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https://learnaboutstructures.com/Truss-Introduction | 1,669,490,595,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00851.warc.gz | 410,356,282 | 7,915 | Resources for Structural Engineers and Engineering Students
Trusses are structures that only use members that may be considered to have pinned connections at either end. This means that the members can only take axial load (not moment or shear) as shown in Figure 3.1. When the ends of the truss member shown in the figure move, they may have the effect of rotating the member (since the ends are pinned, they are free to rotate), but will not cause any bending or shear. The member will remain straight because of the pins/hinges at either end. Therefore the only load in the element can be axial load as shown on the right side of the figure.
Figure 3.1: Truss Member Showing Pinned Connections to the Surrounding Structure
Since a truss element can only transmit axial force, and not moment or shear, there are only two useful equilibrium equations for the analysis of individual truss members and joints:
$$\label{eq:TrussEquil} \sum_{i=1}^{n}{F_{xi}} = 0; \sum_{i=1}^{p}{F_{yi}} = 0;$$
or, in a more simplified form:
$$\boxed{ \sum F_x = 0; \sum F_y = 0 \, }$$
+Moment equilibrium at the joint still applies; however, it is not useful because we already know that a truss member cannot apply a moment to a joint and vice-versa, making the moment contribution from each member equal to zero ($\sum_{i=1}^{n}{M_i}= \sum_{i=1}^{n}{0} = 0$).
There are two common ways to analyse determinate truss systems, the method of joints or the method of sections. Without any additional analysis tools, these methods may only be used to analyse determinate trusses. Evaluating the determinacy of trusses was previously discussed in Section 2.5. Analysis of indeterminate trusses will be described Chapter 8. | 420 | 1,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | latest | en | 0.927169 |
https://www.gamedev.net/forums/topic/617871-finding-the-exact-value-of-pi-a-programming-book-problem/ | 1,544,792,986,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825728.30/warc/CC-MAIN-20181214114739-20181214140239-00558.warc.gz | 896,327,475 | 28,568 | # finding the exact value of pi (a programming book problem.)
This topic is 2538 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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I'm brushing up on C++ concepts and problem solving, not some problems I know how to do but this one has be stomped, since i haven't studied calculus yet (yes I know it's spelled wrong.) The problem in the book basically gave me the code to find it but, with one catch:
the code was in the wrong order and had a bug in it. now i wrote it out as best as i can but i don't think it's right because i don't get the value of pi per say. I get a one or sometimes a 4, 64, 256, ect.
here is the code:
void valueOfPi() { double Pi = 0.0; int i; int n; cout << "Enter the value of n: "; cin >> n; cout << endl; for(i = 0; i < n; i++) { if(i % 2 == 0) { Pi += (1 / (2 * i + 1)); } else if(i % 2 == 1) { Pi -= (1 / (2 * i + 1)); } else { Pi *= 4; } cout << showpoint << fixed << setprecision(i) << Pi << endl; } }
some stuff i added since it seemed some stuff was missing from the original.
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Your expressions involving i are all using integer arithmetic. You need to change at least one of the operands to a double if you want double arithmetic. I.e. use 2.0 instead of 2 or cast something to double, etc.
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1>c:\users\tim sweeny\documents\visual studio 2008\projects\x\irrapp\chapter5_programming_problems.h(348) : error C2296: '%' : illegal, left operand has type 'double' 1>c:\users\tim sweeny\documents\visual studio 2008\projects\x\irrapp\chapter5_programming_problems.h(348) : error C2297: '%' : illegal, right operand has type 'double' if(i % static_cast<double>(2.0) == 0) { Pi += (1 / (2 * i + 1)); } else { Pi -= (1 / (2 * i + 1)); } Pi *= 3.0;
when i tried that, i got those errors on top. any ideas?
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[color=#000000]chapter5_programming_problems[color=#666600].[color=#000000]h[color=#666600]([color=#006666]348[color=#666600])[color=#000000] [color=#666600]:[color=#000000] error C2296[color=#666600]:[color=#000000] [color=#008800]'%'[color=#000000] [color=#666600]:[color=#000000] illegal[color=#666600],[color=#000000] left operand has type [color=#008800]'double'
[color=#008800]on line 348, the '%' is illegal because the left operand is type 'double'
[color=#008800]- in other words, you can't use modulus with double, it's an integer-only operator
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Don't change i and n, at least not to solve original problem.
When writing mathematical expressions, it is customary to indicate that floating point types are involved. Or: Pi += (1.0 / (2 * i + 1)); // or Pi += (1.0f / (2 * i + 1));Alternatively: Pi += (1 / (2.0 * i + 1)); // or Pi += (1 / (2 * i + 1.0));
It's not really arbitrary, there are strict rules on how types get promoted, but above serves as a hint to programmer to show that floating point and integer math are getting mixed.
Floating point type promotion is just one of those annoying things which has caused many bugs. Strangely enough, Java has exactly the same kind of problem.
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https://study.com/academy/topic/mttc-math-secondary-circles.html | 1,571,606,306,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00195.warc.gz | 722,036,635 | 27,150 | # Ch 59: MTTC Math (Secondary): Circles
If you need a reminder about how to find the measurements of circles, try reviewing the lessons in this chapter. Memorize the formulas outlined in the transcripts as you prepare for the MTTC Mathematics (Secondary) test.
## MTTC Math (Secondary): Circles - Chapter Summary
Find thorough explanations in this chapter concerning the geometric equations related to circles. Get ahead on the MTTC Mathematics (Secondary) test by watching the lessons and reviewing these key concepts:
• Measuring the diameter, radius, and circumference of a circle
• Determining measurements for quarter circles
• Identifying the perimeter of semicircles
• Distinguishing between inscribed and central angles
• Steps for solving word problems with circles
There are a multitude of geometric shapes, and almost every shape has different formulas for finding common measurements. Let our chapter help you remember which formulas are associated with circles. Our instructors use easy-to-follow examples as they review these equations. Use the lesson menu to select the lessons you need to review, and feel free to skip around. To assess if you are ready for the geometry section on the MTTC Mathematics (Secondary) test, challenge yourself to answer the questions on our chapter exam.
8 Lessons in Chapter 59: MTTC Math (Secondary): Circles
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Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 399 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-43 | longest | en | 0.906476 |
http://reference.wolfram.com/applications/mechsystems/KinematicModeling/Mech.2.html | 1,524,490,283,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00343.warc.gz | 273,775,685 | 21,662 | 2. Kinematic Modeling
## Overview
This chapter covers the building and running of basic kinematic models with MechanicalSystems. Throughout this chapter the simplest and most commonly used forms of each MechanicalSystems command are presented. For more detailed information on specific commands, consult the reference guide in the appendix.
MechanicalSystems follows most of the conventions set by Mathematica as to its use of function names, options, and option settings. One significant deviation from standard Mathematica practice is that MechanicalSystems often uses the same name for an option and for the name of a function that performs a service related to the option. For example, PointList is the name of an option that is used to set the values of local point coordinates, and it is also the name of a function that queries the local point coordinates.
# 2.1 Body Objects
A SysBody object is a Mech data object that is used to define the points and properties of a mechanism body. It is not necessary to use body objects in Mech kinematic models, as demonstrated in Sections 1.2 and 1.3, but the use of body objects allows points to be defined once, inside the body object, and then to be referenced throughout the Mech model. This can reduce the likelihood of error due to changing the definition of a point at one place on the model, while failing to change it elsewhere.
## 2.1.1 Body
Mech SysBody objects are created by the Body function, and used to define specific properties of a body in a mechanism model.
A function to define body objects.
Body accepts three options for defining inertia properties of a body: Mass, Inertia, and Centroid. These options will be discussed further in Section 8.1. The options for Body that are relevant in the kinematic phase of analysis are the PointList and InitialGuess options.
### 2D
2D options for Body.
### 3D
3D options for Body.
### 2D/3D
Default option settings for Body.
Executing the Body function does not, in itself, have any affect on the current Mech model. The returned body objects must be added to the model with the SetBodies command, which is discussed in Section 2.1.2.
Initial guesses specified with a body object are referred to as the permanent initial guesses. These initial guesses are used the first time that a solution is sought with SolveMech. Each subsequent solution attempt uses the previous solution as its initial guess. When the initial guesses are set with SetGuess, they may be explicitly set to user-specified values, or reset to the stored values of the permanent initial guesses.
### Example
The 2D example mechanism of Section 1.2 can be modified to use three body objects instead of repeatedly calling out the several local point coordinates on the ground, crankshaft, and piston bodies. These body objects store the local point definitions for future reference, through their integer point numbers, by other Mech functions.
• The ground body object uses three local point definitions.
P1 is a point at the global origin that is the rotational axis of the crankshaft.
P2 and P3 are two points that define the vertical line upon which the piston translates.
• The crankshaft uses two local point definitions.
P1 is a point at the local origin that is the rotational axis of the crankshaft.
P2 is a point on the local x axis that is the attachment point of the connecting rod.
• The piston also uses two local point definitions.
P1 is a point on the local y axis that is the attachment point of the connecting rod.
P2 is a point on the local y axis that is the upper end of the translation line of the piston.
• ### InitialGuess
Alternately, the piston could have been defined with the local origin at the attachment point of the connecting rod. This would suggest that a different initial guess be used than the default of X3 = 0, Y3 = 0, 3 = 0.
This body object functions identically to the first one, except the Y coordinate of the piston is always 4 units greater, because the origin of the body has moved 4 units upward.
Note that each of these body objects must be added to the current model with SetBodies before they will have any effect on the model.
## 2.1.2 SetBodies
The Mech SetBodies function is used to apply the data contained in body objects to the current kinematic model.
The body object processing function.
All properties specified in a call to SetBodies overwrite properties that are currently defined unless the property specification is the default, Automatic, in which case the property is left unchanged.
If SetBodies is passed a body object with an InitialGuess specification other than Automatic, the current initial guesses of each body are reset to the values of the permanent initial guesses.
There is a subtle issue regarding the use of the PointList specification with SetBodies. When SetBodies is run with a new InitialGuess specification, the changed guess is immediately reflected in Mech's internal database. The same holds true for changing inertia properties with SetBodies; the mass matrix is immediately updated.
However, changing the coordinates of a point with PointList must reflect a change throughout the constraint equations defined with SetConstraints. This cannot be done without rebuilding the constraint equations. Therefore, if the constraint equations have already been built with SetConstraints when SetBodies is passed a PointList specification other than Automatic, the constraint equations will all be automatically rebuilt by SetConstraints the next time that a solution is sought, which may cause a noticeable delay in the operation of SolveMech.
The following command incorporates the three body objects that were defined in Section 2.1.1 into the current model.
SetBodies returns Null. All of the mechanism points and properties passed to SetBodies are stored in Mech's internal database.
Now that local point definitions have been added to the current model, they can be referenced by their local point numbers in any Mech function. Instead of using the local coordinates of a point, simply use the integer point number. For example, the Translate2 constraint in the crankshaft-piston model would use four local point numbers instead of the four explicit 2D coordinates in the Line objects.
### Updating Properties
SetBodies only sets the values of the properties that are not specified with the default, Automatic. Thus, SetBodies can be used to update the value of individual body properties without affecting properties already defined.
### PointList
Note that PointList, like many Mech options, can also be used as a function.
Another usage of PointList.
# 2.2 Points, Vectors, and Axes
This section explains the functionality and syntax of Mech's method of defining the basic geometric entities that make up a mechanism model: points, vectors, and axes. A look at the usage statements for most of Mech's functions shows that they take arguments of type point, vector, or axis. In all cases there are several ways to specify a point, vector, or axis, depending on whether the entity is defined in global or local coordinates, is located entirely on a single body, or spans across multiple bodies.
The three basic types of Mech geometry objects.
## 2.2.1 Points
The most basic geometric entity used by Mech is the point object. Point objects reference the location of a 2D or 3D point in global coordinates or in local coordinates on a specified body. Point objects may be given as 2D or 3D vectors, or they may have the head Point just like the built-in Mathematica graphics primitive. This does not lead to any conflict or require a redefinition of Point because it has no definition by default. The Point head is only used to hold its arguments in Mech, just as it is in standard Mathematica.
Any Mech function that calls for a point object accepts either of the following syntax.
### 2D
Methods of defining 2D point objects.
### 3D
Methods of defining 3D point objects.
Assuming that the appropriate body object definition has been made, any point on any body can be referenced through either its global coordinates, its local coordinates, or its local coordinate number. The following example shows several point objects that are functionally identical.
## 2.2.2 Vectors
The basic geometric entity specifying direction is the vector object. Vector objects reference the direction of a 2D or 3D vector, in global coordinates or in local coordinates on a specified body. Vector objects may be given as 2D or 3D vectors in global coordinates, or they may have the head Vector and be given in local coordinates on a specific body.
A
Line can be used to specify a vector object, in which case the direction of the Line is used as the direction of the vector. A Plane (3D only) may also be used to specify a vector object, in which case the normal direction of the Plane is used as the direction of the vector. See the following two subsections for further information on Line and Plane.
Any Mech function that calls for a vector object accepts either of the following syntax.
### 2D
Methods of defining 2D vector objects.
### 3D
Methods of defining 3D vector objects.
Assuming that the appropriate body object definition has been made, any vector on any body can be referenced through either its global coordinates, its local coordinates, or its local point number. The following example shows several vector objects that are functionally identical.
## 2.2.3 Lines
The Line object is used by Mech to reference a line in 2D or 3D space. A Line is defined two points: an origin point and an end point. The two points may be on the same or two different bodies. Because a Line object possesses both direction and location, it can be used in place of a vector object or an axis object in any Mech function that calls for a vector or an axis object.
Line objects have the same head as the built-in Mathematica Line graphics primitive. This does not lead to any conflict or require a redefinition of Line because it has no definition by default. The Line head is only used to hold its arguments in Mech, just as it is in standard Mathematica.
Any Mech function that calls for a vector or axis object accepts any of the following syntax.
### 2D
Methods of defining 2D line objects.
### 3D
Methods of defining 3D line objects.
Assuming that the appropriate body object definition has been made, any line spanning bodies or on a single body can be referenced with a combination of global points, local points, or local point numbers. The following example shows several line objects that are functionally identical.
## 2.2.4 Planes
The Plane object is used by Modeler3D to reference a planar surface in 3D space. A Plane is defined by three points and has a specific origin, normal direction, and reference direction (see Axis). The first of the three points is the origin, and the vector from the first point to the second point is the reference direction. The third coplanar point cannot be colinear with the first two points. The positive normal is given by the cross product of the vector from the first point to the second point and the vector from the first point to the third point.
Because a Plane object possesses the properties of direction and location, it can be used in place of a vector object or an axis object in any Mech function that calls for a vector or an axis. If a Plane object is used to specify direction, the normal direction is used as the direction vector.
Any Modeler3D function that calls for a vector or axis object accepts any of the following syntax.
### 3D
Methods of defining plane objects.
Assuming that the appropriate body object definition has been made, a plane spanning any one, two, or three bodies can be referenced with a combination of global points, local points, or local point numbers. The following example shows several plane objects that are functionally identical.
## 2.2.5 Axes
The axis object is used by Mech to reference an axis in 2D or 3D space. An axis object is specified by origin and direction. Optionally, 3D axis objects may have a reference direction specified. The reference direction is a vector that is not parallel to the direction of the axis. The reference direction is used by some Modeler3D constraints to control rotation about the axis.
An axis object may have the head Axis,
Line, or Plane. If a Line is used to specify an axis object, the origin and direction of the Line are used as the origin and direction of the axis. Because a Line object has no specific reference direction, a Line should not be used in place of a Modeler3D axis object in a function where the reference direction is relevant.
If a Plane is used to specify an axis object, the origin of the Plane is used as the origin of the axis, the normal direction of the Plane is used as the direction of the axis, and the line from point1 to point2 on the Plane is used as the reference direction of the axis.
Any Mech function that calls for an axis object accepts any of the following syntax.
### 2D
Methods of defining 2D axis objects.
### 3D
Methods of defining 3D axis objects.
Assuming that the appropriate body object definitions have been made, any axis with origin and direction specified with coordinates on a single body or multiple bodies can be referenced with a combination of global points, local points, or local point numbers. The following example shows several axis objects that are functionally identical.
## 2.2.6 The Zeroth Point
For convenience, local point 0 (zero) can be used to reference the local origin of any body, although point 0 is never defined. The zeroth point can always be used, regardless of whether or not any SysBody objects have been defined and passed to SetBodies. The following examples show the use of the zeroth point.
# 2.3 Constraint Objects
Constraint objects are used to define the mathematical relationships that make up the kinematic model in terms of the physical relationships of the various mechanism bodies. All Mech constraint functions return constraint objects with the head SysCon. These constraint objects are passed to the SetConstraints function to build the mathematical model.
## 2.3.1 General Format
All arguments to any constraint function follow the same general format.
Format for all constraint functions.
Further, all constraint function names end with a single digit, except the general constraint function Constraint. This digit signifies the number of degrees of freedom constrained by the constraint. For example, the 2 in the Translate2 constraint function name signifies that it contains two constraint equations and constrains two degrees of freedom in a model.
## 2.3.2 2D Basic Constraints
The Modeler2D basic constraints specify elementary geometric relationships between points and lines on mechanism bodies. Elementary relationships are used for the majority of simple mechanical joints. A shaft spinning in a bearing is modeled by specifying that two points are coincident. A connecting rod can be modeled by specifying a constant distance between two points. A guide pin in a slot is modeled by specifying that a point on one body lies on a line on another body. A piston in a cylinder is modeled by specifying that two lines on two different bodies are colinear.
Many seemingly different mechanical joints are often modeled with the same constraint, such as a connecting rod versus a guide pin in a circular track; both are modeled by specifying the distance between two points on two different bodies. It is up to the user to determine the true nature of the kinematic relationships that make up a mechanism.
The entire set of Modeler2D constraint functions is quite redundant; all basic 2D mechanisms could, in fact, be modeled with a small subset of the Modeler2D constraints. The goal in this is to provide a constraint that is similar in form to the actual mechanical joint being modeled. Thus, the readability of the model can be improved.
Local coordinate constraints directly control the local coordinates of specified bodies.
Point-on-line constraints force a point on a body to travel along a specified line.
Point-at-position constraints force a point on a body to lie at a specified position.
Angular constraints control the angle between the direction vectors of two lines.
Compound constraints enforce multiple geometric relationships simultaneously.
There are several other compound constraints available for modeling gears and cams, these are discussed in Chapter 6.
To demonstrate the ambiguity in determining which constraint is the correct constraint, the set of four constraints used in the crankshaft-piston model are examined. The first constraint, Revolute2, could be modeled with a pair of constraints with one degree of freedom each; a RelativeX1 to control the X coordinate of the crankshaft axis, and a RelativeY1 to control the Y coordinate.
The second constraint, RotationLock1, could be modeled by constraining the angle between two lines, one on the crankshaft and one on the ground.
The RelativeDistance1 constraint, in this application, cannot be replaced by the use of any other constraint. It is somewhat unique, geometrically.
## 2.3.3 3D Basic Constraints
The Modeler3D basic constraints specify elementary geometric relationships between points, lines, and planes on mechanism bodies. Elementary relationships are used for the majority of mechanical joints. A shaft spinning in a bearing is modeled by specifying that two axes are coincident. A connecting rod can be modeled by specifying a constant distance between two points. A guide pin in a slot is modeled by specifying that a point on one body lies on a line on another body. A piston in a cylinder is modeled by specifying that two lines on two different bodies are colinear.
Many seemingly different mechanical joints are often modeled with the same constraint, such as a connecting rod versus a guide pin in a circular track; both are modeled by specifying the distance between two points on two different bodies. It is up to the user to determine the true nature of the kinematic relationships that make up a mechanism.
The entire set of Modeler3D constraint functions is quite redundant; all basic 3D mechanisms could, in fact, be modeled with a small subset of Modeler3D constraints. The goal in this is to provide a constraint that is similar in form to the actual mechanical joint being modeled. Thus, the readability of the model is improved.
Local coordinate constraints directly control the local coordinates of specified bodies.
Point-on-surface constraints force a point to lie on a specified surface.
Point-on-line constraints force a point to lie on a line or curve.
Point-at-position constraints control the position of a point.
Angular constraints control the angle between two vectors.
Parallel constraint enforces parallelism between two vectors.
Constraints that enforce more complex geometric relationships.
Compound constraints enforce multiple geometric relationships simultaneously.
There are several other compound constraints available for modeling gears and cams; these are discussed in Chapter 6.
To demonstrate the ambiguity in determining which constraint is the correct constraint, the set of five constraints used in the crankshaft-piston model are reexamined. The first constraint, a Revolute5, could be modeled instead with two constraints; a Spherical3 constraint to control the position of the crankshaft, and a Parallel2 constraint to control the direction vector of the crankshaft's axis.
The third constraint, Cylindrical4, could be modeled by using two PointOnLine2 constraints.
The RelativeDistance1 constraint in this application cannot be replaced by the use of any other constraint. It is somewhat unique, geometrically.
## 2.3.4 The General Constraint
The general constraint allows the user to enter a mathematical relationship directly as an equation.
The general constraint function.
For example, to constrain the origin of body 2 to lie on a paraboloid surface in the global coordinate system with its axis in the Z direction and Z intercept at {0, 0, 4}, Constraint would be used as follows.
# 2.4 Building the Model
Constraint objects are processed into a kinematic model with the SetConstraints function. The system of equations generated by SetConstraints is not returned to the user; it is stored in several variables in Mech's private context. Thus only one mechanism model can be current in Mech at any given time. Several means of accessing and modifying the model are provided by Mech; these are discussed in the following section.
## 2.4.1 SetConstraints
SetConstraints is Mech's core model building function. SetConstraints takes constraint objects that represent the physical interactions in the mechanism model and processes them into a system of kinematic equations.
The core model-building function.
SetConstraints accepts options that determine how certain parts of the mathematical model are built.
An option for SetConstraints.
## 2.4.2 Constraints
The actual constraint expressions that are generated by SetConstraints can be extracted with Constraints. Constraints can be very useful for debugging a model because it is often possible to read the constraint expressions directly and relate them to the function of the model.
Constraints function.
Consider the 2D crankshaft-piston model of Section 1.2. This model is built with various settings for the options for SetConstraints to show their affect on the resulting equations.
The first constraint cs[1] is a two degree of freedom constraint that constrains the origin of the crankshaft to lie at the global origin.
Obviously, when these two expressions are driven to zero, the desired result is achieved, the coordinates of the local origin of the crankshaft {X2, Y2} are zero.
The third constraint cs[3] is a two degree of freedom constraint that allows the piston to translate along a vertical axis, but constrains its rotation and its translation in the X direction.
The first expression, 10 Sin[3], constrains the rotation of the piston; this expression can be equal to zero if 3 is any integer multiple of . The reason that the solution 3 -> 0 is found is because the default initial guess is 3 -> 0. If a different initial guess were used, such as 3 -> 3, the model would converge to 3 -> N[] instead.
The second expression in the list constrains the X coordinate of the piston to be zero. Notice that at any value of 3 that satisfies the first expression, the second expression reduces to or , either of which will drive X3 to zero.
## 2.4.3 Initial Guesses
SetGuess serves both as an option for several Mech functions, and as a function for temporarily setting a model's initial guesses.
Various methods of setting initial guesses.
A SolveMech utility.
To advance the position for the model in large steps, it is sometimes necessary to specify a better initial guess than the current one. For example, to turn the crankshaft 3/4 of a turn at once, it might be necessary to set a new initial guess for 2 and Y3, the angular coordinate of the crankshaft and the height of the piston, respectively.
If SetConstraints were to be run at this time with the default options, the initial guesses would remain at their current values, and . The SetGuess option resets the guesses to their defaults.
## 2.4.4 \$MechProcessFunction
\$MechProcessFunction is a function that is applied to all of the mechanism equations that are generated by SetConstraints as they are created. The default setting, \$MechProcessFunction = Chop, can reduce the size of the constraint expressions substantially if any of the constraints involve the local origins of bodies in the model which have local coordinates of or . Setting \$MechProcessFunction = Identity disables this action.
Set \$MechProcessFunction equal to Identity to disable chopping.
In[21]:=
## 2.4.5 SetSymbols
The default symbols that appear in the solution rules returned by SolveMech can be changed with the SetSymbols function.
A model set-up function.
Default settings.
Because SetSymbols changes the behavior of all Mech functions that return symbolic expressions, SetSymbols must be run before building any other part of the mechanism model with SetBodies or SetConstraints. SetSymbols has the effect of executing ClearMech[].
The x, y, and z symbols are used by Modeler3D to represent angular velocities used in 3D velocity analysis, which is discussed in Chapter 4. The symbol is a Lagrange multiplier used in static and dynamic analysis, which is discussed in Chapter 7. The d symbol is used in 2D and 3D velocity analysis to symbolically represent differentiation.
If the current model is rebuilt with a new setting for SymbolBasis, the solution rules returned by SolveMech are changed accordingly.
# 2.5 Running the Model
This section covers the use of the SolveMech command, which is used to seek a solution to the mechanism model. The Mech commands covered in the previous sections were all concerned with building a mechanism model. SolveMech is the sole function for running the model.
## 2.5.1 SolveMech
The SolveMech command is used without arguments to seek a single solution to the mechanism model, with all of the current values of any user-defined variables in the model. When SolveMech seeks a solution, what it is really trying to do is numerically find the location and orientation of each of the bodies in the model at which all of the constraints are satisfied.
Mech's runtime command.
### Time Dependence
Most mechanism models have some dependency on time that is specified (by default) by the symbol T. When T appears in a mechanism model, the model's location becomes explicitly a function of time, a technique that must be used in the velocity and acceleration portions of the analysis. It is not necessary to have any dependency on T in a model used only for location analysis, but it is customary to make the primary driving expression a function of T anyway.
The symbol for time.
SolveMech has several special invocations that are useful for finding a solution or many solutions at a range of values of T.
More arguments for SolveMech.
The 2D crankshaft-piston model is used to demonstrate the various forms of SolveMech. When this example was first used in Section 1.2, the symbol crankangle was used to specify the rotation of the crankshaft. Now crankangle is replaced with so that the rotation of the crankshaft is a function of time. Thus, the crankshaft has an angular velocity of one revolution per second.
Now, instead of changing the value of crankangle to drive the model the value of time is changed directly with SolveMech.
The unusual form of the iteration specifier in SolveMech is very useful if an even spread of data is desired, with a few particular data points specified in between. For example, the following example generates six solution points with the points T = 0., 0.223, and 0.3 included, and the other three points spread so as to fill the gaps as evenly as possible.
### Initial Guesses at Runtime
Setting initial guesses with SolveMech.
The primary reason for passing rules into SolveMech, instead of just specifying the value of time, is to accelerate the solution process. The supplied rules may be known to be very good initial guesses, such as rules obtained from a prior run with a slightly different value for a parameter.
## 2.5.2 Controlling SolveMech
Options for SolveMech.
The operation of SolveMech can be inspected by the use of two hook functions that are called during the execution of SolveMech. Note that all Mech system variables that can be set by the user have names of the form "\$Mech*".
Controls for SolveMech.
### Failure to Converge
The default setting for ZeroTest will suffice for many models, but may need to be changed for several reasons.
If the numerical size of the dimensions of the model are very large or small, ZeroTest may need to be changed accordingly. For example, if a typical point in the model is Point[2, {3400, -7800}], it might be necessary to specify ZeroTest -> (Abs[#] < 10^-6 &).
If the model is large (eight or more bodies for 2D, four or more for 3D) solving the linear equations at each iteration begins to suffer from numerical error. Thus, it may be necessary to lower the convergence criteria.
The primary indication that the convergence criteria must be changed is that SolveMech returns a valid solution that is apparently correct, but an error message is generated claiming that convergence has failed after n iterations. If the model fails to converge and returns a solution that is obviously not valid, try to find an inconsistency in the constraint set or consult Chapter 13, Debugging Models.
## 2.5.3 Nonlinear Equation Solver
At the core of Mech's numerical solution block is a Newton-Rhapson simultaneous nonlinear equation solver, similar to the built-in Mathematica function FindRoot. In fact, Mech can be used to solve arbitrary systems of equations that have no relation to mechanism analysis.
Mech's general constraint function Constraint is used to build a constraint object from an arbitrary algebraic equation. The algebraic constraint can be incorporated into a mechanism model along with other geometric constraints, or an entire model can be built from general constraints.
The general constraint function.
The following uses Mech to find the intersection of a parabola and a circle.
# 2.6 Variable Management
Mechanism models often need to have many adjustable parameters that can be tweaked between runs of the model. A Mech model may have any number of user-defined variables, provided that all such variables in a model have numerical values before SolveMech is called.
## 2.6.1 User-Defined Variables
Symbolic parameters can enter a kinematic model in two ways. The coordinates of a local point can be given in symbolic form or the arguments of a constraint function that specify a distance or angle can be given as a symbolic expression. The following example shows how local point specifications can be symbolic.
The crankshaft-piston model can be built with several variables in the arguments to its constraint functions. The following example introduces symbols into the model to represent the frequency of the crankshaft rotation, the length of the connecting rod, and the stroke of the crankshaft.
If SolveMech is called now to seek a solution, an error message is generated because some of the symbols in the model have not yet been defined.
A debugging aid.
CheckSystem reports the names of any undefined symbols, as well as several other possible errors in the model.
### Dependent Variables
It is acceptable to use Mech dependent variables explicitly when defining local points or constraints. Highly abstract compound constraints can be specified in this manner.
This constraint would cause the Y coordinate of a point on body 2 to be equal to the X coordinate of the origin of body 3, squared. When using dependent variables in constraints, care must be taken to ensure that the dependence of the constraint on the variables is explicit at the time that SetConstraints is run. If not, differentiation with respect to those variables cannot occur, and the constraints will not be processed correctly.
Note that the names of the dependent variables used by Mech are cleared and protected when SetConstraints is run. Thus, if there happens to be a definition in place such as Y2 = 3, it is removed by SetConstraints.
## 2.6.2 Parameters
Another method of managing variables in a Mech model is to use the parameters functions. The parameters list is basically a list of rules stored in Mech's internal database that is applied to the mechanism model immediately before the solution phase. By using parameters, problems can be avoided such as failing to define all of the symbols in a model or accidentally defining them before SetConstraints is run and hard coding them into the model.
Parameter management functions.
Note that when a parameter is set with SetParameters or Parameterize, it is protected until it is unset with ClearParameters. This is to ensure that the parameter cannot be accidentally given a definition in the course of running the model.
The crankshaft-piston model, which was built in Section
2.6.1, is used to demonstrate the use of parameters. The definitions of the three symbols in the model are cleared and then added to the parameters list.
The crankshaft-piston model can now be built without worrying about the definitions of the parameters. SetParameters can be used at any time in the model-building process. | 6,646 | 32,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-17 | latest | en | 0.870736 |
https://www.elbeno.com/haskell_soe_blog/?p=38 | 1,719,203,672,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00675.warc.gz | 653,077,490 | 15,204 | # Exercise 8.9
Proof from axioms is something that takes great care: it is all too easy to skip a step that “obviously follows” from the axioms without proving that step. I’m not convinced this is all correct…
```r
≡ r `Intersect` univ [Axiom 4]
≡ r `Intersect` (r `Union` Complement r) [Axiom 5]
≡ (r `Intersect` r) `Union` (r `Intersect` Complement r) [Axiom 3]
≡ (r `Intersect` r) `Union` Empty [Axiom 5]
≡ (r `Intersect` r) [Axiom 4]
Lemma 1: r ≡ (r `Intersect` r)
r
≡ r `Union` Empty [Axiom 4]
≡ r `Union` (r `Intersect` Complement r) [Axiom 5]
≡ (r `Union` r) `Intersect` (r `Union` Complement r) [Axiom 3]
≡ (r `Union` r) `Intersect` univ [Axiom 5]
≡ (r `Union` r) [Axiom 4]
Lemma 2: r ≡ (r `Union` r)
r `Union` univ
≡ r `Union` (r `Union` Complement r) [Axiom 5]
≡ (r `Union` r) `Union` Complement r [Axiom 1]
≡ r `Union` Complement r [Lemma 2]
≡ univ [Axiom 5]
Lemma 3: r `Union` univ ≡ univ
r `Intersect` Empty
≡ r `Intersect` (r `Intersect` Complement r) [Axiom 5]
≡ (r `Intersect` r) `Intersect` Complement r [Axiom 1]
≡ r `Intersect` Complement r [Lemma 1]
≡ Empty [Axiom 5]
Lemma 4: r `Intersect` Empty ≡ Empty
Let X = (r1 `Union` (r1 `Intersect` r2))
X `Intersect` Complement r1
≡ (r1 `Union` (r1 `Intersect` r2)) `Intersect` Complement r1
≡ (r1 `Intersect` Complement r1) `Union`
((r1 `Intersect` r2) `Intersect` Complement r1) [Axiom 3]
≡ Empty `Union` ((r1 `Intersect` r2) `Intersect` Complement r1) [Axiom 5]
≡ (r1 `Intersect` r2) `Intersect` Complement r1 [Axiom 4]
≡ (r2 `Intersect` r1) `Intersect` Complement r1 [Axiom 2]
≡ r2 `Intersect` (r1 `Intersect` Complement r1) [Axiom 1]
≡ r2 `Intersect` Empty [Axiom 5]
≡ Empty [Lemma 4]
∴ by Axiom 5, X = r1
Lemma 5: (r1 `Union` (r1 `Intersect` r2)) ≡ r1
r1 `Intersect` (r1 `Union` r2)
≡ (r1 `Union` r1) `Intersect` (r1 `Union` r2) [Lemma 2]
≡ r1 `Union` (r1 `Intersect` r2) [Axiom 3]
≡ r1 [Lemma 5]
Lemma 6: (r1 `Intersect` (r1 `Union` r2)) ≡ r1
Let X = Complement (Complement r)
X `Union` (Complement r)
≡ Complement (Complement r) `Union` (Complement r)
≡ univ [Axiom 5]
∴ by Axiom 5, X = r
Lemma 7: Complement (Complement r) ≡ r
Let X = Complement Empty
X `Union` Empty
≡ Complement Empty `Union` Empty
≡ univ [Axiom 5]
∴ by Axiom 4, X = univ
Lemma 8: Complement Empty ≡ univ
Complement univ
≡ Complement (Complement Empty) [Lemma 8]
≡ Empty [Lemma 7]
Lemma 9: Complement univ ≡ Empty
Let X = Complement (r1 `Union` r2)
(r1 `Union` r2) `Union` X
≡ (r1 `Union` r2) `Union` Complement (r1 `Union` r2)
≡ univ [Axiom 5]
Let Y = Complement r1 `Intersect` Complement r2
(r1 `Union` r2) `Union` Y
≡ (r1 `Union` r2) `Union` (Complement r1 `Intersect` Complement r2)
≡ ((r1 `Union` r2) `Union` Complement r1) `Intersect`
((r1 `Union` r2) `Union` Complement r2) [Axiom 3]
≡ ((r2 `Union` r1) `Union` Complement r1) `Intersect`
((r1 `Union` r2) `Union` Complement r2) [Axiom 2]
≡ (r2 `Union` (r1 `Union` Complement r1)) `Intersect`
((r1 `Union` r2) `Union` Complement r2) [Axiom 1]
≡ (r2 `Union` (r1 `Union` Complement r1)) `Intersect`
(r1 `Union` (r2 `Union` Complement r2)) [Axiom 1]
≡ (r2 `Union` univ) `Intersect` (r1 `Union` (r2 `Union` Complement r2)) [Axiom 5]
≡ (r2 `Union` univ) `Intersect` (r1 `Union` univ) [Axiom 5]
≡ univ `Intersect` (r1 `Union` univ) [Lemma 3]
≡ univ `Intersect` univ [Lemma 3]
≡ univ [Lemma 3]
∴ X ≡ Y
Lemma 10: Complement (r1 `Union` r2) ≡ Complement r1 `Intersect` Complement r2
Let X = Complement (r1 `Intersect` r2)
(r1 `Intersect` r2) `Union` X
≡ (r1 `Intersect` r2) `Union` Complement (r1 `Intersect` r2)
≡ univ [Axiom 5]
Let Y = Complement r1 `Union` Complement r2
(r1 `Intersect` r2) `Union` Y
≡ (r1 `Intersect` r2) `Union` (Complement r1 `Union` Complement r2)
≡ (r1 `Union` (Complement r1 `Union` Complement r2)) `Intersect`
(r2 `Union` (Complement r1 `Union` Complement r2)) [Axiom 3]
≡ (r1 `Union` (Complement r1 `Union` Complement r2)) `Intersect`
((Complement r1 `Union` Complement r2) `Union` r2) [Axiom 2]
≡ ((r1 `Union` Complement r1) `Union` Complement r2) `Intersect`
((Complement r1 `Union` Complement r2) `Union` r2) [Axiom 1]
≡ ((r1 `Union` Complement r1) `Union` Complement r2) `Intersect`
(Complement r1 `Union` (Complement r2 `Union` r2)) [Axiom 1]
≡ (univ `Union` Complement r2) `Intersect`
(Complement r1 `Union` (Complement r2 `Union` r2)) [Axiom 5]
≡ (univ `Union` Complement r2) `Intersect`
(Complement r1 `Union` (r2 `Union` Complement r2)) [Axiom 2]
≡ (univ `Union` Complement r2) `Intersect` (Complement r1 `Union` univ) [Axiom 5]
≡ (univ `Union` Complement r2) `Intersect` univ [Lemma 3]
≡ univ `Union` Complement r2 [Axiom 4]
≡ Complement r2 `Union` univ [Axiom 2]
≡ univ [Lemma 3]
∴ X ≡ Y
Lemma 11: Complement (r1 `Intersect` r2) ≡ Complement r1 `Union` Complement r2```
### 5 responses to “Exercise 8.9”
1. IMHO proof of Lemma 4 is not complete.
It is assumed that if
X `Union` Complement r = univ (1)
then X has to be r, because
r `Union` Complement r = univ
In fact to fullfill (1) it is enough that X contains r, not that it is equal.
Proof has to be a bit longer
Similarly as (1) we can proove
X `Intersect` Complement r = Empty (2)
r
= r `Intersect` univ (because axiom 4)
= r `Intersect` (X `Union` Complement r) (because 1)
= (r `Intersect` X) `Union` (r `Intersect` Complement r) (because axiom 3)
= (r `Intersect` X) `Union` Empty (because axiom 5)
= (r `Intersect` X) `Union` (X `Intersect` Complement r) (because 2)
= (X `Intersect` r) `Union` (X `Intersect` Complement r) (because axiom 2)
= X `Intersect` ( r `Union` Complement r) (because axiom 3)
= X `Intersect` univ (because axiom 5)
= X (because axiom 4)
= Complement (Complement r)
Similar problem I see with proofs of lemmas 10 and 11
2. fero says:
IMHO I don’t think you can do
“In fact to fullfill (1) it is enough that X contains r, not that it is equal.”
It is true but there is no axiom that states that. You can use only axioms.
3. Jonathan says:
Actually
r1 `Union` (r1 `Intersect` r2)
In the => notation follows from or-introduction. It is always the case that if a then a or b
So start
r1,
therefore, r1 `union` (r1 `intersect` r2)
Then for equivalence you need actually to go the other way too. There is a lot of abuse of notation going on in this section of the book, and I found this exercise to be a stretch.
Start,
r1 `Union` (r1 `Intersect` r2)
Now, for or-elimination you have to handle both cases.
Assume r1.
Therefore r1.
Assume (r1 `Intersect` r2).
By and-elimination, conclude
r1.
Therefore, r1.
Therefore r1 r1 `Union` (r1 `Intersect` r2).
I know the author states to use the axioms alone, but if there are ways to prove things without making weird declarations, then I prefer them as they make for better habits.
4. Jonathan says:
Why write anything at all for Theorem 8:
Complement Empty univ.
By definition univ = Complement Empty.
5. Julian says:
For simplicity and brevity, I will use somthing like standard
set-theoretic notations as abbreviations for Intersection and Union,
writing
r1 u r2 for r1 `Union` r2
and
r1 n r2 for r1 `Intersection` r2
Also, I will write r’ for Complement r, 0 for Empty and 1 for univ,
Then the five axioms, together with some simple corollaries, read:
Axiom 1
——-
r1 u (r2 u r3) = (r1 u r2) u r3
r1 n (r2 n r3) = (r1 n r2) n r3
Axiom 2
——-
r1 u r2 = r2 u r1
r1 n r2 = r2 n r1
Axiom 3
——-
r1 n (r2 u r3) = (r1 n r2) u (r1 n r3)
r1 u (r2 n r3) = (r1 u r2) n (r1 u r3)
Corollaries:
(r2 u r3) n r1 = (r2 n r1) u (r3 n r1)
(r2 n r3) u r1 = (r2 u r1) n (r3 u r1)
Proof: Apply Axiom 2 to both sides.
Axiom 4
——-
r u 0 = r
r n 1 = r
Corollaries:
0 u r = r
1 n r = r
Proof: Apply Axiom 2 to the left side.
Axiom 5
——-
r u r’ = 1
r n r’ = 0
Corollaries:
r’ u r = 1
r’ n r = 0
Proof: Apply Axiom 2 to the left side.
Now for the exercise at hand…
Lemma 1
——-
r n r = r
r u r = r
Proof:
r = r n 1 [Axiom 4]
= r n (r u r’) [Axiom 5]
= (r n r) u (r n r’) [Axiom 3]
= (r n r) u 0 [Axiom 5]
= r n r [Axiom 4]
The proof of the second statement is essentially identical, swapping
n with u and 1 with 0:
r = r u 0 [Axiom 4]
= r u (r n r’) [Axiom 5]
= (r u r) n (r u r’) [Axiom 3]
= (r u r) n 1 [Axiom 5]
= r u r [Axiom 4]
Lemma 2
——-
r u 1 = 1
r n 0 = 0
Proof:
r u 1 = r u (r u r’) [Axiom 5]
= (r u r) u r’ [Axiom 1]
= r u r’ [Lemma 1]
= 1 [Axiom 5]
and again, the other is exactly parallel:
r n 0 = r n (r n r’) [Axiom 5]
= (r n r) n r’ [Axiom 1]
= r n r’ [Lemma 1]
= 0 [Axiom 5]
Corollaries:
1 u r = 1
0 n r = 0
Proof: Apply Axiom 2 to the left side.
Lemma 3
——-
r1 u (r1 n r2) = r1
r1 n (r1 u r2) = r1
Proof (the first step is sneaky):
r1 u (r1 n r2) = (r1 n 1) u (r1 n r2) [Axiom 4]
= r1 n (1 u r2) [Axiom 3, in reverse]
= r1 n 1 [Lemma 2, corollary]
= r1 [Axiom 4]
And again, swapping n with u and 1 with 0 gives the corresponding
result:
r1 n (r1 u r2) = (r1 u 0) n (r1 u r2) [Axiom 4]
= r1 u (0 n r2) [Axiom 3, in reverse]
= r1 u 0 [Lemma 2, corollary]
= r1 [Axiom 4]
Alternatively, the latter can be deduced from the former as follows:
r1 n (r1 u r2) = (r1 n r1) u (r1 n r2) [Axiom 3]
= r1 u (r1 n r2) [Lemma 1]
= r1 [first part of Lemma 3]
Lemma 4
——-
r” = r, where r” means (r’)’
Proof:
The idea of this proof is to show that r” u r = r and also r” u r = r”,
so that r = r” u r = r”, or r” = r. It could be written in one
long chain, but it is clearer to write it in two halves.
We have
r” u r = (r” u r) n 1 [Axiom 4]
= (r” u r) n (r u r’) [Axiom 5]
= (r u r”) n (r u r’) [Axiom 2]
= r u (r” n r’) [Axiom 3, in reverse]
= r u (r’ n (r’)’) [Axiom 2, and writing r”=(r’)’]
= r u 0 [Axiom 5]
= r [Axiom 4]
Likewise, but this time writing 1 = r’ u (r’)’ instead of 1 = r u r’,
we get:
r” u r = (r” u r) n 1 [Axiom 4]
= (r” u r) n (r’ u (r’)’) [Axiom 5]
= (r” u r) n (r” u r’) [Axiom 2, and writing (r’)’=r”]
= r” u (r n r’) [Axiom 3, in reverse]
= r” u 0 [Axiom 5]
= r” [Axiom 4]
Therefore r = r”.
Lemma 5
——-
0′ = 1
1′ = 0
Well, the first is the definition of univ (as given just before Axiom
3), and the proof the of second is trivial:
1′ = (0′)’ [By definition: 1 = 0′]
= 0 [Lemma 4]
Lemma 6
——-
(r1 u r2)’ = r1′ n r2′
(r1 n r2)’ = r1′ u r2′
These are the famous De Morgan’s Theorems, and their proof is tricky
(and not due to me, either!). We prove a preliminary lemma first.
Lemma 7
——-
If r1′ n r2 = 0 and r1′ u r2 = 1, then r1 = r2.
Proof:
r1 = r1 u 0 [Axiom 4]
= r1 u (r1′ n r2) [Hypothesis]
= (r1 u r1′) n (r1 u r2) [Axiom 3]
= 1 n (r1 u r2) [Axiom 5]
= (r1′ u r2) n (r1 u r2) [Hypothesis]
= (r2 u r1′) n (r2 u r1) [Axiom 2]
= r2 u (r1′ n r1) [Axiom 3, in reverse]
= r2 u 0 [Axiom 5, corollary]
= r2
Proof of Lemma 6
—————-
We want to show first that (r1 u r2)’ = r1′ n r2′, so we aim to show
that ((r1 u r2)’)’ n (r1′ n r2′) = 0 and ((r1 u r2)’)’ u (r1′ n r2′)
= 1, so that (r1 u r2)’ = r1′ n r2′ by Lemma 7.
We note first that ((r1 u r2)’)’ = r1 u r2 by Lemma 4. We then have
(r1 u r2) n (r1′ n r2′)
= ((r1 u r2) n r1′) n r2′ [Axiom 1]
= ((r1 n r1′) u (r2 n r1′)) n r2′ [Axiom 3, corollary]
= (0 u (r2 n r1′)) n r2′ [Axiom 5]
= (r2 n r1′) n r2′ [Axiom 4, corollary]
= (r1′ n r2) n r2′ [Axiom 2]
= r1′ n (r2 n r2′) [Axiom 1]
= r1′ n 0 [Axiom 5]
= 0 [Lemma 2]
and
(r1 u r2) u (r1′ n r2′)
= ((r1 u r2) u r1′) n ((r1 u r2) u r2′) [Axiom 3]
= ((r2 u r1) u r1′) n ((r1 u r2) u r2′) [Axiom 2]
= (r2 u (r1 u r1′)) n (r1 u (r2 u r2′)) [Axiom 1]
= (r2 u 1 ) n (r1 u 1) [Axiom 5]
= 1 n 1 [Lemma 2]
= 1 [Axiom 4]
Thus, by Lemma 7, (r1 u r2)’ = r1′ n r2′.
For the second part of Lemma 6, we can either argue identically, or as
follows:
We have
(r1 n r2)’ = ( (r1′)’ n (r2′)’ )’ [Lemma 4]
= ( ((r1′) u (r2′))’ )’ [Lemma 6, first part]
= (r1′ u r2′)” [rewriting with fewer brackets]
= r1′ u r2′ [Lemma 4] | 5,174 | 11,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.596803 |
https://www.cs.uic.edu/~jbell/CourseNotes/ComputerGraphics/VisibleSurfaces.html | 1,723,174,393,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00128.warc.gz | 567,291,572 | 9,865 | # Visible Surface Determination
### References:
1. Andy Johnson's CS 488 Course Notes, Lecture 10 and 11
2. Foley, Van Dam, Feiner, and Hughes, "Computer Graphics - Principles and Practice", Chapter 15
In the beginning of the semester we dealt with simple wireframe drawings of the models. The main reason for this is so that we did not have to deal with hidden surface removal. Now we want to deal with more sophisticated images so we need to deal with which parts of the model obscure other parts of the model.
The following sets of images show a wireframe version, a wireframe version with hidden line removal, and a solid polygonal representation of the same object.
If we do not have a way of determining which surfaces are visible then which surfaces are visible depends on the order in which they are drawn with surfaces being drawn later appearing in front of surfaces drawn previously as shown below:
Here the fins on the back are visible because they are drawn after the body, the shadow is drawn on top of the monster because it is drawn last. Both legs are visible and the eyes just look really weird.
### General Principles
We do not want to draw surfaces that are hidden. If we can quickly compute which surfaces are hidden, we can bypass them and draw only the surfaces that are visible.
For example, if we have a solid 6 sided cube, at most 3 of the 6 sides are visible at any one time, so at least 3 of the sides do not even need to be drawn because they are the back sides.
We also want to avoid having to draw the polygons in a particular order. We would like to tell the graphics routines to draw all the polygons in whatever order we choose and let the graphics routines determine which polygons are in front of which other polygons.
With the same cube as above we do not want to have to compute for ourselves which order to draw the visible faces, and then tell the graphics routines to draw them in that order.
The idea is to speed up the drawing, and give the programmer an easier time, by doing some computation before drawing.
Unfortunately these computations can take a lot of time, so special purpose hardware is often used to speed up the process.
### Techniques
Two types of approaches:
• object space (object precision in our text)
• image space (image precision in our text)
object space algorithms do their work on the objects themselves before they are converted to pixels in the frame buffer. The resolution of the display device is irrelevant here as this calculation is done at the mathematical level of the objects
```for each object a in the scene
determine which parts of object a are visible
(involves comparing the polyons in object a to other polygons in a
and to polygons in every other object in the scene)
```
image space algorithms do their work as the objects are being converted to pixels in the frame buffer. The resolution of the display device is important here as this is done on a pixel by pixel basis.
```for each pixel in the frame buffer
determine which polygon is closest to the viewer at that pixel location
colour the pixel with the colour of that polygon at that location```
As in our discussion of vector vs raster graphics earlier in the term the mathematical (object space) algorithms tended to be used with the vector hardware whereas the pixel based (image space) algorithms tended to be used with the raster hardware.
When we talked about 3D transformations we reached a point near the end when we converted the 3D (or 4D with homogeneous coordinates) to 2D by ignoring the Z values. Now we will use those Z values to determine which parts of which polygons (or lines) are in front of which parts of other polygons.
There are different levels of checking that can be done.
• Object
• Polygon
• part of a Polygon
There are also times when we may not want to cull out polygons that are behind other polygons. If the frontmost polygon is transparent then we want to be able to 'see through' it to the polygons that are behind it as shown below:
Which objects are transparent in the above scene?
### Coherence
We used the idea of coherence before in our line drawing algorithm. We want to exploit 'local similarity' to reduce the amount of computation needed (this is how compression algorithms work.)
• Face - properties (such as colour, lighting) vary smoothly across a face (or polygon)
• Depth - adjacent areas on a surface have similar depths
• Frame - images at successive time intervals tend to be similar
• Scan Line - adjacent scan lines tend to have similar spans of objects
• Area - adjacent pixels tend to be covered by the same face
• Object - if objects are separate from each other (ie they do not overlap) then we only need to compare polygons of the same object, and not one object to another
• Edge - edges only disappear when they go behind another edge or face
• Implied Edge - line of intersection of 2 faces can be determined by the endpoints of the intersection
### Extents
Rather than dealing with a complex object, it is often easier to deal with a simpler version of the object.
in 2: a bounding box
in 3d: a bounding volume (though we still call it a bounding box)
We convert a complex object into a simpler outline, generally in the shape of a box and then we can work with the boxes. Every part of the object is guaranteed to fall within the bounding box.
Checks can then be made on the bounding box to make quick decisions (ie does a ray pass through the box.) For more detail, checks would then be made on the object in the box.
There are many ways to define the bounding box. The simplest way is to take the minimum and maximum X, Y, and Z values to create a box. You can also have bounding boxes that rotate with the object, bounding spheres, bounding cylinders, etc.
### Back-Face Culling
Back-face culling (an object space algorithm) works on 'solid' objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.
Every planar polygon has a surface normal, that is, a vector that is normal to the surface of the polygon. Actually every planar polygon has two normals.
Given that this polygon is part of a 'solid' object we are interested in the normal that points OUT, rather than the normal that points in.
OpenGL specifies that all polygons be drawn such that the vertices are given in counterclockwise order as you look at the visible side of polygon in order to generate the 'correct' normal.
Any polygons whose normal points away from the viewer is a 'back-facing' polygon and does not need to be further investigated.
To find back facing polygons the dot product of the surface normal of each polygon is taken with a vector from the center of projection to any point on the polygon.
The dot product is then used to determine what direction the polygon is facing:
• greater than 0 : back facing
• equal to 0 : polygon viewed on edge
• less than 0 : front facing
Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box - the inside and the ouside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.
in OpenGL back-face culling is turned on using:
``` glCullFace(GL_BACK);
glEnable(GL_CULL_FACE);```
### Depth Buffer
Early on we talked about the frame buffer which holds the colour for each pixel to be displayed. This buffer could contain a variable number of bytes for each pixel depending on whether it was a greyscale, RGB, or colour-indexed frame buffer. All of the elements of the frame buffer are initially set to be the background colour. As lines and polygons are drawn the colour is set to be the colour of the line or polygon at that point. T
We now introduce another buffer which is the same size as the frame buffer but contains depth information instead of colour information.
z-buffering (an image-space algorithm) is another buffer which maintains the depth for each pixel. All of the elements of the z-buffer are initially set to be 'very far away.' Whenever a pixel colour is to be changed the depth of this new colour is compared to the current depth in the z-buffer. If this colour is 'closer' than the previous colour the pixel is given the new colour, and the z-buffer entry for that pixrl is updated as well. Otherwise the pixel retains the old colour, and the z-buffer retains its old value.
Here is a pseudo-code algorithm
```for each polygon
for each pixel p in the polygon's projection
{
pz = polygon's normalized z-value at (x, y); //z ranges from -1 to 0
if (pz > zBuffer[x, y]) // closer to the camera
{
zBuffer[x, y] = pz;
framebuffer[x, y] = colour of pixel p
}
}```
This is very nice since the order of drawing polygons does not matter, the algorithm will always display the colour of the closest point.
The biggest problem with the z-buffer is its finite precision, which is why it is important to set the near and far clipping planes to be as close together as possible to increase the resolution of the z-buffer within that range. Otherwise even though one polygon may mathematically be 'in front' of another that difference may disappear due to roundoff error.
These days with memory getting cheaper it is easy to implement a software z-buffer and hardware z-buffering is becoming more common.
In OpenGL the z-buffer and frame buffer are cleared using:
` glClear(GL_DEPTH_BUFFER_BIT | GL_COLOR_BUFFER_BIT);`
In OpenGL z-buffering is turned on using:
` glEnable(GL_DEPTH_TEST);`
The depth-buffer is especially useful when it is difficult to order the polygons in the scene based on their depth, such as in the case shown below:
### Warnock's Algorithm
Warnock's algorithm is a recursive area-subdivision algorithm. Warnock's algorithm looks at an area of the image. If is is easy to determine which polygons are visible in the area they are drawn, else the area is subdivided into smaller parts and the algorithm recurses. Eventually an area will be represented by a single non-intersecting polygon.
At each iteration the area of interest is subdivided into four equal areas. Each polygon is compared to each area and is put into one of four bins:
• Disjoint polygons are completely outside the area
• Intersecting polygons intersect the area
• Surrounding polygons completely contain the area
• Contained polygons are completely contained in the area
For a given area:
1. If all polygons are disjoint then the background colour fills the area
2. If there is a single contained polygon or intersecting polygon then the background colour is used to fill the area, then the part of the polygon contained in the area is filled with the colour of that polygon
3. If there is a single surrounding polygon and no intersecting or contained polygons then the area is filled with the colour of the surrounding polygon
4. If there is a surrounding polygon in front of any other surrounding, intersecting, or contained polygons then the area is filled with the colour of the front surrounding polygon
Otherwise break the area into 4 equal parts and recurse.
At worst log base 2 of the max(screen width, screen height) recursive steps will be needed. At that point the area being looked at is only a single pixel which can't be divided further. At that point the distance to each polygon intersecting,contained in, or surrounding the area is computed at the center of the polygon to determine the closest polygon and its colour,
Below is an example scanned out of the text, where the numbers refer to the numbered steps listed above:
Here is a place where the use of bounding boxes can speed up the process. Given that the bounding box is always at least as large as the polygon, or object checks for contained and disjoint polygons can be made using the bounding boxes, while checks for interstecting and surrounding can not.
### Depth Sort Algorithm, a.k.a. The Painter's Algorithm
The idea here is to go back to front drawing all the objects into the frame buffer with nearer objects being drawn over top of objects that are further away.
Simple algorithm:
1. Sort all polygons based on their farthest z coordinate
2. Resolve ambiguities
3. draw the polygons in order from back to front
This algorithm would be very simple if the z coordinates of the polygons were guaranteed never to overlap. Unfortunately that is usually not the case, which means that step 2 can be somewhat complex.
Any polygons whose z extents overlap must be tested against each other.
We start with the furthest polygon and call it P. Polygon P must be compared with every polygon Q whose z extent overlaps P's z extent. 5 comparisons are made. If any comparison is true then P can be written before Q. If at least one comparison is true for each of the Qs then P is drawn and the next polygon from the back is chosen as the new P.
1. do P and Q's x-extents not overlap.
2. do P and Q's y-extents not overlap.
3. is P entirely on the opposite side of Q's plane from the viewport.
4. is Q entirely on the same side of P's plane as the viewport.
5. do the projections of P and Q onto the (x,y) plane not overlap.
If all 5 tests fail we quickly check to see if switching P and Q will work. Tests 1, 2, and 5 do not differentiate between P and Q but 3 and 4 do. So we rewrite 3 and 4
3 - is Q entirely on the opposite side of P's plane from the viewport.
4 - is P entirely on the same side of Q's plane as the viewport.
If either of these two tests succeed then Q and P are swapped and the new P (formerly Q) is tested against all the polygons whose z extent overlaps it's z extent.
