url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://stdlib.io/docs/api/latest/@stdlib/stats/incr/rmse | 1,721,331,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00127.warc.gz | 457,060,408 | 14,106 | # incrrmse
Compute the root mean squared error (RMSE) incrementally.
The root mean squared error (also known as the root mean square error (RMSE) and root mean square deviation (RMSD)) is defined as
## Usage
var incrrmse = require( '@stdlib/stats/incr/rmse' );
#### incrrmse()
Returns an accumulator function which incrementally computes the root mean squared error.
var accumulator = incrrmse();
#### accumulator( [x, y] )
If provided input values x and y, the accumulator function returns an updated root mean squared error. If not provided input values x and y, the accumulator function returns the current root mean squared error.
var accumulator = incrrmse();
var r = accumulator( 2.0, 3.0 );
// returns 1.0
r = accumulator( -1.0, -4.0 );
// returns ~2.24
r = accumulator( -3.0, 5.0 );
// returns ~4.97
r = accumulator();
// returns ~4.97
## Notes
• Input values are not type checked. If provided NaN or a value which, when used in computations, results in NaN, the accumulated value is NaN for all future invocations. If non-numeric inputs are possible, you are advised to type check and handle accordingly before passing the value to the accumulator function.
## Examples
var randu = require( '@stdlib/random/base/randu' );
var incrrmse = require( '@stdlib/stats/incr/rmse' );
var accumulator;
var v1;
var v2;
var i;
// Initialize an accumulator:
accumulator = incrrmse();
// For each simulated datum, update the root mean squared error...
for ( i = 0; i < 100; i++ ) {
v1 = ( randu()*100.0 ) - 50.0;
v2 = ( randu()*100.0 ) - 50.0;
accumulator( v1, v2 );
}
console.log( accumulator() ); | 427 | 1,617 | {"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} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.562768 |
https://oneclass.com/study-guides/ca/york/adms/adms-3330/142696-3330ch11doc.en.html | 1,571,097,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655554.2/warc/CC-MAIN-20191014223147-20191015010647-00522.warc.gz | 644,656,513 | 97,775 | Study Guides (380,000)
CA (150,000)
York (10,000)
# ADMS 3330 Study Guide - Linear Programming Relaxation, Sensitivity Analysis, Feasible Region
Department
Course Code
Professor
hassanq
This preview shows pages 1-3. to view the full 15 pages of the document.
11
Integer Linear Programming
MULTIPLE CHOICE
1. Which of the following is the most useful contribution of integer programming?
a. finding whole number solutions where fractional solutions would not be appropriate
b. using 0-1 variables for modeling flexibility
c. increased ease of solution
d. provision for solution procedures for transportation and assignment problems
TOPIC: Introduction
2. In a model, x1 > 0 and integer, x2 > 0, and x3 = 0, 1. Which solution would not be feasible?
a. x1 = 5, x2 = 3, x3 = 0
b. x1 = 4, x2 = .389, x3 = 1
c. x1 = 2, x2 = 3, x3 = .578
d. x1 = 0, x2 = 8, x3 = 0
TOPIC: Introduction
3. Rounded solutions to linear programs must be evaluated for
a. feasibility and optimality.
b. sensitivity and duality.
c. relaxation and boundedness.
d. each of the above is true.
TOPIC: LP relaxation
4. Rounding the solution of an LP Relaxation to the nearest integer values provides
a. a feasible but not necessarily optimal integer solution.
b. an integer solution that is optimal.
c. an integer solution that might be neither feasible nor optimal.
d. an infeasible solution.
TOPIC: Graphical solution
1
Only pages 1-3 are available for preview. Some parts have been intentionally blurred.
2Chapter 11 Integer Linear Programming
5. The solution to the LP Relaxation of a maximization integer linear program provides
a. an upper bound for the value of the objective function.
b. a lower bound for the value of the objective function.
c. an upper bound for the value of the decision variables
d. a lower bound for the value of the decision variables
TOPIC: Graphical solution
6. The graph of a problem that requires x1 and x2 to be integer has a feasible region
a. the same as its LP relaxation.
b. of dots.
c. of horizontal stripes.
d. of vertical stripes.
TOPIC: Graphical solution
7. The 0-1 variables in the fixed cost models correspond to
a. a process for which a fixed cost occurs.
b. the number of products produced.
c. the number of units produced.
d. the actual value of the fixed cost.
TOPIC: Fixed costs
8. Sensitivity analysis for integer linear programming
a. can be provided only by computer.
b. has precisely the same interpretation as that from linear programming.
c. does not have the same interpretation and should be disregarded.
d. is most useful for 0 - 1 models.
TOPIC: Sensitivity analysis
9. Let x1 and x2 be 0 - 1 variables whose values indicate whether projects 1 and 2 are not done or are done.
Which answer below indicates that project 2 can be done only if project 1 is done?
a. x1 + x2 = 1
b. x1 + x2 = 2
c. x1 - x2 < 0
d. x1 - x2 > 0
TOPIC: Conditional and corequisite constraints
10. Let x1 , x2 , and x3 be 0 - 1 variables whose values indicate whether the projects are not done (0) or are
done (1). Which answer below indicates that at least two of the projects must be done?
a. x1 + x2 + x3 > 2
b. x1 + x2 + x3 < 2
c. x1 + x2 + x3 = 2
d. x1 - x2 = 0
TOPIC: k out of n alternatives constraints
11. If the acceptance of project A is conditional on the acceptance of project B, and vice versa, the appropriate
constraint to use is a
a. multiple-choice constraint.
b. k out of n alternatives constraint.
c. mutually exclusive constraint.
d. corequisite constraint.
TOPIC: Modeling flexibility provided by 0-1 integer variables
Only pages 1-3 are available for preview. Some parts have been intentionally blurred.
Chapter 11 Integer Linear Programming 3
12. In an all-integer linear program,
a. all objective function coefficients must be integer.
b. all right-hand side values must be integer.
c. all variables must be integer.
d. all objective function coefficients and right-hand side values must be integer.
TOPIC: Types of integer linear programming models
13. To perform sensitivity analysis involving an integer linear program, it is recommended to
a. use the dual prices very cautiously.
b. make multiple computer runs.
c. use the same approach as you would for a linear program.
d. use LP relaxation.
TOPIC: A cautionary note about sensitivity analysis
14. Modeling a fixed cost problem as an integer linear program requires
a. adding the fixed costs to the corresponding variable costs in the objective function.
b. using 0-1 variables.
c. using multiple-choice constraints.
d. using LP relaxation.
TOPIC: Applications involving 0-1 variables
15. Most practical applications of integer linear programming involve
a. only 0-1 integer variables and not ordinary integer variables.
b. mostly ordinary integer variables and a small number of 0-1 integer variables.
c. only ordinary integer variables.
d. a near equal number of ordinary integer variables and 0-1 integer variables.
TOPIC: Applications involving 0-1 variables
TRUE/FALSE
1. The LP Relaxation contains the objective function and constraints of the IP problem, but drops all integer
restrictions.
TOPIC: LP relaxation
2. In general, rounding large values of decision variables to the nearest integer value causes fewer problems
than rounding small values.
TOPIC: LP relaxation
3. The solution to the LP Relaxation of a minimization problem will always be less than or equal to the value
of the integer program minimization problem.
TOPIC: Graphical solution
4. If the optimal solution to the LP relaxation problem is integer, it is the optimal solution to the integer
linear program. | 1,393 | 5,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.817983 |
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http://docs.imsidesign.com/projects/DoubleCAD-V6-User-Guide-Publication/DoubleCAD-V6-User-Guide-Publication/Inserting-2D-Objects/Circle-Ellipse/ | 1,632,540,180,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00337.warc.gz | 16,093,398 | 9,581 | # Circle Ellipse
Circle-Ellipse
Default UI menu: Draw/Circle , Draw/Ellipse
Tools for drawing circles and ellipses. You can display the Circle/Ellipse toolbar by right-clicking in any toolbar area and selecting Circle/Ellipse.
### Circle - Center and Point
Default UI menu: Draw/Circle/Center and Point
Creates a circle by defining its center and a point on its circumference.
1. Select the circle center point. Move the cursor to size the circle.
1. Click to create the circle, or enter the radius, diameter, or circumference in the Inspector Bar.
Isocircle
The Isocircle option is available in the Circle - Center and Point tool when an is active. It allows you to draw true isometic cicles instead of ellipses. The tool draws circles using the standard procedure.
### Circle-Concentric
Creates circles that share a center point.
1. Select a centerpoint for the circles.
2. Size the first circle, or enter the radius, diameter, or circumference in the Inspector Bar.
1. Create the second circle the same way.
1. Create more circles as needed.
1. Finish by selecting Finish from the local menu or Inspector Bar, or press Alt+F.
### Circle-Double Point
Creates a circle by defining two endpoints of its diameter.
1. Select a point on the circle circumference.
1. Select the point on the opposite end of the diameter line, or enter the radius, diameter or circumference, and angle of the diameter line in the Inspector Bar.
### Circle-Tangent to 3 Arcs
Default UI menu: Draw/Circle/Tangent to 3 Arcs
Creates a circle tangent to three arcs or circles. You can specify whether the existing arcs will be inside or outside the new circle.
1. Select the first tangent object. Click slightly outside the object to keep it outside the new circle. Click inside if you want the tangent object to be inside the new circle.
1. If the object is clicked slightly outside, an outward-pointing arrow appears.
2. Select the second tangent object. If the object is clicked slightly inside, an inward-pointing arrow appears.
1. Select the third object. The tangent circle is created.
### Circle-Tangent to Arc
Default UI menu: Draw/Circle/Tangent to Arc or Curve
Creates a circle tangent to an arc, circle, or ellipse.
1. Select the existing arc, circle, or ellipse to which you want the circle to be tangent.
1. Select the centerpoint of the circle, or enter the radius, diameter, or circumference in the Inspector Bar. (If you use the Inspector Bar and then press Enter, you will not be able to reposition the circle.) The circle will be created from the point you selected on the tangent object.
1. Move the mouse to position the circle, either inside or outside the tangent object.
Tangent Circle of Fixed Size
By default, Tangent to Entities creates a circle that touches three objects. To fix the size of the circle, enter the radius, diameter, or circumference in the Inspector Bar and lock it, but do not press Enter. When you move the cursor, the circle size remains constant.
Select the second tangent object, close to the point of tangency.
The fixed-sized tangent circle is created.
### Circle-Tangent to Line
Default UI menu: Draw/Circle/Tangent to Line
Creates a circle tangent to a line.
1 .Select the existing line to which you want the circle to be tangent.
1. Select the centerpoint of the circle, or enter the radius, diameter, or circumference in the Inspector Bar. (If you use the Inspector Bar and then press Enter, you will not be able to reposition the circle.) The circle will be created from the point you selected on the line.
1. Move the mouse to position the circle, on either side of the line.
Circle-Triple Point
Creates a circle that passes through three points.
1. Select the first point on the circumference.
1. Select the second point.
1. Select the third point.
Best Fit Circle
Default UI Menu: Draw/Circle/Best Fit Circle
### The Best Circle tool draws a circle that is best-fitted within a set of user-defined points. Points can be extracted from a specified location or from a specific object.
To draw this, select “Best Fit Circle” and define at least 3 points on the work plane. Select Finish and the circle which fits best within the points is drawn.
To draw a best fit circle:
1. Draw first point
2. Draw second point
3. Draw third point and/or other points
4. Select Finish
### Ellipse
Creates an ellipse by defining a bounding rectangle for the ellipse. The axes of the ellipse will be orthogonal.
1. Select the first corner of the bounding rectangle.
2. Define the diagonally opposed corner of the bounding rectangle, or enter the major and minor axis lengths in the Inspector Bar.
### Ellipse Fixed Ratio
Default UI menu: Draw/Ellipse/Ellipse Fixed Ratio
Creates an ellipse by specifying its aspect ratio - the ratio of the major axis to the minor axis. The axes of the ellipse are orthogonal.
1. Enter the aspect ratio in the a:b ratio field on the Inspector Bar. (Lock this value if you want to repeat it later, otherwise the field reverts back to the default.)
2. Select the centerpoint of the ellipse.
3. Move the mouse to size the ellipse.
a:b = 0.5 a:b = 2.0
Tip: The fixed ratio for an "IsoCircle" (an isometric circle on a cube) is 1.73 (the square root of 3).
### Rotated Ellipse
Creates a tilted ellipse.
1. Select the centerpoint of the ellipse.
2. Move the mouse to set the length and angle of the major axis, or enter the major axis length and angle in the Inspector Bar.
3. Set the length of the minor axis, which is always perpendicular to the major axis. You can also enter the axis length in the Inspector Bar.
Circle_2 Points
Creates a circle by defining two endpoints of its diameter.
• Select a point on the circle circumference.
• Select the point on the opposite end of the diameter line, or enter the radius, diameter or circumference, and angle of the diameter line in the Inspector Bar.
Circle_3 Points
Creates a circle that passes through three points.
• Select the first point on the circumference.
• Select the second point.
• Select the third point.
Creates a circle by defining its center and a point on its circumference.
• Select the circle centerpoint. Move the cursor to size the circle.
• Click to create the circle, or enter the radius, diameter, or circumference in the Inspector Bar.
Local menu options: The local menu contains options to switch to any of the other circle tools.
Circle_Concentric
Creates circles that share a centerpoint.
• Select a centerpoint for the circles.
• Size the first circle, or enter the radius, diameter, or circumference in the Inspector Bar.
• Create the second circle the same way.
• Create more circles as needed.
• Finish by selecting Finish from the local menu or Inspector Bar, or press Alt+F.
Creates a circle tangent to two objects, with a fixed radius.
• Select the first tangent object.
• Select the second tangent object.
• Move the mouse to create the tangent circle, or enter the Radius in the Inspector Bar.
### Circle_Tan, Tan, Tan
Creates a circle tangent to three objects (rectangles, polygons, arcs, etc.)
• Select the first tangent object.
• Select the second tangent object.
• Select the third tangent object. The tangent circle is created.
### Circle_Tangent to 3 Arcs
Creates a circle tangent to three arcs or circles. You can specify whether the existing arcs will be inside or outside the new circle.
• Select the first tangent object. Click slightly outside the object to keep it outside the new circle. Click inside if you want the tangent object to be inside the new circle.
If the object is clicked slightly outside, an outward-pointing arrow appears.
• Select the second tangent object.
If the object is clicked slightly inside, an inward-pointing arrow appears.
• Select the third object. The tangent circle is created.
### Circle_Tangent to Arc or Curve
Creates a circle tangent to an arc, circle, or ellipse.
• Select the existing arc, circle, or ellipse to which you want the circle to be tangent.
• Select the centerpoint of the circle, or enter the radius, diameter, or circumference in the Inspector Bar. (If you use the Inspector Bar and then press Enter, you will not be able to reposition the circle.)
• The circle will be created from the point you selected on the tangent object.
• Move the mouse to position the circle, either inside or outside the tangent object.
### Circle_Tangent to Line
Creates a circle tangent to a line.
• Select the existing line to which you want the circle to be tangent.
• Select the centerpoint of the circle, or enter the radius, diameter, or circumference in the Inspector Bar. (If you use the Inspector Bar and then press Enter, you will not be able to reposition the circle.)
• The circle will be created from the point you selected on the line.
• Move the mouse to position the circle, on either side of the line.
### Ellipse_Axis, End
Creates an ellipse by the endpoints of the major axis, and an endpoint of the minor axis.
• Select the endpoints of the major axis, or enter the major axis length and angle in the Inspector Bar.
• Set the length of the minor axis, which is always perpendicular to the major axis. You can also enter the axis length in the Inspector Bar.
Local menu option: Center: Enables you to set the center of the ellipse, which is the same as using the Ellipse: Center tool.
### Ellipse_Center
Creates an ellipse by its centerpoint, major axis and minor axis.
• Select the centerpoint of the ellipse.
• Move the mouse to set the length and angle of the major axis, or enter the major axis length and angle in the Inspector Bar.
• Set the length of the minor axis, which is always perpendicular to the major axis. You can also enter the axis length in the Inspector Bar.
Local menu option: Rotation: Sets the minor axis by rotation from the horizontal axis, rather than perpendicular from the major axis.
### Ellipse_Ellipse (Standard)
Creates a tilted ellipse.
• Select the centerpoint of the ellipse.
• Move the mouse to set the length and angle of the major axis, or enter the major axis length and angle in the Inspector Bar.
• Set the length of the minor axis, which is always perpendicular to the major axis. You can also enter the axis length in the Inspector Bar.
### Ellipse_Fixed Ratio
Creates an ellipse by specifying its aspect ratio - the ratio of the major axis to the minor axis. The axes of the ellipse are orthogonal.
• Enter the aspect ratio in the a:b ratio field on the Inspector Bar. (Lock this value if you want to repeat it later, otherwise the field reverts back to the default.)
• Select the centerpoint of the ellipse.
• Move the mouse to size the ellipse.
Tip: The fixed ratio for an "IsoCircle" (an isometric circle on a cube) is 1.73 (the square root of 3).
### Ellipse, Elliptical Arc
Creates an elliptical arc by defining the major and minor axes.
• Click the endpoints of the major axis, or enter the major axis length and angle in the Inspector Bar.
• Set the length of the minor axis, which is always perpendicular to the major axis. You can also enter the axis length in the Inspector Bar.
• A dotted line appears from the centerpoint. Move the cursor to set the arc start angle, or enter the angle in the Inspector Bar.
• Move the cursor counterclockwise to draw the arc, or enter the end angle or arc length in the Inspector Bar. You can also enter the included angle.
Local menu options: Center: Enables you to define the major axis from the center. Rotation: Enables you to define the minor axis at an angle from the major axis.
### Ellipse, Elliptical Arc, Fixed Ratio
Creates an elliptical arc by specifying its aspect ratio - the ratio of the major axis to the minor axis. The axes of the ellipse are orthogonal.
• Enter the aspect ratio in the a:b ratio field on the Inspector Bar. (If you want to repeat this value later then lock it, otherwise the field reverts back to the default.)
• Select the centerpoint of the ellipse.
• Move the mouse to size the ellipse.
• A dotted line appears from the centerpoint. Move the cursor to set the arc start angle, or enter the angle in the Inspector Bar.
• Move the cursor counterclockwise to draw the arc, or enter the end angle or arc length in the Inspector Bar.
### Ellipse, Rotated Elliptical Arc
Creates a tilted elliptical arc.
• Select the centerpoint of the ellipse.
• Move the mouse to set the length and angle of the major axis, or enter the major axis length and angle in the Inspector Bar.
• Set the length of the minor axis, which is always perpendicular to the major axis. You can also enter the axis length in the Inspector Bar.
• A dotted line appears from the centerpoint. Move the cursor to set the arc start angle, or enter the angle in the Inspector Bar.
• Move the cursor counterclockwise to draw the arc, or enter the end angle or arc length in the Inspector Bar. | 2,836 | 12,918 | {"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-2021-39 | latest | en | 0.768722 |
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# NCTM Annual Meeting and Convention
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### NCTM Annual Meeting and Convention
1. 1. NCTM Annual Meeting and ConventionsFriday, 19 April 2013 Denver COGrades 3 – 8 Gallery WorkshopSession 337 Capitol Ballroom 5-7 Hyatt RegencyExploring Reasoning and Communication Problems fromwithSingapore Classrooms Yeap Ban Har, Ph.D. yeapbanhar@gmail.com Marshall Cavendish Institute, Singapore Presentation is available at www.facebook.com/MCISingapore Ban Har is the Principal at Marshall Cavendish Institute, a global professionaldevelopment institute based in Singapore. Concurrently, he is the Director ofCurriculum and Professional Development at Pathlight School, a Singapore K-10school offering mainstream curriculum to learners with autism. He taughtmathematics education at National Institute of Education, Nanyang TechnologicalUniversity – the sole teacher education institute in Singapore – for more than tenyears prior to his current appointments.1|P ag e NCTM2013 Session 337
2. 2. Problem 1 Reference: Math in Focus Singapore Math by Marshall Cavendish 4BDraw a square such that the area is 1 square unit.The corners of the square must be at where lines meet.Draw a square such that the area is 2 square units. Is it possible?Draw a square such that the area is 3 square units. Is it possible?Draw a square such that the area is 4 square units. Is it possible?How many different squares of area not more than 16 square units are possible?2|P ag e NCTM2013 Session 337
3. 3. Problem 2Nine identical rectangles measuring 7 cm by 2 cm are arranged as shown below.What is the perimeter of the figure? Reference: Henry Park Primary SchoolProblem 3John had 1.5 m of copper wire. He cut some of the wire to bend into the shapeshown in the figure below. In the figure, there are 6 equilateral triangles and thelength of XY is 19 cm. How much of the copper wire was left? Reference: Primary School Leaving Examination3|P ag e NCTM2013 Session 337
4. 4. Problem 4 A rectangular sheet of paper is folded as shown in the figure. Given the angles marked , explain if it is possible to calculate the angle indicated by the arrow?Problem 5The figure below, not drawn to scale, shows a circle overlapping a larger semi-circle. If the ratio of the shaded area to the unshaded area is 5 : 8, find the ratio ofthe radius of the circle to the radius of the larger circle. Reference: Red Swastika School4|P ag e NCTM2013 Session 337 | 886 | 2,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-39 | latest | en | 0.840999 |
https://socratic.org/questions/what-is-the-mole-fraction-of-solute-in-a-3-89-m-aqueous-solution | 1,571,473,220,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00282.warc.gz | 675,677,517 | 5,737 | # What is the mole fraction of solute in a 3.89 m aqueous solution?
Dec 13, 2016
$\text{Molality}$ $=$ $\text{Moles of solute"/"Kilograms of solvent}$
$\chi = \frac{3.89 \cdot m o l}{3.89 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.067$
#### Explanation:
And $\chi ,$ $\text{mole fraction}$ $=$ $\text{Moles of solute"/"Moles of solute and moles of solvent}$
We have a $3.89$ $\text{molal}$ solution (at least I think we do); i.e. $\text{molality}$ $=$ $3.89 \cdot m o l \cdot k {g}^{-} 1$.
And thus $\chi = \frac{3.89 \cdot m o l}{3.89 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.067$ | 264 | 652 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.420647 |
https://number.academy/1053123 | 1,720,911,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00681.warc.gz | 390,370,631 | 11,167 | # Number 1053123 facts
The odd number 1,053,123 is spelled 🔊, and written in words: one million, fifty-three thousand, one hundred and twenty-three, approximately 1.1 million. The ordinal number 1053123rd is said 🔊 and written as: one million, fifty-three thousand, one hundred and twenty-third. The meaning of the number 1053123 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1053123. What is 1053123 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,053,123 in other units
The decimal (Arabic) number 1053123 converted to a Roman number is (M)(L)MMMCXXIII. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
1053123 seconds equals to 1 week, 5 days, 4 hours, 32 minutes, 3 seconds
1053123 minutes equals to 2 years, 2 months, 3 days, 8 hours, 3 minutes
### Codes and images of the number 1053123
Number 1053123 morse code: .---- ----- ..... ...-- .---- ..--- ...--
Sign language for number 1053123:
Number 1053123 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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#### Is Prime?
The number 1053123 is not a prime number. The closest prime numbers are 1053103, 1053179.
#### Factorization and factors (dividers)
The prime factors of 1053123 are 3 * 351041
The factors of 1053123 are 1, 3, 351041, 1053123.
Total factors 4.
Sum of factors 1404168 (351045).
#### Powers
The second power of 10531232 is 1.109.068.053.129.
The third power of 10531233 is 1.167.985.075.315.371.776.
#### Roots
The square root √1053123 is 1026,217813.
The cube root of 31053123 is 101,740304.
#### Logarithms
The natural logarithm of No. ln 1053123 = loge 1053123 = 13,867271.
The logarithm to base 10 of No. log10 1053123 = 6,022479.
The Napierian logarithm of No. log1/e 1053123 = -13,867271.
### Trigonometric functions
The cosine of 1053123 is -0,118263.
The sine of 1053123 is -0,992982.
The tangent of 1053123 is 8,396417.
## Number 1053123 in Computer Science
Code typeCode value
1053123 Number of bytes1.0MB
Unix timeUnix time 1053123 is equal to Tuesday Jan. 13, 1970, 4:32:03 a.m. GMT
IPv4, IPv6Number 1053123 internet address in dotted format v4 0.16.17.195, v6 ::10:11c3
1053123 Decimal = 100000001000111000011 Binary
1053123 Decimal = 1222111121120 Ternary
1053123 Decimal = 4010703 Octal
1053123 Decimal = 1011C3 Hexadecimal (0x1011c3 hex)
1053123 BASE64MTA1MzEyMw==
1053123 MD564c57c296ff7f0588efa323b56b14076
1053123 SHA1cc2ea60aafe865e1b42a2c9372b1df03cfeff46f
1053123 SHA224a015f79dc5673c2302acab5c84ef5ba9a696a4236cf46489ccf4aaf6
1053123 SHA25689d08214c55e6b70471cb255cce0e2fdfbbe369085bf432a18fb759a96376524
1053123 SHA384ab42e1a0915dd8cac0ff8701a50cffd2db399a687612e1b180b38881269035753a50ca06424cbe42344830ea742ca79d
More SHA codes related to the number 1053123 ...
If you know something interesting about the 1053123 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1053123
### Character frequency in the number 1053123
Character (importance) frequency for numerology.
Character: Frequency: 1 2 0 1 5 1 3 2 2 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1053123, the numbers 1+0+5+3+1+2+3 = 1+5 = 6 are added and the meaning of the number 6 is sought.
## № 1,053,123 in other languages
How to say or write the number one million, fifty-three thousand, one hundred and twenty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.053.123) un millón cincuenta y tres mil ciento veintitrés German: 🔊 (Nummer 1.053.123) eine Million dreiundfünfzigtausendeinhundertdreiundzwanzig French: 🔊 (nombre 1 053 123) un million cinquante-trois mille cent vingt-trois Portuguese: 🔊 (número 1 053 123) um milhão e cinquenta e três mil, cento e vinte e três Hindi: 🔊 (संख्या 1 053 123) दस लाख, तिरेपन हज़ार, एक सौ, तेईस Chinese: 🔊 (数 1 053 123) 一百零五万三千一百二十三 Arabian: 🔊 (عدد 1,053,123) مليون و ثلاثة و خمسون ألفاً و مائة و ثلاثة و عشرون Czech: 🔊 (číslo 1 053 123) milion padesát tři tisíce sto dvacet tři Korean: 🔊 (번호 1,053,123) 백오만 삼천백이십삼 Danish: 🔊 (nummer 1 053 123) en millioner treoghalvtredstusinde og ethundrede og treogtyve Hebrew: (מספר 1,053,123) מיליון חמישים ושלושה אלף מאה עשרים ושלוש Dutch: 🔊 (nummer 1 053 123) een miljoen drieënvijftigduizendhonderddrieëntwintig Japanese: 🔊 (数 1,053,123) 百五万三千百二十三 Indonesian: 🔊 (jumlah 1.053.123) satu juta lima puluh tiga ribu seratus dua puluh tiga Italian: 🔊 (numero 1 053 123) un milione e cinquantatremilacentoventitre Norwegian: 🔊 (nummer 1 053 123) en million femtitre tusen en hundre og tjuetre Polish: 🔊 (liczba 1 053 123) milion pięćdziesiąt trzy tysiące sto dwadzieścia trzy Russian: 🔊 (номер 1 053 123) один миллион пятьдесят три тысячи сто двадцать три Turkish: 🔊 (numara 1,053,123) birmilyonelliüçbinyüzyirmiüç Thai: 🔊 (จำนวน 1 053 123) หนึ่งล้านห้าหมื่นสามพันหนึ่งร้อยยี่สิบสาม Ukrainian: 🔊 (номер 1 053 123) один мільйон п'ятдесят три тисячі сто двадцять три Vietnamese: 🔊 (con số 1.053.123) một triệu năm mươi ba nghìn một trăm hai mươi ba Other languages ...
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If you know something interesting about the number 1053123 or any other natural number (positive integer), please write to us here or on Facebook.
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,028 | 5,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.78457 |
http://www.vbforums.com/showthread.php?858769-Two-Particle-motion-quot-approximation-quot-questions&p=5260535 | 1,534,531,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212768.50/warc/CC-MAIN-20180817182657-20180817202657-00537.warc.gz | 596,879,445 | 18,515 | # Thread: Two Particle motion "approximation" questions
1. ## Two Particle motion "approximation" questions
Not sure if anyone has interest in this type of problem. My Physics textbook (which I am self-studying) has a section for "Computer Exercises", where the intent is not that you work the problems out by hand, but write code to approximate the answer for you. Unfortunately, the answers for these specific questions aren't provided in the book. I've written the code to do the approximations, summing values over minute changes in time, which I'm fairly confident is correct. Just looking to see if someone else can verify my answers.
Problem 1: The acceleration of a particle varies as a function of it's position, as described in the following equation:
a = a0 * e ^ (-b * x)
Where a0 = 3 (units of m/s^2) and b = 1 (units of m^-1)
Use numerical integration to find the position and velocity of the particle at 2.37 seconds.
Edit to add: it is given that x(0) = 0 m and v(0) = 0 m/s
I get v(2.37) ~ 2.435 m/s
x(2.37) ~ 4.425 m
Problem 2: The acceleration (in m/s^2) of a particle varies with time (in s), as described in the following equation:
a = (3 + t^3) ^ 0.5
Use numerical integration to find the position and velocity of the particle at 5.7 seconds.
Edit to add: it is given that x(0) = 0 m and v(0) = 0 m/s
I get v(5.7) ~ 33.97 m/s
x(5.7) ~ 62.83 m
2. ## Re: Two Particle motion "approximation" questions
I should note that I'm somewhat familiar with Wolfram Alpha and verified that my v(5.7) for Problem 2 is correct. But I'm not sure how to use it to get the answers for Problem 1 or x(5.7) of Problem 2.
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Featured | 502 | 1,793 | {"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} | 3.546875 | 4 | CC-MAIN-2018-34 | longest | en | 0.923371 |
https://kmath.cn/math/detail.aspx?id=6316 | 1,686,448,194,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00494.warc.gz | 394,900,008 | 3,290 | \begin{aligned} \text { 原式 } & =(7-1)(7+1)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)+1 \\ & =\left(7^2-1\right)\left(7^2+1\right) /\left(7^4+1\right)\left(7^8+1\right)+1 \\ & =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)+1 \\ & =\left(7^8-1\right)\left(7^8+1\right)+1 \\ & =7^{16}-1+1 \\ & =7^{16} \end{aligned} | 181 | 333 | {"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} | 3.734375 | 4 | CC-MAIN-2023-23 | longest | en | 0.296876 |
http://www.jiskha.com/members/profile/posts.cgi?name=darren&page=9 | 1,369,476,998,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705926946/warc/CC-MAIN-20130516120526-00014-ip-10-60-113-184.ec2.internal.warc.gz | 550,559,293 | 3,290 | Saturday
May 25, 2013
# Posts by darren
Total # Posts: 90
Intro to Accounting
Les Fleurs, a boutique in Paris, France, had the following accounts in its accounting records at December 31, 20X2 (amounts in Euros, denoted as "E") Purchases………………... E250,000 Freight In……………… E8,000 Sales...
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PRONOUN REFERENCES. EMPLOYEES ARE BEGINING TO TAKE ADVANTAGE OF THE COMPANY'S ATHLETIC FACILITIES. THEY OFFER SQUASH AND TENNIS COURTS, A SMALL TRACK, AND SERVERAL TRAMPOLINES.
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solve for x. 1.25x^2=13 2.2(x-5)^2=3
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Write your answer with only positive exponents. y^5y^10
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Scientists debate about the use of fossils as a tool to determine records of various events, such as the vanishing of the dinosaurs or the origin of our ancestors.discuss the role of fossils in helping scientists develop accurate records of the geologic history of the Earth.
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I don't know how to find the GCF if the variables are different. For example, 48a^3 and 16b^5.
math right triangle
the hypotenuse of a right triangle is 1 inch longer than one leg and 8 inches longer than the other. Find the length of each side of the triangle. Let l1, l2 be the legs. l1^2 + l2^2 = h^2 but h= l1+1 and h= l2 + 8 solve these for l1, l2, put them in the first equation and sol...
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FEATURES | 455 | 1,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-20 | longest | en | 0.893846 |
https://www.akaqa.com/question/q19191820556-Why-is-anything-to-the-zero-power-equal-to-one | 1,726,276,826,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00017.warc.gz | 592,364,119 | 12,690 | why is anything to the zero power equal to one
0 Views: 321 Answers: 1 Posted: 12 years ago
Its true that "Any number [ except 0] raised to the power zero is 1"
Here is the explanation for you
Consider, 2 ^ 0
= 2 ^ (1-1) [ 1 -1 = 0]
= 2^1 × 2^ ( -1) [ Multiplication law of indices]
= 2^1/ 2^1 [ Division law of indices]
= 1
But for 0 ^ 0
= = 0 ^ (1-1) [ 1 -1 = 0]
= 0^1 × 0^ ( -1) [ Multiplication law of indices]
= 0^1/ 0^1 [ Division law of indices]
= 0/0 [ indeterminate form]
= infinity [ value is undefined]
Hope this helped you to some extent. Now share the idea with your friends too.
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721224 | 400 | 1,064 | {"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} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.789438 |
https://m.wikihow.com/Calculate-Body-Fat-Percentage-Accurately | 1,582,186,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144708.87/warc/CC-MAIN-20200220070221-20200220100221-00121.warc.gz | 471,728,725 | 61,629 | # How to Calculate Body Fat Percentage Accurately
Co-authored by Claudia Carberry, RD, MS
Updated: December 23, 2019
Body fat percentage is the mass of fat that your body holds divided by its total mass, which includes the weight of everything else (muscle, bone, water, etc.). Body fat percentage can be a pretty good indicator of disease risk. For example, the higher your percentage of body fat (especially if it's concentrated around your abdomen), the higher your risk of cardiovascular disease, diabetes, osteoarthritis, and certain types of cancer.[1] There are many different ways of measuring body fat percentage, ranging from old-school methods (such as calipers) to high-tech body scans. Calculating body fat at home can give you a very good approximation, but the most accurate methods rely on expensive equipment operated by healthcare professionals.
