this research-based fourth- through sixth-grade fraction and decimal curriculum, covering thro...
Views: 0Favorites: 0
• Resource Type: Document
## Mathematics Common Core Unpacked: Grade Five
Grades: TK/K to 6
Subjects: Mathematics, Professional Development
This document provides descriptions and examples of what each Mathematics Common Core standard means a Grade Five student will k...
Views: 0Favorites: 0
• Resource Type: Website / HyperLink
## Make Your Own Fractions Worksheet
Grades: TK/K to 6
Subjects: Parts of a Whole (Fractions, Decimals & Percent), Multiplication & Division, Mathematics, Professional Development and 2 additional..
This page, from The Teacher's Corner, allows educators to quickly make worksheets for students to practice addition, subtraction...
Views: 0Favorites: 0
• Resource Type: Website / HyperLink
## Fraction Tutorial
Grades: PS/Pre-K to 12
Subjects: Parts of a Whole (Fractions, Decimals & Percent), Multiplication & Division, Mathematics, Addition & Subtraction, Number Sense
This site provides an introduction to fractions, including the concept of equivalency, and shows how to reduce, decompose, multi...
Views: 0Favorites: 0
• Resource Type: Website / HyperLink
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# FP (function programming): notation
The language used here is FP.
I don't understand the difference between the use of `<>` and `[]`.
For example:
``````2 : <3,4,5> -> 4
``````
But
``````+ o [1,2] : <2,3> -> 5
``````
Why is the first sequence written with `[]` instead of `<>` ?
Thanks!
-
What language in particular? Haskell? Functional languages don't necessarily share syntax. – tjameson Mar 18 '13 at 2:35
Ah. I honestly didn't know about that one. Thought you were using `FP` for the generic idea of functional programming. My bad. – tjameson Mar 18 '13 at 2:39
The answer to your question is actually in the wiki link you provided. While `<...>` is used as notation for lists (e.g., `<3,4,5>` is a list with elements `3`, `4`, and `5`), `[...]` is just syntax for a functional that is already provided by FP. It is called construction and defined by
``````[f1, ..., fn]:x = <f1:x, ..., fn:x>
``````
If you are familiar with Haskell: it is similar to
``````map (\f -> f x) [f1, ..., fn] = [f1 x, ..., fn x]
``````
that is to say, that `[...]` is some kind of map function, see also Higher order function to apply many functions to one argument.
``````2:<3, 4, 5>
``````
selects the second element of the given list, and `+ o [1, 2]:<2, 3>` can be "evaluated" as follows (where `o` is function composition):
``````+ o [1, 2]:<2, 3> => (definition of composition)
+:([1, 2]:<2, 3>) => (definition of construction)
+:<1:<2, 3>, 2:<2, 3>> => (select list elements)
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https://www.causal.app/formulae/cumipmt-google-sheets
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Google Sheets
## How do you use CUMIPMT in Google Sheets?
The CUMIPMT function in Google Sheets calculates the cumulative interest paid on a loan or investment over a period of time. You can use this function to track payments and interest over time, and see the total amount paid on the loan or investment. To use the CUMIPMT function, you'll need to input the loan amount, the interest rate, and the number of payments. You can also include a starting payment date if you'd like to calculate interest for a specific period of time.
## What is the syntax of CUMIPMT in Google Sheets?
The syntax for CUMIPMT in Google Sheets is as follows: =CUMIPMT(rate,nper,pmt,pv,fv,type) rate - The interest rate per period. nper - The number of periods. pmt - The payment per period. pv - The present value. fv - The future value. type - The type of payment. 0 - Annuity due. 1 - Annuity due. 2 - Annuity due. 3 - Annuity due. 4 - Annuity due. 5 - Annuity due. 6 - Annuity due. 7 - Annuity due. 8 - Annuity due. 9 - Annuity due. 10 - Annuity due. 11 - Annuity due. 12 - Annuity due. 13 - Annuity due. 14 - Annuity due. 15 - Annuity due. 16 - Annuity due. 17 - Annuity due. 18 - Annuity due. 19 - Annuity due. 20 - Annuity due. 21 - Annuity due. 22 - Annuity due. 23 - Annuity due. 24 - Annuity due. 25 - Annuity due. 26 - Annuity due. 27 - Annuity due. 28 - Annuity due. 29 - Annuity due. 30 - Annuity due.