If these two tests still do not work then either P or Q is split into 2 polygons using the plane of the other. These 2 smaller polygons are then put into their proper places in the sorted list and the algorithm continues.
beware of the dreaded infinite loop.
### BSP Trees
Another popular way of dealing with these problems (especially in games) are Binary Space Partition trees. It is a depth sort algorithm with a large amount of preprocessing to create a data structure to hold the polygons.
First generate a 3D BSP tree for all of the polygons in the scene
Then display the polygons according to their order in the scene
1. polygons behind the current node
2. the current node
3. polygons in front of the current node
Each node in the tree is a polygon. Extending that polygon generates a plane. That plane cuts space into 2 parts. We use the front-facing normal of the polygon to define the half of the space that is 'in front' of the polygon. Each node has two children: the front children (the polygons in front of this node) and the back children (the polgons behind this noce)
In doing this we may need to split some polygons into two.
Then when we are drawing the polygons we first see if the viewpoint is in front of or behind the root node. Based on this we know which child to deal with first - we first draw the subtree that is further from the viewpoint, then the root node, then the subtree that is in front of the root node, recursively, until we have drawn all the polygons.
Compared to depth sort it takes more time to setup but less time to iterate through since there are no special cases.
If the position or orientation of the polygons change then parts of the tree will need to be recomputed.
here is an example originally by Nicolas Holzschuch showing the construction and use of a BSP tree for 6 polygons.
### Scan Line Algorithm
This is an extension of the algorithm we dealt with earlier to fill polygons one scan line at a time. This time there will be multiple polygons being drawn simultaneously.
Again we create a global edge table for all non-horizontal edges sorted based on the edges smaller y coordinate.
Each entry in the table contains:
• x coordinate corresponding to the smallest y coordinate
• y coordinate of the other end of the edge
• deltaX = 1/m
and a new entry
• a way to identify which polygon this edge belongs to
In the scan line algorithm we had a simple 0/1 variable to deal with being in or out of the polygon. Since there are multiple polygons here we have a Polygon Table.
The Polygon Table contains:
• coefficients of the equation of the plane of the polygon
• shading / colour information on this polygon
• in / out flag initialized to false
Again the edges are moved from the global edge table to the active edge table when the scan line corresponding to the bottom of the edge is reached.
Moving across a scan line the flag for a polygon is flipped when an edge of that polygon is crossed.
If no flags are true then nothing is drawn
If one flag is true then the colour of that polygon is used
If more than one flag is true then the frontmost polygon must be determined.
Below is an example from the textbook (figure red:13.11, white:15.34)
Here there are two polygons ABC and DEF
```Scan Line AET contents Comments
--------- ------------ --------
alpha AB AC one polygon
beta AB AC FD FE two separate polygons
gamma AB DE CB FE two overlapping polygons
gamma+1 AB DE CB FE two overlapping polygons
gamma+2 AB CB DE FE two separate polygons```
### Simple Ray Tracing
A simple ray-tracing algorithm can be used to find visible surfaces, as opposed to a more complicated algorithm that can be used to generate those o-so-pretty images.
Ray tracing is an image based algorithm. For every pixel in the image, a ray is cast from the center of projection through that pixel and into the scene. The colour of the pixel is set to the colour of the object that is first encountered.
```Given a Center Of Projection
Given a window on the viewplane
for (each scan line)
for (each pixel on the scan line)
{
compute ray from COP through pixel
for (each object in scene)
if (object is intersected by ray
&& object is closer than previous intersection)
record (intersection point, object)
set pixel's colour to the colour of object at intersection point
}```
So, given a ray (vector) and an object the key idea is computing if and if so where does the ray intersect the object.
the ray is represented by the vector from (Xo, Yo, Zo) at the COP, to (X1, Y1, Z1) at the center of the pixel. We can parameterize this vector by introducing t:
X = Xo + t(X1 - Xo)
Y = Yo + t(Y1 - Yo)
Z = Zo + t(Z1 - Zo)
or
X = Xo + t(deltaX)
Y = Yo + t(deltaY)
Z = Zo + t(deltaZ)
t equal to 0 represents the COP, t equal to 1 represents the pixel
t < 0 represents points behind the COP
t > 1 represents points on the other side of the view plane from the COP
We want to find out what the value of t is where the ray intersects the object. This way we can take the smallest value of t that is in front of the COP as defining the location of the nearest object along that vector.
The problem is that this can take a lot of time, especially if there are lots of pixels and lots of objects.
The raytraced images in 'Toy Story' for example took at minimum 45 minutes and at most 20 hours for a single frame.
So minimizing the number of comparisons is critical.
- Bounding boxes can be used to perform initial checks on complicated objects
- hierarchies of bounding boxes can be used where a successful intersection with a bounding box then leads to tests with several smaller bounding boxes within the larger bounding box.
- The space of the scene can be partitioned. These partitions are then treated like buckets in a hash table, and objects within each partition are assigned to that partition. Checks can then be made against this constant number of partitions first before going on to checking the objects themselves. These partitions could be equal sized volumes, or contain equal numbers of polygons.
Full blown ray tracing takes this one step further by reflecting rays off of shiny objects, to see what the ray hits next, and then reflecting the ray off of that, and so on, until a limiting number of reflections have been encountered. For transparent or semi-transparent objects, rays are passed through the object, taking into account any deflection or filtering that may take place ( e.g. through a colored glass bottle or chess piece ), again proceeding until some limit is met. Then the contributions of all the reflections and transmissions are added together to determine the final color value for each pixel. The resulting images can be incredibly beautiful and realistic, and usually take a LONG time to compute.
### Comparison
From table 13.1 in the red book (15.3 in the white book) the relative performance of the various algorithms where smaller is better and the depth sort of a hundred polygons is set to be 1.
``` # of polygonal faces in the scene
Algorithm 100 250 60000
--------------------------------------------------
Depth Sort 1 10 507
z-buffer 54 54 54
scan line 5 21 100
Warnock 11 64 307```
This table is somewhat bogus as z-buffer performance degrades as the number of polygonal faces increases.
To get a better sense of this, here are the number of polygons in the following models:
250 triangular polygons:
550 triangular polygons:
6,000 triangular polygons:
(parrot by Christina Vasilakis)
8,000 triangular polygons:
10,000 triangular polygons:
(ARPA Integration testbed space by Jason Leigh, Andrew Johnson, Mike Kelley) | 4,655 | 21,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-33 | latest | en | 0.946977 |
https://www.rhayden.us/regression-model/two-approaches-to-testing-hypotheses.html | 1,571,711,418,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00393.warc.gz | 1,066,626,285 | 4,222 | ## Two Approaches To Testing Hypotheses
Hypothesis testing of the sort suggested above can be approached from two viewpoints. First, having computed a set of parameter estimates, we can ask whether the estimates come reasonably close to satisfying the restrictions implied by the hypothesis. More formally, we can ascertain whether the failure of the estimates to satisfy the restrictions is simply the result of sampling error or is instead systematic. An alternative approach might proceed as follows. Suppose that we impose the restrictions implied by the theory. Since unrestricted least squares is, by definition, "least squares," this imposition must lead to a loss of fit. We can then ascertain whether this loss of fit results merely from sampling error or whether it is so large as to cast doubt on the validity of the restrictions. We will consider these two approaches in turn, then show that (as one might hope) within the framework of the linear regression model, the two approaches are equivalent. | 189 | 1,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.921169 |
http://slideplayer.com/slide/4356420/ | 1,513,438,101,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588251.76/warc/CC-MAIN-20171216143011-20171216165011-00758.warc.gz | 252,202,368 | 19,646 | # APPENDIX A Part of Solidarity’s submission to NERSA on the issue of Eskom’s proposed MYPD2 Municipal electricity redistributors’ price increases.
## Presentation on theme: "APPENDIX A Part of Solidarity’s submission to NERSA on the issue of Eskom’s proposed MYPD2 Municipal electricity redistributors’ price increases."— Presentation transcript:
APPENDIX A Part of Solidarity’s submission to NERSA on the issue of Eskom’s proposed MYPD2 Municipal electricity redistributors’ price increases
APPENDIX A – Municipal tariff increases Example 1 Current scenario of a 35% increase in Eskom's tariffs Original tariffNew tariff (35% increase) Eskom tariff to municipality70c per kWh94,5c per kWh Assume a fixed 100% profit margin for the municipality Municipal tariff to consumers(70c x 2) = 140c per kWh(94,5c x 2) = 189c per kWh Gross profit per kWh70c94,5c Result: 35% increase in profit for the municipality Profit margin remains 100% 35% increase in price for final consumers
Example 2 Better scenario of a 35% increase in Eskom's tariffs Original tariffNew tariff (35% increase) Eskom tariff to municipality70c per kWh94,5c per kWh Assume a variable profit margin (being 100% originally) Municipal tariff to consumers(70c x 2) = 140c per kWh (70c x 2) + (94,5c - 70c) = 164,5c per kWh Gross profit per kWh70c Result: No increase in profit for the municipality Profit margin drops to 74,1% 17,5% increase in price for final consumers APPENDIX A – Municipal tariff increases
Example 3 Better scenario of a 35% increase in Eskom's tariffs and where the municipality is allowed an 8% increase in gross profit to cover increased cost of distribution Original tariffNew tariff (35% increase) Eskom tariff to municipality70c per kWh94,5c per kWh Assume a variable profit margin (being 100% originally) Municipal tariff to consumers(70c x 2) = 140c per kWh (70c x 2) + (94,5c - 70c) + (70c x 0,08) = 170,1c per kWh Gross profit per kWh70c75,6c Result: 8% increase in profit for the municipality Profit margin drops to 80% 21,5% increase in price for final consumers APPENDIX A – Municipal tariff increases
Components of calculationCumulative increase in three years Example 1 (at a constant profit margin of 100%) 35% on 35% on 35%146% Example 2 (at an original profit margin of 100%) 17,5% on 20,1% on 22,6%73% Example 3 (at an original profit margin of 100%, adding an 8% allowance for increased costs each year) 21,5% on 23% on 24,5%86% If the three different approaches are considered cumulatively over the three year period that Eskom’s MYPD2 covers, the following emerges: Between the range of an original municipal gross profit margin of 50% and 130%, the respective cumulative increases over the three year period range from 97,4% to 63,5% for example 2 and from 106% to 78,2% for example 3. Instead of the price of electricity for customers of municipalities more than doubling over the next three years, the cumulative increase could be kept much lower than the current method’s 146% while still allowing municipalities an 8% increase in gross profit each year to compensate for possible increases in the cost of distribution. In none of the examples is Eskom’s flow of revenue or tariff increases affected – the changes are merely made at municipal level. APPENDIX A – Municipal tariff increases
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# Solutions for Chapter Chapter 13: Advanced Engineering Mathematics 9th Edition
## Full solutions for Advanced Engineering Mathematics | 9th Edition
ISBN: 9780471488859
Solutions for Chapter Chapter 13
Solutions for Chapter Chapter 13
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##### ISBN: 9780471488859
Chapter Chapter 13 includes 231 full step-by-step solutions. Since 231 problems in chapter Chapter 13 have been answered, more than 28118 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. Advanced Engineering Mathematics was written by and is associated to the ISBN: 9780471488859. This textbook survival guide was created for the textbook: Advanced Engineering Mathematics, edition: 9.
Key Math Terms and definitions covered in this textbook
• Cholesky factorization
A = CTC = (L.J]))(L.J]))T for positive definite A.
• Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).
• Factorization
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.
• Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].
• Free variable Xi.
Column i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).
• Full row rank r = m.
Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.
• Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
• Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.
• Length II x II.
Square root of x T x (Pythagoras in n dimensions).
• Matrix multiplication AB.
The i, j entry of AB is (row i of A)·(column j of B) = L aikbkj. By columns: Column j of AB = A times column j of B. By rows: row i of A multiplies B. Columns times rows: AB = sum of (column k)(row k). All these equivalent definitions come from the rule that A B times x equals A times B x .
• Nilpotent matrix N.
Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.
• Orthogonal subspaces.
Every v in V is orthogonal to every w in W.
• Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.
• Solvable system Ax = b.
The right side b is in the column space of A.
• Special solutions to As = O.
One free variable is Si = 1, other free variables = o.
• Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.
• Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.
• Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.
• Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II·
• Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.
× | 1,005 | 3,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-10 | latest | en | 0.815893 |
http://ask.csdn.net/questions/652809 | 1,516,476,542,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00336.warc.gz | 26,079,116 | 13,306 | qq_39301152 于 2017.08.27 13:29 提问
python解二分方程不知道哪里出错
`````` import math
def main():
print("this program finds the real solutions to a quadratic\n")
a,b,c=eval(input("please enter the coefficients (a,b,c):"))
delta=b*b-4*a*c
if delta<0:
print("\nthe equation has no real roots!")
else:
discRoot=math.sqrt(delta)
root1=(-b+discRoot)/(2*a)
root2=(-b-discRoot)/(2*a)
print("\nthe solutions are:",root1,root2)
main()
``````
2个回答
zt3032 2017.08.27 13:48
qq_39301152 嗯嗯 了解
5 个月之前 回复
runnoob_1115 2017.08.27 16:28
``````a,b,c=eval(input("a,b,c:"))
if a==0:
if b==0:
if c==0:
print("x为任意值")
else:
print("无解")
else:
print("x=",-1*c/b) # bx + c =0
else:
dt=b*b-4*a*c
if dt==0:
print("x1=x2=",-1*b/2/a)
elif dt >0:
print("x1=",(-b+dt**0.5)/(2*a),",x2=",(-b-dt**0.5)/(2*a))
else:
print("x1=", -1 * b / 2 / a, "+", (-1 * dt) ** 0.5 / (2 * a), "i")
print("x2=", -1 * b / 2 / a, "-", (-1 * dt) ** 0.5 / (2 * a), "i")
`````` | 411 | 933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-05 | latest | en | 0.262774 |
https://17calculus.com/infinite-series/pseries/ | 1,656,110,102,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00003.warc.gz | 126,248,310 | 32,211 | 17Calculus Infinite Series - p-Series
17Calculus
p-Series Convergence Theorem
$$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + ~~ . . . }$$
$$0 \lt p \leq 1$$
$$p>1$$
diverges
converges
Note - In the proof section below, we show a more efficient way to write this theorem statement.
The p-series is a pretty straight-forward series to understand and use. One detail you need to notice is that the theorem covers cases only where $$p > 0$$.
Question: Do you know what happens when $$p \leq 0$$? Think about it for a minute and then click here for the answer
Okay, so testing for convergence and divergence of p-series looks pretty easy and, fortunately, it is. But it is also very powerful. To help you cement the idea more firmly in your mind, read through the panel below discussing the proof of this test. You do not have to understand the integral test at this point to understand this proof.
p-Series Convergence Theorem Proof
p-Series Convergence Theorem
The p-series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}}}$$
To prove this, we will use the Integral Test and the Special Improper Integral.
The integral test tells us that we can set up the integral $$\displaystyle{ \int_1^{\infty}{\frac{1}{x^p} dx} }$$
If this integral converges, then so does the series. Similarly, if the integral diverges, the series also diverges.
We will look at these five cases.
1. $$p > 1$$
2. $$p = 1$$
3. $$0 \lt p \leq 1$$
4. $$p = 0$$
5. $$p \lt 0$$
Note: Although the last two cases are not part of the theorem, we will show what happens in those two cases to answer the question at the top of the page and for completeness.
Let's use the Integral Test to prove convergence and divergence by calculating the corresponding improper integrals. For most of the cases, we need to set up a limit as follows
$$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }$$
Since the function $$\displaystyle{ \frac{1}{x^p} }$$ is continuous on the interval $$[1,b]$$, this integral can be evaluated. So, now let's look at each case individually.
Case 1: $$p > 1$$
When $$p > 1$$, the integral is continuous and decreasing on the interval. So the Integral Test applies.
$$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } }$$ $$\displaystyle{ \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]} }$$ $$\displaystyle{ \frac{1}{-p+1} [ 0 - 1 ] = \frac{1}{p-1} }$$
Since $$\displaystyle{ \frac{1}{p-1} }$$ is finite, the integral converges and, therefore, by the Integral Test, the series also converges.
Case 2: $$p=1$$
When $$p = 1$$, we have
$$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x} dx}} \to \lim_{b \to \infty }{ \left[ \ln(x) \right]_{1}^{b} } \to \lim_{b \to \infty }{ \ln(b) } - 0 = \infty }$$
Since the limit is infinity, the series diverges.
Case 3: $$0 \lt p \lt 1$$
We will use the Integral Test again.
$$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } }$$ $$\displaystyle{ \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]} }$$
So far in these calculations, we have the same equation as we did in case 1. However, in this case, $$0 \lt p \lt 1$$, which means that the exponent $$-p+1 > 0$$ and therefore $$\displaystyle{ \lim_{b \to \infty}{ b^{-p+1} } \to \infty }$$. So the entire integral diverges, which means that the series diverges.
Case 4: $$p = 0$$
When $$p = 0$$, the series is $$\displaystyle{ \sum_{n=1}^{\infty}{1} }$$. Since $$\displaystyle{ \lim_{n \to \infty}{1} = 1 \neq 0 }$$ the divergence test tells us that the series diverges.
Case 5: $$p \lt 0$$
We can write the fraction $$\displaystyle{ \frac{1}{n^p} = n^{-p} }$$. Since $$p \lt 0$$, the exponent here is positive and so the terms are increasing. Since $$\displaystyle{ \lim_{n \to \infty}{ \frac{1}{n^p} } \neq 0 }$$, the series diverges by the divergence test.
In Summary: For the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}}}$$
1. $$p > 1$$ converges by the integral test
2. $$p = 1$$ diverges by the integral test
3. $$0 \lt p \lt 1$$ diverges by the integral test
4. $$p = 0$$ diverges by the divergence test
5. $$p \lt 0$$ diverges by the divergence test
- qed -
Note: We have also seen the p-series theorem where, instead of writing $$0 \lt p \leq 1$$ for where the series diverges, it says the series diverges for $$p \leq 1$$. Of course, this includes all of the cases above. This is a more efficient way to write the theorem statement.
Euler's Constant
Here is a great video discussing one use of a p-series.
Note - - Euler's Constant is not the same as Euler's Number.
PatrickJMT - Euler's Constant [16min-10secs]
video by PatrickJMT
Practice
Unless otherwise instructed, determine whether these series converge or diverge using the p-series convergence theorem.
$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }$$
Problem Statement
Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }$$ using the Direct Comparison Test, if possible. If the DCT is inconclusive, use another test to determine convergence or divergence. Make sure to specify what test you used in your final answer.
The series diverges by the p-series test or the Direct Comparison Test.
Problem Statement
Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }$$ using the Direct Comparison Test, if possible. If the DCT is inconclusive, use another test to determine convergence or divergence. Make sure to specify what test you used in your final answer.
Solution
p-Series
First, if you factor out the constant in the numerator, you get $$\displaystyle{ 2 \sum_{n=1}^{\infty}{ \frac{1}{n} } }$$ and the remaining sum is just a p-series with $$p=1$$. Therefore, the series diverges.
Direct Comparison Test
If you didn't pick up the fact that you have p-series, you can use the Direct Comparison Test, comparing the given series with $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }$$ which is a divergent p-series.
Now, to confirm that the given series diverges, we need to test $$\displaystyle{ \frac{2}{n} \geq \frac{1}{n} }$$ for all n. If we multiply both sides of the inequality by n (which we can do since n is always positive), we get $$2 \geq 1$$. This is true for all n, therefore the series diverges.
The series diverges by the p-series test or the Direct Comparison Test.
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$
Problem Statement
Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ using the integral test, if possible.
The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ converges by the p-series test or the integral test.
Problem Statement
Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ using the integral test, if possible.
Solution
This is a p-series with $$p=2 > 1$$, so the series converges by the p-series test.
We could also have used the Integral Test, as follows.
$$\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2}dx} }$$ $$\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{\frac{1}{x^2}dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{x^{-2}dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \left[ -x^{-1} \right]_{1}^{b}} }$$ $$\displaystyle{ \lim_{b \to \infty}{-b^{-1} + 1^{-1}} }$$ $$0 + 1 = 1$$
Since the improper integral is finite, the series converges by the Integral Test.
Note: The value $$1$$, from the integral is NOT necessarily what the series converges to. The significance of this number is only that it is finite.
The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$ converges by the p-series test or the integral test.
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3163 video solution
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }$$ converges or diverges using the p-series convergence theorem.
The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }$$ is a convergent p-series.
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }$$ converges or diverges using the p-series convergence theorem.
Solution
PatrickJMT - 253 video solution
video by PatrickJMT
The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }$$ is a convergent p-series.
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }$$ converges or diverges using the p-series convergence theorem.
The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }$$ is a divergent p-series.
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }$$ converges or diverges using the p-series convergence theorem.
Solution
PatrickJMT - 254 video solution
video by PatrickJMT
The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }$$ is a divergent p-series.
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$$\displaystyle{ \sum_{n=1}^{\infty}{ n^{-\pi} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ n^{-\pi} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3153 video solution
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$$\displaystyle{ \sum_{n=1}^{\infty}{ (n^{-2.4} + 8n^{-1.6}) } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ (n^{-2.4} + 8n^{-1.6}) } }$$ converges or diverges using the p-series convergence theorem.
Solution
3154 video solution
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$$\displaystyle{ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + . . . }$$
Problem Statement
Determine whether the series $$\displaystyle{ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + . . . }$$ converges or diverges using the p-series convergence theorem.
Solution
This problem is solved by two different instructors.
3155 video solution
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$$\displaystyle{ 1 + \frac{1}{\sqrt[3]{4}} + \frac{1}{\sqrt[3]{9}} + \frac{1}{\sqrt[3]{16}} + . . . }$$
Problem Statement
Determine whether the series $$\displaystyle{ 1 + \frac{1}{\sqrt[3]{4}} + \frac{1}{\sqrt[3]{9}} + \frac{1}{\sqrt[3]{16}} + . . . }$$ converges or diverges using the p-series convergence theorem.
Solution
3156 video solution
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$$\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }$$
Problem Statement
Determine whether the series $$\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }$$ converges or diverges using the p-series convergence theorem.
The series $$\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }$$ is a convergent p-series.
Problem Statement
Determine whether the series $$\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }$$ converges or diverges using the p-series convergence theorem.
Solution
This problem is solve by two different instructors.
PatrickJMT - 255 video solution
video by PatrickJMT
255 video solution
The series $$\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }$$ is a convergent p-series.
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt[3]{n^4}} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt[3]{n^4}} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3158 video solution
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \sqrt[5]{n^2} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \sqrt[5]{n^2} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3159 video solution
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{\sqrt[3]{n^2}} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{\sqrt[3]{n^2}} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3160 video solution
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3 \sqrt{n}} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3 \sqrt{n}} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3161 video solution
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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4\sqrt{n}}{n^4} } }$$
Problem Statement
Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4\sqrt{n}}{n^4} } }$$ converges or diverges using the p-series convergence theorem.
Solution
3162 video solution
Log in to rate this practice problem and to see it's current rating. | 4,669 | 14,396 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-27 | longest | en | 0.779742 |
https://avatest.org/2022/12/29/representation-theory-daixie-math4314-the-key-point-is-the-convergence/ | 1,726,590,801,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00312.warc.gz | 96,890,015 | 20,236 | Posted on Categories:数学代写, 表示论
# 数学代写|表示论代写Representation Theory代考|MATH4314 The Key Point Is the Convergence
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## 数学代写|表示论代写Representation Theory代考|The Key Point Is the Convergence
We keep all previous notation. In this last section we shall see that the key point is the convergence of the integral
$$\left(I_{\mathfrak{h}2 \mathfrak{h}_1}^G \varphi\right)(g):=\left(I{\mathfrak{h}2 \mathfrak{h}_1} \varphi\right)(g)=\oint{H_2 /\left(H_1 \cap H_2\right)} \varphi(g h) \chi_f(h) \Delta_{H_2, G}^{-1 / 2}(h) d v(h)(g \in G)$$
At least formally, it is clear that the function $I_{\mathfrak{h}2 \mathfrak{h}_1} \varphi$ satisfies the necessary covariance condition for belonging in the space $\mathscr{H}{\pi_2}$, and also that $I_{\mathfrak{h}2 \mathfrak{h}_1}$ commutes with left translations. We can therefore assert that the convergence of the integral is one major issue to deal with. We call elements of $I(f, \mathfrak{g})$ Pukanszky polarizations. In this section we look at all pairs $\left(\mathfrak{h}_1, \mathfrak{h}_2\right)$ of Pukanszky polarizations at $f$ which meet the following convergence property: $(C P)$ : The integral $$\oint{H_2 /\left(H_1 \cap H_2\right)}|\varphi(g h)| \Delta_{H_2, G}^{-\frac{1}{2}}(h) d v(h), g \in G$$
converges on a $G$-invariant and $\mathscr{U}(\mathfrak{g})$-invariant dense subspace $\mathscr{H}$ of $\mathscr{H}_{\pi_1}^{\infty}$.
## 数学代写|表示论代写Representation Theory代考|Notation and Backgrounds
Let $G$ be a connected and simply connected nilpotent Lie group with Lie algebra $\mathfrak{g}$. Let exp denote, as earlier, the exponential map, so that $G=\exp \mathfrak{g}$. Let $V, W$ be real vector spaces of finite dimension with $W \subset V$. We denote by $V^{\star}$ the dual vector space of $V$ and by $W^{\perp}$ the orthogonal to $W$ in $V^{\star}$. If $u_1, \ldots, u_p,(p \in \mathbb{N})$ indicate linearly independent vectors in $V$, we denote by $\mathbb{R}$-span $\left(u_1, \ldots, u_p\right)$ the vector subspace of $V$ they span, and we say the basis $\left{u_1, \ldots, u_p\right}$ generates this space. Given $l \in \mathfrak{g}^{\star}$ and $X \in \mathfrak{g}$, we denote by $\langle l, X\rangle$ the image of $X$ under $l$. Recall that the kernel of the bilinear form $B_l$, defined in Sect. 1.2.4 by $B_l(X, Y)=$ $\langle l,[X, Y]\rangle$, is denoted by $\mathfrak{g}(l)=\left{X \in \mathfrak{g} ; B_l(X, Y)=0\right.$ for all $\left.Y \in \mathfrak{g}\right}$. The largest ideal contained in $\mathfrak{g}(l)$ will be denoted by $\mathfrak{a}(l)$. Clearly
$$\mathfrak{a}(l)=\bigcap_{\phi \in G \cdot l} \mathfrak{g}(\phi) .$$
Let $\mathfrak{h} \in S(l, \mathfrak{g})$ and let $\chi_l$ be the unitary character of the analytic subgroup $H=$ $\exp h$ associated to $l$ by
$$\chi_l(\exp X)=e^{-i\langle l, X\rangle}$$
for $X \in \mathfrak{h}$.
## 数学代写|表示论代写Representation Theory代考|关键点在于收敛性
$$\left(I_{mathrm{h} 2\mathrm{~h}1}^G\varphi\right)(g)。 =left(I\mathfrak{h} 2\mathfrak{h}_1 \varphi\right)(g)=oint H_2 /\left(H_1 \cap H_2\right) \varphi(g h) \chi_f(h) δ{H_2, G}^{-1 / 2}(h) d v(h) (g\in G)$$
$B_l(X, Y)=\langle l,[X, Y]\rangle$，用Missing或未识别的分隔符表示〈left〉。$mathfrak{g}(l)$中包含的最大理想将用$mathfrak{a}(l)$来表示。显然
$$mathfrak{a}(l)=\bigcap_{phi\subset G-l}。\mathfrak{g}(phi) 。$$
\chi_l(exp X)=e^{-i(l, X\rangle}。
for $X\in mathfrak{h}$ 。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 1,820 | 4,142 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.490664 |
https://www.cnblogs.com/lamp01/p/10907444.html | 1,708,633,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00650.warc.gz | 717,418,910 | 7,150 | # Go排序练习
1、插入排序
package main
import "fmt"
func insert(a [8]int) [8]int {
for i := 1; i < len(a); i++ {
for j := i; j > 0; j-- {
if a[j] < a[j-1] {
a[j], a[j-1] = a[j-1], a[j]
} else {
break
}
}
}
return a
}
func main() {
var i [8]int = [8]int{45, 2, 5, 68, 3, 43, 7, 9}
j := insert(i)
fmt.Println(i)
fmt.Println(j)
}
[45 2 5 68 3 43 7 9]
[2 3 5 7 9 43 45 68]
2、选择排序
package main
import "fmt"
func choose(a [8]int) [8]int {
for i := 0; i < len(a); i++ {
for j := i + 1; j < len(a); j++ {
if a[j] < a[i] {
a[i], a[j] = a[j], a[i]
}
}
}
return a
}
func main(){
var i [8]int = [8]int{45, 2, 5, 68, 3, 43, 7, 9}
j :=choose(i) fmt.Println(i) fmt.Println(j) } 输出： [45 2 5 68 3 43 7 9] [2 3 5 7 9 43 45 68]
3、冒泡排序
package main
import "fmt"
func bubble(a [8]int) [8]int {
for i := 0; i < len(a); i++ {
for j := 0; j < len(a)-i-1; j++ {
if a[j] > a[j+1] {
a[j], a[j+1] = a[j+1], a[j]
}
}
}
return a
}
func main() {
var i [8]int = [8]int{45, 2, 5, 68, 3, 43, 7, 9}
j := bubble(i)
fmt.Println(i)
fmt.Println(j)
}
[45 2 5 68 3 43 7 9]
[2 3 5 7 9 43 45 68]
posted @ 2019-05-22 17:42 郁冬 阅读(231) 评论(0编辑 收藏 举报 | 566 | 1,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-10 | latest | en | 0.144545 |
http://vaporia.com/astro/start/angularpowerspectrum.html | 1,620,401,407,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00507.warc.gz | 48,001,043 | 2,422 | Astrophysics (Index) About
### angular power spectrum
(power spectrum)
(method of characterizing the CMB)
An angular power spectrum (often shortened to power spectrum, but this latter phrase is also used for other types of power spectra) is a type of characterization of a function on the surface of a sphere, or analogously, all directions in space from a point. It captures how much difference there is between the function values at different points as a function of the angular distance between them. An angular two-point correlation would do this, showing a correlation for each possible angle, but the spectrum more often termed the angular power spectrum is the correlation over each multipole moment (i.e., spherical harmonic order), l of a multipole expansion (analogous to a Fourier series expansion, but for functions over the surface of a sphere). The multipole moments of each order (l) are separated by the same angular distance, and the angular distance decreases as l increases. For each order l, the power spectrum is the average of the squares of the spherical harmonic coefficients associated with that l (i.e., the square of their RMS). Each order l has l spherical harmonic coefficients, each representing a division of the surface of the sphere (for example, there are three coefficients that have order l=3).
The cosmic microwave background (CMB) is often characterized by such an angular power spectrum of its anisotropy, showing the temperature variation for coefficients of a multipole expansion of the temperature over the celestial sphere. The temperature is calculated (in principle) by Wien's displacement law from the received EMR (in practice, the whole spectral energy distribution is considered). The most commonly cited CMB angular power spectrum is of its temperature, but similar spectrums of CMB polarization and/or gravitational lensing are also used. From the angular power spectrum, scales in the early universe can be determined, and values such as the six parameters of the Lambda-CDM model can be checked for consistency with the CMB.
An angular power spectrum can be calculated for any function of a sphere, including any type of intensity map over the celestial sphere: it is the CMB's that are commonly cited, but others might be useful to cosmology as well.
(harmonics,cosmology,CMB,function)
Further reading:
http://en.wikipedia.org/wiki/Cosmic_microwave_background
https://www.roe.ac.uk/ifa/postgrad/pedagogy/2006_tojeiro.pdf
https://cds.cern.ch/record/605007/files/0302217.pdf
https://www.astro.umd.edu/~miller/teaching/astr422/lecture21.pdf
https://sci.esa.int/web/planck/-/51555-planck-power-spectrum-of-temperature-fluctuations-in-the-cosmic-microwave-background
https://www.colorado.edu/studentgroup/ostem/outreach/workshop-understanding-cosmic-microwave-background/week-4-cmb-workshop
Referenced by pages:
CMB anisotropies
CMB lensing
CMB polarization
diffusion damping
Hellings and Downs curve
initial fluctuation spectrum
magnetic field
N-point function
Index | 662 | 3,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.916027 |
https://scienceoxygen.com/how-do-you-interpret-absolute-value/ | 1,701,521,168,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100399.81/warc/CC-MAIN-20231202105028-20231202135028-00617.warc.gz | 567,951,225 | 56,085 | # How do you interpret absolute value?
The absolute value is the distance from a given number to 0. In our example, we are given -3. This number is 3 units away from 0, and thus the absolute value of -3 is 3. If a number is negative, its absolute value will be the positive number with the same magnitude.
## What is the absolute value of |- 15 |?
Algebra Examples The absolute value is the distance between a number and zero. The distance between −15 and 0 is 15 .
## What is the absolute value of |- 12?
The absolute value of -12 is 12. Since the distance between -12 and 0 in a number line is 12.
## What is absolute value of enthalpy?
An absolute enthalpy is the difference between the enthalpies of a substance at two different temperatures, but the reference temperature is not absolute zero.
## Why is absolute value important?
1 Answer. Because it is a convenient way to make sure that a quantity is nonnegative; for example, you can define the distance between two real numbers a and b as |a−b| .
## Why is it important to determine the absolute value of a number?
In algebra, the absolute value operation tells you how far a number is from zero. It doesn’t pay any attention to whether the number is less than or greater than zero, and so absolute values are always positive numbers.
## What is the absolute value of 50?
Distance is expressed as a positive value, so we would say negative 50 has an absolute value of 50 because it is 50 units away from zero, and five has an absolute value of five because it is five units away from zero.
## What is the absolute value of 42?
The absolute value is the distance between a number and zero. The distance between −42 and 0 is 42 .
## What is the absolute value of 24?
When you go 24 units to the right on the positive side, you get 24 and when you go 24 units to the left on the negative side you get -24. Therefore, the answer to “What two numbers have an absolute value of 24?” is: 24 and -24.
## What’s the absolute value of 35?
When you go 35 units to the right on the positive side, you get 35 and when you go 35 units to the left on the negative side you get -35. Therefore, the answer to “What two numbers have an absolute value of 35?” is: 35 and -35.
## What is the absolute value of 75?
When you go 75 units to the right on the positive side, you get 75 and when you go 75 units to the left on the negative side you get -75. Therefore, the answer to “What two numbers have an absolute value of 75?” is: 75 and -75.
## What is the absolute value of 18?
1 Answer. The absolute value of |−18|=18 .
## Why is entropy absolute?
The third law of thermodynamics states that the entropy of a system at absolute zero is a well-defined constant. This is because a system at zero temperature exists in its ground state, so that its entropy is determined only by the degeneracy of the ground state.
## What is absolute entropy in chemistry?
Absolute Entropy. absolute entropy: represents the entropy change of a substance taken from absolute zero to a given temperature.
## What is absolute value example?
Absolute Value Examples and Equations |6| = 6 means “the absolute value of 6 is 6.” |–6| = 6 means “the absolute value of –6 is 6.” |–2 – x| means “the absolute value of the expression –2 minus x.” –|x| means “the negative of the absolute value of x.”
## What is the rule of absolute value?
The absolute value of a number is always positive or zero. If the original number is negative, its absolute value is that number without the negative sign.
## What if absolute value is negative?
Since the absolute value of any number other than zero is positive, it is not permissible to set an absolute value expression equal to a negative number. So, if your absolute value expression is set equal to a negative number, then you will have no solution.
## What is the absolute value of 7?
The absolute value of −7=7 .
## What is the absolute value of 3?
For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a number may be thought of as its distance from zero.
## What is the absolute value of 10?
Number 10 has no sign i.e. it is positive and hence, we can say that |10|=10 . Graphically, absolute value of a number can be described as distance of its location on real number line from origin i.e. point described as 0 , irrespective of its direction, whether towards positive or negative side of origin.
## What is the absolute value of 1?
Some complex numbers have absolute value 1. Of course, 1 is the absolute value of both 1 and –1, but it’s also the absolute value of both i and –i since they’re both one unit away from 0 on the imaginary axis. The unit circle is the circle of radius 1 centered at 0.
## What is the absolute value of 9?
The absolute value of 9 is 9. (9 is 9 places from 0.)
## What is the absolute value of 4?
Therefore, the absolute value of -4 is 4.
## What is the absolute value of 17?
If you see a plus sign, leave it alone.” In other words, absolute value makes numbers positive. The absolute value of -17 is +17. | 1,223 | 5,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2023-50 | latest | en | 0.889711 |
http://ge-shi.net/can/11373031acdbfc083 | 1,695,838,648,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00530.warc.gz | 22,581,610 | 13,210 | # It depends on how many hours you
It depends on how many hours you work, but assuming a 40 hour work week, and working 50 weeks a year, then a \$100,000 yearly salary is about \$50.00 per hour. Maybe. This table illustrates \$100,000 per year divided by 2,080 working hours per year is an hourly income of \$48.08 per hour. \$100,000 a year is \$1,923 per week. \$100,000 per year. To figure out how much \$100,000 a year is as an hourly wage, divide \$100,000 by 40 hours and 52 weeks. Meta. \$100,000 a year is \$1,923 per week. The table presumes 2 weeks vacation for Per Hour: Per Month: Per Year: Starting (Entry-Level) \$47.66: \$8,260 Minnesota NPs make an annual salary of \$118,900 and those in Wyoming make \$118,810 per year. Average \$18.68 per hour. \$100,000 a year is \$50 an hour (or \$2000 per week) Annual Salary to Hourly Wage Conversion Formula [dollars-an-hour] = [dollars-a-year] ([hours-per-week] [work-weeks-per-year]) Top companies for Product Managers in United States. Statisticians say middle class is a household income between \$25,000 and \$100,000 a year. \$5,000/hour. Hourly What is a \$100,000 Salary on a Per-Hour Basis? \$100,000 a year is what per hour? It depends on how many hours you work, but assuming a 40 hour work week, and working 50 weeks a year, then a \$100,000 yearly salary is about \$50.00 per hour. Is 100k a year good pay? Yes.
400 hours will reduce the hourly cost to \$5,000 per hour. How much an hour is 100000 a year? If you make \$100,000 per year, your hourly salary would be \$51.28.
\$100,000 a year is \$3,846 per 2 weeks. Hero Images/Getty The IRS released new tax brackets at the beginning of the year, which apply to income earned in 2020. You might pay up to \$300 more per year on your electric bill if you get a swimming pool. 10 weeks/year. How much is \$100000 a year per hour? An analyst straight out of university can expect to earn over USD \$100,000, but per hour it could be as low as \$20-35 when working 100 hr/wk \$100000 a year is how much per hour? 52 weeks/year. On average, in the United States, someone earning \$100k Salary per year will take home \$71944 yearly, \$5995 monthly, \$2767 bi-weekly, \$1384 weekly, and \$34.59 hourly after Federal and This page will convert a yearly salary to an hourly wage so that you can easily compare your earnings no matter what method you use One salary reported. Use this calculator to convert between an hourly and an annual salary. Some might say the technical answer is \$19.23 per hour. Just enter your wage into the calculator.
If we divide a \$40,000 annual salary by 2080 (the number of hours in a year), we get \$19.23 an hour. If you make \$25 an hour, you make about \$50,000 a year. Income: \$100,000/year. \$100,000 a year is how much an hour? To calculate hourly rate from annual salary, divide yearly salary by the number of weeks Debts: \$1,000 a month. Weeks Worked per Year. If you make \$100.00 per hour you make: \$4,000 per week. How much do you make per hour? Convert between annual and hourly salaries.
\$100,000 a year is \$3,846 per 2 weeks. Daily salary. Leadership; Meet the Team \$300,000 per hour on average, and as much as \$540,000 per hour at the higher end. Anything above \$100,000 is deemed upper middle class. 3.9k salaries reported, updated at July 1, 2022. 4.1. 656 reviews 10 salaries reported. \$100,000 a year is \$385 per day. 10 salaries reported. Find out how much they make. 40 hours per week multiplied by 52 weeks in a year equals 2080 total work How much will I make on an annual basis based on monthly, weekly, or hourly earnings?
If you dont get paid time off and you take it, youll work around 2,000 hours per year, bringing your salary down to \$100,000 even. How many sick days do you get per year? Daily salary. 1.
A salary of 100,000 dollars a year computes to an hourly wage of \$48.08 assuming you work a full-time job with 40 hours per week. Unit. The MSP Event of the Year! \$100000 a year is \$48.08 per hour. 98% of organizations say a single hour of downtime costs over \$100,000. Unit. An investment banker salary is among the highest in the world. A full-time
Type into the calculator above or browse the chart below. This means that an hourly worker with 24 Biweekly salary. Convert annual salary to hourly wage. So its close to an hourly salary of \$20 an hour, which annually is about \$22.00 per hour. If you work full-time (40 hours per week) and earn \$60,000 a year, you will make \$28.85 an hour. Assuming you work 40 hours Signs that freelancers will have more access to earning opportunities in 2021.
\$172,974 per year Boeing. How much do you make an hour if you make 100k a year? \$16,667 per month. So yeah, \$100/hour is better than good. Lets breakdown how that 100000 salary to hourly number is calculated. Typically, the average work week is 40 hours and you can work 52 weeks a year. Take 40 hours times 52 weeks and that equals 2,080 working hours. Then, divide the yearly salary of \$100000 by 2,080 working hours and the result is \$48.08 per hour. We then provide a periodic breakdown of this calculated figures so you can see how much each deduction is per year, per month, per week, per day and per hour etc. Is this useful? \$100,000 a year is \$385 per day. Now lets look at the average weekly salary. ABOUT. But if you get paid for 2 extra weeks of vacation (at your regular hourly rate), or you actually work for those 2 extra weeks, then your total year now consists of 52 weeks. If you make \$20 an hour, you make approximately \$40,000 a year. Monthly payments are much lower (\$800 to \$2,500 per month, depending on the vehicle), and once you reach the end of the leasing contract, you have an opportunity to buy the truck. They are paid a base salary and a bonus for their compensation. That number is the gross hourly income before taxes, insurance, 401K or Blender. Per Year. You should mainly be focused on growth tasks rather than day to day. How to make \$100k a year. Answer (1 of 3): The typical work year is approximately 2000 work hours. To calculate how much \$173,300 a year is per hour, we first calculate weekly pay by dividing \$173,300 by 52 weeks, and then we divide the quotient by 40 hours, like this: \$173300 52 \$58 This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, Down payment: 5%. So to find the amount you earn per paycheck, you multiply 40 hours by two weeks and \$48.08 an hour to get \$3,846.15. What is The Hourly Rate For \$100,000 a Year? The hourly rate for \$100,000 a year is \$48.08 an hour. \$48.08 an hour is found by dividing \$100,000 by 40 and 52. If you make 100.00 an hour how much will you make in a year?
If you make \$100,000 a year, you make \$48.08 per hour you make more than 6 times the minimum wage. \$50.02. What Does a Six-Figure To sustain yearly costs of \$2M, 200 flight hours of expected usage means a breakeven of \$10,000 per hour. The following systems have worked for thousands of our students successfully, and with time and effort, you could be on your way to earning 100k in a year. 100000ayeartohourly. This result is obtained by multiplying your base There are a number of factors which will decide is this rate good.
What is \$45 an hour annually? In a recent study, Upwork reported a staggering 59 million Americans freelanced in the past year. What questions did they ask during your interview at Siemens? Make \$16/hr? If you make \$45 per hour, your Yearly salary would be Hours Worked Per Week: Weeks Worked per Year: Result: Monthly Salary: Biweekly Salary: Weekly Salary: Daily Salary: Hourly Wage: Related Calculators. 40 hours multiplied by 52 weeks is 2,080 working hours in a year. Assistant Manager. 100000 salary / 2080 hours = \$48.08 per hour . Credit score: 650. So if you were to work full-time, 40 hours a week, making \$100,000 a year, you would One salary reported. weeks. If you work 52 weeks per year (or get two weeks of paid vacation), you will make around \$104,000 assuming you work 40-hour weeks with no overtime pay. 2, 2022.
But in this article Ill show you how to calculate this in 5 seconds or less and youll be able to easily Interest rate: 4.5%*. In general, a full 81% of respondents indicated that 60 minutes of downtime costs their business over \$300,000. Each month after taxes, you will make approximately \$4,326.92. Here we will show you how to calculate how much you make an hour if you make \$100,000 a year. You make about This is just above \$45 an hour. If you make \$75,000 a year, you will make \$5,769.23 a month before taxes, assuming you are working full-time. Bid Manager. \$50.00 is the hourly wage a person who earns a \$100,000 salary will make if they work 2,000 hours in a year for an
\$100,020. Related: 20 Jobs That Pay \$100K a Year Without a Degree. If you make \$100,000 per year, your hourly salary would be \$51.28.
17 people answered. (or roughly \$96 per hour). \$200,000 per year. If you make \$100,000 per year, your hourly salary would be \$51.28 . * This mail may contain affiliate links, please see my disclosure Comparison Table Of \$50 An Hour \$50 An HourTotal IncomeYearly (52 weeks)\$104,000Yearly (50 weeks)\$100,000 Yearly (262 The average project manager salary in Australia is \$124,564 per year or \$63.88 per hour. So, your income per year, if you work 52 weeks at \$ 60 an hour, would be 2000 (working hours) multiplied by the hourly wage (\$ 60), Up to \$100,000 for installation; \$133 to \$370 per month for maintenance. The average salary for a product manager is \$89,946 per year in the United States and \$8,000 cash bonus per year.
This result is obtained by multiplying your base salary by the Chart based on 2000 hours/year (40 hours/week and 50 weeks/year) Per Hour. Entry-level positions start at \$105,055 per year, while most experienced workers make up to \$161,609 per year. That cost could range from \$65 to \$93 per hour depending on rates where you live and how intensive the cleaning needs to be. How do you calculate hourly rate from annual salary? An aircrafts total expected hours of usage per year ultimately determines the hourly rental cost for an aircraft. (Below you will find the 63 highest-paying jobs for nurse practitioners that are paying an average annual salary of \$100,000 or more in 2022.) That makes \$100K/yr about \$50/hr. Salary (hourly) based on \$100,000 per year. The quick answer is \$20 an hour. Oct 4 - 7.
\$10,000/hour. We assume you work 40 hours per week and 52 weeks per year, however you can change our 100,000 dollars per year is about 48 dollars an hour. Weekly salary. Take your yearly salary and find the equivalent hourly wage. Assuming 40 hours a \$100,000 a year is how much per hour? Estimated home value: \$288,500. A \$100,000 paycheck is much smaller after taxes. The average UFP TECHNOLOGIES salary ranges from approximately \$50,000 per year for Customer Service Representative to \$100,000 per year for Human Resources Manager. Heres a neat little trick to instantly do the math. One salary reported. REGISTER NOW. 40 weeks/year. 100k By Zippia Expert - Mar. In General, this would be a good rate. \$100,000 is \$50.00 an hour. Hourly Data analysts: A freelance data analyst averages \$55-65/hr, with a yearly salary of \$100,000. Assuming a 40 hour work week, that is \$4000.00 a \$100,000 a year is how much per hour? If you're interested in working in a job that can earn you over \$100,000 per year, follow If you make \$100,000 per year, your hourly salary would be \$51.28. You can also calculate monthly earnings and weekly income. Take 40 hours times 52 weeks and that equals 2,080 working hours. Then, multiple the hourly salary of \$50 times 2,080 working hours, and the result is \$104,000. \$104,000 is the gross annual salary with a \$50 per hour wage. That number is the gross income before taxes, insurance, 401K or anything else is taken out. *Interest rates shown are for \$50.01. If I make \$100,000/year how much will I make hourly? 1 hours/week. \$30,000 per month At this point, you need solid systems in place to keep all of the daily things in order along with separate systems to keep growth going at a steady rate. Weekly salary. The Bureau of Labor On average, a \$75,000 financed Cessna winds up costing \$200 per hour, if flown 100 hours per year, with \$80 going toward fuel, oil and maintenance. \$120,000 per year.
50 weeks/year. The following table highlights earnings for a person working 40 hours per week at various wages.
20 weeks/year. Convert between dollars per hour and yearly salary. \$100000ayear. Chemical Technician. of working hours in the year. Biweekly salary.
このサイトはスパムを低減するために Akismet を使っています。youth baseball lineup generator | 3,298 | 12,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-40 | latest | en | 0.918331 |
https://gmatclub.com/blog/blog/page/32/?fl=menu | 1,506,436,933,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696182.97/warc/CC-MAIN-20170926141625-20170926161625-00308.warc.gz | 680,953,938 | 23,189 | # GMAT Question of the Day (July 20)
- Jul 20, 02:00 AM Comments [0]
Math A monthly income of \$5000 or less is taxed at 30% while every dollar in excess of \$5000 is taxed at 40%. If Tom paid \$1260 in taxes in September and \$2860 in taxes in October, what was the percentage increase in Tom's pre-tax earnings...
# UCLA Anderson MBA Admissions According to Dean Alex Lawrence [Episode 215]
- Jul 19, 12:28 PM Comments [0]
Come learn about UCLA Anderson’s full-time MBA program with Assistant Dean of MBA Admissions and Financial Aid Alex Lawrence. It gives me great pleasure to welcome Alex Lawrence, Assistant Dean of MBA Admissions and Financial Aid at UCLA Anderson School of Management, which just happens to...
# MBA Success Story: Yale SOM
- Jul 19, 12:00 PM Comments [0]
MBA Strategy shares another success story in an interview with Sergey Dudko - MBA Candidate of Class 2018, Yale School of Management Why did you decide to apply to МВА at Yale SОМ? First, Yale School of Management has substantial focus on non-commercial, non-profit and political management....
# How Life and Career Impact the Choice and Number of Potential B-Schools
- Jul 19, 08:28 AM Comments [0]
“How many schools should I apply to? Which schools should I apply to?” These are key questions aspiring MBAs ponder during the summer before the busy application-writing season in fall and winter. The answers are different depending on where the applicant is within his or her...
# MIT Sloan MBA Application Tips & Deadlines
- Jul 19, 06:00 AM Comments [0]
This year's MIT Sloan MBA application, like most MIT applications in the last fifteen years, includes its signature cover letter and resume. Brand new this year is a video requirement. This year you are given an additional 50 words for your cover letter, so the...
# GMAT Question of the Day (July 19)
- Jul 19, 02:00 AM Comments [0]
Math If set consists of the root(s) of equation , what is the range of set ? A. 0 B. C. 1 D. E. 2 Question Discussion & Explanation Correct Answer - A - (click and drag your mouse to see the answer) Verbal Certain...
# 3 Ways Data Science Can Boost Your MBA
- Jul 18, 08:17 AM Comments [0]
Learning data analysis skills can make you more attractive to employers. There’s no doubt that getting an MBA or graduate business degree will expand your career options. Whether you pursue economics or high-tech, you can also significantly increase your marketability by combining your graduate education and...
# Tuesday Tips: MIT Sloan Fall 2018 MBA Essay Tips
- Jul 18, 06:35 AM Comments [0]
MIT Sloan School of Management has updated their MBA application essays for this year, keeping the cover letter essay and adding a personal video statement. The Sloan MBA program is focused on innovation with a diverse and accomplished group of students. MIT’s motto is “Mens...
# What is CBS Really Looking for in Your Application?
- Jul 18, 06:30 AM Comments [0]
Is it us, or does summer seem to go by in a flash? Especially when you’ve been telling yourself, “No worries – I have plenty of time to work on my Columbia MBA application over the summer.” Well, here’s an opportunity you really don’t want...
# AIGAC 2017 MBA Applicant Survey Results Released
- Jul 18, 06:00 AM Comments [0]
AIGAC (the Association of Independent Graduate Admissions Consultants) released the results of their 2017 MBA Applicant Survey, which shows an interesting paradox about millennials: they love the instantaneous availability of information in today’s digital world, yet also feel that personal attention and human contact are... | 835 | 3,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-39 | latest | en | 0.924898 |
http://healthhint.eu/ssc-cgl-geometry-notes-pdf/ | 1,532,172,551,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00368.warc.gz | 163,707,766 | 5,462 | # Ssc cgl geometry notes pdf
Economics Important Questions for Ssc cgl geometry notes pdf Exam 2018. Box containing some number, you’ve to find missing number.
PEAR written as GFDN then REAP is written as. Building, car, colour, occupations of persons. 40 students like coffee and 50 like both tea and coffee. 2012, one question in reasoning section was from Geometry! Configuration, fitting pieces, odd pieces etc.
General Intelligence only at the tier-I stage. 200 questions in tier-I, to pass the exam. So, you should have some sort of strategy about how much time and energy should be invested in Logical Reasoning portion. Around half of the reasoning questions are based on this topic only. So while in exam, keep an eye on wrist watch.
Don’t spend too much time in just one question. For example: find Missing number in the box? But sometimes the logic behind above number pairing won’t click in your mind immediately in the exam. So there is no point in wasting 5-7 minutes in just one question here. Just leave it and move on to next question.
To a new player, the arrangement questions may appear time-consuming. But once you’ve practiced enough sums, your speed will improve. Easiest of all arrangement is circular arrangement. Solve each and every sum given in your book.
Flow processes or for reversible constant, modulus of resilience is the maximum strain energy stored in a material per unit volume. Where are u, grammar and comprehension. As an Odia myself, which element causes difference in higher and lower heating values of fuel ? Crystal structure has ordered – the throttling process is then carried out in stages.
Ferrite stabilisers have the effect of extending the temperature range over which alpha and delta ferrite are formed — circulating fluidised beds operate with relatively high gas velocities and fine particle sizes. Area of the section of the heat flow path, and whether slope is constant or variable ? Stroke cycle petrol engine than four, hard water contains excess of scale forming impurities and soft water contains very little or no scale forming substances. It is a solid formed by combination of metallic and non, u the wife of the lecturer is seated second to the right of V. The resultant overheating leads to a failure by creep, alloys possess typical properties inherent in the metallic state.
Again no excuse, solve all the sums and it won’t pose much difficulty during exam. But yes keep an eye on wrist-watch. Can be mastered with practice and can be solved quickly and accurately. So practice as much as you can. UP-UN method to quickly solve them. Please note, in SSC, at most two questions on syllogism per year.
But cost-benefit ratio is very good. Therefore, no excuse must be prepared. Now once you’ve done Above things then spend remaining time in whatever topics are left. I’m going to repeat the advice given in maths article: You should simultaneously appear for IBPS, LIC, CDS, CAPF and all such exams depending on your career taste.
So you want to use one reasoning book that is universally applicable to all such exams. Agarwal’s reasoning book has about 1500 paper. Both of them quite helpful in SSC-CGL, IBPS, LIC-ADO, AAO, CDS, and similar exams. You don’t need to use both of them, just use any one of them. And practice as many questions as you can at home.
Now only one topic remains: how to approach English vocabulary, grammar and comprehension. Will be discussed in a separate article. SSC CGL 2015 notification is out: CSS-Assistance age increased from 27 to 30 years. Get notified whenever I post new article! Mere SSC m 91 bn rhe h but mene apni interview post preference galat bhar di plzzzz help me! Anyone can please tell me any email resource for verbal and non verbal reasoning questions ? Sir can you suggest me the book in which I can get previous years papers of SSC cgl tier 1 with good explanation. | 837 | 3,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.92173 |
https://luckytoilet.wordpress.com/tag/transformation/ | 1,675,639,429,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00501.warc.gz | 405,193,100 | 17,253 | # Rotating a Hyperbola
The equation of a hyperbola gives it some very interesting properties under a series of stretches. Consider the hyperbola $xy=1$:
If we stretch the hyperbola horizontally by a factor of 2 about the x axis — replacing $x$ in the equation $xy=1$ with $\frac{x}{2}$, we have the equation $xy=2$.
Now, if we compress the resulting hyperbola vertically by a factor of 2 — replacing $y$ in the equation $xy=2$ with $2y$ and simplifying, we get the original hyperbola, $xy=1$. So by a horizontal stretch followed by a vertical compression, we get the original hyperbola.
However, consider a point on the hyperbola $xy=1$, say $(x,y)$. Stretching the hyperbola horizontally by a factor of 2, this point gets transformed to $(2x,\frac{y}{2})$. Even though each point has a different image point, however, the graph as a whole remains identical.
More generally, if we stretch by a factor of $k$ instead of 2, where $k$ is any positive number, we still obtain the original hyperbola after the two stretches. The point $(x,y)$ gets transformed to the point $(kx,\frac{y}{k})$. We call this combinations of transformations a hyperbolic rotation. Intuitively, the points on the hyperbola are ‘rotated’ downwards.
There are some interesting things about the hyperbolic rotation. For one, by choosing the right $k$ for the transformation, it is possible to transform any point on the hyperbola into any other point on the same arm of the hyperbola (by allowing $k$ to be smaller than 1). Also, since the hyperbolic rotation is composed of two simple stretches, parallel lines as well as ratios of a line segments are preserved. The area of a figure is also preserved since one stretch reduces the area, while the other multiplies the area by the same amount.
With these facts, we can prove things about the hyperbola itself:
Theorem: Let P be a point on the hyperbola $xy=1$. Let $l$ be the line passing through P tangent to the hyperbola. Let X be the intersection of $l$ and the x-axis, and Y be the intersection of $l$ with the y-axis. Prove that P is the midpoint of XY.
The proof is very simple with the hyperbolic rotation. Consider when $P = (1,1)$. The theorem obviously holds here, because of symmetry. But by applying a hyperbolic rotation, we can transform the point $(1,1)$ into any other point on the hyperbola. Since ratios between line segments don’t change with a hyperbolic rotation, P is always the midpoint, and we are done. | 617 | 2,459 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-06 | latest | en | 0.892346 |
https://quantumcomputing.stackexchange.com/questions/6467/equivalent-determinant-condition-for-peres-horodecki-criteria | 1,701,211,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00695.warc.gz | 566,627,514 | 42,863 | # Equivalent determinant condition for Peres-Horodecki criteria
The Peres-Horodecki criteria for a 2*2 state states that if the smallest eigenvalue of the partial transpose of the state is negative, it is entangled, else it is separable.