### Part 1 of 2: Calculating Body Fat and BMI at Home
1. 1
Measure your waist with a tape measure. Measuring waist circumference with a tape measure also helps screen for possible health risks (mentioned above) that come with being overweight or obese.[2] More specifically, if most of your fat is around your waist (termed belly fat) rather than lower at your hips, then you’re at a higher risk for cardiovascular and various other diseases. To correctly measure your waist, stand up wearing just your underwear and place a tape measure around your lower abdomen, below your bellybutton and just above your hip bones. Breathe in and then measure your waist just after you fully breathe out.
• When measuring waist circumference, apply the tape so that it makes contact with the skin and conforms to the body, but does not compress the underlying soft tissues.
• Waist sizes greater than 35 inches for women and greater than 40 inches for men represent greater risk of disease.
• The U.S. Navy method incorporates waist, hip, and neck circumference along with height and weight to determine an estimation of body density and fat percentage.[3]
2. 2
Use calipers to measure body fat. The caliper method (also called skinfold or pinch test) involves pulling your subcutaneous fat away from your muscles at certain points and pinching it with measuring calipers. These measurements are then converted to an estimated body fat percentage by an equation — some formulas require only three body measurements, others require as many as seven. Although the caliper method doesn't give an accurate reading of actual body fat percentage, it's a reliable measure of body composition change over time if the test is done by the same person and technique (only a 3% error). Although, measurement error is higher for very lean and obese people. You can buy calipers and have a friend or family member measure you, or get the test done at a fitness club, health clinic or hospital.
• It's very important when doing the caliper test to use fixed pressure on all the points you measure.
• Ideally, have a trained professional do your skin caliper measurements to ensure accuracy.
• Skinfold-based body fat estimates depend somewhat on the type of caliper used and technique. As well, it measures only one type of fat: subcutaneous adipose tissue (fat under the skin).[4]
3. 3
Measure your bioelectrical impedance. Bioelectrical impedance is a method of measuring the fat composition of your body, in comparison to other tissues, by its resistance to electricity.[5] Fat tissue does not conduct electricity, whereas muscle and bone tissue do (albeit poorly). As such, you're measuring how low levels of electricity flow through your fat tissue versus the other tissues in your body. Bioelectrical impedance is reportedly about 95% accurate depending on your body's water content, which fluctuates with exercise, diet, sweating, hydration and use of alcohol or drugs. This method does not require specialized personnel and the equipment is not expensive to buy — most gyms and physical therapy offices have them to use for free.
• You can either stand barefoot on metal plates that send an electrical current through your body (it looks similar to a regular weight scale), or grip a hand-held device (with both hands) that does the same thing.
• To get the most accurate results, don't eat or drink for 4 hours prior to testing; don't exercise vigorously within 12 hours; and no alcohol or diuretic (caffeine) consumption within 48 hours.
4. 4
Calculate your body mass index (BMI). BMI is a useful measure to determine if you're overweight or obese and at risk for heart disease, high blood pressure, type 2 diabetes, and other health problems.[6] However, BMI is not the same as body fat percentage. It's calculated from your height and total body weight, so it's only a general estimate of disease risk. To obtain your BMI number, divide your weight (converted to kilograms) by your height (converted to meters). Higher numbers represent greater risk of disease. Normal BMI measurements range from 18.5 – 24.9; a BMI between 25 – 29.9 is considered overweight, whereas 30 and above is considered obese and at high risk.
### Part 2 of 2: Calculating Body Fat Percentage More Accurately
1. 1
Get a DEXA scan. For a very accurate determination of your body fat percentage, visit a facility that has a dual-energy x-ray absorptiometry (DEXA) scanner. A DEXA scan involves x-ray technology that's used to estimate muscle tissue, bone mineral density and fat tissue in all regions of the body with a high degree of accuracy.[7] It uses a combination of two x-rays to calculate body composition in various body segments, so you can see which part of your body holds the highest percentage of fat (or muscle). The scan imparts about as much radiation to your body as the TSA body imaging devices at airports, which is not very much. A DEXA scan is considered the gold standard of determining fat percentage in your body as a whole, as well as regional parts such as arms and legs.
• Unlike MRI or CT scans, a DEXA scan doesn't involve lying inside a claustrophobic tunnel or enclosure. Instead, you lie on your back on an open table and an X-ray scanner passes slowly over your body — the procedure usually takes about five minutes, although it depends what part of your body is being scanned.
• Most major universities (exercise physiology labs) and many healthcare facilities have DEXA scanners. Ask your doctor for a referral to one in your area. They were originally developed to measure bone mineral density. The cost ranges from \$100-200 USD out of pocket if your health insurance plan doesn't cover it.
EXPERT TIP
Master's Degree, Nutrition, University of Tennessee Knoxville
Claudia Carberry is a Registered Dietitian specializing in kidney transplants and counseling patients for weight loss at the University of Arkansas for Medical Sciences. She is a member of the Arkansas Academy of Nutrition and Dietetics. Claudia received her MS in Nutrition from the University of Tennessee Knoxville in 2010.
Claudia Carberry, RD, MS
Master's Degree, Nutrition, University of Tennessee Knoxville
The average man should have 18-24% body fat. Women should aim for higher levels, such as 25-31%. Women need at least 10-13% body fat to have a regular menstrual cycle.
2. 2
Get weighed underwater. Since muscle tissue and bone is much denser than fat tissue, determining body density is helpful for understanding body composition. With underwater weighing, you are submerged in a water tank and the amount of water that's displaced is measured, which is used to calculate tissue density and full-body composition of fat.[8] The more water you displace, the more bone and muscle tissue you're assumed to have, so the lower your fat percentage. Underwater (or hydrostatic) weighing is a very accurate measure of body fat percentage — its error is only 1.5% if the test is performed according to guidelines.
• The main downside to this method of measuring fat percentage is that you have to get wet and be submerged underwater for a few seconds once you completely exhale your breath.
• Athletes often have denser bone and muscle tissue than non-athletes, so their measurements can underestimate body fat percentage using this method.
• Ask your doctor or do an internet search about which medical or research facilities in your area do hydrostatic weighing — there may not be many. You may have to travel out of your area. The cost should be comparable to getting a DEXA scan.
3. 3
Get a near-infrared interactance (NRI) reading. This method of measuring body fat is based on the principles of light absorption, reflectance, and near-infrared spectroscopy.[9] To estimate body fat composition, a computerized spectrophotometer with a hand-held fiber optic probe is used. The probe is pushed against a body part (often the biceps muscle) and emits an infrared light, which passes through fat and muscle tissue towards bone and is then reflected back to the probe. Density measurements are obtained and incorporated into predictive equations (also taking into account your height, weight, and body type) to give an estimate for overall body fat percentage. This method is not as accurate as DEXA scanning or hydrostatic weighing, but it's likely a more accurate assessment of body fat percentage than you can get at home with calipers or bioelectrical impedance scales.
• NRI tends to be less accurate with people who are either extremely lean (<8% body fat) or obese (>30% body fat).
• The amount of pressure applied to the fiber optic probe, skin color and hydration levels may cause results to differ and be inaccurate.
• NRI devices are widely available at many gyms, health clubs and weight-loss centers for a minimal fee, or sometimes free. Your doctor's or physiotherapist's office may also have an NRI device.
## Community Q&A
Search
• Question
Can you determine body fat based on BMI?
Claudia Carberry, RD, MS
Master's Degree, Nutrition, University of Tennessee Knoxville
Claudia Carberry is a Registered Dietitian specializing in kidney transplants and counseling patients for weight loss at the University of Arkansas for Medical Sciences. She is a member of the Arkansas Academy of Nutrition and Dietetics. Claudia received her MS in Nutrition from the University of Tennessee Knoxville in 2010.
Master's Degree, Nutrition, University of Tennessee Knoxville
No, body mass index is used to predict long-term risk for chronic disease. It does not measure body fat specifically.
• Question
What does your body fat percentage have to be to see abs?
Claudia Carberry, RD, MS
Master's Degree, Nutrition, University of Tennessee Knoxville
Claudia Carberry is a Registered Dietitian specializing in kidney transplants and counseling patients for weight loss at the University of Arkansas for Medical Sciences. She is a member of the Arkansas Academy of Nutrition and Dietetics. Claudia received her MS in Nutrition from the University of Tennessee Knoxville in 2010.
Master's Degree, Nutrition, University of Tennessee Knoxville
There is not a specific number in order to see abdominal muscles. You may need to tone your core if you are hoping to see your abs.
• Question
How do you measure body fat percentage?
Claudia Carberry, RD, MS
Master's Degree, Nutrition, University of Tennessee Knoxville
Claudia Carberry is a Registered Dietitian specializing in kidney transplants and counseling patients for weight loss at the University of Arkansas for Medical Sciences. She is a member of the Arkansas Academy of Nutrition and Dietetics. Claudia received her MS in Nutrition from the University of Tennessee Knoxville in 2010.
Master's Degree, Nutrition, University of Tennessee Knoxville
There are a few ways to measure body fat. Skin calipers are an accessible way whereas a DEXA scan may be more difficult to get access to.
• Question
Correct me if I'm wrong, but for the underwater test, wouldn't more water displacement for a given mass mean the person is less dense?
You are right. The article left out the weight part of density, which is pounds per cubic inch or kilograms per liter. For the same weight, fat displaces more water volume than muscle and is therefore less dense than muscle. The article has it backwards.
200 characters left
## Tips
• Some research laboratories and professional athletic facilities use Bod Pods to assess body composition by measuring air displacement. Similar to hydrostatic weighing (but without the water) this method is pretty accurate and may be best for measuring the body fat composition of the elderly and obese and disabled people, but facilities using Bod Pods are difficult to come across.
Thanks!
Thanks!
Master's Degree, Nutrition, University of Tennessee Knoxville
This article was co-authored by Claudia Carberry, RD, MS. Claudia Carberry is a Registered Dietitian specializing in kidney transplants and counseling patients for weight loss at the University of Arkansas for Medical Sciences. She is a member of the Arkansas Academy of Nutrition and Dietetics. Claudia received her MS in Nutrition from the University of Tennessee Knoxville in 2010.
Co-authors: 19
Updated: December 23, 2019
Views: 156,393
Categories: Body Mass Index (BMI)
Article SummaryX
You can accurately calculate your body fat percentage to get a clear idea of how your weight is distributed and what your lifestyle goals should be. For an accurate determination of your body fat percentage, visit a facility that has a DEXA scanner, which can estimate your muscle tissue, bone mineral density, and fat tissue in all areas of your body. However, you can measure certain things from home, like your waist circumference, to get a rough idea of your body fat percentage. Waist sizes greater than 35 inches for women and 40 inches for men represent a greater risk of disease. You can also figure out your body mass index by dividing your weight in kilograms by your height in meters. Normal BMI measurements range from 18.5 to 24.9, but higher numbers equal a greater chance for disease. For more tips from our Medical co-author, including how to use calipers to measure body fat, keep reading!
Thanks to all authors for creating a page that has been read 156,393 times. | 2,960 | 14,145 | {"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-2020-10 | latest | en | 0.938316 |
http://gameknot.com/comment.pl?t=2&k=5382&pg=-1 | 1,368,877,396,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382360/warc/CC-MAIN-20130516092622-00002-ip-10-60-113-184.ec2.internal.warc.gz | 111,158,657 | 5,321 | # chess online
## CHESS PUZZLE #5382
Added by: fikzter Added on: 25-Jul-08 Description: Grewe, Karge & Voigt, 1940; Open NK Chess Puzzle 25-7-08 Difficulty: online chess puzzle #5382 Attempts: 1114 Solved: 236 (21%)
White to move, mate in 2
From Comment
kingdawar14-Apr-11, 00:54
» Report abuse
Authored by
Grewe, Karge & Voigt, 1940
blahamen
15-Nov-11, 21:10
» Report abuse
pretty good!
smithel
17-Nov-11, 01:34
Comment deleted on 17-Nov-11, 01:35.
smithel
17-Nov-11, 01:37
» Report abuse
doesn't king to f6 prevent the mate in two?
17-Nov-11, 02:31
» Report abuse
after Kf6 follows Qe7#
houthakker17-Nov-11, 03:24
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Nice !!
Very Nice !
arandino
17-Nov-11, 07:29
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It took me some analysis!
blahamen
17-Nov-11, 07:44
» Report abuse
Yeah, it's a good one to keep in your back pocket, very handy!
tauheed
17-Nov-11, 09:03
» Report abuse
Chess puzzle
Imaginative with a high degree of precision! It took me 15 minutes to crack it.
17-Nov-11, 10:05
» Report abuse
I crack it in less than 1 minutes, but it was luck! I only saw 2 mates and I thought, that must be the move!
blahamen
17-Nov-11, 11:07
» Report abuse
No, you must be a genius!
terndrop
17-Nov-11, 15:36
» Report abuse
I thought: 1st move has to be with the queen and second move has to be either Qh6 (if Kh4) or Qe3 (if Kf4), but then I realised the 1st move after some time
phonybenoni
17-Nov-11, 17:02
» Report abuse
This is an example of the chess problem theme known as "star-flights", in which Black's king can move to any diagonal square. Here, it is the squares f4, f6, h4, h6, forming a sort of star. Once I recognized that possibility, it was just a matter of finding a square for the queen that would allow Black's king all four moves, but still set up mates. | 587 | 1,773 | {"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-2013-20 | latest | en | 0.888038 |
http://discourse.iapct.org/t/translating-keynes/11757 | 1,721,702,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00694.warc.gz | 11,496,413 | 5,954 | # Translating Keynes
[From Bill Powers (2003.02.11.1902 MST)]
Bill Williams (2003.02.11) --
Between the beginning of Keynes' Chapter 6 and the point where he reaches
the conclusion that income equals consumption plus investment, there are
only 11 pages. But they are very dense, and I'm not sure they are entirely
consistent. I hope you will help me out here.
Here is the start I have made toward putting Keynes' ideas into the form of
a working model.
We start with his symbol A. "During any period of time, an entrepreneur
will have sold finished output to consumers or to other entrepreneurs for a
certain sum which we will designate as A." (p. 52).
Also defined is A1, a sum which has been spent "on purchasing finished
output from other entrepreneurs." Then:
"And he will end up with a capital equipment, which term includes both his
stocks of unfinished goods or working capital and his finished goods,
having a value G."
It is not clear what the components of G are. Does G consist only of
unfinished goods and finished output from other entrepreneurs, plus
whatever stock of manufactured goods is on hand? Are A, A1, and G three
different and disjoint quantities? The very next sentence, starting a new
paragraph, is not very helpful, since it introduces a composite quantity
that has not, as far as I can see, been previously mentioned:
"Some part, however, of A + G - A1 will be attributable, not to the
activities of the period in question, but to the capital equipment which he
had at the beginning of the period."
Why do we add A and G and then subtract A1? Does this quantity have a name?
I will come back to this question.
This is the statement I have referred to as his struggle with the concept
of an integral. What he appears to be trying to say is that at the start of
the period, there was a quantity to which has been added something to
produce a new value designated as A + G - A1. "We must, therefore, in
order to arrive at what we mean by the _income_ of the current period,
deduct from A + G - A1 a certain sum, to represent that part of its value
which has been (in some sense) contributed by the equipment inherited from
the previous period. The problem of defining income will be solved as soon
as we have found a satisfactory method for calculating this deduction."
One of the deductions he comes up with is "Factor cost" F, which is "the
amount paid out by the entrepereneur to the other factors of production in
return for their services, which from their point of view is income" [note
the seeds of the idea of circular flow]. In other words, as I interpret
this, factor cost consists of the sum of wages and capital distributions.
Another is the fictitious "user cost" U which is a hypothetical cost that I
will ignore by giving it a value of zero until shown a reason to do
otherwise. Still another is a _supplementary cost_ symbolized V, which is
exemplified by physical deterioration of machinery requiring expenditures
just to keep the machinery working properly, or to replace it when it
becomes obsolescent.
If we start with A + G - A1 and deduct these costs, we end up with a value
(to which Keynes doesn't seem to assign a symbol, so I will use X)
X = A + G - A1 - F - U - V.
This would represent the value of X at the beginning of the period. The
value of X at the end of the period is given as A + G - A1, and the change
in value would be - F - U - V. That, of course, makes no sense if this is
supposed to be a calculation of income: it makes income the sum of three costs.
Since a literal interpretation of Keynes' words leads to nonsense, we must
try to find one that makes sense. Clearly, the terms A and G represent
gains in value of X, A being proceeds from sales, and G being an inventory
of manufactured but not yet sold goods that were produced during this
period. G also includes unfinished goods, but since they are never sold we
can absorb them into the costs of production by an alteration in whatever
measure of productivity or efficiency of labor we use.
To approach the problem of income from the front door instead of the back
porch as Keynes does, what we need to do is define a starting condition,
and then define processes which, occurring at different rates, change the
value X from the starting value Xo to the final value Xt. The difference Xt
- X0 will be the income. The starting value will be the ending value from
the previous period. We will first write the differential equations
describing how the variables change, and then convert them to integral
equations to allow calculating changes over a finite interval.
dA/dt = Sg, where Sg stands for a rate of sale of the goods inventory, all
quantities here and below being converted to dollars per unit time.
d(A1)/dt = Pf, where Pf stands for rate of purchasing finished goods.
dG/dt := N*E, where N stands for number of workers, and E is the Efficiency
of the workplace in goods/worker-day. "Day" means time unit.
dU/dt := 0;
dV/dt := D*Vk, where Vk is the current value of capital equipment and D is
the fraction of that value lost per unit time.
dF/dt := N*(W+K), where N is the size of the workforce, W is the wage per
unit time and K is the capital distribution per unit time paid to the
workforce.
The change in value of X is
dX/dt := dA/dt +dG/dt - d(A1)/dt - dU/dt - dF/dt - dV/dt
Income over the accounting period is therefore
t = T
I = integral[(dX/dt)*dt] - Xo
t = 0
This is a good stopping point. I may well have left out things (beside U)
or got things wrong. Corrections or agreements are welcome, as are changes
Best,
Bill P.
···
from my symbols to more conventional ones for the same quantities. | 1,361 | 5,664 | {"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-2024-30 | latest | en | 0.977401 |
https://blog.fosstrading.com/2015/11/overnight-spy-anomaly.html | 1,586,490,230,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00162.warc.gz | 379,640,406 | 33,589 | ## Monday, November 16, 2015
### Comment on Overnight SPY Anomaly
This post is in response to Michael Harris' Price Action Lab post, where he uses some simple R code to evaluate the asymmetry of returns from the day's close to the following day's open. I'd like to respond to his 3 notes, which I've included below.
1. The R backtest assumes fractional shares. This means that equity is fully invested at each new position. This is important because it affects drawdown calculations.
2. When calculating the Sharpe ratio, the “geometric = FALSE” option must be used otherwise the result may not be correct. It took some time to figure that out.
3. The profit factor result in R does not reconcile with results from other platforms or even from excel. PF in R is shown as 1.23 but the correct value is 1.17. Actually, the profit factor is calculated on a per share basis in R, although returns are geometric.
I completely agree with the first point. I'm not sure Mike considers the output of SharpeRatio.annualized with geometric=TRUE to be suspect (he doesn't elaborate). The overnightRets are calculated as arithmetic returns, so it's proper to aggregate them using geometric chaining (i.e. multiplication).
I also agree with the third point, because the R code used to calculate profit factor is wrong. My main impetus to write this post was to provide a corrected profit factor calculation. The calculation (with slightly modified syntax) in Mike's post is:
require(quantmod)
getSymbols('SPY', from = '1900-01-01')
overnightRets <- na.omit(Op(SPY)/lag(Cl(SPY)) - 1)
posRet <- overnightRets > 0
profitFactor <- -sum(overnightRets[posRet])/sum(overnightRets[!posRet])
Note that profit factor in the code above is calculated by summing positive and negative returns, when it should be calculated using positive and negative P&L. In order to do that, we need to calculate the equity curve and then take its first difference to get P&L. The corrected calculation is below, and it provides the correct result Mike expected.
grossEquity <- cumprod(overnightRets+1)
grossPnL <- diff(grossEquity)
grossProfit <- sum(grossPnL[grossPnL > 0])
grossLoss <- sum(grossPnL[grossPnL < 0])
profitFactor <- grossProfit / abs(grossLoss)
I'd also like to respond to Mike's comment:
Since in the past I have identified serious flaws in commercially available backtesting platforms, I would not be surprised if some of the R libraries have some flaws.
I'm certain all of the backtesting R packages have flaws/bugs. All software has bugs because all software is written by fallible humans. One nice thing about (most) R packages is that they're open source, which means anyone/everyone can check the code for bugs, and fix any bugs that are found. With closed-source software, commercial or not, you depend on the vendor to deliver a patched version at their discretion and in their timing.
Now, I'm not making an argument that open source software is inherently better. I simply wanted to point out this one difference. As much as I love open source software, there are times where commercial vendor-supported software presents a more appealing set of tradeoffs than using open source software. Each situation is different. | 765 | 3,224 | {"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-2020-16 | longest | en | 0.924571 |
https://nxttime.wordpress.com/2009/11/21/freerover-steering-ackermann/ | 1,521,954,104,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651820.82/warc/CC-MAIN-20180325044627-20180325064627-00719.warc.gz | 642,426,360 | 14,832 | My steering geometry was square. The steering wheels were always parallel to each other. However, if a car is cornering the outside wheel is making a large circle, the inside wheel is making a small circle. So, the steering angle of the outside wheel should be less than that of the inside wheel. This can be achieved with a steering geometry called Ackerman steering geometry. See the image I took from Wikipedia.
I wanted Ackermann steering. This forced me to redesign the steering mechanism and to recalculate some of the cars properties. The biggest challenge turned was to make the design sturdy. On the run I also managed to get rid of some of my previous design errors. Although I introduced some new ones.
As both front wheel have a different angle this could further complicate calculations. To overcome these complications I now base all calculations on a fictional front wheel that is placed exactly between the two real ones and that has an angle that is the average of the to real ones.
Here are two new pictures of FreeRover. One showing the steering mechanism. The other showing FreeRover in its current state. You might notice the red handle bars. These are intentionally red as the remind me to use them while handling the car. Before that I used to grab the driving motors while carrying FreeRover. Sometimes this would cause the cables to disconnect.
I had to compromise on sturdyness and cornering. It turns more slowly than it used to. Was it worth it? I don’t know. | 304 | 1,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-13 | latest | en | 0.970469 |
https://www.livewebtutors.com/view-solution/if-there-are-3-items-with-the-following-weights-and-valuesweight-8-4-5value20-10-11and-the-capacity-of-the-knapsack-i/165 | 1,553,409,779,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203378.92/warc/CC-MAIN-20190324063449-20190324085449-00264.warc.gz | 809,628,192 | 18,116 | Call Back
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Write a program that uses the brute-force approach to count the number of inversions in the array.
Question Preview:
If there are 3 items with the following weights and values:
weight: 8 4 5
value: 20 10 11
and the capacity of the knapsack is 9, your program should then calculate the total weight and the
total value of each subset in the power set:
...
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Question Preview:
If there are 3 items with the following weights and values:
weight: 8 4 5
value: 20 10 11
and the capacity of the knapsack is 9, your program should then calculate the total weight and the
total value of each subset in the power set:
total weight of subset: 0, 8, 4, 12, 5, 13, 9, 17
total value of subset: 0, 20, 10, 30, 11, 31, 21, 41
The largest value that fits into the knapsack: 21.
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Solution preview
Let a[0..n-1] be an array of n distinct integers. A pair (a[i], a[j]) is said to
be an inversion if these numbers are out of order, i.e., i < j but a[i] > a[j].
For example: if array a contains the following numbers:
9, 8, 4, 5
then the number of inversions is 5.
(inversions are 9 > 8, 9 > 4, 9 > 5, 8 > 4, 8 > 5)
Write a program that uses the brute-force approach to count the number of inversions
in the array.
Get solution
WhatsApp Order WhatsApp Order | 579 | 2,091 | {"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-2019-13 | longest | en | 0.82835 |
http://www.reddit.com/r/gaming/comments/1gyn3j/oh_australia/ | 1,408,675,422,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500822407.51/warc/CC-MAIN-20140820021342-00118-ip-10-180-136-8.ec2.internal.warc.gz | 557,243,731 | 28,339 | [–] 5 points6 points (3 children)
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Am I missing something here?
[–][deleted] (1 child)
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[deleted]
[–] 55 points56 points (42 children)
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Their minimum wage is 16\$ an hour, I thought we've been over this before?
Some math: 550\$/16\$ = 34.375h
Average work day is about 8 hours, 34.375h/8h = 4.3 eight hour days of work.
Get a job.
EDIT: Fuck your rules /u/DollarSignBot !
EDIT EDIT: America's minimum wage is 7.25\$ (Yes I'm putting the dollar sign after the number, because I can't be tamed) and the PS4's price here is 399\$, so, math time.
399 -> 400 for simplicity, divided by 7.25\$ = 55.2h
55.2h/8 hour average workday = 6.9 eight hour days of work.
IT'S CHEAPER FOR THEM.
You can stop the Australia circlejerk now.
[–] 12 points13 points (2 children)
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This. So tired of people complaining about the price difference saying they pay more. No. They don't.
[–] 11 points12 points (0 children)
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You don't pay with money; you pay with time. People don't seem to get that.
[–] -3 points-2 points (0 children)
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Yes they do. Just because they make more dosn't mean they arn't paying more.
[–] 1 point2 points (0 children)
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I can see the logic. For those who are interested I did the maths with the european price of 400€.
Everybody is complaining that we shouldn't have to pay that much since euro is worth more than USD.
When you do the exact same maths with France's minimum wage (7,38€ an hour after taxes) you obtain 54.2 hours of work (6,7 days of work)
Damn, you totally baffled me with your voodoo maths.
[–] 0 points1 point (0 children)
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Most people get a little (or a lot) more than minimum wage at their jobs too.
[–] 0 points1 point (0 children)
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Thank you! Im tired of these fucking Aussies complaining everything is "more expensive".
[–] -2 points-1 points (2 children)
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[–] 3 points4 points (1 child)
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[–] 1 point2 points (0 children)
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smell ya later, Julia
[–] 0 points1 point (22 children)
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Why do people always use the minimum wage in these?
[–] 9 points10 points (7 children)
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Because if people don't have a job, they shouldn't be complaining. If they do have a job, the lowest that they could make is minimum wage.
[–] 1 point2 points (6 children)
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Not in all cases. I know plenty of people that don't make minimum wage. Realistically not every business make enough profit to pay minimum wage and keep going. We are in a recession after all and wage cuts are preferable over job losses.
[–] 2 points3 points (4 children)
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One has to assume that places are obeying the law.
[–] 0 points1 point (0 children)
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In the place where I live in Canada, if you don't pay minimum wage you can get shut down. Many small businesses have been because of this.
[–] 5 points6 points (8 children)
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Because it applies to the largest demographic generally.
[–] 0 points1 point (4 children)
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I think because teenagers (like me) who have jobs make around minimum wage. A lot of 17-20 year old people work to get the console and work low wage jobs part time to get cash there parents don't want to spend on games.
[–] 2 points3 points (3 children)
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The demographic you just described(minimum wage, underage living at home) also doesn't typically have bills to pay. So your income is more disposable than someone trying to pay rent and stuff on minimum wage
[–] 0 points1 point (2 children)
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Exactly. For me it is disposable but I still be need to put in the time to get the money to buy a new system. \$100 makes a difference in a few days of work.
[–] 1 point2 points (1 child)
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My point being it matters less to you than it does for a working man who has bills to pay. It'll take you X amount of days worth of work to earn that console where someone who has bills to pay may have to work a lot more.
For example, after all of my bills, I only get to keep about 200\$ It'll take me a few months to save up for it versus my friend who lives at home and his parents pay for everything (yet he works 30 hour weeks). He could easily have that console 10x faster than me.
[–] 0 points1 point (0 children)
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Oh yeah this is perfectly understandable. I'm going to be in the same boat as you while I go to collage and after. my parents will be there but I'll have to pay for my own things. My job is part time at a greenhouse so getting the money takes a tad amount of time and it's hot in the summer. But like I said, your point makes perfect sense to me.
[–] 0 points1 point (2 children)
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The mimimum wage for an ADULT is just under \$16/hour at the moment... it changes 1 July 2013... but for a 15 year old it's a lot less: http://www.fairwork.gov.au/pay/national-minimum-wage/Pages/default.aspx
There are tons of reasons for product price discrepancy and to be fair... no-one has a simple explanation. but Australian television show "The Checkout" does a very good job - look for the video on this page called "High Tunes": http://www.abc.net.au/tv/thecheckout/clips/
[–] 0 points1 point (0 children)
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and you think that is different in the US?
[–] 0 points1 point (0 children)
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Well, you proved that it's harder to pay for things for a teenager...
[–] -1 points0 points (0 children)
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Exactly, the reason it's so high is because of the crazy minimum wage in Australia. Things like shipping, owning a store, and such would cost a lot more to operate since they, on average, have to pay their employees twice as much as in the US
[–] -1 points0 points (0 children)
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First Australia has an AWARD system of wages, which means that some roles are below \$16 an hour, whilst others are higher.
Second, there is a significantly larger proportion of minimum wage jobs in Australia than in the US.
Third, "minimum wage pricing" is not applied in other regions with lower minimum wages, such as Russia or Latin America.
Last, the \$ symbol comes before the amount, not after.
Cheerio.
[–] -2 points-1 points (0 children)
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It's not cheaper, thats like saying I make \$100,00o a year while you only make \$50,000 so everything is cheaper for me.
[–] 2 points3 points (0 children)
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Bitch, please. In Israel, new gen consoles START from 1000\$+. Why? Because they can and no one complains..
[–] 2 points3 points (10 children)
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\$503. It's \$573 in Norway. We still don't complain, because it's nothing to complain over.
[–] 4 points5 points (8 children)
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Plus, you probably get paid more than the average American.
[–] 1 point2 points (2 children)
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Yes, probaby. However, the Australian middleclass is richer than the Norwegian middleclass.
[–] 0 points1 point (1 child)
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Pro babies?
[–] 0 points1 point (0 children)
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Huh?
[–] 0 points1 point (2 children)
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Wait.. Did you mean that the Australians get more paid, or me, as a Norwegian?
[–] 0 points1 point (1 child)
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I believe both of you are paid more than Americans.
"The federal government mandates a nationwide minimum wage level of \$7.25 per hour" -Wiki
[–] 0 points1 point (0 children)
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Yes. That's the argument I'm trying to get out. There is no reason to complain. Some people have to pay more or less, obviously.
[–] 0 points1 point (1 child)
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I'd say your ordinary job as a cashier is around 25-30\$/hour. When i worked as a 17 year old for a bank scanning some documents i earned like 20\$/hour.
[–] 0 points1 point (0 children)
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then again the launchprice for the PS4 will probably be like 5000 NOK aka 833\$
[–] 4 points5 points (0 children)
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Oh bu-fucking-hu. You guys make about 3x as much compared to where Im from and the price only differs for 200\$. Annual minimum wage in Australia is 32.508, while its 12.056 in Slovenia. Tell me again how expensive it is for you, please. I'd love to hear more.
[–] 1 point2 points (0 children)
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£30 cheaper than here in the UK and the OZ minimum wage is almost twice as much as ours.
[–] 3 points4 points (6 children)
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Welcome to Australia! Where the cost of basically everything is way too high.
It's getting a bit better now that a lot of people are importing things, but considering the average (not minimum) wage of Australia and the USA are pretty close together, it's criminal what we pay for so many things.
Especially when all they do is ship the exact same product down to us with no additional costs. Several people have tried asking big companies why they do this to us and are given the response of "Because fuck you, that's why".
I'm paraphrasing there.. but not much.
[–] 0 points1 point (1 child)
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How much does it cost for a jar of vegemite?
[–] 0 points1 point (0 children)
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Confession: I hate vegemite...
[–] -4 points-3 points (3 children)
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Yeah, because your \$16 dollar an hour minimum wage is totally the same as our \$7.25 minimum wage. Most people in the US don't even make \$16 dollars an hour. It's been said tons of times before, you work for the same amount of time as us to be able to afford these things so quit bitching. Your dollar is worth less than ours, which is why your minimum wage is so much higher and everything is "more expensive".
[–] 0 points1 point (2 children)
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Sigh. Saying “Our minimum wage is this and yours is that, ECONOMY EXPLAINED!” is so woefully inaccurate.
A minimum wage is calculated so that people on that wage, working a full week, are able to support themselves. So they can pay for food/rent/other living expenses and get transport from their home to their work. If they’re careful about how they spend they can afford a hobby and some social activities, but nothing super expensive. You don’t get to buy designer clothes or eat out every day and you certainly don’t get to buy new generation consoles on launch day unless you spend a long time saving up and going without beforehand.
Let me give you the tip: this sucks for everyone who does it. In the USA and Australia. That’s why people try and get out of minimum wage and on to something better. So you can live comfortably and afford nice things, such as games consoles.
So the very fact that the USA minimum wage is less than half what the Australian one is should tell you something: it’s more expensive to live here than there. If it wasn’t, the wages would be the same. Because that’s how minimum wage is calculated. To be the minimum someone needs to stay out of poverty.
Now we have established that, let’s move on to the next statistic. The fact that the average wages in the USA and Australia are basically the same. This was figured out by very smart people who factored in all the kinds of things that make up an economy. Don’t argue with them, you’re wrong. Go google it if you want.
So basically, the average American and the average Australian have the same income. Which means that due to everything in Australia costing significantly more than the USA, those in the USA get to enjoy a lot more than us for their money.
TLDR: You’re wrong.
[–] 0 points1 point (1 child)
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Is full time in Australia considered 40 hours a week? I'm not gonna pretend I know what it's like in Australia, and i'm certainly not gonna say I know a ton about economies, but math shows me that 40 hours a week at your minimum wage= \$576 after a 10% tax. which is enough to buy the ps4 at your price. Here, 40 hours a week gets us \$261 after a ten percent tax. So in Australia, if you can save 1 week worth of income, you have enough for the system. 2 Weeks gets you enough for most of the release games. Here, 2 Weeks gets us enough for the console and maybe a couple games.