## What is an example of how to use CUMIPMT in Google Sheets?
The CUMIPMT function calculates the cumulative interest paid on a loan or investment over a given number of payments. In Google Sheets, you can use the CUMIPMT function to calculate the total interest paid on a loan over its lifetime. For example, if you have a loan with a principal of \$10,000 and a interest rate of 5%, the CUMIPMT function will calculate the total interest paid on the loan over its lifetime. To use the CUMIPMT function in Google Sheets, you can enter the following formula: =CUMIPMT(rate,nper,pv,fv,type) The rate argument is the interest rate for the loan. The nper argument is the number of payments for the loan. The pv argument is the principal of the loan. The fv argument is the final value of the loan. The type argument is the payment type. If the type argument is 0 (the default), the function will calculate the cumulative interest paid on the loan. If the type argument is 1, the function will calculate the cumulative interest paid on the investment.
## When should you not use CUMIPMT in Google Sheets?
There are a few occasions when you should not use the CUMIPMT function in Google Sheets. One such example is when you are trying to calculate the interest for a loan that does not have a fixed interest rate. In this case, you would need to use the PV function to calculate the present value of the loan, and then the FV function to calculate the future value. Another instance where you should not use the CUMIPMT function is when you are trying to calculate the interest for a loan that does not have a fixed length. In this case, you would need to use the NPER function to calculate the number of periods, and then the PMT function to calculate the monthly payment.
## What are some similar formulae to CUMIPMT in Google Sheets?
There are a few similar formulae to CUMIPMT in Google Sheets. One is CUMPRICE, which calculates the cumulative price of a list of items. Another is CUMIPMR, which calculates the cumulative interest paid on a list of items. Finally, there is CUMPRINC, which calculates the cumulative principle paid on a list of items.
### Google Sheets
Get started with Causal today.
Build models effortlessly, connect them directly to your data, and share them with interactive dashboards and beautiful visuals.
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## Weak Acids
#### Learning Objective
• Solve acid-base equilibrium problems for weak acids.
#### Key Points
• The dissociation of weak acids, which are the most popular type of acid, can be calculated mathematically and applied in experimental work.
• If the concentration and Ka of a weak acid are known, the pH of the entire solution can be calculated. The exact method of calculation varies according to what assumptions and simplifications can be made.
• Weak acids and weak bases are essential for preparing buffer solutions, which have important experimental uses.
#### Terms
• weak acidone that dissociates incompletely, donating only some of its hydrogen ions into solution
• conjugate basethe species created after donating a proton.
• conjugate acidthe species created when a base accepts a proton
A weak acid is one that does not dissociate completely in solution; this means that a weak acid does not donate all of its hydrogen ions (H+) in a solution. Weak acids have very small values for Ka (and therefore higher values for pKa) compared to strong acids, which have very large Ka values (and slightly negative pKa values).
The majority of acids are weak. On average, only about 1 percent of a weak acid solution dissociates in water in a 0.1 mol/L solution. Therefore, the concentration of H+ ions in a weak acid solution is always less than the concentration of the undissociated species, HA. Examples of weak acids include acetic acid (CH3COOH), which is found in vinegar, and oxalic acid (H2C2O4), which is found in some vegetables.