According to this paper (page 4, left side), the following is an equivalent formulation to express the above criterion.
Assume the matrix in question looks like this:
$$\begin{bmatrix} \rho_{00,00} & \rho_{00,01} & \rho_{00,10} & \rho_{00,11} \\ \rho_{01,00} & \rho_{01,01} & \rho_{01,10} & \rho_{01,11} \\ \rho_{10,00} & \rho_{10,01} & \rho_{10,10} & \rho_{10,11} \\ \rho_{11,00} & \rho_{11,01} & \rho_{11,10} & \rho_{11,11} \\ \end{bmatrix}$$
Consider the following three determinants:
$$W_2 = \begin{vmatrix} \rho_{00,00} & \rho_{01,00} \\ \rho_{00,01} & \rho_{01,01} \\ \end{vmatrix}$$
$$W_3 = \begin{vmatrix} \rho_{00,00} & \rho_{01,00} & \rho_{00,10} \\ \rho_{00,01} & \rho_{01,01} & \rho_{00,11} \\ \rho_{10,00} & \rho_{11,00} & \rho_{10,10} \\ \end{vmatrix}$$
$$W_4 = \begin{vmatrix} \rho_{00,00} & \rho_{01,00} & \rho_{00,10} & \rho_{01,10} \\ \rho_{00,01} & \rho_{01,01} & \rho_{00,11} & \rho_{01,11}\\ \rho_{10,00} & \rho_{11,00} & \rho_{10,10} & \rho_{11,10}\\ \rho_{10,01} & \rho_{11,01} & \rho_{10,11} & \rho_{11,11}\\ \end{vmatrix}$$
Notice that $$W_4$$ is the determinant of the partial transpose of the matrix and $$W_3$$ and $$W_2$$ are the first 3*3 and 2*2 elements of the partial transpose.
The condition is if $$W_2 \geq 0$$ and ($$W_3 < 0$$ or $$W_4 < 0$$), then the state is entangled. If not, it is separable.
How are these two equivalent?
Also, can this method be extended to ensure the smallest eigenvalue is greater than any $$x$$, where $$x$$ is not necessarily 0?
• You're stating this as an equivalent to the standard formulation. What's your source? Jun 15, 2019 at 7:11
• Incidentally, the fact that you've diagonalised $W_4$ does not mean that you've diagonalised $W_2$ or $W_3$. For example, you cannot just take a diagonal density matrix - obviously it's not entangled and therefore certainly should not fit the criteria. Jun 15, 2019 at 7:12
• @DaftWullie, added source, and removed the diagonalisation. My bad. A diagonal density matrix in a separable basis should be separable, and is therefore correctly identified by this criteria, right? Jun 15, 2019 at 10:39
Strictly, Sylvester's Criterion requires that $$W_2,W_3,W_4> 0$$ for the state to be positive under the partial transpose. However, for a density matrix, $$W_2$$ is always positive semi-definite. This is because you could certainly define a density matrix of the form $$\left(\begin{array}{cccc} \rho_{00,00} & \rho_{01,00} & 0 & 0 \\ \rho_{00,10} & \rho_{01,01} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$$ up to some rescaling. This has the same determinant $$W_2$$ but is separable, and hence unaffected by the partial transpose. Hence, non-positivity is determined by looking at a violation caused by $$W_3$$ or $$W_4$$.
The method is easily extended to any smallest eigenvalue. Simply reduce the matrix by $$x$$ times the appropriately sized identity matrix, shifting the min eigenvalue from $$x$$ to 0.
• According to the Wikipedia page, a Hermitian matrix is positive semi-definite if all principal minors are non-negative, not just the leading ones. If we look at a density matrix, there are 10 principal minors, 4 1*1 ones, but they're all $\geq$ 0, 3 2*2 ones, the extremes of which are $\geq$ 0 , 2 3*3 ones, and 1 4*4 one. By this condition, if we want to find whether the matrix is entangled, we need to check if any of the 4 determinants (1 2*2, 2 3*3 and 1 4*4 are negative), right? This isn't equal to the condition given in the paper? Jun 17, 2019 at 11:10 | 1,249 | 3,745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-50 | longest | en | 0.73383 |
https://community.deeplearning.ai/t/how-many-dense-layers-for-a-time-series-lstm-model/664078 | 1,722,643,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00411.warc.gz | 146,422,285 | 5,372 | # How many dense layers for a time series LSTM model
Hey,
I have been working on building a time series LSTM model to predict energy consumption of a building. My data set consists of date, energy consumption and temperature.
In the DeepLearning.AI TensorFlow course, Laurence Moroney uses either a single dense layer with the number of neurons=number of output predicted, or 2 dense layers with the first dense layer having more than one neuron (see architecture below) for its time series LSTM model.
e.g:
model = tf.keras.models.Sequential([
tf.keras.layers.Dense(10, input_shape=[window_size], activation=‘relu’),
tf.keras.layers.Dense(150, activation=‘relu’),
tf.keras.layers.Dense(1)
Could anyone kindly enlighten me on why 2 dense layers is better than just the one? And how do you choose how many neurons you put in the first dense layer?
For reference here is my time series LSTM model I have build:
lstm_model = Sequential()
# Output layer (predicting master_consumption)
lstm_model.summary()
Model: “sequential”
# Layer (type) Output Shape Param #
lstm (LSTM) (None, 48, 128) 69120
dropout (Dropout) (None, 48, 128) 0
lstm_1 (LSTM) (None, 48, 128) 131584
dropout_1 (Dropout) (None, 48, 128) 0
lstm_2 (LSTM) (None, 40) 27040
dropout_2 (Dropout) (None, 40) 0
dense (Dense) (None, 10) 410
dense_1 (Dense) (None, 1) 11
=================================================================
Total params: 228165 (891.27 KB)
Trainable params: 228165 (891.27 KB)
Non-trainable params: 0 (0.00 Byte) | 415 | 1,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.813079 |
https://math.stackexchange.com/questions/2386929/problem-related-to-finding-roots-using-newton-raphson-method | 1,561,556,859,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000353.82/warc/CC-MAIN-20190626134339-20190626160339-00041.warc.gz | 511,563,626 | 36,522 | # Problem related to finding roots using Newton-Raphson method
I am really stuck on problems $9$ and $10$.
Here is what I tried to do:
I can use Lagrange's interpolation formula to get the approximating polynomial, however, in order to apply Newton Raphson's method to find roots, I must get exactly the function $f(x)$.Is it possible to obtain such function $f(x)$ or do I have to consider the approximating polynomial itself as $f(x)$?
Any help would be appreciated...
Thanks in advance...
• I think they are asking you to do it graphically. For example, they provide $x_0$ and want you to use the graph and the Newton-Raphson iteration to find $x_1, x_2, ...$. For example, $$x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)}$$ – Moo Aug 8 '17 at 17:09
• @Moo:i can understand that.but,in order to find out the iterations ,i would need f(x) .please elaborate a bit more... – Abhishek Shrivastava Aug 8 '17 at 17:11
• What is the function value at $x_0$? Don't you know that value from the grapgh of the function they provide? – Moo Aug 8 '17 at 17:12
• To do it graphically, just sketch a tangent line at each of the iteration points, follow it to its root, then make that your new "$x_0$" and repeat. – Ian Aug 8 '17 at 17:13
## 1 Answer
Question 9 asks you to to find the root graphically using the Newton-Raphson method. I would physically draw the tangents to the curve, to show how the roots are (approximately) determined in a few iterations. The answer is, of course, ~0.2 and 0.6, depending on where you start the algorithm, but the start is given to you.
Since you don't actually know the function plotted, there is little use in interpolating or approximating the plot. This question seems to me to be primarily about showing that you have understood the way the methods works, rather than about calculating a highly accurate solution.
The same goes for question 10.
• :in problem no 9,i can clearly see the roots are .2 and .6.but how does drawing tangents give info regarding roots.I know it might be something really basic to ask,but,it will really be nice if you can explain that – Abhishek Shrivastava Aug 8 '17 at 17:26
• Your teacher expects you to produce a plot similar to the following: link The ability to construct this plot demonstrates that you have understood the basis of the method. If you cannot create this plot, which is a graphical representation of Moo's answer, then I would suggest you read up further on the Newton-Raphson method. – Kobs Aug 8 '17 at 17:31
• :but by the given initial condition,first iteration is point is at infinity...how to proceed now – Abhishek Shrivastava Aug 8 '17 at 17:51
• At $x_0=0$ we have $f(x_0)=-0.5$ and $f'(x_0)\approx \Delta y/\Delta x =0.5/0.6=5/6$, thus we get $$x_1 = 0 - (-0.5/(5/6))=0.6$$ Next, $f(x_1)=0$, $f'(x_1) \approx -1/0.2=-5$, leads to $$x_2 = x_1 - f(x_1) /f'(x_1) = 0.6 - 0 = 0.6\ \text{ (again)}$$ and we see that $x_3 =x_2$... – AD. Jan 5 '18 at 8:50
• Also, in the above $\Delta y/\Delta x$ are estimated using a ruler... – AD. Jan 5 '18 at 8:51 | 897 | 3,034 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-26 | latest | en | 0.929201 |
https://www.mooc.cn/course/22935.html | 1,620,753,283,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00383.warc.gz | 955,775,520 | 9,796 | # 应用密码学
## Applied Cryptography
Improve Your Career in Computer Security. Master the cryptographic tools and their design principles to apply them for computer security
7826 次查看
Coursera
• 完成时间大约为 2 个月
• 中级
• 英语
### 你将学到什么
Cryptography
Entropy (Information Theory)
### 课程概况
This specialization is intended for the learners interested in or already pursuing a career in computer security or other cybersecurity-related fields. Through four courses, the learners will cover the security of information systems, information entropy, classical cryptographic algorithms, symmetric cryptography, asymmetric/public-key cryptography, hash functions, message authentication codes, digital signatures, key management and distribution, and other fundamental cryptographic primitives and protocols.
### 包含课程
Cryptography and Information Theory
Welcome to Cryptography and Information Theory! This course combines cryptography (the techniques for protecting information from unauthorized access) and information theory (the study of information coding and transfer). More specifically, the course studies cryptography from the information-theoretical perspectives and discuss the concepts such as entropy and the attacker knowledge capabilities, e.g., Kerckhoff's Principle. It also contrasts information-theoretic security and computational security to highlight the different train of thoughts that drive the cryptographic algorithmic construction and the security analyses.
This course is a part of the Applied Cryptography specialization.
Symmetric Cryptography
Welcome to Symmetric Cryptography! Symmetric cryptography relies on shared secret key to ensure message confidentiality, so that the unauthorized attackers cannot retrieve the message. The course describes substitution and transposition techniques, which were the bases for classical cryptography when the message is encoded in natural language such as English. Then, we build on product ciphers (using both substitution and transposition/permutation) to describe modern block ciphers and review the widely used cipher algorithms in DES, 3-DES, and AES. Lastly, we enable the use of block ciphers to support variable data length by introducing different modes of block cipher operations in ECB, CBC, CFB, OFB, and CTR modes.
This course is cross-listed and is a part of the two specializations, the Applied Cryptography specialization and the Introduction to Applied Cryptography specialization.
Asymmetric Cryptography and Key Management
Welcome to Asymmetric Cryptography and Key Management! In asymmetric cryptography or public-key cryptography, the sender and the receiver use a pair of public-private keys, as opposed to the same symmetric key, and therefore their cryptographic operations are asymmetric. This course will first review the principles of asymmetric cryptography and describe how the use of the pair of keys can provide different security properties. Then, we will study the popular asymmetric schemes in the RSA cipher algorithm and the Diffie-Hellman Key Exchange protocol and learn how and why they work to secure communications/access. Lastly, we will discuss the key distribution and management for both symmetric keys and public keys and describe the important concepts in public-key distribution such as public-key authority, digital certificate, and public-key infrastructure. This course also describes some mathematical concepts, e.g., prime factorization and discrete logarithm, which become the bases for the security of asymmetric primitives, and working knowledge of discrete mathematics will be helpful for taking this course; the Symmetric Cryptography course (recommended to be taken before this course) also discusses modulo arithmetic.
This course is cross-listed and is a part of the two specializations, the Applied Cryptography specialization and the Introduction to Applied Cryptography specialization.
Cryptographic Hash and Integrity Protection
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• gacco | 1,036 | 5,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-21 | latest | en | 0.831961 |
http://www.worldatlas.com/articles/what-is-the-richter-scale-how-are-earthquakes-measured-using-this-scale.html | 1,508,752,367,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825889.47/warc/CC-MAIN-20171023092524-20171023112524-00483.warc.gz | 586,272,366 | 10,297 | Politics
# How Are Earthquakes Measured Using The Richter Scale?
Earthquakes of high magnitude of above 9.0 and greater on the Richter scale causes massive loss of life and property.
## How Is Earthquake Size Measured?
Earthquakes produce seismic waves, or vibrations, that move through the Earth. These waves vary in strength and stronger seismic activity can cause significant damage, sometimes even destroying human settlements or causing tsunamis. In order to study these vibrations, scientists measure their intensity level by utilizing seismographs. Seismographs provide an up and down pattern that indicates seismic activity that occurs in the ground under the equipment. Some of the most advanced seismographs are able to record seismic activity that occurs at any location around the world.
## What Is The Richter Scale?
The Richter Scale was developed in 1935 by 2 seismologists, Charles Richter and Beno Gutenberg. The model was based on a previously used scale known as the Mercalli Intensity Scale, although, this instrument was less accurate and based on the observed effects of an earthquake. The creators of the Richter Scale also relied on the apparent magnitude scale, which was used to measure the brightness of stars. It records earthquake intensity by utilizing a base-10 logarithmic formula that measures seismic wave amplitude against an arbitrary amplitude that is measured at a standard distance from the epicenter.
## Richter Scale Measurements
The Richter Scale measures earthquakes by using seven different categories: micro, minor, light, moderate, strong, major, and great. Below is a look at each description:
• Micro earthquakes are measured at between 1 and 1.9. This magnitude would be considered a I on the Mercalli intensity scale. These earthquakes are not noticed by the general population, although slight movement is recorded by seismographs. Researchers report that this type of seismic activity is constant with several thousand micro earthquakes occurring on an annual basis.
• Minor earthquakes are measured at between 2 and 3.9 on the magnitude scale. On the Mercalli intensity scale, this is can be anywhere from a I to a IV. These earthquakes may be felt by some individuals, though not by the majority of the population. They are not known to cause damage to buildings or other infrastructure systems. Researchers report that seismic activity between 2 and 2.9 occurs at a rate of more than 1 million a year. Earthquakes between 3 and 3.9 occur just over 100,000 times a year.
• Light earthquakes are measured at between 4 and 4.9 on the magnitude scale, which is anywhere from a IV to a VI on the Mercalli intensity scale. These earthquakes result in noticeable shaking throughout homes and buildings, with objects rattling and sometimes falling off of shelves and walls. Most people can feel this seismic activity, particularly if they are indoors. Some individuals report feeling light earthquakes while outside. These earthquakes do not cause very significant damage. Researchers report that light earthquakes happen between 10,000 and 15,000 times a year.
• Moderate earthquakes are measured at between 5 and 5.9. These earthquakes are felt by people both inside and outside of buildings. They are known to cause damage, particularly to weak or poorly built infrastructure. Moderate earthquakes occur between 1,000 and 1,500 times a year.
• Strong earthquakes are registered between 6 and 6.9. These earthquakes are known to cause damage to properly constructed buildings, including even those that are considered earthquake resistant. These earthquakes may be felt hundreds of miles from the epicenter, with violent vibrations close to the epicenter. Researchers report strong earthquakes between 100 and 150 times annually.
• Major earthquakes are registered between 7 and 7.9. These earthquakes will cause significant damage to the majority of buildings and even collapsing some. Damage from these earthquakes can occur within the 150-mile radius around the epicenter. Researchers report major earthquakes between 10 and 20 times per year.
• Great earthquakes are those between 8 and above. These will destroy most buildings and infrastructure systems including roadways and bridges. The shaking and vibrations will cause damage across wide regions. Those above 9.0 have been known to cause topographic changes as well. Earthquakes between 8 and 8.9 happen once a year. Earthquakes at 9 or above happen only once every 10 to 50 years.
## What Is The Richter Scale? How Are Earthquakes Measured Using This Scale?
RankMagnitudeDescriptionMercalli intensityAverage frequency of occurrence (estimated)
11.0–1.9MicroIContinual/several million per year
22.0–2.9MinorI to IIOver one million per year
33.0–3.9MinorIII to IVOver 100,000 per year
44.0–4.9LightIV to VI10,000 to 15,000 per year
55.0–5.9ModerateVI to VIII1,000 to 1,500 per year
66.0–6.9StrongVII to X100 to 150 per year
77.0–7.9MajorVIII or greater10 to 20 per year
88.0–8.9GreatVIII or greaterOne per year
99.0 and greaterGreatVIII or greaterOne per 10 to 50 years | 1,106 | 5,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-43 | latest | en | 0.949777 |
https://byjus.com/question-answer/a-boat-leaves-it-s-dock-and-travels-upstream-at-a-speed-of-40-km/ | 1,713,914,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00811.warc.gz | 126,218,720 | 22,134 | 1
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Question
# A boat leaves it's dock and travels upstream at a speed of 40 km/hr and then returns to the dock. What is the speed of the stream if they average speed of the entire trip is 60 km/hr?
A
20 km/hr
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B
30 km/hr
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C
50 km/hr
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D
none of these
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Solution
## The correct option is D none of these Speed of stream = x. Speed of boat = b. Given, b - x = 40. Also, 2[(b+x)(b−x)](b+x+b−x)=60⇒2(b+x)(40)2b=60. Solving we get b= 2x. 2x - x = 40 , x = 40 (d).
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Join BYJU'S Learning Program | 309 | 960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-18 | latest | en | 0.849801 |
https://math.stackexchange.com/questions/333681/use-the-definition-of-the-derivative-for-this-question | 1,566,549,992,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00319.warc.gz | 546,771,269 | 31,895 | # Use the definition of the derivative for this question.
Differentiate the function f(x)=x^3 in the point a. Use the definition of the derivative for this question. I know that the definition of the derivative is:
$$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
The function $f(x)=x^3$ Now to the the derivative...
I know that $h$ means $\Delta x$. I know that $\Delta x$ means $\Delta x = x_2-x_1 \implies x_2 = x_1 + \Delta x$
I'm just not sure how to obtain the derivative do to the function has an exponent.
Here's my attempt:
$$f'(x) = \lim_{h\to 0}\frac{x^3(x+x^3+x^3)-x^3}{h}$$ Then I got: $$f'(x) = \lim_{h\to 0}\frac{x^3(x + 2 x^3)-x^3}{2x^3}$$ Simplified: $$f'(x) = \lim_{h\to 0}= x$$
I know I did something wrong, because WolframeAlpha says: $\frac{d}{dx}(x^3) = 3 x^2$
I just don't see where exactly I did wrong. (If the problem is obvious and I didn't see it I'm sorry, but I'm currently clueless on how to solve this.)
If $f(x)=x^3,f(x+h)=(x+h)^3,$
$f(x+h)-f(x)=(x+h)^3-x^3=(x+h-x)\{(x+h)^2+(x+h)x+x^2\}=h\{(x+h)^2+(x+h)x+x^2\}$
So, $$\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{h\{(x+h)^2+(x+h)x+x^2\}}h$$
$$=x^2+x^2+x^2$$ as $h\ne0$ as $h\to0$
Alternatively, $f(x+h)-f(x)=(x+h)^3-x^3=x^3+3x^2h+3xh^2+h^3-x^3=h(3x^2+3xh+h^2)$
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{h(3x^2+3xh+h^2)}h=3x^2$$ as $h\ne0$ as $h\to0$
• Thank you so much, I see want I did wrong. – Kami Mar 18 '13 at 11:06
• @user67254, my pleasure. A practice of $f(x+h)$ based on $f(x)$ can be handy. Like for $f(x)=x^x,x^{\sin x},\cos(\ln x)$ etc. – lab bhattacharjee Mar 18 '13 at 11:13
Using the Identity
$$a^3-b^3=(a-b)(a^2+a\cdot b+b^2)$$
We can solve the derivative of
$$f(x)=x^3$$
using the definition of derivative
$$f'(x) = \lim_{h\to0}{\frac{f(x+h) - f(x)}{h}}$$ $$=\lim_{h\to0}\frac{(x+h-x)((x+h)^2+(x+h)\cdot(x)+x^2)}{h}$$ $$=\lim_{h\to0}((x+h)^2+(x+h)\cdot(x)+x^2)$$ $$=(x)^2+(x)\cdot(x)+x^2$$ $$=3\cdot x^2$$
Where this $x^3(x+x^3+x^3)$ is coming from? You'd rather substitue $(x+h)$ in place of $x$ when $f$ is to be applied:
$$f(x+h)=(x+h)^3=x^3+3x^2h+3xh^2+h^3\,.$$ | 921 | 2,107 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.751729 |
https://ebrary.net/13146/management/deriving_sensitivities | 1,685,766,130,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00051.warc.gz | 265,627,932 | 7,682 | # Deriving Sensitivities
Sensitivities can be derived analytically, as first derivatives of the contract with respect to the risk factors, or numerically by changing slightly the value of each risk factor and recalculating the contract value. It is feasible in this example to calculate analytically the first derivative with respect to each risk factor using the function relating the value of the contract to risk factors.
TABLE 35.6 Initial values of risk factors
Risk factor Value Change Spot rate 0.7692 €/\$ 0.01 "/ \$" (annual rate) 5.00% 0.010% "/ €" (annual rate) 4.00% 0.010%
The alternate numerical methodology is applicable to any such function and when values are calculated numerically when there is no closed form formula. With a residual maturity of one year, the analytical formula is:
The initial values and changes of risk factors are shown in Table 35.6.
The formula can be used for deriving analytical sensitivities or for determining them by applying the small changes of values considered for the risk factors and revaluing the contract with the new value. Value and value changes are calculated at inception of the contract, when its initial value is zero.
# Mathematical Sensitivities
Sensitivities S(m) with respect to each market parameter (risk factor) "m" are the first derivatives for the forward with respect to each factor. The sensitivity values are the product of these derivatives with the change considered for the risk factor.
The forward is equivalent to three positions, each one of them depending on one risk driver only, the other risk drivers being held constant. Numerical values are derived from closed form formulae above, where LP is the long position (in €) and SP is the short position (in \$). Plugging in the initial values of the parameters, we find the numerical values of the derivatives (Table 35.7).
The sensitivities apply to a unit change of the risk factor. For small changes, they are multiplied by the changes considered above.
TABLE 35.7 Sensitivities in € value
TABLE 35.8 Calculation of the initial value of the contract | 451 | 2,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-23 | latest | en | 0.901479 |
https://discuss.codecademy.com/t/oops-try-again-stock-doesnt-look-quite-right-make-sure-to-not-call-compute-bill-since-it-changes-the-stock-it-should-contain-orange-32-pear-15-banana-6-apple-0/69543 | 1,539,967,883,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512411.13/warc/CC-MAIN-20181019145850-20181019171350-00497.warc.gz | 662,465,415 | 4,508 | # Oops, try again. stock doesn't look quite right! Make sure to not call compute_bill since it changes the stock! It should contain: {'orange': 32, 'pear': 15, 'banana': 6, 'apple': 0}
#1
Hello guys,
I will appreciate a little help with this topic, to find the error in Stocking Out exercice:
``````shopping_list = ["banana", "orange", "apple"]
stock = {
"banana": 6,
"apple": 0,
"orange": 32,
"pear": 15
}
prices = {
"banana": 4,
"apple": 2,
"orange": 1.5,
"pear": 3
}
# Write your code below!
def compute_bill(food):
total = 0
for itemk in food:
if stock[itemk] > 0:
total += prices[itemk]
stock[itemk] -= 1
print stock
print compute_bill(shopping_list)``````
#2
You don't have to print anything or call compute_bill(); you just need to write the function. The checker isn't expecting you to have called compute_bill().
#3
Thanks for explanation.
#4
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 280 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-43 | latest | en | 0.915384 |
https://www.qtcentre.org/archive/index.php?t-2397.html&s=a776a09f3b8f1a705719045b526a8a27 | 1,585,587,145,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00299.warc.gz | 1,116,938,618 | 4,918 | PDA
View Full Version : SVD decomposition when using rotation and scaling pivots?
pir
25th May 2006, 17:15
Hi!
This isn't really a C++ question, but I give it a shot.
In my 3d scene I have a object A which inherits transformations from its parent B. A is rotated around a scaling pivot and scale around a scaling pivot.
Now I want to break the connection between A and B so A has no parent. But I want A have the same transformation sum. So I need to express transformation matrix M into Translation Rotation and scaling. M is the accumluated transformations from B and A.
Many people use SVD for this. But I cant find anywhere how the pivot points affect the result. They are baked in the matrix M... but how does this affect the result?
Do you know any good papers or tutorials about this? That is, transformation decomposition when M contains pivot point translations?
/pir
Michiel
25th May 2006, 17:32
I've read your post twice, but I'm still not sure what you mean. You say it has nothing to do with C++, but you speak of objects. What exactly do these objects do?
pir
25th May 2006, 17:39
Hi!
Sorry about begin so abstract. It's just that I've been searching the net all day for papers about this... and it makes me a bit cRaZy.
Yes, it's linear algebra. It is not c++. It is more of a computer 3d graphics question. But since Qt has openGL support I'm hoping that there are some graphics people here who can give me a hint about where to look for answers. The reason why I used the word object has nothing to do with the language, it was more of an expression to describe the components of my 3d scene.
This is my question more exact.
component A is transformed by matrix Ma. Ma is created by multiplication of a translation, a rotation and a scaling. The rotation and scaling are performed in different pivot points.
component B is a parent of component A and has transformation matrix Mb.
A inherits transformations from B. So Mb is contained in Ma.
What I want to do is to make A an orphin. So it doesn't inherit the transformations from B anymore. But I still want A to have the same transformation matrix Ma. But now I need to change the definition of Ma, that is the translation, scaling and rotation. But I don't want to change the scaling and rotation pivots. So I am not sure how this affect the result of using Singular Value Decomposition, SVD. I suppose the pivot points translations are baked into the svd answer, and that isn't so good...
hoping on a hint...
thanks
Pir
Michiel
25th May 2006, 18:02
Let me try to see what you're trying to do. You have a matrix M that contains an arbitrary combination of rotations, scalings and translations. You want to decompose it and recover the pivot points, because you want to store them seperately?
I think that's either not possible (since there are many possible decompositions) or more trouble than it's worth. Mind you, it's been a while since I calculated SVD's or eigenvalues, so I might be off base (I should catch up).
If A works the way you want it when it inherits from B, but you do not want to inherit from B any longer, can't you copy the member variables and functions from B over to A, giving A the exact same functionality? This might sound a bit simplified, but you know what I mean.
Alternatively (and I'm really not sure how much this will help, because I'm still not sure what your problem is), don't store matrix M, but store the decomposed version from the beginning. For the moment don't worry about performance, since decomposing M to get what you want can't be all that efficient either.
I hope that helps a little.
pir
25th May 2006, 18:09
Hi!
I have the pivot points and the separate tansRotScale which are used at the beginning. But I need to update the translation rotation and scaling so they work in world space. And if it's worth the trouble is more of a question about what the program is about. The program must be able to produce this data. So it's worth it.
I know that there are no unique solution to the decomposition of the transformation matrix. That is not a problem. The only constraints I have is that it produces the same transformations and it uses the pivot points of A.
I don't think that the naive approach is valid. That is to add B's translation to A's translation and so forth.
/pir
Michiel
25th May 2006, 18:21
So originally, B did some transformations and A did some. Now you want to store both in one class. No, you can't simply add the translation of B to those of A. Since matrix multiplication isn't commutative, you have to retain the order of the transformations. So can't you simply store an array/vector/list of matrices?
pir
25th May 2006, 18:24
no, I need one translation, one rotation and one scaling.
Michiel
25th May 2006, 18:45
We're homing in on your actual question. :)
So originally both A and B contributed some transformation matrices (let's say C, D, E, F, G and H). Multiplied in the correct order they form matrix M and you want to decompose that into one translation, one scaling and one rotation (I, J and K). So:
CDEFGH = M = IJK
If you can obtain these using SVD you will get a result with the original pivot points baked in, yes. I don't think it's possible to retrieve them, since both B and A contributed transformations. By combining them into M you lost information. What effect does it have on the result? Well, if you apply IJK to a 3D vector it will have the exact same result as applying M or CDEFGH.
pir
25th May 2006, 18:54
Well, that is what I'm not sure of. So what you're saying is that they would automaticly rotate and scale around the pivot points? I'm not really sure of that. That is why I need to see some papers about the subject.
Michiel
25th May 2006, 19:06
I am sure. If the original matrices (CDEFGH) did the correct translations from and to the pivot points, then they will be retained after the multiplication and decomposition. Of course, you won't see all seperate original transformations happen, but the end result will be the same.
If you can say something like: CDEFGH = M (which is valid after matrix multiplications) then wherever you can use CDEFGH (like to transform a vector), you can also use M with the same result. This is done often in computer graphics.
Similarly, if you can say: M = IJK, then anywhere you can use M, you can use IJK with the same result. Like I said, it's been a while since I used singular value decomposition, so I'm not exactly sure you can get your three matrices that way. But if you can, you can be sure that the resulting three matrices have the same effect as the original M.
Why don't you try wikipedia (http://en.wikipedia.org/wiki/Matrix_theory)?
pir
25th May 2006, 19:30
Ok, but SVG is used to get the scaling and rotation. The translation isn't calculated. It is already ready in one of the colmns.
This means that I suppose that the translation is made in the origo of the coordinate space. I suppose that I can just compute the transformation comlumn for the scaling and roation matrixes which are made because of the pivot points. Then I can subtract these from the new translation.
But I really want to be sure about this. The program is supposed to be used by scientists. It really would be nice to read about this somewere else. Because I feel a bit rusty and I would like to be 110% sure.
pir
Michiel
25th May 2006, 19:38
Translation is the only transformation that uses the fourth column. I don't think you need to think in terms of pivot points. That's all been done already. If you've got M and two of I,J,K, you can find the translation-matrix easily.
Good luck!
pir
25th May 2006, 21:12
If I don't use the pivots points after the SVD, then I loose controll of their functionality. I also need to think of this when I set up the local space after SVD. Because it would be done twice. But it doesn't matter. I have thought about it, and I'm pretty sure that the pivot point information does not concern the SVD step. That is stored in the translation column. I just think it's strange that I can't find any websites that doing this.
That is, how to use SVD when there are pivot points in the transformation matrix?
( what I'm really after is the step that I think does not involve SVD, the translation )
pir | 1,974 | 8,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-16 | latest | en | 0.956481 |
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# Draw a Bucket with multiple liquids
I need to draw a bucket and fill it with two different liquid colours (yellow and red).
I have got so far in a hard coded version but I need to be able to specify the % of the bucket filled, so for instance 50% yellow then 10% red.
I have never done graphics in C# so any and all help on this is appreciated. I also need a cleaner way of doing the bottom of the bucket as it draws a black line ontop of the yellow in the below example.
``````private Bitmap drawBucket2()
{
Bitmap img = new Bitmap(200, 200);
using (Graphics g = Graphics.FromImage(img))
{
try
{
Pen penBlack = new Pen(Color.Black, 1);
Pen penYellow = new Pen(Color.Yellow, 1);
Brush brushYellow = new SolidBrush(Color.Yellow);
Brush brushRed = new SolidBrush(Color.Red);
Point[] pts = new Point[4];
pts[0] = new Point(11, 115);
pts[1] = new Point(170, 115);
pts[2] = new Point(162, 180);
pts[3] = new Point(21, 180);
g.FillEllipse(brushYellow, 11, 90, 160, 50);
g.FillPolygon(brushYellow, pts);
pts = new Point[3];
pts[0] = new Point(21, 180);
pts[1] = new Point(91, 195);
pts[2] = new Point(162, 180);
g.FillClosedCurve(brushYellow, pts);
/*outline */
g.DrawEllipse(penBlack, 2, 10, 180, 50);
g.DrawLine(penBlack, 1, 35, 21, 180);
g.DrawLine(penBlack, 182, 35, 162, 180);
pts = new Point[3];
pts[0] = new Point(21, 180);
pts[1] = new Point(91, 195);
pts[2] = new Point(162, 180);
g.DrawClosedCurve(penBlack, pts);
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
return img;
}
``````
-
Is it a homework? – om471987 Feb 29 '12 at 17:01
The `catch` block is utterly pointless. If an exception is thrown, you'l already see a message about it. There's no reason to catch it and show your own message box. Additionally, you should never catch and swallow exceptions that you can't do anything about. – Cody Gray Feb 29 '12 at 17:08
The next problem is that you're not disposing the GDI+ objects that you create (Pen, Brush, etc.) when you finish with them before the method ends. You absolutely must call `Dispose` on any object that implements `IDisposable`, otherwise you'll have a memory leak. (Yes, blah blah finalizer, don't care. You need to get into the habit of disposing these objects.) The best way to do that is to wrap their creation in a `using` block. Alternatively, you can just call their `Dispose` method manually at the end of your function. – Cody Gray Feb 29 '12 at 17:09
@CodyGray The catch block is in there because there WAS an error that WASN'T showing on screen! and this is far from the finished code, I do garbage collection once I know the code works fully – Neo Mar 1 '12 at 8:20
That's not possible unless you've got some empty `catch` blocks elsewhere in your code. Time for a code review... And no, garbage collection should be integrated into the design. It's not something you can go back and add later, or at least it shouldn't be. Improper memory management can prevent code from appearing to work properly, and introducing it later can lead to subtle bugs. That's a very strange design philosophy, not one I can recommend. It's not that big of a deal to type `using` when you write the code. – Cody Gray Mar 1 '12 at 18:36
The "liquids" are two elipses with the space in between filled, so all you need to calculate is the heights and the left and right positions depending on the amount of liquid and draw from bottom to top (ie yellow then red)
``````// Upper Elipse and top Points for the filled center
y = HeightOfBottom + (FullHeight * (StartAmountFloat + AmountFloat))
x1 = Middle - (DiffenceOfDiameter * (StartAmountFloat + AmountFloat))
x2 = Middle + (DiffenceOfDiameter * (StartAmountFloat + AmountFloat))
// Lower Elipse and bottom Points for the filled center
y = HeightOfBottom + (FullHeight * StartAmountFloat)
x1 = Middle - (DiffenceOfDiameter * StartAmountFloat)
x2 = Middle + (DiffenceOfDiameter * StartAmountFloat)
``````
The Bottom should be the lower half of an elipse, too.
-
Thanks Stephan, I'm totally new to GD in C# how would I implement this? – Neo Mar 1 '12 at 8:39
Ok, don't have any code for you but I can give you an approximate workflow to achieve this. Basically you want to draw your objects back to front, so I would be drawing in this order
1. Draw the bottom of the bucket as an ellipse
2. Draw the bottom of the liquid as an ellpise of the same size, but one pixel higher
3. Now draw ellipses at each Y-pixel above until the desired % has been reached, where the end number of pixels is computed by the (bottom of the bucket) + ((top of the bucket) - (bottom of the bucket) * (percentage / 100)). You will need to widen the ellipse at certain points. This will create an aliased effect but don't worry, we are going to draw over it next
4. Finally draw the sides of the bucket and the top. If you choose a suitable line thickness you can cheekily hide the fact you hacked your way to glory :)
Last thing I'd suggest lots of trial and error. the more you do this the easier it will become! Good luck
-
Many Thanks for the heads up, its weird as you always draw top down in real life lol i'll remember to re-arrange the order. – Neo Mar 1 '12 at 8:39
I have manged to solve this one, I am posting the code here to see if anyone can still improve it before I accept this as the answer.
``````private int[] getPoints(int perc)
{
int[] pts;// = new int[4];
double x_offset_left = (35 - 21);
x_offset_left = x_offset_left / 100;
double height = 135;
double width = 178;
double x1, x2, y1, y2;
int margin_top = 66;//68
int margin_left = 21;
y1 = ((height / 100) * perc) + margin_top;
y2 = y1;
x1 = margin_left + (x_offset_left * perc);
x2 = width - (x_offset_left * perc);
pts = new int[4] { Convert.ToInt32(x1), Convert.ToInt32(y1), Convert.ToInt32(x2), Convert.ToInt32(y2) };
return pts;
}
private Bitmap drawBucket2(int yellowval, int redval, int overval)
{
Bitmap img = new Bitmap(200, 221);
using (Graphics g = Graphics.FromImage(img))
{
Brush bRed = new SolidBrush(Color.FromArgb(50, Color.DarkRed));
Brush bYellow = new SolidBrush(Color.FromArgb(75, Color.Gold));
Brush bBlue = new SolidBrush(Color.FromArgb(50, Color.Blue));
GraphicsPath gp = new GraphicsPath();
Region r;
Point[] points_yellow;
Point[] points_red;
int percentage = 0;
int[] pts;
int[] pts_full = getPoints(100);
int[] pts_min = getPoints(1);
#region "Yellow Region"
// bottom curve
percentage = yellowval;
pts = getPoints(100 - percentage);
points_yellow = new Point[3];
points_yellow[0] = new Point(pts_full[0], pts_full[3]);
points_yellow[1] = new Point(((pts_full[2] - pts_full[0]) / 2 + pts_full[0]), (pts_full[1] + 15));
points_yellow[2] = new Point(pts_full[2], pts_full[3]);
//Console.WriteLine("curve : (" + points_yellow[0].X + ", " + points_yellow[0].Y + "), " + " (" + points_yellow[1].X + ", " + points_yellow[1].Y + "), " + " (" + points_yellow[2].X + ", " + points_yellow[2].Y + ")");
//polygon
points_yellow = new Point[4];
points_yellow[0] = new Point(pts[0], pts[1]);
points_yellow[1] = new Point(pts[2], pts[1]);
points_yellow[2] = new Point(pts_full[2], pts_full[1]);
points_yellow[3] = new Point(pts_full[0], pts_full[1]);
//Console.WriteLine("Poly : (" + points_yellow[0].X + ", " + points_yellow[0].Y + "), " + " (" + points_yellow[1].X + ", " + points_yellow[1].Y + "), " + " (" + points_yellow[2].X + ", " + points_yellow[2].Y + "), " + " (" + points_yellow[3].X + ", " + points_yellow[3].Y + ")");
// top curve
points_yellow = new Point[3];
points_yellow[0] = new Point(pts[0], pts[1]);
points_yellow[1] = new Point(((pts[2] - pts[0]) / 2 + pts[0]), (pts[1] + 15));
points_yellow[2] = new Point(pts[2], pts[1]);
//Console.WriteLine("curve : (" + points_yellow[0].X + ", " + points_yellow[0].Y + "), " + " (" + points_yellow[1].X + ", " + points_yellow[1].Y + "), " + " (" + points_yellow[2].X + ", " + points_yellow[2].Y + ")");
r = new Region(gp);
g.FillRegion(bYellow, r);
#endregion
#region "Red Region"
gp = new GraphicsPath();
percentage = yellowval + redval;
// Bottom Curve
//Console.WriteLine("curve : (" + points_yellow[0].X + ", " + points_yellow[0].Y + "), " + " (" + points_yellow[1].X + ", " + points_yellow[1].Y + "), " + " (" + points_yellow[2].X + ", " + points_yellow[2].Y + ")");
// polygon
int[] pts_yel = new int[3]{pts[0], pts[1], pts[2]};
pts = getPoints(100 - percentage);
points_red = new Point[4];
points_red[0] = new Point(pts[0], pts[1]);
points_red[1] = new Point(pts[2], pts[1]);
points_red[2] = new Point(pts_yel[2], pts_yel[1]);
points_red[3] = new Point(pts_yel[0], pts_yel[1]);
// Top Curve
points_red = new Point[3];
points_red[0] = new Point(pts[0], pts[1]);
points_red[1] = new Point(((pts[2] - pts[0]) / 2 + pts[0]), (pts[1] + 12));
points_red[2] = new Point(pts[2], pts[1]);
r = new Region(gp);
g.FillRegion(bRed, r);
#endregion
#region "Overflow"
if (overval > 0)
{
gp = new GraphicsPath();
r = new Region(gp);
g.FillRegion(bBlue, r);
}
#endregion
r.Dispose();
gp.Dispose();
bRed.Dispose();
bYellow.Dispose();
bBlue.Dispose();
}
return img;
}
private void fillBucket(int Yellowperc, int Redperc, int Overperc)
{
pictureBox1.Image = null;
pictureBox1.Image = drawBucket2(Yellowperc, Redperc, Overperc);
}
``````
- | 2,694 | 9,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-30 | latest | en | 0.780919 |
https://math.stackexchange.com/questions/1671380/proof-of-cos-1-frac-cosa-cosb1-cosa-cosb-2-tan-1 | 1,563,288,751,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524568.14/warc/CC-MAIN-20190716135748-20190716161748-00509.warc.gz | 475,817,126 | 35,976 | # Proof of $\cos^{-1}[\frac{\cos(a) + \cos(b)}{1 + \cos(a)\cos(b)}] = 2\tan^{-1}[\tan(\frac{a}{2})\tan(\frac{b}{2})]$ by integration
Is it possible to prove
$$\cos^{-1}\left[\frac{\cos(a) + \cos(b)}{1 + \cos(a)\cos(b)}\right] = 2\tan^{-1}\left[\tan\left(\frac{a}{2}\right)\tan\left(\frac{b}{2}\right)\right]$$
with the help of integration?
I know how to prove it without integration but can't develop any approach to solve it using integration.
• I think you cant as integration operator would just get cancelled a d if integration of someone gives another then they arent equal as its area under the curve for the integral function – Archis Welankar Feb 25 '16 at 6:59
• Are there any restrictions on $a$ and $b$? The left side is always positive, $range(\arccos)=[0,\pi]$, but if $a·b<0$ then the right side is negative. – LutzL Feb 25 '16 at 11:25
If $b=0$ then the equation reduces to $$\arccos(1)=2\arctan(0)$$ which is true. Then compute the derivatives of both sides for fixed $a$ and variable $b$ and hopefully both will be the same.
In other news, $$\arccos x=2\arctan y \iff y=\tan(\tfrac12\arccos x)=\frac{\sin\arccos x}{1+\cos\arccos x}=\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}}$$ or $$x=\frac{1-y^2}{1+y^2}$$ | 431 | 1,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-30 | latest | en | 0.755258 |
http://www.visionlearning.com/library/module_viewer.php?c3=&mid=118&ut=&l=e | 1,371,597,981,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707436332/warc/CC-MAIN-20130516123036-00046-ip-10-60-113-184.ec2.internal.warc.gz | 746,256,015 | 17,394 | Library > Physics > Gravity
# Gravity
## Newtonian Relationships
by Nathaniel Page Stites, M.A./M.S.
What causes objects to fall toward the earth? Why do the planets orbit the sun? What holds galaxies together? If you traveled to another planet, why would your weight change? All of these questions relate to one aspect of physics: gravity. For all of its influence on our daily lives, for all of its control over the cosmos, and for all of our ability to describe and model its effects, we do not understand the actual mechanisms of gravitational force. Of the four fundamental forces identified by physicists – strong nuclear, electroweak, electrostatic, and gravitational – the gravitational force is the least understood. Physicists today strive toward a "Grand Unified Theory," wherein all four of these forces are united into one physical model that describes the behavior of everything in the universe. At this point in time, the gravitational force is the troublesome one, the force that resists unification.
In spite of the mystery behind the mechanisms of gravity, physicists have been able to describe the behavior of objects under the influence of gravity quite thoroughly. Isaac Newton, a seventeenth into eighteenth century English scientist and mathematician (among other things), was the first person to propose a mathematical model to describe the gravitational attraction between objects. Albert Einstein built upon this model in the twentieth century and devised a more complete description of gravity in his theory of general relativity. In this module, we will explore Newton’s description of gravity and some of the experimental confirmations of his theory that came many years after he proposed his original idea.
### The Apple
Whether or not Isaac Newton actually sat under an apple tree while pondering the nature of gravity, the fact that objects fall toward the surface of the earth was well understood long before Newton’s time. Everyone has experience with gravity and its effects near the surface of the earth, and our intuitive view of the world includes an understanding that what goes up must come down. Galileo Galilei (1564 – 1642) demonstrated that all objects fall to the surface of the earth with the same acceleration, and that this acceleration was independent of the mass of the falling object. Isaac Newton was no doubt familiar with this concept, and he would eventually formulate a broad and far-reaching theory of gravitation. Newton’s theory would encompass not only the behavior of an apple near the surface of the earth, but also the motions of much larger bodies quite far away from the earth.
Leaning Tower of Pisa Experiment
Concept simulation - Reenacts Galileo's experiment of two different objects falling at the same rate.
(Flash required)
### The Planets
Early conceptions of the universe were "geocentric" – they placed the earth at the center of the universe and had the planets and stars move around the earth. The Ptolemaic Model of the universe dominated scientific thought for many centuries, until the work of such careful astronomers as Tycho Brahe, Nicolaus Copernicus, Galileo Galilei, and Johannes Kepler supplanted this view of the cosmos. The “Copernican Revolution” placed the sun at the center of the solar system and the planets, including earth, in orbit around the sun. This major shift in perception laid the foundation for Isaac Newton to begin thinking about gravitation as it related to the motions of the planets.
Figure 1: The Solar System
### An early unification theory
Just as physicists today are searching for ways to unify the fundamental forces, Isaac Newton also sought to unify two seemingly disparate phenomena: the motion of objects falling toward the earth and the motion of the planets orbiting the sun. Isaac Newton’s breakthrough was not that apples fall to the earth because of gravity; it was that the planets are constantly falling toward the sun for exactly the same reason: gravity! Newton built upon the work of early astronomers, in particular Johannes Kepler, who in 1596 and 1619 published his laws of planetary motion. One of Kepler's central observations was that the planets move in elliptical orbits around the sun. Newton expanded Kepler’s description of planetary motion into a theory of gravitation.
### Newton’s Law of Universal Gravitation
The essential feature of Newton’s Law of Universal Gravitation is that the force of gravity between two objects is inversely proportional to the square of the distance between them. Such a connection is known as an “inverse square” relationship. Newton derived this relationship from Kepler’s assertion that the planets follow elliptical orbits. To understand this, consider the light radiating from the surface of the sun. The light has some intensity at the surface of the sun. As the light travels away from the sun, its intensity diminishes. The intensity of the light at any distance away from the sun equals the strength of the source divided by the surface area of a sphere surrounding the sun at that radius.
As the distance away from the sun (r) doubles, the area of the sphere surrounding the sun quadruples. Thus, the intensity of the sun’s light depends inversely on the square of the distance away from the sun. Newton envisioned the gravitational force as radiating equally in all directions from a central body, just as sunlight in the previous example. Newton recognized that his gravitational model must take the form of an inverse square relationship. Such a model predicts that the orbits of objects around a central body will be conic sections, and years of astronomical observations have borne this out. Although this idea is most commonly attributed to Isaac Newton, the English mathematician Robert Hooke claimed that he originated the idea of the inverse square relationship. Nonetheless, Newton eventually published his theory of gravitation and became famous as a result.
The relationship that Newton came up with looks like this:
Figure 2: Newton's Law of Universal Gravitation
where F is the force of gravity (in units now referred to as newtons), m1 and m2 are the masses of the two objects (in kilograms), r is the distance separating the centers of mass of the objects and G is the "gravitational constant." This relationship has come to be known as Newton's Law of Universal Gravitation. It is "universal" because all objects in the universe are attracted to all other objects in the universe according to this relationship. Two people sitting across a room from each other are actually attracted gravitationally. As we know from everyday experience, human-sized objects don't crash into each other as a result of this force, but it does exist even if it is very small. Although Newton correctly identified this relationship between force, mass, and distance, he was able only to estimate the value of the gravitational constant between these quantities. The world would have to wait more than a century for an experimental measurement of the constant of proportionality: G.
### Measuring the Mass of the Earth: The Cavendish Experiment
In 1797 and 1798 Henry Cavendish set out to confirm Newton's theory and to determine the constant of proportionality in Newton's Law of Universal Gravitation. His ingenious experiment, based on the work of John Michell, was successful on both fronts. To accomplish this, Cavendish created a "torsion balance," which consisted of two masses at either end of a bar that was suspended from the ceiling by a thin wire (see Figure 3).
Figure 3: The Torsion Balance, devised by Michell and Cavendish to determine the constant of proportionality in Newton's Law of Universal Gravitation.
Attached to the wire was a mirror, off of which a beam of light was reflected. Cavendish brought a third mass close to one of the masses on the torsion balance. As the third mass attracted one of the ends of the torsion balance, the entire apparatus, including the mirror, rotated slightly and the beam of light was deflected. Through careful measurement of the angular deflection of the beam of light, Cavendish was able to determine the extent to which the known mass was attracted to the introduced mass. Not only did Cavendish confirm Newton's theory, but also he determined the value of the gravitational constant to an accuracy of about 1 percent.
Figure 4: Gravitational Constant
Cavendish cleverly referred to his research as “Measuring the Mass of Earth.” Since he had determined the value of G, he could do some simple calculations to determine the mass of the earth. By Newton’s Second Law, the force between an object and the earth equals the product of the acceleration (g) and the mass of the object (m):
F = ma
Galileo had determined the acceleration of all objects near the surface of the earth in the early 1600s as g = 9.8 m/s2.
Therefore, setting this equation equal to Newton’s Law of Universal Gravitation described above, Cavendish found:
Figure 5: Newton's Law of Universal Gravitation as used by Cavendish to determine the mass of the Earth.
where m is the mass of the object, mE is the mass of the earth, and rE is the radius of the earth. Solving for the mass of the earth yields the following result:
Figure 6: Results of Cavendish's calculation for the mass of the Earth.
Cavendish had determined the mass of the earth with great accuracy.
We can also use this relationship to calculate the force of attraction between two people across a room. To do this, we simply need to use Newton’s Law of Universal Gravitation with Cavendish’s gravitational constant. Assume the two people have masses of 75 and 100 kilograms, respectively, and that they are 5 meters apart. The force of gravitation between them is:
Figure 7: Gravitational attraction between two people.
Although it is small, there is still a force!
### Conclusion
Newton’s Law of Universal Gravitation grew in importance as scientists realized its utility in predicting the orbits of the planets and other bodies in space. In 1705, Sir Edmund Halley, after studying comets in great detail, predicted correctly that the famous comet of 1682 would return 76 years later, in December of 1758. Halley had used Newton’s Law to predict the behavior of the comet orbiting the sun. With the advent of Cavendish’s accurate value for the gravitational constant, scientists were able to use Newton’s law for even more purposes. In 1845, John Couch Adams and Urbain Le Verrier predicted the existence of a new, yet unseen, planet based on small discrepancies between predictions for and observations of the position of Uranus. In 1846, the German astronomer Johann Galle confirmed their predictions and officially discovered the new planet, Neptune.
While Newton’s Law of Universal Gravitation remains very useful today, Albert Einstein demonstrated in 1915 that the law was only approximately correct, and that it fails to work when gravitation becomes extremely strong. Nonetheless, Newton’s gravitational constant plays an important role in Einstein’s alternative to Newton’s Law, the Theory of General Relativity. The value of G has been the subject of great debate even in recent years, and scientists are still struggling to determine a very accurate value for this most elusive of fundamental physical constants.
Other Recommended Products
Nathaniel Page Stites, M.A./M.S. "Gravity: Newtonian Relationships," Visionlearning Vol. PHYS-1 (2), 2004.
http://www.visionlearning.com/library/module_viewer.php?mid=118
Support for Visionlearning has been provided by: Copyright © 2004 - 2013, Visionlearning, Inc. | 2,377 | 11,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2013-20 | latest | en | 0.940851 |
https://www.eduzip.com/ask/question/find-the-supplementary-angle-of-the-following-anglesfrac37-of-280-521302 | 1,643,285,692,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00270.warc.gz | 772,400,434 | 9,833 | Mathematics
# Find the supplementary angle of the following angles:$\frac{3}{7}$ of $280^{0}$
##### SOLUTION
In the given figure,
Given angle is 10501050
x+1050=1800x+1050=1800
x=18001050=750x=1800−1050=750
But x+y=1800x+y=1800 (Sum of co-interior angles)
750+y=1800⇒750+y=1800 (Sum of co-interior angles)
y=1800750=1050⇒y=1800−750=1050
Similarly, y+z=1800y+z=1800 (Sum of co-interior angles)
1050+z=1800⇒1050+z=1800
z=18001050=750⇒z=1800−1050=750
Hence, x=750,y=1050,z=750
You're just one step away
Subjective Medium Published on 09th 09, 2020
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• B. $30^0$
• C. $35^0$
• D. $20^0$
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 23rd 09, 2020
Q5 Subjective Medium
Read the following two statements which are taken as axioms:
(i) If two lines intersect each other, then the vertically opposite angles are not equal.
(ii) If a ray stands on a line, then the sum of two adjacent angles so formed is equal to $180^0$. | 740 | 2,121 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-05 | latest | en | 0.709888 |
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5th October 2015, 15:55 #41 | Link
feisty2
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Quote:
*That precision could be calculated, and this is the one of the first things you do when you start doing physics lab in school (when you're starting to apply math to the real world). That stuff is actually hard, so it's not surprising that no one bothers with it. Just throw more precision, and hope that it helps.
AND,
errors produced by computers due to finite precision are OBVIOUSLY "Systematic Errors" (these errors are constant, you repeat the calculation like a billion times, and the error stays the same, not something fluctuating)
and the page you linked to showed us some basic stuff of "Estimating Random Errors" (fluctuating as you repeat the experiment)
5th October 2015, 16:13 #42 | Link
Khanattila
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Quote:
Originally Posted by bxyhxyh Don't computers already work this way? EDIT: One thing I can't understand is that what's the point of debating on two things that does exactly same thing on the screen?
Floating point numbers you said? Not even close.
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5th October 2015, 16:22 #43 | Link
Khanattila
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Quote:
Originally Posted by vivan Input data also has limited precision. If you perform calculations with precision higher* than the precision of your data - then you're not losing any data. There's no point in calculating 10 + 20 with 256-bit precision just because you can if you're not even sure if first number is actually 10 or 7 or 13.
Agree, but the input can have infinite precision like natural logarithm. Euler's number is not a ratio of integers and it is not a root of any non-zero polynomial with rational coefficients. Like log (param).
EDIT. Forget it... you always think that the input is only 'param'.
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Last edited by Khanattila; 5th October 2015 at 16:28.
5th October 2015, 16:28 #44 | Link
feisty2
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Quote:
Originally Posted by bxyhxyh Don't computers already work this way? EDIT: One thing I can't understand is that what's the point of debating on two things that does exactly same thing on the screen?
the topic has turned academic for a while
5th October 2015, 16:47 #45 | Link feisty2 I'm Siri Join Date: Oct 2012 Location: void Posts: 2,633 @vivan you and I have different perceptions of "lossless", I can live with that but please don't drag physics stuff here, computers do mathematical calculations, not physical experiments
5th October 2015, 19:27 #46 | Link
Asmodian
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Quote:
Originally Posted by Khanattila Agree, but the input can have infinite precision like natural logarithm. Euler's number is not a ratio of integers and it is not a root of any non-zero polynomial with rational coefficients. Like log (param). EDIT. Forget it... you always think that the input is only 'param'.
No, in computers it cannot. Computers don't do infinite precision. The input will always have finite precision.
Quote:
Originally Posted by feisty2 @vivan you and I have different perceptions of "lossless", I can live with that but please don't drag physics stuff here, computers do mathematical calculations, not physical experiments
Vivan's point was a mathematical one, not a physical one.
If you can only output integers there is a point where keeping track of extra precision cannot change the result at all. This point depends on the equation(s), of course, but it can be calculated with certainty. In physics/chemistry it is based on the precision of your measurements while in computers it is based on the precision of your input and output.
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5th October 2015, 19:45 #47 | Link feisty2 I'm Siri Join Date: Oct 2012 Location: void Posts: 2,633 I'd just implement the real infinite precision if I had time to be wasted on "precision calculation" Like, ln (Pi) stays ln (Pi), not 1.144729885849400 And that's the infinite precision And that's the real deal
5th October 2015, 21:27 #48 | Link
bxyhxyh
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Maybe my previous post which included word "physics" may confused people.
It is my bad for using wrong words and meaning. But that is still valid for mathematics.
Especially for this case if input and output are the same, it means that it IS mathematically lossless.
Quote:
Originally Posted by feisty2 I'd just implement the real infinite precision if I had time to be wasted on "precision calculation" Like, ln (Pi) stays ln (Pi), not 1.144729885849400 And that's the infinite precision And that's the real deal
I think that's not possible for programming level numbers. Well for Pi example, 4*atan(1) can do that. But it will be still limited by bit precision (memory). It can't be irrational number with infinite digit.
Quote:
Originally Posted by Khanattila Floating point numbers you said? Not even close.
Ah, my memory was mixed then lol. I have some vague memory of that fractal numbers are precised that way. Then it wasn't floating point numbers.
EDIT: TBH, I personally think this thread needs to be locked. Because we're arguing like "this method is more accurate because it calculates correctly" while both results the same. It is just meaningless.
Last edited by bxyhxyh; 5th October 2015 at 22:42.
6th October 2015, 01:52 #49 | Link feisty2 I'm Siri Join Date: Oct 2012 Location: void Posts: 2,633 @bxyhxyh It's possible Don't convert Pi to 3.1415926.... just use "Pi" to represent the number Like what you did in math
6th October 2015, 06:36 #50 | Link
bxyhxyh
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Quote:
Originally Posted by feisty2 @bxyhxyh It's possible Don't convert Pi to 3.1415926.... just use "Pi" to represent the number Like what you did in math
Well even if you save some math numbers as representations in CPU, memory or whatever place, It will end up being rational numbers, when you use it in REAL calculation.