[–] 0 points1 point (0 children)
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Yes math does show that.. but it doesn’t show actual income tax, cost of food/rent (or mortgage)/clothing/transport/power and everything else in life that you have to pay for in order to live. You have to subtract all of those ‘essentials’ first. Then you subtract things like education/phone/internet bills as they have a higher priority than a games system. What you have left is your spare cash and almost all of those things are more expensive in Australia. Also we are taxed differently than the USA which needs to be factored in.
It also doesn’t address my original point – that the average wages in the USA are very similar to those in Australia. Calculating a luxury item like a brand new console at its launch using minimum wages is pointless, because those on minimum wage in the USA and those on minimum wage in Australia are equally not able to afford one unless they are very frugal and save for a long time. Even then, few of them would ever spend that amount of cash because of what else they could do with it. I know people who are eagerly awaiting the launch of the new consoles just so they can finally afford the old generation – you can buy a second hand Xbox 360 or PS3 and every title you want that come out in the last 7 years for less than what a new console, extra controller and a single game is going to cost.
I mean have you ever lived off minimum wage? And I mean lived and had to pay all the bills yourself, not worked for spare cash while you lived with your parents or whatever. You can’t afford cutting edge gaming systems of any kind. If you do have a system it’ll be a PC because you can use that for working/study/whatever else as well. Or you use your phone. Or you get a games console long after its launch that you bought from a friend for cheap or whatever.
Because seriously, if people in the USA living off minimum wage can afford next gen systems as soon as they launch without making some major sacrifices then you have absolutely nothing to complain about.
[–] 2 points3 points (5 children)
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Just get it off amazon, \$399
[–] 1 point2 points (0 children)
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Amazon doesn't ship worldwide.
[–] 0 points1 point (2 children)
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AUD =/= USD
[–] 1 point2 points (1 child)
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AUD are actually worth more than USD, bring some AUD to America and convert it to dollars and buy an Xbox One/PS4. Problem solved. And before you say anything about plane tickets, be a real man and swim goddamnit.
[–] 0 points1 point (0 children)
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The American dollar is worth more than the Australian dollar at the moment.
[–] 0 points1 point (0 children)
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I'm not sure if this is advanced humour or idiocy...
[–] 0 points1 point (0 children)
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Paying the same as us Brits.
[–] 0 points1 point (2 children)
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Ok, I understand that the console is the same and we can all buy it from Amazon... But what about the power difference here down under? From what I've been told (shakey way to start, I know) US power output is 120 (amps/volts-can never remember) and here in Australia it's 240 If we buy overseas will it be able to the power of Australian providers (If the animals aren't trying to kill us, our own power companies will kill our dreams of cheaper systems)
[–] 0 points1 point (0 children)
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The PS4 uses 240 amps regardless of regional design.
[–] 0 points1 point (0 children)
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I got an American Asus TF300T from somewhere, and while it came with an NZ-friendly charger, I use a travel adapter with the original 'cause it's so much nicer. Plus it operates between 110 and 240V.
Which brings me to the point that maybe (hopefully, for some) the PS4 will have a PSU that takes both 110 and 240V input. >_>
[–] 0 points1 point (0 children)
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It's 650 in new zealand :(. BUT! We have the option to trade in old PS3 games to pay off the fees for the PS4.
[–] 0 points1 point (0 children)
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\$522, Oh Europe...
[–] 0 points1 point (0 children)
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It's still the exact same product. Australia actually gets jipped with all the 'Xbox tv' rubbish that is pretty much useless to them. It's not a matter of complaining about not being able to afford it, rather, asking why the price is so much more for an identical product? The biggest gripe comes with games/media that are distributed digitally.
[–] 0 points1 point (0 children)
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How many motha fuckin australian posts is there gunna be before everybody on reddit realizes there minimum wage is more? Like seriosly theres a new one every day and nobody gives a shit.
[–] 0 points1 point (0 children)
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Well as a brit I'd save £20 on this.
[–] 0 points1 point (0 children)
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In America it would take 55 hours to get \$400 at \$7.25 minimum wage. In Australia it would take 34 Hours to get \$550 at \$15.96 minimum wage. Please remember the fact that some countries currency is worth less/more than Americas. \$1 USD is equivalent to \$1.08 AUD
[–] 0 points1 point (0 children)
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Jokes on them, it's not region locked and I doubt it's more than \$50 to ship one from the states
[–] 0 points1 point (0 children)
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That's 550 AUD. It's a bit under the 400 Euro that everyone pays, except for North America, who have it cheaper, and the UK, who have it more expensive. Also, the PS3, XBox 360 and XBox One pricing work the same way across regions. No idea about the Wii [U] pricing.
[–] -1 points0 points (1 child)
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That sucks.
[–] 1 point2 points (0 children)
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Yeah for Americans because it's more expensive in the US than in Australia since their minimum wage more than doubles the US. It's cheaper for them.
[–] -2 points-1 points (0 children)
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I don't get it... You can't afford a PS4? Which is half the price of what the PS3 was at launch? What are you complaining about?
[–] -1 points0 points (0 children)
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https://school.careers360.com/ncert/ncert-exemplar-class-11-physics-solutions-chapter-12-thermodynamics | 1,726,111,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00249.warc.gz | 468,329,310 | 100,093 | NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics
Edited By Safeer PP | Updated on Aug 09, 2022 11:14 AM IST
NCERT Exemplar Class 11 Physics solutions chapter 12 - Thermodynamics is one of the most sought topics in physics for both CBSE exams and entrance exams. This chapter of NCERT Class 11 Physics Syllabus explains various topics that can be turned into a question for the students. In this NCERT exemplar Class 11 Physics chapter 12 solutions, the students will learn about the relation between energy, matter, heat, and work. This entire Class 11 Physics NCERT exemplar solutions chapter 12 is based on the process of turning heat into energy and how this entire process affects matters.
Also, check NCERT Solutions for Class 11 other subjects.
NCERT Exemplar Class 11 Physics Solutions Chapter 12 MCQI
Question:1
The answer is the option (c) 4 is the isobaric process, 1 is isochoric. Out of 3 and 2, 3 has a smaller slope hence is isothermal. Remaining process 2 is adiabatic.
Question:2
If an average person jogs, he produces This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires l for evaporation) is
(a) 0.025 kg
(b) 2.25 kg
(c) 0.05 kg
(d) 0.20 kg
The answer is the option (a) 0.025 kg
$\text{Amount of sweat produced}=\frac{14.5\times10^3}{580\times10^3}=0.025Kg$
Question:3
Consider P-V diagram for an ideal gas shown in Figure.
Out of the following diagrams (Figure), which represents the T-P diagram?
(a) (iv)
(b) (ii)
(c) (iii)
(d) (i)
The answer is the option (c)
Question:4
The answer is the option (b). Since the direction of arrows is anticlockwise, so work done is negative equal to the area of the loop
$=-2P_{0}V_{0}$
Question:5
Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is
a) $2^{\gamma-1}$
b) $\left (\frac{1}{2} \right )^{\gamma-1}$
c) $\left (\frac{1}{1-\gamma} \right )^{2}$
d) $\left (\frac{1}{\gamma-1} \right )^{2}$
The answer is the option (a). Let us consider a P-V diagram for container A and B. Compression of gas is involved in both the cases. For the isothermal process (gas A) during
Question:6
The answer is the option (b) Let the equilibrium temperature of the system=T
Let
As there is no loss to the surroundings.
Heat lost by =Heat gain by +Heat gain by
$M_{3}s(T_{3}-T)=M_{1}s(T-T_{1})+M_{2}s(T-T_{2})$
$M_3sT_3-M_3sT=M_1sT-M_1sT_1+M_2sT-M_2sT_2$
$T(M_{3}+M_{1}+M_2)=M_3T_3+M_1T_1+M_2T_2$
$T=\frac{M_1T_1+M_2T_2+M_{3}T_{3}}{M_{1}+M_2+M_3}$
Hence (b) is correct.
NCERT Exemplar Class 11 Physics Solutions Chapter 12 MCQII
Question:7
Which of the processes described below are irreversible?
(a) The increase in temperature of an iron rod by hammering it.
(b) A gas in a small container at a temperature T1 is brought in contact with a big reservoir at a higher temperature T2 which increases the temperature of the gas.
(c) A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless piston.
(d) An ideal gas is enclosed in a piston-cylinder arrangement with adiabatic walls. A weight W is added to the piston, resulting in compression of the gas.
The answer is the option (a), (b), (d)
(a) During hammering the work done on the rod in hammering converts into heat, raises the temperature of the rod but this heat energy cannot be converted into work, so the process is not reversible.
(b) Heat flows from higher temperature to lower. When the containers are brought in contact, the heat of the bigger one gets transferred to a smaller container until the temperature in both the containers become equal, which is an average of both. But the heat cannot flow from smaller to larger.
(d) Adding weight to the piston increases the pressure and decreases the volume. It cannot be reversed back itself.
Question:8
An ideal gas undergoes isothermal process from some initial state I to final state f. Choose the correct alternatives
a) dU = 0
b) dQ = 0
c) dQ = dU
d) dQ = dW
The answer is the option (a) and (d).
Since the process is isothermal or T is constant
For an ideal gas, dU=Change in internal energy=nCvdT
dT=0 ;Thus dU=0
dQ=dU+dW
dQ=dW
Hence, a and d are correct.
Question:9
Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
(a) Change in internal energy is same in IV and III cases, but not in I and II.
(b) Change in internal energy is same in all the four cases.
(c) Work done is maximum in case I
(d) Work done is minimum in case II.
The answer is the option (b) and (c). dU is independent of the path followed in a P-V diagram. It depends only on the initial and final position. Work done is the area enclosed with the V-axis.
The initial and final position is the same for different parts I, II, III, IV. So change in U is same. Hence b is correct.
As area enclosed by the path I is maximum with V-axis, so work done during path I is maximum and minimum in III.
Hence, c is correct.
Question:10
Consider a cycle followed by an engine
1 to 2 is isothermal
Such a process does not exist because
a) heat is completely converted to mechanical energy in such a process, which is not possible
b) mechanical energy is completely converted to heat in this process, which is not possible
c) curves representing two adiabatic processes don’t intersect
d) curves representing an adiabatic process and an isothermal process don’t intersect
The answer is the option (a) and (c).
The given process is cyclic, which starts at 1 and ends at 1.
So, dU=0
And dQ= dU+dW
dQ=dW
The heat energy supplied to the system gets converted to mechanical work which isn't possible by the second law of thermodynamics.
Question:11
The answer is the option (a) and (c)
(From above figure)
Thus, if both and are positive.
And if both are negative.
Hence, a and c are correct.
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Very Short Answer
Question:12
Can a system be heated, and its temperature remains constant?
A system can be heated, but with a constant temperature when it does work against the surrounding to compensate for the heat supplied.
$\Delta Q=\Delta W$
Question:13
For path 1 Q1=1000J
through path 1
through path 2
(Change in internal energy by both paths are same)
$\Delta U=Q_{1}-W_{1}=Q_{2}-W_{2}\\ 1000-W_{1}=Q_{2}-(W_1-100)$
Question:14
If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
When the refrigerator door is kept open, the amount of heat absorbed from inside the refrigerator and the work is done on it by electricity both will be rejected by the refrigerator in the room. This will make the room hotter.
Question:15
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
In adiabatic compression, the temperature of gas increases while no heat is added to the system.
dQ=0
dQ=dU+dW
dU=-dW
So in compression work done on the system is negative. Thus, dU is positive, increasing the temperature of the system. As the internal energy of gas increases, its temperature decreases.
Question:16
Air pressure in a car tyre increases during driving. Explain.
During driving, a reaction force is applied due to the force on tyres. The temperature of gas increases due to this reaction force causing the gas inside the tyre to expand while the volume inside the tyre remains constant. In accordance with the Charles Law, the temperature of the car tyre increases as well as the pressure
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Short Answer
Question:17
Consider a Carnot’s cycle operating between and producing 1 k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
The efficiency of Carnot's engine $\mu =1-T_2/T_1$
(Temp of source)=500K
Temp of sink=300K
Question:18
A person of mass 60 kg wants to lose 5kg by going up and down a 10m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?
The energy produced by 1 kg fat=7000 k cal
Energy produced by 5 kg fat=35000 k cal=35×106 cal
Energy consumed to go up one time=mgh
Energy consumed to go up and down one time=mgh+0.5mgh
Let us assume that he goes up and down n times, then
Question:19
Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump )
of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P1 to P2?
Air is transferred into tyre adiabatically.
Let the initial volume of air be V and after pumping it becomes V+dV and pressure P+dP
$P_1 V_1^\gamma =P_2V_{2}^{\gamma }$
$P(V+dV)^\gamma =(P+dP)V^\gamma$
$PV^\gamma \left [1+\frac{dV}{V} \right ]^\gamma =P\left [1+\frac{dP}{P} \right ]V^\gamma$
$PV^\gamma \left [1+\gamma \frac{dV}{V} \right ]=PV^\gamma\left [ 1+\frac{dP}{P} \right ]$
On expanding by binomial theorem and neglecting the higher terms
Question:22
The initial state of a certain gas is It undergoes expansion till its volume becomes Consider the following two cases:
a) the expansion takes place at constant temperature
b) the expansion takes place at constant pressure
Plot the P-V diagram for each case. In which of the two cases, is the work done by the gas more?
The expansion from Vi to Vf at a constant temperature in (a) is an isothermal expansion.i.e.
at a constant temp.
The expansion at constant pressure in (b) is the isobaric process wherein the graph of P-V will be parallel to V axis until its volume becomes
The area enclosed by graph (a) is less than the graph (b). So, work done by the process (b) is more than (a).
NCERT Exemplar Class 11 Physics Solutions Chapter 12 Long Answer
Question:23
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure.
a) find the work done when the gas is taken from state 1 to state 2
b) what is the ratio of temperature if $V_{2}=2V_{1}$
c) given the internal energy for one mole of gas at temperature T is , find the heat supplied to the gas when it is taken from state 1 to 2 with .
Let $P_{1}V_{1}^{\frac{1}{2}}=P_BV_B^{\frac{1}{2}}=K$
And
Work done for process 1 to 2;
WD from V1to V2=
1. Equation of ideal gas PV=nRT
and
(c )
$dW=2P_1V_1(\sqrt{2}-1) =2nRT_{1}(\sqrt{2}-1)$
$n=1$
Question:24
A cycle followed by an engine is shown in the figure.
A to B: volume constant
C to D: volume constant
a) in which part of the cycle heat is supplied to the engine from outside?
b) in which part of the cycle heat is being given to the surrounding by the engine?
c) what is the work done by the engine in one cycle in terms of ?
d) what is the efficiency of the engine?
(a) Heat is supplied to the engine in part AB where dV=0
dW=PdV=P(0)
dW=0
dQ=dU+dW(First law of thermodynamics)
dQ=dU
$P=\frac{nRT}{V}$
V=constant
(b) Heat is given out by the system in a part CD where dV=0 and pressure and temperature decreases
(c ) WD by system=
Total
Net WD=
Question:25
A cycle followed by an engine is shown in the figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle considering .
AB: constant volume
BC: constant pressure
DA: constant pressure
Ans.
(a) For , dV=0
dQ=dU
(b)
1. In diagram
Compression of gas at constant pressure takes place. Therefore, heat exchange is similar to part (b)
Question:26
Consider that an ideal gas is expanding in a process given by P = f(V), which passes through a point . Show that the gas is absorbing heat at if the slope of the curve P = f(V) is larger than the slope of the adiabat passing through .
Slope of graph at
$(V_0,P_0)=\frac{dP}{dV}_{(P_0,V_0)}$
$P=f(V)$ for adiabatic process $PV^{\gamma }=\text{constant }K$
$P=\frac{K}{V^{\gamma}}$
$\frac{dP}{dV}=-\frac{\gamma P}{}V$
$(\frac{dP}{dV})_{(P_0,V_0)}=-\gamma \frac{P_0}{V_0 }$
$P=f(V)$
$dQ=dU+dW$
$dQ=nC_vdT+PdV$
$PV=nRT$
$T=\frac{PV}{nR}=\frac{V}{nR}f(V)$
$\frac{dT}{dV}=\frac{1}{nR}[f(V)+Vf'(V)]$
$\frac{dQ}{dV}=nC_v\left (\frac{dT}{dV} \right )+P\left (\frac{dV}{dV} \right )=\frac{nC_v}{nR}[f(V)+V(f'(V))+P]$
$\frac{dQ}{dV}_{(V=V_0)}=\frac{C_v}{}R\left [f(V_0)+V_0(f'(V_0))+f(V_0) \right ]$
$=f(V_0)\left [\frac{C_v}{}R+1 \right ]+V_0f'(V_0)\frac{C_v}{}R$
$C_P-C_v=R\Rightarrow \frac{C_P}{C_V}-1=\frac{R}{C_v}$
$\gamma -1=\frac{R}{C_v}\Rightarrow C_v=\frac{R}{\gamma -1}\Rightarrow \frac{C_v}{}R=\frac{1}{\gamma -1}$
$\frac{dQ}{dV}_{(V=V_0)}=\frac{1}{r-1}\left [\gamma P_0+V_0f'{(V_0)} \right ]$
$\frac{dQ}{dV}_{(V=V_0)}>1 and \gamma >1 so\frac{1}{\gamma -1}>0$
$\gamma P_0+V_0f'(V_0)>0$
$V_0f'(V_0)>-\gamma P_0$
$f'V_0>-\frac{\gamma P_0}{V_0}$
Question:27
Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached. A spring is attached to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from V0 to V1.
a) what is the initial pressure of the system?
b) what is the final pressure of the system?
c) using the first law of thermodynamics, write down the relation between Q, Pa, V, V0, and k.
1. Piston is considered massless and balanced by atmospheric pressure. So the initial
pressure of system inside cylinder is Pa
1. On supplying heat volume of gas increases from
Increase in Volume= =Area of base×height=
$x=\frac{V_{1}-V_{0}}{A}$
(Force exerted by spring)
$F=Kx=\frac{K(V_1-V_0)}{A}$
(Force due to spring on unit area)
(Final total pressure on gas)
1. (First law of thermodynamics)
T=final temperature of the gas
T0=initial temperature of the gas
n=1
$T_y=T=\frac{P_fV_f}{R}=\frac{\left [P_a+K(V_1-V_0) \right ]}{R}$
(Increase in potential energy of spring)
One of the most crucial topics covered is thermodynamic equilibrium, which explains when a system is isolated from the surrounding, the properties do not change. Other than this, various other topics are covered like entropy, enthalpy, laws of thermodynamics, thermodynamic processes, types of system, etc. are covered in Class 11 Physics NCERT exemplar solutions chapter 12.
We have solved all the questions that the NCERT book has after the chapter. These questions play a major role in understanding the chapter in detail, and also in answering questions in exams. Students can easily check their answers against the ones that our experts have solved. Also, one can download the NCERT exemplar Class 11 Physics solutions chapter 12 pdf download for offline access using the download link.
Also, Read NCERT Solution subject wise -
Also, check NCERT Notes subject wise -
Topics Covered in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics
• 1. Introduction
• 2. Thermal equilibrium
• 3. Zeroth law of thermodynamics
• 4. Heat, internal energy, and work
• 5. The first law of thermodynamics
• 6. Specific heat capacity
• 7. Thermodynamic state variables and equation of state
• 8. Thermodynamic processes
• 9. Heat engines
• 10. Refrigerators and heat pumps
• 11. The second law of thermodynamics
• 12. Reversible and irreversible processes
• 13. Carnot engine
What Will The student learn in NCERT Exemplar Class 11 Physics Solutions Chapter 12 Thermodynamics?
• In this chapter, several topics revolve around heat and energy and interconversion between them. In thermodynamics, the rate of the change or how the change happens is not accounted for. Instead, it takes into account the final and initial stage of a process in bulk and not at microscopic levels.
• In NCERT exemplar Class 11 Physics chapter 12 solutions, the learners will delve into various thermodynamic systems like closed system, open systems, and isolated systems.
• The thermodynamic process like isothermal, isobaric, isochoric, and adiabatic process is explained in detail in Class 11 Physics NCERT exemplar solutions chapter 12.
• Two of the major topics are explained, that is enthalpy and entropy. Enthalpy is the measurement of energy, and entropy is the measurement of disorder in a system. Other than these factors, 4 laws of thermodynamics and state of equilibriums are topics that hold the most importance.
• Also, in the NCERT exemplar Class 11 Physics solutions chapter 12, various applications of thermodynamics are added like heat engines, refrigerators, and Carnot engines.
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NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
Chapter 2 Units and Measurement Chapter 3 Motion in a Straight Line Chapter 4 Motion in a Plane Chapter 5 Laws of Motion Chapter 6 Work, Energy, and Power Chapter 7 Systems of Particles and Rotational Motion Chapter 8 Gravitation Chapter 9 Mechanical Properties of Solids Chapter 10 Mechanical Properties of Fluids Chapter 11 Thermal Properties of Matter Chapter 13 Kinetic Theory Chapter 14 Oscillations Chapter 15 Waves
Important Topics To Cover From NCERT Exemplar Solutions for Class 11 Physics Chapter 12
· One of the very things that students should learn from this chapter and NCERT exemplar Class 11 Physics solutions chapter 12 is the factors, terminology, processes, and variables. Enthalpy, entropy, specific heat, thermodynamic processes, etc., are some topics that will help in higher education and a better understanding of the chapter.
· Secondly, students should focus on all the laws of thermodynamics. There are in total four laws which hold importance for exams, entrance exams, and from an application point of view in the field of engineering and science.
· Lastly, the students should focus on the real-world applications of thermodynamics with the help of NCERT exemplar Class 11 Physics chapter 12 solutions. One should learn in detail about the heat engines, refrigerators, and the Carnot engine. Focus on the working process, the importance, and its application.
NCERT Exemplar Class 11 Solutions
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Check Class 11 Physics Chapter-wise Solutions
Chapter 1 Physical world Chapter 2 Units and Measurement Chapter 3 Motion in a straight line Chapter 4 Motion in a Plane Chapter 5 Laws of Motion Chapter 6 Work, Energy and Power Chapter 7 System of Particles and Rotational motion Chapter 8 Gravitation Chapter 9 Mechanical Properties of Solids Chapter 10 Mechanical Properties of Fluids Chapter 11 Thermal Properties of Matter Chapter 12 Thermodynamics Chapter 13 Kinetic Theory Chapter 14 Oscillations Chapter 15 Waves
Also Check NCERT Books and NCERT Syllabus here
1. Is this chapter and crucial for NEET?
Yes, this chapter is important for NEET and the medical entrance exam. The physics papers tend to have 1-2 questions from this chapter every year.
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All 27 questions given in the exercise of the physics NCERT exemplar book is solved with utmost detail. All the NCERT exemplar solutions for class 11 physics, chapter 12 are in simple language and easy to understand.
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Our teams of experienced teachers who have industry and teaching knowledge for years have solved these questions keeping in mind the student’s level of understanding.
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Get answers from students and experts
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9 | 6,294 | 22,983 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 47, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.85198 |
https://brilliant.org/problems/is-it-a-5th-degree-polynomial/ | 1,627,207,628,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00338.warc.gz | 163,554,239 | 8,790 | # Is it a 5th degree Polynomial?
Algebra Level 4
\large \begin{aligned} a + 4b + 9c + 16d + 25e & = 1 \\ 4a + 9b + 16c + 25d + 36e & = 8 \\ 9a + 16b + 25c+ 36d + 49e & = 23 \end{aligned}
Real numbers $a$, $b$, $c$, $d$, and $e$ satisfy the system of equations above. Find $a + b + c+ d + e$.
× | 137 | 297 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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} | 3.5625 | 4 | CC-MAIN-2021-31 | latest | en | 0.523401 |
https://www.systutorials.com/docs/linux/man/docs/linux/man/l-slasq2/ | 1,721,615,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00877.warc.gz | 846,496,305 | 4,193 | # slasq2 (l) - Linux Manuals
## NAME
SLASQ2 - computes all the eigenvalues of the symmetric positive definite tridiagonal matrix associated with the qd array Z to high relative accuracy are computed to high relative accuracy, in the absence of denormalization, underflow and overflow
## SYNOPSIS
SUBROUTINE SLASQ2(
N, Z, INFO )
INTEGER INFO, N
REAL Z( * )
## PURPOSE
SLASQ2 computes all the eigenvalues of the symmetric positive definite tridiagonal matrix associated with the qd array Z to high relative accuracy are computed to high relative accuracy, in the absence of denormalization, underflow and overflow. To see the relation of Z to the tridiagonal matrix, let L be a unit lower bidiagonal matrix with subdiagonals Z(2,4,6,,..) and let U be an upper bidiagonal matrix with 1aqs above and diagonal Z(1,3,5,,..). The tridiagonal is L*U or, if you prefer, the symmetric tridiagonal to which it is similar.
Note : SLASQ2 defines a logical variable, IEEE, which is true on machines which follow ieee-754 floating-point standard in their handling of infinities and NaNs, and false otherwise. This variable is passed to SLASQ3.
## ARGUMENTS
N (input) INTEGER
The number of rows and columns in the matrix. N >= 0.
Z (input/output) REAL array, dimension ( 4*N )
On entry Z holds the qd array. On exit, entries 1 to N hold the eigenvalues in decreasing order, Z( 2*N+1 ) holds the trace, and Z( 2*N+2 ) holds the sum of the eigenvalues. If N > 2, then Z( 2*N+3 ) holds the iteration count, Z( 2*N+4 ) holds NDIVS/NIN^2, and Z( 2*N+5 ) holds the percentage of shifts that failed.
INFO (output) INTEGER
= 0: successful exit
< 0: if the i-th argument is a scalar and had an illegal value, then INFO = -i, if the i-th argument is an array and the j-entry had an illegal value, then INFO = -(i*100+j) > 0: the algorithm failed = 1, a split was marked by a positive value in E = 2, current block of Z not diagonalized after 30*N iterations (in inner while loop) = 3, termination criterion of outer while loop not met (program created more than N unreduced blocks)
## FURTHER DETAILS
The shifts are accumulated in SIGMA. Iteration count is in ITER. Ping-pong is controlled by PP (alternates between 0 and 1). | 600 | 2,211 | {"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-2024-30 | latest | en | 0.809921 |
http://slideplayer.com/slide/5360660/ | 1,571,121,983,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00016.warc.gz | 169,474,279 | 17,782 | # The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 1 Mathematical Modeling.
## Presentation on theme: "The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 1 Mathematical Modeling."— Presentation transcript:
The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Numerical Analysis ECIV 3306 Chapter 1 Mathematical Modeling
Chapter 1: Mathematical Modeling Mathematical Model A formulation or equation that expresses the essential features of a physical system or process in mathematical terms. Generally, it can be represented as a functional relationship of the form
Mathematical Modeling
Simple Mathematical Model Example: Newton’s Second Law (The time rate of change of momentum of a body is equal to the resultant force acting on it) a = acceleration (m/s 2 ) ….the dependent variable m = mass of the object (kg) ….the parameter representing a property of the system. f = force acting on the body (N)
Complex Mathematical Model Example: Newton’s Second Law Where: c = drag coefficient (kg/s), v = falling velocity (m/s)
Complex Mathematical Model At rest: (v = 0 at t = 0), Calculus can be used to solve the equation
Analytical solution to Newton's Second Law.
Numerical Solution to Newton's Second Law Numerical solution: approximates the exact solution by arithmetic operations. Suppose
Numerical Solution to Newton's Second Law.
.
Comparison between Analytical vs. Numerical Solution
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https://codeforces.com/blog/entry/46075 | 1,590,391,465,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388012.14/warc/CC-MAIN-20200525063708-20200525093708-00569.warc.gz | 297,483,575 | 23,070 | ### Edvard's blog
By Edvard, history, 4 years ago, translation, ,
### 691A - Fashion in Berland
The problem was suggested and prepared by Arthur Jaworski KingArthur.
In this problem you should simply check the conditions from the problem statement.
С++ solution
Complexity: O(n).
### 691B - s-palindrome
The problem was suggested by Nikita Melnikov nickmeller.
In this problem you should simply find the symmetric letters by picture and also observe that the pairs (b, d) and (p, q) is the symmteric reflections.
C++ solution
Complexity: O(n).
### 691C - Exponential notation
The problem was suggsted by user I_Had_A_Great_Time.
This is an implementation problem. You should do exactly what is written in the problem statement. On my mind the simplest way is to find the position of the first not zero digit and the position of the dot. The difference between that positions is the value of b (if the value is positive you should also decrease it by one).
C++ solution
Complexity: O(n).
### 691D - Swaps in Permutation
The problem was suggested by Zi Song Yeoh zscoder.
Consider a graph with n vertices whose edges is the pairs from the input. It's possible to swap any two values with the positions in some connected component in that graph. So we can sort the values from any component in decreasing order. Easy to see that after sorting the values of each component we will get the lexicographically maximal permutation.
C++ solution
Complexity: O(n + m).
### 691E - Xor-sequences
The problem was suggested by Zi Song Yeoh zscoder.
Let z ij be the number of xor-sequences of length i with the last element equal to a j. Let g ij be equal to one if contains the number of ones in binary presentation that is multiple of three. Otherwise let g ij be equal to zero. Consider a vectors z i = {z ij}, z i - 1 = {z i - 1, j} and a matrix G = {g ij}. Easy to see that z i = G × z i - 1. So z n = G n z 0. Let's use the associative property of matrix multiplication: at first let's calculate G n with binary matrix exponentiation and then multiply it to the vector z 0.
C++ solution
Complexity: O(n 3 logk).
### 691F - Couple Cover
The problem was suggested by Michael Kirsche mkirsche.
Let's count the number of pairs with multiple less than p. To get the number of not less pairs we should sumply subtract from n·(n - 1) the number of less pairs. Let cnt i be the number of values in a equal to i and z j be the number of pairs from a with the multiple equal to j. To calculate the values from z we can use something like Eratosthenes sieve: let's iterate over the first multiplier a and the multiple of it b = ka and increment z b by the value cnt a·cnt k. After calculating the array z we should calculate the array of its partial sums and find the number of less pairs in O(1) time.
C++ solution
Complexity: O(n + XlogX), where X is the maximal value in p.
• +47
» 4 years ago, # | 0 Can 691D - Swaps in Permutation be solved using merge sort ? If we allow swaps only if its one of valid swaps given in the input.
» 4 years ago, # | ← Rev. 2 → 0 For problem D,Though not necessary , even has a larger complexity and even dsu can be used in a better way to lower complexity I always wanted to use some kind of heuristic in my code 19211367. Got WA in contest due to some minor major mistake( A mistake that is so minor but changes everything you are doing :P ).
» 4 years ago, # | 0 In case someone faced the same issue. For problem E, my O(n3logk) solution in C# was getting TLE, so I parallelized matrix multiplication and got AC. 19309384
» 4 years ago, # | 0 Apologies all for what is probably a silly question.The solution to 691D includes the line: const int N = 1200300; pti a[N];I can't find any documentation for a pti datatype. I assume this is an abbreviation commonly used in codeforces solutions? If so can someone link me to an explanation of what this is? Or a solution where the abbreviation is defined.
• » » 4 years ago, # ^ | 0 I had never seen this abbreviation before, but looking through Edvard's submissions, for example, 19489538 you can see it means pair .
» 3 years ago, # | +3 I think that problem 691D - Swaps in Permutation have the same concept as problem 500B - New Year Permutation. I use dsu to solve both of these two problems. Anyway, D is still very good problems I think so. Thanks zscoder.
» 23 months ago, # | 0 I think the complexity of problem D is wrong. did you mention the sorting complexity? | 1,166 | 4,489 | {"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} | 3.875 | 4 | CC-MAIN-2020-24 | longest | en | 0.880004 |
https://www.mathsisfun.com/definitions/micro-.html | 1,582,148,300,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00070.warc.gz | 779,913,541 | 2,451 | Definition of
# Micro-
A prefix meaning one-millionth (1/1,000,000 or 10-6)
Examples:
• a micrometer is one-millionth of a meter, which is about the size of a small bacterium!
• a microsecond is one-millionth of a second
Symbol is µ
Example: 12 µm = 12 micrometers = 12 millionths of a meter (0.000012 m) | 95 | 309 | {"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} | 2.859375 | 3 | CC-MAIN-2020-10 | latest | en | 0.829854 |
http://fiftysixtysoftware.com/mean-square/root-mean-square-error-matlab.html | 1,516,369,491,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00323.warc.gz | 114,594,221 | 6,612 | # fiftysixtysoftware.com
Home > Mean Square > Root Mean Square Error Matlab
# Root Mean Square Error Matlab
## Contents
CicioraMetin Parçacığı görünümü - 2004Modern Cable Television Technology: Video, Voice, and Data CommunicationsWalter S. For single carrier modulations, Preference is, by convention, the power of the outermost (highest power) point in the reference signal constellation. If you're desperate to speed up this particular operation, you could write a MEX file in C and call it from Matlab. –mtrw Nov 30 '11 at 8:14 a Moon Dust Is it possible to return an object of type T by reference from a lambda without using trailing return type syntax? check over here
The repetition of these three steps as more data becomes available leads to an iterative estimation algorithm. Thus, we may have C Z = 0 {\displaystyle C_ σ 4=0} , because as long as A C X A T {\displaystyle AC_ σ 2A^ σ 1} is positive definite, The system returned: (22) Invalid argument The remote host or network may be down. When the observations are scalar quantities, one possible way of avoiding such re-computation is to first concatenate the entire sequence of observations and then apply the standard estimation formula as done https://www.mathworks.com/matlabcentral/answers/4064-rmse-root-mean-square-error
## Root Mean Square Error Matlab
EVM or Error vector magnitude provides insight into quality of the modulated signal/symbol. The orthogonality principle: When x {\displaystyle x} is a scalar, an estimator constrained to be of certain form x ^ = g ( y ) {\displaystyle {\hat − 4}=g(y)} is an Another approach to estimation from sequential observations is to simply update an old estimate as additional data becomes available, leading to finer estimates. Also, I presume you know that you're calculating the RSS not RMS.