## Dissociation
Weak acids ionize in a water solution only to a very moderate extent. The generalized dissociation reaction is given by:
$HA(aq) \rightleftharpoons H^+ (aq) + A^- (aq)$
where HA is the undissociated species and A is the conjugate base of the acid. The strength of a weak acid is represented as either an equilibrium constant or a percent dissociation. The equilibrium concentrations of reactants and products are related by the acid dissociation constant expression, Ka:
$K_a = \frac{[H^+][A^-]}{[HA]}$
The greater the value of Ka, the more favored the H+ formation, which makes the solution more acidic; therefore, a high Ka value indicates a lower pH for a solution. The Ka of weak acids varies between 1.8×10−16 and 55.5. Acids with a Ka less than 1.8×10−16 are weaker acids than water.
If acids are polyprotic, each proton will have a unique Ka. For example, H2CO3 has two Ka values because it has two acidic protons. The first Ka refers to the first dissociation step:
$H_2CO_3 + H_2O \rightarrow HCO_3^{-} + H_3O^+$
This Ka value is 4.46×10−7 (pKa1 = 6.351). The second Ka is 4.69×10−11 (pKa2 = 10.329) and refers to the second dissociation step:
$HCO_3^- + H_2O \rightarrow CO_3^{2- } + H_3O^+$
## Calculating the pH of a Weak Acid Solution
The Ka of acetic acid is $1.8\times 10^{-5}$. What is the pH of a solution of 1 M acetic acid?
In this case, you can find the pH by solving for concentration of H+ (x) using the acid’s concentration (F) and Ka. Assume that the concentration of H+ in this simple case is equal to the concentration of A, since the two dissociate in a 1:1 mole ratio:
$K_a = \frac{[H^+][C_2H_3O_2^-]}{[HA]} = \frac{x^2}{(F-x)}$
This quadratic equation can be manipulated and solved. A common assumption is that x is small; we can justify assuming this for calculations involving weak acids and bases, because we know that these compounds only dissociate to a very small extent. Therefore, our above equation simplifies to:
$K_a=1.8\times 10^{-5}=\frac{x^2}{F-x}\approx \frac{x^2}{F}=\frac{x^2}{\text{1 M}}$
$1.8\times 10^{-5}=x^2$
$x=3.9 \times 10^{-3}\text{ M}$
$pH=-log[H^+]=-log(3.9\times 10^{-3})=2.4$
Although it is only a weak acid, a concentrated enough solution of acetic acid can still be quite acidic.
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Maple TA HW 15.1
Page 1 of 2 1, 2
Maple TA HW 15.1
Here is where you can add and discuss 122 HW's
Posts : 22
Join date : 2008-09-17
mapleta?
Is mapleTA down for anyone? I have not been able to open it in a few hours.
Guest1
Guest
Question 3 / Forum
Where did the forum that was posted for 15.1 go ? Is there anyone who knows how to do question 3 I cant figure it out? I hate this new format.
sjames
Guest
Re: Maple TA HW 15.1
you would think that if they had a semester to screw up, that it'd be ready for spring....
a
Guest
Q6
img519.imageshack.us/img519/57/151q6dz1.jpg
don't include a www
anon
Guest
Question 5
Can someone please help with question 5 I have all the other questions if anyone needs them I just need number 5 please?
sjames
Guest
question 5
I cant figure out 5 either anyone know how to do that one?
hwilson
Guest
q
can anyone help with 4 and 6?
helpme
Guest
question5
yes if u can post up 5 i will post 4 and 6 for u
hwilson
Guest
#5
all i need is the expression for number 5 so if anybody has that I will post up all my answers once I am done please help???????
craziech
Guest
Re: Maple TA HW 15.1
I'll list them in order of blanks for most questions, and those involving the tasks I will paste my task in case they differ from yours, and show you how to do it so it can be applied to your own questions
1.) protons, a deficiency, an excess, 2, 1 , electron, 0, 1.6E-19 , huge compared to
2.) I don't seem to understand what they actually want in this question so I'll just tell you what I got...
An object carries a charge of +1.802 . How many elementary electron charges is this ?"