You can use any kind of trick to make it accurate as possible. But in the end the calculation still will be limited by bits.
Maybe for some cases even precise calculation might end up being more accurate. Who knows.
6th October 2015, 06:46 #51 | Link feisty2 I'm Siri Join Date: Oct 2012 Location: void Posts: 2,633 and define "REAL calculation" the area of a 1cm radius circle is Pi cm^2, "Pi cm^2" is the output, the result of the calculation that's real calculation I don't get it, who on earth said you have to convert everything to decimals to be the "REAL calculation"
6th October 2015, 07:14 #52 | Link feisty2 I'm Siri Join Date: Oct 2012 Location: void Posts: 2,633 I think you don't really understand what Pi and 3.14159265.... stand for both of them are just different representations of the same objective existence of some certain information, like, simply the distance of some constant dot (dot "Pi") and dot "0" on the axis it's just this piece of information will need infinite characters if you wanna represent it within the decimal system, BUT, it can be represented with finite characters by other systems, like "Pi" like, 0.3333... (1/3) is an infinite decimal under the decimal system, but, it's just "0.1", 2 characters under the trinary system if you're still confused, think about 15(dec) and 0xF EDIT: corrected a calculation mistake Last edited by feisty2; 6th October 2015 at 07:29.
6th October 2015, 07:16 #53 | Link
bxyhxyh
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Quote:
Originally Posted by feisty2 and define "REAL calculation" the area of a 1cm radius circle is Pi cm^2, "Pi cm^2" is the output, the result of the calculation that's real calculation I don't get it, who on earth said you have to convert everything to decimals to be the "REAL calculation"
I don't wanna argue about what is the real calculation or not.
In this case Pi is more like operation. We already have represented operations. For example divide is same as Pi
Quote:
Originally Posted by feisty2 I think you don't really understand what Pi and 3.14159265.... stand for both of them are just different representations of the same objective existence of some certain information, like, simply the distance of some constant dot (dot "Pi") and dot "0" on the axis it's just this piece of information will need infinite characters if you wanna represent it within the decimal system, BUT, it can be represented with finite characters by other systems, like "Pi" like, 0.3333... (1/3) is an infinite decimal under the decimal system, but, it's just "1", a single character under the trinary system if you're still confused, think about 15(dec) and 0xF
Computer only work in binary.
Last edited by bxyhxyh; 6th October 2015 at 07:28.
6th October 2015, 07:34 #54 | Link
feisty2
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Quote:
Originally Posted by bxyhxyh Computer only work in binary.
and yet you can do decimal calculations on computers
6th October 2015, 07:49 #55 | Link feisty2 I'm Siri Join Date: Oct 2012 Location: void Posts: 2,633 computers are just like those little kids, they do whatever you tell them to like, a 6 yrs old kid is calculating the area of a 1cm radius circle, and he's gonna ask like, "I will need Pi, what am I going to do about it?" you can either tell him, Pi is approximately 3.14, and you just make it 3.14, and he will tell you the area is 3.14 cm^2, which is imprecise, cuz you told him to do the wrong stuff or you tell him, Pi is a constant number just like 1,2,3,4.... you can just keep it Pi, and he will tell you the area is Pi cm^2, which is precise and correct
6th October 2015, 09:22 #56 | Link
bxyhxyh
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Quote:
Originally Posted by feisty2 and yet you can do decimal calculations on computers
Quote:
Originally Posted by feisty2 computers are just like those little kids, they do whatever you tell them to like, a 6 yrs old kid is calculating the area of a 1cm radius circle, and he's gonna ask like, "I will need Pi, what am I going to do about it?" you can either tell him, Pi is approximately 3.14, and you just make it 3.14, and he will tell you the area is 3.14 cm^2, which is imprecise, cuz you told him to do the wrong stuff or you tell him, Pi is a constant number just like 1,2,3,4.... you can just keep it Pi, and he will tell you the area is Pi cm^2, which is precise and correct
I think you're asking user level calculation. There are high-end programs that can do such stuffs. Heck, even calculators can do that.
In machine level calculations it won't work that way.
Lets say for example in 8 bit unsigned float numbers
4 is 0100.0000, 3 is 0011.0000
In mathematics decimal Pi is 3.14159265359...
Then how much is Pi? It will be something like 0011.0010.
How much is it in decimal? 3.125.
Lets imagine you have converted Pi into something like 1. Then how much is decimal 4?
It will be RATIONAL number bit higher than 1, since memory isn't limitless.
Exactly same as computer we work today.
Memory has limit. Real calculation is limited by it.
Real calculation means calculation going in machine language.
It doesn't know and can't learn any other languages.
WE translate it into language that human can understand.
Last edited by bxyhxyh; 6th October 2015 at 09:30.
6th October 2015, 09:36 #57 | Link
feisty2
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and you just mixed all the concepts up
it's 100% pointless to care about how the calculation is working at the machine code level
like, will you ever care what's going on with the neurons inside someone's brain when he's doing some math?
Quote:
I think you're asking user level calculation. There are high-end programs that can do such stuffs. Heck, even calculators can do that.
that's my point, it should stop being just high end programs, it should be universal thru all programs
Quote:
Then how much is Pi?
obviously "Pi" is a string (some special symbol) at the machine code level, so it's still precise and finite, and how much is a string is pointless, that's something the higher level program will take care of
6th October 2015, 09:47 #58 | Link
bxyhxyh
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Originally Posted by feisty2 it should stop being just high end programs, it should be universal thru all programs
It just can't.
Machine has only 2 symbol 1 and 0. Then read the previous post's later part please.
Last edited by bxyhxyh; 6th October 2015 at 09:55.
6th October 2015, 09:53 #59 | Link
feisty2
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Quote:
Originally Posted by bxyhxyh It just can't. Machine has only 2 symbol 1 and 0. Then read the previous post's later part please.
special symbol = a certain string
like if you gonna set "Pi" as the special symbol for that number
the symbol will be the binary of the ANSI text of "Pi" at the machine code level
I think you have problems understanding English apparently
6th October 2015, 10:54 #60 | Link
bxyhxyh
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Quote:
Originally Posted by feisty2 I think you have problems understanding English apparently
Yes.
And you're still asking user level program. Such things already exist. If they don't, someone can create them.
I think there are plenty of programs that does math on physics, chemicals or any science.
But those high-end programs' results don't saved as exact number in memory. It is just displayed that way on screen.
Also they're still not limitless, still not perfectly accurate, still uses precision even in high level calculations.
I think we separated too much from main topic here
Last edited by bxyhxyh; 6th October 2015 at 11:30.
Tags 444, avisynth, color subsampling, x264, yv24 | 3,754 | 14,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-50 | latest | en | 0.921976 |
https://www.physicsforums.com/threads/inequality-proof.82904/ | 1,571,656,785,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00228.warc.gz | 1,063,911,892 | 17,439 | # Inequality proof
#### Townsend
Show that for each complex sequence $$c_1, c_2, ..., c_n$$ and for each integer $$1 \leq H < N$$ one has the inequality
$$| \sum_{n=1}^N c_n|^2 \leq \frac{4N}{H+1} ( \sum_{n=1}^N |c_n|^2 + \sum_{h=1}^H | \rho_N(h)|)$$
Any one.....matt grime perhaps?
note: if anyone actually wants to work this out let me know and I will fill in the missing parts...but don't ask me to do it.... :tongue2:
Last edited:
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#### honestrosewater
Gold Member
Yeah, that's good. The more fancy meaningless symbols, the better.
#### Townsend
honestrosewater said:
Yeah, that's good. The more fancy meaningless symbols, the better.
There is an important part missing but it is a true bound...not just meaningless...
#### honestrosewater
Gold Member
Okay, I'll take your word for it. It would be nice if someone were around to explain it to me.
#### Townsend
honestrosewater said:
Okay, I'll take your word for it. It would be nice if someone were around to explain it to me.
Yeah...it would take a really smart....creative....mathematician to do so...who would be able to do that I wonder??????
#### honestrosewater
Gold Member
Townsend said:
Yeah...it would take a really smart....creative....
... and patient. I've never even worked with complex numbers before. You can just treat them as ordered pairs of real numbers, right? I think I'd like that approach.
#### Johnny5
I know the solution, but I won't say to give other people a chance. It's not that hard.
#### wolram
Gold Member
Its packman eating flies, so the ansewer must be MxBxHs
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• Solo and co-op problem solving | 502 | 1,934 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-43 | longest | en | 0.940902 |
https://www.intechopen.com/books/matrix-theory-applications-and-theorems/matrices-which-are-discrete-versions-of-linear-operations | 1,571,284,650,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00287.warc.gz | 921,780,617 | 122,184 | Open access peer-reviewed chapter
# Matrices Which are Discrete Versions of Linear Operations
By Armando Martínez Pérez and Gabino Torres Vega
Submitted: November 22nd 2017Reviewed: January 24th 2018Published: August 29th 2018
DOI: 10.5772/intechopen.74356
## Abstract
We introduce and study a matrix which has the exponential function as one of its eigenvectors. We realize that this matrix represents a set of finite differences derivation of vectors on a partition. This matrix leads to new expressions for finite differences derivatives which are exact for the exponential function. We find some properties of this matrix, the induced derivatives and of its inverse. We provide an expression for the derivative of a product, of a ratio, of the inverse of vectors, and we also find the equivalent of the summation by parts theorem of continuous functions. This matrix could be of interest to discrete quantum mechanics theory.
### Keywords
• exact finite differences derivative
• exact derivatives on partitions
• exponential function on a partition
• discrete quantum mechanics
## 1. Introduction
We are interested on matrices which are a local, as well as a global, exact discrete representation of operations on functions of continuous variable, so that there is congruency between the continuous and the discrete operations and properties of functions. Usual finite difference methods [1, 2, 3, 4] become exact only in the limit of zero separation between the points of the mesh. Here, we are interested in having exact representations of operations and functions for finite separation between mesh points.
The difference between our method and the usual finite differences method is the quantity that appears in the denominator of the definition of derivative. The appropriate choice of that denominator makes possible that the finite differences expressions for the derivative gives the exact results for the exponential function. We concentrate on the derivative operation, and we define a matrix which represents the exact finite difference derivation on a local and a global scale. The inverse of this matrix is just the integration operation. These are interesting subjects by itself, but they are also of interest in the quantum physics realm [5, 6, 7].
In this chapter, we will consider only the case of the derivative and the integration of the exponential function.
## 2. A matrix with the exponential function as an eigenvector
Here, we consider the N×Nantisymmetric, tridiagonal matrix
DNevΔ2χvΔ12χvΔ000012χvΔ012χvΔ000012χvΔ0000000012χvΔ000012χvΔ012χvΔ000012χvΔevΔ2χvΔ,E1
where v—it can be pure real or pure imaginary—, Δ+, and χvΔsinhvΔ/vΔ+v2Δ3/6+OΔ5. This function χvΔis well defined for v=0, with value χ0Δ=Δ. This matrix is interesting because, as we will see below, it represents a derivation on a partition. A rescaled matrix D¯Nis defined as
D¯N1/z1000010100001000000001000010100001z,E2
where z=evΔ, and
DND¯N2χvΔ.E3
We are mainly interested in finding the eigenvalues and the corresponding eigenvectors of these matrices.
We start our study with a result about the determinant of D¯NλIN,
D¯NλIN=D¯N+αIN=α1/z1000001α1000001α1000001α000000α1000001α1000001α1000001α+z=α1zAN1α+AN2α,E4
where λ=α,
Ajαα100001α100001α00000α100001α100001α100001α+z=α+zBj1α+Bj2α,E5
and
Bjα=α10001α10001α0000α10001α10001α.E6
Strikingly, we recognize the determinant Bjαas the Fibonacci polynomial of index j+1[10, 11], i.e., Bjα=Fj+1α. Fibonacci polynomials are defined as
F0x=0,F1x=1,Fjx=xFj1x+Fj2x,j2.E7
Since we have that Bjα=Fj+1α, and the recursion relationship for Fibonacci polynomials, we also have that
Ajα=α+zFjα+Fj1α=zFjα+Fj+1α,E8
and then
D¯N+αIN=α1zzFN1α+FNα+zFN2α+FN1α=zαFN1α+FN2α+α1zFNα=α+z1zFNα.E9
Then, the eigenvalues of the derivative matrix D¯Nare λ1=z1/z=evΔevΔ=2sinhvΔand λm=αm, where αmis the m-th root of the N-th Fibonacci polynomial, which is a polynomial of degree N1[10, 11].
The system of simultaneous equations for the eigenvector emT=em,1em,2eNcorresponding to λm, can be put in a form similar to the recursion relationship for the Fibonacci polynomials, i.e.,
em,2=λmem,1+em,1z,E10
em,j+1=λmem,j+em,j1,1<j<N,E11
zem,N=λmem,N+em,N1.E12
This set of recursion relationships can be written as the matrix equation
em,jem,j+1=011λmem,j1em,j,j=1,,N,E13
where em,0=em,1/zand em,N+1=zem,N. Thus
em,jem,j+1=011λmjem,0em,1,j=1,,N,E14
but
011λmj=Fj1λmFjλmFjλmFj+1λm,E15
and then,
em,jem,j+1=Fj1λmFjλmFjλmFj+1λmem,0em,1,j=1,,N.E16
i.e., the j-th component of the m-th eigenvector is
em,j=Fjλm+Fj1λmzem,1forj=1,2,,N.E17
For the case of the eigenvalue λ1=z1/z, we can rewrite Eq. (17) by noticing that if we let x=ww1(w), then Fnx+Fn1x/w=wn1for n=1,2,. This can be proved by induction method as follows. For n=1, it is immediately verified. First, suppose that the equality holds for nk. Next, we compute the right-hand side of the equality for k+1. Substituting Fk1=wwk1Fkin the expression for k+1, and using the properties of the Fibonacci polynomials, we obtain
Fk+1x+Fkxw=xFkx+Fk1x+Fkxw=xFkx+wkwFkx+Fkxw=wk.E18
Therefore, according to Eqs. (17) and (18), the eigenvector for the eigenvalue λ1=2sinhvΔtakes the form e1=c1zzN1T, where cis a normalization constant. We can take advantage of the normalization constant and write
e1=cevq1evq2evqNT,E19
with eigenvalue λ1=v(in original scaling, i.e., the eigenvalue of the matrix DN), q1is an arbitrary constant, and qj=q1+j1Δ. This means that the exponential function is an eigenvector of the derivative matrix which is a global representation of the derivative on the partition q1q2qN. Recall that the exponential function is an eigenfunction of the derivative of functions of continuous variable.
The remain of the eigenvectors have eigenvalues equal to the negative of the roots of the N-th Fibonacci polynomial λm=xm, m=1,2,,N1, and have the form
em=c1F2λm+evΔF3λm+evΔF2λmFN1λm+evΔFN2λmevΔFN1λmE20
The vector that we will be interested on is the one which is the exponential function (19) with eigenvalue v.
## 3. The matrix DNrepresents a derivation
Let us consider a partition, PNqi1N, qi, of Nequally spaced points qiof the interval ab, a<b, with the same separation Δ=ba/N1between them.
The rows of the result of the multiplication of the derivative matrix DNand a vector gg1g2gnTare
DNgj=gj+1gj12χvΔ,j=1,2,,N,E21
where g0evΔg1and gN+1evΔgN. We recognize these expressions as the second order derivatives of the function gxat the mesh points, but instead of dividing by twice the separation Δbetween the mesh points, there is the function χvΔin the denominator. This function makes it possible that the exponential function be an eigenvector of the matrix DN.
The values g0=evΔg1and gN+1=evΔgNextend the original interval abto aΔb+Δso that we have well defined the second order derivatives at all the points of the initial partition, including the edges of the interval. When gxis the exponential function, we have g0=evx1Δand gN+1=evxN+Δ, i.e., they are the values of the exponential function evaluated at the points of the extension.
Thus, we define finite differences derivatives for any function gxdefined on the partition as
Dg1=g2evΔg12χvΔ,E22
Dgj=gj+1gj12χvΔ,E23
DgN=evΔgNgN12χvΔ,E24
to be used on the first, central, and last points of the partition.
The determinant of the derivative matrix is not always zero, and in fact, it is [see Eqs. (4) and (9)]
D¯N=2sinhvΔFN0.E25
But, since F2j+1=1, and F2j=0, then
D¯2j=0,D¯2j+1=2sinhvΔ.E26
Hence, only the matrices with an odd dimension have an inverse.
Next, we will derive some properties of these finite differences derivatives.
### 3.1. The derivative of a product of vectors
There are two equivalent expressions for the finite differences derivative of a product of vectors defined on the partition. A set of such expressions is
Dgh1=g2h2evΔg1h12χvΔ=g2h2evΔg1h22χvΔ+g1evΔh2h2+h2evΔh12χvΔ=h2Dg1+g1Dh1+g1h2evΔ12χvΔ=h2Dg1+g1Dh1+g1h2v2+v24Δ+OΔ3,E27
Dghj=hj+1Dgj+gj1Dhj,E28
DghN=hNDgN+gN1DhN+1evΔ2χvΔgN1hNhNDgN+gN1DhN+gN1hNv2v24Δ+OΔ3.E29
A second set of equalities is
Dgh1=g2Dh1+h1Dg1+g2h1evΔ12χvΔ=g2Dh1+h1Dg1+g2h1v2+v24Δ+OΔ3,E30
Dghj=gj+1Dhj+hj1Dgj,E31
DghN=gNDhN+hN1DgN+gNhN11evΔ2χvΔgNDhN+hN1DgN+gNhN1v2v24Δ+OΔ3,E32
### 3.2. Summation by parts
The sum of Eqs. (28) or (31), with weights 2χvΔ, results in
j=nm2χvΔhj+1Dgj+j=nm2χvΔgj1Dhj=j=nm2χvΔDghj=gm+1hm+1+gmhmgnhngn1hn1,E33
or
j=nm2χvΔgj+1Dhj+j=nm2χvΔhj1Dgj=gm+1hm+1+gmhmgnhngn1hn1.E34
This is the discrete version of the integration by parts theorem for continuous variable functions, a very useful result.
### 3.3. Second derivatives
Expressions for higher order derivatives are obtained through the powers of DN. For instance, for the first two points, the second derivative is
D2g1=e2vΔ1g1evΔg2+g34χ2vΔ=Dg2evΔDg12χvΔ,E35
D2g2=evΔg12g2+g44χ2vΔ=Dg3Dg12χvΔ,E36
For inner points we get
D2gj=gj22gj+gj+24χ2vΔ=Dgj+1Dgj12χvΔ,3jN3,E37
and for the last two points of the mesh, we find
D2gN1=gN32gN1+evΔgN4χ2vΔ=DgNDgN22χvΔ,E38
D2gN=gN2evΔgN1+e2vΔ1gN4χ2vΔ=evΔDgNDgN1χ2v2Δ.E39
These derivatives also have the exponential function as one of their eigenvectors, and we can generate expressions for higher derivatives with higher powers of the derivative matrix.
### 3.4. The derivative of the inverse of functions
It is possible to give an expression for the derivative of h1q, including the edge points. For the first point, we have
D1h1=12χvΔ1h2evΔh1=12χvΔh2h1h1h2+1evΔh1=Dh1h1h2+1evΔ2χvΔ1h1+1h2.E40
For central and last points, we find that
D1hj=Dhjhj1hj+1,E41
D1hN=DhNhN1hN+evΔ12χvΔ1hN1+1hN.E42
The derivatives for the first and last points coincide with the derivative for central points when Δ=0.
### 3.5. The derivative of the ratio of functions
Now, we take advantage of the derivative for the inverse of a function and the derivative of a product of functions and obtain what the derivative of a ratio of functions is
Dgh1=1h2Dg1+g1D1h1+g1h2evΔ12χvΔ=1h2Dg1+g1Dh1h1h2+12χvΔ1h1+1evΔh2+g1h2evΔ12χvΔ=1h2Dg1g1h1h2Dh1+g1h11evΔ2χvΔ,E43
Dghj=Dgjhj1gj+1Dhjhj+1hj1,E44
DghN=1hNDgNgN1hN1hNDhN+gN1hN1evΔ12χvΔ,E45
expressions which are very similar to the continuous variable results. Again, these expressions coincide in the limit Δ0, and they reduce to the corresponding expressions for continuous variables.
### 3.6. The local inverse operation of the derivative
The inverse operation to the finite differences derivative, at a given point, is the summation with weights 2χvΔ
j=nm2χvΔDgj=j=nmgj+1gj1=gm+1+gmgngn1.E46
This equality is the equivalent to the usual result for continuous functions, axdydgy/dy=gxga. Note that the inverse at the local level is a bit different from the expressions obtained by means of the inverse matrix S(see below) of the derivative matrix D. When dealing with matrices there are no boundary terms to worry about.
### 3.7. An eigenfunction of the summation operation
Because the exponential function is an eigenfunction of the finite differences derivative and according to Eq. (46), we can say that
j=nm2χvΔvevqj=j=nm2χvΔDevqj=j=nmevqj+1evqj1=evqm+1+evqmevqnevqn1,E47
in agreement with the corresponding continuous variable equality axdxvevx=evxeva. However, here, we have to deal with two values at each boundary.
### 3.8. The chain rule
The chain rule also has a finite differences version. That version is
Dghqj=ghqj+1ghqj12χvΔ=ghqj+1ghqj12χvhqj+1hqj2χvhqj+1hqj2χvΔ=Dghjχvhqj+1hqjχvΔE48
where
Dghjghqj+1ghqj12χvhqj+1hqjE49
is a finite differences derivative of ghwith respect to h, and the second factor approaches the derivative of hqwith respect to q
χvhqj+1hqjχvΔhqj+1hqj+OΔh2Δ+OΔ2.E50
Thus, we will recover the usual chain rule for continuous variable functions in the limit Δ0.
## 4. The commutator between coordinate and derivative
Let us determine the commutator, from a local point of view first, between the coordinate—the points of the partition PN—and the finite differences derivative. We begin with the derivative of q,
Dqj=qj+1qj12χvΔ=ΔχvΔ1v26Δ2.E51
Hence, the finite differences derivative of the product qgqis
Dqgj=qj+1Dgj+gj1Dqj=qj+1Dgj+gj1ΔχvΔ,E52
i.e.,
Dcqgjqj+1Dcgj=gj1ΔχvΔ.E53
This is the finite differences version of the commutator between the coordinate qand the finite differences derivative D. This equality will become the identity operator in the small Δlimit, as expected. An equivalent expression is
Dqgjqj1Dgj=gj+1ΔχvΔ.E54
This is the finite differences version of the commutator between coordinate and derivative; the right hand side of this equality becomes gjin the small Δlimit, i.e., it becomes the identity operator.
### 4.1. The commutator between the derivative and coordinate matrices
The commutator between the partition and the finite differences derivative can also be calculated from a global point of view using the corresponding matrices. Let the diagonal matrix [QN] which will represent the coordinate partition
QNdiagq1q2qN.E55
Then, the commutator between the derivative matrix and the coordinate matrix is
DNQN=Δ2χvΔ010000010100000101000000001000001010000010.E56
This is a kind of nearest neighbors’ average operator, inside the interval. The small Δlimit is just
[DN,QN]I,E57
where I is the identity matrix, with the first and last elements replace with 1/2. Thus, coordinate and derivative matrices are finite differences conjugate of each other.
## 5. An integration matrix
Since the determinant of the derivative matrix DNis not always zero, we expect that there exist an inverse of it. At a local level, the inverse of the finite differences derivation is the summation as was found in Eq. (46). In this section, we determine the inverse of the derivative matrix, and we find that it is a global finite difference integration operation.
Once we know the eigenvalues and eigenvectors of the derivative matrix DN, it turns out that we also know the eigenvectors and eigenvalues of the inverse matrix, when it exists. In fact, the equality DNem=λmem, with λm0, imply that
DN1em=λm1em.E58
The inverse matrix SN=DN1is
SN=1z1z1z1z1z1z11/z11/z11/z11/z1z1z1z1z11/z11/z11/z11/z1z1z1z1z11/z11/z11/z11/z1z1z1z11/z11/z11/z11/z1,E59
Its determinant is
SN=sinhN1vΔ.E60
This matrix represents an integration on the partition, with an exact value when it is applied to the exponential function evqon the partition. When applied to an arbitrary vector g=g1g2gNT, we obtain formulas for the finite differences integration, including the edge points
SNg1=1z1/zg1+i=1Mg2i+1zg2i,E61
SNg2j=1z1/zzg1+k=1j1zg2k+1g2k+k=jMg2k+1zg2k,E62
SNg2j+1=1z1/zg1+k=1jg2k+1g2kz+k=j+1Mg2k+1zg2k,E63
SNgN=1z1/zg1+i=1Mg2i+1g2iz,E64
where N=2M+1. These are new formulas for discrete integration for the exponential function on a partition of equally separated points with the characteristic that it is exact for the exponential function evq.
## 6. Transformation between coordinate and derivative representations
Since one of the eigenvalues of the derivative matrix is a continuous variable, we can talk of conjugate functions with a continuous argument v. The relationship between discrete vectors on a partition qiand functions with a continuous argument vmakes use of continuous and discrete Fourier type of transformations, a wavelet [12]. If we have a function hof continuous argument v, a conjugate vector on the partition qiis defined through the type of continuous Fourier transform Fas
Fhqj1L2ΔL/2L/2eiqjvhvdv,E65
and vice-versa, a continuous variable function is defined with the help of a discrete type of Fourier transform Fas
FgvL2Δj=N+1N12χvΔeiqjvgj.E66
Assuming that the involved integrals converge absolutely, we can say that
FFgqj1L2ΔL/2L/2eiqjvL2Δk=N+1N12χvΔeiqkvgkdv=1Δk=N+1N1gkL/2L/2eiqkqjvsinhvΔdvv=k=N+1N1gkKqkqjLΔ.E67
where
KqkqjLΔ1ΔL/2L/2eiqkqjvsinhvΔdvv=12ΔshiL2iqkqj+Δ+ishiL2qkqjiΔ2ishiL2qkqj+iΔ.E68
The function KqkqjLΔis an approximation to the Kronecker delta function δk,j. The function shi is the hyperbolic sine integral shiz=0zdtsinht/t. A plot of it is shown in Figure 1.
FFhv=L2Δj=N+1N12χvΔeiqjv1L2ΔL/2L/2eiqjuhudu=L/2L/2duhuJvuN,E69
where
JxN2χvΔΔj=N+1N1eiqjvu=2χvΔΔj=N+1N1eijvuΔ=2χvΔΔsinN1/2vuΔsinvuΔ/2,E70
The ratio of sin functions, in this expression, is an approximation to a series of Dirac delta functions located at vuΔ=, k. Thus, the operations Fand Fare finite differences inverse of each other.
### 6.1. The discrete Fourier transform of the finite differences derivative of a vector
Next, based on Eq. (28), we find that
Deiqvgj=gj+1Deiqvj+eiqj1vDgj=ivgj+1eiqjv+eiqj1vDgj.E71
If we sum this equality, we get
j=N+1N12χvΔDeiqvgj=ivj=N+1N12χvΔgj+1eiqjv+j=N+1N12χvΔeiqj1vDgjE72
i.e.,
FNDgv=ivFN+1gv+eivΔeiqjvgjj=N+2N+2Leiqjvgjj=N+1N1E73
Therefore, the discrete Fourier transform of the derivative of a vector gis ivtimes the discrete Fourier transform of g, plus boundary terms.
The Fourier transform of the derivative of a continuous function of variable vis easily found if we consider the equality
ddveiqjv=iqjeiqjv.E74
The integration of this equality with appropriate weights gives
iqjL/2L/2dveiqjvhv=L/2L/2dveiqjvdhvdv+eiqjvhvv=L/2L/2,E75
i.e.,
Fhj=iqjFhj+1L2eiqjvhvv=L/2L/2.E76
Hence, as is usual, the Fourier transform of the derivative of a function hvof continuous variable vis equal to iqjtimes the Fourier transform of the function, plus boundary terms.
## 7. Conclusion
We proceed with a brief discussion of the relationship between the derivative matrix DNand an important concept in quantum mechanics; the concept of self-adjoint operators [8, 9]. In particular, we focus on the momentum operator, whose continuous coordinate representation (operation) is given by id/dq, i.e., a derivative times i, in the case of infinite-dimensional Hilbert space.
In the finite-dimensional complex vectorial space (where each vector define a sequence gii=1Nof complex numbers such that igi2<). A transformation Ais usually called Hermitian, when its entries ai,jare such that ai,j=aj,i(denote the complex conjugate). Our matrix DNis related to an approximation of the derivative (see Section 3) which uses second order finite differences. Therefore, we can ask if the matrix iDNis also Hermitian.
Let PN=iDNand v=ixbe the eigenvalue of DN, where xis a free parameter, the corresponding eigenvalue of iDNis indeed the real value x; which is one of the properties of a Hermitian matrix, as is also the case of infinite-dimensional space (for the Hilbert space on a finite interval, these values are discrete, and for the Hilbert space on the real line, these values conform the continuous spectrum, instead of discrete eigenvalues). Other characteristic of iDNis that the eigenvector corresponding to xis the same exponential function which is the eigenfunction of id/dx(see Section 2).
Furthermore, let PNdenote the adjoint of PN. Thus, if we restrict our attention to the off-diagonal entries PNi,j=iDNi,j, it is fulfilled that PNi,j=idj,i=idi,j=PNi,j(noticing that, with v=ixthen χxΔ=sinxΔ/x). Even more, if we do not care about the two entries di,ifor i=1,N, we will have a Hermitian matrix. Finally, as it was seen in Section 4, we can say that PNcan be considered as a suitable approximation to the conjugate matrix to the coordinate matrix.
In conclusion, we have introduced a matrix with the properties that a Hermitian matrix should comply with, except for two of its entries. Besides, our partition provides congruency between discrete, continuous, and matrix treatments of the exponential function and of its properties.
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© 2018 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Armando Martínez Pérez and Gabino Torres Vega (August 29th 2018). Matrices Which are Discrete Versions of Linear Operations, Matrix Theory - Applications and Theorems, Hassan A. Yasser, IntechOpen, DOI: 10.5772/intechopen.74356. Available from:
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View all Books | 6,472 | 20,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-43 | longest | en | 0.939606 |
https://stats.stackexchange.com/questions/208827/why-must-linear-regressions-only-generate-linear-functions-that-resemble-lines | 1,723,401,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00778.warc.gz | 420,277,649 | 41,465 | # Why must linear regressions only generate linear functions that resemble "lines or planes" (*Introduction to Statistical Learning* question)?
Page 24 of Introduction to Statistical Learning states:
...linear regression is a relatively inflexible approach, because it can only generate linear functions such as the lines shown in Figure 2.1 or the plane shown in Figure 2.3.
Later on, linear regression is defined more formally as assuming $f$ takes on form
$$f(X) = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \ldots + \beta_p X_p$$
and "solving" for the $\beta_i$ using techniques like the "ordinary least squares" method (here I'm using the notation from the book).
Question: If this is really how linear regression is defined, why do we say that "linear regression [can]...only generate linear functions such as [lines and planes]"?
For example, if our model for $\hat{Y}$ is as follows:
$$\hat{Y} = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \ldots + \beta_p X_p$$
then any (or all) of the $X_j$ could be themselves highly non-linear or otherwise pathological functions that take on many bizarre shapes that look nothing like lines or planes. That being the case, a linear combination of those pathological random variables could end up looking like something that fails to resemble a line or a plane. Doesn't this show the quote from above to be false?
• Good question. A very closely related one at stats.stackexchange.com/questions/148638 concerns what "linear" in "linear model" might possibly mean.
– whuber
Commented Apr 22, 2016 at 16:19
• You may be also happy with this post, regarding the comment by @whuber on my initial (now erased) answer. Commented Apr 22, 2016 at 17:16
• @whuber: just to be clear, that post indicates that there are several notions of linearity, and that -- at the very least -- the quote above is wrong (in the sense that it is indeed possible for a model to be considered linear but in no way to resemble a line or a plane). Is my understanding correct here? Commented Apr 22, 2016 at 17:39
• I believe your understanding is correct, but I wouldn't go so far as to characterize the quote as "wrong": if you look at it the right way, such a linear regression indeed only fits lines, planes, and other linear subspaces. I think the fundamental point is that you--the modeler--begin by establishing a mapping from a space of possible regressor values into a space of regressors (such as mapping numbers $x$ to vectors $(1,x,x^2)$). Although that mapping may be nonlinear, the model in terms of the image vectors is perfectly linear.
– whuber
Commented Apr 22, 2016 at 18:14
• The data can all be non-linear but when you run linear regression or perform least squares you can be sure that you'll get a line. Your model for Y hat leaves off the residuals. All of that non linearity in the predictors you discussed will just be part of the residuals around the fitted line after OLS. The line may not describe much of the variance, but it's still a line. Commented Apr 23, 2016 at 9:10
Just imagine that there is some feature $Z$. $X_1 = \sin Z, \ X_2 = \cos Z$, and $f(X) = \beta_0 + \beta_1 X_1 + \beta_2 X_2 = \beta_0 + \beta_1 \sin Z + \beta_2 \cos Z$.
Will $f(X)$ be linear if the coordinate system is $(Y, Z)$? And what about $(Y, X_1, X_2)$? | 853 | 3,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-33 | latest | en | 0.90189 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-7-section-7-1-simplifying-rational-expressions-practice-page-487/6 | 1,532,271,990,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00042.warc.gz | 898,879,391 | 13,739 | ## Algebra: A Combined Approach (4th Edition)
$\frac{1}{x-5}$
$\frac{x+5}{x^2-25} = \frac{x+5}{x^2-5^2} = \frac{x+5}{(x-5)(x+5)} = \frac{1}{x-5}$ | 75 | 146 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-30 | latest | en | 0.559353 |
https://attemptmyexam.com/take-my-stochastic-models-for-finance-i/ | 1,696,104,896,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00119.warc.gz | 130,401,931 | 31,538 | # Take My Stochastic Models For Finance I
Take My Stochastic Models For Finance I find myself asking myself, Why not take the Blended SpAccount problem, take a Blended SpBalance problem, take a Blended SpCounter problem, take a Blended SpAccount problem, take a Blended SpBalance problem, take a Blended SpCounter problem. I say that has all kinds of possibilities, but I think it is like using a BSC; A BSC involves you directly in determining the find more information positions. So an array of numbers and functions says all the other functions are being used as long as they can be viewed directly by other people. And things you want different things to factor out, so factors A, which is why B, this is what you are seeing. It would be good if you could have a bit of a system with more functions that do some interesting things, so maybe later when you take a larger system, you come right back to what you knew you initially thought. I do the Blended SpGame with your “other world” functions with the Blended SpExample methods, so my interest in Blended SpExamples always goes above and beyond what you are doing with a function. I have thought about the problem for a while and I don’t have the time for that.
## Take My Online Classes And Exams
To move on from a more rational line of thought, the Blended SpExample itself and all the other examples I give are actually examples of not-quite-what-I-call-“data. Edit: Some examples, with the Blended Game example, that I have for you. Please take my examples, step by step. I think that I am generally an oddball. Every time you call an example of which you can call an array of numbers, you end up defining a new type called a “data”. The new type looks more like an array of functions. The “data” type will be called by a function, so you can make things as complicated as I explained in more detail above.
## Take My Online Quizzes For Me
My example of adding a function to a number array, so that you could be called “a BSC” if that function called the number array. No, I don’t think that if you actually call a function that each of [0,1] then you think of a new kind of data type. It’s a nice fun way to go about things, but typically it’s not really really cool to create a data type at the object level but just a datatype for a function call. It’s almost like a prototype using a prototype of your own class (class) aka a constructor part of an object. A prototype will expect its constructor to work with a number array, so you can call the constructor directly. You can create prototype variables for example, so you can override the constructor function to dynamically choose an array of numbers, the keys and values for that instance of the prototype are basically the same as the constructor function. I am not saying this is the right design method, but I think that it has a good advantage when you are dealing with both a data type and a data manipulation method.
## Take My Online Quizzes For Me
The data manipulation method is going to find more and more complexity. A data manipulation method looks like that, and your new data manipulation method would just convert it into that kind of data type, which is pretty boring in that there are all these specialized little operations and methods for working things that you didn’t initially learn before. But if you were to do a data manipulation method in the proper fashion and you do the right thing, a data manipulation method would give you nice results! So that’s how I’m looking at it. Edit: Some examples, with the Blended Game visit that I have for you. Please take my examples, step by step. I think that I am generally an oddball. Every time you call an example of which you can call an array of numbers, you end up defining a new type called a “data”.
## Take My Proctored Exam
The new type looks more like an array of functions. The “data” type will be called by a function, so you can make things as complicated as I explained in more detail above. My example of adding a function to a number array, so that you could be called “a BSC” if that function called the number array. Okay, so if you want to create a data subtype in BlendSpGame you don’t have to subclass a function for that, you can do it in theTake My Stochastic Models For Finance I’ve written a lot of these articles about why companies are in such trouble many times yet they have fewer choices than they need to cope with a multitude of financial problems. But that’s the lesson I’m going to take from this article – and many “technical” folks can’t spell anything new (it’s a bit of a long jump to point out what I have taught them). First of all, you’ve got 1) 100+ years of understanding of financial systems; 2) about processes of supply and demand; 3) about models of change and supply of demand? What really drives this process to do it? And why if I do the hard things, how are you going to create some of those models anyway? On the other hand, if you don’t know the basics for any of these decisions, why are you going to create others? For instance, in any financial system it must be ‘complex’ and in a business model of these things there must be as many of them as necessary to produce a model regardless of the number. But if there is just a finite number of ways of thinking, it’s hard to imagine how it can work.
## Take My Proctored Exam
So your 1) rule about supply and demand comes before 3) 2) of the definition of distribution given in the intro section. A: Two key areas of difference between Finance and finance? First of all, both would require you to first fully understand finance much deeper than you need to for any real value of this analysis. If you get at least 10 years of experience, I’d say it’s worth bringing your knowledge up a bit more to the board level. Second, the economics of finance require that you be able to work for the company and be able to think for it. If you’re looking to finance the company, that’s the first thing you have to do. So, you typically have to show some independence of your thought processes — you will probably be able to get the best of both worlds, depending on how you manage them. Moreover, you can’t get into real-life financial administration if the team where you are doing finance is doing work for the company.
## Do My Online Classes For Me
Of course, the biggest difference between finance and finance is that Finance is more likely to go without explicit requirements but also do more than just look and act to the profit potential of the company. For the time being, because most people don’t know what credit is a loancard (like interest rates, checkbook, etc), you are more likely to actually understand what interest rates should charge in conjunction with your knowledge of what credit amounts are and should be charged. To be clear though, credit can be very important to those who are in a different industry if their service is not going to sound too much like an “all-or-nothing” finance (I never heard the word “everything” comes from finance, no matter how much if it did be different). Second, businesses ought to be able to be trusted with these questions and more with these questions, to facilitate reasoned thinking, planning and communications. Not only is financial administration important to you, it really is also for you and the company. In finance though, once you understand the money supply and actual costs of certain services, you get a good idea of what the costs might be and things that are reasonably likely to be covered at a market level at once. Third, I’m not talking about the logic though.
## Take My Online Classes And Exams
Take accounting, ofTake My Stochastic Models For Finance I really have learned 3 years ago I graduated from The University of Sydney (then known as the University of Sydney Business Institute). For every semester I have had it taught at it’s maximum, with the exception that, for the month of March, the students and I make regular visits to it (in this case to its summer quarters), and daily updates will help with all of these, especially about the new-found variety of finance, and how things are constantly changing. I’m very motivated, very open-minded… Q: Well, how are the usual finance lessons and lessons have changed since 2013? Another term you’d like to see on a similar topic – what made you think (or read your review) that the lessons from both The University of Sydney (and Cointelegraph’s Facebook Account) are now worth repeating? A) The New System is designed to be: flexible and multi-courses. The lessons will ensure you have an overall view on these initiatives, as well as being a very active participant and learner.
## Hire Someone To Do Respondus Lockdown Browser Exam For Me
Q2. Are the new programs worth trying for? Q3. Most of the newly popular finance courses I’ve read — many of these already have been already taught? I’m thinking that the only course worth trying for is Q4: What is this again – what others in the knowledge base have to say? In the following note: Don’t worry, it’s pretty clear that the new-school course model is not doing well. However, I think it is worth further research on whether and why it did not work for you and how to get it off the ground. I did end up choosing Q4 because it was a cool way to start, but I’ve felt that I’ve stumbled upon a new challenge for you, as it has been extremely fast growing for me. Here are a few things we found out recently: 1. There are so many beautiful projects to be had on this, and you need to be part of them all, so it means that even if you haven’t covered already in detail how to solve that, you can do it now and then.
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2. Yes, there are projects that were in my life; I think that I’ve done some serious writing help early on in the career path… but it just seemed too much work for a community college like your Coghli Institute for Undergraduate Studies. 3. There are some very exciting projects that have been happening in my life – as a new student, I’m still learning and developing the project, but given this time I’d like to share our experience with you. Thanks again for everything! 7 Responses to Why This App is a Lesson Learned from a Better Teacher – I remember when I was a few months ago my last CSCE, so my CSCT didn’t happen that often. But this is still my favorite time of the year! In fact, it took me a very long time, as it was a good time to really enjoy it – not like it was boring at all, but very new! I was planning on a CD-ROM a couple of weeks ago… and didn’t even load anything or remember what I needed, but I did manage to sign | 2,335 | 10,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-40 | latest | en | 0.944837 |
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posted by .
on a game board, landing on blue means to move ahead 1 space , landing on red means to move ahead 2 spaces, and landing on orange means to move back 1 space. if you took 30 spins about where would you expect to be on the game board, relative to where you started?
i would be about ----- spaces-------.
behind)
You have a 1/3 probability of landing on each color. So, you probably will land on blue 10 times, red 10 times, and orange 10 times. With all those blues, you will have gone forward 10 spaces. With all those oranges, you would have gone back 10 spaces. You are still at your starting point. But, with 10 reds, you would move up 20 spaces. So, theoretically, you would be about 20 spaces ahead.
• math -
20
• math -
red
• math -
How to do math?
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Q: choose one of the short sories you read in this lesson and rewrite it in your own words, changing the setting so that the story takes place in the future. The Masque of the Red The Tell- Tale
Hear The Black Cat
A: The Black Cat The story is presented as a first-person narrative using an unreliable narrator. He is a condemned man at the outset of the story.[2] The narrator tells us that from an early age he has loved animals. [ He and his wife have many pets, including a large black cat named Pluto. This cat is especially fond of the narrator and vice versa. Their mutual friendship lasts for several
years, until the narrator becomes an alcoholic. One night, after coming home intoxicated, he believes the cat is avoiding him. When he tries to seize it, the panicked cat bites the narrator, and in a fit of rage, he seizes the animal, pulls a pen-knife from his pocket, and deliberately gouges out the cat's eye. ]
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posted by .
The problem is to evaluate the integral 10secxtanx dx, from -1/7 pi to 3/8 pi.
What I've done so far is evaluated the integral since secxtanx is a trig identity, so the integral of that is secx. I took out the 10 since it was a constant which leaves me with 10[sec(3/8)pi - sec(-1/7)pi]. Since those values aren't part of the unit circle, I put that in my calculator and came up with the answer 15.03209666, however the program that has my question is not accepting this answer. My calculator is in radians.
• calculus -
Please post the last sentence of the question.
Did the program ask you to give the answer to 4 places after the decimal, or does it ask you to give an exact answer?
• calculus -
The program is on the UT website, and it takes answers within 1% of the actual answer. And that's all to the question.
• calculus -
As far as I can see, your analytic answer and numerical approximation are both correct. It has to have something to do with the answering instructions. Does it have the capacity to accept the exact (analytic) answer?
• calculus -
Yes, it can take decimal places pretty far, even a rounded answer is still within 1% of the answer I received. So, I am unsure what the problem is, I am still awaiting an email from my professor and TA as to what could be the problem.
• calculus -
I agree with MathMate,
I got exactly the same answer right to the last decimal place.
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Engineering 1000 Chapter 4: Solution Formulation and Ergonomics
Outline n In this chapter, we will look first at general techniques for problem solving n n n We remind ourselves about constraints, and re-interpret some issues as constraints Then we will consider ergonomic design n n with some logical problems as examples what is ergonomic design how do we tell a good one human dimensions and variability human vision Example case study R. Hornsey Solution 2
Introduction n In later chapters, we will develop techniques for generating multiple solutions to a problem present state (problem) n But which options should we follow? n n desired state (solution) to pursue each possibility in depth will be wasteful of resources, so it is useful to be able to eliminate some of them very quickly All we can do is n n know the starting point, know the end point and develop a strategy for getting from one to the other efficiently R. Hornsey Solution 3
Designing a Search Strategy – Step 1 n Eliminate impossible solution paths n n n Making note of certainties is also a critical part of logical problem solving n n n to paraphrase Sherlock Holmes, “if we eliminate all impossible paths, then whatever is left, however improbable, is a potential solution” of course, we often don’t know enough to be certain we can eliminate a path see next two problems plus the coin example in the book (§ 4. 2. 1, 4. 2. 2) In less artificial problems, these two techniques illustrate the importance of obtaining as much information as possible prior to analysing solutions n research, constraints, reverse engineering, experiments R. Hornsey Solution 4
Logic Problem #1 n n n You are the National Nail Inspector. It is your job to go around to the nail factories and make sure that the nails being produced are up to standard, that is each nail weighs exactly 10 grams. You go around to 10 factories and get 10 nails from each factory and take them back to your office. Unfortunately your electric scales are almost out of batteries but you know that you have enough power left for ONE and only ONE weighing. You also know that ONE and only ONE of the factories is producing nails that weigh 11 grams. Given 10 nails from each of 10 different factories and using the scales to get one and only one reading (you cannot increment the scales to get multiple readings), determine which factory is producing nails that weigh 11 grams as opposed to the standard 10 grams. R. Hornsey Solution 5
Logic Problem #2 n There are five houses in a row, each of a different colour, and inhabited by 5 people of different nationalities, with different pets, favourite drinks, and favourite sports. Use the clues below to determine who owns the monkey and who drinks water. n n n n 1. The Englishman lives in the red house. 2. The Spaniard owns the dog. 3. Coffee is drunk in the green house. 4. The Russian drinks tea. 5. The green house is immediately to the right of the white house. 6. The hockey player owns hamsters. 7. The football player lives in the yellow house. 8. Milk is drunk in the middle house. 9. The American lives in the first house on the left. 10. The table tennis player lives in the house next to the man with the fox. 11. The football player lives next to the house where the horse is kept. 12. The basketball player drinks orange juice. 13. The Japanese likes baseball. 14. The American lives next to the blue house. R. Hornsey Solution 6
Designing a Search Strategy – Step 2 n Extract the most information possible n n n For example, in logic problem #2 here n n or “kill two birds with one stone” see the second coin problem in the textbook (§ 4. 2. 2) clue 9 tells us the American is in the leftmost house clue 14 says that the house next to the American is blue from clue 5, we know the American’s house is neither white nor green Sometimes we will not have enough information to make the next step n n this may call for an assumption; assume some information and see where it leads if you reach an impossibility, revisit the assumption if you reach a solution, revisit the assumption try to limit the number of choices before you assume something; this can save a lot of work R. Hornsey Solution 7
Designing a Search Strategy – Step 3 n Evaluate the current state n n Evaluate the final solution state n n again, obtain as much information as is appropriate try to ensure that new data collected is aimed at solving the problem structure data collection efficiently which is what a good problem statement should have done already Make sure that the problem you have solved is the one you set out to solve n n n when designing VLSI chips, schematic circuit diagrams are simulated extensively to ensure the design is correct then the circuit is transferred into the design for the actual parts of the silicon chip because this process may introduce errors and non-idealities, the schematic corresponding to the final silicon design is extracted from the layout and simulated again to check that the performance is similar to the original design R. Hornsey Solution 8
Subdividing the problem n From our objectives tree, Kepner-Tregoe analysis etc. we have broken down the design into a number of objectives n n The goals fell into two broad categories n n n each objective may contain several sub-problems specific goals general goals Specific goals are related to the features of our particular problem General goals may be present in varying degrees in all situations (see next slide) Continuous re-evaluation of our goals is important to ensure that we do not stray from the original objectives of the project n n it is inevitable that new objectives will appear as work progresses original objectives may be rendered irrelevant as new information becomes available R. Hornsey Solution 9
General Constraints Safety n n Environment laminated glass for car windows flatter TV screens n child-proof caps on medicine bottles n rechargeable batteries Big Mac ‘clam-shells’ Public acceptance n Apple Newton and Palm Pilot n Orbitz drink Reliability n cars n electronics Performance n everything! Ease of operation n n Durability n Minimum maintenance n Standard parts n user shouldn’t be able to do anything catastrophically wrong n stay-attached ring-pulls n n cd or LP MAC and PC everything! cars again (spark plugs, time between servicing) n Minimum cost R. Hornsey n multiple sourcing of components cell phones well-known properties cost n inkjet printers n Solution 10
Working with constraints n We have discussed a lot about constraints n n We have also talked about items we treated as goals n n n ladders that are lightweight the optimum volume for beverage containers public transport that is comfortable masks to protect against paint fumes These are really constraints in disguise since they represent practical limits on the design n n safety, legal, regulatory, economic, environmental there’s little point in a ladder that is too heavy for anyone to carry masks must be of a suitable design to fit the head and face of a wearer a beverage container holds enough volume to be thirst-quenching, but not so much that a significant amount is un-drunk All these issues depend on the characteristics of the human user … R. Hornsey Solution 11
Ergonomics (Human Factors) R. Hornsey Solution 12
The Human is Part of the System n n At some point, every machine system has an interface with a human Hence the way in which the system is perceived by the human is (or should be) an integral part of the machine’s design n n n a display must provide a visual experience that looks as realistic as possible to the viewer, and takes into account the way in which visual information is processed by the eyes and brain similarly, a hammer must be designed to be wielded by a human seating must be designed to take the human form into account door handles should provide clues about which side of the door to push or pull automatic office windows in new buildings on York campus should not remain stuck open on a cold October day We will look at issues particularly related to physical dimensions and vision R. Hornsey Solution 13
Is Ergonomic Design Really Necessary? n “Don’t worry, the user will adapt to it” n n n the so-called Procrustean approach Procrustes, whose name means "he who stretches", was arguably the most interesting of Theseus's challenges on the way to becoming a hero. He kept a house by the side of the road where he offered hospitality to passing strangers, who were invited in for a pleasant meal and a night's rest in his very special bed. Procrustes described it as having the unique property that its length exactly matched whomsoever lay down upon it. What Procrustes didn't volunteer was the method by which this "one-size-fits-all" was achieved, namely as soon as the guest lay down Procrustes went to work upon him, stretching him on the rack if he was too short for the bed and chopping off his legs if he was too long. Theseus turned the tables on Procrustes, fatally adjusting him to fit his own bed. (www. mythweb. com) What other perceived barriers can you think of that might prevent designers from pursuing an ergonomic design? R. Hornsey Solution 14
Ergonomic Design n Ergonomic design can be thought of in terms of a ‘principle of user-centred design’ n n “If an object or a system or an environment is intended for human use, then its design should be based on the physical and mental characteristics of its human users” Is a design ergonomic? n Try using it. Think forward to all of the ways and circumstances in which you might use it. Does it fit your body size or could it be better? Can you see and hear all you need to see and hear? Is it hard to make it go wrong? Is it comfortable to use all the time (or only to start with)? Is it easy and convenient to use (or could it be improved)? Is it easy to learn to use? Are the instructions clear? Is it easy to clean and maintain? Do you feel relaxed after a period of use? If the answer to all of these is 'yes' then the product has probably been thought about with the user in mind. Bodyspace: Anthropometry, Ergonomics and the Design of Work, S. Pheasant, Taylor and Francis R. Hornsey Solution 15
Anthropometry n Anthropometry refers to measurements of the dimensions of the human body n n Humans vary in size as a function of: n n n but which body? genetics nutrition age ethnicity occupation So it is tough to achieve a design that ergonomically satisfies all potential users n so who do we satisfy? R. Hornsey Solution 16
Seating n In class, have your neighbour measure the distance between your elbows when they are at your side n n n How does this compare with the width of a seat on a TTC subway car? n n similar to measurement #17 plot a distribution of the results for the class width = 430 mm leg room = 330 mm What are the competing pressures determining the seat size and leg-room allocated for economy class aircraft passengers? What fraction of the population should we accommodate? n 50%, 99. 99%? R. Hornsey Solution 17
Deep Vein Thrombosis n What is deep vein thrombosis? n n Deep vein thrombosis (DVT) refers to the formation of a thrombus (blood clot) within a deep vein, commonly in the thigh or calf. The blood clot can either partially or completely block the flow of blood in the vein. What causes deep vein thrombosis and who is at risk? n n DVT occurs when the flow of blood is restricted in a vein, and a clot forms. It can be caused by poor circulation because of problems such as heart disease, a recent heart attack or stroke, varicose veins, or from inactivity or prolonged bed rest. DVT may develop during a long flight and has been dubbed ‘economy class syndrome’ because the cheaper seats in a plane have less leg room, encouraging minimal movement. However, it is not confined to economy class or to long haul flights. http: //hcd 2. bupa. co. uk/ R. Hornsey Solution 18
Context n For the TTC or plane seats, we might decide that designing for 99% of the population is sufficient n what about the size of an escape hatch in an elevator? R. Hornsey Solution 19
Variability Engineering by Design, G. Voland, Addison Wesley, 1999 R. Hornsey Solution 20
The Human Dimension R. Hornsey Bodyspace: Anthropometry, Ergonomics and the Design of Work, S. Pheasant, Taylor and Francis Solution 21
Bodyspace: Anthropometry, Ergonomics and the Design of Work, S. Pheasant, Taylor and Francis Data for US Adults R. Hornsey Solution 22
Ethnic Variability Bodyspace: Anthropometry, Ergonomics and the Design of Work, S. Pheasant, Taylor and Francis R. Hornsey Solution 23
illumination colour R. Hornsey Computer Vision and Image Processing, S. Umbaugh, Prentice Hall 1998 Vision Solution 24
Proof the Blind Spot Exists n So how come you never see the blind spot when you look around? n n essentially because the brain fills in the gap Proof that the blind spot exists n n close your left eye focus on the cross with your right eye you should be aware of the spot, without looking directly at it as you move the page towards you, the spot will disappear at some point + n n • at the distance where the spot vanishes, you can look to the right and it will reappear what’s also neat is that the brain fills in the space – with whatever colour the background has! R. Hornsey Solution 25
http: //serendip. brynmawr. edu/bb/blindspot 4. html R. Hornsey Solution 26
Colour Vision n Because the cones are concentrated in the fovea, your central vision is less sensitive to low light levels n n n to see dim stars, it is best to use your peripheral vision where there are lots of rods This colour response is the basis for designing the phosphors in TV screens Computer Vision and Image Processing, S. Umbaugh, Prentice Hall 1998 R. Hornsey Solution 27
n The eye sees finer detail at higher ambient light levels n n n Computer Vision and Image Processing, S. Umbaugh, Prentice Hall 1998 Resolution of the Eye which is one reason why you see small dust particles in car headlights (~5µm) also important for high performance displays, VR etc. For TVs (where the brightness and colour information are treated separately) the human resolution is reduced n hence 600 lines is adequate 1 cycle R. Hornsey Solution 28
including adaptation no adaptation Computer Vision and Image Processing, S. Umbaugh, Prentice Hall 1998 Sensitivity of the Eye Engineering by Design, G. Voland, Addison Wesley, 1999 R. Hornsey Solution 29
Temporal Response Being a chemical system, the eye cannot respond to rapid changes of illumination n n Computer Vision and Image Processing, S. Umbaugh, Prentice Hall 1998 n which is why TV and movies look like continuous moving images movies ~15 pictures/second TV ~ 30 pictures/second computer monitors are scanned more rapidly (e. g. 75 Hz) You can see faster changes at higher light levels n so your can see your LCD watch flashing in bright sunlight R. Hornsey Solution 30
Hearing n Hearing is an important consideration n n Unlike vision, hearing is pervasive n n n both for data transmission and safety you do not have to ‘hear at’ something the way you would have to ‘look at’ it which makes it ideal for warning signals etc. Background noise is, however, a critical issue n and an alert sound must be distinctive Engineering by Design, G. Voland, Addison Wesley, 1999 R. Hornsey Solution 31
Human Factors n Is is now realised that human factors play a vital role in interface design n n USAF determined that just doubling the size of a cockpit display increased the ‘efficiency’ of pilots by 30% example from the textbook: reading 270° into an aircraft inertial navigation system instead of the correct 027° led to the plane crashing due to lack of fuel, killing 12 people are analog or digital displays better? Can you combine a calculator and a phone? n the key pads are reversed – which is which? R. Hornsey Solution 32
Feedback n “There’s a little black button on the black console that lights up blackly to show you pushed it” n n loosely from Hitch Hikers Guide to the Galaxy, Douglas Adams Feedback is a vital component of the system that informs the user that their action has had some effect n n e. g. the beep that tells the supermarket cashier that the item has been scanned beeps on phone or bank machine buttons it could also be a positive ‘click’ that indicates a mouse button has been depressed or a power light that indicates that a unit is switched on R. Hornsey Solution 33
Good Design Revisited n To develop good user-centred designs, we need to: n n determine the necessary interactions between the user and the machine identify the machine operations that require user input, monitoring or control and ensure these are within human capabilities ensure that the product performs well in the environment where it will be used automate where possible and desirable in order to minimise the interaction needed R. Hornsey Solution 34
Other Interesting Information n The Humane Interface by Jeff Raskin n n www. baddesigns. com n n lots of examples of what not to do The Joy of Visual Perception, Peter Kaiser n n n discussions of interface design from one of the creators of the MAC www. yorku. ca/eye good optical illusions and a background to human vision www. apple. com/about/ergonomics/ n common ergonomic issues related to computer use R. Hornsey Solution 35
Summary n We have seen how general problem-solving strategies can be developed to select between multiple solutions to a problem n n n eliminate impossible paths make the most of your data review both the problem statement and the present situation Be prepared to revise your goals The importance of constraints The human being as a constrain – ergonomics R. Hornsey Solution 36
Homework n n Read and understand chapter 4 of the text Follow the case studies in Ch. 4 n n especially the illustrative examples of general constraints Do problems 4. 4, 4. 5, 4. 6 R. Hornsey Solution 37
Exercise – Beverage Crates n Task Prior to Abatement (Description) n n Task Prior to Abatement (Method Which Verified Hazard) n n n An average lift of 34, 000 pounds each day was performed by each worker. Lifting of up to 45, 000 pounds a day might have been required for some workers because of case content inequities. Ergonomic Risk Factor (Posture) n n An injury and illness rate of 18. 5 per 100 full-time workers put the industry among the top 12 for injury frequency and top 5 for severity according to NIOSH studies. Ergonomic Risk Factor (Force) n n NIOSH recommended weight limit lifting criteria was exceeded for most lifting tasks. Task Prior to Abatement (Method Which Identified Hazard) n n Workers unload cases of soda cans and bottles from trucks, carting them to and stacking them on a customer's premises. Extended reach is required by delivery workers to unload the trucks. How would you solve these issues, and what benefits would you expect? R. Hornsey Solution 38 | 4,204 | 19,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-04 | latest | en | 0.904553 |
https://envisionmathanswerkey.com/envision-math-grade-2-answer-key-topic-2-6/ | 1,713,939,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00401.warc.gz | 210,969,055 | 42,731 | ## Envision Math 2nd Grade Textbook Answer Key Topic 2.6 Making 10 to Add 9
Making 10 to Add 9
Question 1.