Thus, we can combine the two sounds as y = w 1 y 1 + w 2 y 2 {\displaystyle y=w_{1}y_{1}+w_{2}y_{2}} where the i-th weight is given as w i = EVM, as conventionally defined for single carrier modulations, is a ratio of a mean power to a peak power. The "RMS Error Vector Spectrum" trace is the same as averaging multiple Error Vector Spectrum trace measurements, one Error Vector Spectrum measurement is made per average count. Normalized Root Mean Square Error Matlab Since some error is always present due to finite sampling and the particular polling methodology adopted, the first pollster declares their estimate to have an error z 1 {\displaystyle z_{1}} with
Every new measurement simply provides additional information which may modify our original estimate. Matlab Rms Function ciltStructural Health Monitoring and Intelligent Infrastructure: Proceedings of the First International Conference on Structural Health Monitoring and Intelligent Infrastructure, 13-15 November 2003, Tokyo, Japan, Zhishen Wu, ISBN 9058096475, 9789058096470EditörlerZhishen Wu, Masato Alternative form An alternative form of expression can be obtained by using the matrix identity C X A T ( A C X A T + C Z ) − 1 In particular, when C X − 1 = 0 {\displaystyle C_ − 6^{-1}=0} , corresponding to infinite variance of the apriori information concerning x {\displaystyle x} , the result W =
## How To Calculate Mean Square Error In Matlab
M. (1993). check my blog Browse other questions tagged search matlab vector find or ask your own question. Please try the request again. Studies have shown that dynamic EVM with a 50% duty cycle square wave applied to PA Enable to be worse than the static EVM (PA Enable with 100% duty cycle).[2] See Root Mean Square Error Formula
We can describe the process by a linear equation y = 1 x + z {\displaystyle y=1x+z} , where 1 = [ 1 , 1 , … , 1 ] T A signal sent by an ideal transmitter or received by a receiver would have all constellation points precisely at the ideal locations, however various imperfections in the implementation (such as carrier CicioraMorgan Kaufmann, 2004 - 1053 sayfa 0 Eleştirilerhttps://books.google.com.tr/books/about/Modern_Cable_Television_Technology.html?hl=tr&id=tvUoQJXEwNECFully updated, revised, and expanded, this second edition of Modern Cable Television Technology addresses the significant changes undergone by cable since 1999--including, most notably, this content This is in contrast to the non-Bayesian approach like minimum-variance unbiased estimator (MVUE) where absolutely nothing is assumed to be known about the parameter in advance and which does not account
x ^ = W y + b . {\displaystyle \min _ − 4\mathrm − 3 \qquad \mathrm − 2 \qquad {\hat − 1}=Wy+b.} One advantage of such linear MMSE estimator is Rmse Interpretation ISBN978-0521592710. Play games and win prizes!
## the dimension of y {\displaystyle y} ) need not be at least as large as the number of unknowns, n, (i.e.
It is an average.sqrt(sum(Dates-Scores).^2)./Dates Thus, you have written what could be described as a "normalized sum of the squared errors", but it is NOT an RMSE. In your case, I guess it would be A(ones(1, size(B,1)),:) - B. The average power of the error vector, normalized to signal power, is the EVM. have a peek at these guys Intended as a day-to-day reference for cable engineers, this book illuminates all the technologies involved in building and maintaining a cable system.
Luenberger, D.G. (1969). "Chapter 4, Least-squares estimation". Thus we can re-write the estimator as x ^ = W ( y − y ¯ ) + x ¯ {\displaystyle {\hat − 4}=W(y-{\bar − 3})+{\bar − 2}} and the expression But this can be very tedious because as the number of observation increases so does the size of the matrices that need to be inverted and multiplied grow. It is required that the MMSE estimator be unbiased.
In terms of the terminology developed in the previous sections, for this problem we have the observation vector y = [ z 1 , z 2 , z 3 ] T An Error Occurred Unable to complete the action because of changes made to the page. From the figure it is imperative that M and Φ are magnitude and phase errors respectively between two constellation points. We can model our uncertainty of x {\displaystyle x} by an aprior uniform distribution over an interval [ − x 0 , x 0 ] {\displaystyle [-x_{0},x_{0}]} , and thus x
Prentice Hall. Moreover, if the components of z {\displaystyle z} are uncorrelated and have equal variance such that C Z = σ 2 I , {\displaystyle C_ ¯ 4=\sigma ^ ¯ 3I,} where I need to calculate the RMSE between every point. Subtracting y ^ {\displaystyle {\hat σ 4}} from y {\displaystyle y} , we obtain y ~ = y − y ^ = A ( x − x ^ 1 ) +
These methods bypass the need for covariance matrices. Here the left hand side term is E { ( x ^ − x ) ( y − y ¯ ) T } = E { ( W ( y − The system returned: (22) Invalid argument The remote host or network may be down. Also, this method is difficult to extend to the case of vector observations. | 1,663 | 6,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-05 | latest | en | 0.843227 |
https://pt.scribd.com/document/36676646/Percentages | 1,582,613,749,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00107.warc.gz | 502,242,386 | 63,953 | Você está na página 1de 1
# Percentages are used all the time. They are used just like fractions.
## Just like fractions, they
are used to find part of a value which may be an increase or decrease of the original value.
## A percentage is a number out of 100, so:
1% is equal to 1/100
23% is equal to 23/100
50% is equal to 50/100 or 1/2
When using percentages on a calculator, I find that converting the percentage into a fraction
or decimal is much more simpler than using the percentage button. Knowing how to convert a
percentage would make this process easier.
## Finding percentages of values includes decreases and increases:
• Decreases
A decrease of 10% means 90% of the value ( 100% - 10% = 90% or 0.90 ). To find
the answer you work out 0.90 × number.
• Increases
An increace of 10% means your value and an extra 10% added to it. This means it is
110% ( 100% + 10% = 110% or 1.10 ) of the value. To find the answer you work out
1.10 × number. | 270 | 962 | {"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} | 3.984375 | 4 | CC-MAIN-2020-10 | latest | en | 0.893768 |
https://forums.examsbook.com/topic/439/one-card-is-drawn-from-a-pack-of-52-cards-each-of-the-52-cards-being-equally-likely-to-be-drawn-find-the-probability-that-the-card-drawn-is-neither-a-spade-nor-a-king | 1,603,254,553,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00265.warc.gz | 339,932,675 | 7,580 | Answer : 2 9/13 Explanation : Answer: B) 9/13 Explanation: There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits Thus n(E) = 13 3 = 16 Hence nE¯=52-16=36 Therefore, PE=3652=913 | 80 | 221 | {"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-2020-45 | longest | en | 0.622883 |
http://stackoverflow.com/questions/3507922/excel-sum-values-in-a-column-based-on-date/3511137 | 1,406,711,660,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510270313.12/warc/CC-MAIN-20140728011750-00115-ip-10-146-231-18.ec2.internal.warc.gz | 276,309,623 | 17,153 | # Excel: sum values in a column based on date
I have written this function that will give me a monthly sum for two columns: one has the date of each order, one has the cost of each order.
``````=SUMIF(\$C\$1:\$C\$1000,">="&DATE(2010,6,1),\$D\$1:\$D\$1000)-SUMIF(\$C\$1:\$C\$1000,">="&DATE(2010,7,1),\$D\$1:\$D\$1000)
``````
Using data like this:
``````8/16/10 17:00 7.99
8/16/10 14:25 7.99
8/15/10 22:42 7.99
``````
I end up with a table like this:
``````May 998
June 968.28
July 1239.76
August 514.96
``````
However, now I would like to do daily sumas, and using my way, I have to hand edit each row. How can I do this better in Excel?
-
Use a column to let each date be shown as month number; another column for day number:
`````` A B C D
----- ----- ----------- --------
1 8 6 8/6/2010 12.70
2 8 7 8/7/2010 10.50
3 8 7 8/7/2010 7.10
4 8 9 8/9/2010 10.50
5 8 10 8/10/2010 15.00
``````
The formula for A1 is =Month(C1).
The formula for B1 is =Day(C1)
For Month sums, put the month number next to each month:
`````` E F G
----- ----- -------------
1 7 July \$1,000,010
2 8 Aug \$1,200,300
``````
The formula for G1 is =SumIf(\$A\$1:\$A\$100, E1, \$D\$1:\$D\$100). This is a portable formula; just copy it down.
Total for the day will be be a bit more complicated, but you can probably see how to do it.
-
Use pivot tables, it will definitely save you time. If you are using excel 2007+ use tables (structured references) to keep your table dynamic. However if you insist on using functions, go with Smandoli's suggestion. Again, if you are on 2007+ use SUMIFS, it's faster compared to SUMIF.
-
I would like to accept your answer as well, because I think it is also valid. Best I can do is upvote though :) – Andrew Johnson Aug 26 '10 at 1:03
+1 Yes, this is a great application for pivot tables. – Smandoli Aug 26 '10 at 3:32
In all honesty, I must add I don't like pivot tables although I use them sometimes. I can't offer any valid reasons for disliking them -- it's probably from trying to use Excel's chart wizards (the early versions, pre-Y2K). Also, I use OpenOffice more than Excel -- it has a pivot-table feature (I believe), but I imagine it is less mature; I stay on the simple paths. – Smandoli Aug 26 '10 at 18:42
+1 Used a pivot table to solve a problem identical to OP's. Thank you. – Rhyuk Oct 24 '13 at 16:34
Following up on Niketya's answer, there's a good explanation of Pivot Tables here: http://peltiertech.com/WordPress/grouping-by-date-in-a-pivot-table/
For Excel 2007 you'd create the Pivot Table, make your Date column a Row Label, your Amount column a value. You'd then right click on one of the row labels (ie a date), right click and select Group. You'd then get the option to group by day, month, etc.
Personally that's the way I'd go.
If you prefer formulae, Smandoli's answer would get you most of the way there. To be able to use Sumif by day, you'd add a column with a formula like:
``````=DATE(YEAR(C1), MONTH(C1), DAY(C1))
``````
where column C contains your datetimes.
You can then use this in your sumif.
-
Also would like to accept your answer as part of this... upvoted. – Andrew Johnson Aug 26 '10 at 1:03
If the second row has the same pattern as the first row, you just need edit first row manually, then you position your mouse pointer to the bottom-right corner, in the mean time, press ctrl key to drag the cell down. the pattern should be copied automatically.
-
The pattern of the dates does not copy down... only the row/column pattern. I suppress the row/column carrying down too, with the & signs. – Andrew Johnson Aug 18 '10 at 0:11 | 1,149 | 3,753 | {"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-2014-23 | latest | en | 0.905391 |
https://brilliant.org/problems/symmetric-sum-of-altitudes/ | 1,657,198,281,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00755.warc.gz | 193,339,675 | 8,262 | # Symmetric sum of altitudes
Triangle $ABC$ has area 15 and perimeter 20. Furthermore, the product of the 3 side lengths is 255. If the three altitudes of the triangle have lengths $d, e$, and $f$, then the value of $de+ef+fd$ can be written as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. What is $m+n$?
× | 97 | 329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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-2022-27 | latest | en | 0.853042 |
https://www.beatthegmat.com/a-project-requires-a-rectangular-sheet-of-cardboard-satisfying-the-following-requirement-when-the-sheet-is-cut-into-ide-t323781.html?sid=96017d8cff7fdaf150439d60a65b4a6c | 1,624,592,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488567696.99/warc/CC-MAIN-20210625023840-20210625053840-00138.warc.gz | 597,276,855 | 13,423 | ## A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into ide
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### A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into ide
by M7MBA » Fri May 14, 2021 2:13 am
00:00
A
B
C
D
E
## Global Stats
A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into identical rectangular halves, each of the resulting rectangles has the same ratio of length to width as the original sheet. Which of the following sheets comes closest to satisfying the requirement?
(A) A sheet measuring 7 inches by 10 inches
(B) A sheet measuring 8 inches by 14 inches
(C) A sheet measuring 10 inches by 13 inches
(D) A sheet measuring 3 feet by 5 feet
(E) A sheet measuring 5 feet by 8 feet
Source: Manhattan GMAT
### GMAT/MBA Expert
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### Re: A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into
by [email protected] » Fri May 14, 2021 5:29 am
M7MBA wrote:
Fri May 14, 2021 2:13 am
A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into identical rectangular halves, each of the resulting rectangles has the same ratio of length to width as the original sheet. Which of the following sheets comes closest to satisfying the requirement?
(A) A sheet measuring 7 inches by 10 inches
(B) A sheet measuring 8 inches by 14 inches
(C) A sheet measuring 10 inches by 13 inches
(D) A sheet measuring 3 feet by 5 feet
(E) A sheet measuring 5 feet by 8 feet
Source: Manhattan GMAT
Here's an algebraic solution:
Let x be length of the LONG side of the original rectangle
Let y be length of the SHORT side of the original rectangle
Then cut the rectangle into two pieces
We want the resulting rectangles to have the same ratio of length to width as the original sheet.
In other words, we want x/y = y/(x/2)
Cross multiply to get: x²/2 = y²
Multiply both sides by 2 to get: x² = 2y²
Divide both sides by y² to get: x²/y² = 2
Take square root of both sides to get: x/y = √2
IMPORTANT: For the GMAT, everyone should know the following APPROXIMATIONS: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2
So, we know that x/y ≈ 1.4
In other words, the ratio (LONG side)/(SHORT side) ≈ 1.4
(A) 10/7 = 1 3/7 ≈ 1.4 LOOKS GOOD!
(B) 14/8 = 1 6/8 = 1.75 ELIMINATE
(C) 13/10 = 1.3 ELIMINATE
(D) 5/3 = 1 2/3 ≈ 1.66 ELIMINATE
(E) 8/5 = 1.6 ELIMINATE | 798 | 2,751 | {"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} | 3.796875 | 4 | CC-MAIN-2021-25 | latest | en | 0.892801 |
https://www.physicsforums.com/threads/show-that-the-nested-intervals-property-fails-for-the-rational-numbers.389722/ | 1,701,919,120,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00274.warc.gz | 1,000,171,553 | 15,244 | # Show that the nested intervals property fails for the rational numbers
• Kb13
In summary, the conversation discusses constructing a nest of closed, bounded intervals in the field of formal rational functions to show that the Nested Intervals property fails in this field. This involves using an irrational number, such as sqrt2, as the center of the intervals and considering what happens as the intervals get smaller. The conversation also touches on the topology of the field and the subtlety of irrational numbers not being included in the intervals.
#### Kb13
In the field of formal rational functions, construct a nest of closed, bounded intervals whose intersection is empty. (That is, show that the Nested Intervals property fails in this field)
I know it has to involve radical 2 but because that is the only number we know is irrational but other than that I have no idea.
Try 'centering' your intervals around an irrational (you should know more irrationals than that, but sqrt2 is still a good choice). For example, one interval can be [sqrt2 - 1/2, sqrt2 + 1/2]. Turn this into a nested sequence and consider what happens as the length of the intervals get very small.
Kb13 said:
In the field of formal rational functions, construct a nest of closed, bounded intervals whose intersection is empty. (That is, show that the Nested Intervals property fails in this field)
Is there a topology assumed for this field? If so, what is it? I'm not sure what an "interval" looks like in this field.
I suspected he meant the set of rational numbers (since this is a common problem and he didn't know any other irrationals so odds are he doesn't know what a topology is).
Good point. I would like to point out one subtlety that he might otherwise overlook, namely that [sqrt2 - 1/2, sqrt2 + 1/2] in the rational field does NOT contain the endpoints since sqrt2 +/- 1/2 is irrational. It is nonetheless a closed (AND an open) interval. | 430 | 1,946 | {"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} | 3.578125 | 4 | CC-MAIN-2023-50 | latest | en | 0.944126 |
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# What would be the boiling point of methylene chloride at 670 mm Hg?
Updated: 8/9/2023
Wiki User
9y ago
53o C
It will be somewhat lower than 329 K (57oC), the boiling point at standard 760 mmHg, but it is not calculable. Some suggest -3 K, lower, so about 54oC but this couldn't be verified.
Wiki User
11y ago
Wiki User
14y ago
Hexane's boiling point at 760mm HG is 69. A correction is needed to go from 760mmHG to 670mm Hg. The correction is about 3.1. You subtract 3.1 from 69 to get 65.9 degrees celcius at 670 mm Hg
Wiki User
9y ago
The boiling point of acetone at 760 mmHg is 133.3 degrees Fahrenheit. Its freezing point is -138 degrees Fahrenheit, and the chemical formula is CH3COCH3.
Wiki User
15y ago
Hexane is part of the alkane series, which contains 6 carbon atoms. Its boiling point at 670 mmHg is 65.9 degrees Celsius or 160.62 degrees Fahrenheit.
Wiki User
9y ago
The boiling point of toluene at 670 mm Hg is 106.3 degrees Celsius. This is equal to 223.3 degrees Fahrenheit.
Wiki User
7y ago
The boiling point will be approx. 58 0C.
Jake
Lvl 7
2y ago
It is around 36.2*C
Wiki User
7y ago
61.2 deg C.
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Q: What would be the boiling point of methylene chloride at 670 mm Hg?
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The boiling point of the solution is lower, the boiling point is higher. | 842 | 3,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.832567 |
https://de.zxc.wiki/wiki/Gewichtskraft | 1,709,208,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00178.warc.gz | 192,061,452 | 10,586 | # Weight force
The force of gravity, and weight is caused by the effect of a gravitational field caused power to a body . In the rotating reference system of a celestial body (such as that of the earth), this gravitational field is made up of a gravitational component and a small centrifugal component. The gravitational force is perpendicular to the bottom directed what in the gravitational field of the earth almost, but not exactly the direction to the center of the Earth corresponds.
Usually or is used as a symbol . The SI unit for weight is the Newton (N). ${\ displaystyle {\ vec {F}} _ {\ text {G}}}$${\ displaystyle {\ vec {G}}}$
## Amount, direction and point of attack
The weight force can be calculated as the product of the mass with the gravitational acceleration : ${\ displaystyle {\ vec {F}} _ {\ text {G}}}$ ${\ displaystyle m}$ ${\ displaystyle {\ vec {g}}}$
${\ displaystyle {\ vec {F}} _ {\ text {G}} = m \, {\ vec {g}}}$
Apart from minor irregularities, the weight of a body is always directed towards the center of the celestial body on which it is located, since the gravitational field is a radial field as a good approximation . In most applications, however, sufficient accuracy can be achieved if the gravitational field is viewed as a homogeneous field , namely when all dimensions are much smaller than the radius of the celestial body. In this case the weight has the same direction and strength at every location.
The trajectory of a moving rigid body runs exactly as if the entire weight force was attacking the center of gravity (center of gravity) of the body. This also applies to the movement of the center of gravity of a system of several bodies. In a homogeneous gravitational field, the center of gravity coincides with the center of mass . If the weight is the only force acting, the body or the multi-body system is in a state of free fall . Since the inertia of a body depends on the mass in the same way as the weight, the accelerations of all freely falling bodies are the same. The acceleration due to gravity does not depend on the mass or other properties of the body, but at most on its location.
## Weight on earth
The approximate value can be used for the acceleration due to gravity on the earth's surface . ${\ displaystyle g = 9 {,} 81 \, \ mathrm {\ tfrac {m} {s ^ {2}}}}$
However, if you want to determine the weight on the earth more precisely, the location dependency of the gravitational acceleration ( at the equator or at the poles) must be taken into account using gravity formulas, for example the Somigliana formula . There are various causes for this location dependency: ${\ displaystyle g = 9 {,} 78 \, \ mathrm {\ tfrac {m} {s ^ {2}}}}$${\ displaystyle 9 {,} 83 \, \ mathrm {\ tfrac {m} {s ^ {2}}}}$
1. the centrifugal acceleration caused by the rotation of the earth ,
2. the different strength of gravity due to the flattening of the earth ,
3. local gravitational anomalies .
On the earth's surface, the first two effects depend on the geographical latitude : the first, because the latitude determines the distance of the location from the earth's axis; the second, because the latitude determines the distance from the center of the earth and the exact distribution of the earth's mass in relation to the location. In addition, there is a dependence on the height of the location above the earth's surface.
## Weight and mass
In everyday language , the weight of a body is spoken of without distinguishing whether it is its mass or its weight. However, the physical terms are very different:
• The mass is a measure of how strongly a body is generally influenced by gravitational fields and how much it resists accelerations ( inertia ).
• The weight, on the other hand, indicates how strongly a body is specifically attracted to the earth or the celestial body on which it is located.
The mass is therefore an inherent property of the body, while the weight force is the result of an external influence on the body.
As a result, the mass of a body is always the same , regardless of where it is (earth, moon , weightlessness , ...), while the weight acting on it depends on the acceleration of gravity. (On the moon the weight is only about one sixth of that on earth. A body with a mass of 6 kg is therefore only as heavy on the moon as a body with a mass of 1 kg on earth)
Until 1960 it was customary to give the force in the unit kilopond (kp). This was defined in such a way that the weight on earth, measured in kiloponds, had the same measure as the mass in kilograms ( ). After that, the kilopond in the SI system of units was replaced by the unit Newton (1 kp = 9.80665 N ≈ 1 daN ). Since then, the mass and the weight force have dimensions that are approximately around the above. Differentiating factor . ${\ displaystyle 1 \, \ mathrm {kp} = g \ cdot 1 \, \ mathrm {kg}}$${\ displaystyle 9 {,} 81}$
## Measurement
Measuring devices for the direct determination of a weight force are force gauges , for example spring balances . However, the static buoyancy falsifies the result, which is particularly noticeable with bodies of low density .
The weight force can also be determined indirectly by weighing and then converting the weight value . If you take a closer look at the functionality of a balance, you will find that the actual directly recorded measured variable is the weight force falsified by the buoyancy, even if a mass is displayed as the weighing value. For example, B. a simple beam balance the forces that the two masses exert on their respective pan.
## literature
Weight force is explained in many books that introduce mechanics. Examples are:
## Individual evidence
1. James H. Allen: Statics for mechanical engineers for dummies . John Wiley & Sons , 2012, ISBN 978-3-527-70761-4 , pp. 158 ( Google Books ).
2. Eberhard Brommundt, Gottfried Sachs , Delf Sachau: Technical Mechanics: An Introduction . 4th edition. Oldenbourg Verlag , Munich 2007, ISBN 978-3-486-58111-9 , p. 70 ( Google Books ).
3. Karlheinz Kabus: Mechanics and strength theory . 8th edition. Carl Hanser Verlag, Munich 2017, ISBN 978-3-446-45320-3 , pp. 121 ( Google Books ). | 1,424 | 6,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "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} | 3.984375 | 4 | CC-MAIN-2024-10 | latest | en | 0.917419 |
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# 1.3 - Math 2940 Solutions Fall 2011 Section 1.3 3 The...
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Math 2940 Solutions, Fall 2011 Section 1 . 3 3 ) The equations are y 1 = B 1 , y 1 + y 2 = B 2 , y 1 + y 2 + y 3 = B 3 so substituting, we see y 2 = B 2 - B 1 and y 3 = B 3 - y 1 - y 2 = B 3 - B 1 - ( B 2 - B 1 ) = B 3 - B 2 . So y 1 y 2 y 3 = 1 0 0 - 1 1 0 0 - 1 1 B 1 B 2 B 3 . 4 ) We have to solve x 1 w 1 + x 2 w 2 + x 3 w 3 = 0. The equations are x 1 + 4 x 2 + 7 x 3 = 0 , 2 x 1 + 5 x 2 + 8 x 3 = 0 , 3 x 1 + 6 x 2 + 9 x 3 = 0 . We eliminate the varible x 1 from the last trwo equations by adding - 2 and - 3 times the first equation to these two to get - 3 x 2 - 6 x 3 = 0 , - 6 x 2 - 12 x 3 = 0 . These last two equations are multiples of one another, so the solutions to one are exactly the solutions to the other. We’ll take x 2 = 2 and x 3 = - 1. Plugging back into the first equation gives x 1 = - 1. The three vectors lie in a plane . 6 ) Clearly the first two columns of 1 3 5 1 2 4 1 1 c are independent. We need to consider x 1 1 1 1 + x 2 3 2 1 = 5 4 c .
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{[ snackBarMessage ]} | 542 | 1,353 | {"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} | 4.375 | 4 | CC-MAIN-2018-17 | latest | en | 0.75001 |
https://www.hindawi.com/journals/jam/2012/835495/ | 1,544,536,286,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2018-51/segments/1544376823621.10/warc/CC-MAIN-20181211125831-20181211151331-00622.warc.gz | 930,684,229 | 90,357 | `Journal of Applied MathematicsVolumeΒ 2012, Article IDΒ 835495, 21 pageshttp://dx.doi.org/10.1155/2012/835495`
Research Article
## Existence of Solutions for the Evolution π(π₯)-Laplacian Equation Not in Divergence Form
Department of Mathematics, Jilin University, Changchun 130012, China
Received 31 October 2011; Accepted 6 December 2011
Copyright Β© 2012 Changchun Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
The existence of weak solutions is studied to the initial Dirichlet problem of the equation , with inf . We adopt the method of parabolic regularization. After establishing some necessary uniform estimates on the approximate solutions, we prove the existence of weak solutions.
#### 1. Introduction
In this paper, we investigate the existence of solutions for the -Laplacian equation The equation is supplemented the boundary condition and the initial condition where , , is a bounded domain with smooth boundary and .
In the case when is a constant, there have been many results about the existence, localization and extendibility and of weak solutions. We refer the readers to the bibiography given in [1β5] and the references therein.
A new interesting kind of fluids of prominent technological interest has recently emerged, the so-called electrorheological fluids. This model includes parabolic equations which are nonlinear with respect to the gradient of the thought solution, and with variable exponents of nonlinearity. The typical case is the so-called evolution -Laplace equation with exponent as a function of the external electromagnetic field (see [6β12] and the references therein). In [6], the authors studied the regularity for the parabolic systems related to a class of non-Newtonian fluids, and the equations involved are nondegenerated.
On the other hand, there are also many results to the corresponding elliptic -Laplace equations [13β15].
In the present work, we will study the existence of the solutions to problem (1.1)β(1.3). As we know, when is a constant, the nondegenerate problems have classical solutions, and hence the weak solutions exist. But in the case of -Laplace type, there are no results to the corresponding non-degenerate problems. Since (1.1) degenerates whenever and , we need to regularize the problem in two aspects corresponding to two different degeneracy: the first is the initial and boundary value and the second is the equation. We will first consider the non-degenerate problems. Based on the uniform Schauder estimates and using the method of continuity, we obtain the existence of classical solutions for non-degenerate problems. After establishing some necessary uniform estimates on the approximate solutions, we prove the existence of weak solutions.
This paper is arranged as follows. We first state some auxiliary lemmas in Section 2, and then we study a general quasilinear equation in Section 3. Subsequently, we discuss the existence of weak solutions in Section 4.
#### 2. Preliminaries
Denote that
Throughout the paper, we assume that where are given constants.
To study our problems, we need to introduce some new function spaces. Denote that We use to denote the closure of in .
Remark 2.1. In [16, 17], Zhikov showed Hence, the property of the space is different from the case when is a constant. This will bring us some difficulties in taking the limit of the weak solutions. Luckily, our approximating solutions are in , and hence the limit function is also in which avoids the above difficulties.
We now give the definition of the solutions to our problem.
Definition 2.2. A nonnegative function , , and is said to be a weak solution of (1.1)β(1.3), if for all satisfies the following:
In the following, we state some of the properties of the function spaces introduced as above.
Proposition 2.3 (see [15, 18]). (i)The space is a separable, uniform convex Banach space, and its conjugate space is , where . For any and , we have (ii)If for any , then and the imbedding continuous.(iii)There is a constant , such that This implies that and are equivalent norms of .(iv)We have .
Proposition 2.4 (see [18]). If we denote then(i), (ii), , (iii).
Lemma 2.5 (see [4]). Let . Then where is a positive constant depending on .
#### 3. A General Quasilinear Equation
Here, we will consider the general quasilinear equations where .
Proposition 3.1 (see [19, Theoremββ2.9 of Chapter I]). Let be a classical solution of (3.1) in . Suppose that the functions and take finite value for any finite , and , and that for and arbitrary where and are nonnegative constants. Then where depends only on , and .
We suppose that for and arbitrary the functions are continuous in , continuously differentiable with respect to , and , and satisfy the inequalities where is a nonnegative continuous function that tends to zero for and is an arbitrary function.
Lemma 3.2. Let be a classical solution of (3.1) in . Suppose that the conditions of Proposition 3.1 hold and satisfy (3.5) with a sufficiently small determined by the numbers , , , , and Then
The proof of Lemma 3.2 is quite similar to the Theoremββ4.1, chapter VI of [19]; one only has to replace with and remark that the constants in the proof are depending only on and ; we omit the details.
Theorem 3.3. Suppose that the following conditions are fulfilled.(a)For and arbitrary either conditions (3.3) are fulfilled.(b)For , (where is taken from estimate (3.4)) and arbitrary , the functions and are continuous and differentiable with respect to , , and and satisfy inequalities (3.5) with a sufficiently small determined by the numbers , , , , and (c)For and (where is taken from estimate (3.7)), the functions and are continuously differentiable with respect to all of their arguments.(d)The boundary condition (3.2) is given by a function belonging to and satisfying on (3.1), that is, (in other words, the compatibility conditions of zero and first orders are assumed to be fulfilled).(e)We have .
Then there exists a unique solution of problem (3.1) and (3.2) in the space . This solution has derivatives from .
Proof. We consider problem (3.1) and (3.2) along with a one-parameter family of problems of the same type Define the Banach space For any , let . Using Schauder theory, the linear problem admits a unique solution . Let , clearly , and define the map such that . By [19], we know that is continuous and compact. By Proposition 3.1, Lemma 3.2, and the Leray-Schauder fixed point principle, the operator has a fixed point .
#### 4. Existence
In this section, we are going to prove the existence of solutions of the problem (1.1)β(1.3).
Theorem 4.1. Assume that , and . Then the problem (1.1)β(1.3) admits a weak solution .
Consider the following problem: where , , and . Roughly speaking, here we use to regularize the initial-boundary value and use to regularize the equation. Thus, we have to carry out two limit processes, that is, first let (along a certain subsequence) and then let .
We first change (4.1) into the form where It is easily seen that (4.4) satisfies (3.3) and (3.5), where instead of . By Theorem 3.3, we know that (4.1)-(4.2) has a classical solution .
Proposition 4.2. We have
Proof. By the maximum principle, we know that .
A simple calculation shows that where and are defined as (4.4).
It is easy to prove that Hence, we have Let . Using the mean value theorem, we have Similarly, we get where , and are bounded functions. Hence, we see that Since on , by comparison principle of linear parabolic equation, we have .
Lemma 4.3. For all and , there hold
Proof. Multiplying (4.1) by , integrating both sides of the equality over and integrating by parts, we derive where denotes the outward normal to . Since from (4.5), , we have on . Hence where .
Using and Youngβs inequality, we have
Combining (4.14) with (4.15) yields Hence, Multiplying (4.1) by , integrating both sides of the equality over and integrating by parts and noticing that on , we derive
Equation (4.5), Lemma 4.3, and Proposition 2.3 imply that, for any , there exists a subsequence of , denoted by , and a function ββ, such that, as ,
Lemma 4.4. As , we have
Proof. Observe that . Multiplying (4.1) by , integrating both sides of the equality over and integrating by parts, we derive By HΓΆlder inequality and Lemma 4.3, we obtain Hence, We divide the integral in (4.25) in the following way: From (4.20), we see that Using Lemma 4.3, we have Now we estimate . If , then . Using () gives If , we obtain By (4.25), (4.26), and , we obtain Again by Lemma 2.5, we get Letting , we obtain (1). Again noticing that by Proposition 2.4, we see that (2) holds. To prove (3), we have Using Lemma 4.3 and (2), we see that (3) holds.
Finally, we prove (4). We have Equation (4.19) implies that To estimate , notice that As , we have , By HΓΆlder inequality, we have If , we have Hence, Thus (4) is proved, and the proof of Lemma 4.4 is complete.
Proposition 4.5. We obtain that is a weak solution of the problem then where is independent of .
Proof. Obviously, for all , . By Proposition 4.2 and (4.19)β(4.21), we know that (4.43) holds. (4.44) follows from (4.5), (4.19)β(4.21), and Lemma 4.3. To prove that satisfies the integral equality in Definition 2.2, we multiply (4.1) by , integrate both sides of the equality on , and integrate by parts to derive Letting to pass to limit and using (4.19) and Lemma 4.4 show that Applying Lemma 4.3, we derive where is independent of and . Hence,
From (4.43), we see that is bounded and increasing in , which implies the existence of a function , such that, as ,
Lemma 4.6. As , we have
Proof. We may take , in the integral equality satisfied by . Then it is easy to see that Hence, by (4.43), we have Notice that So Hence, By (4.43), (4.44), and (4.49), we see that as , Therefore, By Lemma 2.5, Hence, that is, Applying , (), and , we have Equation (4.50) and (1) imply that the right side tends to zero as . Since in , (2) is proved. (3) is an immediate consequence of (2).
Lemma 4.7. For any , we have where is independent of .
Proof. From Lemmas 4.3 and 4.4, it is easily seen that Using , (4.66), and Proposition 4.5, we have The proof of Lemma 4.7 is completed by combining (4.67) with (4.65).
Lemma 4.8. As , we have
Proof. Let and be the characteristic functions of and , respectively. Then Taking in Lemma 4.7, we obtain that . Since (), for any , we can choose such that . For fixed | 2,689 | 10,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.901032 |
http://sas-and-r.blogspot.com/2010/03/example-730-simulate-censored-survival.html | 1,469,455,014,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00010-ip-10-185-27-174.ec2.internal.warc.gz | 234,300,750 | 30,290 | ## Tuesday, March 30, 2010
### Example 7.30: Simulate censored survival data
To simulate survival data with censoring, we need to model the hazard functions for both time to event and time to censoring.
We simulate both event times from a Weibull distribution with a scale parameter of 1 (this is equivalent to an exponential random variable). The event time has a Weibull shape parameter of 0.002 times a linear predictor, while the censoring time has a Weibull shape parameter of 0.004. A scale of 1 implies a constant (exponential) baseline hazard, but this can be modified by specifying other scale parameters for the Weibull random variables.