Input variable given = Q
Output variable wanted = N
Value of input variable in SI units = 1.802E-6 (just convert micro-coulombs to coulombs)
SI unit of output variable is = none
The object has more protons then electrons
if someone can help me figure out what is meant by "Give the expression for the output variable, The value of the additional variable in S.I. Units is, and The value of the output variable is " that would be great. I don't want answers...if there is an equation to use just let me know that so I can work it out myself. I looked on ch.15 sheet 4 like it says but there is no expressions or anything
3.) G, masses, r, G, k, m, q , along , attractive, attractive, repulsive, 6.7E-11, much smaller, 9.0E9
A 2.157 negative electric charge is placed at a distance of 23.857 cm from a positive electric
charge and experiences a force of 5.983 N. What is the magnitude of the positive charge ?
Fill in the blanks: q_1, r, F, k, q_2, k*((q_1*q_2)/(r^2)) , (F*r^2)/(k*q_1), multiply negative electric charge by 10^-6 for this answer, convert your radius in cm to meters, Force is already given in newtons , 9.0E9 , solve value of input variable by plugging in numbers into the (F*r^2)/(k*q_1) equation, giving you the q_2 value in Coulombs . The two charges attract each other since they are opposite
5.) m, r, G, F so far
I have to go to class but I'll post the rest of 5 and 6 when I figure it out. Once again if anyone can help me figure out whats wanted for #2 that'd be great. Also - is anyone else having a problem with #1 erasing your "Click list" answers everytime you go back to it? It keeps happening to me even after I save it and I don't want to submit the quiz later and have it show that I didn't put answers in there...
Thanks!
kat
Guest
MapleTA
How are you guys getting access to quiz 15_1? It's not under the course website when clicking Link to the course workshops.
coolson
Guest
Q 5
It's under assignments on blackboard I also need just the expression for 5 I have the rest so if someone can share that I am willing 2 share what I have which is 1 thru 6
Craziech
Guest
#2
What problems are you having with number two? I have the answer so ill post it but it seems as though you have all the answers
Guest 87
Guest
#2
OoooO ..... for the output it is Q/q ..... that is all that is being asked
Guest 87
Guest
Number 2
Im pretty lost with number 2. I cant figure out the expression for the output variable and the input of the variable in S.I. units whihc ultimately means I dont have the value of the output variable. I also need help with Question 5 the Gravitational Force in those variables alsong with the value of the output variable in S.I. Units. I have all the other answers so ill post them if you need. I just need help with those questions. Thanks to alll!!
shanMone
Guest
Q5
SOMEONE PLZ HELP WITH Q5 THE EXPRESSION?????
sck105
Guest
#4
what is the Unit for #4
maria
Guest
Re: Maple TA HW 15.1
Units for #4 are C standing for Coulomb
Can you guys stop doing the whole.."Oh i have all the answers but ill only post mine if you post yours" nonsense? This website is here so we can all help each other out...you don't need conditions for trading answers. If we all just sit here and do that, no ones going to get anything done. If you have answers or methods that can help someone else out, quit being a baby and just do it. we're in college here...
Kay
Guest
I got the Expression for Gravitational Force for Question 5.
You use F= G*(m_1*m_2/r^2) but since m values are the same, you can just do m^2.
Therefore, the mapleTA syntax for this expression and the answer is: G*((m^2)/(r^2))
You plug your answers for mass (in kg) and radius (in meters) into that to solve for your output variable (F). The SI units will be in Newtons..
Kathleen
Posts : 42
Join date : 2008-12-18
Question 6
Fill in the blanks: q, r , k , F, Expression for electrostatic force= k*(q^2/r^2), 1.6E-19, number given in task above in meters, 9.0E9, output variable is found by plugging all your knowns into the equation and solving for electrostatic force, SI units is N and gravitational force is "tiny compared to" electrostatic force.
goodddd luck!