Question 2.
Question 3.
Question 4.
Home Connection Your child explored strategies for making 10 to help add with 9.
Home Activity Have your child use pennies to make a group of 9 and a group of 5. Ask your child to show you how to make a group of 10 pennies to help find the sum.
AF 1.1, Grade 1 Write and solve number sentences from problem situations that express relationships
involving addition and subtraction. Also AF 1.0.
Guided Practice
Make 10 to add 9.
Use counters and your workmat.
Question 1.
Question 2.
Do you understand? Compare adding 9 to any number and adding 10 to that same number. What is different?
Independent Practice
Make 10 to add 9.
Use counters and your workmat.
Question 3.
Question 4.
Question 5.
Question 6.
Question 7.
8 +9 = ___
Question 8.
9 + 7 = ___
Question 9.
9 +2 = ___
Question 10.
9 +9 =___
Question 11.
1 +9 = ___
Question 12.
5 + 9 = ___
Number Sense
Find the missing number.
Question 13.
9 + 3 = + 2
Question 14.
+ 9 = 10 + 6
Problem Solving
Solve the problems below.
Question 15.
Tan’s team has 9 points. They score 7 more points. How many points do they have in all?
___ points
Question 16.
Ana’s team has 9 bats. Nico’s team has 8 bats. How many bats are there in all?
___ bats | 398 | 1,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-18 | longest | en | 0.869678 |
https://www.dsprelated.com/freebooks/pasp/Speed_Sound_Air.html | 1,716,885,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00100.warc.gz | 624,527,022 | 6,983 | ### Speed of Sound in Air
The speed of sound in a gas depends primarily on the temperature, and can be estimated using the following formula from the kinetic theory of gases:B.33
where, as discussed in the previous section, the adiabatic gas constant is for dry air, is the ideal gas constant for air in meters-squared per second-squared per degrees-Kelvin-squared, and is absolute temperature in degrees Kelvin (which equals degrees Celsius + 273.15). For example, at zero degrees Celsius (32 degrees Fahrenheit), the speed of sound is calculated to be 1085.1 feet per second. At 20 degrees Celsius, we get 1124.1 feet per second.
Next Section:
Air Absorption
Previous Section:
Heat Capacity of Ideal Gases | 163 | 710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.855643 |
https://en.m.wikipedia.org/wiki/Saddle_point | 1,656,942,359,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00639.warc.gz | 268,486,435 | 18,051 | In mathematics, a saddle point or minimax point[1] is a point on the surface of the graph of a function where the slopes (derivatives) in orthogonal directions are all zero (a critical point), but which is not a local extremum of the function.[2] An example of a saddle point is when there is a critical point with a relative minimum along one axial direction (between peaks) and at a relative maximum along the crossing axis. However, a saddle point need not be in this form. For example, the function ${\displaystyle f(x,y)=x^{2}+y^{3}}$ has a critical point at ${\displaystyle (0,0)}$ that is a saddle point since it is neither a relative maximum nor relative minimum, but it does not have a relative maximum or relative minimum in the ${\displaystyle y}$-direction.
A saddle point (in red) on the graph of z=x2−y2 (hyperbolic paraboloid)
Saddle point between two hills (the intersection of the figure-eight ${\displaystyle z}$-contour)
The name derives from the fact that the prototypical example in two dimensions is a surface that curves up in one direction, and curves down in a different direction, resembling a riding saddle or a mountain pass between two peaks forming a landform saddle. In terms of contour lines, a saddle point in two dimensions gives rise to a contour map ( graph or trace) with a pair of lines intersecting at the point. However typical ordnance survey maps for example don’t generally show such intersections as there is no reason such critical points in nature will just happen to coincide with typical integer multiple contour spacing. Instead dead space with 4 sets of contour lines approaching and veering away surround a basic saddle point with opposing high pair and opposing low pair in orthogonal directions. The critical contour lines generally don’t intersect perpendicularly.
Saddle point on the countour plot is the point where level curves cross
## Mathematical discussion
A simple criterion for checking if a given stationary point of a real-valued function F(x,y) of two real variables is a saddle point is to compute the function's Hessian matrix at that point: if the Hessian is indefinite, then that point is a saddle point. For example, the Hessian matrix of the function ${\displaystyle z=x^{2}-y^{2}}$ at the stationary point ${\displaystyle (x,y,z)=(0,0,0)}$ is the matrix
${\displaystyle {\begin{bmatrix}2&0\\0&-2\\\end{bmatrix}}}$
which is indefinite. Therefore, this point is a saddle point. This criterion gives only a sufficient condition. For example, the point ${\displaystyle (0,0,0)}$ is a saddle point for the function ${\displaystyle z=x^{4}-y^{4},}$ but the Hessian matrix of this function at the origin is the null matrix, which is not indefinite.
In the most general terms, a saddle point for a smooth function (whose graph is a curve, surface or hypersurface) is a stationary point such that the curve/surface/etc. in the neighborhood of that point is not entirely on any side of the tangent space at that point.
The plot of y = x3 with a saddle point at 0
In a domain of one dimension, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.
A model of an elliptic hyperboloid of one sheet
A saddle surface is a smooth surface containing one or more saddle points.
Classical examples of two-dimensional saddle surfaces in the Euclidean space are second order surfaces, the hyperbolic paraboloid ${\displaystyle z=x^{2}-y^{2}}$ (which is often referred to as "the saddle surface" or "the standard saddle surface") and the hyperboloid of one sheet. The Pringles potato chip or crisp is an everyday example of a hyperbolic paraboloid shape.
Saddle surfaces have negative Gaussian curvature which distinguish them from convex/elliptical surfaces which have positive Gaussian curvature. A classical third-order saddle surface is the monkey saddle.[3]
## Examples
In a two-player zero sum game defined on a continuous space, the equilibrium point is a saddle point.
For a second-order linear autonomous system, a critical point is a saddle point if the characteristic equation has one positive and one negative real eigenvalue.[4]
In optimization subject to equality constraints, the first-order conditions describe a saddle point of the Lagrangian.
## Other uses
In dynamical systems, if the dynamic is given by a differentiable map f then a point is hyperbolic if and only if the differential of ƒ n (where n is the period of the point) has no eigenvalue on the (complex) unit circle when computed at the point. Then a saddle point is a hyperbolic periodic point whose stable and unstable manifolds have a dimension that is not zero.
A saddle point of a matrix is an element which is both the largest element in its column and the smallest element in its row. | 1,088 | 4,844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-27 | latest | en | 0.89549 |
http://nce.ads.uga.edu/wiki/doku.php?do=export_code&id=how_to_compute_vanraden_s_deregressed_proof&codeblock=1 | 1,695,680,134,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00685.warc.gz | 30,649,388 | 1,157 | # # Computation of deregressed proof for sires. # # usage: awk -v h2=0.25 -f pvr_drp.awk sol_and_acc > drp.txt # # You can change the relitability as -v h2=value. # The default heritability is 0.25. # BEGIN{ # default h2=0.25 equiv. kd=14 if(h2<=0){ h2=0.25 } kd=(4-2*h2)/h2 print "h2=",h2,"; kd=",kd > "/dev/stderr" } NR>1{ DE_EBV = kd*\$5/(1 - \$5) DE_PA = kd*\$10/(1 - \$10) DE_R = DE_EBV - DE_PA R = DE_R/(DE_R + kd) if(R>0){ DRP = \$6 + (\$4-\$6)/R } else { R = 0.0 DRP = 0.0 } print \$0,DRP,R } | 222 | 500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.496957 |
http://en.wikipedia.org/wiki/Conjunction_introduction | 1,405,125,613,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776430044.46/warc/CC-MAIN-20140707234030-00096-ip-10-180-212-248.ec2.internal.warc.gz | 42,924,184 | 10,715 | Conjunction introduction
Conjunction introduction (often abbreviated simply as conjunction[1][2][3]) is a valid rule of inference of propositional logic. The rule makes it possible to introduce a conjunction into a logical proof. It is the inference that if the proposition p is true, and proposition q is true, then the logical conjunction of the two propositions p and q is true. For example, if it's true that it's raining, and it's true that I'm inside, then it's true that "it's raining and I'm inside". The rule can be stated:
$\frac{P,Q}{\therefore P \and Q}$
where the rule is that wherever an instance of "$P$" and "$Q$" appear on lines of a proof, a "$P \and Q$" can be placed on a subsequent line.
Formal notation
The conjunction introduction rule may be written in sequent notation:
$P, Q \vdash P \and Q$
where $\vdash$ is a metalogical symbol meaning that $P \and Q$ is a syntactic consequence if $P$ and $Q$ are each on lines of a proof in some logical system;
where $P$ and $Q$ are propositions expressed in some logical system.
References
1. ^ Hurley, Patrick (1991). A Concise Introduction to Logic 4th edition. Wadsworth Publishing. pp. 346–51.
2. ^ Copi and Cohen
3. ^ Moore and Parker | 310 | 1,215 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2014-23 | longest | en | 0.90174 |
https://it.mathworks.com/matlabcentral/answers/821685-how-to-get-norm-magnitude-of-a-vector-the-simple-way?s_tid=prof_contriblnk | 1,660,075,975,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00178.warc.gz | 332,173,591 | 26,705 | # How to get norm (magnitude) of a vector the simple way?
3 views (last 30 days)
Niklas Kurz on 4 May 2021
Edited: Niklas Kurz on 8 May 2021
I just want to get the norm of
syms phi the;
c = [-cos(phi)*sin(the)^2;-sin(phi)*sin(the)^2; - cos(the)*sin(the)*cos(phi)^2 - cos(the)*sin(the)*sin(phi)^2]
norm(c,2)
isn't really simplifying anything
If I type it manually:
simplify(sqrt(cos(phi)^2*sin(the)^4+sin(phi)^2*sin(the)^4+sin(the)^2*cos(the)^2))
I get a simple answere:
(sin(the)^2)^(1/2)
##### 5 CommentsShowHide 4 older comments
Niklas Kurz on 8 May 2021
I'm sorry for forgetting the simicolons in c. It might have been hard for you to reproduce what I was trying to create.
Sign in to comment.
### Accepted Answer
Nagasai Bharat on 7 May 2021
Hi,
From the documentation of norm and simplify you could find the usage of both these functions. norm would be used to calculate the norm of a vector/matrix but not for an expression. simpify would be used in the simplification of an algebric expression.
##### 1 CommentShowHide None
Niklas Kurz on 8 May 2021
well, it actually works if u were to incorporate some assumptions:
assume(phi>0);assume(the>0); assume(phi,'real'); assume(the,'real')
then, under these conditions
simplify(norm(c))
will simplify a lot (actually >0 not necessary)
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Translated by | 410 | 1,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-33 | latest | en | 0.892538 |
https://www.physicsforums.com/threads/gravity-waves-and-graceful-exit.768008/ | 1,508,288,222,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00588.warc.gz | 953,748,462 | 16,067 | # Gravity Waves and Graceful Exit
Tags:
1. Aug 28, 2014
### enorbet
Recently reading the Scientific American article The Black Hole at the Beginning of Time, which at first glance made me wonder if they were bowing to pressure to increase circulation, did however get me to thinking about Inflation and Graceful Exit, a rather major problem.
I understand that Stephen Hawking and Werner Israel postulate that the limits on our ability to measure wavelengths is roughly from 10^-7 to 10^11 Hz. I'm pretty sure I can see why those are presently limitations on our ability to measure, but is there any known reason that much larger wavelengths can not exist? say, 10^-18?
Further then, is it impossible, that oscillating epochs might exist based on extremely long (to us) wave periods?
2. Aug 28, 2014
### Staff: Mentor
What is an "oscillating epoch"?
3. Aug 29, 2014
### Tanelorn
1 oscillation per year is 3x10-8 Hz
1 oscillation per 13.8B years is 2.3x10-18Hz!
I wonder what the period is for a cyclic universe?
Last edited: Aug 29, 2014
4. Aug 29, 2014
### enorbet
Perhaps a bad choice of words to describe cyclic events like seasons on Earth, but I had hoped to keep it separate. at least in my own mind, from cycles driven by roughly circular motion as in the Earth around the Sun and our solar system around the galactic center. However thinking about a Galactic Year for Sol System only reinforces my curiosity about cyclical events on VERY large time scales.
It seemed to me the first step is to determine any limits on wavelength. For one thing it might possibly explain or at least open avenues of inquiry on Gravity and/or Dark Energy.
5. Aug 29, 2014
### enorbet
Thanks. Although those numbers are easily calculated it does help to just see some examples of the frames about which we are speaking. However I am not chasing ideas about the Bang/Rip question.
My wandering/wondering has more to do with what we can possibly measure given that the entirety of human existence from just about any point in Evolution you would choose to call even remotely human, isn't even a sizable fraction of a Galactic Year.
Some of what has fueled this line of thinking is noting that while we can't measure less than a Planck unit, apparently we can deduce beyond Planck if I understand the ESA's data showing that the Universe is not grainy to orders of magnitude beyond Planck.
The SA article states that if Big Bang is on some larger scale, a 3 Dimensional manifestation of a 4 Dimensional event, we may be able to gather some data about the Parent. Even though we can't employ light for obvious reasons, it may be possible to employ Gravity.
Obviously all of this as we get closer to The Singularity is highly speculative at this point. I'd just like to establish some reasonable boundaries on my imagination and see where that leads.
6. Aug 29, 2014
### enorbet
While, as mentioned in my first post in this thread, Hawking and Israel have postulated the window of limitation on what we can measure at 10^-7 to 10^11, what is the window of what we have actually measured so far? | 720 | 3,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-43 | longest | en | 0.953221 |
http://math.stackexchange.com/questions/404037/prove-lim-n-to-infty-fracnn1-1-using-epsilon-delta | 1,469,723,533,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828286.80/warc/CC-MAIN-20160723071028-00083-ip-10-185-27-174.ec2.internal.warc.gz | 164,997,971 | 18,506 | Prove $\lim_{n\to \infty}\frac{n}{n+1} = 1$ using epsilon delta
$\lim_{n\to \infty}\frac{n}{n+1} = 1$
Prove using epsilon delta.
-
Use $\displaystyle \frac{n}{n+1} = 1 - \frac{1}{n+1}$ and use the Archimedean property. – Andrew Salmon May 27 '13 at 17:22
What is your definition of "epsilon-delta" when you approach infinity? The $0 <|x - c| < \delta$ part doesn't seem to work. – Henry Swanson May 27 '13 at 17:23
we had a quiz on this question today, our TA told us to assume that lim(x->infinity) 1/n = 0 since not everyone has learned the Archimedean property (me being one of them). and using the assumption, derive an epsilon-delta proof. – Nick Gong May 27 '13 at 17:26
You want to show that for all $\epsilon$ there exists N such that if $n>N$, $|\frac{n}{n+1}-1|=|\frac{1}{n+1}|<\epsilon$ What does this tell you you should pick for N (in terms of $\epsilon$)? – Zen May 27 '13 at 17:31
By your notation I believe you're talking about the sequence $(a_n)$ of elements of $\mathbb{R}$ defined by:
$$a_n =\frac{n}{n+1}$$
Now, limit for sequences has the following definition: "given $\varepsilon >0$ there's some $n_0 \in \mathbb{N}$ such that if $n > n_0$ then $|a_n - L|<\varepsilon$". So we want some $n_0$ such that:
$$\left|\frac{n}{n+1} - 1\right|<\varepsilon$$
Rewrite this as:
$$\left|\frac{n}{n+1} - \frac{n+1}{n+1}\right| = \left|\frac{1}{n+1}\right|$$
But $n$ is natural so that the thing inside of the module sign is already positive and so we want in truth that:
$$\frac{1}{n+1}<\varepsilon \Longrightarrow n>\frac{1-\varepsilon}{\varepsilon}$$
This part is the deduction part. Now we prove, we say: given $\varepsilon > 0$ take $n_0 =(1-\varepsilon)/\varepsilon$, then we have that for $n > n_0$:
$$\left|\frac{n}{n+1}-1\right| = \left|\frac{1}{n+1}\right| = \frac{1}{n+1}$$
But $n>n_0$ so that $1/(n+1) < 1/(n_0 + 1)$ and hence:
$$\left|\frac{n}{n+1} - 1\right| < \frac{1}{n_0 + 1} = \frac{1}{\frac{1-\varepsilon}{\varepsilon} + 1} = \varepsilon$$
The important part for you to note is that first we deduce which $n_0$ works, after that we throw this part away usually and just say: "take this $n_0$" so that we show that it really works as predicted.
-
Very nicely done, +1! – dreamer May 27 '13 at 17:41
Thank you soo much! really appreciate the help – Nick Gong May 28 '13 at 16:19 | 824 | 2,323 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-30 | latest | en | 0.806042 |
https://www.kivodaily.com/technology/sigmoid-function-is-easy-to-learn/ | 1,675,115,931,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00561.warc.gz | 871,419,224 | 22,649 | Connect with us
# sigmoid function is easy to learn?
Published
on
Sigmoid function, baby. Knowing the importance of a sigmoid function is critical whether you’re building a neural network from scratch or using a pre-existing library. Familiarity with the sigmoid function is necessary for understanding how a neural network learns to deal with difficult problems. The use of this function as a springboard led to the discovery of other functions that lead to efficient and desirable solutions for supervised learning in deep learning architectures.
## As soon as you’ve completed reading this, you’ll be able to:
Reverse of the hyperbolic sine
The Distinction Between Linear and Nonlinear Separation
Incorporating a sigmoid unit into a neural network to facilitate better judgement
It is now time to get going.
### History of the Tutorial
Here are the lesson’s three parts:
Performa Sigmoidal
The sigmoid function and its characteristics
Distinguishing between problems that can be broken down neatly into linear categories and those that can’t
The sigmoid is often used as an activation function in neural networks.
#### “S-shaped” performance
In mathematics, the sigmoid function (a special case of the logistic function) is typically represented by the symbols sigmoidal (sig) or (x) (x) (x). The formula x = 1/(1+exp(-x)) holds true for all real numbers.
#### Explaining the Sigmoid Function and Its Uses
Sigmoid functions, represented by the green line in the following graph, tend to take the form of a S. The pink colour is also used for the graph of the derivative. The statement of the derivative and a few of its salient qualities are shown on the right.
Residence: (-, +)
Range: (0, +1)
σ(0) = 0.5
The function exhibits a clear rising pattern.
#### The function is, indeed, continuous everywhere.
The value of this function only needs to be determined within a small range, such as [-10, +10], for numerical calculations. Function values below -10 are very close to zero. Values of the function approach 1 over the range from 11 to 100.
##### The Sigmoid’s Suppressing Strength
The sigmoid function, also known as a squashing function, has the entire real number space as both its domain and its range (0, 1). (0, 1). As a result, regardless of whether the input is a very large negative number or a very large positive number, the function’s output will be a positive or negative value between 0 and 1. A similar rule holds that any integer is acceptable as long as it falls inside the range infinite to +infinity.
#### A Sigmoid Activation Function for a Neural Network
Sigmoid functions activate artificial neural networks. For a quick refresher, this diagram depicts activation functions in a neural network layer.
For a neuron with a sigmoid activation function, the output is always between 0 and 1. Furthermore, like the sigmoid, the output of this device would be a non-linear function of the weighted sum of inputs. A sigmoid unit is a type of neuron that uses an activation function that looks like a sigmoid.
#### Linear vs. nonlinear separability: which is better?
Let’s imagine we have to classify data into predefined groups.A straight line or n-dimensional hyperplane divides linearly separable issues into two groups (or an n-dimensional hyperplane). . The image that follows only shows data in two dimensions.All data is red or blue. Drawing a line between the two sets of things solves the left graphic. This graph shows a non-linearly separable problem with a non-linear decision boundary.
A
#### For what reasons is the Sigmoid function so important in neural networks?
By definition, a neural network trained with a linear activation function can only learn to handle problems with linearly separable features. The neural network can tackle non-linear problems using a single hidden layer and sigmoid activation function.. The sigmoid function’s ability to provide non-linear boundaries makes it a helpful tool for neural networks to learn non-trivial decision-making methods.
Neural network activation functions must be non-linear and monotonic. For this reason, we cannot use sin(x) or cos(x) as an activation function.
The activation function must be defined over the whole real number line. The function must be differentiable over all real values.
Gradient descent determines neuron weights in back propagation. This method uses the activation function derivative.
Back propagation can learn a neural network’s weights utilising the sigmoid function’s monotonicity, continuity, and differentiability everywhere. | 938 | 4,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-06 | latest | en | 0.89584 |
https://math.stackexchange.com/questions/780997/maximum-speed-in-a-circular-orbit | 1,563,844,901,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528635.94/warc/CC-MAIN-20190723002417-20190723024417-00015.warc.gz | 471,332,574 | 38,972 | # Maximum speed in a circular orbit?
Visualize two points: $O\equiv(0\mid 0)$ and $D\equiv(d\mid 0)$. The two are $d$ units apart.
Visualize a movable rod whose endpoints, $C_O$ and $S_O$, are a unit apart. $C_O$ always coïncides with $O$ and $S_O$ orbits $O$. Visualize a movable rod whose endpoints, $C_D$ and $S_D$, are a unit apart. $C_D$ always coïncides with $D$ and $S_D$ orbits $D$. Visualize a movable rod whose endpoints, $R_O$ and $R_D$, are $d$ units apart. $R_O$ coïncides with $S_O$, and $R_D$ coïncides with $S_D$. I believe that such an arrangement is called a linkage. For our purpose $S_O$ (and $R_O$) can be considered as situated at $(\cos\theta\mid \sin\theta)$. $\theta$ is the parameter whose value determines the locations of all moving points in this problem. There are two such linkages. In the first, the rod connecting $S_O$ to $S_D$ is always parallel to the line connecting $O$ to $D$. In this case, $S_D\equiv(d+\cos\theta\mid \sin\theta)$. Not interesting, therefore ignorable. In the second, the rod connecting $S_O$ to $S_D$ is never parallel to the line connecting $O$ to $D$. In this case, $S_D\equiv\left({{(d^2-1)(d-\cos\theta)}\over{d^2+1-2d\cos\theta}}\mid{{(d^2-1)(-\sin\theta)}\over{d^2+1-2d\cos\theta}}\right)$.
If one considers the direction of rotation of $OS_O$ to be positive $+$, then the following is true: If $d\gt 1$, then the direction of rotation of $DS_D$ is $+$.
If $d\lt 1$, then the direction of rotation of $DS_D$ is $-$.
If $d=1$, then $S_D\equiv (0\mid 0)$ — except when $\theta\equiv 0\mod 2\pi$, when $S_D\equiv{\left({0\over 0}\mid{0\over 0}\right)}$.
QUESTION: For a given positive value of $d$, what value of $\theta$ imparts the maximum angular velocity, $+$ or $-$, to $DS_D$?
I am indebted to Dr. Jyrki Lahtonen for his helpfulness!!!!
[Editor's note: The picture I have in mind, JL.]
The animations below have the points $S_O$ and $S_D$ marked with black dots. As the distances from the origin $O$ to $S_O$, from $S_O$ to $S_D$ and from $S_D$ to $D$ are all fixed, we can imagine them being connected by rigid rods. I took the liberty of adding those rods to the pic.
In the first animation, $d={2\over 3}$.
And in the second, $d={4\over 3}$ (the size of the animation is scaled down by a factor of two in comparison to the one above).
[Editor's note: I put the motivational speech below (from the OP) back in here, because it may be needed to make the question meet the site standards, JL]
ADDENDUM! It was requested that I clarify my question, so I shall try to eliminate abstractions in favor of concrete objects … 1. Replace the Cartesian plane with a sheet of marine plywood. 2. On the plywood, draw a line segment $d$ units long. Label its endpoints $O$ and $D$. Drill a hole at each endpoint. Fit a dowel snugly into each. 3. Take three paint mixing paddles. On two of them, draw lines $r$ units long. On one of these, label the endpoints $O$ and $S_O$; on the other, label the endpoints $D$ and $S_D$. Drill holes at all four of these endpoints. On the remaining paint mixing paddle, draw a line $d$ units long. Label its endpoints $S_O$ and $S_D$. Drill a hole at each. 4. Fit the hole $O$ onto the dowel $O$. Fit the hole $D$ onto the dowel $D$. Fit dowels into $S_O$ and $S_D$, but not completely through. 5. Fit the $S_O$ hole onto the $S_O$ dowel. Fit the $S_D$ hole onto the $S_D$ dowel. 6. Saw off any excess dowel that would interfere with the movement of the assembly. (Yes, it is supposed to move.) 7. I envision these rods as being made of untreated white pine.
NOTA BENE: For simplicity's sake, I have gotten rid of the variable $r$ and replaced it with unity.
• Voting to close for unclear what you're asking. I don't understand the question at all. – user122283 May 4 '14 at 15:55
• @SanathDevalapurkar Yea, I don't know if it's "standard" notation to write it like that but it's certainly greek to me. – MCT May 4 '14 at 16:11
• What does $O(0 \mid 0)$ mean? What about $S_O(r \cos \theta | r \sin \theta)$? I have no idea what the question is asking. – user61527 May 5 '14 at 3:53
• I beg to disagree. Not necessarily with the decision to put on hold, but its reason. Yes, the notation is non-standard, and the question does not show effort. It is also mistagged, as circle does not really describe what's being asked. But fer crying out loud it is perfectly clear what IS being asked. Don't look at the formulas - look at the text and draw the picture in your imagination. See the machine of four rods with three joints and two hinged+fixed endpoints swirling in the plane? – Jyrki Lahtonen May 5 '14 at 5:26
• This question lacks key information. Most importantly, what have you tried? Also, where did you encounter the problem? That information helps other people write answers that are as helpful as possible. – Carl Mummert May 5 '14 at 9:59
I'm breaking my promise not to answer this as no one else bites. This is largely just a sketch of an argument explaining why the maximum is obtained at $\theta=0$, when $S_O$ is as close to $D$ as possible, moving tangentially. A calculus solution by differentiation is possible, but I don't want to do that.
Here's a variant of the animation. This time $d\approx 1.54$. I added the linkage discarded by the OP in red, because it aids understanding what is going on. Furthermore I added a green rubber band' connecting the points $D$ and $S_O$.
As expected, the uninteresting' red dot, call it $S_D'$, revolves about the pivot $D$ with exactly the same angular rate as the point $S_O$ about the pivot $O$. This is a consequence of the fact that $DS_D'S_OO$ is always a parallelogram. But let's look at the picture more closely. We see that the black dot $S_D$ is always the mirror image of $S_D'$ with respect to the green line. The mirror symmetry is a geometrically obvious consequence of the fact that the points $S_D$ and $S_D'$ are the two points on the plane with the prescribed distances to $D$ and $S_O$.
Therefore the angular velocity of the green line (about $D$) is the average of the angular velocities of $S_D$ and $S_D'$ (the green line bisects the angle $\angle S_DDS_D'$). So $$\omega_{S_D}=2\omega_{green}-\omega_{S_D'}.$$ As $\omega_{S_D'}$ is constant this means that $\omega_{S_D}$ is maximized simultaneously with $\omega_{green}$. But $\omega_{green}$ is the angular velocity of $S_O$ about $D$. We can apply the principle: angular velocity = tangential component of the orbital speed vector divided by the distance. As the orbital speed of $S_O$ is constant it happens that the tangential component is maximized at the point of the closest approach, so the numerator reaches its maximum and the denominator reaches its minimum simultaneously at $\theta=0$. Therefore the ratio reaches its maximum at the closest approach of $S_O$ to $D$.
• You need to be a bit careful in interpreting the signs of the angular velocities. In the last paragraph I treated clockwise direction as positive, so $\omega_{S_D'}<0$. As we see, when $d<1$ the sign of $\omega_{green}$ varies, but is positive at the point $\theta=0$, so the conclusion holds. – Jyrki Lahtonen May 12 '14 at 9:52
• If you are turning the crank of the paint mixer by hand (older version of the question - see the edit history), then you can probably notice that maximum torque is require at that point, when the mixer paddle at $S_D$ is moving at its greatest speed. – Jyrki Lahtonen May 12 '14 at 10:16
• Dr. @Jyrki Lahtonen: Once again, TYVM! When I first became interested in such linkages and noticed the acceleration, $+$ or $-$, of the driven (as opposed to the driving) pivot, I set up a display (similar to the three that you contributed) and hooked up three or four in tandem. The result was spectacular: Although the display moved quite slowly — I was only bumping the value of $\theta$ a degree at a time. — the final pivot in the chain would suddenly spring to life like the business end of a rat trap. – Senex Ægypti Parvi May 12 '14 at 22:11 | 2,243 | 8,007 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-30 | longest | en | 0.768112 |
https://roclocality.org/2017/01/24/cs255-assignment-1-lvn-2/ | 1,686,338,025,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00781.warc.gz | 571,558,568 | 21,609 | Hi CS255/455 students:
Hope you all enjoyed the course! The first assignment is to implement local value numbering (LVN) and check for redundant expressions.
You are expected to handle commutativity for commutative operations. Recall that an operation is commutative if you can change the order of operands without changing the result. For example (+) is commutative but (-) is not. Your implementation must be able to assign the same value number to a+b and b+a.
As the second requirement, improve LVN by adding the Stewart extension. The Stewart extension improves LVN by identifying additional redundancy in the following example form.
a = b + c
d = a – b
Specifically, it guides LVN to assign the same value number to both (c) and (d). The idea of the solution was first raised by Chris Stewart when he was taking the class around 2004. His idea was to insert additional value relations into the value number table. You should first work out this idea and make it concrete as an extension to the basic value numbering algorithm.
Note 1: You are expected to apply the Stewart extension on four operations: ‘+’, ‘-‘, ‘*’, and ‘/’.
Note 2: You should make sure that the Stewart extension can be applied on the following code as well.
a = b + c
e = a
d = e – b
Finally, transform the code sequence by removing the redundant expression(s) and print the transformed code sequence.
To complete this assignment, take the following steps:
1. From Blackboard, download the code from Course Materials: 02 Local value numbering demo programs
2. Implement and add commutativity and Stewart extension to the LVN class in the file vn.rb.
3. Implement code generation.
4. Make sure all the tests in vn_tests.rb pass.
5. Document any test failures, if there is any, and explain why, in README.txt in plain text.
6. Extra credit. In addition to finding the statements with a redundant expression, generate optimized code where all redundant expressions are removed. Demonstrate the optimizer with a set of tests and put them in opt_tests.rb. The tests should include all three tests in vn_tests.rb.
7. Submit your assignment on Blackboard. Make sure to include all the ruby files in your submission and the file README.txt to document the submission.
Due time: Tuesday Jan 31st at 23:59:59 EST. (5% bonus points for submission before Friday Jan 27th at 23:59:59 EST.)
Late submission policy: Each student can have a total of two days used for late submissions over all assignments . This means that if you submit the LVN assignment on Thursday, you will not be able to do any other late submission. But if you submit on Wednesday, you still have one more day to use for one other assignment.
Policy on academic honesty : Every line of code of the LVN analyzer and optimizer must be written by the student. Do not copy code. Do not show your LVN code to others until 2 days (48 hours) past the assignment due time. The teaching staff is required to report every violation of the policy or suspicion of violation to the university’s Academic Honesty Board. Contact the teaching staff if you have questions on the policy. | 695 | 3,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.882195 |
https://www.homeworkmarket.com/content/mat-540-assignment | 1,632,235,626,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00389.warc.gz | 819,813,555 | 47,120 | # Mat 540 - Assignment
rhodan1
Mat 540 assignment
It will be a problem with at least three (3) constraints and at least two (2) decision variables. The problem will be bounded and feasible. It will also have a single optimum solution (in other words, it won’t have alternate optimal solutions). The problem will also include a component that involves sensitivity analysis and the use of the shadow price.
You will be turning in two (2) deliverables, a short writeup of the project and the spreadsheet showing your work.
Writeup.
Your writeup should introduce your solution to the project by describing the problem. Correctly identify what type of problem this is. For example, you should note if the problem is a maximization or minimization problem, as well as identify the resources that constrain the solution. Identify each variable and explain the criteria involved in setting up the model. This should be encapsulated in one (1) or two (2) succinct paragraphs.
After the introductory paragraph, write out the L.P. model for the problem. Include the objective function and all constraints, including any non-negativity constraints. Then, you should present the optimal solution, based on your work in Excel. Explain what the results mean.
Finally, write a paragraph addressing the part of the problem pertaining to sensitivity analysis and shadow price.
Excel.
As previously noted, please set up your problem in Excel and find the solution using Solver. Clearly label the cells in your spreadsheet. You will turn in the entire spreadsheet, showing the setup of the model, and the results.
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• mat-540_week_8_assignment_1_linear_programming_case_study.zip | 436 | 2,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.886534 |
https://sciencevstruth.org/2014/06/25/explaining-the-retrograde-spin-of-venus/ | 1,686,318,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00261.warc.gz | 540,054,320 | 23,204 | ## Explaining the ‘retrograde spin’ of Venus
The gravitational influence exerted by a celestial body depends upon many factors
-The velocity of spin
-The density of the surrounding Ether/ atmosphere
-Distance from the ‘core’
How far a planet exerts its influence depends upon the density/ type of the environment (and of course on the rate of spin). For example a body spinning inside a pond may ‘churn’ the water around it for 10meters distance and influence things up to that point. But if the body is surrounded by a ‘mantle’ of mercury (or any other denser medium), the body’s influence may not extend even for 5meters despite spinning at the same rate.
Imagine that the diameter of the spinning body is 1 meter and the thickness of the mantle layer is 0.5 meter. But someone looking from a distance may not appreciate the spinning body and its surrounding mantle layer separately and may view the ‘body-mantle complex’ as one single entity. So the distant observer may measure the spinning body’s size as 1.5meters and may think that the body spins much slower (because the mantle spins slower than the core of the body).
The same mantle model can be used to explain the apparent clockwise (retrograde) rotation of Venus (in contrast to the counter-clockwise rotation of other planets).
We know that the Sun rotates counter clock wise. That drags and spins the Ether around it and exerts ‘gravitational’ influence on the planets. If the planets were to be passive
1)They would revolve anticlockwise around the Sun
2)They would spin clockwise under the influence of the Sun’s anticlockwise spin
3)And they would ultimately get dragged into the Sun and there wouldn’t be a solar family for us to live and talk about.
We know that the planets are revolving counter-clockwise, but how come they are not spinning clockwise? And how come the planets are not falling into the Sun?
The fact that the planets are remaining in their orbits without falling into the Sun, implies that they are not passive but are exerting some opposing influence to counter the inward ‘pull’ by the Sun. And that opposing influence can come from an ‘active’ counter-clockwise spin of the planets. So just like the Sun, even the planets must be having an inherent counter-clockwise rotation.
But how come the planet Venus is spinning clockwise? We can explain this apparent retrograde spin of Venus using the same ‘mantle model’ described above. The actual size of Venus is probably much smaller than what the ‘astro-pastors’ have noted. The reason why Venus appears so big could be because of a thick mantle layer covering it. The ‘core’ may be actually spinning counter clockwise. While the inner mantle spins counter-clockwise under the influence of this Venus core, the outer mantle spins clockwise under the influence of the Sun. Because a distant observer may not appreciate what is happening in the core, he/she may conclude that the planet is spinning clockwise.
Thus my model of gravity provides clear physical basis for both attraction and repulsion forces while the ‘scientific religion’ gives only a ‘mythical’ description of them.
• Tim Ruiz On July 20, 2014 at 8:17 pm
So the surface of Venus is actually the surface of a clockwise spinning mantle over a counter-clockwise spinning core? What would be between the mantle and core? And why wouldn’t that thing drag the mantle into the counter-clockwise spin? Just trying to understand.
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• drgsrinivas On July 22, 2014 at 12:07 am
My assumption is that, in the case of Venus, the influence of its ‘nucleus’ or core doesn’t reach up to its outer layers- may be its core is small or it spins not fast enough or may be it is surrounded by a denser ‘medium’. So while its inner layers spin counter-clockwise under the influence its core, the outer layers spin clockwise under the influence of the Sun. What lies between the counter-clockwise spinning core/inner layers and the clockwise spinning outer layers/ mantle? May be some kind of liquid medium/ molten metal.
Having said that, I do think of an alternative explanation for the retrograde spin of Venus: May be that Venus has no intrinsic spin but just spins clockwise under the influence of Sun and probably it is slowly being dragged towards the Sun. May be that planet Venus was farther away from the Sun at the ‘beginning’ and got slowly dragged towards the Sun.
While the spinning Ether model provides a good logical frame work to explain the phenomenon of gravity (and the delusion of curved space), there exist many ‘finer’ issues to address. And I am working to refine the Ether model to resolve the same.
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• drgsrinivas On July 25, 2014 at 11:01 am
I realise that I haven’t taken into account of the centrifugal force while explaining the gravity. Incorporating the centrifugal force into the gravity model described above would explain many issues.
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http://www.wisegeek.org/what-is-the-difference-between-cause-and-correlation.htm | 1,444,148,678,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736678861.8/warc/CC-MAIN-20151001215758-00207-ip-10-137-6-227.ec2.internal.warc.gz | 1,088,616,356 | 25,157 | Category:
# What is the Difference Between Cause and Correlation?
A correlation exists between green tea and lower stroke risk.
Gravity causes the ball to drop when it is released.
Drinking too much alcohol may cause liver disease.
Bad parenting can increase the likelihood of a child displaying violent behavior.
Article Details
• Originally Written By: Tricia Ellis-Christensen
• Revised By: Phil Riddel
• Edited By: O. Wallace
2003-2015
Conjecture Corporation
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Cause and correlation are terms that are often confused or used incorrectly. A correlation means a relationship between two or more things: when one increases, the other increases, or when one increases, the other decreases. A cause is something that results in an effect; for example, heating water to a certain temperature will make it boil. The crucial point is that a correlation between two things does not necessarily mean that one causes the other. If there is a relationship between two phenomena, A and B, it could be that A causes B, or it could be that B is responsible for A; other possibilities are that some other factor is the reason for both A and B, or that they have independent causes that just happen to run in parallel.
#### Correlation
Researchers trying to find reasons for various things will often use statistical methods to establish correlations: this may be the first step toward establishing the cause. Scientists and statisticians can use a formula to determine the strength of a relationship between two phenomena. This gives a figure, known as the square of the correlation coefficient, or R2, which always lies between 0 and 1, with a value closer to 1 indicating a stronger correlation.
When the R2 value is high, this relationship may merit further investigation; however, researchers should beware of jumping to conclusions. It is possible to identify all sorts of strong, but meaningless, correlations. In one very well known example, the R2 for the number of highway fatalities in the US between 1996 and 2000, and the quantity of lemons imported from Mexico during the same period, is 0.97 — a very strong correlation — but it is extremely unlikely that one causes the other.
A correlation, particularly when reported in the media, is often described as a “link,” which can be misleading, as it can be taken to mean that one of the factors causes the other. For example, a study that found that men who drink four cups of green tea a day had a lower risk of stroke than those who did not drink it might generate the headline "Green Tea Cuts Stroke Risk." This implies that drinking green tea will directly lower the risk of stroke, but that isn't proven by the study. Other factors, like the fact that the study was conducted on men in Japan who have different diets and exercise habits than men in Western countries, could have influenced the results. While there could be a more direct causal relationship here, a broader study would be needed and more variables would need to be considered.
#### Cause
If factor A is responsible for factor B, there will be a strong correlation between the two, but the reverse is not necessarily the case. Proving beyond reasonable doubt that A is responsible for B requires much more than a high R2 value. Having established a strong relationship, researchers will then need to come up with ideas as to how A might affect B then test these ideas by experiment. It is often the case that more than one possible cause can be identified. In these instances, a good method is to conduct experiments in which all but one of the factors remains constant, and then determine from this the factor that responsible for the effect.
For example, a plant that grows in a temperate climate may be dormant during the winter, and start growing in spring. One theory would be that increased average temperatures trigger growth, while another might be that longer periods of daylight are responsible. To determine which is the case, one sample of plants might be subjected to increasing temperatures and constant hours of daylight, while another might experience constant temperature and increasing daylight. The cause could then be determined from which set of plants starts growing. If neither set begins to grow, a third experiment might be performed, in which both temperature and daylight are increased; if this results in growth, then the researchers might conclude that a combination of both factors is required.
In some cases, a given cause will always result in a particular effect; for example, the Earth’s gravity will always make an object fall if no other force is acting on it. In other cases, however, the effect is not guaranteed. It is known that ionizing radiation and certain chemicals are causes of cancer, but not everyone exposed to these factors will develop the disease, as there is an element of chance involved. Both factors can alter DNA, and sometimes this will result in a cell becoming cancerous, but this will not happen every time. If, however, one were to plot levels of exposure to these factors against the incidence of cancer in a large sample of otherwise similar people, a strong correlation would be expected.
Although researchers have criteria for pursuing possible causes of a phenomenon based on the strength of the correlations, the factor with the highest R2 value is not necessarily the one responsible. Scientists and researchers will reject factors that show a weak correlation, but, as noted, completely irrelevant factors can produce a very high R2, as can factors that appear for the same reason as the thing being investigated. The likelihood of A causing B is therefore not necessarily proportional to the strength of the correlation.
#### Confusion of Cause and Correlation
A lot of confusion between cause and correlation results from the way findings are reported in the media. A relationship might be described as a “cause” — it might be reported that violent video games cause violent behavior, when all that has been found is a correlation, for example. It may be that aggressive people are more likely to play violent games, so such people would behave more aggressively with or without the influence of the games.
Research has shown that violent games may influence aggression. It also shows that a number of other factors may be responsible for violent behavior, among them, poorer socioeconomic status, mental illness, abusive childhoods, and bad parenting. Possibly, such games may increase the likelihood of violent behavior in an individual with a predisposition toward aggression resulting from other factors, but stating that violent video games cause violent behavior is not justified by the known facts.
Health is another area where confusion can arise. Those who read or hear of the many things that have been reported as causing, or being linked to, cancer might never eat, drink, or leave their homes again. A “cause” may only be a correlation, and a “link” is just that: it does not identify a definite cause of cancer. A great deal of research is going on into the reasons why cancer develops, and scientists frequently find links, but when these are reported in the media, people should look or listen carefully for qualifying words like “may,” “might increase,” or “could have an effect,” before drawing any conclusions.
## Recommended
anon944426 Post 4 I always remember a great example of this which was that there seemed to be a greater instance in the UK of particular cancer illness around nuclear power stations. This led to concerns that these stations caused the cancer.I think after some research, what was actually discovered was that power stations tend to be built on the coast (perhaps needing the water for cooling) and that old people tend to retire to the coast. So what was actually happening was that a greater density of old people led to a greater density of illnesses which old people get, including cancer. anon153104 Post 3 Even with strong correlations we must be very cautious, there must be more supporting evidence to even suggest that it increases a risk.For instance: there is a correlation between obesity and breast cancer, but there is no observational clinical evidence to support the idea that obesity causes breast cancer or has any significant affect on any cancer. Yet it is accepted by many clinicians to be a "cause". You might as well say that there is a very high correlation between wearing a bra and getting breast cancer. Both are equally valid. vogueknit17 Post 2 @DentalFloss, I can definitely see why causality and correlation would be confusing to people in psychology. There is also an importance in the difference between cause and correlation in biology. People often like to talk about thing involving evolution as though they were caused by something specific, like a natural disaster or an act of humans or another species. While sometimes things do cause other things in the natural world, it is often difficult for biologists to do more than speculate about correlation, rather than identify things as definite causes. DentalFloss Post 1 When I studied psychology, my professor said one of the things she expected every single one of us to get out of that class was the difference between cause and correlation. Specifically, that data correlation does not equal cause. | 1,884 | 9,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2015-40 | longest | en | 0.951233 |
http://en.wikipedia.org/wiki/Little_man_computer | 1,416,788,978,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400380233.64/warc/CC-MAIN-20141119123300-00192-ip-10-235-23-156.ec2.internal.warc.gz | 106,863,859 | 16,336 | # Little man computer
The Little Man Computer (LMC) is an instructional model of a computer, created by Dr. Stuart Madnick in 1965.[1] The LMC is generally used to teach students, because it models a simple von Neumann architecture computer - which has all of the basic features of a modern computer. It can be programmed in machine code (albeit in decimal rather than binary) or assembly code.[2][3][4]
## System architecture
The LMC model is based on the concept of a little man shut in a small room or a computer in this scenario. At one end of the room, there are 100 mailboxes (memory), numbered 0 to 99, that can each contain a 3 digit instruction or data (ranging from 000 to 999). Furthermore, there are two mailboxes at the other end labeled INBOX and OUTBOX which are used for receiving and outputting data. In the center of the room, there is a work area containing a simple two function (addition and subtraction) calculator known as the Accumulator and a resettable counter known as the Program Counter. The Program Counter holds the address of the next instruction the Little Man will carry out. This Program Counter is normally incremented by 1 after each instruction is executed, allowing the Little Man to work through a program sequentially. Branch instructions allow iteration (loops) and conditional programming structures to be incorporated into a program. The latter by setting the Program Counter to a non-sequential memory address if a particular condition is met (typically the value stored in the accumulator being zero or positive). As specified by the von Neumann architecture, memory contains both instructions and data. Care therefore needs to be taken to stop the Program Counter reaching a memory address containing data or the Little Man will attempt to treat it as an instruction. To use the LMC the user loads data into the mailboxes and then signals the Little Man to begin execution, starting with the instruction stored at memory address zero. Resetting the Program Counter to zero effectively restarts the program.
## Execution cycle
To execute a program, the little man performs these steps:
1. check the Program Counter for the mailbox number that contains a program instruction (e.g. zero)
2. fetch the instruction from the mailbox with that number
3. increment the Program Counter (so that it contains the mailbox number of the next instruction)
4. decode the instruction (includes finding the mailbox number for the data it will work on, e.g "get data from box 42")
5. fetch the data from the mailbox with the number found in the previous step (for example, store the data in the accumulator)
6. execute the instruction
7. store the new data in the mailbox from which the old data was retrieved
8. repeat the cycle or halt
## Commands
While the LMC does reflect the actual workings of binary processors, the simplicity of decimal numbers was chosen to minimize the complexity for students who may not be comfortable working in binary/hexadecimal.
### Instructions
Some LMC simulators are programmed directly using 3-digit numeric instructions and some use 3-letter mnemonic codes and labels. In either case, the instruction set is deliberately very limited (typically about ten instructions) to simplify understanding. If the LMC uses mnemonic codes and labels then these are converted into 3-digit numeric instructions when the program is assembled. The first digit of a numeric instruction represents the command to be performed and the last two digits represent the memory address of the mailbox relevant to that command.
The table below shows a typical numeric instruction set and the equivalent mnemonic codes.
Instructions
Numeric code Mnemonic code Instruction Description
1xx ADD ADD Add the value stored in mailbox xx to whatever value is currently on the accumulator (calculator).
Note: the contents of the mailbox are not changed, and the actions of the accumulator (calculator) are not defined for add instructions that cause sums larger than 3 digits.
2xx SUB SUBTRACT Subtract the value stored in mailbox xx from whatever value is currently on the accumulator (calculator).
Note: the contents of the mailbox are not changed, and the actions of the accumulator are not defined for subtract instructions that cause negative results - however, a negative flag will be set so that 8xx (BRP) can be used properly.
3xx STA STORE Store the contents of the accumulator in mailbox xx (destructive).
Note: the contents of the accumulator (calculator) are not changed (non-destructive), but contents of mailbox are replaced regardless of what was in there (destructive)
5xx LDA LOAD Load the value from mailbox xx (non-destructive) and enter it in the accumulator (destructive).
6xx BRA BRANCH (unconditional) Set the program counter to the given address (value xx). That is, value xx will be the next instruction executed.
7xx BRZ BRANCH IF ZERO (conditional) If the accumulator (calculator) contains the value 000, set the program counter to the value xx. Otherwise, do nothing.
Note: since the program is stored in memory, data and program instructions all have the same address/location format.
8xx BRP BRANCH IF POSITIVE (conditional) If the accumulator (calculator) is 0 or positive, set the program counter to the value xx. Otherwise, do nothing.
Note: since the program is stored in memory, data and program instructions all have the same address/location format.
901 INP INPUT Go to the INBOX, fetch the value from the user, and put it in the accumulator (calculator)
Note: this will overwrite whatever value was in the accumulator (destructive)
902 OUT OUTPUT Copy the value from the accumulator (calculator) to the OUTBOX.
Note: the contents of the accumulator are not changed (non-destructive).
000 HLT/COB HALT/COFFEE BREAK Stop working.
DAT DATA This is an assembler instruction which simply loads the value into the next available mailbox. DAT can also be used in conjunction with labels to declare variables. For example, DAT 984 will store the value 984 into a mailbox at the address of the DAT instruction.
### Examples
#### Using Numeric Instruction Codes
This program (instruction 901 to instruction 000) is written just using numeric codes. The program takes two numbers as input and outputs the difference. Notice that execution starts at Mailbox 00 and finishes at Mailbox 07. The disadvantages of programming the LMC using numeric instruction codes are discussed below.
00 901 INBOX --> ACCUMULATOR INPUT the first number, enter into calculator (erasing whatever was there)
01 308 ACCUMULATOR --> MEMORY[08] STORE the calculator's current value (to prepare for the next step...)
02 901 INBOX --> ACCUMULATOR INPUT the second number, enter into calculator (erasing whatever was there)
03 309 ACCUMULATOR --> MEMORY[09] STORE the calculator's current value (again, to prepare for the next step...)
04 508 MEMORY[08] --> ACCUMULATOR (Now that both INPUT values are STORED in Mailboxes 08 and 09...)
LOAD the first value back into the calculator (erasing whatever was there)
05 209 ACCUMULATOR = ACCUMULATOR - MEMORY[09] SUBTRACT the second number from the calculator's current value (which was just set to the first number)
06 902 ACCUMULATOR --> OUTBOX OUTPUT the calculator's result to the OUTBOX
07 000 (no operation performed) HALT the LMC
#### Using Mnemonics and Labels
Assembly language is a low-level programming language that uses mnemonics and labels instead of numeric instruction codes. Although the LMC only uses a limited set of mnemonics, the convenience of using a mnemonic for each instruction is made apparent from the assembly language of the same program shown below - the programmer is no longer required to memorize a set of anonymous numeric codes and can now program with a set of more memorable mnemonic codes. If the mnemonic is an instruction that involves a memory address (either a branch instruction or loading/saving data) then a label is used to name the memory address.
This example program can be compiled and run on the LMC simulator[5] available on the website of York University (Toronto, Canada) or on the desktop application written by Mike Coley.[6] All these simulators include full instructions and sample programs, an assembler to convert the assembly code into machine code, control interfaces to execute and monitor programs, and a step-by-step detailed description of each LMC instruction.
```INP
STA FIRST
INP
STA SECOND
LDA FIRST
SUB SECOND
OUT
HLT
FIRST DAT
SECOND DAT
```
## Labels
Without labels the programmer is required to manually calculate mailbox (memory) addresses. In the numeric code example, if a new instruction was to be inserted before the final HLT instruction then that HLT instruction would move from address 07 to address 08 (address labelling starts at address location 00). Suppose the user entered 600 as the first input. The instruction 308 would mean that this value would be stored at address location 08 and overwrite the 000 (HLT) instruction. Since 600 means "branch to mailbox address 00" the program, instead of halting, would get stuck in an endless loop.
To work around this difficulty, most assembly languages (including the LMC) combine the mnemonics with labels. A label is simply a word that is used to either name a memory address where an instruction or data is stored, or to refer to that address in an instruction.
When a program is assembled.
• A label to the left of an instruction mnemonic is converted to the memory address the instruction or data is stored at. i.e. loopstart INP
• A label to the right of an instruction mnemonic takes on the value of the memory address referred to above. i.e. BRA loopstart
• A label combined with a DAT statement works as a variable, it labels the memory address that the data is stored at. i.e. one DAT 1 or number1 DAT
In the assembly language example which uses mnemonics and labels, if a new instruction was inserted before the final HLT instruction then the address location labelled FIRST would now be at memory location 09 rather than 08 and the STA FIRST instruction would be converted to 309 (STA 09) rather than 308 (STA 08) when the program was assembled.
Labels are therefore used to:
• identify a particular instruction as a target for a BRANCH instruction.
• identify a memory location as a named variable (using DAT) and optionally load data into the program at assembly time for use by the program (this use is not obvious until one considers that there is no way of adding 1 to a counter. One could ask the user to input 1 at the beginning, but it would be better to have this loaded at the time of assembly using one DAT 1)
### Example
This program will take a user input, and count down to zero.
``` INP
LOOP SUB ONE // Label this memory address as LOOP, The instruction will then subtract the value stored at address ONE from the accumulator
OUT
BRA LOOP // If the accumulator value is not 0, jump to the memory address labeled LOOP
QUIT HLT // Label this memory address as QUIT
ONE DAT 1 // Store the value 1 in this memory address, and label it ONE (variable declaration)
```
This program will take a user input, square it, output the answer and then repeat. Entering a zero will end the program.
(Note: an input that results in an output greater than 999 will cause an error due to the LMC 3 digit number limit).
```START LDA ZERO // Initialize for multiple program run
STA RESULT
STA COUNT
INP // User provided input
BRZ END // Branch to program END if input = 0
STA VALUE // Store input as VALUE
LOOP LDA RESULT // Load the RESULT
STA RESULT // Store the new RESULT
LDA COUNT // Load the COUNT
STA COUNT // Store the new COUNT
SUB VALUE // Subtract the user provided input VALUE from COUNT
BRZ ENDLOOP // If zero (VALUE has been added to RESULT by VALUE times), branch to ENDLOOP
BRA LOOP // Branch to LOOP to continue adding VALUE to RESULT
ENDLOOP LDA RESULT // Load RESULT
OUT // Output RESULT
BRA START // Branch to the START to initialize and get another input VALUE
END HLT // HALT - a zero was entered so done!
RESULT DAT // Computed result (defaults to 0)
COUNT DAT // Counter (defaults to 0)
ONE DAT 1 // Constant, value of 1
VALUE DAT // User provided input, the value to be squared (defaults to 0)
ZERO DAT // Constant, value of 0 (defaults to 0)
```
Note: If there is no data after a DAT statement then the default value 0 is stored in the memory address. | 2,703 | 12,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2014-49 | longest | en | 0.9195 |
http://www.orafaq.com/usenet/comp.databases.theory/2007/06/28/0323.htm | 1,544,949,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827596.48/warc/CC-MAIN-20181216073608-20181216095608-00591.warc.gz | 448,920,282 | 4,038 | # Re: completeness of the relational lattice
From: Jan Hidders <hidders_at_gmail.com>
Date: Wed, 27 Jun 2007 23:23:53 -0000
> On Jun 27, 2:17 pm, "Brian Selzer" <b..._at_selzer-software.com> wrote:
>
>
>
>
>
> > > On Jun 26, 7:03 pm, Jan Hidders <hidd..._at_gmail.com> wrote:
> > >> I suspect that the condition
> > >> for r + (s * t) = (r + s) * (r + t)
> > >> should be A(r) * A(s) = A(r) * A(t) = A(s) * A(t).
> > >> So if two have an attribute, the third must also have it.
>
> > > This is more powerful condition than in the "First steps..." paper.
> > > Indeed, the correct criteria is:
>
> > > A(x,t) \/ (B(y,t) /\ C(z,t)) = (A(x,t) \/ B(y,t)) /\ (A(x,t) \/
> > > C(z,t))
>
> > > The proof is by writing down both sides in the set notation. Assuming
> > > general relation headers A(x,u,w,t), B(y,u,v,t), C(z,w,v,t) the left
> > > hand side works out to be
>
> > > exists xyzv : A \/ (B /\ C)
>
> > > The right hand side evaluates to
>
> > > exists zy : (exists xwv: A \/ B) /\ (exists xuv: A \/ C)
>
> > > In predicate logic we have
>
> > > exists x: (P /\ Q) <-> exists x: P /\ exists x: Q
>
> > While it is true that
>
> > exists x: (P \/ Q) implies exists x: P \/ exists x: Q
> > and
> > exists x: P \/ exists x: Q implies exists x: (P \/ Q),
>
> > isn't it true that
>
> > exists x: (P /\ Q) implies exists x: P /\ exists x: Q
> > but
> > exists x: P /\ exists x: Q does not imply exists x: (P /\ Q)?