First we'll simulate the data, then we'll fit a Cox proportional hazards regression model (section 4.3.1) to see the results.
Simulation is relatively straightforward, and is helpful in concretizing the notation often used in discussion survival data. After setting some parameters, we generate some covariate values, then simply draw an event time and a censoring time. The minimum of these is "observed" and we record whether it was the event time or the censoring time.
SAS
`data simcox; beta1 = 2; beta2 = -1; lambdat = 0.002; *baseline hazard; lambdac = 0.004; *censoring hazard; do i = 1 to 10000; x1 = normal(0); x2 = normal(0); linpred = exp(-beta1*x1 - beta2*x2); t = rand("WEIBULL", 1, lambdaT * linpred); * time of event; c = rand("WEIBULL", 1, lambdaC); * time of censoring; time = min(t, c); * which came first?; censored = (c lt t); output; end;run;`
The phreg procedure (section 4.3.1) will show us the effects of the censoring as well as the results of fitting the regression model. We use the ODS system to reduce the output.
`ods select censoredsummary parameterestimates;proc phreg data=simcox; model time*censored(1) = x1 x2;run;The PHREG ProcedureSummary of the Number of Event and Censored Values Percent Total Event Censored Censored 10000 5971 4029 40.29 Analysis of Maximum Likelihood Estimates Parameter StandardParameter DF Estimate Error Chi-Square Pr > ChiSqx1 1 1.98628 0.02213 8059.0716 <.0001x2 1 -1.01310 0.01583 4098.0277 <.0001Analysis of Maximum Likelihood Estimates HazardParameter Ratiox1 7.288x2 0.363`
R
`n = 10000beta1 = 2; beta2 = -1lambdaT = .002 # baseline hazardlambdaC = .004 # hazard of censoringx1 = rnorm(n,0)x2 = rnorm(n,0)# true event timeT = rweibull(n, shape=1, scale=lambdaT*exp(-beta1*x1-beta2*x2)) C = rweibull(n, shape=1, scale=lambdaC) #censoring timetime = pmin(T,C) #observed time is min of censored and trueevent = time==T # set to 1 if event is observed`
Having generated the data, we assess the effects of censoring with the table() function (section 2.2.1) and load the survival() library to fit the Cox model.
`> table(event)eventFALSE TRUE 4083 5917 `
`> library(survival)> coxph(Surv(time, event)~ x1 + x2, method="breslow")Call:coxph(formula = Surv(time, event) ~ x1 + x2, method = "breslow") coef exp(coef) se(coef) z px1 1.98 7.236 0.0222 89.2 0x2 -1.02 0.359 0.0160 -64.2 0Likelihood ratio test=11369 on 2 df, p=0 n= 10000 `
These parameters result in data where approximately 40% of the observations are censored. The parameter estimates are similar to the true parameter values.
Anonymous said...
proc lifereg data =simcox;
model t*censored(1) =x1 x2/dist =weibull;
run;
I find that the estiamted beta1 and beta2 are not mathced to those in simulation. i think i missed something. do you have any comments?
Ken Kleinman said...
I see two potential problems. First, you have "t" in the model statement, rather than "time." This is the uncensored version. You need either
model t =x1 x2 / dist =weibull;
or
model time*censored(1) =x1 x2 / dist =weibull;
otherwise the model won't fit very well.
The other issue is that the models are interpreted very differently. The parameter estimate from PHREG is the log relative hazard associated with a one-unit increase in the covariate, while the estimate from LIFEREG is the log relative change in time to event associated with the same one-unit increase.
This means that when we see a positive est estimate in PHREG (increased hazard of event) we should expect a _negative_ estimate in LIFEREG (sooner occurrence of event).
Anonymous said...
Cool. I got it. Maybe the following question is more basic but sort of confused to me:
linpred = exp(-beta1*x1 - beta2*x2);
I don’t see intercept and measurement error here. On the other hand, in the lifereg model, we indeed have an intercept. Can you illustrate what the estimated intercept stand for?
Thanks.
Ken Kleinman said...
The intercept ought to reflect here exactly what it does in any regression: the expected outcome when the covariates are all set to 0. There's no measurement error, but there's Weibull variability injected directly.
Szilard said...
Could you think a similar simulation scenario for Aalen Additive Hazards model?
I did try everything i could come up,but so far got nowhere.
Anonymous said...
I used the same simulation program, but I got different figures for the event, beta1 and beta2, is it normal to have different figures when the same simulation program is repeated a number of times?
Ken Kleinman said...
Very typical. Unless you set the seed for the pseudo-random number generator, you get a different stream of numbers every time, and the parameter estimates differ. This is one of the great strengths of simulation.
Wen said...
Would you please advise how to check if the estimates for scale and shape are correct? I used the code to generate a weibull distribution with shape=2 and scale=0.0008*(-beta1*x1-beta2*x2). And using the generated dataset to run a survreg model. The transformed shape and beta's are correct, but I have no idea how to check "scale"? What the scale estimate we should expect from the regression model?Is it the original lamdaT or scale=0.0008*(-beta1*x1-beta2*x2)?
Kennedy said...
This comment has been removed by the author.
Kennedy said...
Dear Ken. Your blog is helpful. In your frailty model simulation, time is generated by Weibull i.e t = rand("WEIBULL", 1, lambdaT * linpred); Would you advise on how one can simulate this time so that its roughly say every 3 months i.e baseline, month 3, month 6 etc. Kennedy | 1,731 | 6,512 | {"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-2016-30 | longest | en | 0.806021 |
http://priorart.ip.com/IPCOM/000088226 | 1,516,282,503,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887414.4/warc/CC-MAIN-20180118131245-20180118151245-00176.warc.gz | 288,865,761 | 7,661 | Browse Prior Art Database
# Arithmetic Algorithm for Determining Computer Program Execution Sequences
IP.com Disclosure Number: IPCOM000088226D
Original Publication Date: 1977-May-01
Included in the Prior Art Database: 2005-Mar-04
Document File: 1 page(s) / 12K
IBM
## Related People
Thomas, DR: AUTHOR
## Abstract
This algorithm "records", via a simple arithmetic computation, the logical path followed by a computer program. Hence, numerical data concerning program sequences may be obtained (e.g., the order in which modules are "called" during input/output operations), and faults detected (if the failure alters the sequences).
This text was extracted from a PDF file.
This is the abbreviated version, containing approximately 65% of the total text.
Page 1 of 1
Arithmetic Algorithm for Determining Computer Program Execution Sequences
This algorithm "records", via a simple arithmetic computation, the logical path followed by a computer program. Hence, numerical data concerning program sequences may be obtained (e.g., the order in which modules are "called" during input/output operations), and faults detected (if the failure alters the sequences).
More precisely, suppose that there are n events (e.g., module "calls" or procedure "openings") which may or may not occur during program execution. An algorithm will be given which determines after the fact, the order in which these events actually occurred (possibly with some occurring more than once or not at all).
With each event, "associate" a unique positive integer (e.g., when the events are procedure openings, association could be via a table to be used by the executive program). Let m be any integer strictly greater than all of the aforementioned integers (e.g., if the events are denoted by E(1), E(2),..., E(n), associate i with E(i) and let m = n+1). An integer P shall now be computed based upon the order of occurrence of some of these events.
Initially, let P = 0. Upon the occurrence of an event, replace P by Pm+a, where "a" is the integer previously associated with that event. Notice that after k> or = 1 events have occurred, P is given by the following expression: (1) P = a(k) + a(k-1)m + a(k-2)m/2/ + ... + a(i)m/-i/ + ... + a(1)m/k-1/. where a(i) denotes the integer... | 522 | 2,271 | {"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-2018-05 | latest | en | 0.913272 |
https://forum.bionicturtle.com/members/david-harper-cfa-frm.10/ | 1,726,787,742,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00017.warc.gz | 233,295,264 | 18,057 | Reactions
## Profile posts Latest activity Postings About
• I see that bionic turtle is now using VitalSource. However, unfortunately I did not receive any email from VitalSource with the new login.
Can you please send to me the login details ASAP? I really only have today to start studying and am already very behind. Also, I unfortunately did not receive any notification of this change until I logged in.
Your prompt response is much appreciated.
Hi David. I have a question. May sound silly, but I want some clarity. Risk Free Value = Risky Value + CVA. As the name indicates, Risk Free value is for assuming Zero risk. Not sure how can it have a risky component. Can you please advice? Thanks.
hey David,
I have just found a response to my question ,the spot rates in thee mapping exemple are just an assumption.
Hey David,
Could you please help me to understand how did you extract the spot rates in the exemple of
the Summarize how to map a fixed income portfolio into positions of standard instruments.
Instructional Video: Jorion, Chapter 11: VaR Mapping
Sincerely,
Soraya
Hello David,
I would like to thank you for this platform I am 19 years old, and in my 2nd year in college. I took financial accounting and corporate finance, and am a math major (I have done calc 3). Is it common for students like me to take the FRM? Cause I love this stuff (I spend upwards of 8 hours daily working through the material) but I'm so, so overwhelmed.... I plan to take the Oct/Nov exams. Any advice?
Hi David... I am still stuck with the straddle puzzle.... could I send you an email?
Hello David, can you help me solve this question?
Consider a ten-year mortgage loan secured by residential real estate, with an EAD of Euro 250,000 and an estimated LGD of 10 %. Supposing the customer’s PD is 2%, compute the capital requirement under the Basel II IRB approach, the standardized approach of Basel II, and under Basel I? Please explain step by step
Hi David, Kindly advise how many observations need to calculate daily VaR?
Nicole Seaman
Hello @Hassan2016. Please see the private message that I responded to. We do not provide one-on-one support in private messages or on profile pages like this. All of our support is provided in our public forum so it can also help other members. Thank you.
Hi David - in the AR(1) model Xt+1 = a + bXt + et+1 what is a? the intercept? why is a/1-b equal to the LRM?
Hi, I am surprised there is no BSM in Part 3 of FRM 1 (at least not in the videos and the first 3 option Hull chapters). Should we know this for the exam? Thanks in advance.
2. i am supposed to find review questions after each chapter i finish, i couldn't find that ? is there any depository for all practice questions related to each chapter?
Nicole Seaman
Hello I will answer this because David is very busy answering content questions in the forum. Our study planner is organized according to the GARP syllabus, not the GARP books. If you are referring to chapters in the GARP books, that is not how our study planner is laid out. It is organized by Part, Topic and Reading. You will find the individual materials under each Reading in the study planner.
Hello David, i'm a new member and have a couple of Q's as i find the website is pretty scrambled and the experience was below my expectation for the money paid.
1. The chapters' study notes are repeated under different authors (for ex. in Quant. we have chapter 4,5,6 under miller and watson and Diebold) which one in this case we should study from? also for their respective question set.
Nicole Seaman
The study planner is extremely organized (not scrambled), as it follows the exact GARP syllabus (NOT the GARP books). Please email me or send a private forum message if you have more questions about how our website is organized. Posting on our profiles in the forum is not the best way to ask questions. Thank you.
David Harper CFA FRM
@eldakrory GARP's syllabus assigns different authors !? ... we're helping you with authors THEY assigned .... I don't even know what to say
done with forum support today, just getting too tired and I need to get on to other work, I'll be back tomorrow (Monday). Don't stress, if you are practicing, you'll be find. And good luck
spent all morning in the forum catching up. Need to write practice questions and prep video, back tomorrow. Have a good Sunday!
I've been in the forum all morning, I need to record a video now. I will check back tomorrow AM (Sunday) to respond to new questions. Have a good weekend!
I'm back from 3-day vacation to Austin, TX ... catching up on forum questions!
I won't be actively replying to FRM type questions until Monday June 4th: Nicole & I are working on admin and the 2nd semester content plan
Hi David I found 1 error question in 2018 FRM Part 1 exam conducted on Saturday..Please let me know when i can notify you for the same | 1,133 | 4,872 | {"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-2024-38 | latest | en | 0.944464 |
https://ppqty.com/moves-only-king-left/ | 1,721,931,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00646.warc.gz | 409,845,611 | 21,145 | # How Many Moves When Only a King Is Left? (Chess)
What happens when only a king is left on the board?
How many moves can be made in this situation?
When only a king is left on the board for one player and there are no other pieces or pawns for either player, the game is in a specific endgame scenario.
Here’s what happens:
1. Stalemate: If it’s the turn of the player with the lone king and the king has no legal moves, then the game is a stalemate. In a stalemate, the game ends in a draw, meaning neither player wins.
2. Insufficient Material: If both players have only their kings left on the board, the game is immediately drawn due to insufficient material to checkmate. Neither side can deliver checkmate with just a king.
3. Checkmate: If one player has just a king, but enough material to checkmate to opposing king, the game will continue until checkmate, stalemate, three-fold repetition, or 50-move rule (no capture or pawn movement after 50 moves).
Below we look at the concept of stalemates and the 50-move rule in chess, shedding light on the possibilities and limitations when only a king remains.
## Understanding Stalemates
Stalemate is a unique situation in chess where a player’s king is not in check, but they have no legal moves to make.
In other words, the king is not under immediate threat, but there are no available moves to escape from the current position.
When a stalemate occurs, the game ends in a draw, regardless of the material advantage one player may have had.
Stalemates often arise when a player with a lone king is trying to avoid checkmate by maneuvering their king into a safe position.
However, if the player is not careful, they may unintentionally reach a stalemate position, resulting in a draw instead of a victory or loss.
### Example
Let’s consider a scenario where White has only a king left, and Black has a king and a rook.
White’s king is in check, and the only available move is to move the king to an adjacent square.
However, all adjacent squares are also under attack from Black’s rook.
In this case, White’s king is in a stalemate position, as there are no legal moves to make.
## The 50-Move Rule
The 50-move rule is a regulation in chess that states if no capture has been made and no pawn has been moved in the last 50 moves by each player, a draw can be claimed.
This rule prevents games from dragging on indefinitely when there is no progress or significant material change on the board.
The 50-move rule is particularly relevant when only a king is left on the board.
Since the king is unable to capture any pieces, and pawns cannot move, it is crucial to understand the limitations imposed by this rule.
### Example
Imagine a scenario where both players have only a king remaining, and no captures or pawn moves have occurred for the last 50 moves.
According to the 50-move rule, either player can claim a draw. This rule ensures that the game does not continue indefinitely without any progress or decisive moves.
## FAQs – How Many Moves in Chess When Only a King Is Left?
### 1. Can a king move anywhere on the board?
No, a king can only move one square in any direction: horizontally, vertically, or diagonally.
### 2. Can a king capture other pieces?
Yes, a king can capture other pieces by moving to a square occupied by an opponent’s piece.
However, in the scenario where only a king is left on the board, there are no pieces to capture.
### 3. What happens if a king is in checkmate?
If a king is in checkmate, the game ends, and the player whose king is checkmated loses.
Checkmate occurs when a player’s king is under attack and has no legal moves to escape capture.
### 4. Can a stalemate occur with more than just a king on the board?
Yes, a stalemate can occur with more than just a king on the board.
It can happen when a player’s king is not in check, but they have no legal moves to make with any of their pieces.
### 5. Can a stalemate be avoided when only a king is left?
Yes, a stalemate can be avoided when only a king is left by ensuring that the king always has at least one legal move available.
This requires careful maneuvering and avoiding positions where the king has no escape.
### 6. How does the 50-move rule affect games with only a king left?
The 50-move rule can lead to a draw in games with only a king left if no captures or pawn moves have occurred in the last 50 moves.
This rule prevents games from dragging on indefinitely without any progress.
### 7. Can the 50-move rule be overridden in certain situations?
No, the 50-move rule is a standard regulation in chess and cannot be overridden.
It applies to all games, regardless of the number of pieces left on the board.
### 8. Is there a way to force a checkmate with only a king?
No, it is not possible to force a checkmate with only a king.
Checkmate requires coordination between multiple pieces to trap the opponent’s king, which is not possible with just one king.
### 9. Can a player claim a draw if they are in a losing position with only a king left?
No, a player cannot claim a draw solely based on being in a losing position with only a king left.
The game continues until checkmate, stalemate, or the 50-move rule comes into effect.
### 10. Are stalemates and the 50-move rule commonly encountered in professional chess games?
Stalemates and the 50-move rule are relatively rare in professional chess games.
They are more likely to occur in amateur or casual games where players may not be as experienced or skilled.
## Summary – How Many Moves in Chess When Only a King Is Left?
When only a king is left on the chessboard, the number of moves that can be made is limited.
Stalemates can occur when a player’s king has no legal moves but is not in check.
The game ends in a draw in such cases.
The 50-move rule, on the other hand, allows for a draw to be claimed if no captures or pawn moves have occurred in the last 50 moves.
This rule prevents games from dragging on indefinitely without any progress.
Understanding these concepts is essential for chess players to navigate endgame scenarios effectively and make informed decisions. | 1,366 | 6,163 | {"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-2024-30 | latest | en | 0.944404 |
http://angolodeisaporifirenze.it/ryuz/infinite-square-well-expectation-value.html | 1,591,423,532,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00308.warc.gz | 6,930,691 | 19,044 | # Infinite Square Well Expectation Value
The expectation value of the position operator squared is. Find the commutator of the parity operator and the kinetic energy operator. Consider two eigenfunctions ψ 1 and ψ 2 of an operator Oˆ with corresponding eigen-values λ 1 and λ 2 respectively. Reconcile your answer with the fact that the KE of the particle in this level is 9p 2 hbar 2 /2ML 2. Particle in Finite-Walled Box One way to estimate the ground state energy of a finite potential well is to use the infinite well energy to produce a trial attenuation factor α. Examination of this problem enables us to understand the origin of many features of such systems, such as the appearance of discrete energy levels and the important concept of boundary conditions [ 3 ]. Calculating the expectation value of position and momentum. 3 Infinite Square-Well Potential 6. Wick's theorem must be applied for the evaluation of that function (see Ap- pendix A for details about these well-known techniques and references to earlier work). Itisnatural toidentify t R withthe timescalethat controls the eventual escape from the quasi-steady state, hence the approach to thermal equilibrium. %***** % Program 3: Matrix representation of differential operators, % Solving for Eigenvectors & Eigenvalues of Infinite Square Well %*****. 490L ≤ x ≤ 0. b) Find $\Psi(x,t)$. Energy Levels 4. This is call the Expectation Value. It is independent of n! Well is symmetric, so particle does not prefer one sid f ll h h hide of well to the other, no matter what state n it is in. 6 Simple Harmonic Oscillator 6. Use the v=0 and v=1 harmonic oscillator wavefunctions given below which are normalized such that ⌡⌠-∞ +∞ Ψ (x) 2dx = 1. Only the bound states are shown in this applet. 3 Infinite Square-Well Potential 6. A deuteron is bound state of proton and neutron (mp ~ mn~m~939 MeV/c2). Inside the box, δ²ψ/δx² + 2m (E)ψ /h²=0 (1) Put 2mE/h² = K². For any wavefunction ψ(q) the expectation value of gˆ for that wavefunction is defined as ψgˆψ≡∫ψ∗(q)gˆψ(q)dq Since ψ(q) 2 dq is the probability density, the expectation value can be considered to be the usual statistical notion of expectation value. Quantum Mechanics: Ground States for 2 Charged Particles in the 1D Infinite Square Well. (a) Normalize Ψ(𝑥𝑥,0) Graph it. 6 Simple Harmonic Oscillator 6. Compute the expectation value of the 𝑥𝑥 component of the momentum of a particle of mass 𝑛𝑛 in the 𝑛𝑛= 3 level of a one-dimensional infinite square well of width 𝐿𝐿. However, this holds when the random variables are independent: Theorem 5 For any two independent random variables, X1 and X2, E[X1 X2] = E[X1] E[X2]:. Given a wave function and an observable operator, calculate that operator's expectation value. I'm working in the infinite square well, and I have the wavefunction: I'm asked to calculate the expectation value of the particles position $\left\langle x\right. Reconcile your answer with the fact that the kinetic energy of the particle in this level is 𝐸𝐸. Examples are to predict the future course of the national economy or the path of a rocket. If the system is initially in an eigenstate of an operator Ĝ, then the expectation value of that operator is time independent. In an infinite system we have Π = 1 above p c and Π = 0 below p c. 5 Three-Dimensional Infinite- Potential Well 5. 4: Determine the time-independent expectation values for a two-state superposition. c) Calculate the uncertainty and explain your results. 2 Scattering from a 1D Potential Well *. What is the mass current at x= a=2? Problem14. onality of the in nite-square-well energy eigenfunctions in Gri ths or almost any other quantum mechanics textbook. For the position x, the expectation value is defined as Can be interpreted as the average value of x that we expect to obtain from a large number of measurements. In the momentum representation, wavefunctions are the Fourier transforms of the equivalent real-space wavefunctions, and dynamical variables are represented by different operators. 5 Three-Dimensional Infinite-Potential Well 6. 00 g marble is constrained to roll inside a tube of length L= 1:00cm. Quantum Mechanics Homework #6 1. The default wave function is a two-state superposition of infinite square well states. The expectation value of an observable A in the state ψ Infinite square well. front of it. 2 Expectation Value Consider a QM operator gˆ. The model is mainly used as a hypothetical example to illustrate the differences between classical and quantum systems. In quantum physics, if you are given the wave equation for a particle in an infinite square well, you may be asked to normalize the wave function. Expectation Value, Operators and Some Tricks (in Hindi) 9:49 mins. For the ground state, that is n=1 the energy is. have established a well-defined Hermitian operator in the infinite-dimensional space 'R. Calculating the expectation value of position and momentum. I'm working in the infinite square well, and I have the wavefunction:$$\psi(x,t=0)=A\left( i\sqrt{2}\phi_{1}+\sqrt{3}\phi_{2} \right). fined to an infinite one-dimensional square-well potential whose volume (width) is V. For any state n, the expectation value of the momentum of the particle is. Check that the uncertainty principle is satisfied. In quantum mechanics this model is referred to as. What is the expectation value of the energy?. It represents the expected average if we were to make many many measurements. In contrast, the most probable value is where we are MOST LIKELY to observe the particle. Proba bil ity, Exp ectat io n V al ue s, and U nce rtai n ties As indi cated earli er, on e of the re mark ab le featu res of the p h ysical w or ld is that rand om n ess is in carn ate, irred ucibl e. An Infinite Series for Resistor Grids. Expectation Values To relate a quantum mechanical calculation to something you can observe in the laboratory, the "expectation value" of the measurable parameter is calculated. Schrodinger's wave equation. 20) In addition, we know that such an initial waveform must be normalized:R. inside the well and (x) = 0 outside. In quantum mechanics the average value of an observable A whose operator is  is called the expectation value and is written like so: It is calculated from = ∫ ψ ψdτ / ∫ ψ 2 dτ (14. It is shown that this force apart from a minus sign is equal to the expectation value of the. To do this, we will solve for the expectation values of x, p, x^2, and p^2 for a wave function in a SINGLE basis state 'n. Standard Deviation. 7 Barriers and Tunneling Erwin Schrödinger (1887-1961) Homework due next Wednesday Oct. 2 Expectation Values 6. Now we know that the Schrodinger equation in general form-δ²ψ /δx²+ 2m (E-V)ψ /h²=0. Problem 1: A 3-D Spherical Well(10 Points) For this problem, consider a particle of mass min a three-dimensional spherical potential well, V(r), given as, V = 0 r≤ a/2 V = W r>a/2. INFINITE SQUARE WELL Lecture 6 (that naked mn is the motivation for the funny factor of q 2 a in (6. Reconcile your answer with the fact that the KE of the particle in this level is 9p 2 hbar 2 /2ML 2. the eigenfunctions and eigenvalues for the infinite square well Hamiltonian. to in nity, but care is. Infinite potential well A particle at t =0 is known to be in the right half of an infinite square well with a probability density that is uniform in the right half of the well. The tube is capped at both ends. We often refer to the expected value as the mean, and denote E(X) by µ for short. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0. 7(b)] Calculate the percentage change in a given energy level of a particle in a cubic box when the length of the edge of the cube is decreased by 10 percent in each direction. If the system is initially in an eigenstate of an operator Ĝ, then the expectation value of that operator is time independent. To do this, we will solve for the expectation values of x, p, x^2, and p^2 for a wave function in a SINGLE basis state 'n. Reconcile your answer with the fact that the KE of the particle in this level is 9p 2 hbar 2 /2ML 2. In Quantum Mechanics, everything is probabilistic (e. evaluated as an expectation value of a product of four Fermi operators. 2 Expectation Values 6. Application of Quantum Mechanics to a Macroscopic Object Problem 5. Mendes 2 6. But a theory may be mathematically rigorous yet physically irrelevant. Angular momentum operator 4. The potential is 0 inside a rectangle with diagonal points of the origin and (L x,L y) and infinite outside the rectangle. Bound States of a Semi-Infinite Potential Well. Energy levels. Quantum Mechanics: Ground States for 2 Charged Particles in the 1D Infinite Square Well. The collector current versus stopping voltage has minima for each energy value of the Hg atom. The mean value of x is thus the first moment of its distribution, while the fact that the probability distribution is normalized means that the zeroth moment is always 1. At t= 0, the walls are suddenly removed. In Quantum Mechanics, everything is probabilistic (e. A particle in an infinite square well has the initial wave function Ψ(x,0) = Ax(a− x). Determine A, find psi(x, t), and calculate (x) as a function of time. We are often interested in the expected value of a sum of random variables. The default wave function is a two-state superposition of infinite square well states. The ground state for an infinite square well of width ais 1 = r 2 a sin ˇx a (1) Clearly this isn't an easy sum, but we can calculate the expectation value of the Hamiltonian directly as an integral to get the result. Question: A particle in an infinite square well potential has an initial wave function {eq}\psi (x,t=0)=Ax(L-x) {/eq}. Find the second-order correction to the energies for the above potential. PROBLEMS FROM THE The time-dependent operator A(t) is defined through the expectation value, as Consider an electron in the infinite square well Suppose the electron is known to be in the first excited state for t 0. (c) What If?. Phase velocity and group velocity. The first two period behaviors of a quantum wave packet in an infinite square well potential is studied. Finding expected value of sum of random variables. Application of Quantum Mechanics to a Macroscopic Object Problem 5. the infinite square well potential V (a:) = O if O < < a and V = otherwise. The wavefunction of the election is said to contain all the information we can gather about the. Check that the uncertainty principle is satisfied. Choose all of the following statements that are correct at a given time t>0. The QM Momentum Expectation Value program displays the time evolution of the position-space wave function and the associated momentum expectation value. In classical systems, for example, a particle trapped inside a. The solutions are obtained by solving the time-independent Schrödinger equation in each region, and requiring continuity of both the wavefunction and its first derivative. [3] If all of the projectors act on different qudits, then this expectation value simply factors R ( ker H ) = ∏ i = 1 M ( 1 − Π i ) ¯ = ( 1 − p ) M , where p = Π i ¯ is the relative dimension of Π i. (a) Determine the expectation value of x. 2 Expectation Values 6. Application of Quantum Mechanics to a Macroscopic Object Problem 5. In addition to emphasizing the appearance of wave packet revivals, i. We now examine the nite square well, de ned as. In this video you will learn how to calculate expectation values for momentum or position when given a wavefunction. Was it possible programmatically to manipulate the volume as well as the pitch on computers with. The jth central moment about x o, in turn, may be defined as the expectation value of the quantity x minus x o, this quantity to the jth power,. The width of the well and the field direction and strength are adjustable. The proof of the sine Basel conjecture (PI)^2/2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 depends on the Newtonian Infinite Series formulae which are the ABC summation 1 + Ax + Bx^2 + Cx^3 = (1+ax)(1+ bx)(1 +cx) the ABC Alternating 1 -Ax + Bx^2 - Cx^3. Physics 452 Quantum mechanics II Winter 2012 Karine Chesnel Physics 452 Quantum mechanics II Winter 2012 Homework Phys 452 Thursday Feb 9 Assignment # 8: 7. Hypothesis HO. Problem 1 A particle of mass m is in the ground state (n=1) of the infinite square well: Suddenly the well expands to twice its original size -the right wall moving from a to 2a leaving the wave function (momentarily) undisturbed. 3 Bound States of a 1D Potential Well. So If your wave function for the nth state is. expected value calculation for squared normal distribution. Addition of angular momentum 4. A particle is in the ground state of an infinite square well potential given by, The probability to find the particle in the interval between and is (a) (b) (c) (d) Q46. 1st Read Chapters 4 and 5 of Kane Chapter 4: 2, 3, 5, 13, 20 Chapter 5: 3, 4, 5, 7, 8. A particle in the in nite square well has the initial wave function (x;0) = Ax(a x) (0 x a) for some constant A. We square this matrix to construct the energy levels and use the energy theorem of Fourier analysis to establish the wave-matrix connection. A particle in the infinite square well has initial wavefunction: a) Plot$\Psi(x,0)$and determine the constant$A$. ' Let us start with the x and p values below:. 1 Bound problems 4. Mendes 2 6. What is the mass current at x= a=2? Problem14. The configura-tions used for this estimate are chosen so that the probability p,of the p,th state appearing is related to the importance of its contribution to the expec-tation value. We use the notion of site percolation [2, 10] here i. Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexatio. However, these past couple of years have seen an incredible upswing in Korean music. The average of any quantity is defined (with some example value C_1 and C_2) as:. Schrodinger equation in spherical coordinates 4. The expectation value is de ned. Tricks to Find Expectation Value of Momentum and Position (in Hindi) Infinite Square Well Potential (in Hindi) 10:37 mins. However, this holds when the random variables are independent: Theorem 5 For any two independent random variables, X1 and X2, E[X1 X2] = E[X1] E[X2]:. When the wave function reaches one boundary, it is reflected back. Make the range of the wave function in the well clearly visible, show with a dot where the wave function vanishes. Coupled Well Pair: this is two square wells with a wall between them. The Infinite Square Well Potential. While results may vary (sometimes drastically), we can always find some average value to expect. The expectation value of the x - component of the orbital angular momentum in the state (where are the eigenfunctions in usual notation), is (a). Given Ψ(x, t) as one of the eigenstates of ĤΨ = EΨ, what is the expectation value of the Hamiltonian-squared? A) E B) an infinite square well of width a (0 a. b) Calculate the expectation of energy E. We are not going to discuss the consistency of the theory,. To calculate the expectation / average value for quantum operators, let us revisit the general definition of average values. L φ( x ,2 ) x Just for kicks, plot the n=2. Topics Fall 2018 Prof. Modified square well potential: Consider the following potential (a variation of the infinite square well): V(x) = ˆ −V 0 if 0 L For a particle in this potential, the normalized energy eigenfunctions are ψ n(x) = r 2 L sin nπx L. A particle, which is confined to an infinite square well of width L, has a wavefunction given by, lþ(x) = — Sin x) a) Calculate the expectation value of position x and momentum p. (Hint: use normalization and. Make the range of the wave function in the well clearly visible, show with a dot where the wave function vanishes. 2 Expectation Values 6. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0. Expectation value. 3: Infinite Square-Well Potential The simplest such system is that of a particle trapped in a box with infinitely hard walls that the particle cannot penetrate. Delta function potential as a shallow well. The Green function in I1 which has zero I I I I I I I " - r, - Figure 2. If the E is not much below V0, then the difference between the infinite and finite well solutions is larger. This implies that the operators representing physical variables have some spe-cial properties. 5 Three-Dimensional Infinite-Potential Well 6. A particle in an infinite square well, V(x) = 0 for 0 < x < L, V(x) = ∞ otherwise, has the time independent wavefunction: (a) By exploiting the orthonormality of the expansion functions, find the value of the normalization factor A. An electron in a 2D infinite potential well needs to absorb electromagnetic wave with wavelength 4040 nm (IR radiation) to be excited from lowest excited state to next higher energy state. The Algebra of an Infinite Grid of Resistors. At the boundaries, the wave function has to be continuous. Gea-Banacloche, "A quantum bouncing ball". 5 Three-Dimensional Infinite- Potential Well 5. This means that if you ran a probability experiment over and over, keeping track of the results, the expected value is the average of all the values obtained. Graduate Quantum Mechanics – Final Exam Problem 1) A particle is moving in one dimension (along the x-axis) and is confined to a box of length L (the potential V(x) is infinite for x < 0 and x > L and 0 for 0 ≤ x ≤ L). Physics 452 Quantum mechanics II Winter 2012 Karine Chesnel Physics 452 Quantum mechanics II Winter 2012 Homework Phys 452 Thursday Feb 9 Assignment # 8: 7. Expectation Values To relate a quantum mechanical calculation to something you can observe in the laboratory, the "expectation value" of the measurable parameter is calculated. Assuming that the system can be described by a square well of depth V0 and width R, show that to a good approximation V0 R2 = (\u3c0 2 )2 (\ufffd2 M ) 3. Angular momentum operator 4. Now we know that the Schrodinger equation in general form-δ²ψ /δx²+ 2m (E-V)ψ /h²=0. A particle in an infinitely deep square well has a wave function given by for 0 ≤ x ≤ L and zero otherwise. 2 A complete set of solutions is. (C) Is the uncertainty principle satisfied? For which state is the product Ar. Suppose we have an initial state vector lv(t -0) results from Mclntrye without re-deriving them, and you may use a computer for your math as long as you include your code in your solution A(3E1) 4iE2)). Two possible eigenfunctions for a particle moving freely in a region of length a, but strictly confined to that region, are shown in the figure below. The configura-tions used for this estimate are chosen so that the probability p,of the p,th state appearing is related to the importance of its contribution to the expec-tation value. , situations where a spreading wave packet reforms with close to its initial shape and width, we also examine in detail the approach to the collapsed phase where the position-space probability density is almost. I first needed to normalise the energy eigenfunction to determine A, I got something similar to the example here:. Expectation value of Hamiltonian. 