Kathleen
Posts : 42
Join date : 2008-12-18
#2
Hi
for #2 I am getting The value of the output variable wrong
even though i divided Q/q
what did u guys get for an answer
what was the powe u used?
any heeelp?
sarah
Guest
Q5 and 6
5.) m, r, g,F, G*((m^2)/(r^2)), SI units at the end are in N
6.) q, r, k, F, k*(q^2/r^2) , SI units are also N
Guest 87
Guest
Re: Maple TA HW 15.1
YUP I AM GETTING 2 WRONG AS WELL
WAT DID U GUYS GET FOR 2
The value of the output variable ?
#2
Guest
did any of you get 5.88 out of 6?
i got every Q right, even checked with "how did i do"...
WTH is going on?
uhhhh
Guest
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# Search by Topic
#### Resources tagged with Practical Activity similar to Tiles in the Garden:
Filter by: Content type:
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### There are 183 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Practical Activity
### Cover the Tray
##### Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
### Square Corners
##### Stage: 2 Challenge Level:
What is the greatest number of counters you can place on the grid below without four of them lying at the corners of a square?
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Sticks and Triangles
##### Stage: 2 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### Four Colours
##### Stage: 1 and 2 Challenge Level:
Kate has eight multilink cubes. She has two red ones, two yellow, two green and two blue. She wants to fit them together to make a cube so that each colour shows on each face just once.
### Tri.'s
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How many triangles can you make on the 3 by 3 pegboard?
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What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
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This activity investigates how you might make squares and pentominoes from Polydron.
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Are all the possible combinations of two shapes included in this set of 27 cards? How do you know?
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How can you put five cereal packets together to make different shapes if you must put them face-to-face?
### Map Folding
##### Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### Three Sets of Cubes, Two Surfaces
##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
### Order the Changes
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Can you order pictures of the development of a frog from frogspawn and of a bean seed growing into a plant?
### Egyptian Rope
##### Stage: 2 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Putting Two and Two Together
##### Stage: 2 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
### Two by One
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An activity making various patterns with 2 x 1 rectangular tiles.
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Cutting Corners
##### Stage: 2 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Escher Tessellations
##### Stage: 2 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Little Boxes
##### Stage: 2 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### The Numbers Give the Design
##### Stage: 2 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Sort Them Out (2)
##### Stage: 2 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
### It's a Fence!
##### Stage: 1 and 2 Challenge Level:
In this challenge, you will work in a group to investigate circular fences enclosing trees that are planted in square or triangular arrangements.
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### Pyramid Numbers
##### Stage: 2 Challenge Level:
What are the next three numbers in this sequence? Can you explain why are they called pyramid numbers?
### Advent Calendar 2006
##### Stage: 2 Challenge Level:
NRICH December 2006 advent calendar - a new tangram for each day in the run-up to Christmas.
### Making Maths: Birds from an Egg
##### Stage: 2 Challenge Level:
Can you make the birds from the egg tangram?
### Double Your Popcorn, Double Your Pleasure
##### Stage: 2 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
### Making Maths: Happy Families
##### Stage: 1 and 2 Challenge Level:
Here is a version of the game 'Happy Families' for you to make and play.
### Advent Calendar 2008
##### Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Square Tangram
##### Stage: 2 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### Four Layers
##### Stage: 1 and 2 Challenge Level:
Can you create more models that follow these rules?
### Domino Sets
##### Stage: 2 Challenge Level:
How do you know if your set of dominoes is complete?
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### Car Journey
##### Stage: 2 Challenge Level:
This practical activity involves measuring length/distance.
### Making Maths: Indian Window Screen
##### Stage: 2 Challenge Level:
Can you recreate this Indian screen pattern? Can you make up similar patterns of your own?
### Counting Counters
##### Stage: 2 Challenge Level:
Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
### Triangle Shapes
##### Stage: 1 and 2 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### Cuisenaire Rods
##### Stage: 2 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
### A Patchwork Piece
##### Stage: 2 Challenge Level:
Follow the diagrams to make this patchwork piece, based on an octagon in a square.