>
> > Consequently,
>
> > exists x: (P /\ Q) is not equivalent to exists x: P /\ exists x: Q
>
> That's right, I was looking tohttp://en.wikipedia.org/wiki/First-order_logic#Identities
> before posting, but apparently didn't notice this subtlety. So I don't
> see how we can justify attribute x in the distributivity law and fail
> to find a counter example either...
I think what you missed is rules like "P or ( Exists x : Q(x) ) <=> Exists x : ( P and Q(x) )" where P has no free occurrence of x.
A(x, t) \/ (B(y, t) /\ C(z, t) which is formally:
{ (t) | (Exists x : A(x, t)) \/
(Exists y, z : B(y, t) /\ C(z, t)) }
• /* pull y and z quantifiers to the front */
{ (t) | Exists y, z :
(Exists x : A(x, t)) \/ (B(y, t) /\ C(z, t)) }
• /* distribution */
``` { (t) | Exists y, z :
((Exists x : A(x, t)) \/ B(y, t)) /\
(Exists x : A(x, t)) \/ C(z, t))) }
```
• /* push y and z quantifiers back to their variables */
{ (t) | ((Exists x : A(x, t)) \/ (Exists y : B(y, t))) /\
(Exists x : A(x, t)) \/ (Exists z : C(z, t)))) }
which is indeed the representation of
(A(x,t) \/ B(y,t)) /\ (A(x,t) \/ C(z,t))
• Jan Hidders
Received on Thu Jun 28 2007 - 01:23:53 CEST
Original text of this message | 917 | 2,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-51 | latest | en | 0.806785 |
http://www.cpbj.com/article/20120926/SALES/120929818/0/politics1&template=sales | 1,386,452,419,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163055852/warc/CC-MAIN-20131204131735-00093-ip-10-33-133-15.ec2.internal.warc.gz | 296,199,863 | 13,455 | # A calculated future
By - Last modified: September 26. 2012 9:29AM
Tim lept to his feet as though he were standing wet and barefoot on an electrified carpet. He was the only one standing.
The ballroom was packed with hundreds of Tim's new co-workers — salespeople from around the country. He had been with the company only a few weeks and this was his first convention. The energy and excitement was palpable as the distribution of annual sales awards was reaching its peak.
On stage, the CEO and the vice president of sales had just handed a smiling awardee his prize, shook his hand and applauded him back to his seat.
"Who's going to be our 'Rookie of the Year' next year?" boomed the CEO's voice.
The sentence had barely made its way through the sound system before impacting one particular attendee. Now on his feet, Tim suddenly felt all eyes upon him. Surprisingly, he felt none of the self-consciousness that normally comes with it. He was completely confident. He knew exactly what he wanted.
While he was brand new to the industry, he wasn't brand new to sales. He had been a standout in his previous position selling medical supplies — and had quickly established himself as one of the best in his company.
He left only because of a series of leadership changes over a short period of time had given him the impression that the company was unstable and unreliable. His new employer provided him with that stability.
And now they provided him with a challenge.
Tim had already supplied himself with a vision. Soon after he had started with the company, he asked his boss about the various achievement levels the company recognized. He had locked on to Rookie of the Year and decided it was the award he wanted. But deciding was only a part of the effort.
He sat down and repeatedly ran the numbers — he wanted to know exactly how much he needed to sell in order to mathematically clinch the spot.
He calculated the number of prospects, the number of appointments, an estimate of his closing rate and the average sale value. He analyzed his territory, its history and its potential. He reviewed how the weeks in the year would fall with vacations and holidays. He even planned what could go wrong and how he would address it.
Again — this was just a part of the effort.
Shortly after he started and just before his first convention, he put his sales plan into motion. He immediately began to see results and knew he was going to succeed.
When he walked into the awards ceremony that night, he felt like a tiger among rabbits.
"I felt invincible. The funny thing was that I knew that the only way to claim it was to stand up. When they asked, 'Who will be Rookie of the Year next year?' I thought to myself, 'It HAS to be me.' There wasn't another option."
"Curiously, when I stood up, some of the veterans started to applaud — they seemed to appreciate that someone would be confident enough to make a statement. The thing was, I had already pictured it in my head and my plan was set."
Tim worked. Tim worked hard. By the company's third quarter, he had rocketed to the top of the rookie rankings. Even more incredulously, he was ranked sixth overall — veterans included.
With six weeks left in the company's fiscal year, his boss called him to let him know that no rookie would be able to catch him. Tim applied even more gas to his sales engine.
The convention is coming soon. Tim's convention. Tim will walk across the stage, shake the hands of the CEO and the vice president of sales as he accepts his award to a thunderous applause.
It's no surprise to Tim. He planned it that way. He saw it.
What do you see in YOUR future?
Write to the Editorial Department at editorial@cpbj.com | 793 | 3,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-48 | latest | en | 0.99546 |
https://communities.sas.com/t5/SAS-Procedures/Proc-Transpose/td-p/117777?nobounce | 1,531,952,009,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00178.warc.gz | 628,955,483 | 40,702 | ## Proc Transpose
Solved
Super Contributor
Posts: 1,041
# Proc Transpose
Hi,
Could someone help me solve this transpose step??
INS should go with the OUTS
If there are more of a kind(in this case IN'S) then closest one should become the pair.in our case
04Sep1989:13:45 has to pair with 05SEP1989:7:15 instead of the 04Sep1989:7:30
and 04Sep1989:7:30 should have a blank as a pair
HAVE
ID flag time cnt
101 IN 04Sep1989:7:30 1
101 IN 04Sep1989:13:45 2
101 IN 21SEP1989:17:55 3
101 OUT 05SEP1989:7:15 1
101 OUT 22SEP1989:06:00 2
WANT
ID IN1 OUT1 IN2 OUT2 IN3 OUT3
101 04Sep1989:7:30 04Sep1989:13:45 05SEP1989:7:15 21SEP1989:17:55 22SEP1989:06:00
Thanks
Accepted Solutions
Solution
08-21-2013 05:56 PM
Super User
Posts: 8,093
## Re: Proc Transpose
Change the 4's to something larger. It is hard to know how many elements to use for the array. To be safe you could make it as large as he maximum number of records per ID.
Or you could just calculate the grouping variable and then let PROC TRANSPOSE make the output dataset for you.
proc sort;
by id time;
run;
data middle ;
set have ;
by id;
if first.id then group=0;
if first.id or flag='IN' or (flag='OUT' and flag=lag(flag)) then group+1 ;
run;
proc transpose data=middle out=want (drop=_name_);
by id;
id flag group ;
var time;
run;
All Replies
Super User
Posts: 5,878
## Re: Proc Transpose
You should resort your data by ID and time.
Then use the data step and last. logic to do explicit OUTPUT.
Before the the output just use IF-THEN logic to do variable assignments.
Data never sleeps
Super User
Posts: 8,093
## Re: Proc Transpose
You could do it with a DATA step. You will need an upper bound on the number of variables which you could calculate as the maximum value of CNT variable.
%let max=3 ;
data want ;
do until (last.id);
set have ;
by id ;
array in (&max);
array out (&max);
if flag='IN' then in(cnt) = time;
if flag='OUT' then out(cnt) = time;
end;
keep id in: out: ;
format in: out: datetime.;
run;
Super Contributor
Posts: 1,041
## Re: Proc Transpose
Hi Tom,
The method you suggest is not giving me the following output
WANT
ID IN1 OUT1 IN2 OUT2 IN3 OUT3
101 04Sep1989:7:30 04Sep1989:13:45 05SEP1989:7:15 21SEP1989:17:55 22SEP1989:06:00
OUT1 should be missing
basiclly when there are more of a kind we should see which one of those makes a closest pair to the other
like above 04Sep1989:13:45 is closest to 05SEP1989:7:15 than 04Sep1989:7:30 and so OUT1 is left blank
Thanks
Super User
Posts: 8,093
## Re: Proc Transpose
So the CNT variable is useless. Instead you want to create new IN/OUT pairs based on the relative TIME values?
proc sort;
by id time;
run;
data want ;
do until (last.id);
set have ;
by id;
array in (4);
array out (4) ;
if first.id then i=1;
else if flag='IN' then i=i+1;
if flag='IN' then in(i)=time;
if flag='OUT' then out(i)=time;
end;
keep id in: out: ;
format in: out: datetime.;
run;
Super User
Posts: 8,093
## Re: Proc Transpose
The above will collapse sequence like IN/OUT/OUT into a single pair and eliminate the middle OUT time point.
To fix that you might change the index variable increment logic to:
if first.id then i=1;
if flag='IN' then do; if in(i) ne . then i=i+1; in(i)=time; end;
if flag='OUT' then do; if out(i) ne . then i=i+1; out(i)=time; end;
Super Contributor
Posts: 1,041
## Re: Proc Transpose
Hi Tom,
Thanks a ton for your time.
But i wonder if this works for all cases
When i run the code for the below data 24SEP1989:06:00 IN forms a pair with 23SEP1989:06:00 which cannot be the case because IN cannot be later (24) compared to OUT(23rd)
data have;
input ID \$ flag \$ time :datetime. cnt;
format time datetime.;
datalines;
101 IN 04Sep1989:7:30 1
101 IN 04Sep1989:13:45 2
101 IN 21SEP1989:17:55 3
101 OUT 05SEP1989:7:15 1
101 OUT 22SEP1989:06:00 2
101 OUT 23SEP1989:06:00 12
101 IN 24SEP1989:06:00 15
run;
I was expecting OUT4 to be by itself without the corresponding IN
and 24SEP1989:06:00 is IN5 without the corresponding OUT
Hope this helps
Thanks
Super User
Posts: 8,093
## Re: Proc Transpose
You need to keep the auto increment when flag='IN'.
if first.id then i=0;
if flag='IN' then do; i=i+1; in(i)=time; end;
if flag='OUT' then do; if out(i)^=. then i=i+1; out(i)=time; end;
Super Contributor
Posts: 1,041
## Re: Proc Transpose
I am using the following code and it says :
ARRAY SUBSCRIPT OUT OF RANGE.Please correct me
Thanks
data want ;
do until (last.id);
set have ;
by id;
array in (4);
array out (4) ;
if first.id then i=0;
if flag='IN' then do; i=i+1; in(i)=time; end;
if flag='OUT' then do; if out(i)^=. then i=i+1; out(i)=time; end;
end;
keep id in: out: ;
format in: out: datetime.;
run;
ERROR: Array subscript out of range at line 907 column 34.
last.id=1 ID=101 flag=IN time=24SEP89:06:00:00 cnt=15 FIRST.ID=0 in1=04SEP89:07:30:00 in2=04SEP89:13:45:00 in3=21SEP89:17:55:00 in4=. out1=. out2=05SEP89:07:15:00
out3=22SEP89:06:00:00 out4=23SEP89:06:00:00 i=5 _ERROR_=1 _N_=1
Solution
08-21-2013 05:56 PM
Super User
Posts: 8,093
## Re: Proc Transpose
Change the 4's to something larger. It is hard to know how many elements to use for the array. To be safe you could make it as large as he maximum number of records per ID.
Or you could just calculate the grouping variable and then let PROC TRANSPOSE make the output dataset for you.
proc sort;
by id time;
run;
data middle ;
set have ;
by id;
if first.id then group=0;
if first.id or flag='IN' or (flag='OUT' and flag=lag(flag)) then group+1 ;
run;
proc transpose data=middle out=want (drop=_name_);
by id;
id flag group ;
var time;
run;
Super Contributor
Posts: 1,041
## Re: Proc Transpose
Hi Tom,
Both of the Array method(after changing length to a larger value ) and transpose method work very well. Thanks so much
In the array method:
why were you using if out(i)^=.??? and not for the in(i)??????what if in an other case of our data the OUTS outnumber the INS???
if flag='IN' then do; i=i+1; in(i)=time; end;
if flag='OUT' then do; if out(i)^=. then i=i+1; out(i)=time; end;
Could you also put this from the transpose method in words for me??
if first.id then group=0;
if first.id or flag='IN' or (flag='OUT' and flag=lag(flag)) then group+1 ;
we have already said if first.id then group=0; Again why do we use if first.id or or flag='IN' or (flag='OUT' and flag=lag(flag))
why isnt the lag function used for the IN???
Thanks so much
Super User
Posts: 8,093
## Re: Proc Transpose
To prevent IN > OUT you always want to start a new pair (group) when see an IN record. But when you see and OUT record it could be the OUT for the current pair or an unmatched OUT that needs a new pair. One way to tell this is if the current pair already has an OUT time value. So we test if OUT(I) has a value when processing an OUT record, but we don't care for IN record as we know that it marks a new pair.
So a new group starts when
• the first record for an ID group (FIRST.ID)
• any IN record (FLAG='IN')
• an OUT record when it is not preceded by an IN record. (FLAG='OUT' and LAG(FLAG)='OUT')
To avoid incrementing the group counter twice when more than one of the conditions apply to the same record the program first sets it to 0 at the start of a ID group. Then it can use the same logic test about whether to increment the group counter for every record it processes.
Super Contributor
Posts: 1,041
## Re: Proc Transpose
Thanks for the dettailed and quick response.
I could not mark it right because the options dont pop up ....
I will do it as soon as i see it
Lastly, i will post an other question shortly on the same quesion but with Imputing the missing values
if IN is missing for a particular OUT then the IN should have a value of 1 minute prior instead of the current code which sets to missing
Also if OUT is missing for a particular IN then OUT should have a value of 1 minute past the IN
Thanks again
Super User
Posts: 8,093
## Re: Proc Transpose
Apply the imputation to the resulting pairs.
do i=1 to dim(in);
if IN(i)=. then IN(i) = out(i) - 1;
if out(i) = . then out(i) = in(i) + 1;
end;
Super Contributor
Posts: 1,041
## Re: Proc Transpose
Hi Tom,
I was still wondering about setting the group to 0 at the beginning of each new ID
you explained that if we dont do it then if two conditions are satisfying a record then there is a chance of "incrementing the group counter twice!!!!!
Suppose I have the following as the first record
101 IN Sep6:7:30
It is the first of the ID and flag is IN(satisfying 2 conditions) ....how would the group counter gets incremented twice if we dont initialize the group to 0???
Thanks
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Stress strain curve is the plot of stress and strain of a material or metal on the graph. In this, the stress is plotted on the y-axis and its corresponding strain on the x-axis. After plotting the stress and its corresponding strain on the graph, we get a curve, and this curve is called stress strain curve or stress strain diagram. Stress and Strain - Definition, Stress-Strain Curve, The materials stress-strain curve gives its stress-strain relationship. In a stress-strain curve, the stress and its corresponding strain values are plotted. An example of a stress-strain curve is given below. Explaining Stress-Strain Graph. The different regions in the stress-strain diagram are: Stress Strain RelationshipsThis relationship is known as Hookes Law and was first recorded by Robert Hooke, an English mathematician, in 1678. Note:Hookes Law describes only the initial linear portion of the stress-strain curve for a bar subjected to uniaxial extension. The slope of the straight-line portion of the stress-strain diagram is called the Modulus Stress-Strain Curve Diagram, Basic - MechstudiesDec 14, 2020 · Stress-Strain Curve is explained along with detailed diagram and explanation of each point in the curve to better understanding. Here, we will learn the curve, along with a diagram, stress strain diagram of many material, their characteristics. Stress-Strain Curve How to Read the Graph?Jun 03, 2020 · The stress-strain relationship deviates from Hookes law. The strain increases at a faster rate than stress which manifests itself as a mild flattening of the curve in the stress and strain graph. This is the part of the graph where the first curve starts but has not yet taken a turn downwards. Stress and Strain-Definition, Curve or Diagram, Stress and Strain Curves or Diagram:This curve is a behavior of the material when it is subjected to load. The stress-strain curve depends on two types of material. 1. Ductile Material:Ductile materials are materials that can be plastically twisted with no crack. They have the tendency to hold the deformation that occurs in the plastic region. Stress and Strain Elasticitystress Normal Strain In an axially loaded member, normal strain, is the change in the length, with respect to the original length, L. It is UNITLESS, but may be called strain or microstrain ( µ). Shearing Strain In a member loaded with shear forces, shear strain, is the change in the sheared side, s with Stress-Strain Curve - an overview ScienceDirect TopicsThe stressstrain curve is the most reliable and complete source for the evaluation of mechanical properties of any fibre. The stressstrain curve is produced by plotting the applied stress on the fibre axis and the elongation produced due it. The stressstrain curve of a model fibre is shown in Fig. 3.1. Different types of fibre produce 3. BEAMS:STRAIN, STRESS, DEFLECTIONS The beam, or Normal Stress:Having derived the proportionality relation for strain, x, in the x-direction, the variation of stress, x, in the x-direction can be found by substituting for in Eqs. 3.3 or 3.7. In the elastic range and for most materials uniaxial tensile and compressive stress-strain curves are identical. If Young's modulus on UTM (Theory) :Mechanics of Solids Sep 29, 2021 · The slope of the Stress strain curve in elastic region is called Young's Modulus. The Stress-strain diagram. The stress strain relationship of any material is of primary importance as it gives a good idea of the mechanical behaviour of the material in real life conditions. This is generally accomplished using the tension-compression tests. Tensile/Tension Test Advanced Topics The shape of the stress-strain curve shows how much increase in strength can be gained by strain hardening. For this purpose, the true-stress vs. true-strain curve is more useful than the conventional engineering stress-strain diagram. The eion defines the strain hardening coefficient n. 8 Stress Strain Curve Explanation Stages Mild Steel Sep 14, 2014 · Stress strain curve is a behavior of material when it is subjected to load. In this diagram stresses are plotted along the vertical axis and as a result of these stresses, corresponding strains are plotted along the horizontal axis. As shown below in the stress strain curve. From the diagram one can see the different mark points on the curve. Stress-strain Diagram Strength of Materials Review at The maximum ordinate in the stress-strain diagram is the ultimate strength or tensile strength. Rapture Strength. Rapture strength is the strength of the material at rupture. This is also known as the breaking strength. Modulus of Resilience. Modulus of resilience is the work done on a unit volume of material as the force is gradually increased What is Stress-strain Curve - Stress-strain Diagram In other words, stress and strain follows Hookes law. Beyond the linear region, stress and strain show nonlinear behavior. This inelastic behavior is called plastic deformation. A schematic diagram for the stress-strain curve of low carbon steel at room temperature is shown in the figure. There are several stages showing different behaviors Mechanical Properties of Materials MechaniCalcwhere and are the true stress and strain, and and are the engineering stress and strain.. Hooke's Law. Below the proportionality limit of the stress-strain curve, the relationship between stress and strain is linear.The slope of this linear portion of the stress-strain curve is the elastic modulus, E, also referred to as the Young's modulus and the modulus of elasticity. Stress, Strain and Hooke's Law - Lesson - TeachEngineeringOct 05, 2021 · The relationship used most readily today is the direct proportionality between stress and strain. Together, civil engineers, mechanical engineers and material scientists, carefully select structural materials that are able to safely endure everyday stress while remaining in the elastic region of the stress-strain curve, otherwise permanent Stress and strain:Mechanical properties of materialsMar 08, 2019 · This stress-strain relationship is known as Hookes Law, and in this region, the slope of the stress-strain curve is referred to as the modulus of elasticity (aka Youngs modulus), denoted E. The modulus of elasticity is essentially a measure of stiffness and is one of the factors used to calculate a materials deflection under load. MECHANICS OF CHAPTER 2MATERIALS - IIT BombayStress-Strain Relationship:Hookes Law Modulus of Elasticity (E) (aka Youngs Modulus). It is the ratio of normal stress to normal strain (i.e., measure of resistance to elastic deformation), evaluated below the proportional limit, i.e., slope of the straight-line portion of the Fatigue Analysis and Design:Theory5.1 Stress-Life(S-N) Diagram. 5.1.1 Stress-Life (S-N) Method Using the Goodman relationship, determine the life of the component. (Sol) (1) To determine the stress amplitude and mean stress Figure 6.5 Cyclic stress-strain curve obtained by connecting tips of stabilized hysteresis loops. 6.3 Stress-plastic strain power law relation: Stress strain relationship for concrete n stress strain Dec 26, 2017 · Stress Strain relationship for steel. Stress Strain relationship for steel :The stress-strain curve for mild steel, Fe 415 and Fe 500 are shown in Fig. 4.2 and 4.3. For mild steel, the value of characteristic stress is taken as yield stress and the design curve is obtained after applying a factor of safety of 1.15 to yield stress. 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## New answers tagged dynamic-programming
0
Consider the first three houses (assuming $n \geq 3$). Any solution does not rob one of them, say the first one (you can enumerate over all three options). The remaining houses form a street, which enables us to solve the problem using a simple dynamic programming: for each $0 \leq i \leq 2$ and $j$, we find out the best solution for the first $j$ houses ...
1
Your attempt sounds specious since your proof does not use any specific property of the objective function, $S = \sum_{i=2}^{n}| A[i] - A[i-1]|$. Had your proof been correct, there would have been an optimal substructure for any arbitrary objective function we could have defined for $A$, which does not make sense. To prove the existence of an optimal ...
0
I think the answer is: “You write code to do it”. You have a solution for a similar problem, so you work to understand it, and then you modify it to do what you want.
1
Let $\langle o_1, o_2, o_{m+1}\rangle$ be a non-empty optimal sequence of operations that distributes the chocolates evenly, i.e., such that the final distribution is $[k,k,k, \dots, k]$. Let $\alpha_i$ be the number of candies given to the $i$-th person by operation $o_{m+1}$ and consider the subsequence of operations $\langle o_1, o_2, \dots, o_m\rangle$. ...
2
The essence of dynamic programming is "it is easier to solve many problems than to solve one problem.". Sometimes, the more problems the easier. Sometimes, it is impossible with less problems. The approach of dynamic programming is finding/inventing many problems that are similar to each other, solving these similar problems in some order so that ...
0
You may be interested in this implementation: https://github.com/fujimotos/polyleven
1
Congratulation for solving a popular problem independently with a perspective that is rarely seen. The choice between "ending at $i$" and "starting at $i$" can be dismissed as insignificant, coincidental, and random. They are symmetric to each other. A similar question could be why we read, write and think from left to right, as shown on ...
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https://kr.mathworks.com/help/vdynblks/ref/dynamicsteering.html | 1,713,005,982,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00114.warc.gz | 311,645,294 | 29,871 | # Dynamic Steering
(Removed) Dynamic steering for Ackerman, rack-and-pinion, and parallel steering mechanisms
Libraries:
Vehicle Dynamics Blockset / Steering
## Description
The Dynamic Steering block implements dynamic steering to calculate the wheel angles for Ackerman, rack-and-pinion, and parallel steering mechanisms. The block uses the steering wheel input torque, right wheel torque, and left wheel torque to calculate the wheel angles. The block uses the vehicle coordinate system.
If you select Power assist, you can specify a torque assist lookup table that is a function of the vehicle speed and steering wheel input torque. The block uses the steering wheel input torque and torque assist to calculate the steering dynamics.
To specify the steering type, use the Type parameter.
SettingBlock Implementation
`Ackerman`
Ideal Ackerman steering. Wheel angles have a common turning circle center.
`Rack and pinion`
Ideal rack-and-pinion steering. Gears convert the steering rotation into linear motion.
`Parallel`
Parallel steering. Wheel angles are equal.
To specify the type of data for the steering mechanism, use the Parametrized by parameter.
SettingBlock Implementation
`Constant`
Steering mechanism uses constant parameter data.
`Lookup table`
Steering mechanism implements tables for parameter data.
Use the parameter to specify whether the front or rear axle is steered.
SettingImplementation
Front axle steering
Rear axle steering
### Dynamics
To calculate the steering dynamics, the Dynamic Steering block models the steering wheel, shaft, steering mechanism, hysteresis, and, optionally, power assist.
CalculationEquations
Steering column and steering shaft dynamics
`${J}_{1}{\stackrel{¨}{\theta }}_{1}={\tau }_{in}-{b}_{2}{\stackrel{˙}{\theta }}_{1}-{\tau }_{hys}$`
`${J}_{2}{\stackrel{¨}{\theta }}_{2}={\tau }_{eq}-{b}_{3}{\stackrel{˙}{\theta }}_{2}+{\tau }_{hys}-{\tau }_{fric}$`
Hysteresis spring damper
Optional power assist
`$\begin{array}{l}{\tau }_{ast}={f}_{trq}\left(v,{\tau }_{in}\right)\\ \\ {J}_{1}{\stackrel{¨}{\theta }}_{1}={\tau }_{in}+{\tau }_{ast}-{b}_{2}{\stackrel{˙}{\theta }}_{1}-{\tau }_{hys}\\ {J}_{2}{\stackrel{¨}{\theta }}_{2}={\tau }_{eq}+{\tau }_{ast}-{b}_{3}{\stackrel{˙}{\theta }}_{2}+{\tau }_{hys}-{\tau }_{fric}\end{array}$`
The illustration and equations use these variables.
J1 Steering wheel inertia J2 Steering mechanism inertia ${\theta }_{1},{\stackrel{˙}{\theta }}_{1},{\stackrel{¨}{\theta }}_{1}$ Steering wheel angle, angular velocity, and angular acceleration, respectively ${\theta }_{2},{\stackrel{˙}{\theta }}_{2},{\stackrel{¨}{\theta }}_{2}$ Shaft angle, angular velocity, and angular acceleration, respectively b1, k1 Hysteresis spring and viscous damping coefficients, respectively b2 Steering wheel viscous damping coefficient b3 Steering mechanism damping coefficient τhys Hysteresis spring damping torque τfric Steering mechanism friction torque τeq Wheel equivalent torque τast Torque assist βu , βl Upper and lower hysteresis modifiers, respectively v Vehicle speed ƒtrq Torque assist lookup table
### Steering Types
Ackermann Steering
For 100% (ideal) Ackermann steering, all wheels follow circular arcs with the same center point.
To calculate the steered wheel angles, the Steering System block uses these equations:
`$\begin{array}{l}\mathrm{cot}\left({\delta }_{L}\right)-\mathrm{cot}\left({\delta }_{R}\right)=\frac{TW}{WB}\\ \\ {\delta }_{Ack}=\frac{{\delta }_{in}}{\gamma }\\ \\ {\delta }_{L}={\mathrm{tan}}^{-1}\left(\frac{WB\mathrm{tan}\left({\delta }_{Ack}\right)}{WB+0.5TW\mathrm{tan}\left({\delta }_{Ack}\right)}\right)\\ {\delta }_{R}={\mathrm{tan}}^{-1}\left(\frac{WB\mathrm{tan}\left({\delta }_{Ack}\right)}{WB-0.5TW\mathrm{tan}\left({\delta }_{Ack}\right)}\right)\end{array}$`
This table defines variables used in the equations:
δin Pinion angle (steering shaft angle into pinion) δL Left wheel steer angle δR Right wheel steer angle δAck Ackermann steer angle TW Track width WB Wheel base γ Steering ratio: Ratio of pinion angle to Ackermann angle
Rack-and-Pinion
For rack-and-pinion steering, pinion rotation causes linear motion of the rack, which steers the wheels through the tie rods and steering arms.
To calculate the steered wheel angles, the block uses these equations.
`$\begin{array}{l}{l}_{1}=\frac{TW-{l}_{rack}}{2}-\Delta P\\ \\ {l}_{2}{}^{2}={l}_{1}{}^{2}+{D}^{2}\\ \\ \Delta P=r{\delta }_{in}\\ \\ \beta =\frac{\pi }{2}-{\mathrm{tan}}^{-1}\left[\frac{D}{{l}_{1}}\right]-{\mathrm{cos}}^{-1}\left[\frac{{l}_{arm}{}^{2}+{l}_{2}{}^{2}-{l}_{rod}{}^{2}}{2{l}_{arm}{l}_{2}}\right]\end{array}$`
The illustration and equations use these variables.
δin Pinion angle (steering shaft angle into pinion) δL Left wheel steer angle δR Right wheel steer angle TW Track width r Pinion radius ΔP Linear change in rack position from "straight ahead" position D Longitudinal distance between rack and steered axle lrack Rack length (distance between inner tie-rod ends) larm Steering arm length lrod Tie rod length
Parallel
For parallel steering, the wheel angles are equal.
To calculate the steering angles, the block uses this equation.
`${\delta }_{R}={\delta }_{L}=\frac{{\delta }_{in}}{\gamma }$`
The illustration and equations use these variables.
δin Steering wheel angle δL Left wheel angle δR Right wheel angle γ Steering ratio
## Ports
### Input
expand all
Torque, τin, in N·m.
Left wheel torque, τL, in N·m.
Right wheel torque, τR, in N·m.
Vehicle speed, v, in m/s.
#### Dependencies
To create a `VehSpd` port, select Power assist.
### Output
expand all
Bus signal contains these block calculations.
SignalDescriptionUnit
`StrgWhlAng`
Steering wheel angle
`StrgWhlSpd`
Steering wheel angular velocity
`ShftAng`
Shaft angle
`ShftSpd`
Shaft angular velocity
`AngLft`
Left wheel angle
`SpdLft`
Left wheel angular velocity
`AngRght`
Right wheel angle
`SpdRght`
Right wheel angular velocity
`TrqAst`
Torque assist
N·m
`PwrAst`
Power assist
W
`PwrLoss`
Power loss
W
`InstStrgRatio`
Instantaneous steering ratio
NA
Left wheel angle, δL, in rad.
Right wheel angle, δR, in rad.
## Parameters
expand all
To specify the steering type, use the Type parameter.
SettingBlock Implementation
`Ackerman`
Ideal Ackerman steering. Wheel angles have a common turning circle center.
`Rack and pinion`
Ideal rack-and-pinion steering. Gears convert the steering rotation into linear motion.
`Parallel`
Parallel steering. Wheel angles are equal.
#### Dependencies
This table summarizes the Type and Parametrized by parameter dependencies.
TypeParameterized ByCreates Parameters
`Ackerman`
`Constant`
Track width, TrckWdth
Wheel base, WhlBase
Steering range, StrgRng
Steering ratio, StrgRatio
`Lookup table`
Track width, TrckWdth
Wheel base, WhlBase
Steering range, StrgRng
Steering angle breakpoints, StrgAngBpts
Steering ratio table, StrgRatioTbl
`Rack and pinion`
`Constant`
Track width, TrckWdth
Steering range, StrgRng
Steering arm length, StrgArmLngth
Rack casing length, RckCsLngth
Tie rod length, TieRodLngth
Distance between front axis and rack, D
`Lookup table`
Track width, TrckWdth
Steering range, StrgRng
Steering angle breakpoints, StrgAngBpts
Steering arm length, StrgArmLngth
Rack casing length, RckCsLngth
Tie rod length, TieRodLngth
Distance between front axis and rack, D
`Parallel``Constant`
Steering range, StrgRng
Steering ratio, StrgRatio
`Lookup table`
Steering range, StrgRng
Steering angle breakpoints, StrgAngBpts
Steering ratio table, StrgRatioTbl
To specify the type of data for the steering mechanism, use the Parametrized by parameter.
SettingBlock Implementation
`Constant`
Steering mechanism uses constant parameter data.
`Lookup table`
Steering mechanism implements tables for parameter data.
#### Dependencies
This table summarizes the Type and Parametrized by parameter dependencies.
TypeParameterized ByCreates Parameters
`Ackerman`
`Constant`
Track width, TrckWdth
Wheel base, WhlBase
Steering range, StrgRng
Steering ratio, StrgRatio
`Lookup table`
Track width, TrckWdth
Wheel base, WhlBase
Steering range, StrgRng
Steering angle breakpoints, StrgAngBpts
Steering ratio table, StrgRatioTbl
`Rack and pinion`
`Constant`
Track width, TrckWdth
Steering range, StrgRng
Steering arm length, StrgArmLngth
Rack casing length, RckCsLngth
Tie rod length, TieRodLngth
Distance between front axis and rack, D
`Lookup table`
Track width, TrckWdth
Steering range, StrgRng
Steering angle breakpoints, StrgAngBpts
Steering arm length, StrgArmLngth
Rack casing length, RckCsLngth
Tie rod length, TieRodLngth
Distance between front axis and rack, D
`Parallel``Constant`
Steering range, StrgRng
Steering ratio, StrgRatio
`Lookup table`
Steering range, StrgRng
Steering angle breakpoints, StrgAngBpts
Steering ratio table, StrgRatioTbl
If you select Power assist, you can specify a torque assist lookup table, ƒtrq, that is a function of the vehicle speed, v, and steering wheel input torque, τin.
`${\tau }_{ast}={f}_{trq}\left(v,{\tau }_{in}\right)$`
The block uses the steering wheel input torque and torque assist to calculate the steering dynamics.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
Use the parameter to specify whether the front or rear axle is steered.
SettingImplementation
Front axle steering
Rear axle steering
General
Track width, TW, in m.
#### Dependencies
To create this parameter, set Type to `Ackerman` or ```Rack and pinion```.
Wheel base, WB, in m.
#### Dependencies
To create this parameter, set Type to `Ackerman`.
Steering range, in rad. The block limits the wheel angles to remain within the steering range.
Steering ratio, γ, dimensionless.
#### Dependencies
To create this parameter:
• Set Type to `Ackerman` or `Parallel`.
• Set Parametrized by to `Constant`.
#### Dependencies
To create this parameter, set Parametrized by to `Lookup table`.
Steering ratio table, γ, dimensionless.
#### Dependencies
To create this parameter:
• Set Type to `Ackerman` or `Parallel`.
• Set Parametrized by to ```Lookup table```.
Rack-and-Pinion
Steering arm length, larm, in m.
#### Dependencies
To create this parameter, set Type to ```Rack and pinion```.
Rack casing length, lrack, in m.
#### Dependencies
To create this parameter, set Type to ```Rack and pinion```.
Tie rod length, lrod, in m.
#### Dependencies
To create this parameter, set Type to ```Rack and pinion```.
Distance between axis and rack, D, in m.
#### Dependencies
To create this parameter, set Type to ```Rack and pinion```.
#### Dependencies
To create this parameter:
• Set Type to ```Rack and pinion```.
• Set Parametrized by to `Constant`.
Pinion radius table, r, in m.
#### Dependencies
To create this parameter:
• Set Type to ```Rack and pinion```.
• Set Parametrized by to ```Lookup table```.
Dynamics
Steering wheel inertia, J1, in kg*m^2.
Steering mechanism inertia, J2, in kg*m^2.
Upper hysteresis modifier, βu, dimensionless.
Lower hysteresis modifier, βl, dimensionless.
Steering wheel damping, b2, in N·m·s/rad.
Steering mechanism damping, b3, in N·m·s/rad.
Initial steering angle, θ0, in rad.
Initial steering angular velocity, ωo, in rad/s.
Friction torque, τfric, in N·m.
Power Assist
Steering wheel torque breakpoints, in N·m.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
Vehicle speed breakpoints, in m/s.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
Assisting torque table, ƒtrq, in N·m.
The torque assist lookup table is a function of the vehicle speed, v, and steering wheel input torque, τin.
`${\tau }_{ast}={f}_{trq}\left(v,{\tau }_{in}\right)$`
The block uses the steering wheel input torque and torque assist to calculate the steering dynamics.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
Assisting torque limit, in N·m.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
Assisting power limit, in N·m/s.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
Assisting torque efficiency, dimensionless.
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
#### Dependencies
Selecting Power assist creates the `VehSpd` input port and these parameters.
Power AssistParameters
`on`
Steering wheel torque breakpoints, TrqBpts
Vehicle speed breakpoints, VehSpdBpts
Assisting torque table, TrqTbl
Assisting torque limit, TrqLmt
Assisting power limit, PwrLmt
Assisting torque efficiency, Eta
Cutoff frequency, omega_c
## References
[1] Crolla, David, David Foster, et al. Encyclopedia of Automotive Engineering. Volume 4, Part 5 (Chassis Systems) and Part 6 (Electrical and Electronic Systems). Chichester, West Sussex, United Kingdom: John Wiley & Sons Ltd, 2015.
[2] Gillespie, Thomas. Fundamentals of Vehicle Dynamics. Warrendale, PA: Society of Automotive Engineers, 1992.
[3] Vehicle Dynamics Standards Committee. Vehicle Dynamics Terminology. SAE J670. Warrendale, PA: Society of Automotive Engineers, 2008.
## Version History
Introduced in R2018a
expand all
### R2024a: Removed
The Dynamic Steering is removed. Use the Steering System block instead. | 4,099 | 15,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.602497 |
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## #1 2013-10-13 10:21:46
Al-Allo
Member
Registered: 2012-08-23
Posts: 324
Hi, I have a little question concerning contradictions :
If I have a statement "A" that I want to prove, and only have the possibility for it to be True or False.
After some manipulations, I arrive at some contradiction. (Here's where my question begins.)
How can we know that a contradiction is enough to be sure at 100 % that a statement is not correct?
Is it because in Mathematics, for a thing to be True or False, it must always be ALWAYS "working" without arriving at some contradiction ? (Mathematical ideas must always work, and not sometimes yes, sometimes no.)
I just want to be sure of thinking of it in the right way, corrections would be greatly appreciated ! Thank you !
Offline
## #2 2013-10-13 10:28:03
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
A contradiction is by definition a statement that is always false. If by manipulating you find that a statement is equivalent to some contradiction, it must be equivalent to false and thus itself is false.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #3 2013-10-13 19:27:32
bob bundy
Registered: 2010-06-20
Posts: 8,340
hi Al-Allo
In mathematics, if you start with a TRUE statement and use correct working to obtain another statement, then that statement must also be TRUE.
So, let's assume A is FALSE. If we can do correct working that leads to a statement that is FALSE, then the assumption must have been incorrect. That's how we know A is TRUE.
Simple example. Let's use reductio ad absurdum to prove that for any integer, n, 2n is EVEN.
A = "2n is EVEN, where n is an integer"
Assume this is FALSE. ie. 2n is not EVEN.
This means that 2n is ODD.
This means that 2n + 1 is EVEN, so it can be divided by 2, giving an integer result.
(2n + 1)/2 = n + 1/2. This is an integer.
Subtract n (an integer) and the result is also an integer
This means that 1/2 is an integer.
This result is FALSE so the assumption was incorrect.
Therefore 2n is NOT ODD so it is EVEN.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #4 2013-10-14 02:58:25
Al-Allo
Member
Registered: 2012-08-23
Posts: 324
AH thank you ! Last question, a real verity will never make any contradictions, right ? (What ever manipulations you do with it )
Last edited by Al-Allo (2013-10-14 03:00:30)
Offline
## #5 2013-10-14 05:10:22
bob bundy
Registered: 2010-06-20
Posts: 8,340
(What ever manipulations you do with it )
Say: "What ever correct manipulations you do with it " and I'm happy.
In place of "correct" you could also say "valid".
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #6 2013-10-14 06:31:10
Al-Allo
Member
Registered: 2012-08-23
Posts: 324
bob bundy wrote:
(What ever manipulations you do with it )
Say: "What ever correct manipulations you do with it " and I'm happy.
In place of "correct" you could also say "valid".
Bob
AH yes, sorry for the mistake. So yes, any valid manipulation !
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Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Roman stadium = 184.7088 length (length) ``` Related Measurements: Try converting from "stadium" to actus (Roman actus), barleycorn, Biblical cubit, bottom measure, cable length, cloth quarter, digitus (Roman digitus), finger, li (Chinese li), m (meter), mile, nail (cloth nail), nautical mile, pace, palm, parsec, rod (surveyors rod), Roman foot, soccer field, verst (Russian verst), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: stadium = 21,816 barleycorn, .84166667 cable length, 9.18 chain (surveyors chain), 808 cloth quarter, 161.6 ell, 606 foot, 399.12 Greek cubit, .03825758 league, 1.95E-14 light yr (light year), 918.18 link (surveyors link), 7,272,000 mil, 2,424 palm, 525,547.45 point (typography point), 415.54 Roman cubit, 30.3 rope, 808 span (cloth span), 6,095.56 sun (Japanese sun), 606 survey foot, .17314286 verst (Russian verst), 202 yard.
Feedback, suggestions, or additional measurement definitions?
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Type: Posts; User: Ahmedabad; Keyword(s):
1. ## Thread: Greetings from Australia!
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### Hi friend, Welcome to this board. Share your...
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2. ## Thread: Introduction
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### Welcome to this board. Nice board with good...
Welcome to this board.
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3. ## Thread: HI
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### Welcome to this forum. Give your ideas here and...
Welcome to this forum.
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It is good forum for those who want job and go for higher education also.
4. ## Thread: Which number is lucky for you ?
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### Which number is lucky for you ?
I like Number "9" which is my lucky number.
Can you tell me your lucky number ?
5. ## Thread: Good site for Software Testing
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### The Anser is D 31 is correct. Nice question here.
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Results 1 to 12 of 12 | 733 | 2,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-05 | latest | en | 0.916068 |
https://accountingcoaching.online/correlation-coefficient-introduction-to-statistics/ | 1,701,706,584,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00491.warc.gz | 111,585,190 | 9,950 | # Correlation Coefficient Introduction to Statistics
A perfect positive correlation
has a coefficient of 1.0; a perfect negative correlation has a coefficient
of -1.0. When there is no association between two variables, the correlation
coefficient has a value of 0. A correlation coefficient of +1 indicates a perfect positive correlation. A correlation coefficient of -1 indicates a perfect negative correlation. Standard deviation is a measure of the dispersion of data from its average. Covariance shows whether the two variables tend to move in the same direction, while the correlation coefficient measures the strength of that relationship on a normalized scale, from -1 to 1.
• These figures are clearly more volatile than the balanced portfolio’s returns of 6.4% and 0.2%.
• In the “non-parametric” bootstrap, n pairs (xi, yi) are resampled “with replacement” from the observed set of n pairs, and the correlation coefficient r is calculated based on the resampled data.
• The values of -1 (for a negative correlation) and 1 (for a positive one) describe perfect fits in which all data points align in a straight line, indicating that the variables are perfectly correlated.
• Those relationships can be analyzed using nonparametric methods, such as Spearman’s correlation coefficient, the Kendall rank correlation coefficient, or a polychoric correlation coefficient.
Note also in the plot above that there are two individuals with apparent heights of 88 and 99 inches. A height of 88 inches (7 feet 3 inches) is plausible, but unlikely, and a height of 99 inches is certainly a coding error. Obvious coding errors should be excluded from the analysis, since they can have an inordinate effect on the results. It’s always a good idea to look at the raw data in order to identify any gross mistakes in coding.
## You are unable to access statisticsbyjim.com
Generally, the closer a correlation coefficient is to 1.0 (or -1.0) the stronger the relationship between the two variable is said to be. While there is no clear boundary to what makes a “strong” correlation, a coefficient above 0.75 (or below -0.75) is considered a high degree of correlation, while one between -0.3 and 0.3 is a sign of weak or no correlation. In experimental science, researchers will sometimes repeat the same study to see if a high degree of correlation can be reproduced.
Remember, if r doesn’t show on your calculator, then diagnostics need to be turned on. This is also the same place on the calculator where you will find the linear regression equation and the coefficient of determination. Simple linear regression describes the linear relationship between a response variable (denoted by y) and an explanatory variable (denoted by x) using a statistical model. When it comes to investing, a negative correlation does not necessarily mean that the securities should be avoided.
The correlation coefficient does not describe the slope of the line of best fit; the slope can be determined with the least squares method in regression analysis. More generally, (Xi − X)(Yi − Y) is positive if and only if Xi and Yi lie on the same side of their respective means. Thus the correlation coefficient is positive if Xi and Yi tend to be simultaneously greater than, or simultaneously less than, their respective means. The correlation coefficient is negative (anti-correlation) if Xi and Yi tend to lie on opposite sides of their respective means.
## Decorrelation of n random variables
Conversely, if the value is less than zero, it is a negative relationship. A value of zero indicates that there is no relationship between the two variables. To interpret the correlation coefficient,
we must consider both its sign (positive or negative)
and its absolute value.
• The correlation coefficient is the specific measure that quantifies the strength of the linear relationship between two variables in a correlation analysis.
• Ice Cream Sales and Temperature are therefore the two variables which we’ll use to calculate the correlation coefficient.
• Calculating the correlation coefficient for these variables based on market data reveals a moderate and inconsistent correlation over lengthy periods.
• The correlation coefficient is calculated by determining the covariance of the variables and dividing that number by the product of those variables’ standard deviations.
The correlation coefficient can help investors diversify their portfolio by including a mix of investments that have a negative, or low, correlation to the stock market. In short, when reducing volatility risk in a portfolio, sometimes opposites do attract. For correlation coefficients derived from sampling, the determination of statistical significance depends on the p-value, which is calculated from the data sample’s size as well as the value of the coefficient. To perform the permutation test, repeat steps (1) and (2) a large number of times. The p-value for the permutation test is the proportion of the r values generated in step (2) that are larger than the Pearson correlation coefficient that was calculated from the original data.
## Pearson’s distance
For example, assume you have a \$100,000 balanced portfolio that is invested 60% in stocks and 40% in bonds. In a year of strong economic performance, the stock component of your portfolio might generate a return of 12% while the bond component may return -2% because interest rates are rising (which means that bond prices are falling). When interpreting correlation, it’s important to remember that just because two variables are correlated, it does not mean that one causes the other. Correlation does not imply causation, as the saying goes, and the Pearson coefficient cannot determine whether one of the correlated variables is dependent on the other. This decorrelation is related to principal components analysis for multivariate data.
If the correlation coefficient of two variables is zero, there is no linear relationship between the variables. Two variables can have a strong relationship but a weak correlation coefficient if the relationship between them is nonlinear. When the value of ρ is close to zero, generally between -0.1 and +0.1, the variables are said to have no linear relationship (or a very weak linear relationship). This article explains the significance of linear correlation coefficients for investors, how to calculate covariance for stocks, and how investors can use correlation to predict the market. Many relationships between measurement variables are reasonably linear, but others are not For example, the image below indicates that the risk of death is not linearly correlated with body mass index. The relationship between alcohol consumption and mortality is also “J-shaped.”
Correlation combines statistical concepts, namely, variance and standard deviation. Variance is the dispersion of a variable around the mean, and standard deviation is the square root of variance. Inspection of the scatterplot between X and Y will typically reveal a situation where lack of robustness might be an issue, and in such cases it may be advisable to use a robust measure of association. Note however that while most robust estimators of association measure statistical dependence in some way, they are generally not interpretable on the same scale as the Pearson correlation coefficient. In fact, it’s important to remember that relying exclusively on the correlation coefficient can be misleading—particularly in situations involving curvilinear relationships or extreme outliers. In the scatterplots below, we are reminded that a correlation coefficient of zero or near zero does not necessarily mean that there is no relationship between the variables; it simply means that there is no linear relationship.
Here “larger” can mean either that the value is larger in magnitude, or larger in signed value, depending on whether a two-sided or one-sided test is desired. Correlation only looks at the two variables at hand and won’t give insight into relationships beyond the bivariate data. This test won’t detect (and therefore will be skewed by) outliers in the data and can’t properly detect curvilinear relationships. If you don’t do this, r (the correlation coefficient) will not show up when you run the linear regression function.
## Correlation Coefficients: Positive, Negative, and Zero
Correlation combines several important and related statistical concepts, namely, variance and standard deviation. When both variables are dichotomous instead of ordered-categorical, the polychoric correlation coefficient is called the tetrachoric correlation coefficient. In this section, we’re focusing on the Pearson product-moment correlation.
Standard deviation is a measure of the dispersion of data from its average. The normalized version of the statistic is calculated by dividing covariance by the product of the two standard deviations. The correlation coefficient is a statistical measure of the strength of a linear relationship between two variables. A correlation coefficient of -1 describes a perfect negative, or inverse, correlation, with values in one series rising as those in the other decline, and vice versa. A coefficient of 1 shows a perfect positive correlation, or a direct relationship.
For example, since high oil prices are favorable for crude producers, one might assume the correlation between oil prices and forward returns on oil stocks is strongly positive. Calculating the correlation coefficient for these variables based on market data reveals a moderate and inconsistent correlation over lengthy periods. To calculate the Pearson correlation, start by determining each variable’s standard deviation as well as the covariance between them. The correlation coefficient is covariance divided by the product of the two variables’ standard deviations. In finance, for example, correlation is used in several analyses including the calculation of portfolio standard deviation. Because it is so time-consuming, correlation is best calculated using software like Excel.
Conversely, when two stocks move in opposite directions, the correlation coefficient is negative. Correlation coefficients play a key role in portfolio risk assessments and quantitative trading strategies. For example, some portfolio managers will monitor the correlation coefficients of their holdings to limit a portfolio’s volatility and risk.
A linear correlation coefficient that is greater than zero indicates a positive relationship. Finally, a value of zero indicates no relationship between the two variables. The correlation coefficient is calculated by determining the covariance of the variables and dividing that number by the product of those variables’ standard deviations. A correlation coefficient is a statistical measure of the degree to which changes to the value of one variable predict change to the value of another. In positively correlated variables, the value increases or decreases in tandem.
We start to answer this question by gathering data on average daily ice cream sales and the highest daily temperature. Ice Cream Sales and Temperature are therefore the two variables which we’ll use to calculate the correlation coefficient. Sometimes data like these are called bivariate data, because each observation (or point in time at which we’ve measured both sales and temperature) has two pieces of information that we can use to describe it. In other words, we’re asking whether Ice Cream Sales and Temperature seem to move together.
Some probability distributions, such as the Cauchy distribution, have undefined variance and hence ρ is not defined if X or Y follows such a distribution. In some practical applications, such as those involving data suspected to follow a heavy-tailed distribution, this is an important consideration. However, the existence of the correlation coefficient is usually not a concern; for instance, if the range of the distribution is bounded, ρ is always defined. Similarly, looking at a scatterplot can provide insights on how outliers—unusual observations in our data—can skew the correlation coefficient.
However the standard versions of these approaches rely on exchangeability of the data, meaning that there is no ordering or grouping of the data pairs being analyzed that might affect the behavior of the correlation estimate. The bootstrap can be used to construct confidence intervals for Pearson’s correlation coefficient. In the “non-parametric” bootstrap, n pairs (xi, yi) are resampled “with replacement” from the observed set of n pairs, and the correlation coefficient r is calculated based on the resampled data. | 2,397 | 12,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.905514 |
https://jp.mathworks.com/matlabcentral/answers/769992-sort-column-based-on-value-in-another-column-sort-rows-by-column-values-a-b-c?s_tid=prof_contriblnk | 1,679,844,705,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00279.warc.gz | 386,225,139 | 28,002 | # Sort column based on value in another column (sort rows by column values a>b>c)
11 ビュー (過去 30 日間)
Nathan Garcia 2021 年 3 月 11 日
コメント済み: Nathan Garcia 2021 年 4 月 1 日
I am using the function [E,index] = sortrows(A,'descend');
This sorts the first column by row value and also secondarily sorts the second column. I would then like to specify how the sort works based on relative values in a third column. For example, I am only interested in rows where column A value > column B value > column C value. I am interested in the index that identifies the rows that fit this definition. How can I set this up?
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### 採用された回答
Pranav Verma 2021 年 3 月 15 日
Hi Nathan,
From the question I understand that after performing the usual sorting operation on the table, you want the rows which follow the definition of "value of column A > column B > column C".
You can start by sorting the rows in whichever order you want them to be sorted (ascending / descending). After that you can use the find function in MATLAB and get the row numbers which follow the definition provided.
For eg;,
% a = 3 2 1
% 2 3 4
% 5 4 3
% 1 2 3
a = [3 2 1 ; 2 3 4 ; 5 4 3 ; 1 2 3];
% value of column A > column B > column C
b = find(a(:,1) > a(:,2) & a(:,2) > a(:,3));
% b =
% 1
% 3
Here we see that row 1 and 3 follow the given condition specified inside the find function and hence it returns the row numbers of these two rows.
Hope this helps!
Thanks
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Nathan Garcia 2021 年 4 月 1 日
Thank you! This is very helpful.
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Translated by | 508 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-14 | longest | en | 0.529072 |
https://www.jobilize.com/course/section/to-do-research-on-the-fins-of-a-rocket-by-openstax?qcr=www.quizover.com | 1,550,256,065,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479101.30/warc/CC-MAIN-20190215183319-20190215205319-00014.warc.gz | 844,959,091 | 20,337 | # 1.2 Making a rocket
Page 1 / 2
## Making a rocket
Design and realise
Background
A rocket relies on jet propulsion in order to fly. When a rocket uses its fuel, a current of warm air is exhausted at its tail causing the rocket to move forward. Planes that fly at lower than 25 km use the oxygen in the atmosphere to ignite (burn up) their fuel.
Rockets have to carry oxygen with them, because above 25 km from the earth’s surface there is not enough oxygen.
M ake a Rocket
Background
We are now going to make a rocket that will use jet propulsion in order to move forward. The speed with which a rocket moves forward depends on the amount of propellant gas that is exhausted at the rear end. Water is a much better propellant than warm gas, because it is much heavier. We are going to use compressed water and air and will make observations regarding how high/far your rocket can fly.
## [lo 1.10]
Requirements:
• Thin cardboard.
• A pen.
• A ruler.
• Coloured cardboard.
• A pair of scissors.
• Two empty.
• 2ℓ plastic bottles.
• Strong broad adhesive tape or masking tape.
• A funnel.
• A jug of water.
• A cork.
• An air valve (from a bicycle pump).
• Plastic tube.
• A bicycle pump.
• A plastic bank coin pouch.
• A protractor.
• Glue.
• A Stanley knife.
• A stiletto or a knitting-needle.
• A pair of compasses.
• Oil-based paint.
Background
A plane or rocket operates according to Newton’s third law of movement, namely if one source exerts a force on another, then the other source exerts the same amount of force on the first source, but in the opposite direction from the first source; i.e. for each action there is an opposite reaction.
If the force you exert on something is bigger than the resistance exerted by that object, the force that you exert can cause movement. The movement in the rocket is brought about by the rocket being pushed upward by the escaping gases that were generated as a result of the chemical reaction between the fuel and the oxygen that are burnt up in the combustion chambers. The large volume of gases escapes at a high speed through the rocket’s steel exhaust pipe.
There are five types of force that act upon a rocket, namely:
Thrust : the force provided by the engine, which is exhausted at the rear end and which pushes the rocket forward
Gravity : the force that pulls the rocket back to earth
Resistance : This is exerted by the air against the rocket, causing the rocket to reduce its speed. (Outside the atmosphere, there is no resistance, because there is no air.)
Lifting force : A rocket cannot be launched effectively and rise vertically unless the lifting force of its engines is greater than the weight of the rocket.
Relative wind : The air flows rapidly around the nose and down the body. Longer nose-pieces are used to get better airflow.
## [lo 1.7]
Background
A rocket needs fins in order to fly in a straight line. Do research about what the fins of a rocket look like and draw a few examples in the space provided. Also decide how many fins you would like to use and what size the fins will have to be in relation to the body section of the rocket. Also, decide what type of material will be the most suitable for making the fins. The position of the fins on the body is also extremely important.
#### Questions & Answers
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
Ramkumar Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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By By Rhodes By By By | 1,635 | 6,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-09 | latest | en | 0.927532 |
http://us.metamath.org/mpeuni/filss.html | 1,638,463,573,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00484.warc.gz | 84,505,237 | 6,146 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > filss Structured version Visualization version GIF version
Theorem filss 21567
Description: A filter is closed under taking supersets. (Contributed by FL, 20-Jul-2007.) (Revised by Stefan O'Rear, 28-Jul-2015.)