7 Barriers and Tunneling Erwin Schrödinger (1887-1961). Suppose we have an initial state vector lv(t -0) results from Mclntrye without re-deriving them, and you may use a computer for your math as long as you include your code in your solution A(3E1) 4iE2)). to in nity, but care is. 1st Read Chapters 4 and 5 of Kane Chapter 4: 2, 3, 5, 13, 20 Chapter 5: 3, 4, 5, 7, 8. 67 x 10-27 Kg. Expected value of a product In general, the expected value of the product of two random variables need not be equal to the product of their expectations. The jth central moment about x o, in turn, may be defined as the expectation value of the quantity x minus x o, this quantity to the jth power,. Because the energy is a simple sum of energies for the , and directions, the wave function will be a product of wave function forms for the one-dimensional box, and in order to satisfy the first three of the boundary conditions, we can take the functions:. Proba bil ity, Exp ectat io n V al ue s, and U nce rtai n ties As indi cated earli er, on e of the re mark ab le featu res of the p h ysical w or ld is that rand om n ess values of ev ery ph ysical prop ert y at some in stan t in time , to un limited precis ion. Infinite square well We now turn to the most straightforward (and therefore educational) non-zero potentials. The wave functions in are sometimes referred to as the "states of definite energy. So at some point, someone just made one up, and designated it by the letter i (which stands for "imaginary"): i 2 = 1, by definition. This is quantum mechanics! Who knows? Separation of Variables and Stationary States. Using the ground state solution, we take the position and. (3 marks) B): For a spherical symmetric state of a hydrogen atom, the Schrodinger equation in spherical coordinates is h2 2 du kee2 2m dr r dr. Without such loading, "expected value of a random variable taking countably infinite values" doesn't have plausible meaing due to Riemann Rearrangement Thm, and irresistant to change of the terms in the series itself. Quantum Mechanics: Ground States for 2 Charged Particles in the 1D Infinite Square Well. You can see the first two wave functions plotted in the following figure. An Interpretation of Quantum Mechanics One- Dimensional Wave Functions & Expectation Values The Particle Under Boundary Conditions The Schrodinger Equation The particle in a BOX A particle in a Well of finite Height Tunneling Through a Potential Energy Barrier The Scanning Tunneling Microscope The Simple Harmonic Oscillator Text Book PHYSICS for Scientists and. 8 A particle in the infinite square well has the initial wave function. You can use the wave function to calculate the "expectation value A perfect example of this is the "particle in a box" group of solutions where the particle is assumed to be in an infinite square potential well in one dimension, so there is zero potential (i. square well, radius rs, depth V. 4 Calculate the expectation values of x and x 2 for a particle in the state n = 2 in a square-well potential. A physical variable must have real expectation values (and eigenvalues). 6 Simple Harmonic Oscillator 6. wave function outside well. particles in a quantum state Ψ. The first three quantum states (for of a particle in a box are shown in. In quantum mechanics, well compute expectation values. 7 Barriers and Tunneling Erwin Schrödinger (1887-1961) Homework due next Wednesday Oct. 22) ( ) sin 22 2 2 sin cos 0. To see how a result matches with Classical Mechanics, we can use the concept of an "Expectation Value". (a) Find the possible values of the energy, that is, the energies E n. (b) Compute hxi, hpi and hHi,att=0. Finding expected value. The infinite square well potential and the evolution operator method for the purpose of overcoming misconceptions in quantum mechanics Article (PDF Available) in European Journal of Physics 35(2. In probability theory, an expected value is the theoretical mean value of a numerical experiment over many repetitions of the experiment. , 1-D infinite square well), find the eigenvectors and eigenvalues for the energy operator. That is, the allowed wavelengths are just slightly longer than if it were an infinite well. Hence the name isosurface - the value of the function is the same at all points on the surface. 2 Expectation Values 6. The average or expectation value of the energy of a particle in an infinite square well can be worked out either by using the series solution in the form hHi=å n jc nj 2 E n (1) or directly using an integral, using H=p2=2mand p=(¯h=i )(d=dx): hHi= h¯ 2 2m a 0 Y(x;t) d dx2 Y(x;t)dx (2) Since Y(x;t) in the general case is a sum over. onality of the in nite-square-well energy eigenfunctions in Gri ths or almost any other quantum mechanics textbook. is Planck's constant. Schrodinger equation in spherical coordinates 4. Finite square well 4. Modelling this as a one-dimensional in nite square well, determine the value of the quantum number nif the marble is initially given an energy of 1. Barbaroux1 and A. Now we can answer the question as to the probability that a measurement of the energy will yield the value E1? The energy levels of an infinite square well is given as. Inflnite potential energy constitute an impenetrable barrier. The model is mainly used as a hypothetical example to illustrate the differences between classical and quantum systems. An electron is confined to a box of width 0. The Quantum 1D Infinite Square Well (ISW) The only “simple” problem in quantum mechanics is the infinite square well. 7 - An electron with kinetic energy 2. 3 Infinite Square-Well Potential 6. 23, 2013 Dr. In this article author has developed computer simulation using Microsoft Excel 2007 ® to graphically illustrate to the students the superposition principle of wave functions in one dimensional infinite square well potential. Continuity of the first derivative of the wave function and boundary conditions. (e) By symmetry considerations alone and on the basis of the infinitely deep square potential well model, what will be the expectation value of the position x. In this program, We can: 1. The infinite square well is a weird animal if analysed by itself. Infinite square well We now turn to the most straightforward (and therefore educational) non-zero potentials. It is independent of n! Well is symmetric, so particle does not prefer one sid f ll h h hide of well to the other, no matter what state n it is in. uncertain; b. 1) We assume that the width of the well is initially V = V1 and that the initial energy of the system is a fixed. Actually it's quite simple to comprehend - a finite square well is a one-dimensional function V(x) which has a constant value V 0 everywhere except where |x| < L, when it drops to zero (Figure). 2 Expectation Values 6. The expected value can really be thought of as the mean of a random variable. Superposition of energy eigenstates in the one-dimensional infinite square well. Schrodinger's Equation, Current Density, Continuity Equation (in Hindi) 8:30 mins. Infinite potential well A particle at t =0 is known to be in the right half of an infinite square well with a probability density that is uniform in the right half of the well. 3 Infinite Square-Well Potential 6. property A is estimated by its expectation value over a relatively small sample of the total collection of states of the system. For the position x, the expectation value is defined as. Indeterminacy in expectation value. This model also deals with nanoscale physical phenomena, such as a nanoparticle trapped in a low electric potential bounded by high-potential barriers. 6 Simple Harmonic Oscillator 6. Which state comes closest to the uncertainty limit? 4. A particle, which is confined to an infinite square well of width L, has a wavefunction given by, lþ(x) = — Sin x) a) Calculate the expectation value of position x and momentum p. Generic 3-Level Quantum System Ket Representation Matrix Representation Graph Representation. A): A quantum particle is in an infinite deep square well has a wave function l/f(x) — — sin —x for 0 x L and zero otherwise. 4 Finite Square-Well Potential 6. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0. (b) Compute hxi, hpi and hHi,att=0. Answer to: A particle in an infinite square well potential has an initial wave function psi (x,t=0)=Ax(L-x). 3 Infinite Square-Well Potential 5. 2 Expectation Values 6. 5 Three-Dimensional Infinite-Potential Well 6. It is independent of n! Well is symmetric, so particle does not prefer one sid f ll h h hide of well to the other, no matter what state n it is in. , for the nth stationary state of the infinite square well. This Demonstration considers a Gaussian wavepacket , in the position and momentum representations, respectively. Outside the well, of course, = 0. In the position domain, this is equivalent to an infinite square-well potential, or particle-in-a-box. What is the position expectation value as a function of time? Solution (a) The given initial conditions have already been expanded in the basis of energy eigenfunctions of. 3 Infinite Square-Well Potential 6. Next: Expectation Values and Variances Up: Fundamentals of Quantum Mechanics Previous: Schrödinger's Equation Normalization of the Wavefunction Now, a probability is a real number between 0 and 1. Quantum Mechanics in 3D: Angular momentum 4. 6 Simple Harmonic Oscillator 6. (a) Normalize Ψ(x,0). Question: A particle in an infinite square well potential has an initial wave function {eq}\psi (x,t=0)=Ax(L-x) {/eq}. 67, 776-782 (1999). The infinite square‐well potential describes a one‐dimensional problem where a particle of mass m bounces back and forth in a “box” described by the potential, V(x), which is zero for x between 0 and a and infinite when x is either smaller than 0 or larger than a. It's like asking you what is the area under a curve on just this line. 23 expectation value of x expectation value of p px. In quantum physics, if you are given the wave equation for a particle in an infinite square well, you may be asked to normalize the wave function. The momentum operator in position space is given by. What is the expectation value of the energy?. For example, start with the following wave equation: The wave function is a sine wave, going to zero at x = 0 and x = a. Infinite square well, particle in a finite well; barrier penetration, reflection 3. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0. (b) Calculate the wavelengths of the emitted photons when the electron makes transitions between the fourth and the second excited states, between the second excited state and the ground state, and between the third and the second excited states. Schrodinger's wave equation. At the height of this COVID-19 pandemic, the government of Kano State, in its infinite wisdom, decided to "repatriate" Nigerian citizens (almajirai) back to their "states of indigeneity". Examples are to predict the future course of the national economy or the path of a rocket. Example III–1. 2 Expectation Values 6.$\begingroup$This example ignores the loading of absolute-summability in the def'n of expected value of a random variable taking countably infinite values. Choose all of the following statements that are correct at a given time t>0. Calculate the expectation value for position and momentum operator. A particle, which is confined to an infinite square well of width L, has a wavefunction given by, lþ(x) = — Sin x) a) Calculate the expectation value of position x and momentum p. (a) Determine the expectation value of x. What is the expectation value of the Posted 2 years ago. 2 Expectation Values 5. An electron trapped in a one-dimensional infinite square potential well of width $L$ obeys the time-independent Schrodinger equation (TISE). Expectation values in the infinite square well. Robinett, “Visualizing the Collapse and Revival of Wave Packets in the Infinite Square Well Using Expectation Values,” Am. Quantum Mechanics 1 (TN2304) Geüpload door. A physical variable must have real expectation values (and eigenvalues). Find (a) the wave function at a later time, (b) the probabilities of energy measurements, and (c) the expectation value of the energy. I’ll let you work out a few special cases in the homework. Also studied in the article is a statistic ~x2 , which is a function of a random variable x =()x() () ( )0 , x 1 ,, x M −1 created from discrete random signal samples. An electron energy of 4. Quantum Mechanics Homework #6 1. The momentum and Hamil-tonian operators. expectation value of the position operator squared. A physical variable must have real expectation values (and eigenvalues). So at some point, someone just made one up, and designated it by the letter i (which stands for "imaginary"): i 2 = 1, by definition. 23 expectation value of x expectation value of p px. PHYS 3313 - Section 001 Lecture #13 Wednesday, Oct. Expectation Value, Operators and Some Tricks (in Hindi) 8:00 mins. The expected value of a random variable is essentially a weighted average of possible outcomes. [The time independent Schrodinger's equation for a particle in an in nite square well is h 2 2m d dx2 = E Substitution of the. Answer to: A particle in an infinite square well potential has an initial wave function psi (x,t=0)=Ax(L-x). So the expectation value of the momentum of a particle in an infinite square well is zero? Of course it is! The allowed energy levels in a well can be thought of as the standing waves that "fit" in the well. Griffiths, Pearson Education, Inc. An Interpretation of Quantum Mechanics One- Dimensional Wave Functions & Expectation Values The Particle Under Boundary Conditions The Schrodinger Equation The particle in a BOX A particle in a Well of finite Height Tunneling Through a Potential Energy Barrier The Scanning Tunneling Microscope The Simple Harmonic Oscillator Text Book PHYSICS for Scientists and. 6 Simple Harmonic Oscillator 5. Consider two eigenfunctions ψ 1 and ψ 2 of an operator Oˆ with corresponding eigen-values λ 1 and λ 2 respectively. In quantum physics, if you are given the wave equation for a particle in an infinite square well, you may be asked to normalize the wave function. So, the deviation is a spread of the quantity under consideration. Particle in an infinite square well potential. Reconcile your answer with the fact that the KE of the particle in this level is 9p 2 hbar 2 /2ML 2. As William Feller notes on p. The Infinite Square Well Potential. 7 - Atoms in a crystal lattice vibrate in simple Ch. For example, start with the following wave equation: The wave function is a sine wave, going to zero at x = 0 and x = a. We usually combine equation 9 with the normalization condition to write Z a 0 m(x) n(x)dx= mn; (11) where mnis an abbreviation called the Kronecker delta symbol, de ned. have established a well-defined Hermitian operator in the infinite-dimensional space 'R. (a) Normalize Ψ(x,0). We review the history, mathematical properties, and visualization of these models, their. For brevity, we omit the commands setting the parameters L,N,x,and dx. 5 3-D Finite Square Well 6. Finite square well 4. The figure below shows two isosurfaces of the wave function. In an infinite system we have Π = 1 above p c and Π = 0 below p c. Show that Emust exceed the minimum value of V(x), for every normalizable solution to the time independent schrodinger equation h2 2m d2 dx2 + V = E 2 In nite square well 3. Boundary conditions. A particle, which is confined to an infinite square well of width L, has a wavefunction given by, lþ(x) = — Sin x) a) Calculate the expectation value of position x and momentum p. 5: Normalization and Expectation Values of Given Wavefunction Problem from Introduction to Quantum Mechanics, 2nd edition, by David J. On page 2 of SC2 there is a finite 1D well with U 0 = 17 eV and L = 0. Example III–1. U= ∞ U= ∞ 0 L x E n n=1 n=2 n=3 The idea here is that the photon is absorbed by the electron, which gains all of the photon's energy (similar to the photoelectric effect). 7 Barriers and Tunneling I think it is safe to say that no one understands quantum mechanics. 23, 2013 Dr. V(x) is called the potential function and it determines behavior of the quantum particle. infinite square well are orthogonal: i. Graduate Quantum Mechanics – Final Exam Problem 1) A particle is moving in one dimension (along the x-axis) and is confined to a box of length L (the potential V(x) is infinite for x < 0 and x > L and 0 for 0 ≤ x ≤ L). x>, and similar language can be used for p. Find the expectation value. Barbaroux1 and A. Check that the uncertainty principle is satisfied. (a) Determine the expectation value of x. 6 Simple Harmonic Oscillator 6. the eigenfunctions and eigenvalues for the infinite square well Hamiltonian. 2 Expectation Values 5. The following code finds the square root of a number, it runs fine unless you compile with MinGW gcc: [code] #include #include. An outcome of a measurement which has a probability 0 is an impossible outcome, whereas an outcome which has a probability 1 is a certain outcome. of the ground state is zero. Griffiths, Pearson Education, Inc. Finite 1-D square well: For an electron in a potential well of finite depth we must solve the time-independent Schrödinger equation with appropriate boundary conditions to get the wave functions. 2 A complete set of solutions is. 4 Finite Square-Well Potential 6. Linear Combination. Linear harmonic oscillator (§2. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0. (5 pts) There are an infinite number of bound energy states for the finite potential. Choose any arbitrary initial un-normalized wave function : psi(x,0) Example : Gaussian wave packet 2. A particle in an infinite square well has the initial wave function Ψ(x,0) = Ax(a− x). In the infinite square well, the potential energy is very simple, and has a graph that kind of looks like—well, a big, square, well. Find the commutator of the parity operator and the kinetic energy operator. Energy in Square infinite well (particle in a box) 4. In the last century, the urban development of Hangzhou concentrated and grew around the single center of the West Lake. 1 The case of an infinite square well whose right wall expands at a constant velocity (v) can be solved exactly. 7(b)] Calculate the percentage change in a given energy level of a particle in a cubic box when the length of the edge of the cube is decreased by 10 percent in each direction. In quantum mechanics, well compute expectation values. Particle in Finite-Walled Box One way to estimate the ground state energy of a finite potential well is to use the infinite well energy to produce a trial attenuation factor α. The expectation value, in particular as presented in the section "Formalism in quantum mechanics", is covered in most elementary textbooks on quantum mechanics. 68, 410-420 (2000). It can be shown that the expectation values of position and momentum are related like the classical position and. Calculate x, x^2, p, p^2, ?x and ?p for the nth stationary state of the infinite square well. Probability theory - Probability theory - Conditional expectation and least squares prediction: An important problem of probability theory is to predict the value of a future observation Y given knowledge of a related observation X (or, more generally, given several related observations X1, X2,…). 7 Barriers and Tunneling Erwin Schrödinger (1887-1961) Homework due next Wednesday Oct. Finite square well 4. Bound States of a Semi-Infinite Potential Well. Check that the uncertainty principle is satisfied. Position expectation Position expectation value value for for infinite square well This result means that average of many measurements of the position would be at x=L/2. That is, the allowed wavelengths are just slightly longer than if it were an infinite well. 5 Three-Dimensional Infinite-Potential Well 6. Median: value where we half the population has a higher value and half the population has a lower value. Some of the properties associated with a particle are continuous, like position, while others are. As an example of program , we use the time evolution of a wave packet. 22) ( ) sin 22 2 2 sin cos 0. However, this holds when the random variables are independent: Theorem 5 For any two independent random variables, X1 and X2, E[X1 X2] = E[X1] E[X2]:. As William Feller notes on p. Of course, these are theoretical idealizations, but it gives a basic idea of how you solve the Schrodinger equation without accounting for many of the complications that exist in nature. The tube is capped at both ends. wave function outside well. A particle in an infinite square well has an initial wave. Itisnatural toidentify t R withthe timescalethat controls the eventual escape from the quasi-steady state, hence the approach to thermal equilibrium. lation the statistical-mechanical expectation value (A) of a. Infinite Square Well Potential (in Hindi) 10:37 mins. This paper focuses on common stock returns governed by a formula structure; the APT is a one-period model, in which avoidance of arbitrage over static portfolios of these assets leads to a linear relation between the expected return and its covariance with the factors. 5 Three-Dimensional Infinite-Potential Well 6. 3 Infinite Square-Well Potential 5. %***** % Program 3: Matrix representation of differential operators, % Solving for Eigenvectors & Eigenvalues of Infinite Square Well %*****. If X>0:1, then you are succesful in round 1; if X>0:2, then you are succesful in round 2; if X>0:3, then you are succesful in round 3. So the expectation value of the momentum of a particle in an infinite square well is zero? Of course it is! The allowed energy levels in a well can be thought of as the standing waves that "fit" in the well. (We use here the "alternative origin" rather having the well centered on the origin. Basically you calculate the expectation value of "x^2" and subtract from it the expectation value of x, which is then squared. 3 Infinite Square-Well Potential 6. the infinite square well potential V (a:) = O if O < < a and V = otherwise. The fact that the expectation values satisfy the New. 23 expectation value of x. This is a more involved process, though, so here you'll only be able to see the results rather than run through. For example, if the potential V (x) takes the value V 0 outside the potential well and 0 inside it, the wave function can be determined in the three main regions covered by the problem. 7 Barriers and Tunneling Erwin Schrödinger (1887-1961). Find the expectation value of H. These models frequtly appear in the research literature and are staples in the teaching of quantum they on all levels. fined to an infinite one-dimensional square-well potential whose volume (width) is V. Consider two cases: (a) The infinite well, U(x) = 0 for 0 < x < L, and U(x) infinite. Hyosung Group, under the leadership of Chairman Cho Hyun-joon is set to construct the world's largest liquid hydrogen plant to better vitalize the local hydrogen economy. Their procedure inust be worked entirely in the 4 space, and allows its dimension tend to infinity only after expectation values are calcu-lated in order to become true expectation values. Find the expectation values of the components of in the total angular momentum eigenstate ; that is, J 2 has eigenvalue and J z has eigenvalue. Examination of this problem enables us to understand the origin of many features of such systems, such as the appearance of discrete energy levels and the important concept of boundary conditions [ 3 ]. At the height of this COVID-19 pandemic, the government of Kano State, in its infinite wisdom, decided to "repatriate" Nigerian citizens (almajirai) back to their "states of indigeneity". 1D scattering problem. 6 Simple Harmonic Oscillator 6. Choose all of the following statements that are correct at a given time t>0. x>, and similar language can be used for p. (b) If a measurement of the energy is made, what are the possible results? What is the. The particle is thus bound to a potential well. b) Use the result from part (a) to find the expectation value for X and the expectation value for X^2 for a classical particle in such a well. A particle in the infinite square well has the initial wave function 15 (a) Sketch ψ (x, 0), and determine the constant A. 2: Rank infinite square well energy eigenfunctions; 10. We can choose this energy value to be zero V= 0, 0 < x < L, V , x 0 and x L Particle in a one dimensional Box (infinite square well potential) Particle in a one dimensional Box (infinite square well potential) Page 6 Since the walls are impenetrable, there is zero probability of finding the particle outside the box.$\begingroup$This example ignores the loading of absolute-summability in the def'n of expected value of a random variable taking countably infinite values. In probability theory, an expected value is the theoretical mean value of a numerical experiment over many repetitions of the experiment. At time t=0, the state of a particle in this square well is. Expectation value of Hamiltonian. We review the histy, mathematical properties, and visualization of these models, their many. (Hint: use normalization and. You can do it by straight forward substitution of the appropriate y and A in calculating = or you can use some ingenuity to get the. The Infinite Square Well Potential. However, these past couple of years have seen an incredible upswing in Korean music. Which state. The wavefunction of an electron in a one-dimensional infinite square well of width a, x (0, a), at time t =0 is given by Ψ(x,0)=√2/7 ψ 1 (x) +√5/7 ψ 2 (x), where ψ 1 (x) and ψ 2 (x) are the ground state and first ex-cited stationary. For the ground state, that is n=1 the energy is. Joye1,2 Received June 20, 1997; final November 24, 1997 Let U(t) be the evolution operator of the Schrodinger equation generated by a Hamiltonian of the form H0(t) + W(t), where H0(t) commutes for all t with a. For any wavefunction ψ(q) the expectation value of gˆ for that wavefunction is defined as ψgˆψ≡∫ψ∗(q)gˆψ(q)dq Since ψ(q) 2 dq is the probability density, the expectation value can be considered to be the usual statistical notion of expectation value. We are often interested in the expected value of a sum of random variables. 7 - An electron with kinetic energy 2. (b) Find ψ (x, t) (c) What is the probability that a measurement of energy would yield the value E 1? (d) find the expectation value of the energy. Example 1: Expectation values of momentum? Recall – standing wave has two. 6-2 The Infinite Square Well 243 Just as in the case of the standing-wave function for the vibrating string, we can con-sider this stationary-state wave function to be the superposition of a wave traveling to the right (first term in brackets) and a wave of the same frequency and amplitude trav-eling to the left (second term in brackets). In the Stark effect for a hydrogen atom in its ground state, the energy change D E associated with a small applied electric field is proportional to the square of the field strength , , where a is the polarizability. One version of the Heisenberg Uncertainty Principle is ΔxΔp x=α x. Perturbation remove degeneracy. Mean: average value in limit of infinite number of measurements:. Probability theory - Probability theory - Conditional expectation and least squares prediction: An important problem of probability theory is to predict the value of a future observation Y given knowledge of a related observation X (or, more generally, given several related observations X1, X2,…). When a probability distribution is normal, a plurality of the outcomes will be close to the expected value. Expectation values of of a particle in the infinite well box of width a is given by 33. Let Ψ = a ψ1 + b ψ2 + c ψ3 + d ψ4 a superposition of 4 states <Ε> = expectation value of E ( similar to the average value of E) = E1 x P1 + E2 x P2 + E3 x P3 + E4 x P4. uncertain; b. Derive the equation for scattering we had started in class. INFINITE SQUARE WELL - CHANGE IN WELL SIZE 3 hEi = a 0 1(x) h¯ 2 2m! d dx2 1(x)dx (18) h¯ 2 ma a 0 sin ˇx a d dx2 sin ˇx a dx (19) = ¯h 2ˇ ma3 a 0 sin2 ˇx a dx (20) = h¯ 2ˇ 2ma2 (21) = E 2 (22) If you’re interested, a side effect of this result is that we can evaluate the. In another article we discussed the problem of determining the limiting value of resistance between two diagonally neighboring nodes of an square grid of unit resistors as the size of the grid increases toward infinity. 3 Infinite Square-Well Potential 6. Uncertainty in p Now that we have found the expectation value of momentum and of the momentum squared, we can find the uncertainty in the momentum of the particle using the standard expression (see the. is Planck's constant. To find the expected value of a continuous function, we use integration. We call this the expectation value. 1 IntroductionThe use of sandwich structures has been increasing in recent years because of their lightweight and high stiffness. A particle in the infinite square well has the initial wave function 15 (a) Sketch Ψ(x. Now we know that the Schrodinger equation in general form-δ²ψ /δx²+ 2m (E-V)ψ /h²=0. The Hamiltonian is$H = p^2 / 2m$ inside the potential. Interactive simulation that allows the user to set up different superposition states in a one-dimensional infinite square well, and that depicts the expectation value of position and the position uncertainty. The infinite square well is a weird animal if analysed by itself. These models frequently appear in the research literature and are staples in the teaching of quantum theory on all levels. At t= 0, the walls are suddenly removed. Tricks to Find Expectation Value of Momentum and Position (in Hindi) Infinite Square Well Potential in 2-D (in Hindi) 8:30 mins. A particle in an infinite square well has the initial wave function Ψ(x,0) = Ax(a− x). t), and calculate (x), as a function of time. (That is, find A. So, for instance,. Schrodinger's Equation, Current Density, Continuity Equation (in Hindi) 8:30 mins. However, this holds when the random variables are independent: Theorem 5 For any two independent random variables, X1 and X2, E[X1 X2] = E[X1] E[X2]:. (b) Compute hxi, hpi and hHi,att=0. ***Problem 10. Infinite deep square well A box for which we define positions in the box to correspond to no potential energy while positions outside the box to have infinite potential energy. (Hint: use normalization and. The figure below shows two isosurfaces of the wave function. Suppose on a given measurement we nd energy E 1. (b) Calculate the expectation value for (p) and (p) as a function of n. At the height of this COVID-19 pandemic, the government of Kano State, in its infinite wisdom, decided to "repatriate" Nigerian citizens (almajirai) back to their "states of indigeneity". For brevity, we omit the commands setting the parameters L,N,x,and dx. A particle in an infinite square well has an initial wave. Determine the probability of finding the particle between x=0 and x=L/3. Infinite square well, particle in a finite well; barrier penetration, reflection 3. Next: Expectation Values and Variances Up: Fundamentals of Quantum Mechanics Previous: Schrödinger's Equation Normalization of the Wavefunction Now, a probability is a real number between 0 and 1. How to Verify the Uncertainty Principle for a Quantum Harmonic Oscillator. The infinite square well potential is given by: () ⎩ ⎨ ⎧ ∞ < > ≤ ≤ = x x a x a V x,,, 0 0 0. Expected value and probability. Quantum Mechanics Homework #6 1. 4 Calculate the expectation values of x and x 2 for a particle in the state n = 2 in a square-well potential. 2 Expectation The most basic parameter associated to a random variable is its expected value or mean. 7 Barriers and Tunneling CHAPTER 6 Quantum Mechanics II I think it is safe to say that no one understands quantum mechanics. For a discussion of conceptual aspects, see: Isham, Chris J (1995). Notion of deep and shallow level. Finite 1-D square well: For an electron in a potential well of finite depth we must solve the time-independent Schrödinger equation with appropriate boundary conditions to get the wave functions. This is a more involved process, though, so here you'll only be able to see the results rather than run through. This fast time scale must be put in comparison with the much longer one, t R of Fig. A deuteron is bound state of proton and neutron (mp ~ mn~m~939 MeV/c2). lation the statistical-mechanical expectation value (A) of a. Expectation values in the infinite square well. Question: A particle in an infinite square well potential has an initial wave function {eq}\psi (x,t=0)=Ax(L-x) {/eq}. If the above sum. 7 Consider two noninteracting particles of mass m in the harmonic oscillator potential well. (a) For the infinite square well potential, show that the expectation value of the momentum. The width of the well and the field direction and strength are adjustable. This is very easy, if you exploit the orthonomarlity of ψ1 andψ2. 1 The Schrödinger Wave Equation 6. Consider a quantum mechanical particle, described by the wavefunction$\psi (x)\$, in one dimension. The bottom of the in nite square well was at zero potential energy. A particle is in the ground state of an infinite square well potential given by, The probability to find the particle in the interval between and is (a) (b) (c) (d) Q46. 8 A particle in the infinite square well has the initial wave function. The square well itself constitutes region I, and the constant potential outside, region I1 (figure 2). 2 Scattering from a 1D Potential Well *. What is the expectation value of the Posted 2 years ago. The width of the well is adjustable. , situations where a spreading wave packet reforms with close to its initial shape and width, we also examine in detail the approach to the collapsed phase where the position-space probability density is almost. In an infinite system we have Π = 1 above p c and Π = 0 below p c. It is also called “a particle in a rigid box”, and even though it’s relatively easy, there are many important applications of the solution. You can do it by straight forward substitution of the appropriate y and A in calculating = or you can use some ingenuity to get the. uncertain; b. Expected value is a measure of central tendency; a value for which the results will tend to. Check that the uncertainty principle is satisfied. Physics 48 February 1, 2008 Happy Ground Hog Day (a day early)! • A few remarks about solutions to the SE. 7 Barriers and Tunneling CHAPTER 6 Quantum Mechanics II I think it is safe to say that no one understands quantum mechanics. (b) Determine the probability of finding the particle near L/2, by calculating the probability that the particle lies in the range 0. A comparison has been performed along the lines of Chen (1983). This potential is represented by the dark lines in Fig. 2: What photon energy is required to excite the trapped. The collector current versus stopping voltage has minima for each energy value of the Hg atom. Infinite Well: this is the "particle in a box"; the particle is confined between two walls of infinite potential. Itisnatural toidentify t R withthe timescalethat controls the eventual escape from the quasi-steady state, hence the approach to thermal equilibrium. Only a finite number of the states are shown; increase the resolution to see more states. expectation value of the position operator squared. Infinite Square Well Potential in 2-D (in Hindi) 10:13 mins. ψ = √(2/a)sin(nπx/a), E = n²π²(hbar)²/(2ma²) continuous superposition. Choose all of the following statements that are correct at a given time t>0. an infinite potential well), or a one-dimensional box of base length L.
7c83sm10ewoqgi, y4vixp3egvtv, q71qk2w8cylg9q9, 7wtof5wflla, jmgf6z8u2qnu2e4, z9p2an9omn, zk3aehkxex6o0s0, 57ldn05fmqo, o6u84jjzu9, co1akdyiosxd36x, 8yynka79q0ay, 57xdy4pr96, 8o2fyl0iaif, gcbirjxeqkqcn, oyiug85ota7, 3sgc6q9em8m, 5tf5p95gcq5ye, 79nlaa0mn883, tujwrqwouopf7, 7417li8azt1ozo5, 75jdo9jvl5wepgx, s8x9u4x1w7, 1s3z8spe12rpbs9, 5ibdu93k75, joluux5r9cj, u6xjh7bb8m15eda, lpbhedrxvcn5r, frocv2x0ukfjdlu, kbdxo2rdntg, uqd7ygqnlf, fr0wxwefsnx | 13,509 | 55,842 | {"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": 2, "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-2020-24 | latest | en | 0.848269 |
https://metric-calculator.com/convert-inch-per-second-to-meters-per-second.htm | 1,726,540,106,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00525.warc.gz | 354,877,645 | 7,075 | # Inches Per Second to Meters Per Second Converter
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Convert Meters Per Second to Inches Per Second (m/s to in/s) ▶
## Conversion Table
inches per second to meters per second in/s m/s 10 in/s 0.254 m/s 20 in/s 0.508 m/s 30 in/s 0.762 m/s 40 in/s 1.016 m/s 50 in/s 1.27 m/s 60 in/s 1.524 m/s 70 in/s 1.778 m/s 80 in/s 2.032 m/s 90 in/s 2.286 m/s 100 in/s 2.54 m/s 110 in/s 2.794 m/s 120 in/s 3.048 m/s 130 in/s 3.302 m/s 140 in/s 3.556 m/s 150 in/s 3.81 m/s 160 in/s 4.064 m/s 170 in/s 4.318 m/s 180 in/s 4.572 m/s 190 in/s 4.826 m/s 200 in/s 5.08 m/s
## How to convert
1 inch per second (in/s) = 0.0254 meter per second (m/s). Inch Per Second (in/s) is a unit of Speed used in Standard system. Meter Per Second (m/s) is a unit of Speed used in Metric system.
## Inches per second: A unit of speed
Inches per second ( in/s) is a unit of speed or velocity in the US customary and imperial systems. It measures how fast an object is moving by calculating the distance traveled in inches divided by the time taken in seconds. For example, if a worm travels 2 inches in 1 second, its speed is 2 in/s.
## How to convert inches per second
Inches per second can be converted to other units of speed or velocity by using simple conversion factors. Here are some common units and their conversion factors:
• Meters per second (m/s): To convert from in/s to m/s, multiply by 0.0254. To convert from m/s to in/s, divide by 0.0254. For example, 1 in/s is equal to 0.0254 m/s, and 10 m/s is equal to 393.701 in/s.
• Kilometers per hour (km/h): To convert from in/s to km/h, multiply by 0.09144. To convert from km/h to in/s, divide by 0.09144. For example, 1 in/s is equal to 0.09144 km/h, and 50 km/h is equal to 547.867 in/s.
• Miles per hour (mph): To convert from in/s to mph, multiply by 0.0568182. To convert from mph to in/s, divide by 0.0568182. For example, 1 in/s is equal to 0.0568182 mph, and 30 mph is equal to 528 in/s.
• Knots (kn): To convert from in/s to kn, multiply by 0.0493737. To convert from kn to in/s, divide by 0.0493737. For example, 1 in/s is equal to 0.0493737 kn, and 15 kn is equal to 303.795 in/s.
• Feet per second (ft/s): To convert from in/s to ft/s, divide by 12. To convert from ft/s to in/s, multiply by 12. For example, 1 in/s is equal to 0.0833333 ft/s, and 10 ft/s is equal to 120 in/s .
• Miles per second (mi/s): To convert from in/s to mi/s , multiply by 1.5783x10 -5 . To convert from mi/s to in/s, divide by 1.5783x10 -5 . For example,1 in/s is equal to 1.5783x10 -5 mi/s, and 1 mi/s is equal to 63360 in/s .