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
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# Solutions of $\varphi\left(n^n\sigma(n)\right)=\varphi(n^{n+1})$, where $\sigma(n)$ is the sum of divisors and $\varphi(n)$ the Euler's totient
I would like to know a full characterization of the solutions of next equation involving the sum of divisors function $\sigma(m)=\sum_{d\mid m}d$ and the Euler's totient function denoted as $\varphi(m)$
$$\varphi\left(n^n\sigma(n)\right)=\varphi(n^{n+1}).\tag{1}$$ The equation arises from my experiments with odd perfect numbers, since one has the following claim.
Claim. If there exists an odd perfect number then satisfies $(1)$ (and as a consequence we can get also an statement invoking Euler-Fermat $2^{\varphi(n^n\sigma(n))}\equiv 1\text{ mod }n^{n+1}$, I know the first few solutions of this congruence).
Just a comment is that I was playing with more reasonable powers before this $n^n$.
Proof. The proof is easy from the fact that the Euler's totient function is multiplicative, $\varphi\left(n^n\sigma(n)\right)=\varphi(2n^{n+1})=\varphi(n^{n+1}).$$\square Now, it was fun when I ran a program to know what integers satisfy our equation (1). The sequence starts as 1, 2, 8, 128, that are the terms that I can compute with my program. But due my belief I suspect that if fact every term from the OEIS A058891 satisfy (1). Question. I would like to know a full characterization (I am saying what reasonable work can be done) of solutions of previous equation (1). I know the Claim and I suspect that the integers of the form$$2^{2^{\lambda-1}-1}\tag{2}$$satisfy our equation, where \lambda runs over integers \geq 1. What can you prove about it? Can you prove that (2) satisfies our equation (I know that it should be easy but is required patience)? Are there more solutions? Many thanks. Please if you know the equation (1) from the literature please answer this as a reference request and I try to find and read those facts from the literature. • If you run a program in your computer and you find a sufficiently large odd solution of$$2^{\varphi(n^n\sigma(n))}\equiv 1\text{ mod }n^{n+1},$$please add it as a comment. I am saying an odd integer say us >100. Many thanks. – user243301 Mar 7 '18 at 23:17 • Just a comment if some user was interested in this kind of equations is that it is possible to prove (easily by cases) that each integer m\geq 1 satisfies \varphi(4m\sigma(m))=2\varphi(2m\sigma(m)). And also the fact that if n is an odd perfect number and a\geq 2 an integer then \varphi(2^a n\sigma(n))=2^{a-1}\varphi(n^2). – user243301 Mar 8 '18 at 11:59 ## 1 Answer About 2^{2^{\lambda-1}-1}: Assume 2^a satisfies the equation. Then, since \sigma(2^a)=2^{a+1}-1, we have that$$\varphi\left(2^{a2^a}\left(2^{a+1}-1\right)\right)=\varphi\left(2^{a2^a+a}\right).$$Since \varphi(mn)=\varphi(m)\varphi(n) when \gcd(m,n)=1 and \phi(2^a)=2^{a-1}, we have that then$$2^{a2^a-1}\varphi\left(2^{a+1}-1\right) = 2^{a2^a+a-1}.\varphi\left(2^{a+1}-1\right) = 2^a.$$You wish to show that$$\varphi\left(2^{2^k}-1\right) = 2^{2^k-1}.$$Let$$F_n=2^{2^n}+1.$$Then$$2^{2^n}-1=\left(2^{2^{n-1}}-1\right)\left(2^{2^{n-1}}-1\right),$$so we have inductively that$$2^{2^n}-1 = F_0\cdots F_{n-1}.$$Since all Fermat numbers are coprime, this holds iff we have that$$\prod_{k=0}^{n-1} \varphi\left(2^{2^k}+1\right) = \prod_{k=0}^{n-1} \left(2^{2^k}\right).$$Indeed, since \varphi(x)\leq x-1, with equality holding iff x is prime, we have that$$\prod_{k=0}^{n-1} \varphi\left(2^{2^k}+1\right) \leq \prod_{k=0}^{n-1} \left(2^{2^k}\right),$$with equality holding iff each of the first$n-1$Fermat numbers are prime. Thus, your conjecture is equivalent to that made by Fermat that they are all prime, and thus fails first at$n=6$, and similarly everywhere thereafter (as$F_5\$ is not prime).