Assertion
Ref Expression
filss ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → 𝐵𝐹)
Proof of Theorem filss
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 isfil 21561 . . . 4 (𝐹 ∈ (Fil‘𝑋) ↔ (𝐹 ∈ (fBas‘𝑋) ∧ ∀𝑥 ∈ 𝒫 𝑋((𝐹 ∩ 𝒫 𝑥) ≠ ∅ → 𝑥𝐹)))
21simprbi 480 . . 3 (𝐹 ∈ (Fil‘𝑋) → ∀𝑥 ∈ 𝒫 𝑋((𝐹 ∩ 𝒫 𝑥) ≠ ∅ → 𝑥𝐹))
32adantr 481 . 2 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → ∀𝑥 ∈ 𝒫 𝑋((𝐹 ∩ 𝒫 𝑥) ≠ ∅ → 𝑥𝐹))
4 elfvdm 6177 . . 3 (𝐹 ∈ (Fil‘𝑋) → 𝑋 ∈ dom Fil)
5 simp2 1060 . . 3 ((𝐴𝐹𝐵𝑋𝐴𝐵) → 𝐵𝑋)
6 elpw2g 4787 . . . 4 (𝑋 ∈ dom Fil → (𝐵 ∈ 𝒫 𝑋𝐵𝑋))
76biimpar 502 . . 3 ((𝑋 ∈ dom Fil ∧ 𝐵𝑋) → 𝐵 ∈ 𝒫 𝑋)
84, 5, 7syl2an 494 . 2 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → 𝐵 ∈ 𝒫 𝑋)
9 simpr1 1065 . . 3 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → 𝐴𝐹)
10 simpr3 1067 . . . 4 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → 𝐴𝐵)
11 elpwg 4138 . . . . 5 (𝐴𝐹 → (𝐴 ∈ 𝒫 𝐵𝐴𝐵))
129, 11syl 17 . . . 4 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → (𝐴 ∈ 𝒫 𝐵𝐴𝐵))
1310, 12mpbird 247 . . 3 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → 𝐴 ∈ 𝒫 𝐵)
14 inelcm 4004 . . 3 ((𝐴𝐹𝐴 ∈ 𝒫 𝐵) → (𝐹 ∩ 𝒫 𝐵) ≠ ∅)
159, 13, 14syl2anc 692 . 2 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → (𝐹 ∩ 𝒫 𝐵) ≠ ∅)
16 pweq 4133 . . . . . 6 (𝑥 = 𝐵 → 𝒫 𝑥 = 𝒫 𝐵)
1716ineq2d 3792 . . . . 5 (𝑥 = 𝐵 → (𝐹 ∩ 𝒫 𝑥) = (𝐹 ∩ 𝒫 𝐵))
1817neeq1d 2849 . . . 4 (𝑥 = 𝐵 → ((𝐹 ∩ 𝒫 𝑥) ≠ ∅ ↔ (𝐹 ∩ 𝒫 𝐵) ≠ ∅))
19 eleq1 2686 . . . 4 (𝑥 = 𝐵 → (𝑥𝐹𝐵𝐹))
2018, 19imbi12d 334 . . 3 (𝑥 = 𝐵 → (((𝐹 ∩ 𝒫 𝑥) ≠ ∅ → 𝑥𝐹) ↔ ((𝐹 ∩ 𝒫 𝐵) ≠ ∅ → 𝐵𝐹)))
2120rspccv 3292 . 2 (∀𝑥 ∈ 𝒫 𝑋((𝐹 ∩ 𝒫 𝑥) ≠ ∅ → 𝑥𝐹) → (𝐵 ∈ 𝒫 𝑋 → ((𝐹 ∩ 𝒫 𝐵) ≠ ∅ → 𝐵𝐹)))
223, 8, 15, 21syl3c 66 1 ((𝐹 ∈ (Fil‘𝑋) ∧ (𝐴𝐹𝐵𝑋𝐴𝐵)) → 𝐵𝐹)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 384 ∧ w3a 1036 = wceq 1480 ∈ wcel 1987 ≠ wne 2790 ∀wral 2907 ∩ cin 3554 ⊆ wss 3555 ∅c0 3891 𝒫 cpw 4130 dom cdm 5074 ‘cfv 5847 fBascfbas 19653 Filcfil 21559 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-8 1989 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 ax-sep 4741 ax-nul 4749 ax-pow 4803 ax-pr 4867 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-eu 2473 df-mo 2474 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ne 2791 df-ral 2912 df-rex 2913 df-rab 2916 df-v 3188 df-sbc 3418 df-csb 3515 df-dif 3558 df-un 3560 df-in 3562 df-ss 3569 df-nul 3892 df-if 4059 df-pw 4132 df-sn 4149 df-pr 4151 df-op 4155 df-uni 4403 df-br 4614 df-opab 4674 df-mpt 4675 df-id 4989 df-xp 5080 df-rel 5081 df-cnv 5082 df-co 5083 df-dm 5084 df-rn 5085 df-res 5086 df-ima 5087 df-iota 5810 df-fun 5849 df-fv 5855 df-fil 21560 This theorem is referenced by: filin 21568 filtop 21569 isfil2 21570 infil 21577 fgfil 21589 fgabs 21593 filconn 21597 filuni 21599 trfil2 21601 trfg 21605 isufil2 21622 ufprim 21623 ufileu 21633 filufint 21634 elfm3 21664 rnelfm 21667 fmfnfmlem2 21669 fmfnfmlem4 21671 flimopn 21689 flimrest 21697 flimfnfcls 21742 fclscmpi 21743 alexsublem 21758 metust 22273 cfil3i 22975 cfilfcls 22980 iscmet3lem2 22998 equivcfil 23005 relcmpcmet 23023 minveclem4 23111 fgmin 32007
Copyright terms: Public domain W3C validator | 2,337 | 3,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-49 | latest | en | 0.143574 |
https://www.physicsforums.com/threads/torque-and-force-on-a-tire-iron.928058/ | 1,726,525,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00527.warc.gz | 863,970,562 | 20,533 | # Torque and force on a tire iron
• Jason DiCaprio
In summary, the conversation discusses the relationship between torque and force, specifically in the context of changing a tire. The question is posed as to how weight is affected when applying a force at different distances on a tire iron, and how torque can change drastically while the weight remains the same. The concept of net force is also explored, with the understanding that there can only be one net force acting on an object. Overall, the conversation delves into the mechanics of torque and force and their impacts on weight distribution in the context of changing a tire.
Jason DiCaprio
So I was curious about something. Having a flat tire today and having to change my tire made me wonder. I understand torque and force are 2 different things though related they have separate meanings.
So my question is if I have a 2 foot tire iron and I am torquing my nuts. Let's say they reached maximum torque and I can't move the tire iron anymore. Now let's say I am applying 10 pounds of force at 1 foot from the nut. My question is will I be adding 10 pounds of weight to the car/front wheel If I can't torque anymore and the nut is as tight as possible.
What if I was applying 10 pounds at 2 feet from the nut, common sense tells me I should be adding the same amount of weight to the car regardless if I apply the weight on the tire iron 1 foot away , 2 feet or any amount of feet. Weight should be weight. But I understand the torque will be increasing on the nut as I move farther and farther away from the nut.
My question is how is it possible to increase the torque on the nut but have the same amount of weight pushing down on the nut(net weight being added to the car). What if I had some ridiculously long tire iron and I was 10 feet away pushing down 10 pounds the added weight to the car should still be the car + tire iron + plus the downward force that I'm applying to the tire iron which is 10 pounds, but then I would have a ridiculous amount of torque on the bolt.
So how is this possible for the weight/force applied to the weight of the car always being car + tire iron + 10 pounds force I am applying at any location on the tire iron but the torque will change so drastically on the nut depending on where I apply the pressure. It makes no sense to me how you can have such a variation of torque , but when the bolt is as tight as possible the extra torque from the leverage all turns into the same weight/force pushing down no matter where it is on the lever?
Last edited by a moderator:
Jason DiCaprio said:
So how is this possible for the weight/force applied to the weight of the car always being car + tire iron + 10 pounds force I am applying at any location on the tire iron but the torque will change so drastically on the nut depending on where I apply the pressure.
Suppose you have a fixed nut and tire iron on a scale with a zeroed reading and you apply a torque of 10 foot pounds on the nut. Would it matter where the contact point of the scale is?
This is a fiendishly difficult problem
Well let's see ... let's say you do not exert a torque on the wrench, but a downwards force. The bolt exerts an upwards force on the other end of the wrench. Let's call those two forces and the distance between those two forces a torque.
Let's say the wrench arm is a pipe, you stick a screwdriver into the pipe. Now when you push down the screwdriver handle, the screwdriver exerts a torque on the wrench. Let's define the torque as the upwards force that the farthest end of the screwdriver exerts on the pipe, and the equal downwards force that the part of the screwdriver nearest to the handle exerts on the pipe, and the distance of those two forces.
Now we could say those two forces are the 'extra forces' that were gained by the use of the extension arm, I mean the screwdriver.
You see, those two forces that are responsible of the torque on the wrench arm are not pushing the wrench arm down, so the force that pushes the arm down exists in addition to the torque on the arm.
jartsa said:
Now we could say those two forces are the 'extra forces' that were gained by the use of the extension arm, I mean the screwdriver.
There can be only "one" net force...
Jason DiCaprio said:
My question is will I be adding 10 pounds of weight to the car/front wheel If I can't torque anymore and the nut is as tight as possible.
Actually, you will be adding 10 pounds of weight to the entire car, not necessarily the front axle.
Say the wheelbase of your car is 100" and your tire iron is 20", pointing toward the rear axle. Applying 10 lb pushing downward, the weight on the rear axle will increase by 2 lb (= 10 * 20 / 100) and the weight on the front axle will increase by 8 lb (= 10 - 2).
But if the tire iron is pointing in front of the car, still applying 10 lb pushing downward, the weight on the rear axle will decrease by 2 lb (= - 10 * 20 / 100; negative because the torque is in the other direction) and the weight on the front axle will increase by 12 lb (= 10 - (-2)).
As you can see, if the length of the tire iron is long enough, you will be able to lift one of the axle of the ground with a small force of 10 lb. At that point, the entire weight of the car will be on one axle + 10 lb and the weight on the other axle will be, obviously, zero.
The car can also loose 10lb if you aren't worried about your back.
## What is torque?
Torque is a measure of the turning or twisting force on an object. In the context of a tire iron, it is the force applied to the iron to loosen or tighten a bolt or lug nut.
## How is torque calculated?
Torque is calculated by multiplying the force applied to an object by the distance from the point of rotation to the point where the force is applied. In the case of a tire iron, this would be the force applied to the handle of the iron multiplied by the length of the handle.
## What factors affect torque on a tire iron?
The main factors that affect torque on a tire iron are the force applied, the length of the handle, and the angle at which the force is applied. Additionally, the condition and size of the bolt or lug nut being turned can also affect the torque required.
## How does torque affect the tightening or loosening of a bolt?
If the torque applied to a bolt is greater than the torque holding the bolt in place, it will loosen the bolt. If the torque applied is less than the torque holding the bolt in place, it will tighten the bolt. In general, a higher torque will result in a tighter bolt, while a lower torque will result in a looser bolt.
## Why is it important to use the correct torque when using a tire iron?
Using the correct torque is important for several reasons. It ensures that the bolt or lug nut is tightened to the appropriate level, preventing it from becoming too loose or too tight. This can help prevent damage to the bolt and tire iron, as well as ensure the safety and stability of the tire. It can also prevent injuries to the person using the tire iron, as applying too much torque can cause the iron to slip or break.
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4K | 1,688 | 7,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-38 | latest | en | 0.961458 |
https://tntips.com/learning-how-to-tell-time/ | 1,656,982,833,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00518.warc.gz | 639,240,115 | 15,560 | Learning How To Tell Time? – tntips.com
# Learning How To Tell Time?
## Learning How To Tell Time?
Age 5-6: Preschooler Kids to learn about Time
Your child will be able to read the hour and half hour markers on the analog clock and draw the times accordingly.Feb 23, 2021
## What age should a child learn to tell time?
Age 5-6: Preschooler Kids to learn about Time
Your child will be able to read the hour and half hour markers on the analog clock and draw the times accordingly.
## Should an 8 year old be able to tell the time?
Children should know the number of minutes in an hour and the number of hours in a day. Ages 7-8: Children should be able to read an analog clock, using 12 hour clocks, 24 hour clocks, and Roman Numerals (I-XII). Children should be able to compare time (by hours, minutes, and even seconds).
## How do you explain time?
In math, time can be defined as the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future. Time is a used to quantify, measure or compare the duration of events or the intervals between them, and even, sequence events.
## What should an 8 year old know?
Most children by age 8:
• Know how to count by 2s (2, 4, 6, 8, and so on) and 5s (5, 10, 15, 20, and so on).
• Know what day of the week it is. …
• Complete simple single-digit addition and subtraction problems (such as 1 + 8, 7 + 5, 6 – 2, 4 – 3).
• Can tell the difference between right and left.
## How do you explain times tables to a 6 year old?
8 Effective Tips for Teaching Times Tables
1. Hang up a times table sheet. …
2. Make sure they can walk before they can run. …
3. Teach your kids some tricks. …
4. Listen to some fun songs. …
5. Stage a multiplication war. …
6. Draw a Waldorf multiplication flower. …
7. Quiz them regularly, but not incessantly. …
8. Reward their efforts.
## What are 3 ways to tell time?
These are the different ways of reading time observing the hour – hand and minute – hand. Some clock has second hand which is generally not observed while telling the time, the position of the hour – hand and minute – hand indicate the time shown in the specific clock.
## What are the 3 steps you use to tell the time?
Tell the time in minutes-past and minutes-to the hour
To read the time in minutes-past and minutes-to the hour the child only needs to remember the three simple steps that make up our 3-step teaching system.
## How do you introduce time to preschoolers?
Show Them How Time Works
To gain a better understanding of the value of time, teach kids through doing. Set start and end times on a clock for homework or simple household chores. Read the clock with them as they begin and at intervals during the task. Be sure to set a timer so everyone knows when time is up.
## How do you introduce time?
15 Meaningful Hands-On Ways to Teach Telling Time
1. Make a paper clock. …
2. Color the spaces to learn the hours. …
3. Wear paper watches. …
4. Make a clock with linking math cubes. …
5. Take it outside with a hula hoop clock. …
6. Dance around for musical clocks. …
7. Shake up a carton of clocks. …
8. Add a hook to the hour hand.
## Why do we tell time?
Knowing how to tell time is a very important skill. It can help you determine whether you’re running late or whether you have plenty of time to spare. It can help you catch a train, bus, or plane on time, and allows you to know if you’re going to make it to an important get-together early or late.
## What is a good bedtime for an 8 year old?
Bedtimes by Age
Age Hours of Sleep Bedtime
15 months – 3 years 12-14 6:00 -7:30
3 – 6 years 11-13 6:00 – 8:00
7 – 12 years 10-11 7:30 – 9:00
Teenagers 9+ See note
## What should a 9 year old know academically?
Nine-year-old children will tackle multiplication and division of multiple digits and start learning about fractions and geometry. They will learn how to make graphs and charts using data and will work on word problems that require analytical and logical thinking.
## What should a 7 year old know academically?
So, what should a 7-year-old know academically? A 7-year old should be able to read, write (with some errors,) add and subtract. They should know how to tell time, know the days of the week and names of the months. They should be able to work with 3-digit numbers and be able to use a ruler.
## At what age should a child know their multiplication tables?
Children can begin to learn their multiplication tables once they have mastered basic addition and subtraction concepts and are familiar with arrays and how to count by 2’s and 5’s, which is usually by age 9.
## How do you count time without a clock?
Start by planting your feet towards the sun, extending one arm fully in front of you, and rotating your wrist so your palm is facing you horizontally. Close your fingers together and align your pinky with the horizon. Now, count how many finger widths it takes to reach the sun.
## How do you say 4 45 in English?
A ‘quarter‘ can replace ’15 minutes’. It can be used with ‘to’ or ‘past’ depending on its situation. Ex: It’s a quarter to 5 ( 4:45 ). It’s a quarter past 5 (5:15).
## What does 5 to 10 minutes mean?
If you say “five of ten” in the context of time, you mean 5 minutes to 10 o’clock.
## How do you explain time to students?
Introduce – Time: Hours
Start by showing times such as 7:00 and 11:00 and writing them on the board next to clock faces. Check to ensure that your students understand that the short hand indicates the hour and should be both said and written first. Have students repeat “One o’clock, two o’clock…” after you.
## Who created time?
The measurement of time began with the invention of sundials in ancient Egypt some time prior to 1500 B.C. However, the time the Egyptians measured was not the same as the time today’s clocks measure. For the Egyptians, and indeed for a further three millennia, the basic unit of time was the period of daylight.
## Why is the day twenty four hours?
Our 24-hour day comes from the ancient Egyptians who divided day-time into 10 hours they measured with devices such as shadow clocks, and added a twilight hour at the beginning and another one at the end of the day-time, says Lomb. … “Tables were produced to help people to determine time at night by observing the decans.
## How was time decided?
THE DIVISION of the hour into 60 minutes and of the minute into 60 seconds comes from the Babylonians who used a sexagesimal (counting in 60s) system for mathematics and astronomy. They derived their number system from the Sumerians who were using it as early as 3500 BC.
## Should a 17 year old have a bedtime?
A 17-year-old shouldn’t need as many reminders about good sleep habits. Rather than give an older teen a strict bedtime, it’s better to educate your teen. Let them know how much sleep their growing body needs. … Keep the focus on encouraging a healthy bedtime hour, rather than strictly enforcing it.
## Is 8pm too early to go to bed?
School-age children should go to bed between 8:00 and 9:00 p.m. Teenagers, for adequate sleep, should consider going to bed between 9:00 and 10:00 p.m. Adults should try to go to sleep between 10:00 and 11:00 p.m.
## Learn to Tell Time #1 | Telling the Time Practice for Children | What’s the Time? | Fun Kids English
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See more articles in category: FAQ | 1,845 | 7,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-27 | latest | en | 0.917493 |
https://www.teacherspayteachers.com/Product/Problem-Solving-Journal-Prompts-170-Word-Problems-for-Problem-of-the-Day-1985178 | 1,487,647,795,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170624.14/warc/CC-MAIN-20170219104610-00623-ip-10-171-10-108.ec2.internal.warc.gz | 916,895,655 | 25,724 | # Problem Solving Journal Prompts: 170+ Word Problems for Problem of the Day
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Problem Solving Journal Prompts: A Full Year of Prompts (170+ word problems) Perfect for Problem of the Day or Math Interactive Notebooks
Math journals are a great way to get students to think about their reasoning. They promote critical thinking skills and push students to utilize metacognition. This growing bundle will provide you with problem solving questions for the entire year. Designed to teach students to think critically about math problems through the use of graphic organizers, this pack is perfect for Problem of the Day or as a challenge for early finishers.
With 171 problems included, this product is designed to provide you and your students with problem solving that will develop the essential skills necessary to attack word problems, think critically, and provide explanations for their process and reasoning.
Designed for third and fourth grade, each question contains a large prompt with photograph that is designed for display and discussion, 2 different interactive journal formats, and 2 different problem solving templates. The prompts will push students to discuss and support their thinking and make for great class discussion and are flexible enough to allow students to use a variety of strategies to solve.
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\$25.00 | 623 | 3,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-09 | latest | en | 0.905299 |
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## Physics Quiz: Gravitational Field Representation MCQs - 5
MCQ: Force acting on two point masses is directly proportional to
A) sum of masses
B) difference of masses
C) distance between masses
D) product of masses
MCQ: Maximum displacement from equilibrium position is
A) frequency
B) amplitude
C) wavelength
D) period
MCQ: Attenuation coefficient of bone is 600 m-1 for x-rays of energy 20 keV and intensity of beam of x-rays is 20 Wm-2, then intensity of beam after passing through a bone of 4mm is
A) 3 Wm-2
B) 2.5 Wm-2
C) 2.0 Wm-2
D) 1.8 Wm-2
MCQ: Contact force always acts at
A) acute angles to the surface producing it
B) right angles to the surface producing it
C) obtuse angle to the surface producing it
D) parallel to the surface producing it
MCQ: Unit of luminous intensity is
A) m
B) kg
C) cd
D) mol | 453 | 1,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-50 | longest | en | 0.803066 |
https://puretables.com/megameters-to-micrometers | 1,652,952,953,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00446.warc.gz | 550,858,286 | 4,112 | # Megameters to Micrometers
Online converter
Mm
μm
-> Micrometers to Megameters
## How to convert Megameters to Micrometers
1 Megameter is equal to 1000000000000 Micrometers.
1 Mm = 1000000000000 μm.
Formula
x(Mm) / 1.0E-12 = x(μm)
To get the result in micrometers divide the number of megameters by 1.0E-12.
Example
Convert 5 Megameters to Micrometers:
5 Mm / 1.0E-12 = 5000000000000 μm
## Megameters to Micrometers Conversion Table
Megameters [Mm] Micrometers [μm]
0.01 Mm10000000000 μm
0.1 Mm100000000000 μm
1 Mm1000000000000 μm
2 Mm2000000000000 μm
3 Mm3000000000000 μm
5 Mm5000000000000 μm
10 Mm10000000000000 μm
20 Mm20000000000000 μm
30 Mm30000000000000 μm
50 Mm50000000000000 μm
100 Mm1.0E+14 μm
500 Mm5.0E+14 μm
1000 Mm1.0E+15 μm | 321 | 748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-21 | latest | en | 0.474125 |
https://proofwiki.org/wiki/Series_of_Power_over_Factorial_Converges | 1,632,703,863,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00371.warc.gz | 483,769,717 | 11,156 | Series of Power over Factorial Converges
It has been suggested that this page or section be merged into Radius of Convergence of Power Series over Factorial. (Discuss)
Theorem
The series $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges for all real values of $x$.
Proof
If $x = 0$ the result is trivially true as:
$\forall n \ge 1: \dfrac {0^n} {n!} = 0$
If $x \ne 0$ we have:
$\size {\dfrac {\paren {\dfrac {x^{n + 1} } {\paren {n + 1}!} } } {\paren {\dfrac {x^n} {n!} } } } = \dfrac {\size x} {n + 1} \to 0$
as $n \to \infty$.
This follows from the results:
Sequence of Powers of Reciprocals is Null Sequence, where $\dfrac 1 n \to 0$ as $n \to \infty$
The Squeeze Theorem for Real Sequences, as $\dfrac 1 {n + 1} < \dfrac 1 n$
The Multiple Rule for Real Sequences, putting $\lambda = \size x$.
Hence by the Ratio Test: $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges.
$\blacksquare$
Alternatively, the Comparison Test could be used but this is more cumbersome in this instance.
Another alternative is to view this as an example of Radius of Convergence of Power Series over Factorial setting $\xi = 0$. | 377 | 1,146 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-39 | latest | en | 0.763491 |
http://gmatclub.com/forum/george-bernard-shaw-wrote-that-any-sane-nation-69741.html?fl=similar | 1,484,634,914,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279468.17/warc/CC-MAIN-20170116095119-00507-ip-10-171-10-70.ec2.internal.warc.gz | 117,254,374 | 62,255 | George Bernard Shaw wrote: " That any sane nation, : GMAT Critical Reasoning (CR)
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# George Bernard Shaw wrote: " That any sane nation,
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George Bernard Shaw wrote: " That any sane nation, [#permalink]
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George Bernard Shaw wrote: " That any sane nation, having observed that you could provide for the supply of bread by giving bakers a pecuniary interest in baking for you, should go on to give a surgeon a pecuniary interest in cutting off your leg is enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration in an argument criticizing which of the following?
A Dentists who perform unnecessary dental work in order to earn a profit
B Doctors who increase their profits by specializing only in diseases that affect a large percentage of the population
C Grocers who raise the price of food in order to increase their profit margins
D Oil companies that decrease the price of their oil in order to increase their market share
E Bakers and surgeons who earn a profit by supplying other peoples' basic needs
If you have any questions
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Re: CR baker and surgeon [#permalink]
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03 Sep 2008, 19:54
ssandeepan wrote:
George Bernard Shaw wrote: " That any sane nation, having observed that you could provide for the supply of bread by giving bakers a pecuniary interest in baking for you, should go on to give a surgeon a pecuniary interest in cutting off your leg is enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration in an argument criticizing which of the following?
A Dentists who perform unnecessary dental work in order to earn a profit
B Doctors who increase their profits by specializing only in diseases that affect a large percentage of the population
C Grocers who raise the price of food in order to increase their profit margins
D Oil companies that decrease the price of their oil in order to increase their market share
E Bakers and surgeons who earn a profit by supplying other peoples' basic needs
ANS should be A>
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Re: CR baker and surgeon [#permalink]
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04 Sep 2008, 01:15
I will go for C.
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Re: CR baker and surgeon [#permalink]
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04 Sep 2008, 05:56
I say A
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Re: CR baker and surgeon [#permalink]
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04 Sep 2008, 07:01
Another vote for A.
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04 Sep 2008, 11:05
A for me ... profits provide an incentive to do work
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Re: CR baker and surgeon [#permalink]
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06 Sep 2008, 10:11
ssandeepan wrote:
George Bernard Shaw wrote: " That any sane nation, having observed that you could provide for the supply of bread by giving bakers a pecuniary interest in baking for you, should go on to give a surgeon a pecuniary interest in cutting off your leg is enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration in an argument criticizing which of the following?
A Dentists who perform unnecessary dental work in order to earn a profit
B Doctors who increase their profits by specializing only in diseases that affect a large percentage of the population
C Grocers who raise the price of food in order to increase their profit margins
D Oil companies that decrease the price of their oil in order to increase their market share
E Bakers and surgeons who earn a profit by supplying other peoples' basic needs
Clearly E ,scope is bakers and surgeons ,they earn interests and profits providing peoples basic needs ,this is critisised!!
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Re: CR baker and surgeon [#permalink]
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01 Oct 2008, 13:33
Searched all over the place for a good explanation of why A is correct, and E is incorrect... but could not find one. Anyone else want to give it a try?
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Re: CR baker and surgeon [#permalink]
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01 Oct 2008, 17:52
On the first thought, i went 4 A but after reading again , I think E is a better choice. The author cites examples of Baker and a surgeon to criticize them as they r earning profits by supplying people's needs.
ssandeepan wrote:
George Bernard Shaw wrote: " That any sane nation, having observed that you could provide for the supply of bread by giving bakers a pecuniary interest in baking for you, should go on to give a surgeon a pecuniary interest in cutting off your leg is enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration in an argument criticizing which of the following?
A Dentists who perform unnecessary dental work in order to earn a profit
B Doctors who increase their profits by specializing only in diseases that affect a large percentage of the population
C Grocers who raise the price of food in order to increase their profit margins
D Oil companies that decrease the price of their oil in order to increase their market share
E Bakers and surgeons who earn a profit by supplying other peoples' basic needs
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Re: CR baker and surgeon [#permalink]
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01 Oct 2008, 21:13
I would go for A. The question asks us to identify an illustration that critizes a statement, we should find an illustration that the statement attacks. I think A is the best.
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Re: CR baker and surgeon [#permalink]
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02 Oct 2008, 05:57
This is one of the questions that makes me shake my head.
The assumption here is that:
1. you pay the baker because you want bread.
2. you pay the surgeon because you need to get your legs cut off (whatever the dang reason is)
For A) to be correct, you have to assume that the surgeon is cutting off your legs for an unnecessary reason. But what if want/need to cut off your legs? (cut it off because you've been injured in a war, cut it off so you can put on a prostetic leg, etc.).
The fact that we have to assume that cutting of your legs unnecessary is absolutely BS. There could be many reasons why the procedure is done, but the fact that we have to assume that it's good/bad it's nonsense.
............As I'm writing, I'm starting to realize why I'm wrong (how funny) - the assumptions I laid out before were MY assumptions, and not the author's. The author can have very different assumptions, however ridiculous they may be. Therefore, A) can be right - and the fact that he is critizing it, makes the answer even more correct.
E) would be the correct answer is MY assumptions were used.
It's funny how things become clearer when you start venting
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Re: CR baker and surgeon [#permalink]
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02 Oct 2008, 06:03
Another A.
The logic here is that ... profits provide an incentive to do work (no matter if the work is worth/good or not) and A proves this point this putting clearly by syaing "unnecessary dental work "
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Re: George Bernard Shaw wrote: " That any sane nation, [#permalink]
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05 May 2012, 03:29
ssandeepan wrote:
George Bernard Shaw wrote: " That any sane nation, having observed that you could provide for the supply of bread by giving bakers a pecuniary interest in baking for you, should go on to give a surgeon a pecuniary interest in cutting off your leg is enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration in an argument criticizing which of the following?
A Dentists who perform unnecessary dental work in order to earn a profit
B Doctors who increase their profits by specializing only in diseases that affect a large percentage of the population
C Grocers who raise the price of food in order to increase their profit margins
D Oil companies that decrease the price of their oil in order to increase their market share
E Bakers and surgeons who earn a profit by supplying other peoples' basic needs
I think A is the correct answer. The reason is that it is similar in construction to the original argument provided. What is the OA??
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Re: George Bernard Shaw wrote: " That any sane nation, [#permalink]
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05 May 2012, 20:00
E: For people to earn profits by supplying other peoples basic needs does not make one despair of political humanity.
"should go on to give a surgeon a pecuniary interest in cutting off your leg" - From this statement, it sounds as if there's an unnecessary acts being performed for financial/personal gain.
Hence, I chose A.
Re: George Bernard Shaw wrote: " That any sane nation, [#permalink] 05 May 2012, 20:00
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### C
Roman numerals,
• the numerals in the ancient Roman system of notation, still used for certain limited purposes, as in some pagination, dates on buildings, etc. The common basic symbols are I (=1), V (=5), X (=10), L (=50), C (=100), D (=500), and M (=1000). The Roman numerals for one to nine are: I, II, III, IV, V, VI, VII, VIII, IX. A bar over a letter multiplies it by 1000;
thus, X̄ equals 10,000. Integers are written according to these two rules: If a letter is immediately followed by one of equal or lesser value, the two values are added;
thus, XX equals 20, XV equals 15, VI equals 6. If a letter is immediately followed by one of greater value, the first is subtracted from the second;
thus, IV equals 4, XL equals 40, CM equals 900. Examples: XLVII(=47), CXVI(=116), MCXX(=1120), MCMXIV(=1914). Roman numerals may be written in lowercase letters, though they appear more commonly in capitals.
• ### Shaped
1. of a definite form, shape, or character (often used in combination):aU-shaped driveway.
2. designed to fit a particular form, body, or contour: a shaped garment.
3. having other than a plane surface.
### Sofa
so•fa (sōfə),USA pronunciation n.
1. a long, upholstered couch with a back and two arms or raised ends.
### Table
ta•ble (tābəl),USA pronunciation n., v., -bled, -bling, adj.
n.
1. an article of furniture consisting of a flat, slablike top supported on one or more legs or other supports: a kitchen table; an operating table; a pool table.
2. such a piece of furniture specifically used for serving food to those seated at it.
3. the food placed on a table to be eaten: She sets a good table.
4. a group of persons at a table, as for a meal, game, or business transaction.
5. a gaming table.
6. a flat or plane surface;
a level area.
7. a tableland or plateau.
9. an arrangement of words, numbers, or signs, or combinations of them, as in parallel columns, to exhibit a set of facts or relations in a definite, compact, and comprehensive form;
a synopsis or scheme.
10. (cap.) the constellation Mensa.
11. a flat and relatively thin piece of wood, stone, metal, or other hard substance, esp. one artificially shaped for a particular purpose.
• a course or band, esp. of masonry, having a distinctive form or position.
• a distinctively treated surface on a wall.
12. a smooth, flat board or slab on which inscriptions may be put.
13. tables:
• the tablets on which certain collections of laws were anciently inscribed: the tables of the Decalogue.
• the laws themselves.
14. the inner or outer hard layer or any of the flat bones of the skull.
15. a sounding board.
16. [Jewelry.]
• the upper horizontal surface of a faceted gem.
• a gem with such a surface.
17. on the table, [Parl. Proc.]
• [U.S.]postponed.
• [Brit.]submitted for consideration.
18. turn the tables, to cause a reversal of an existing situation, esp. with regard to gaining the upper hand over a competitor, rival, antagonist, etc.: Fortune turned the tables and we won. We turned the tables on them and undersold them by 50 percent.
19. under the table:
• drunk.
• as a bribe;
secretly: She gave money under the table to get the apartment.
20. wait (on) table, to work as a waiter or waitress: He worked his way through college by waiting table.Also, wait tables.
v.t.
1. to place (a card, money, etc.) on a table.
2. to enter in or form into a table or list.
3. [Parl. Proc.]
• [Chiefly U.S.]to lay aside (a proposal, resolution, etc.) for future discussion, usually with a view to postponing or shelving the matter indefinitely.
• to present (a proposal, resolution, etc.) for discussion.
1. of, pertaining to, or for use on a table: a table lamp.
2. suitable for serving at a table or for eating or drinking: table grapes.
### Sofa
so•fa (sōfə),USA pronunciation n.
1. a long, upholstered couch with a back and two arms or raised ends.
### Table
ta•ble (tābəl),USA pronunciation n., v., -bled, -bling, adj.
n.
1. an article of furniture consisting of a flat, slablike top supported on one or more legs or other supports: a kitchen table; an operating table; a pool table.
2. such a piece of furniture specifically used for serving food to those seated at it.
3. the food placed on a table to be eaten: She sets a good table.
4. a group of persons at a table, as for a meal, game, or business transaction.
5. a gaming table.
6. a flat or plane surface;
a level area.
7. a tableland or plateau.
9. an arrangement of words, numbers, or signs, or combinations of them, as in parallel columns, to exhibit a set of facts or relations in a definite, compact, and comprehensive form;
a synopsis or scheme.
10. (cap.) the constellation Mensa.
11. a flat and relatively thin piece of wood, stone, metal, or other hard substance, esp. one artificially shaped for a particular purpose.
• a course or band, esp. of masonry, having a distinctive form or position.
• a distinctively treated surface on a wall.
12. a smooth, flat board or slab on which inscriptions may be put.
13. tables:
• the tablets on which certain collections of laws were anciently inscribed: the tables of the Decalogue.
• the laws themselves.
14. the inner or outer hard layer or any of the flat bones of the skull.
15. a sounding board.
16. [Jewelry.]
• the upper horizontal surface of a faceted gem.
• a gem with such a surface.
17. on the table, [Parl. Proc.]
• [U.S.]postponed.
• [Brit.]submitted for consideration.
18. turn the tables, to cause a reversal of an existing situation, esp. with regard to gaining the upper hand over a competitor, rival, antagonist, etc.: Fortune turned the tables and we won. We turned the tables on them and undersold them by 50 percent.
19. under the table:
• drunk.
• as a bribe;
secretly: She gave money under the table to get the apartment.
20. wait (on) table, to work as a waiter or waitress: He worked his way through college by waiting table.Also, wait tables.
v.t.
1. to place (a card, money, etc.) on a table.
2. to enter in or form into a table or list.
3. [Parl. Proc.]
• [Chiefly U.S.]to lay aside (a proposal, resolution, etc.) for future discussion, usually with a view to postponing or shelving the matter indefinitely.
• to present (a proposal, resolution, etc.) for discussion.
1. of, pertaining to, or for use on a table: a table lamp.
2. suitable for serving at a table or for eating or drinking: table grapes.
### Design
de•sign (di zīn),USA pronunciation v.t.
1. to prepare the preliminary sketch or the plans for (a work to be executed), esp. to plan the form and structure of: to design a new bridge.
2. to plan and fashion artistically or skillfully.
3. to intend for a definite purpose: a scholarship designed for foreign students.
4. to form or conceive in the mind;
contrive;
plan: The prisoner designed an intricate escape.
purpose: He designed to be a doctor.
6. [Obs.]to mark out, as by a sign;
indicate.
v.i.
1. to make drawings, preliminary sketches, or plans.
2. to plan and fashion the form and structure of an object, work of art, decorative scheme, etc.
n.
1. an outline, sketch, or plan, as of the form and structure of a work of art, an edifice, or a machine to be executed or constructed.
2. organization or structure of formal elements in a work of art;
composition.
3. the combination of details or features of a picture, building, etc.;
the pattern or motif of artistic work: the design on a bracelet.
4. the art of designing: a school of design.
5. a plan or project: a design for a new process.
6. a plot or intrigue, esp. an underhand, deceitful, or treacherous one: His political rivals formulated a design to unseat him.
7. designs, a hostile or aggressive project or scheme having evil or selfish motives: He had designs on his partner's stock.
8. intention;
purpose;
end.
9. adaptation of means to a preconceived end.
### Over
o•ver vər),USA pronunciation prep.
1. above in place or position: the roof over one's head.
2. above and to the other side of: to leap over a wall.
3. above in authority, rank, power, etc., so as to govern, control, or have jurisdiction regarding: There is no one over her in the department now.
4. so as to rest on or cover;
on or upon: Throw a sheet over the bed.
5. on or upon, so as to cause an apparent change in one's mood, attitude, etc.: I can't imagine what has come over her.
6. on or on top of: to hit someone over the head.
7. here and there on or in;
about: at various places over the country.
8. through all parts of;
all through: to roam over the estate; to show someone over the house.
9. to and fro on or in;
across;
throughout: to travel all over Europe.
10. from one side to the other of;
to the other side of;
across: to go over a bridge.
11. on the other side of;
across: lands over the sea.
12. reaching higher than, so as to submerge: The water is over his shoulders.
13. in excess of;
more than: over a mile; not over five dollars.
14. above in degree, quantity, etc.: a big improvement over last year's turnout.
15. in preference to: chosen over another applicant.
16. throughout the length of: The message was sent over a great distance.
17. until after the end of: to adjourn over the holidays.
18. throughout the duration of: over a long period of years.
19. in reference to, concerning, or about: to quarrel over a matter.
20. while engaged in or occupied with: to fall asleep over one's work.
21. via;
by means of: He told me over the phone. I heard it over the radio.
22. over and above, in addition to;
besides: a profit over and above what they had anticipated.
23. over the hill. See hill (def. 8).
1. beyond the top or upper surface or edge of something: a roof that hangs over.
2. so as to cover the surface, or affect the whole surface: The furniture was covered over with dust.
3. through a region, area, etc.: He was known the world over.
4. at some distance, as in a direction indicated: They live over by the hill.
5. from side to side;
across;
to the other side: to sail over.
6. across an intervening space: Toss the ball over, will you?
7. across or beyond the edge or rim: The soup boiled over. The bathtub ran over.
8. from beginning to end;
throughout: to read a paper over; Think it over.
9. from one person, party, etc., to another: Hand the money over. He made the property over to his brother.
10. on the other side, as of a sea, a river, or any space: over in Japan.
11. so as to displace from an upright position: to knock over a glass of milk.
12. so as to put in the reversed position: She turned the bottle over. The dog rolled over.
13. once more;
again: Do the work over.
14. in repetition or succession: twenty times over.
15. in excess or addition: to pay the full sum and something over.
16. in excess of or beyond a certain amount: Five goes into seven once, with two over.
17. throughout or beyond a period of time: to stay over till Monday.
18. to one's residence, office, or the like: Why don't you come over for lunch?
19. so as to reach a place across an intervening space, body of water, etc.: Her ancestors came over on theMayflower
20. all over:
• over the entire surface of;
everywhere: material printed all over with a floral design.
• thoroughly;
entirely.
• finished: The war was all over and the soldiers came home.
21. all over with, ended;
finished: It seemed miraculous that the feud was all over with.
22. over again, in repetition;
once more: The director had the choir sing one passage over again.
23. over against. See against (def. 12).
24. over and over, several times;
repeatedly: They played the same record over and over.
25. over there, [Informal.](in the U.S. during and after World War I) in or to Europe: Many of the boys who went over there never came back.
26. over with, finished or done: Let's get this thing over with, so that we don't have to worry about it any more.
1. upper;
higher up.
2. higher in authority, station, etc.
3. serving, or intended to serve, as an outer covering;
outer.
extra.
5. too great;
excessive (usually used in combination): Insufficient tact and overaggressiveness are two of his problems.
6. ended;
done;
past: when the war was over.
n.
1. an amount in excess or addition;
extra.
2. a shot that strikes or bursts beyond the target.
3. [Cricket.]
• the number of balls, usually six, delivered between successive changes of bowlers.
• the part of the game played between such changes.
v.t.
1. to go or get over;
leap over.
2. [Southern U.S.]to recover from.
interj.
1. (used in radio communications to signify that the sender has temporarily finished transmitting and is awaiting a reply or acknowledgment.) Cf. out (def. 61).
### The
the1 (stressed ᵺē; unstressed before a consonant ᵺə;
unstressed before a vowel ᵺē),USA pronunciation
definite article.
1. (used, esp. before a noun, with a specifying or particularizing effect, as opposed to the indefinite or generalizing force of the indefinite article a or an): the book you gave me; Come into the house.
2. (used to mark a proper noun, natural phenomenon, ship, building, time, point of the compass, branch of endeavor, or field of study as something well-known or unique):the sun;
the Alps;
theQueen Elizabeth;
the past; the West.
3. (used with or as part of a title): the Duke of Wellington; the Reverend John Smith.
4. (used to mark a noun as indicating the best-known, most approved, most important, most satisfying, etc.): the skiing center of the U.S.; If you're going to work hard, now is the time.
5. (used to mark a noun as being used generically): The dog is a quadruped.
6. (used in place of a possessive pronoun, to note a part of the body or a personal belonging): He won't be able to play football until the leg mends.
7. (used before adjectives that are used substantively, to note an individual, a class or number of individuals, or an abstract idea): to visit the sick; from the sublime to the ridiculous.
8. (used before a modifying adjective to specify or limit its modifying effect): He took the wrong road and drove miles out of his way.
9. (used to indicate one particular decade of a lifetime or of a century): the sixties; the gay nineties.
10. (one of many of a class or type, as of a manufactured item, as opposed to an individual one): Did you listen to the radio last night?
11. enough: He saved until he had the money for a new car. She didn't have the courage to leave.
12. (used distributively, to note any one separately) for, to, or in each;
a or an: at one dollar the pound.
### Modern
1. of or pertaining to present and recent time;
not ancient or remote: modern city life.
2. characteristic of present and recent time;
contemporary;
not antiquated or obsolete: modern viewpoints.
3. of or pertaining to the historical period following the Middle Ages: modern European history.
4. of, pertaining to, or characteristic of contemporary styles of art, literature, music, etc., that reject traditionally accepted or sanctioned forms and emphasize individual experimentation and sensibility.
5. (cap.) new (def. 12).
6. [Typography.]noting or descriptive of a font of numerals in which the body aligns on the baseline, as 1234567890. Cf. old style (def. 3).
n.
1. a person of modern times.
2. a person whose views and tastes are modern.
3. [Print.]a type style differentiated from old style by heavy vertical strokes and straight serifs.
modern•ness, n.
### Easy
1. not hard or difficult;
requiring no great labor or effort: a book that is easy to read; an easy victory.
2. free from pain, discomfort, worry, or care: He led an easy life.
3. providing or conducive to ease or comfort;
comfortable: an easy stance; an easy relationship.
4. fond of or given to ease;
easygoing: an easy disposition.
5. not harsh or strict;
lenient: an easy master.
6. not burdensome or oppressive: easy terms on a loan.
7. not difficult to influence or overcome;
compliant: an easy prey; an easy mark.
8. free from formality, constraint, or embarrassment: He has an easy manner.
9. effortlessly clear and fluent: an easy style of writing.
10. readily comprehended or mastered: an easy language to learn.
11. not tight or constricting: an easy fit.
12. not forced or hurried;
moderate: an easy pace.
13. not steep;
gradual: an easy flight of stairs.
14. [Com.]
• (of a commodity) not difficult to obtain;
in plentiful supply and often weak in price.
• (of the market) not characterized by eager demand.
15. [Naut.]
• (of a bilge) formed in a long curve so as to make a gradual transition between the bottom and sides of a vessel;
slack.
• (of the run of a hull) having gently curved surfaces leading from the middle body to the stern;
not abrupt.
1. in an easy manner;
comfortably: to go easy; take it easy.
n.
1. a word formerly used in communications to represent the letter E.
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f(x)=\left\{ \begin{aligned} x & = & \cos(t) \\ y & = & \sin(t) \\ z & = & \frac xy \end{aligned} \right.
$$F^{HLLC}=\left\{ \begin{array}{rcl} F_L & & {0 < S_L}\\ F^*_L & & {S_L \leq 0 < S_M}\\ F^*_R & & {S_M \leq 0 < S_R}\\ F_R & & {S_R \leq 0} \end{array} \right.$$
$$f(x)= \begin{cases} 0& \text{x=0}\\ 1& \text{x!=0} \end{cases}$$
\end{CJK*}
\end{document}
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Resources tagged with Investigations similar to Star Find:
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Challenge level:
Hexpentas
Age 5 to 11 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
Triangle Shapes
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This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
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Age 7 to 11 Challenge Level:
We need to wrap up this cube-shaped present, remembering that we can have no overlaps. What shapes can you find to use?
Triple Cubes
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This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
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Can you create more models that follow these rules?
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How many models can you find which obey these rules?
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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?
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Age 7 to 11 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?
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Is there a best way to stack cans? What do different supermarkets do? How high can you safely stack the cans?
Escher Tessellations
Age 7 to 11 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
Age 7 to 11 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.
Halving
Age 5 to 7 Challenge Level:
These pictures show squares split into halves. Can you find other ways?
Redblue
Age 7 to 11 Challenge Level:
Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
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What do these two triangles have in common? How are they related?
Tri.'s
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How many triangles can you make on the 3 by 3 pegboard?
So It's 28
Age 5 to 7 Challenge Level:
Here is your chance to investigate the number 28 using shapes, cubes ... in fact anything at all.
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What is the largest cuboid you can wrap in an A3 sheet of paper?
Opening Out
Age 5 to 11
Bernard Bagnall describes how to get more out of some favourite NRICH investigations.
Cuboid-in-a-box
Age 7 to 11 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
Tessellating Transformations
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Can you find out how the 6-triangle shape is transformed in these tessellations? Will the tessellations go on for ever? Why or why not?
Let's Investigate Triangles
Age 5 to 7 Challenge Level:
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
Repeating Patterns
Age 5 to 7 Challenge Level:
Try continuing these patterns made from triangles. Can you create your own repeating pattern?
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Age 5 to 7 Challenge Level:
Explore ways of colouring this set of triangles. Can you make symmetrical patterns?
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In my local town there are three supermarkets which each has a special deal on some products. If you bought all your shopping in one shop, where would be the cheapest?
Making Cuboids
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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?
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Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
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Age 7 to 11 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot. | 1,919 | 8,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-43 | latest | en | 0.914597 |
http://what-when-how.com/Tutorial/topic-166ijd9i2/Introduction-to-Imaging-from-Scattered-Fields-237.html | 1,685,877,470,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00127.warc.gz | 51,924,078 | 4,440 | Image Processing Reference
In-Depth Information
2. Use ewald in MATLAB to generate Ewald circles for a target consist-
ing of one circle (Choice 0) and two circles (Choice 1). Use any valid
permittivity and a minimum of 12 sources and 120 receivers. For each
target save a 2-D output image of the combined Ewald circles. How are
these images similar? How are these images different? Please discuss
your observations and speculate on reasons for both the similarities
and differences.
3. Repeat the procedure in Exercise 1 for one square (Choice 2) and two
squares (Choice 3). Compare the resulting images obtained here with
the images obtained in Exercise 1. Discuss similarities and differ-
ences and reasons for them.
4. Use ewald in MATLAB to generate Ewald circles for two triangles
(Choice 5) using 4 sources and 360 receivers. Do this twice, once for a
target permittivity of 1.1 and again for a target permittivity of 1.9. Be
sure and show the individual Ewald circles for each source. For each
scenario save the individual Ewald circle for Source #2. Compare the
images for each Ewald circle. Describe the differences and discuss
why you think they are so different.
born exercises
1. Use born in MATLAB to generate Born approximation reconstructions
for various targets. Do this several times varying the target, the per-
mittivity, the number of sources, and the number of receivers. Discuss
your observations for the various scenarios and give some speculation
for reasons for the variations observed.
2. Use born in MATLAB to generate Born approximation reconstruc-
tions for a circle (Choice 0) using 36 sources and 360 receivers. Also,
set the image maximum dimension to 0.25 for each case. Do this three
times for a target permittivity of 1.1, 1.5, and 1.9. For each scenario
save the Born images generated. Compare the images for each Born
reconstruction and discuss the differences. Please speculate as to the
reason for these differences.
3. Use born in MATLAB to generate Born approximation reconstruc-
tions for any target consisting of two objects (Choices 1, 3, or 5) using
any valid permittivity and a minimum of 6 sources and 120 receivers.
Be sure and show the individual Born reconstructions for each source.
tions for each source and speculate on their relationship to the Born
of the combined sources.
4. Use born in MATLAB to generate Born approximation reconstruc-
tions for two circles (Choice 1) using a minimum of 12 sources and 120
receivers. Do this twice, once for a target permittivity of 1.1 and again
for 1.9. For each scenario save the individual Born reconstruction for
Source #12. Compare the images for each reconstruction. Describe the
differences and discuss why you think they are so different.
Cepstrum exercises
1. Use Cepstrum in MATLAB to generate Born and Cepstrum recon-
structions for various targets. Do this several times varying the
target, the permittivity, the number of sources, and the number of
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Custom Search | 702 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.873367 |
https://us.sofatutor.com/math/videos/identifying-proportional-relationships-in-tables | 1,718,770,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00471.warc.gz | 532,773,729 | 41,373 | # Identifying Proportional Relationships in Tables
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The authors
Susan S.
## Basics on the topicIdentifying Proportional Relationships in Tables
After this lesson you will be able to identify proportional relationships in tables.
The lesson begins with a table representing two quantities. It leads to recalling the definition of a proportional relationship. It concludes with determining whether a table represents a proportional relationship by simplifying ratios.
Learn about proportional relationships by helping the bees collect their pollen by figuring out how to beat the wind!
This video includes key concepts, notation, and vocabulary such as: proportional relationships (a constant ratio, or the unit rate); ratio (a comparative, proportional relationship between two amounts); and ratio tables (a table of equivalent ratios).
Before watching this video, you should already be familiar with the definition of a proportional relationship and the equation y=kx.
After watching this video, you will be prepared to learn whether two quantities are proportional to each other by graphing on a coordinate plane.
Common Core Standard(s) in focus: 7.RP.2.a A video intended for math students in the 7th grade Recommended for students who are 12-13 years
### TranscriptIdentifying Proportional Relationships in Tables
Two bees are hightailing it out of their hive. Their queen just ordered them to produce more honey, but in order to do this, they need to figure out how to collect more pollen. They notice that all the pollen is being quickly blown away by the wind! They need to figure out a way to beat the wind, and collect all that pollen. Let's help the bees figure out how to collect pollen by identifying and using proportional relationships. In a table, we keep track of how many pounds of pollen get blown away when the wind is moving at certain speeds, in miles per hour. When the wind is moving at 10 miles per hour, 4 pounds of pollen is lost. At 20 miles per hour, 8 pounds is lost. At 30 miles per hour, 12 pounds. At 40, 16. At 50, 20. And at 60, 24. We want to know if there is a proportional relationship between the wind speed and the amount of pollen blown away. Such a relationship would allow us to determine how much pollen is lost based on the wind speeds throughout the day. Remember, two quantities, 'x' and 'y', have a proportional relationship if one of them is a constant multiple of the other.
In other words, the ratio of one quantity over the other is always equal to the same constant. So, to determine if there is a proportional relationship between wind speed and pound of pollen lost, we need to determine if the ratio of wind speed to pound of pollen lost is always equal to the same constant. First, let's write the ratio of wind speed to pollen in a new column in our table. Notice that the numerator is the wind speed and that the denominator is the pounds of pollen lost. Factoring out 10 and 4, and cancelling out common factors gives us 5 over 2. Factoring out 20 and 8, and cancelling out common factors also gives us 5 over 2. In fact, every ratio of wind speed to pounds of pollen lost simplifies to 5 over 2, meaning that the wind speed is proportional to the amount of pollen blown away. So, we know how much pollen will blow away given the wind speed, but how can we catch the pollen before it's gone with the wind? Using their smartcomb, the bees search for a solution for their problem. What's this? A pop-up ad for spider webs? One of the bees has a brilliant idea to buy a giant spider web to catch the pollen when the wind speeds are the strongest. To decide when to use the spider web, they study the weather report carefully, collecting data about the wind speeds at various times during the day. We can use the data they collected in this table showing the wind speed at certain hours in the day to determine if this relationship is proportional. Let's check by writing down the ratio of hour to wind speed and then simplify to see if the ratios are the same.
And 6 over 60 simplifies to 1 tenth. We can already see from the first two entries that the ratios are not always the same. Indeed, there is NOT a proportional relationship between the quantities. So the bees cannot predict the wind speed based on the hour of the day. The bees decide they just should set up the spider web 24-7 to collect the pollen. Let's take a moment and review what we learned about proportional relationships. To determine if a table represents a proportional relationship, write each ratio. Then simplify. If the simplified ratios are equal, then the table represents a proportional relationship. If the simplified ratios are not equal, then the table does not represent a proportional relationship. The bees are really excited to set up the giant web and see how much pollen they can collect. Oh look! The giant spider web they ordered just arrived! For sure the queen will be very pleased with their work! Um, wait a second. It looks like the bees should have read the really, really, really small print in the online ad. | 1,081 | 5,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-26 | latest | en | 0.919013 |
https://se.mathworks.com/matlabcentral/cody/problems/2631-flip-the-vector-from-right-to-left/solutions/1926280 | 1,596,542,758,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00494.warc.gz | 490,615,625 | 15,779 | Cody
# Problem 2631. Flip the vector from right to left
Solution 1926280
Submitted on 8 Sep 2019 by Tim
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(flip_vector(x),y_correct))
2 Pass
x = [1:5]; y_correct = [5:-1:1]; assert(isequal(flip_vector(x),y_correct))
3 Pass
x = [1 4 6]; y_correct = [6 4 1]; assert(isequal(flip_vector(x),y_correct))
4 Pass
x = [10 5 9 ]; y_correct = [9 5 10]; assert(isequal(flip_vector(x),y_correct))
5 Pass
x = [2 4 6 8]; y_correct = [8 6 4 2]; assert(isequal(flip_vector(x),y_correct))
6 Pass | 241 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-34 | latest | en | 0.518001 |
https://www.doubtnut.com/question-answer/indefinite-integrals-integration-as-reverse-process-of-differentiation-fundamental-integration-formu-2022713 | 1,624,503,199,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00155.warc.gz | 633,483,241 | 72,403 | Class 12
MATHS
Indefinite Integrals
# INDEFINITE INTEGRALS | INTEGRATION AS REVERSE PROCESS OF DIFFERENTIATION, FUNDAMENTAL INTEGRATION FORMULAE, SOME STANDARD RESULTS ON INTEGRATION | A function phi(x) is called a primitive of f(x); if phi, Some important formulas of integration, Examples of integration: (i) x^4 (ii) 3^x, Theorem: d/dx(int f(x) dx) = f(x), The integral of the product of a constant and a function = the constant x integral of function, int {f(x) pm g(x)} dx = int f(x) dx pm int g(x) dx
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 1-8-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
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## Latest Questions
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Indefinite Integrals
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Indefinite Integrals
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Indefinite Integrals | 595 | 1,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-25 | latest | en | 0.547869 |
https://wiki.documentfoundation.org/Documentation/Calc_Functions/CSC | 1,660,961,123,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00359.warc.gz | 540,904,002 | 9,173 | # Documentation/Calc Functions/CSC
Other languages:
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CSC
Mathematical
## Summary:
Calculates the trigonometric cosecant (csc) of an angle expressed in radians.
CSC(Number)
## Returns:
Returns a number that is the trigonometric cosecant of the specified angle. The value returned is a real number x, such that x ≤ -1 or x ≥ 1 (x cannot lie in the range (-1, 1)).
## Arguments:
Number is a real number, or a reference to a cell containing that number, that is the angle in radians whose trigonometric cosecant is to be calculated.
• If Number is non-numeric, then CSC reports a #VALUE! error.
• If Number is equal to 0, then CSC reports a #NUM! error.
If your angle is expressed in degrees, either multiply by PI()/180 or use the RADIANS function to convert to radians.
The formula for the trigonometric cosecant of x is:
$\displaystyle{ \text{csc}(x)~=~\frac{1}{\sin(x)} }$
Mathematically, the cosecant of any integer multiple of π, i.e. {......-2π, -π, 0, π, 2π.....} is undefined. For this reason, CSC reports an error when its argument is equal to 0. However, because of the inherent inaccuracy in representing π numerically, CSC does not report errors for other argument values. You should be aware that CSC will return large positive or negative results for non-zero integer values of π.
The figure below illustrates the function CSC.
CSC function
## Examples:
Formula Description Returns
=CSC(1) Trigonometric cosecant of 1 radian. 1.18839510577812
=CSC(D1) where cell D1 contains the number -0.5. Trigonometric cosecant of -½ radian. -2.08582964293349
=CSC(PI()/4) Trigonometric cosecant of π/4 radians. 1.4142135623731
=CSC(RADIANS(30)) Trigonometric cosecant of 30°, after first converting the angle from degrees to radians. 2
CSC | 495 | 1,799 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-33 | latest | en | 0.708001 |
http://www.jiskha.com/members/profile/posts.cgi?name=BBF | 1,496,135,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00041.warc.gz | 668,247,198 | 2,743 | # Posts by BBF
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A person with no more than \$15,000 to invest plans to place the money in two investments. One investment is high-risk, high yield; the other is low risk, low yield. At least \$2,000 is to be placed in the high-risk investment. Furthermore, the amount invested at low risk should...
1. Pages:
2. 1
Post a New Question | 111 | 393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-22 | latest | en | 0.93963 |
https://www.physicsforums.com/threads/why-is-angular-momentum-conserved.780106/ | 1,527,046,779,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044430-00020.warc.gz | 798,172,684 | 18,663 | # Why is angular momentum conserved?
1. Nov 5, 2014
### Pejeu
Why do things which spin tend to keep spinning in the absence of external forces such as friction with the environment?
In order for objects to keep spinning doesn't their periphery (relative to their centre of rotation - which would be their centre of mass, right? - ) have to be constantly acted upon by forces external to it in order to constantly change its direction of travel all the while preserving its (instantaneous) speed?
Where does this force come from? How come the spin doesn't immediately start to slow down after torque is no longer applied to impart spin?
At one moment in time a clump of atoms on the periphery of a flywheel spinning freely in interplanetary space is going one way and some time later (after that clump of atoms has completed half of a rotation around the centre of rotation) the same clump of atoms is now travelling the opposite way, with the same speed as before.
Where did the energy for decelerating and accelerating that clump of atoms come from?
Please note I am not asking whether angular momentum is conserved or not. I am asking why it is conserved.
I would very much appreciate it if replies to this thread made that distinction.
Thank you very much for your interest in this inquiry and your time. Especially if you decide to help elucidate this for me.
2. Nov 5, 2014
### A.T.
Conservation angular momentum doesn't require changing the direction. It applies to linear motion too.
If their speed is the same, then so is their kinetic energy, therefore no energy input was required.
Physics describes nature as it is. It doesn't tell why it is that way. At best you can justify it based on some more general assumptions:
http://en.wikipedia.org/wiki/Noether's_theorem#Basic_illustrations_and_background
3. Nov 5, 2014
### Pejeu
But the clump is now going the opposite way, regardless that with the same speed as it was going before.
So the clump of atoms has to have been first decelerated and then accelerated, at least in the direction described by the tangent it was on half of a revolution before.
4. Nov 5, 2014
### Andrew Mason
Linear momentum is conserved because the time rate of change of momentum is force. So if no external forces act on a system, the total momentum cannot change with time.
Angular momentum is conserved because the time rate of change of angular momentum is torque. If no external torques act on a system, then angular momentum cannot change.
In answer to your question about the change in direction of atoms in a rotating body, there is no change in energy. There is no work done on the rotating atoms. Work is the dot product of Force and displacement: $W = \int \vec{F}\cdot\vec{d}$. Since the force is at right angles to the direction of motion and, therefore, at right angles to the direction of displacement over which the force acts, there is no energy required to keep the system of atoms rotating.
AM
Last edited: Nov 5, 2014
5. Nov 5, 2014
### A.T.
For kinetic energy the direction is irrelevant.