## Where inches per second are used
Inches per second are mainly used in engineering and manufacturing to measure the speed or velocity of machines and processes.
For example, the cutting speed of a lathe or a milling machine is often given in inches per minute, which is equivalent to inches per second divided by 60.
The rotational speed of a hard disk drive is often given in revolutions per minute, which can be converted to inches per second by multiplying by the circumference of the disk.
The flow rate of a fluid through a pipe or a nozzle is often given in gallons per minute, which can be converted to inches per second by multiplying by the cross-sectional area of the pipe or the nozzle.
## Definition of inches per second
According to the US customary and imperial systems definition, one inch per second is the speed of a body that covers a distance of one inch in a time of one second.
Mathematically, it can be expressed as:
where v is the speed or velocity in inches per second, s is the distance traveled in inches, and t is the time taken in seconds.
## History of inches per second
The concept of speed or velocity has been studied since ancient times by philosophers and scientists such as Aristotle, Galileo, Newton, etc.
The inch was originally derived from the Roman uncia which was one twelfth of a Roman foot or about 0.97 inches .
The second was originally defined in terms of the Earth’s rotation as one eighty-six thousand four hundredth of a mean solar day.
The combination of these two units resulted in the inch per second as a unit of speed or velocity.
The inch per second was officially adopted as part of the US customary and imperial systems in the late 18th century.
## Example conversions of inches per second to other units
Here are some examples of converting inches per second to other units of speed or velocity:
1 in/s = 0.0254 m/s = 0.09144 km/h = 0.0568182 mph = 0.0493737 kn = 0.0833333 ft/s = 1.5783x10 -5 mi/s
2 in/s = 0.0508 m/s = 0.18288 km/h = 0.113636 mph = 0.0987474 kn = 0.166667 ft/s = 3.1566x10 -5 mi/s
5 in/s = 0.127 m/s = 0.4572 km/h = 0.284091 mph = 0.246869 kn = 0.416667 ft/s = 7.8915x10 -5 mi/s
10 in/s = 0.254 m/s = 0.9144 km/h = 0.568182 mph = 0.493737 kn = 0.833333 ft/s = 1.5783x10 -4 mi/s
20 in/s = 0.508 m/s = 1.8288 km/h = 1.13636 mph = 0.987475 kn = 1.66667 ft/s = 3.1566x10 -4 mi/s
50 in/s = 1.27 m/s = 4.572 km/h = 2.84091 mph = 2.46869 kn = 4.16667 ft/s = 7.8915x10 -4 mi/s
100 in/s = 2.54 m/s = 9.144 km/h = 5.68182 mph = 4.93737 kn = 8.33333 ft/s = 1.5783x10 -3 mi/s
Inch per second also can be marked as ips.
## Meters per second: A unit of speed
Meters per second (m/s) is a unit of speed or velocity in the International System of Units (SI). It measures how fast an object is moving by calculating the distance traveled in meters divided by the time taken in seconds. For example, if a car travels 100 meters in 5 seconds, its speed is 20 m/s.
## How to convert meters per second
Meters per second can be converted to other units of speed or velocity by using simple conversion factors. Here are some common units and their conversion factors:
• Kilometers per hour (km/h): To convert from m/s to km/h, multiply by 3.6. To convert from km/h to m/s, divide by 3.6. For example, 20 m/s is equal to 72 km/h, and 50 km/h is equal to 13.89 m/s.
• Miles per hour (mph): To convert from m/s to mph, multiply by 2.2369. To convert from mph to m/s, divide by 2.2369. For example, 20 m/s is equal to 44.74 mph, and 30 mph is equal to 13.41 m/s.
• Knots (kn): To convert from m/s to kn, multiply by 1.9438. To convert from kn to m/s, divide by 1.9438. For example, 20 m/s is equal to 38.88 kn, and 15 kn is equal to 7.72 m/s.
• Feet per second (ft/s): To convert from m/s to ft/s, multiply by 3.2808. To convert from ft/s to m/s, divide by 3.2808. For example, 20 m/s is equal to 65.62 ft/s, and 40 ft/s is equal to 12.19 m/s.
## Where meters per second are used
Meters per second are widely used in science and engineering to measure the speed or velocity of various phenomena, such as sound waves, light waves, wind speed, fluid flow, projectile motion, etc.
Meters per second are also used in some countries as the official unit of speed or velocity for road traffic signs and regulations. For example, in Canada, the maximum speed limit on highways is usually 100 km/h, which is equivalent to 27.78 m/s.
Meters per second are also used in sports and athletics to measure the performance of athletes, such as sprinters, swimmers, cyclists, etc.
## Definition of meters per second
According to the SI definition, one meter per second is the speed of a body that covers a distance of one meter in a time of one second.
Mathematically, it can be expressed as:
where v is the speed or velocity in meters per second, s is the distance traveled in meters, and t is the time taken in seconds.
## History of meters per second
The concept of speed or velocity has been studied since ancient times by philosophers and scientists such as Aristotle, Galileo, Newton, etc.
The meter was originally defined in 1793 by the French Academy of Sciences as one ten-millionth of the distance from the equator to the North Pole along a meridian through Paris.
The second was originally defined in terms of the Earth’s rotation as one eighty-six thousand four hundredth of a mean solar day.
The combination of these two units resulted in the meter per second as a unit of speed or velocity.
The meter per second was officially adopted as part of the SI system in 1960.
## Example conversions of meters per second to other units
Here are some examples of converting meters per second to other units of speed or velocity:
• 10 m/s = 36 km/h = 22.37 mph = 19.44 kn = 32.81 ft/s
• 5 m/s = 18 km/h = 11.18 mph = 9.72 kn = 16.4 ft/s
• 15 m/s = 54 km/h = 33.55 mph = 29.17 kn = 49.21 ft/s
• 25 m/s = 90 km/h = 55.92 mph = 48.61 kn = 82.02 ft/s
• 50 m/s = 180 km/h = 111.85 mph = 97.22 kn = 164.04 ft/s
• 100 m/s = 360 km/h = 223.69 mph = 194.39 kn = 328.08 ft/s
• 200 m/s = 720 km/h = 447.39 mph = 388.79 kn = 656.17 ft/s
Meters per second also can be marked as m/s and metres per second(alternative British English spelling in UK).
Español Russian Français | 2,660 | 8,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.736989 |
https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_11&diff=prev&oldid=30935 | 1,623,988,074,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00462.warc.gz | 113,685,797 | 11,777 | # Difference between revisions of "2009 AIME I Problems/Problem 11"
## Problem
Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$. Find the number of such distinct triangles whose area is a positive integer.
## Solution
Let the two points be $P$ and $Q$; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).$ (Error compiling LaTeX. ! Package amsmath Error: Old form `\matrix' should be \begin{matrix}.)
Since the triangle has half the area of the parallelogram, we just need the determinant to be even.
The determinant is
$$(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))$$
Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are then $25$ even and $25$ odd numbers and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.
2009 AIME I (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions | 522 | 1,507 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "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} | 4.65625 | 5 | CC-MAIN-2021-25 | latest | en | 0.692469 |
https://priorprobability.com/2016/08/31/the-law-of-the-trolley-problem/ | 1,500,712,783,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423927.54/warc/CC-MAIN-20170722082709-20170722102709-00542.warc.gz | 689,984,465 | 40,086 | ## The Law of the Trolley Problem
We recently rediscovered and reread Lon Fuller’s classic “Case of the Speluncean Explorers” (via Peter Suber), and in the process of writing up our own response to Fuller, we noticed a possible parallel between this hypothetical case and the famous “Trolley Problem” in the field of moral philosophy. (There are several variants of the problem (see below); we will refer to the standard version in this post.) By way of background, we summarized the facts of Fuller’s fictional case in a previous post; alternatively, here is a simplified account of the cave case (via Wikipedia):
The case involves five explorers who are caved in following a landslide. They learn via intermittent radio contact that, without food, they are likely to starve to death before they can be rescued. They decide that someone should be killed and eaten so that the others may survive. They decide who should be killed by throwing a pair of dice. After the four [remaining] survivors are rescued, they are charged and found guilty of the murder of the fifth explorer.
By comparison, the trolley problem is a moral dilemma that involves a similar numerical calculus, but instead of five explorers trapped in a cave, there are six hapless workmen (five on one track and one on another) and an innocent bystander standing next to a switch or lever where the track divides into two. We restated the standard version of this moral dilemma in a previous paper; here too is the standard set up of the trolley problem (again, via Wikipedia):
There is a runaway trolley barreling down the railway tracks. Ahead, on the tracks, there are five people tied up and unable to move. The trolley is headed straight for them. You are standing some distance off in the train yard, next to a lever. If you pull this lever, the trolley will switch to a different set of tracks. However, you notice there is one person on the side track. You have two options: (1) Do nothing, and the trolley kills the five people on the main track. (2) Pull the lever, diverting the trolley onto the side track where it will kill one person. Which is the correct choice?
The question we have is this: are the factual differences in both the cave and trolley scenarios morally or legally relevant? In both cases there is an imminent threat or mortal danger to five human lives, and both cases involve sacrifices, but the relevant decision makers are different in each case. In one case, the persons whose own lives are in danger are able to deliberate and ponder their collective fate. Moreover, they initially agree to sacrifice one member of their party through a random mechanism. In the other case, by contrast, someone external to the danger (the person standing next to the lever) has no time to deliberate. He must make a snap decision and that decision (including the decision to do nothing) will result in the loss of either one or five lives.
In short, is the trolley problem relevant to our legal or moral analysis of the case of the explorers? By way of illustration, let’s transpose the trolley problem to Fuller’s fictional land of Newgarth. (Recall the language of the relevant murder statute in Newgarth: “Whoever shall willfully take the life of another shall be punished by death.”) Putting the ethics of the trolley problem aside (assuming it is even possible to separate morality from law), would the person at the switch in the trolley problem be guilty of murder under this statute if he were to divert the trolley onto the side track? Further, does our answer to this question apply or control the result in the case of the explorers?
Trolley problems …
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### 6 Responses to The Law of the Trolley Problem
1. Craig says:
Another possible solution is that you are the fat man: you switch the trolley to the nearest track and jump in front of it yourself, sparing all others at the expense of your own life.
• That’s true (similarly, in the case of the explorers trapped in the cave, one of them could volunteer to be sacrificed), but is there a moral obligation for the fat man to do so?
• Craig says:
Not a moral “obligation” per se — but perhaps one “could not live with oneself” if one pushed someone else into the path, or switched the track to kill a different person — in that case, one might choose self-sacrifice rather than other actions or positive inactions. Situations like this always bring the movie “Sophie’s Choice” to mind.
• I never saw “Sophie’s Choice,” so I will add this movie to my must-see list. | 995 | 4,601 | {"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-2017-30 | longest | en | 0.941819 |
https://www.extremeoptimization.com/QuickStart/IronPython/MultipleRegression.aspx | 1,545,013,424,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00252.warc.gz | 853,582,300 | 7,159 | Data Analysis Mathematics Linear Algebra Statistics
New Version 7.0!
QuickStart Samples
# Multiple Linear Regression QuickStart Sample (IronPython)
Illustrates how to use the LinearRegressionModel class to perform a multiple linear regression in IronPython.
import numerics
from System import Array, Char
from Extreme.Mathematics import *
from Extreme.Statistics import *
# Illustrates building multiple linear regression models using
# the LinearRegressionModel class in the
# Extreme.Statistics namespace of the Extreme
# Optimization Numerical Libraries for .NET.
# Multiple linear regression can be performed using
# the LinearRegressionModel class.
#
# This QuickStart sample uses old economic data about 50 countries
# from Belsley, Kuh and Welsch. The fields are as follows:
# DispInc: Per capita disposable income.
# Growth: Percent rate of change of DispInc.
# Pop15: Percentage of population under 15.
# Pop75: Percentage of population over 75.
# Savings: Aggregate savings divided by disposable income.
#
# We want to investigate the effect of the first four variables
# on the savings ratio.
# First, read the data from a file into an ADO.NET DataTable.
# For the sake of clarity, we put this code in its own method.
#/ Reads the data from a text file into a <see cref="DataTable"/>.
import clr
from System.Data import *
from System.IO import *
data = DataTable("savings")
whitespace = Array[Char]([ ' ', '\t' ])
pos = 0
while True:
while Char.IsWhiteSpace(line[pos]):
pos = pos + 1
pos2 = line.IndexOfAny(whitespace, pos)
if pos2 < 0:
break
else:
pos = pos2
if pos < 0:
break
# Assumes all columns except the first are numerical.
rowData = Array.CreateInstance(object, data.Columns.Count)
while line != None and line.Length > 0:
column = 0
pos = 0
while True:
while Char.IsWhiteSpace(line[pos]):
pos = pos + 1
pos2 = line.IndexOfAny(whitespace, pos)
if pos2 < 0:
field = line.Substring(pos)
else:
field = line.Substring(pos, pos2 - pos)
if column == 0:
rowData[column] = field
else:
rowData[column] = float.Parse(field)
column = column + 1
pos = pos2
if pos < 0 or column >= data.Columns.Count:
break
return data
# Next, create a VariableCollection from the data table:
data = VariableCollection(dataTable)
# Now create the regression model. Parameters are the name
# of the dependent variable, a string array containing
# the names of the independent variables, and the VariableCollection
# containing all variables.
model = LinearRegressionModel(data, "Savings", \
Array[str]([ "Pop15", "Pop75", "DispInc", "Growth"]))
# We can set model options now, such as whether to include a constant:
model.NoIntercept = False
# The Compute method performs the actual regression analysis.
model.Compute()
# The Parameters collection contains information about the regression
# parameters.
print "Variable Value Std.Error t-stat p-Value"
for parameter in model.Parameters:
# Parameter objects have the following properties:
print "{0:20}{1:10.5f}{2:10.5f}{3:8.2f} {4:7.4f}".format( # Name, usually the name of the variable:
parameter.Name, # Estimated value of the parameter:
parameter.Value, # Standard error:
parameter.StandardError, # The value of the t statistic for the hypothesis that the parameter
# is zero.
parameter.Statistic, # Probability corresponding to the t statistic.
parameter.PValue)
print
# In addition to these properties, Parameter objects have a GetConfidenceInterval
# method that returns a confidence interval at a specified confidence level.
# Notice that individual parameters can be accessed using their numeric index.
# Parameter 0 is the intercept, if it was included.
confidenceInterval = model.Parameters[0].GetConfidenceInterval(0.95)
print "95% confidence interval for constant:{0:.4f} - {1:.4f}".format(confidenceInterval.LowerBound, confidenceInterval.UpperBound)
# Parameters can also be accessed by name:
confidenceInterval = model.Parameters["DispInc"].GetConfidenceInterval(0.95)
print "95% confidence interval for Growth: {0:.4f} - {1:.4f}".format(confidenceInterval.LowerBound, confidenceInterval.UpperBound)
print
# There is also a wealth of information about the analysis available
# through various properties of the LinearRegressionModel object:
print "Residual standard error: {0:.3f}".format(model.StandardError)
print "R-Squared: {0:.4f}".format(model.RSquared) | 1,036 | 4,381 | {"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} | 3.375 | 3 | CC-MAIN-2018-51 | longest | en | 0.654289 |
https://us.metamath.org/mpeuni/difundir.html | 1,686,166,098,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654012.67/warc/CC-MAIN-20230607175304-20230607205304-00785.warc.gz | 635,905,247 | 4,272 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > difundir Structured version Visualization version GIF version
Theorem difundir 4107
Description: Distributive law for class difference. (Contributed by NM, 17-Aug-2004.)
Assertion
Ref Expression
difundir ((𝐴𝐵) ∖ 𝐶) = ((𝐴𝐶) ∪ (𝐵𝐶))
Proof of Theorem difundir
StepHypRef Expression
1 indir 4102 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ (V ∖ 𝐶)) ∪ (𝐵 ∩ (V ∖ 𝐶)))
2 invdif 4095 . 2 ((𝐴𝐵) ∩ (V ∖ 𝐶)) = ((𝐴𝐵) ∖ 𝐶)
3 invdif 4095 . . 3 (𝐴 ∩ (V ∖ 𝐶)) = (𝐴𝐶)
4 invdif 4095 . . 3 (𝐵 ∩ (V ∖ 𝐶)) = (𝐵𝐶)
53, 4uneq12i 3988 . 2 ((𝐴 ∩ (V ∖ 𝐶)) ∪ (𝐵 ∩ (V ∖ 𝐶))) = ((𝐴𝐶) ∪ (𝐵𝐶))
61, 2, 53eqtr3i 2810 1 ((𝐴𝐵) ∖ 𝐶) = ((𝐴𝐶) ∪ (𝐵𝐶))
Colors of variables: wff setvar class Syntax hints: = wceq 1601 Vcvv 3398 ∖ cdif 3789 ∪ cun 3790 ∩ cin 3791 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1839 ax-4 1853 ax-5 1953 ax-6 2021 ax-7 2055 ax-9 2116 ax-10 2135 ax-11 2150 ax-12 2163 ax-ext 2754 This theorem depends on definitions: df-bi 199 df-an 387 df-or 837 df-tru 1605 df-ex 1824 df-nf 1828 df-sb 2012 df-clab 2764 df-cleq 2770 df-clel 2774 df-nfc 2921 df-ral 3095 df-rab 3099 df-v 3400 df-dif 3795 df-un 3797 df-in 3799 This theorem is referenced by: dfsymdif3 4119 difun2 4272 diftpsn3 4566 setsfun0 16295 strleun 16368 mreexmrid 16693 mreexexlem2d 16695 mvdco 18252 dprd2da 18832 dmdprdsplit2lem 18835 ablfac1eulem 18862 lbsextlem4 19562 opsrtoslem2 19885 nulmbl2 23744 uniioombllem3 23793 ex-dif 27859 indifundif 29922 imadifxp 29981 ballotlemfp1 31156 ballotlemgun 31189 onint1 33035 lindsadd 34033 lindsenlbs 34035 poimirlem2 34042 poimirlem6 34046 poimirlem7 34047 poimirlem8 34048 poimirlem22 34062 dvmptfprodlem 41097 fourierdlem102 41362 fourierdlem114 41374 caragenuncllem 41663 carageniuncllem1 41672
Copyright terms: Public domain W3C validator | 1,074 | 1,967 | {"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-2023-23 | longest | en | 0.099192 |
https://www.cpc.ncep.noaa.gov/products/predictions/short_range/tools/ppgc.html | 1,701,282,888,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00233.warc.gz | 809,255,322 | 7,669 | Points, Pixels, Grids, and Cells
Terms of Reference
Introduction
Satellite remote sensing has made possible the collection of data over large areas of the Earth. These data are often stored in "grids." Grids are an efficient means of storing data because the location of a value within the grid is implicit--it is not explicitly stored in the grid. The location is also constant, which makes it easy to compare data from different sensors or different time periods.
The purpose of this document is to give a brief introduction to mapping, gridding, and the associated terminology used in the mapx library at NSIDC. Along with comments in the source code, it also describes the format of the map and grid parameter files.
Maps
A map projection is a procedure or mathematical formula to transform a curved surface onto a plane. The curved surface is usually the surface of the Earth and the plane is what we call a "map." The derivation of a map transformation requires a mathematical representation of the surface of the Earth. The surface of constant geopotential referred to as mean sea level or the geoid is quite complicated and its mathematical form is correspondingly complex. When defining map projections then, it is common to approximate the surface of the earth with a simpler surface such as an ellipsoid or sphere.
The ellipsoid is flattened at the poles, relative to the sphere, by about 1 part in 300. For maps of large areas (continents or bigger) using the sphere introduces no significant error. When mapping smaller areas it is not uncommon to choose an ellipsoid specifically for each area; each area is mapped on an ellipsoid with a different radius and eccentricity.
Coordinates for points on the surface of the Earth are given in latitude and longitude. Geodetic latitude is the angle between a vertical ray at the point and the plane of the equator. Geocentric latitude is the angle between a ray from the center of the Earth to the point and the plane of the equator. On a sphere, the geodetic latitude is equivalent to the geocentric latitude. It's more common to use geodetic latitude. There is only one definition of longitude--the angle between a ray from the center of the Earth to the point and the plane of the Greenwich Meridian (an arbitrary reference point). Coordinates in the map plane (x,y) are defined in the usual manner with x horizontal and increasing to the right and y vertical and increasing upward.
Projections
Map transformations are referred to as projections because they are commonly visualized as geometric constructs. That is, we think of the map as a piece of paper wrapped around a globe with lines emanating from points on the globe and "projected" onto the paper. It's important to note that not all map projections can be constructed in this manner.
Map projections are specified as a set of equations giving x and y in terms of latitude and longitude. The manner in which we wrap the paper around the globe determines the form of the equations. Cylindrical projections wrap the paper around in a tube, conical projections use a cone, and azimuthal projections use a flat surface. Several parameters are used in these equations to specify exactly where the surface intersects the globe, the scale or size of the globe, and the translation and rotation of the map on the plane surface.
There is no one best map projection. Each projection has different properties and different "best" uses. Two of the most important characteristics of maps are whether they are conformal or equal-area. No map projection is both, and some are neither. If a map is conformal then angles within a small area are reproduced accurately. This means that shapes are preserved. A small circle on the globe will look like a small circle on the map. In formal terms, at any point on the map the scale h along a meridian of longitude is equal to the scale k along a parallel of latitude. If a map is equal-area then a small circle placed anywhere on the map will always cover the same amount of area on the globe and the product of h and k at any point is one.
For maps that are not equal-area, hk - 1 gives a measure of the areal distortion. For maps that are not conformal, k/h gives the aspect ratio, which is a measure of shape distortion. For example, on the Polar Stereographic map true at 70N (a conformal map) the areal distortion varies from -6% at the pole to +29% at 45N and goes up to +276% at the equator while, by definition, the aspect ratio remains 1:1 everywhere. The Azimuthal Equal-Area map over the pole has, of course, no areal distortion, while the aspect ratio varies from 1:1 at the pole to 1.17:1 at 45N and goes up to 2:1 at the equator.
Tissot's indicatrix combines the two previous measures. Tissot's indicatrix is a small ellipse which shows both the scale distortion and the maximum angular deformation at a particular point on the map. See Maling or Snyder for a complete discussion.
Grids
A grid is a rectangular array of points. Grids record regularly spaced samples over an area. When sampling over the surface of the earth a grid is determined by a map projection, a sampling interval, an origin, and the number of rows and columns. So called "lat/lon" or "equal angle" grids are equivalent to grids sampled on a Cylindrical Equidistant map projection.
A grid coordinate system is defined in the map plane with axes parallel to the rows and columns of the grid and units equal to the sampling interval. The grid sample locations are at the whole integer grid coordinate points. To conform to mathematical array conventions, the grid coordinates (r,s) start at (0,0) in the upper left corner with r increasing to the right, and s increasing downward (this also conforms to digital image processing conventions). A grid cell (i,j) is defined as the area between grid coordinates i-.5 and i+.5, and j-.5 and j+.5. To conform to rounding conventions, the lower bound is included in the cell while the upper bound is not (round up at .5). Note that the r coordinate corresponds to the grid column number j, and the s coordinate corresponds to the grid row number i.
Think of the grid as a map drawn on graph paper. To find a grid cell location, count graph lines over from the left and down from the top, then read the latitude and longitude off the map.
Binning
The process of sampling data on a grid is sometimes referred to as binning, especially when the sampling method is to average all data that falls into the grid cell. Other methods used to sample data are: take the closest value to the grid point (nearest neighbor) or interpolate between surrounding grid points. Also, binning doesn't need to be the average of multiple data points; it could be the minimum, the maximum, the median, or the latest value in the grid cell. The characteristics of the data should determine the sampling method.
Pictures
Pixels are analogous to grid cells. To display gridded data as a picture, define a transformation from grid coordinates to pixel coordinates and then sample the gridded data at the whole integer pixel coordinate points. The most common technique is to map grid cells onto pixels one for one. More sophisticated techniques, often referred to as "resampling", allow for scrolling, zooming, and rotating.
Parameters
Many programs we use at NSIDC depend on the "mapx" module to handle map projections. These include: GISMO, PSQ, regrid, gridomatic, and the EASE-Grid processing software. Most of these tools provide access to predefined maps, but, it is relatively easy to define your own map with a map projection parameters (.mpp) file.
Map Projection Parameters (.mpp) File Format
Each parameter is described by a "keyword: value" pair on a single line. Parameters can be followed by comments, which begin with a semi-colon. Parameters can appear in any order. The default value is used when the corresponding keyword is not found. Unrecognized keywords are silently ignored. This means that if you misspell a keyword you'll get the default value without warning. The following parameters define the map projection.
Keyword Format Default Description
Map Projection string required field projection name (see list below)
Map Reference Latitude decimal degrees required field reference latitude for map projection
Map Reference Longitude decimal degrees required field reference longitude for map projection
Map Second Reference Latitude decimal degrees none used by some projections
Map Rotation decimal degrees 0.0 rotation counter-clockwise
Map Scale float 1.0 map scale factor (radius units per map unit)
Map ISin Justify int 2=even justify flag for ISin map
Map Origin Latitude decimal degrees refrence latitude translated map origin latitude
Map Origin Longitude decimal degrees reference longitude translated map origin longitude
Map False Easting float 0.0 offset map x origin in map units
Map False Northing float 0.0 offset map y origin in map units
Map Eccentricity float Clark 1866 eccentricity of ellipsoid
Map Equatorial Radius float Clark 1866 ellipsoid or equivalent authallic sphere radius of ellipsoid or sphere
The following parameters are used only by programs that actually draw maps.
Keyword Format Default Description
Map Southern Bound decimal degrees 90.00S bottom of map, starting point for latitude graticule
Map Northern Bound decimal degrees 90.00N top of map
Map Western Bound decimal degrees 180.00W left side of map, starting point for longitude graticule
Map Eastern Bound decimal degrees 180.00E right side of map
Map Graticule Latitude Interval decimal degrees 30. graticule spacing
Map Graticule Longitude Interval decimal degrees 30. graticule spacing
Map Graticule Label Latitude decimal degrees 0.00N where to label meridians
Map Graticule Label Longitude decimal degrees 0.00E where to label parallels
Map CIL Detail Level integer 1 level of detail for coastlines, islands, and lakes
Map BDY Detail Level integer 0 level of detail for political boundaries
Map RIV Detail Level integer 0 level of detail for rivers
note: For Integerized Sinusoidal, set scale to size of each row.
Map Projection Names
• Albers Conic Equal-Area
• Azimuthal Equal-Area
• Azimuthal Equal-Area (ellipsoid)
• Cylindrical Equal-Area
• Cylindrical Equal-Area (ellipsoid)
• Cylindrical Equidistant
• Integerized Sinusoidal
• Interupted Homolosine Equal-Area
• Lambert Conic Conformal (ellipsoid)
• Mercator
• Mollweide
• Orthographic
• Polar Stereographic
• Sinusoidal
Keywords must be entered verbatim. Capitalization and white space matter. Unrecognized (or mispelled) keywords are silently ignored. Map names are not case sensitive, white space and dashes are ignored, and most word permutations are acceptable. For example, "Equal-Area Cylindrical" is equivalent to "CYLINDRICAL EQUALAREA".
The reference latitude and longitude specify the original location and orientation of the map projection. The origin of the rectangular map coordinate system can be rotated and translated with the rotation and origin keywords.
For the Cylindrical Equal-Area and Polar Stereographic projections, the second reference latitude specifies the latitude of "true" scale. Conceptually, this is where the projection plane intersects the surface. For the Albers Conic Equal-Area and Lambert Conic Conformal projections, the conic plane is secant to the surface between the reference latitude and the second reference latitude.
The equatorial radius and eccentricity specify the surface to be mapped (usually the Earth's surface). The scale and the equatorial radius determine the map units. For example, if the radius is in kilometers (as is the case with the default) and the scale is 1.0, then the map units will be kilometers. PostScript units are inches, so for PostScript output you would specify the scale in kilometers per inch. For a grid you can define the scale in terms of kilometers per grid cell. The gridding module "grids" also has parameters that specify grid cells per map unit. This allows the same .mpp file to be used for multiple nested grids.
The remaining parameters are used by programs that actually draw maps (e.g. psmap, mapenum). The southern, northern, western, and eastern bounds of the map are the starting point for the graticule and are used to speed up the search in the map outline data base. They cannot be counted on to clip the map accurately. All longitudes should be in the range -180 to +360. West to east should not span more than 360 degrees. West specifies the left side of the map and east the right, not necessarily the minimum and maximum longitudes.
The latitude and longitude graticule intervals specify the spacing between graticule lines. The graticule is the overlay of latitude and longitude lines on the map. By default, a parallel is drawn every 30 degrees starting at the bottom of the map (90.00S) and a meridian is drawn every 30 degrees starting at the left side of the map (180.00W). Label latitude and longitude specify the parallel and meridian along which to draw the graticule line labels.
The last three numbers are used by the database search routine to specify the level of detail for map lines. The level refers to the number and size of features that will be selected, as opposed to resolution of each feature. The higher the number, the more detail will be included. For example, level 1 will include only the largest features, such as whole continents, while level 2 would also include some smaller features like large lakes. In either case the continental outline will appear with the same amount of detail (the same jaggedness).
Grid Parameter Definition (.gpd) File Format
Grids are defined in grid parameter definition (.gpd) files.
Keyword Format Default Description
Grid Width integer required field number of columns in grid
Grid Height integer required field number of rows in grid
Grid Map Origin Column float 0.0 column coordinate of map origin
Grid Map Origin Row float 0.0 row coordinate of map origin
Grid Cells per Map Unit float 1.0 sets both column and row scale
Grid Columns per Map Unit float 1.0 sets column scale if not already set
Grid Rows per Map Unit float 1.0 sets row scale if not already set
Grid Map Units per Cell float 1.0 alternative method to set both column and row scale
Grid Map Units per Column float 1.0 sets column scale if not already set
Grid Map Units per Row float 1.0 sets row scale if not already set
Grid MPP File string none name of .mpp file, map projection parameters can also be specified in the same file with the grid definition parameters
Limitations
The mapx library is very flexible but it does impose certain restrictions on the kinds of grids that can be defined. All grids must be rectangular arrays. Rectangular arrays are easy to use with existing scientific visualization tools, like IDL, but they may be inefficient in terms of storage space for non-rectangular data sets. If some flag value (for example zero) is stored in the unused cells, then this wasted space can be compacted by any good data compression scheme. This same invalid data flag can be used within the grid as well, leading to further storage savings. Another restriction is that each grid must be based on a map projection defined in mapx. Many of the most useful projections have already been implemented and fortunately it is easy, given a mathematical formulation of the forward and inverse transformations, to add a new projection to mapx (see source code for instructions). The mathematical formulations of most common map projections can be found in the references below.
References
American Society of Photogrammetry. 1983. Manual of Remote Sensing. Robert N. Colwell, editor. Second edition. Falls Church, VA.
Gonzalez, Rafael C. and Paul Wintz. 1987.Digital Image Processing.Second edition. Reading, MA: Addison-Wesley.
Maling, D. H. 1992. Coordinate Systems and Map Projections. Second edition. Elmsford, N.Y.: Pergamon Press.
Snyder, John P. 1987.Map Projections, A Working Manual. U. S. Geological Survey Professional Paper 1395. Department of the Interior. Washington, D. C. | 3,464 | 16,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-50 | latest | en | 0.909734 |
http://math.stackexchange.com/questions/18677/speeding-up-gauss-elimination | 1,469,592,404,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825365.1/warc/CC-MAIN-20160723071025-00324-ip-10-185-27-174.ec2.internal.warc.gz | 146,647,964 | 17,701 | # Speeding up Gauss Elimination
I was working on something on Matlab couple of months back and found a tweak to speed up Gauss elimination on Matlab by dividing the original matrix into 4 block matrices and then solving them in a block-wise fashion.
I am however wondering about couple of things.
1. What is the reason for the optimal "split parameter" to be $\approx \frac{2}{3}$? Is there a heuristic or quick argument for this?
2. Why is the number of additions independent of how we block partition the matrix?
-
Volker Strassen amazed the algorithmic world in 1969 by producing an $O(n^{\log_2 7})$ algorithm for matrix multiplication, based on a similar idea but applied recursively. The algorithm can also be adapted to matrix inversion and so to the solving of linear equations.
Since then, researchers have produced even better algorithms. The best known so far is by Coppersmith and Winograd (1987), who achieve $O(n^{2.376})$. Recent work suggests that matrix multiplication is actually possible in $O_\epsilon(n^{2+\epsilon})$ for any $\epsilon > 0$. Usually this is stated as $\omega = 2$, where $\omega$ is the infimum of all exponents $k$ such that matrix multiplication is possible in time $O(n^k)$.
Matrix multiplication is actually equivalent to matrix inversion, in the sense that they both have the same $\omega$. The proof can be found in the well-known CLRS.
All these algorithms - even Strassen's - are impractical (the constants hidden inside the big O are huge), and so never actually used. However, "blocking" is a wide-spread technique, since accessing memory "locally" is more efficient (due to the way caches work).
-
+1 for a very educational answer! – M.B. Jan 24 '11 at 4:22
Yes. Here is an article (lehigh.edu/~gi02/m242/08linstras.pdf) which Moron posted as an answer to another question. This has the proof why Matrix Multiplication cost is same as Matrix Inversion. – user17762 Jan 24 '11 at 4:54
There's some controversy whether Strassen is practical -- rjlipton.wordpress.com/2010/10/23/galactic-algorithms/… – Yaroslav Bulatov Jan 24 '11 at 5:33
Here are Strassen's and Coppersmith/Winograd's articles. – J. M. Apr 11 '11 at 2:08 | 541 | 2,178 | {"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} | 3.34375 | 3 | CC-MAIN-2016-30 | latest | en | 0.921892 |
https://www.jiskha.com/display.cgi?id=1388246613 | 1,501,108,959,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00061.warc.gz | 781,182,052 | 4,045 | # math
posted by .
I'm not understanding and writing the following proportion: 8 is to 64 as 2 is to x, can somebody explain how to solve this problem.