• I am going to study your answer. I can not believe that the same mistake was made again (the same false conjecture that did Fermat). Many thanks.
– user243301
Mar 9 '18 at 8:18
• I'm sure that your answer is right 100% but I have not studied it yet, thus I'm sorry for not accepting it yet.
– user243301
Mar 17 '18 at 16:31
• @user243301 That's fine - and although it's highly unlikely that a complete characterization of the solutions to this can be provided, this answer doesn't fully respond to all you asked, so it's reasonable to want to leave the question open in case anyone else has anything to provide on it. Mar 17 '18 at 23:33
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Commit 0a823652 by Loic Huder
### Added NumPy-vectorised version of cort (dtw is as V1)
parent c924286d
#!/usr/bin/env python3 from functools import partial from runpy import run_path from pathlib import Path import numpy as np import math util = run_path(Path(__file__).absolute().parent.parent / "util.py") def serie_pair_index_generator(number): """ generator for pair index (i, j) such that i < j < number :param number: the upper bound :returns: pairs (lower, greater) :rtype: a generator """ return ( (_idx_greater, _idx_lower) for _idx_greater in range(number) for _idx_lower in range(number) if _idx_lower < _idx_greater ) def DTWDistance(s1, s2): """ Computes the dtw between s1 and s2 with distance the absolute distance :param s1: the first serie (ie an iterable over floats64) :param s2: the second serie (ie an iterable over floats64) :returns: the dtw distance :rtype: float64 """ len_s1 = len(s1) len_s2 = len(s2) _dtw_mat = np.empty([len_s1, len_s2]) _dtw_mat[0, 0] = abs(s1[0] - s2[0]) # two special cases : filling first row and columns for j in range(1, len_s2): dist = abs(s1[0] - s2[j]) _dtw_mat[0, j] = dist + _dtw_mat[0, j - 1] for i in range(1, len_s1): dist = abs(s1[i] - s2[0]) _dtw_mat[i, 0] = dist + _dtw_mat[i - 1, 0] # filling the matrix for i in range(1, len_s1): for j in range(1, len_s2): dist = abs(s1[i] - s2[j]) _dtw_mat[(i, j)] = dist + min( _dtw_mat[i - 1, j], _dtw_mat[i, j - 1], _dtw_mat[i - 1, j - 1] ) return _dtw_mat[len_s1 - 1, len_s2 - 1] def cort(s1, s2): """ Computes the cort between series one and two (assuming they have the same length) :param s1: the first serie (or any iterable over floats64) :param s2: the second serie (or any iterable over floats64) :returns: the cort distance :rtype: float64 """ slope_1 = s1[1:] - s1[:-1] slope_2 = s2[1:] - s2[:-1] num = np.sum(slope_1 * slope_2) sum_square_x = np.sum(slope_1 * slope_1) sum_square_y = np.sum(slope_2 * slope_2) return num / (math.sqrt(sum_square_x * sum_square_y)) def compute(series, nb_series): gen = serie_pair_index_generator(nb_series) _dist_mat_dtw = np.zeros((nb_series, nb_series), dtype=np.float64) _dist_mat_cort = np.zeros((nb_series, nb_series), dtype=np.float64) for t1, t2 in gen: dist_dtw = DTWDistance(series[t1], series[t2]) _dist_mat_dtw[t1, t2] = dist_dtw _dist_mat_dtw[t2, t1] = dist_dtw dist_cort = 0.5 * (1 - cort(series[t1], series[t2])) _dist_mat_cort[t1, t2] = dist_cort _dist_mat_cort[t2, t1] = dist_cort return _dist_mat_dtw, _dist_mat_cort main = partial(util["main"], compute) if __name__ == "__main__": main()
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