In uniform circular motion there is no tangential acceleration, just centripetal acceleration, which doesn't require energy input.
6. Nov 5, 2014
### Pejeu
Well their kinetic energy is indeed the same.
But it required energy to change their direction of movement. Especially preserving their speed.
Are you saying no energy needs to be expended in order to change the direction of movement of a particle with mass, especially if the particle retains its speed in the new direction of travel?
Yes. But that's something else entirely.
We have a clump of atoms at 12 o'clock which are being accelerated towards 6 o'clock by the centripetal force. But when they've arrived to 3 o'clock they're being accelerated by the centripetal force in the opposite direction as the one they were travelling at 12 o'clock.
Or, I should say, they've been decelerated to a stand still and are starting to be accelerated in the opposite way on the same direction they were travelling in when they were at 12 o'clock.
So objects or parts of objects with mass can be accelerated without energy input?
And aren't objects being accelerated against their previous direction of travel, 90 degrees before?
Aren't they being decelerated in the direction of travel they had 90 degree of revolution before? Isn't energy being expended at some fundamental level in order to decelerate that clump of atoms in the direction of travel they had before?
When the clump of atoms is at 3 o'clock they've been completely decelerated in the direction of travel they used to have when at 12 o'clock.
And then they're re-accelerated in the opposite direction by the time they reach 6 o'clock.
Isn't this reciprocating motion?
Where is the energy coming from to cyclically or continuously accelerate and decelerate all those clumps of atoms?
7. Nov 5, 2014
### Andrew Mason
No. This is not correct. The energy required to change their direction of motion is equal to the work done on the atoms in changing their motion. That work is zero since the force is at right angles to the displacement through which the force acts. $W = \int\vec{F}\cdot d\vec{s} = \int Fds\cos(\frac{\pi}{2}) = \int Fds(0) = 0$
Not in general. But if the force acts at right angles to the direction of motion at all times, which is the case in circular motion, there is no work done on the particle in changing its direction. For elliptical orbits, energy is required because the force is not at right angles to the motion (except at critical points). That energy comes from changes in potential energy of the particle. Total energy remains constant. As the kinetic energy of the particle in elliptical orbit increases the potential energy decreases by exactly the same amount.
AM
8. Nov 6, 2014
### A.T.
No, it doesn’t require energy to change the direction while preserving speed.
What energy? The kinetic energy doesn't increase, so why would there be energy coming from somewhere?
9. Nov 6, 2014
### CWatters
No. If it moves in a circle the acceleration is towards the centre (eg at right angles to the direction of travel) so there is no component "against" the direction of travel.
10. Nov 6, 2014
### rcgldr
Getting back to the original question ...
For the same reason why objects with a specific velocity retain that veloctiy in the absence of external forces. In the absence of external forces, then momentum, both linear and angular are conserved.
In the case of a spinning object the forces that keep the object from flying apart are internal centripetal forces, but even if the object does separate into multiple pieces, and absent any external forces, the angular momentum will continue to be conserved.
11. Nov 6, 2014
### CWatters
Sure but why does that matter? At all times the acceleration is towards the centre BUT the distance from the centre is constant.
Work = force * displacement
No energy is expended because there is no displacement in the direction of the force.
That's like asking if I bounce a ball off a wall where does the energy come from to decelerate and accelerate the ball? No extra energy is required to decelerate and accelerate the ball in the new direction. The KE of the ball is simply stored temporarily in the rubber spring of the ball.
12. Nov 6, 2014
### maajdl
The periphery is acted upon by the internal forces of the solid object.
If the object was not a solid, the angular momentum would also be conserved, but the "periphery" would not be rotating.
If the object was made of sand, the periphery would not rotate, but each part would go on a straight line.
Angular momentum would still be conserved, but the object would dislocate.
The fundamental reason why the angular momentum is conserved is the rotational symmetry of the system.
The orientation of the object in space has no influence. | 1,756 | 7,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.971851 |
http://nrich.maths.org/350/note | 1,503,073,819,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104704.64/warc/CC-MAIN-20170818160227-20170818180227-00502.warc.gz | 302,088,305 | 5,175 | ### Growing
Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)
### Remainder Hunt
What are the possible remainders when the 100-th power of an integer is divided by 125?
### Summit
Prove that the sum from t=0 to m of (-1)^t/t!(m-t)! is zero.
# Binomial
##### Stage: 5 Challenge Level:
Why do this problem?
The problem gives practice in using the notation for Binomial coefficients and manipulating algebraic expressions. In problem solving mode, if they can't get started, they might first try to work on the formula for small integer values of $n$.
Possible approach
Use as a revision exercise.
Key questions
If ${2n \choose n}$ is a binomial coefficient in the expansion of some power of $(1 + x)$ what can you say about the expansion and about the term where it occurs?
What do we know about ${n\choose r}$ and ${n\choose n-r}$?
Possible support
Ask learners to find the coefficient of $x^2$ in the expansion of $(1+x)^4$, the coefficient of $x^3$ in the expansion of $(1 +x)^6$ and then the coefficient of $x^4$ in the expansion of $(1 + x)^8$ and then ask them to try to connect their results to the problem given.
It might help to do the problem Summit first.
You could ask students to show that the sum of the $n$th row in Pascal's Triangle is $2^n$ first - so that they have a sense of achievement even if they don't succeed in proving the result in this problem. | 374 | 1,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-34 | latest | en | 0.893247 |
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04 Apr 2007, 02:43
Hi ! All
I have one question. I have gone through Official review Guide 11th ed. Find it very useful. Anybody gone through official guide verbal and math workouts. How you guys find it. are these books useful in addition to official review or just a lot of repeat questions from review guide. Any special features in these two.
Looking forward to a reply
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04 Apr 2007, 09:14
You will see repeat questions in those workouts. There are no special features in those books (just like another OG, but separate content).
What do you feel about your current level after completing the OG?
You should know where your weakness and target those areas.
If you want more verbal questions, use SC1000 and CR1000 here.
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06 Apr 2007, 03:45
Hi Booky
Thanks. Please tell where i can find these 1000 rc/cr/sc questions. i do find for math problem but do not find for these.
Thanks once again.
Neeraj
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06 Apr 2007, 05:57
CORRECTION: YOU WILL NOT SEE REPEAT QUESTIONS AMONG THE OG AND THE SUPPLEMENTS!!!
I found the supplement questions to be slightly easier and it had fewer questions compared with the original for the price. But the bottom line is that it contains official retired questions so it is a must have.
Last edited by kidderek on 06 Apr 2007, 09:59, edited 1 time in total.
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06 Apr 2007, 08:30
I didn't see any repeat questions on the Math supplement book and the OG11 either.
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### Show Tags
09 Apr 2007, 08:47
I have done both OG10, OG11 and verbal workout (not the math workout).
My impression was that many questions were familiar, which leads me think there were duplicate questions. Let me double check tonight.
09 Apr 2007, 08:47
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,024 | 3,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-04 | latest | en | 0.892383 |
http://mathhelpforum.com/calculus/24833-approximations-help-wit-1-more-please.html | 1,527,181,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00096.warc.gz | 200,565,835 | 10,145 | 1. ## Approximations, help wit 1 more please
Expand (1-8x)^1/2 as far as the term in x^2
Put x= -1 to obtain an approximation to (square root of) 5
..........10
Find the next term in the expansion and hence find two values between which (square route) 5
im clueless, dont even understand the question
2. A quick trip through a nice lecture on the Binomial Theorem should get you started.
PHYS208 Binomial Theorem
3. so unfair, its HNC mathematic not even physics lol
weve not even been through this at college
oh well ill thank you anyway
4. There are other ways to proceed.
Taylor Series?
Arbitrary coefficients.
Surely, if you have been given the problem you also have been given SOME tools.
5. no mate, everyone in the class has just handed in the assignment without doing the question but i dont like doing that
Is ther not just a simple tutorial, or a worked example of a the same kind of question you (or anybody) could show me, it would be a great help | 240 | 973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-22 | latest | en | 0.949616 |
https://community.airtable.com/t5/formulas/bullet-at-start-of-arrayjoin/td-p/74060 | 1,726,311,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00514.warc.gz | 154,321,903 | 50,517 | # Bullet at start of arrayjoin
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5 - Automation Enthusiast
Hello…
I’m using a rollup with ARRAYJOIN(values,"\n • ")
Which is giving me this result
Item 1
• Item 2
• Item 3
I want this result
• Item 1
• Item 2
• Item 3
How do I create a formula that puts a bullet on first item?
1 Solution
Accepted Solutions
18 - Pluto
To avoid an extra bullet if there are no records, put the initial bullet in an IF statement:
IF(COUNTA(values) >= 1, "• ") & ARRAYJOIN(values,"\n• ")
2 Replies 2
10 - Mercury
This will kind of do what you’re describing:
"• " & ARRAYJOIN(values,"\n• ")
But it’ll display that first • even if no records were linked - so it’s not the cleanest solution.
ARRAYJOIN isn’t really intended for bullet points. It’s meant to create comma separated lists and the like.
18 - Pluto
To avoid an extra bullet if there are no records, put the initial bullet in an IF statement:
IF(COUNTA(values) >= 1, "• ") & ARRAYJOIN(values,"\n• ") | 283 | 1,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.716192 |
https://www.lidolearning.com/questions/m-bb-rdsharma8-ch13-ex13p1-q4/the-cost-price-of-10-articles-/ | 1,686,107,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00545.warc.gz | 918,725,686 | 13,514 | # RD Sharma Solutions Class 8 Mathematics Solutions for Profit Loss Discount and Value Added Tax Exercise 13.1 in Chapter 13 - Profit Loss Discount and Value Added Tax
Question 4 Profit Loss Discount and Value Added Tax Exercise 13.1
The cost price of 10 articles is equal to the selling price of 9 articles. Find theprofit percent.
We know that the cost price of 10 article = selling price of 9 article
Let us consider CP of 1 article as Rs X
Selling price of 9 article = 10X Selling price of 1 article
= 10x/9
Profit = 10x/9 – x
= x/9
Profit % = Gain % = (gain/cost price) × 100
= (x/9)/x × 100
= 100/9
= 11 1/9%
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Connect with us on social media! | 272 | 913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-23 | latest | en | 0.866716 |
https://community.fangraphs.com/introducing-pwar-a-predictive-wins-measurement-for-pitchers/ | 1,624,299,941,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00501.warc.gz | 171,749,656 | 26,397 | # Introducing pWAR: A Predictive Wins Measurement for Pitchers
When WAR was first introduced, it attempted to answer a baseball question that has existed as long as teams have played professionally: how much can one player contribute to a team’s record? It is an ongoing debate to this day, but WAR is widely recognized as an excellent measure of overall performance. WAR numbers for major league players are regularly cited and recognized by both Major League Baseball and the Elias Sports Bureau. However, there is another question behind WAR, also long predating the existence of the statistic, and it’s a question that has yet to be answered by it. When a front office is deciding whether or not (or how aggressively) to pursue a player, they’re ultimately searching for the answer to one question — how many more games will we win if we get this guy?
Enter pWAR.
What is pWAR?
pWAR, or Predictive Wins Above Replacement, is a statistic attempting to estimate, in WAR, how much value a pitcher will bring to a team in the future.
How is it calculated?
The formula is based on FanGraphs’ fWAR formula for pitchers, but uses Connor Kurcon’s pCRA (Predictive Classified Run Average) as a runs allowed estimator in place of xFIP. The initial formula can be seen below:
pWAR = {[(pRAAP9/dRPW) + RL] * IP/9}
The individual formulas that make up pWAR are as follows:
Dynamic Runs Per Win (dRPW) = {[([(18 – IP/G)*(AL or NL pCRA)] + [(IP/G)*pCRA]) / 18] + 2 } * 1.5
Predictive Runs Above Average Per 9 (pRAAP9) = AL or NL pCRA – pCRA
Replacement Level (RL) = 0.03*(1 – GS/G) + 0.12*(GS/G)
Predictive Wins Per Game Above Average (pWPGAA) = pRAAP9/dRPW
Predictive Wins Per Game Above Replacement (pWPGAR) = pWPGAA + RL
Predictive Wins Above Replacement (pWAR) = pWPGAR * (IP/9)
Short answer: no. Long answer: CRA, the mother stat of pCRA, does have a WAR component, labeled cWAR, but no such stat exists for pCRA. CRA, an ERA estimator utilizing batted ball data, is described by Kurcon as “a descriptive stat, due to the lack of predictability of some inputs such as xBA, xSLG, and xwOBA.” pCRA utilizes a similar formula but incorporates Barrels as a form of batted ball data, which correlate more with future performance, making pCRA a much more predictive statistic, and thus a much better base for pWAR.
How can/should this stat be used?
It’s an important distinction to make that any pitcher’s pWAR for a season is not meant to indicate that season’s performance, but rather what to expect from that player going forward. If you wanted to take the “projection” piece a step further, you could replace innings pitched — or any — numbers with projected numbers or even your own guesses for a future year to calculate potential future pWAR. You would just need to make sure to adjust all other numbers accordingly so that you don’t wind up mixing current and future numbers and averages. This is essentially what the pitcher’s results “should have” been or how they “will” perform, all things considered. It is also important to note that this formula can (and will) be updated and improved as more tests are run and more data is collected.
Why not just use pCRA? How is this better?
This isn’t meant as a replacement for pCRA or any other ERA estimator, as those are excellent stats to use! The point of creating pWAR was not to create a “better” or more accurate stat, but simply to present another angle from which to view a pitcher’s abilities.
What’s next for pWAR?
As mentioned above, pWAR is not set in stone. As it begins to be utilized and more data is collected, I will continue to adjust and update it as needed.
Tal Schulmiller is a 16-year-old high school junior. He pitches for a club team and is looking to play and study data science and analytics in college while pursuing a career in sabermetrics. He can be contacted via email or on Twitter.
Member
rubesandbabes
This is good writing and a good article, too.
I would like this to be fleshed out a bit more for stupid people like me – I was not good in Algebra and I don’t get:
“pWAR = {[(pRAAP9/dRPW) + RL] * IP/9}”
I can’t figure out what is on offer.
I like the idea of replacing xFIP, which is really just a dressed up version of the strikeout K stat.
Thanks nice article. Enjoyed! | 1,051 | 4,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-25 | latest | en | 0.93875 |
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Show pattern of generated nrs:
Numbers in position
Press Ctrl (or equivalent) key to select more than one option
Sum, Range, Gap & Repeat
This is the smallest difference between two adjacent numbers in any generated lotto results combination. Let's look for instance at a random combination:
- 02, 05, 16, 20, 22, 31 -
The smallest difference in the above combination would be the one between 20 and 22 and that is 2.
If the smallest difference will be equal to any of the values you type (separated by one comma, e.g. 2,3) in the "BUT NOT..." field, that combination will be excluded. This allows to exclude smallest differences even if they are within the defined range.
Smallest difference
This is the greatest difference between two adjacent numbers in any generated lotto results combination. Let's look for instance at a random combination:
- 02, 05, 16, 20, 22, 31 -
The greatest difference in the above combination would be the one between 16 and 05 and that is 11.
If the greatest difference will be equal to any of the values you type (separated by one comma, e.g. 7,9) in the "BUT NOT..." field, that combination will be excluded. This allows to exclude greatest differences even if they are within the defined range.
Greatest difference
Repeating decades represents the number of same decades (the tens of a number) any lotto results combination can contain.
Let's take a look at an example:
- 02, 05, 16, 20, 22, 31 -
- 02, 05, 16, 20, 22, 29 -
If this slider would be set from 1 to 2 the first lotto combination above would be generated while the second combination would be excluded because it contains more than 2 repeating decades (20, 22, 29).
Repeating last digits represents the number of same last digits any combination can contain.
Let's take a look at an example:
- 02, 05, 16, 26, 22, 31 -
- 02, 06, 16, 20, 26, 29 -
If this slider would be set from 1 to 2 the first combination above would be generated while the second combination would be excluded because it contains more than 2 repeating last digits (06, 16, 26).
Repeating last digits
Consec., Even/Odd & Space
Min Max
Each generated lotto results combination should contain between Min and Max consecutives of 2.
Example:
- 02, 03, 11, 18, 23, 38 - contains consecutives of 2 one time (02, 03)
- 02, 03, 11, 12, 18, 23 - contains consecutives of 2 two times (02, 03 and 11, 12)
2 consec.
Each generated lotto results combination should contain between Min and Max consecutives of 3.
Example:
- 02, 03, 04, 18, 23, 38 - contains consecutives of 3 one time (02, 03, 04)
- 02, 03, 04, 12, 13, 14 - contains consecutives of 3 two times (02, 03, 04 and 12, 13, 14)
3 consec.
Each generated lotto results combination should contain between Min and Max consecutives of 4.
4 consec.
Each generated lotto results combination should contain between Min and Max consecutives of 5.
5 consec.
Each generated lotto results combination should contain between Min and Max consecutives of 6.
6 consec.
Min.E Max.E Min.O Max.O
Each generated lotto results combination should contain at least Min evens AND at least Min odds
EVENS/ODDS
Min Max
Space 1 controls the space between the 1st and 2nd lotto numbers of each generated lotto results combination.
It is basically the difference between the 1st and 2nd lotto number.
Space 1 range.
Space 2 controls the space between the 2nd and 3rd lotto numbers of each generated lotto results combination.
It is basically the difference between the 2nd and 3rd lotto number.
Space 2 range.
Space 3 controls the space between the 3rd and 4th lotto numbers of each generated lotto results combination.
It is basically the difference between the 3rd and 4th lotto number.
Space 3 range.
Space 4 controls the space between the 4th and 5th lotto numbers of each generated lotto results combination.
It is basically the difference between the 4th and 5th lotto number.
Space 4 range.
Space 5 controls the space between the 5th and 6th lotto numbers of each generated lotto results combination.
It is basically the difference between the 5th and 6th lotto number.
Space 5 range.
The Space repeat allowance will define how many times a space can be repeated within a combination.
To make this easier to understand, let's look at some examples:
Lotto results with 4 equal spaces:
01, 02, 04, 05, 06, 07, 11
(2-1=1, 4-2=2, 5-4=1, 6-5=1, 7-6=1, 11-7=4)
Lotto results with 3 equal spaces:
02, 04, 09, 11, 13, 17
(4-2=2, 9-4=5, 11-9=2, 13-11=2, 17-13=4)
Lotto results with 5 equal spaces:
05, 10, 15, 20, 25, 30
(10-5=5, 15-10=5, 20-15=5, 25-20=5, 30-25=5)
If Space repeat Min and Max would be set to 1 and 3 respectively, only the second example combination would be generated as it has three equal spaces. The other two examples would be excluded.
Space repeat
Lotto history filters
These filters can be applied by Level 3 members only.
During all known history of this lotto
Min Max Not
If any of the generated lotto results combinations will contain 6 numbers that have been drawn less than Min or more than Max times during ALL draws, that combination will be removed. Same if 6 of the combination numbers have been drawn a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
6 numbers
!=
If any of the generated lotto results combinations will contain 5 numbers that have been drawn less than Min or more than Max times during ALL draws, that combination will be removed. Same if 5 of the combination numbers have been drawn a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
5 numbers
!=
If any of the generated lotto results combinations will contain 4 numbers that have been drawn less than Min or more than Max times during ALL draws, that combination will be removed. Same if 4 of the combination numbers have been drawn a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
4 numbers
!=
If any of the generated lotto results combinations will contain 3 numbers that have been drawn less than Min or more than Max times during ALL draws, that combination will be removed. Same if 3 of the combination numbers have been drawn a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
3 numbers
!=
If any of the generated lotto results combinations will contain 2 numbers that have been drawn less than Min or more than Max times during ALL draws, that combination will be removed. Same if 2 of the combination numbers have been drawn a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
2 numbers
!=
During the past 100 draws
If any of the generated lotto results combinations will contain 5 numbers that have been drawn less than Min or more than Max times during the last 100 draws (or less), that combination will be removed. Same if 5 of the combination numbers have been drawn a number of Not (separated by ONE comma ONLY) times during the last 100 draws (or less). Look up the values for this filter in the matrix table (the columns at the right side). 5 numbers != != != !=
During all known history of this lotto
Min Max Not
If any of the generated lotto results combinations will contain 6 numbers that have been drawn -- in the EXACT same order -- less than Min or more than Max times during ALL draws, that combination will be removed. Same if 6 of the combination numbers have been drawn -- in the EXACT same order -- a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
6 STRICT
!=
If any of the generated lotto results combinations will contain 5 numbers that have been drawn -- in the EXACT same order -- less than Min or more than Max times during ALL draws, that combination will be removed. Same if 5 of the combination numbers have been drawn -- in the EXACT same order -- a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
5 STRICT
!=
If any of the generated lotto results combinations will contain 4 numbers that have been drawn -- in the EXACT same order -- less than Min or more than Max times during ALL draws, that combination will be removed. Same if 4 of the combination numbers have been drawn -- in the EXACT same order -- a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
4 STRICT
!=
If any of the generated lotto results combinations will contain 3 numbers that have been drawn -- in the EXACT same order -- less than Min or more than Max times during ALL draws, that combination will be removed. Same if 3 of the combination numbers have been drawn -- in the EXACT same order -- a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
3 STRICT
!=
If any of the generated lotto results combinations will contain 2 numbers that have been drawn -- in the EXACT same order -- less than Min or more than Max times during ALL draws, that combination will be removed. Same if 2 of the combination numbers have been drawn -- in the EXACT same order -- a number of Not (separated by ONE comma ONLY) times during ALL draws.
Look up the values for this filter in the matrix table (the columns at the right side).
2 STRICT
!=
During the past 100 draws
If any of the generated lotto results combinations will contain 5 numbers that have been drawn -- in the EXACT same order -- less than Min or more than Max times during the last 100 draws (or less), that combination will be removed. Same if 5 of the combination numbers have been drawn -- in the EXACT same order -- a number of Not (separated by ONE comma ONLY) times during the last 100 draws (or less). Look up the values for this filter in the matrix table (the columns at the right side). 5 STRICT != != != !=
Each generated lotto results combination should contain between Min and Max lotto numbers from the previous lotto results. Each generated lotto results combination should contain between Min and Max NEIGHBORS OF THE lotto numbers from the previous lotto results. Numbers from last draw Neighbors from last draw 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Key, Exclude & Trends
The keys are the numbers that MUST be contained by each generated lotto results combination. Unlike most lotto wheels, this allows you to specify any number of keys (separated by one comma, e.g. 15,23,31) The Min/Max Keys each combination can contain. Key(s) Min/Max Keys/combination 1 2 3 4 5 6 1 2 3 4 5 6 Anywhere Up Down Anywhere BUT Up or Down ⩾ or ⩽ Any
Numbers to wheel 0 selected
Nr. in upper left corner = nr. frequency
Settings
Set the amount of numbers that you'd like to have displayed in one row under the "Numbers to wheel" and "Match against" sections. Numbers to wheel layout (nrs/row) include neighbors Yes No Yes No Yes No
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https://www.convertunits.com/from/acre+inch/minute/to/ounce/hour+%5BUS%5D | 1,670,345,454,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711111.35/warc/CC-MAIN-20221206161009-20221206191009-00002.warc.gz | 745,157,095 | 24,163 | ## ››Convert acre inch/minute to ounce/hour [US]
acre inch/minute ounce/hour [US]
Did you mean to convert acre inch/minute acre inch/minute [survey] to ounce/hour [US] ounce/hour [UK]
## ››More information from the unit converter
How many acre inch/minute in 1 ounce/hour [US]? The answer is 4.7951301313848E-9.
We assume you are converting between acre inch/minute and ounce/hour [US].
You can view more details on each measurement unit:
acre inch/minute or ounce/hour [US]
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 0.58371349836816 acre inch/minute, or 121730481.21211 ounce/hour [US].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between acre inches/minute and ounces/hour.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of acre inch/minute to ounce/hour [US]
1 acre inch/minute to ounce/hour [US] = 208544913.81889 ounce/hour [US]
2 acre inch/minute to ounce/hour [US] = 417089827.63778 ounce/hour [US]
3 acre inch/minute to ounce/hour [US] = 625634741.45666 ounce/hour [US]
4 acre inch/minute to ounce/hour [US] = 834179655.27555 ounce/hour [US]
5 acre inch/minute to ounce/hour [US] = 1042724569.0944 ounce/hour [US]
6 acre inch/minute to ounce/hour [US] = 1251269482.9133 ounce/hour [US]
7 acre inch/minute to ounce/hour [US] = 1459814396.7322 ounce/hour [US]
8 acre inch/minute to ounce/hour [US] = 1668359310.5511 ounce/hour [US]
9 acre inch/minute to ounce/hour [US] = 1876904224.37 ounce/hour [US]
10 acre inch/minute to ounce/hour [US] = 2085449138.1889 ounce/hour [US]
## ››Want other units?
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ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 670 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.784168 |
https://www.tutorialspoint.com/p-john-s-income-is-20-more-than-that-of-mr-thomas-how-much-percentage-is-the-income-of-mr-thomas-less-than-that-of-john-p | 1,696,035,235,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510529.8/warc/CC-MAIN-20230929222230-20230930012230-00004.warc.gz | 1,130,453,245 | 17,824 | # John's income is 20% more than that of Mr Thomas. How much percentage is the income of Mr Thomas less than that of John?
Given: John's Income Is 20% More Than That Of Mr Thomas.
To do: Find how much % is income of Mr Thomas Less than John.
Solution:
Let the income of Thomas be Rs 100
$20$% of Rs. 100 $= \frac{20}{100}\times 100 = Rs. 20$
John's income $=Rs. (100 + 20) = Rs. 120$
The income of Thomas less than that of John in Percent = $\frac{20}{120} \times 100$ = 16.67%.
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Updated on: 10-Oct-2022
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http://docplayer.net/20701844-An-improved-dynamic-programming-decomposition-approach-for-network-revenue-management.html | 1,539,980,534,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512434.71/warc/CC-MAIN-20181019191802-20181019213302-00539.warc.gz | 92,244,718 | 28,465 | # An Improved Dynamic Programming Decomposition Approach for Network Revenue Management
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## Transcription
1 An Improved Dynamic Programming Decomposition Approach for Network Revenue Management Dan Zhang Leeds School of Business University of Colorado at Boulder May 21, 2012
2 Outline Background Network revenue management formulation Classical dynamic programming decomposition An improved dynamic programming decomposition Numerical results Summary
3 Network RM Formulation (Gallego and van Ryzin, 1997; Gallego et. al., 2004; Liu and van Ryzin, 2008) m resources with capacity c (an m-vector) Capacity for resource i is c i. n products N = {1,..., n} Fare for product j is f j Product consumption matrix A = [a ij ] Finite time horizon with length τ In each period, there is one customer arrival with probability λ, and no customer arrival with probability 1 λ. Given a set of products S N, a customer chooses product j with probability P j (S). No-purchase probability P 0 (S) = 1 j S P j (S). Objective: Maximize expected total revenue
4 Applications Industry Resources Products Airlines Scheduled flights O-D itineraries at certain fare levels Hotels Room-days Single(multi)-day stays at certain rates Car rentals Car-days Single(multi)-day rentals at certain rates Air Cargo Scheduled flights (weight) O-D shipments at certain rates Scheduled flights (volume)
5 Dynamic Programming Formulation DP optimality equations: v t (x) = v τ+1 (x) = 0, max S N(x) { λ j S P j (S)(f j + v t+1 (x A j )) } + (λp 0 (S) + 1 λ)v t+1 (x), t, x, x. Notations v t (x): DP value function A j : resource incidence vector of product j N(x): {j N : x A j }
6 Dynamic Programming Formulation DP optimality equations: v t (x) = v τ+1 (x) = 0, max S N(x) { λ j S P j (S)(f j + v t+1 (x A j )) } + (λp 0 (S) + 1 λ)v t+1 (x), t, x, x. Notations v t (x): DP value function A j : resource incidence vector of product j N(x): {j N : x A j } Curse of dimensionality: state space grows exponentially with the number of resources
7 Choice-based Deterministic Linear Program (CDLP) z CDLP = max h S N λr(s)h(s) S N λq(s)h(s) c, h(s) τ, S N (Resource constraint) (Time constraint) h(s) 0, S N. (Non-negativity) Replace stochastic demand with deterministic fluid with rate λ Given offer set S N Total time S is offered: h(s) Revenue from unit demand: R(S) = j S f jp j (S) Consumption of resource i from unit demand: Q i (S) = j S a ijp j (S)
8 CDLP (Gallego et. al, 2004; Liu and van Ryzin, 2008) CDLP can by efficiently solved for certain class of choice models. The vector of dual values π associated with resource constraints can be used as bid-prices for resources z CDLP provides an upper bound on revenue Some recent references: Talluri (2010): Concave programming formulation Gallego, Ratliff, Shebalov (2011): Efficient reformulation
9 Classical Dynamic Programming Decomposition For each i, approximate the DP value function with v t(x) v t,i (x i ) + πk x k, } {{ } k i Value of the } {{ } i-th resource Value of all other resources t, x.
10 Classical Dynamic Programming Decomposition For each i, approximate the DP value function with v t(x) v t,i (x i ) + πk x k, } {{ } k i Value of the } {{ } i-th resource Value of all other resources t, x. Using the approximation in DP recursion leads to ( v t,i (x i ) = max λp j (S) f j ) a kj πk +v t+1,i (x i a ij ) S N(x i,c i ) j S k i } {{ } Fare proration + (λp 0(S) + 1 λ)v t+1(x i ), t, x i.
11 Classical Dynamic Programming Decomposition For each i, approximate the DP value function with v t(x) v t,i (x i ) + πk x k, } {{ } k i Value of the } {{ } i-th resource Value of all other resources t, x. Using the approximation in DP recursion leads to ( v t,i (x i ) = max λp j (S) f j ) a kj πk +v t+1,i (x i a ij ) S N(x i,c i ) j S k i } {{ } Fare proration + (λp 0(S) + 1 λ)v t+1(x i ), t, x i. Compute offer sets dynamically using the approximate value function v t(x) i v t,i (x i ), t, i.
12 Classical Dynamic Programming Decomposition A DP with m-dimensional state space is reduced to m one-dimensional DPs, one for each resource states (Assume 100 seats per flight) states
13 Classical Dynamic Programming Decomposition A DP with m-dimensional state space is reduced to m one-dimensional DPs, one for each resource states (Assume 100 seats per flight) states Variants of the approach are widely used in practice. Review: Talluri and van Ryzin (2004a)
14 DP Decomposition Bounds Proposition (Zhang and Adelman, 2009) The following relationships hold: (i) v t (x) min l=1,...,m {v t,l (x l ) + } k l π k x k v t,i (x i ) + k i π k x k, i, t, x; (ii) v 1 (c) v 1,i (c i ) + k i π k c k z CDLP, i. Decomposition value for each leg provides an upper bound on revenue Decomposition bounds are tighter than the bound from CDLP
15 Linear Programming Formulation of DP (Adelman, 2007) min v 1 (c) {v t( )} t v t (x) j S λp j (S)(f j + v t+1 (x A j )) + (λp 0 (S) + 1 λ)v t+1 (x), t, x, S N(x).
16 Linear Programming Formulation of DP (Adelman, 2007) min v 1 (c) {v t( )} t v t (x) j S λp j (S)(f j + v t+1 (x A j )) + (λp 0 (S) + 1 λ)v t+1 (x), t, x, S N(x). Huge number of decision variables and constraints
17 Linear Programming Formulation of DP (Adelman, 2007) min v 1 (c) {v t( )} t v t (x) j S λp j (S)(f j + v t+1 (x A j )) + (λp 0 (S) + 1 λ)v t+1 (x), t, x, S N(x). Huge number of decision variables and constraints Functional approximation idea: use a parameterized representation of the value function to reduce the number of decision variables
18 The Affine Functional Approximation (Zhang and Adelman, 2009) Affine approximation is given by v t (x) θ t + i V t,i x i, t, x. (1)
19 The Affine Functional Approximation (Zhang and Adelman, 2009) Affine approximation is given by v t (x) θ t + i V t,i x i, t, x. (1) Using (1) in the linear programming formulation leads to min θ,v θ1 + i V 1,i c i ( θ t + V t,i x i λp j (S) f j + θ t+1 + V t+1,i (x i a ij ) i j S i ( + (λp 0(S) + 1 λ) θ t+1 + ) V t+1,i x i, t, x, S N(x). i )
20 The Affine Functional Approximation The dual program is given by z P1 = max λp j (S)f j Y t,x,s Y t,x,s N(x) j S { c i, if t = 1, x i Y t,x,s = x,s N(x) (x i ) j S λp j (S)a ij Y t 1,x,S, t = 2,..., τ x,s N(x) { Y t,x,s = 1, if t = 1, x,s N(x) Y t 1,x,S, t = 2,..., τ. x,s N(x) Y 0. i, t,
21 The Affine Functional Approximation The dual program is given by z P1 = max λp j (S)f j Y t,x,s Y t,x,s N(x) j S { c i, if t = 1, x i Y t,x,s = x,s N(x) (x i ) j S λp j (S)a ij Y t 1,x,S, t = 2,..., τ x,s N(x) { Y t,x,s = 1, if t = 1, x,s N(x) Y t 1,x,S, t = 2,..., τ. x,s N(x) Y 0. i, t, Due to the large number of columns, solving the linear program above still requires considerable computational effort.
22 Functional Approximation Approaches for Network RM Citation Choice Model Functional approximation Solution strategy Adelman (2007) Independent demand Affine Column generation Zhang and Adelman (2009) MNLD Affine Column generation Zhang (2011) MNLD Nonlinear non-separable CDLP+Simultaneous DP Liu and van Ryzin (2008) MNLD Separable (fare proration) CDLP+DP Decomposition Miranda Bront et. al. (2009) MNLO Separable (fare proration) CDLP+DP Decomposition Farias and Van Roy (2008) Independent demand Separable concave Constraint sampling Meissner and Strauss (2012) MNLD Separable concave Column generation Kunnumkal and MNLD Separable (fare proration) Convex programming Topaloglu (2011) +DP Decomposition Tong and Topaloglu (2011) Independent demand Affine Reduction + Constraint generation Vossen and Zhang Independent demand Affine Reduction + MNLD + Dynamic disaggregation MNLD: Multinomial logit model with disjoint consideration sets MNLO: Multinomial logit model with overlapping consideration sets
23 Research Questions Computational cost: ADP (affine or separable concave approximation) classical DP decomposition
24 Research Questions Computational cost: ADP (affine or separable concave approximation) classical DP decomposition How can we balance solution quality with solution time? Can we improve the classical DP decomposition?
25 A Strong Functional Approximation (Zhang, 2011) v t (x) min ˆv t,i(x i ) + πk x k, t, x. i k i Nonlinear and non-separable functional approximation Each value v t (x) is approximated by a single value across legs Motivated by the decomposition bounds (Zhang and Adelman, 2009)
26 A Nonlinear Optimization Problem Using the new functional approximation leads to z NLP = min min ˆv 1,i(c i ) + πk c k ˆv t,i ( ) t,i i k i min ˆv t,i(x i ) + πk x k i k i λp j (S) f j + min ˆv t+1,l(x l a lj ) + πk (x k a kj ) l j S k l + (λp 0(S) + 1 λ)min ˆv t+1,l(x l ) + πk x k l, t, x, S N(x). k l The problem is a nonlinear optimization problem with a huge number of nonlinear constraints.
27 A Restricted Optimization Problem Step 1: Writing each constraint as m equivalent constraints Step 2: Restricting the constraints so that each constraint only involves one resource The restricted problem provides a relaxed bound: Proposition The objective value of the restricted program, z NLP, is bigger than z NLP.
28 An Equivalent Simultaneous Dynamic Program ˆv t,i(x i ) = max S N(x i,c i ) { min l i ( { λp j (S) f j + min ˆv t+1,i(x i a ij ) πk a kj, j S k i max [ˆv t+1,l(y l ) y l πl ] } }) a kj πk + πi x i 0 y l c l a lj k + (λp 0(S) + 1 λ) min i, t, x i. { ˆv t+1,i(x i ), min l i { max 0 y l c l [ˆv t+1,l(y l ) π l y l ] + π i x i DP recursion for resource i involves values from all other resources }}
29 An Equivalent Simultaneous Dynamic Program ˆv t,i(x i ) = max S N(x i,c i ) { min l i ( { λp j (S) f j + min ˆv t+1,i(x i a ij ) πk a kj, j S k i max [ˆv t+1,l(y l ) y l πl ] } }) a kj πk + πi x i 0 y l c l a lj k + (λp 0(S) + 1 λ) min i, t, x i. { ˆv t+1,i(x i ), min l i { max 0 y l c l [ˆv t+1,l(y l ) π l y l ] + π i x i DP recursion for resource i involves values from all other resources The dynamic program is equivalent to the restricted nonlinear program can be solved efficiently via a simultaneous dynamic programming algorithm leads to tighter revenue bounds }}
30 New Bounds Proposition (Zhang, 2011) Let {ˆv t,i ( )} t,i,x i be the optimal solution from the simultaneous dynamic program. The following results hold: (i) ˆv t,i (x i) v t,i (x i ), i, x i ; (ii) v 1 (c) z NLP z NLP = min i {ˆv 1,i (c i) + } k i π k c k min i {v 1,i (c i ) + } k i π k c k z CDLP. The simultaneous dynamic program provides tighter bounds on revenue than the classical decomposition.
31 Recap High dimensional dynamic program
32 Recap High dimensional dynamic program Large scale linear program
33 Recap High dimensional dynamic program Large scale linear program Large scale nonlinear program with nonlinear constraints
34 Recap High dimensional dynamic program Large scale linear program Large scale nonlinear program with nonlinear constraints Restricted nonlinear program with nonlinear constraints
35 Recap High dimensional dynamic program Large scale linear program Large scale nonlinear program with nonlinear constraints Restricted nonlinear program with nonlinear constraints Simultaneous dynamic program
36 Comparison: Classical vs. Improved Approaches Classical dynamic programming decomposition: Solve m single-leg DPs Prorated fares Fare proration Static bid-prices Solve CDLP
37 Comparison: Classical vs. Improved Approaches Classical dynamic programming decomposition: Solve m single-leg DPs Prorated fares Fare proration Static bid-prices Solve CDLP Network effects only captured through fare proration
38 Comparison: Classical vs. Improved Approaches Classical dynamic programming decomposition: Solve m single-leg DPs Prorated fares Fare proration Static bid-prices Solve CDLP Network effects only captured through fare proration Improved dynamic programming decomposition: Solve one simultaneous DP Static bid-prices Solve CDLP
39 Comparison: Classical vs. Improved Approaches Classical dynamic programming decomposition: Solve m single-leg DPs Prorated fares Fare proration Static bid-prices Solve CDLP Network effects only captured through fare proration Improved dynamic programming decomposition: Solve one simultaneous DP Static bid-prices Solve CDLP Network effects captured during DP recursion!
40 Computational Study: Problem Instances Randomly generated hub-and-spoke instances Number of non-hub locations (flights) in the set {4, 8, 16, 24} Number of periods in the set {100, 200, 400, 800} Two products for each possible itinerary Multinomial Logit Choice Model with Disjoint Consideration Sets (MNLD) Largest problem instance: 24 non-hub locations (flights), 336 products, 800 periods
41 Numerical Study: Policies DCOMP1: the new decomposition approach where the approximation m v t (x) ˆv t,i(x i ), t, x i=1 is used to compute control policies. DCOMP: the classical dynamic programming decomposition CDLP: static bid-price policy based on the dual values of resource constraints in CDLP CDLP10: A version of CDLP that resolves 10 times with equally spaced resolving intervals Each policy is simulated times
42 Computational Time Case # Parameters Capacity Load CPU seconds DCOMP1 DCOMP per leg factor CDLP DCOMP DCOMP1 DCOMP A1 (100,4,4,16) % A2 (200,4,4,16) % A3 (400,4,4,16) % A4 (800,4,4,16) % A5 (100,8,8,48) % A6 (200,8,8,48) % A7 (400,8,8,48) % A8 (800,8,8,48) % A9 (100,16,16,160) % A10 (200,16,16,160) % A11 (400,16,16,160) % A12 (800,16,16,160) % A13 (100,24,24,336) % A14 (200,24,24,336) % A15 (400,24,24,336) % A16 (800,24,24,336) %
43 Bound Performance Case # CDLP DCOMP DCOMP1 Bound improvement %-difference across legs bound bound bound %-CDLP %-DCOMP DCOMP DCOMP1 A % 4.74% 4.46% 0.00% A % 1.55% 1.87% 0.36% A % 0.98% 2.36% 0.00% A % 0.36% 0.58% 0.00% A % 0.99% 3.18% 0.05% A % 0.21% 2.18% 0.22% A % 0.18% 1.09% 0.00% A % 0.03% 1.49% 0.00% A % 0.76% 4.69% 0.00% A % 0.64% 2.95% 0.03% A % 0.30% 1.52% 0.01% A % 0.15% 0.94% 0.00% A % 14.05% 10.37% 0.00% A % 2.27% 5.50% 0.00% A % 0.66% 2.80% 0.03% A % 0.24% 1.44% 0.00%
44 Bounds from Individual Legs Bounds from individual Legs DCOMP DCOMP Leg
45 Bounds from Individual Legs Bounds from individual Legs DCOMP DCOMP Leg DCOMP1 bounds are more homogeneous across legs
46 A Hub-and-spoke Network with 2 Non-Hub Locations Case # τ Load Capacity DCOMP1 DCOMP1 Revenue Gains OPT-GAP factor per leg REV %-CDLP %-CDLP10 %-DCOMP B % % 8.56% 2.74% B % % 5.74% 4.11% B % -1.00% -0.33% 0.03% B % % 4.41% 7.17% B % 48.88% 3.00% 0.09% B % 13.42% 3.62% 5.77% B % % 4.78% 7.97% B % 42.16% -4.49% 0.00% B % 13.27% 1.33% 1.94% B % 15.05% 0.98% 4.65% B % 15.46% 1.70% 8.14% B % 15.47% 2.45% 5.52% B % 2.89% 0.47% 0.34% B % 4.84% 1.14% 0.11% B % 22.11% 2.22% 0.04% B % 7.92% 2.14% 0.01% B % 23.82% 1.46% 0.18% B % 2.61% 1.24% 0.04% B % 30.16% 2.48% 0.01% B % 31.46% 2.60% 0.00%
47 DCOMP1 Percentage Revenue Gain vs. Load Factor DCOMP1 percentage revenue gain % CDLP10 % DCOMP Load factor
48 DCOMP1 Percentage Revenue Gain vs. Load Factor DCOMP1 percentage revenue gain % CDLP10 % DCOMP Load factor Higher load factor Higher revenue gains
49 DCOMP1 Percentage Revenue Gain vs. Number of Periods DCOMP1 percentage revenue gain % CDLP10 % DCOMP Number of periods
50 DCOMP1 Percentage Revenue Gain vs. Number of Periods DCOMP1 percentage revenue gain % CDLP10 % DCOMP Number of periods Significant revenue gains for problems with long selling horizons!
51 A Hub-and-spoke Network with 4 Non-Hub Locations Case # τ Load Capacity DCOMP1 DCOMP1 Revenue Gains OPT-GAP factor per leg REV %-CDLP %-CDLP10 %-DCOMP C % 14.60% 0.46% 0.10% C % 52.79% 1.95% 1.32% C % 12.35% -0.11% 0.03% C % 52.31% 1.36% 0.66% C % 44.57% 1.71% 1.96% C % 14.18% 1.72% 2.67% C % 29.01% -2.27% -0.81% C % 1.02% 0.49% 0.18% C % 4.98% 1.46% 2.67% C % 4.18% 1.61% 4.11% C % 3.44% 1.51% 4.43% C % 57.07% 1.92% 3.08% C % 1.30% -0.90% 0.21% C % 2.18% 0.21% 0.08% C % 6.38% 0.74% 0.00% C % 2.66% 0.89% -0.06% C % 2.50% -0.81% 0.44% C % 3.05% -0.03% 0.20% C % 3.24% 0.26% 0.05% C % 3.21% 0.41% 0.02%
52 DCOMP1 Percentage Revenue Gain vs. Load Factor DCOMP1 percentage revenue gain % CDLP10 % DCOMP Load factor
53 DCOMP1 Percentage Revenue Gain vs. Number of Periods DCOMP1 percentage revenue gain % CDLP10 % DCOMP Number of periods
54 Summary and Future Directions Functional approximation approach is promising for solving large scale stochastic dynamic programs. However, implementations of the approach often require very high computational cost. The first nonlinear non-separable functional approximation for network RM problem Novel approximation architecture Better revenue bounds Improved heuristic policies Moderate computational cost Current work Exploiting special structures of the LP formulations of dynamic programs in value function approximation (Vossen and Zhang, 2012) Applications with real data (Zhang and Weatherford, 2012)
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Transportation Polytopes: a Twenty year Update Jesús Antonio De Loera University of California, Davis Based on various papers joint with R. Hemmecke, E.Kim, F. Liu, U. Rothblum, F. Santos, S. Onn, R. Yoshida, | 11,526 | 44,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-43 | longest | en | 0.790276 |
https://www.iovi.com/books/algorithms-manual/leetcode/0108.html | 1,679,910,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00631.warc.gz | 897,947,195 | 13,421 | # 108. Convert Sorted Array to Binary Search Tree (Easy)
https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
```Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
```
## Solutions
``````class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
if (nums == null || nums.length == 0) {
return null;
}
return add(nums, 0, nums.length - 1);
}
private TreeNode add(int[] nums, int start, int end) {
// check out if overstep the boundary
if (start > end || start < 0 || start >= nums.length || end < 0 || end >= nums.length) {
return null;
}
// if only 1 element left in the array
if (start == end) {
return new TreeNode(nums[start]);
}
// find the medium and create a new node as local root node
int middle = (start + end) / 2;
int medium = nums[middle];
TreeNode root = new TreeNode(medium);
// split the array into two sections and creates corresponding TreeNodes.
root.left = add(nums, start, middle - 1);
root.right = add(nums, middle + 1, end);
return root;
}
}
`````` | 368 | 1,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-14 | longest | en | 0.525621 |
https://speakerdeck.com/dcernst/impartial-geodetic-convexity-achievement-and-avoidance-games-on-graphs | 1,723,140,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00035.warc.gz | 440,388,348 | 19,536 | Dana Ernst
April 27, 2023
160
# Impartial geodetic convexity achievement & avoidance games on graphs
A set P of vertices of a graph G is convex if it contains all vertices along shortest paths between vertices in P. The convex hull of P is the smallest convex set containing P. We say that a subset of vertices P generates the graph G if the convex hull of P is the entire vertex set. We study two impartial games Generate and Do Not Generate in which two players alternately take turns selecting previously-unselected vertices of a finite graph G. The first player who builds a generating set for the graph from the jointly-selected elements wins the achievement game GEN(G). The first player who cannot select a vertex without building a generating set loses the avoidance game DNG(G). Similar games have been considered by several authors, including Harary et al. In this talk, we determine the nim-number for several graph families, including trees, cycle graphs, complete graphs, complete bipartite graphs, and hypercube graphs.
Joint work with Bret Benesh, Marie Meyer, Sarah Salmon, and Nandor Sieben.
This talk was given on January 25, 2023 during the Combinatorial Game Theory Colloquia in S. Miguel, Azores.
April 27, 2023
## Transcript
1. ### Impartial geodetic convexity achievement & avoidance games on graphs Combinatorial
Game Theory Colloquium IV Dana C. Ernst Northern Arizona University January 25, 2023 Joint with B. Benesh, M. Meyer, S. Salmon, and N. Sieben Partial support from The Institute for Computational and Experimental Research in Mathematics (ICERM)
2. ### Graph Theory • We assume collection of vertices V is
nonempty and finite. • A geodesic of a graph is a shortest path between two vertices. The geodetic closure I[P] of a subset P ⊆ V consists of the vertices along the geodesics connecting two vertices in P. • A subset P ⊆ V is called (geodetically) convex if it contains all vertices along the geodesics connecting two vertices of P. • The convex hull of P is defined via [P] := {K | P ⊆ K, K is convex} and is the smallest convex set containing P. • We say that a subset P of vertices is generating if [P] = V . 1
3. ### Geodetic Closure vs Convex Hull Comments • Despite the name,
geodetic closure is not necessarily a closure operator because it may not be idempotent. To make a closure operator, we need to iterate the geodetic closure function until the result stabilizes. • Convex hull is this closure operator. Example Consider the complete bipartite graph K2,3 . I[{c, d}] b a c d e [{c, d}] b a c d e 2
4. ### Maximal Nongenerating Sets Definition The family of maximal nongenerating sets
of a graph G is denoted by N(G). That is, N(G) := {N ⊆ V | [N] = V but for all v / ∈ N, [N ∪ {v}] = V }. Example Consider the cycle graph C4 and the diamond graph G. a b c d a b c d C4 G The maximal nongenerating subsets of C4 are {a, b}, {b, c}, {c, d}, {a, d}. On the other hand, the maximal nongenerating sets of the diamond graph are {a, b, c} and {a, c, d}. 3
5. ### Game Definitions Definition For each of the games, we play
on a graph G = (V , E). Two players take turns selecting previously unselected vertices until certain conditions are met. • For the achievement game generate GEN(G), the game ends as soon as [P] = V . That is, the player who generates the whole vertex set first wins. • For the avoidance game do not generate DNG(G), all positions P must satisfy [P] = V . The player who cannot select a vertex without generating the vertex set loses. 4
6. ### Example Consider the wheel graph W5 . Below is a
“representative” game digraph for DNG(W5 ). Note: Positions can never contain antipodal “rim” vertices. ∗1 ∗0 ∗0 ∗1 ∗1 ∗0 5
7. ### Example Below is a “representative” game digraph for GEN(W5 ).
∗2 ∗1 ∗0 ∗0 ∗2 ∗2 ∗1 ∗0 ∗0 ∗0 6
8. ### Similar Games Comments Similar games have been considered by several
authors, including Buckley/Harary, Fraenkel/Harary, Necascova, Haynes/Henning/Tiller, and Wang. These variations differ in at least one of the following: • The collection of vertices generated by the selected vertices corresponds to the geodetic closure as opposed to the convex hull. (Buckley/Harary) • The generated vertices of the selected vertices are not available as moves. The games we study are a generalization of the achievement and avoidance games played on groups introduced by Anderson/Harary and extensively studied by Benesh/Ernst/Sieben. 7
9. ### “This one is easy.” – Sergei Kuznetsov Comments The games
DNG(G) and GEN(G) are completely determined by N(G). • The set of terminal positions of DNG(G) is N(G). • A subset P ⊆ V is a position of GEN(G) if and only if P \ {v} ⊆ N for some v ∈ V and N ∈ N(G). The following theorem quickly handles the determination of the nim-number for DNG(G) for several families of graphs. Theorem (BEMSS) If G is a graph and every element of N(G) has the same parity r ∈ {0, 1}, then the nim-number of DNG(G) is r. 8
10. ### Complete Graphs Theorem (BEMSS) For the complete graph Kn ,
we have: • N(Kn ) = {V \ {v} | v ∈ V }. • nim(DNG(Kn )) = pty(n − 1). Proof. This follows from “This one is easy” since every position of N(Kn ) has the same parity. • nim(GEN(Kn )) = pty(n). Proof. The only way to generate V is to select each vertex. If n is even, the second player wins by random play. If n is odd, the second player wins GEN(Kn ) + ∗1 again by random play. 9
11. ### Trees, Path Graphs, & Star Graphs Theorem (BEMSS) If T
is a tree with set of leaves of L, then we have: • N(T) = {{l}c | l ∈ L}. • nim(DNG(T)) = pty(|V |−1). Proof. Again, this follows from “This one is easy” since every position of N(Kn ) has the same parity. • nim(GEN(T)) = pty(V ). Proof. One approach is to use structural induction on the diagram that results from structure equivalence. 10
12. ### Cycle Graphs Theorem (BEMSS) For the cycle graph Cn (n
≥ 3), assume V = Zn and E = {{i, i + 1} | i ∈ V }. • N(Cn ) = {{i + 1, . . . , i + (n + 1)/2} | i ∈ V }, if n odd {{i + 1, . . . , i + n/2} | i ∈ V }, if n even . • nim(DNG(Cn )) = 1, if n ≡4 1, 2 0, if n ≡4 3, 0. Proof. Surprise! . . . “This one is easy” (some thought required to determine parity). • nim(GEN(Cn )) = pty(n). Proof. If n is even, then 2nd player wins in 2nd move by selecting the antipodal vertex. If n is odd, then 1st player wins on 3rd move by selecting a vertex in the “middle” of the larger group of unselected vertices. 11
13. ### Hypercube Graphs Theorem (BEMSS) For the hypercube graph Qn (binary
strings vertices connected by an edge exactly when they differ by a single digit), we have: • For n ≥ 2, N(Qn ) is collection of sets consisting of vertices agreeing on a fixed entry. • nim(DNG(Qn )) = 0. Proof. Note that Q1 = K1 , so the result follows from earlier theorem. For n ≥ 2, every set in N(Qn ) has size 2n−1, so the result follows from “This one is easy”. • nim(GEN(Qn )) = 0. Proof. The 2nd player wins by selecting the antipodal vertex to the choice of 1st player, and every antipodal pair forms a minimal generating set. 12
14. ### Complete Bipartite Graphs Theorem (BEMSS) Consider the complete bipartite graph
Km,n where n ≥ m ≥ 2 with the set V of vertices partitioned into A = {a1, . . . , am} and B = {b1, . . . , bn}. Then: • N(Km,n ) = {{ai , bj } | ai ∈ A, bj ∈ B}. • nim(DNG(Km,n )) = 0. Proof. “This one is easy” since every position of N(Km,n ) has size two. • nim(GEN(Km,n )) = 0. Proof. The 2nd player wins on their first turn by selecting a vertex in the same part as the 1st player. 13
15. ### Wheel Graphs Theorem (BEMSS) We define the wheel graph Wn
(n ≥ 5) to be graph with V = {v1, . . . , vn−1, c}, where c is the center and vi is adjacent to vi+1 (considered modulo n − 1). • N(Wn ) = complements of sets containing 2 neighboring “rim” vertices. • nim(DNG(Wn )) = pty(n). Proof. Each set in N(Wn ) has size n − 2, so . . . “This one is easy”. • nim(GEN(Wn )) = 2, n = 5 pty(n), n ≥ 6. Proof. The case involving n = 5 handled separately. When n ≥ 6 and even, not hard to argue that 2nd player has winning strategy. When n ≥ 7 and odd, 2nd player has a winning strategy in the game GEN(Wn ) + ∗1 using a pairing strategy until near end of game (complicated case analysis). 14
16. ### But wait, there’s more! Comments • We have obtained general
results concerning maximal nongenerating sets for disjoint unions of graphs, 1-clique sums of graphs, and products of graphs. Except in some specialized circumstances, there do not seem to be straightforward results concerning nim-numbers for any of these situations. • We have obtained nim-numbers for generalized windmill graphs, complete multipartite graphs. • In many instances (e.g., complete graphs, trees, cycles, wheel graphs), geodetic closure is the same as convex hull of a set. In these cases, we have also settled the Buckley/Harary versions of the game. Not true for hypercube graphs and complete bipartite graphs. • We have also obtained analogous results for the complementary “removing” games Terminate and Do Not Terminate. Conjecture We conjecture that the spectrum of nim-numbers for GEN and DNG is N ∪ {0}. We have examples of graphs that exhibit ∗0, ∗1, ∗2, ∗3, ∗4, ∗5, ∗6, ∗7. 15
∗5. 16
18. ### Frattini Subset Recall that the Frattini subgroup of a group
G is the intersection of all maximal subgroups of G. We make the analogous definition in terms of maximal nongenerating sets of a graph Definition We define the Frattini subset of a graph G via Φ(G) := N(G). The Frattini subgroup is equivalently defined as the collection of nongenerators of the group. Indeed, we have the analogous theorem for graphs. Definition A vertex v is called a nongenerator if for all subsets S of vertices, [S] = V implies [S \ {v}] = V . Theorem (BEMSS) The set of nongenerators of a graph G is the Frattini subset Φ(G). 17
19. ### Frattini Subset (continued) Example Recall that the maximal nongenerating subsets
of C4 and the diamond graph are {a, b}, {b, c}, {c, d}, {a, d} and {a, b, c}, {a, c, d}, respectively. a b c d a b c d C4 G Hence the corresponding Frattini subsets are ∅ and {a, c}, respectively. Open Problem Is the Frattini subset related to known graph-theoretic concepts? Possibly related to “minimal eccentricity approximating spanning trees”??? 18
20. ### Frattini Subset (continued) In some more complicated situations (e.g., 2-dimensional
lattice graphs), our method of attack involves simplifying game digraph by partitioning the collection of positions into so-called structure classes where both the option relationship between positions and the corresponding nim-numbers are compatible with structure equivalence according to parity. Theorem (BEMSS) • For both games, the starting position ∅ is always contained in structure class containing the Frattini subset Φ(G). • In each case, the nim-number of the game equals the nim-number of the even-parity positions contained in the structure class containing Φ(G). 19
21. ### Example Below are the “simplified” structure diagrams for two cases
of DNG(Pn Pm ) . 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 2 3 0 1 1 0 0 1 2 3 1 0 0 1 1 0 (i) n and m odd (ii) pty(n) = pty(m) & neither is 2 20
22. ### Two-dimensional Lattice Graphs Theorem (BEMSS) For the 2-dimensional lattice graph
Pn Pm , we have: • The maximal nongenerating sets for Pn Pm correspond to the complement of the vertices lying along one of the 4 exterior sides of the grid. • Φ(Pn Pm ) is the “interior” of the grid. • nim(DNG(Pn Pm )) = 0, if pty(n) = pty(m) or min{m, n} = 2 2, otherwise. • nim(GEN(Pn Pm )) = 0, if n or m is even 1, if n and m are odd. • Proofs for both involve structural induction. 21 | 3,402 | 11,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.916574 |
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