• math -
A survey indicated that 3 out of doctors used brand x aspirin. if 2000 doctors were surveyed, how many used brand X?
I need help understanding the problem of this math. Can somebody explain this to me.
• math -
Here's the first one.
8/64 = 2/x
Cross multiply.
8x = 128
Solve for x.
x = 128/8
x = 16
You haven't posted enough information to solve the second problem.
• math -
Solve for x 8 is to 64 as 2 is to x
• math -
A survey indicated that 3 out of 5 doctors use Brand X aspirin. If \$2,000 were surveyed, how many used Brand X? | 197 | 711 | {"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} | 3.75 | 4 | CC-MAIN-2017-30 | longest | en | 0.95812 |
https://repl.it/talk/share/ryanmuraliraj-Using-list-comprehension/2840/7321 | 1,586,139,384,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371612531.68/warc/CC-MAIN-20200406004220-20200406034720-00175.warc.gz | 645,635,503 | 30,566 | An Actual Calculator
It is mind blowing how no one has created a calculator where you just type in an expression and get the result. All of the calculators I've seen have you type in one number then press enter then type in +, -, / or * then press enter then type your second number. This calculator takes simple expressions and returns the result.
Examples:
Input:1+1 | Output:2
Input:1000-15 | Output:985
Input:10*10 | Output:100
Input:100/10 | Output:10
Input:2/3 | Output:0.666666
Input:0/0 | Output: Error
*No spaces in expressions
**Can do addition, subtraction, multiplication, and division. Planning to add more operations in the future.
***Has float/int switcher in division. If is not a whole number it is a float, if it is a whole number it is an int.
My Calculator:
https://repl.it/@RyanMuraliraj/Calculator
Some of the calculators I was talking about:
https://repl.it/ibuiltthis/p/my-calculator-1
https://repl.it/ibuiltthis/p/calculator
https://repl.it/@HongjiDai/Simple-Calculator
This calculator does it but uses external modules without actually coding it themself:
https://repl.it/ibuiltthis/p/mega-calculator
Shoutout to the creators of the calculators listed above for the idea and inspiration.
pls upvote
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proofofconcept (21)
@ryanmuraliraj: Using list comprehensions you can do it in one line:
[print("Result: "+str(eval(input("Enter expression: " )))) if x else print("Welcome to the single line calculator") for x in range(2)] | 392 | 1,501 | {"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-2020-16 | latest | en | 0.775105 |
https://www.enotes.com/homework-help/40x-45-3-7-2-7x-40-can-u-solve-this-problem-308989 | 1,485,248,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284376.37/warc/CC-MAIN-20170116095124-00032-ip-10-171-10-70.ec2.internal.warc.gz | 893,283,556 | 13,356 | # -40x+45=3[7-2(7x-40] Can you solve this problem?
txmedteach | High School Teacher | (Level 3) Associate Educator
Posted on
This problem is a standard "Solve for x" problem that we so often see in algebra. I'm going to rewrite it below to get everything in a good font:
`-40x+45 = 3(7-2(7x-40))`
To do this equation most easily, we're going to have to simplify the term on the right. To do this we'll need to use the distributive property (see link). Here's an example of that:
`a(b+c) = ab + ac`
Whenever we're multiplying numbers in parentheses by another number, we're "distributing" it to each term inside the parentheses.
In our equation, we're going to "distribute" the 3 outside the parentheses to the terms inside the parentheses:
`-40x+45 = 3(7-2(7x-40))`
`-40x+45 = 3*7-3*2(7x-40)`
`-40x+45=21-6(7x-40)`
We can do the same with the 6 on the right side seen here (don't forget to distribute the -1! This makes it a + on that third term!):
`-40x+45=21-6*7x+6*40`
Keep in mind, we're only multiplying the 6by the terms inside its adjacent parentheses. We aren't multiplying it by the 21, for this reason.
Let's simplify:
`-40x+45=21-42x+240`
We can combine "like" terms now, meaning we can combine numbers with other numbers on the same side and terms with x with other terms of x on the same side. In our case, the only "like" terms we'll be combining are 21 and -240:
`-40x+45=-42x+261`
Now, we are in a good position to isolate the x and solve the equation. First, we're going to subtract 45 from both sides:
`-40x+45-45=-42x+261-45`
`-40x=-42x+216`
Now, we can add 42x to both sides:
`-40x + 42x = -42x+216+42x`
`2x=216`
Almost there, we can now divide by 2:
`2x/2 = 216/2`
`x=108`
And there's your answer! Now, you can substitute 108 for x in the original equation and check to make sure both sides are equal, which they are. Hope that helps!
Sources: | 595 | 1,893 | {"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} | 4.75 | 5 | CC-MAIN-2017-04 | longest | en | 0.898816 |
https://ezmart4u.com/the-secret-of-odd-strike-price-option/ | 1,628,080,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154805.72/warc/CC-MAIN-20210804111738-20210804141738-00094.warc.gz | 243,130,545 | 22,436 | # The Secret of Odd Strike Price Option
In the options market, a strike price (or exercise price in the Australian options market) refers to a fixed price that options will get bought or sold.
Basically, when someone buys a call option, he holds a right to exercise that option to buy 100 shares of stock at the strike price of that call option. On the other side, the option seller who gets exercised will be obligated to deliver 100 shares at that price.
This is simply a right to do so. It does not mean if chose to exercise; the option owner will make profit.
Strike of options is generally formulated as the following:
– If the stock price is higher than \$100, the strike price is in the intervals of 10, i.e.: \$100, \$110, \$120…etc
– If the stock price ranges from \$25 to \$100, it is in the intervals of 5, i.e. \$25, \$30…, \$50, \$55, \$60…, \$90, \$95.
– If the stock price is lower than \$25, it is in the intervals of 2.5, i.e. \$17.7, \$20 and \$22.5.
Strike also reflects the liquidity of the option itself. That’s why you will see a number of stocks that don’t follow the formula above. For instance, at the point of this writing Microsoft (MSFT) trading at \$28 has \$1 strike price increment. Also Proctor & Gamble (PG) at the point of this writing trading at \$60 has \$1 strike price.
Sometimes, an unexpected event like a stock split at the ratio of 2 to 1 or 3 to 2 can result in odd number like a \$2.5 or \$3. It is because when the underlying stock gets split, the options get split too.
If you own one \$55 strike call options contract and the stock gets split from \$50 to \$25, you will end up owning 2 \$27.5 call options contracts. The value won’t change but your contract size and strike price do.
However, once you start trading and gain some experience, you will know strike prices don’t matter that much. What matters is your risk and rewards analysis, control of emotion and continuation of learning and refining your skills. | 481 | 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} | 2.890625 | 3 | CC-MAIN-2021-31 | latest | en | 0.908221 |
https://math.stackexchange.com/questions/1746092/why-cant-we-keep-adding-axioms-forever?noredirect=1 | 1,571,890,558,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00349.warc.gz | 592,538,826 | 33,823 | # Why can't we keep adding axioms forever?
Let F be a formal system falling prey to Gödel's incompleteness theorems, implyng there is a true but unprovable statement, call it $G_1$. Of course, adding $G_1$ to the axioms of F doesn't solve the problem - there would appear another true-but-unprovable statement, call it $G_2$. Now, adding $G_2$ to F+$G_1$ again doesn't solve the problem. In other words, no matter how many times you try to patch the original formal system, you will always end up having another true-but-unprovable statement.
My question: in light of the fact that most mathematicians accept the absulute infinite, why can't we just solve the whole problem by constructing the formal system F+$\bigwedge_{1}^{\infty}G_i$?
• – Pockets Apr 17 '16 at 8:38
• In what sense are you regarding the unprovable statement as true? Do you mean a statement that is true in some preferred model of the theory? – joriki Apr 17 '16 at 9:11
• @Mikhail Katz: on one hand, as I pointed out, this sense of "true" can also be read by formalists as "provable in PRA". It is common even for formalists to write "true" as if they were platonists, because it is well known how to re-interpret that language in a formalist way. However, it would be silly to go around math forum such as this challenging all speakers on their platonist views, for various reasons. – Carl Mummert Apr 17 '16 at 12:10
• @CarlMummert: Thanks for the explanation. I wasn't aware of this usage. It seems confusing to use something as meaning three different things according to people's taste :-), but I guess in some useful practical sense the differences don't matter too much in many cases. – joriki Apr 17 '16 at 12:21
• @joriki: no problem - the word "challenge" was introduced by someone else's comment anyway. I agree that, for practical purposes, there isn't much difference. My opinion on why the terminology is so common in logic is that it makes naive sense, but also those who want to use a non-naive understanding of "truth" can easily translate sentences written that way into their own preferred viewpoint. There is an analogy with reading a proof in constructive mathematics - a naive reader can pretend it's a classical proof, while experts can interpret the proof in their preferred constructive system. – Carl Mummert Apr 17 '16 at 12:34
The challenge is that the theory has to remain effective. You are asking about adding new axioms in stages. For the first few stages things are clear. We can make $F_2 = F + G_1$, $F_3 = F_2 + G_2$, etc.
So we can make the theory $F + \{ G_i : i \in\mathbb{N}\}$. Let's call that $F_\omega$. That will be an effective theory as well. This means that, just from $F$, we can enumerate the entire sequence of formulas $\{G_i : i \in \mathbb{N}\}$.
So we can then continue, making $F_{\omega + 1} = F_{\omega} + G_{\omega}$, $F_{\omega + 2} = F_{\omega+ 1} + G_{\omega+1}$, etc.
Eventually, we can make $F_{\omega + \omega}$ in the same way. To do this, we need to enumerate the entire sequence of formulas $\{G_\alpha: \alpha < \omega + \omega\}$, but we can do that effectively by enumerating $\{G_i : i \in \mathbb{N}\}$, then $G_\omega$, then $\{G_{\omega + i} : i \in \mathbb{N}\}$. So, as long as we have a good grasp of the overall sequence of extensions we have made, we can enumerate the necessary sequence of Goedel sentences.
For the incompleteness theorem to apply, we need the theory at hand to be effective. The problem, as you can see, is that we have to have a computable way to keep track of the "stages", so that we know which sentences have been added at each stage. There are limits to how many stages can be described in a coherent, computable way. The stages are usually treated as ordinals, re-using a concept from set theory.
Essentially, in order to create the Goedel sentence $G_\alpha$ for some ordinal $\alpha$, we need to have an effective way of describing $\alpha$. Recall that the Goedel sentence for a theory refers to an effective axiomatization of the theory, and to axiomatize one of the theories in our sequence of extensions, we need to know just how much it has been extended.
The easiest limit on how long we can keep going is that the number of stages can be at most countable, especially when the theory at hand is only countable. If there are only countably many sentences overall, then we can't continue adding Goedel sentences an uncountable number of times.
It turns out, however, that if we want the description of the stages to be effective, it cannot be near uncountable. The least bound for the computable ordinal numbers, known as the "Church-Kleene ordinal", $\omega_1^{CK}$, is countable. So our sequence of extensions will eventually have to stop, not because we run out of sentences, but because we run out of effective ways to describe the overall sequence of extensions we have made. | 1,248 | 4,874 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-43 | latest | en | 0.959949 |
https://im.kendallhunt.com/MS/teachers/1/2/11/index.html | 1,726,047,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00316.warc.gz | 292,964,814 | 38,899 | # Lesson 11
Representing Ratios with Tables
## 11.1: How Is It Growing? (10 minutes)
### Warm-up
This warm-up encourages students to look for regularity in how the tiles in the image are growing. Students may use each color to reason about the total, while others may reason about the way the total tiles increase each time. Emphasize both insights as students share their strategies.
### Launch
Arrange students in groups of 2. Display the image for all to see and tell students that the collection of images of green and blue tiles is growing. Ask how many total tiles will be in the 4th, 5th and 10th image if it keeps growing in the same way. Tell students to give a signal when they have an answer and strategy. Give students 3 minutes of quiet think time, and then time to discuss their responses and reasoning with their partner.
### Student Facing
Look for a pattern in the figures.
1. How many total tiles will be in:
1. the 4th figure?
2. the 5th figure?
3. the 10th figure?
2. How do you see it growing?
### Activity Synthesis
Invite students to share their responses and reasoning. Record and display the different ways of thinking for all to see. If possible, record the relevant reasoning on or near the images themselves. After each explanation, ask the class if they agree or disagree and to explain alternative ways of thinking, referring back to what is happening in the images each time.
## 11.2: A Huge Amount of Sparkling Orange Juice (15 minutes)
### Activity
Here, students are asked to find missing values for significantly scaled-up ratios. The activity serves several purposes:
• To uncover a limitation of a double number line (i.e., that it is not always practical to extend it to find significantly scaled-up equivalent ratios),
• To reinforce the multiplicative reasoning needed to find equivalent ratios (especially in cases when drawing diagrams or skip counting is inefficient), and
• To introduce a table as a way to represent equivalent ratios.
To find equivalent ratios involving large values, some students may simply try to squeeze numbers on the extreme right side of the paper, ignoring the previously equal intervals. Others may use multiplication (or division) and write expressions or equations to capture the given scenarios. Notice students’ reasoning processes, especially any struggles with the double number line (e.g., the lines not being long enough, requiring much marking and writing, the numbers being too large, etc.), as these can motivate a need for a more efficient strategy.
### Launch
Give students 2–3 minutes to work on the first two questions and then ask them to pause. As a class, discuss the two approaches students are likely to take: counting multiples of 4 and 5 up to 36 and 45; and multiplicative reasoning (asking “What number times 4 equals 36?”). Also discuss how a double number line like the one below might be used to support reasoning.
Reiterate the multiplicative relationship between equivalent ratios before students move on.
Representation: Internalize Comprehension. Activate or supply background knowledge about equivalent ratios describing situations that happen at the same rate. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
### Student Facing
Noah’s recipe for one batch of sparkling orange juice uses 4 liters of orange juice and 5 liters of soda water.
1. Use the double number line to show how many liters of each ingredient to use for different-sized batches of sparkling orange juice.
2. If someone mixes 36 liters of orange juice and 45 liters of soda water, how many batches would they make?
3. If someone uses 400 liters of orange juice, how much soda water would they need?
4. If someone uses 455 liters of soda water, how much orange juice would they need?
5. Explain the trouble with using a double number line diagram to answer the last two questions.
### Launch
Give students 2–3 minutes to work on the first two questions and then ask them to pause. As a class, discuss the two approaches students are likely to take: counting multiples of 4 and 5 up to 36 and 45; and multiplicative reasoning (asking “What number times 4 equals 36?”). Also discuss how a double number line like the one below might be used to support reasoning.
Reiterate the multiplicative relationship between equivalent ratios before students move on.
Representation: Internalize Comprehension. Activate or supply background knowledge about equivalent ratios describing situations that happen at the same rate. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
### Student Facing
Noah’s recipe for one batch of sparkling orange juice uses 4 liters of orange juice and 5 liters of soda water.
1. Use the double number line to show how many liters of each ingredient to use for different-sized batches of sparkling orange juice.
2. If someone mixes 36 liters of orange juice and 45 liters of soda water, how many batches would they make?
3. If someone uses 400 liters of orange juice, how much soda water would they need?
4. If someone uses 455 liters of soda water, how much orange juice would they need?
5. Explain the trouble with using a double number line diagram to answer the last two questions.
### Anticipated Misconceptions
Students may become frustrated when they “run out of number line,” but remind them of what they know about how to find ratios equivalent to $$4:5$$ (they need to multiply both 4 and 5 by the same number). Consider directing their attention to a definition of equivalent ratios displayed in your room or in a previous lesson, or suggesting they reexamine some of the simpler cases (e.g., the relationship between $$4:5$$ and $$36:45$$). Be on the lookout for students trying to tape on more paper to extend their number lines.
### Activity Synthesis
After students have a chance to share with a partner, select a few to share their reasoning with the class for the last few questions. Start with students who tried to extend the double number line (if anyone did so). Discuss any challenges of using the double number line and merits of alternative methods students might have come up with.
Explain that there is a more appropriate tool—a table—that can be used to represent equivalent ratios. Display for all to see the double number line from the activity above and a table of equivalent ratios. Explain that even though the table is oriented vertically and the double number line is oriented horizontally, the two representations represent the same ratios. Explain what we mean by row and column and demonstrate the use of these words. Fill in the table using the values from the orange-soda ratios and, along the way, compare and contrast how the two representations work. A few other key insights to convey:
• Just as it was important to label the double number line, it is important to label the columns of the table to indicate what the values represent (MP6).
• Each row of a table shows a pair of values from a collection of equivalent ratios. Unlike a number line, distances between values do not matter.
• On each line of a double number line, numbers are shown in order. In each column of a table, order is not important, i.e., pairs of values can be placed in any order that is convenient. When complete, the display should look something like this:
Representing: MLR7 Compare and Connect. Use this routine to help students make connections between specific features of tables and double number lines. Ask students to describe to a partner how multiplication appears in each representation, and then invite listeners to restate or revoice what they heard, back to their partner, using mathematical language (e.g., product, row, column, table, equivalent ratio, etc.). After students have a chance to share with a partner, select a few to share their reasoning with the class.
Design Principle(s): Maximize meta-awareness
## 11.3: Batches of Trail Mix (10 minutes)
### Activity
This task gets students to interact with a table in a way that discourages skip counting. Numbers within each column are deliberately out of order. This is intended to encourage students to multiply the pairs of values from a given ratio by the same number and to emphasize that the order in which pairs of values appear is not a necessary part of the structure of a table. (Order within rows, however, is necessary.) The last question reinforces the definition of equivalent ratios.
Students may use the given values (7 and 5) as the basis for every calculation (e.g., for every row, they think “7 times what . . . ” or “5 times what . . .”). They may also reason with values from another row (e.g., they may see 250 as $$10\boldcdot 25$$ rather than as $$5\boldcdot 50$$). As students work, notice different approaches.
### Launch
Explain that a table is just a list of equivalent ratios. In this case, one column contains amounts of almonds, and the other column contains corresponding amounts of raisins. Each row shows the amount of each ingredients in a particular batch.
Reiterate that multiplying both parts of a ratio by the same non-zero number always creates a ratio that is equivalent to the original ratio.
### Student Facing
A recipe for trail mix says: “Mix 7 ounces of almonds with 5 ounces of raisins.” Here is a table that has been started to show how many ounces of almonds and raisins would be in different-sized batches of this trail mix.
almonds (oz) raisins (oz)
7 5
28
10
3.5
250
56
1. Complete the table so that ratios represented by each row are equivalent.
2. What methods did you use to fill in the table?
3. How do you know that each row shows a ratio that is equivalent
to $$7:5$$? Explain your reasoning.
### Student Facing
#### Are you ready for more?
You have created a best-selling recipe for chocolate chip cookies. The ratio of sugar to flour is $$2:5$$.
Create a table in which each entry represents amounts of sugar and flour that might be used at the same time in your recipe.
• One entry should have amounts where you have fewer than 25 cups of flour.
• One entry should have amounts where you have between 20–30 cups of sugar.
• One entry can have any amounts using more than 500 units of flour.
### Anticipated Misconceptions
Students may make patterns that do not yield equivalent ratios. For example, they may think “7 minus 2 is 5, so for the next row, 28 minus 2 is 26.” Or they may think “7 plus 21 is 28, so then 5 plus 21 is 26.” If so, consider:
• Appealing to what students know about batches of recipes. “The second row represents how many batches of trail mix?” (4, because 28 is $$7\boldcdot 4$$.) “Okay, so to make 4 batches of trail mix, how will we figure out how many raisins?” (Also multiply the 5 by 4.)
• Refreshing what students learned about equivalent ratios. “We need a ratio that is equivalent to the ratio represented in row 1. So what do we need to do to the 7 and the 5?” (Multiply them by the same number.)
Students may be unsure about how to find the missing value in the row with 3.5. Encourage them to reason about it the same way they reasoned about the other rows. “We need a ratio that is equivalent to the ratio represented in row 1. So what do we need to do to the 7 and the 5?” They may have to get there by way of division. 7 divided by 2 is 3.5, so 7 times $$\frac12$$ is 3.5; this means multiplying 5 by $$\frac12$$ as well.
### Activity Synthesis
Invite one or more students who used multiplicative approaches to share their reasoning with the class. Consider displaying the table and using it to facilitate gesturing and arrow-drawing while students explain. Highlight the strategy of multiplying the 7 and 5 values by the same number.
## Lesson Synthesis
### Lesson Synthesis
Sometimes it is easier to use a table rather than a double number line to represent equivalent ratios. Each row contains a ratio that is equivalent to all the other ratios, so if we know one row, we can multiply both of its values by the same number to find another row’s values. | 2,649 | 12,186 | {"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} | 4.625 | 5 | CC-MAIN-2024-38 | latest | en | 0.93227 |
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Oxidation Numbers Watch
1. Hi can somebody please help me with oxidation numbers. I am retaking OCR as chemistry atoms,bonds and groups and really struggling with oxidation numbers!!
I have these questions which say...
Work out the oxidation state of the element in bold.
e.g
NaClO3.
All it says in my text book is
Uncombined element = 0
Combined Oxygen= -2
Combined Hydrogen = +1
Simple Ion = Charge on ion
Combined Fluorine = -1.
Using this i can only work out the 03 part which is -6. But how am I supposed to work out the oxidation numbers of the Na and Cl. The answer says Cl = +5.
Would really appreicate the help.
2. Na is in group 1, so take it as being plus 1
if it asked you for Na, you would take Cl as being -1
3. Generally [but not always the case] the rule is:
group 1 plus 1
group 2 plus 2
group 3 plus 3
group 4 0
group 5 -3
group 6 -2
group 7 -1
-6 + 1 = -5
x + -5= 0
x=5
4. Thankyou
That has really helped, I can now do all the questions!
5. (Original post by Alwee)
Thankyou
That has really helped, I can now do all the questions!
6. (Original post by nazirard)
Generally [but not always the case] the rule is:
group 1 plus 1
group 2 plus 2
group 3 plus 3
These are good to go...
(Original post by nazirard)
group 4 0 (never the case except in the element itself)
group 5 -3
group 6 -2
group 7 -1
These are not...
The 'rules' below are better
--------------------------
Hydrogen is always +1 unless bonded to a metal in a binary salt (i.e. a metal hydride) in which case it's -1
Oxygen is always -2 unless a peroxide,in which case -1
The binary salts with metals have:
group 5 -3
group 6 -2
group 7 -1
BUT otherwise you have to add up.
A compound has a sum of zero and an ion has a sum of the charge on the ion.
Na2SO4
Na (group 1) = +1 , therefore 2 x Na = +2
Oxygen = -2 therefore 4 x O = -8
---------------------------------------------sum
total = -6, therefore sulphur must = +6
7. (Original post by charco)
These are good to go...
These are not...
The 'rules' below are better
--------------------------
Hydrogen is always +1 unless bonded to a metal in a binary salt (i.e. a metal hydride) in which case it's -1
Oxygen is always -2 unless a peroxide,in which case -1
The binary salts with metals have:
group 5 -3
group 6 -2
group 7 -1
BUT otherwise you have to add up.
A compound has a sum of zero and an ion has a sum of the charge on the ion.
Na2SO4
Na (group 1) = +1 , therefore 2 x Na = +2
Oxygen = -2 therefore 4 x O = -8
---------------------------------------------sum
total = -6, therefore sulphur must = +6
Some of what you said i presumed he already knew as it's in the ocr book (the situation with ions).
In the AS spec it says you don't need to know about the oxygen and peroxide situation.
My chem teacher told us this is the best way to know it at AS level.
But thanks for the corrections.
8. This is the list of priority:
Group 1 is always +1
Group 2 is always +2
Group 3 is +3
F is -1
H is +1
O is -2
Cl is -1
You can work out others from this as you tend to have compounds given with at least one of these elements in.
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https://www.leonieclaire.com/other/how-do-you-do-sem-bars-in-excel/ | 1,653,505,341,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00368.warc.gz | 998,852,127 | 13,721 | ## How do you do SEM bars in Excel?
The “Chart Tools” menu should appear at the top of your screen: Now choose the “Layout” tab under the “Chart Tools” menu, and click on “Error Bars.” Select “More Error Bar Options”: Page 2 The “Format Error Bars” box should now appear, as shown below.
## What do error bars in Excel mean?
Error bars in Excel are graphical representations of data variability. They show the precision of a measurement. The bars usually represent standard deviation. It will the estimate standard deviation based on a sample. Whether checks are written to be and allow users to see the margins or errors in the data.
## What is SEM error bars?
Unlike s.d. bars, error bars based on the s.e.m. reflect the uncertainty in the mean and its dependency on the sample size, n (s.e.m. = s.d./√n). Intuitively, s.e.m. bars shrink as we perform more measurements.
## What do SEM bars do?
SEM error bars quantify how precisely you know the mean, taking into account both the SD and sample size. Looking at whether the error bars overlap, therefore, lets you compare the difference between the mean with the precision of those means.
## What do error bars show you?
Error bars are graphical representations of the variability of data and used on graphs to indicate the error or uncertainty in a reported measurement. They give a general idea of how precise a measurement is, or conversely, how far from the reported value the true (error free) value might be.
## Should I show SD or SEM?
SEM quantifies uncertainty in estimate of the mean whereas SD indicates dispersion of the data from mean. As readers are generally interested in knowing the variability within sample, descriptive data should be precisely summarized with SD.
## Should I plot SD or SEM?
In conclusion, SD quantifies the variability, whereas SEM quantifies uncertainty in estimate of the mean. As readers are generally interested in knowing the variability within sample and not proximity of mean to the population mean, data should be precisely summarized with SD and not with SEM.
## Should I use SEM or SD?
It helps present data precisely and draws the meaningful conclusions. SEM quantifies uncertainty in estimate of the mean whereas SD indicates dispersion of the data from mean. As readers are generally interested in knowing the variability within sample, descriptive data should be precisely summarized with SD.
## What does SEM mean in Excel?
standard error of mean
The Standard Error of Mean, also known as SEM is another measure of variability of data. It is an estimate of the deviation of a sample mean from the population mean. SEM is not as popular as standard deviation, and it is sometimes just referred to as “standard error”.
## How to add or remove error bars in Excel?
[Solution] Excel Add Error Bars 1 Click anywhere in the chart. 2 Click the Chart Elements button next to the chart, and then check the Error Bars box. (Clear the box to remove error bars.) 3 To change the error amount shown, click the arrow next to Error Bars, and then pick an option. See More….
## How do I remove the error bars on an Excel scatter chart?
Scatter charts can show both horizontal and vertical error bars. You can remove either of these error bars by selecting them, and then pressing Delete. People often ask how Excel calculates error amounts. Excel uses the following equations to calculate the Standard Error and Standard Deviation amounts that are shown on the chart.
## Why do scientists use error bars in Excel?
In science, uncertainty is the beginning of an investigation. And because scientists love quantifying things, they found a way to quantify uncertainty. For this, they calculate confidence intervals, or margins of error, and display them by using what is known as error bars.
## How is the standard deviation ( SEM ) calculated in Excel?
SEM is calculated by taking the standard deviation and dividing it by the square root of the sample size. In Excel, this is calculated using the following formula: =stdev(range)/sqrt(count(range)) Posted by at 8:47 AM. Email ThisBlogThis!Share to TwitterShare to FacebookShare to Pinterest. | 851 | 4,155 | {"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-2022-21 | latest | en | 0.883146 |
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# Module Five Chapter - Chapter 8 problems Problem 11 a No change in optimal solution there is no upper limit for the range of optimality for the
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Chapter 8 problems: Problem 11 a. No change in optimal solution, there is no upper limit for the range of optimality for the objective coefficient for S. b. No change in the optimal solution, the objective coefficient for M can increase to 6.4. c. The optimal solution does not change Problem 17 a. Produce 1000 units of model DRB and 800 units of model DRW. Total profit contribution \$424,000 b. The dual value for constraint one is 8.80. Each additional pound of steel will increase profit by \$8.80. At \$2 per pound Deegan should purchase the additional 500 pounds. The upper limit on the right hand side range for constraint 1 is approximately 40,909, the dual value of \$8.80 is applicable for an increase of as much as 909 pounds. c. Constraint three has a slack of 4000 hours. Increasing the number of assembly time is not worthwhile. d. The objective coefficient range for model DRB shows a lower limit of \$112. The optimal solution will not change, the value of the optimal solution will be \$175(1000) + \$280(800) = \$399,000
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$a{{x}^{2}}+bx+c=0$ An example of equation reducible to a quadratic is ${{x}^{4}}-8{{x}^{2}}-9=0$ Let ${{x}^{2}}=u$, Then ${{x}^{4}}={{u}^{2}}$. Substitute the values of ${{x}^{2\text{ }}}\text{and }{{x}^{4}}$ into the equation ${{x}^{4}}-8{{x}^{2}}-9=0$ as follows: \begin{align} & {{x}^{4}}-8{{x}^{2}}-9=0 \\ & {{u}^{2}}-8u-9=0 \end{align} Factor ${{u}^{2}}-8u-9=0$, \begin{align} & {{u}^{2}}-8u-9=0 \\ & \left( u-9 \right)\left( u+1 \right)=0 \end{align} Thus, $\text{ }u=9\text{ or }u=-1$ Now, this is incomplete because the solution is being carried out for x and not u. Now replace $u$ with ${{x}^{2}}$. \begin{align} & \text{ }u=9\text{ or }u=-1 \\ & {{x}^{2}}\text{= 9 or }{{x}^{2}}=-1 \\ & x=\pm 3\text{ square can }\!\!'\!\!\text{ t be negative,no solution}\text{. } \\ \end{align} Thus, the final solution is $x=\pm \text{ }3$ and not $\text{ }u=9\text{ or }u=-1$. | 406 | 951 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-21 | latest | en | 0.536623 |
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``` Stock Valuation: First In First Out (FIFO)
So far we have assumed that the cost of stock is constant, that is, it does not change over time:
Q 5
STOCK CARD: Chairs IN OUT BALANCE C V Q C V Q C V 3 200 600 200 1000 8 200 1600 2 200 400 6 200 1200
However, constant prices is not realistic!
Example: Chairs.
Q
1/1 BAL 5 2 3/1 CHQ 4/1 CHQ 5/1 REC
STOCK CARD: Chairs IN OUT BALANCE C V Q C V Q C V
220 230 1100 460 3 3 5 3 5 2 200 200 220 200 220 230 600 600 1100 600 1100 460
What happens when the business sells stock on 5/1?
EG: What happens if 4 chairs are sold on 5/1?
Identified Cost.
When the chairs are purchased, they are tagged or coded in some way to identify their cost price: this would be easy for chairs. Assume the tags or codes identify that the 4 sales were: 1 chair @ \$200, 2 chairs @ \$220, 1 chair @ \$230
STOCK CARD: Chairs IN OUT BALANCE Q C V Q C V Q C V
1/1 BAL 3/1 CHQ 4/1 CHQ 5 2 220 230 1100 460 1 2 1 200 220 230 200 440 230 3 3 5 3 5 2 2 3 1 200 200 220 200 220 230 200 220 230 600 600 1100 600 1100 460 400 660 230
5/1 REC
For stock that is easy to tag or code, such as Chairs, identified cost can be used to assign cost to the stock. What about stock that is not easy to tag or code?
Example:
Purchases of grain (IN)
Sales of grain (OUT)
Because of the nature of the stock, separate purchases cannot be tagged or coded: the stock merges together and loses its individual identity. So how is the stock valued?
First In First Out (FIFO)
FIFO assumes stock is sold on a FIFO basis, that is, the oldest stock is sold first. This method can be used by businesses which cannot tag or code their stock or, if they can, they prefer FIFO because it is administratively easier and cheaper to operate than identified cost. FIFO = last in last out!
Bulk Grain example:
1/1 Balance at start: 90 tonnes @ \$100 tonne 3/1 Purchase: 80 tonnes @ \$110 tonne 5/1 Sales: 120 tonnes.
STOCK CARD: Bulk Grain FIFO IN OUT BALANCE Q C V Q C V Q C V
1/1 BAL 3/1 CHQ 80 5/1 REC 110 8800 90 30 100 110 9000 3300 90 90 80 50 100 100 110 110 9000 9000 8800 5500
Chair example using FIFO
Even though the chair business could use identified cost, it may use FIFO since it would not need to tag or code its stock. FIFO is administratively easier and cheaper to operate than identified cost.
Sold 4 chairs on 5/1. Using FIFO.
STOCK CARD: Chairs: FIFO
Q
1/1 BAL 3/1 CHQ 5
IN C
220
V
1100
OUT Q C V
BALANCE Q C V
3 3 5 3 5 2 200 200 220 200 220 230 220 230 600 600 1100 600 1100 460 880 460
4/1 CHQ
2
230
460 3 1 200 220 600 220
5/1 REC
4 2
Sold 4 units. Using FIFO, oldest stock is sold first, thus 3 @ \$200 then 1 @ \$220.
Other theory!
Identified cost is a 100% accurate method of stock valuation which means: Profit/Loss statement: Cost of sales, gross and net profit are accurate Balance Sheet: Value of stock and owner’s equity (due to profit impact) are accurate Thus identified cost = reliability characteristic of the reports.
More theory!
FIFO is based on an assumption that stock flows in a first in first out basis: this may not match the actual stock flows thus FIFO is unlikely to be a 100% accurate method of stock valuation: Profit/Loss statement: Cost of sales, gross and net profits would reflect this possible inaccuracy (less reliable) Balance Sheet: Value of stock and owner’s equity (due to profit impact) would reflect this possible inaccuracy
Thus FIFO reduces the reliability characteristic of the reports.
… and more theory!
Units 3&4 only require a knowledge of FIFO not identified cost (but good to know for comparison purposes)
Identified cost may not be possible/realistic for some businesses due to their type of stock (grain, fuel, hard to tag/code stock etc) so FIFO is used.
FIFO is often a realistic assumption is many businesses (think, eg, perishable stock) FIFO is also used by many businesses which could use identified cost but choose FIFO since it is easier and cheaper to administer
… and even more theory!
•Usually, overtime, prices tend to rise. •This means that using FIFO stock at end is valued at higher more recent prices (does this contravene conservatism?) and cost of sales is valued at earlier or cheaper prices. Gross and net profit thus tend to be higher using FIFO.
```
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# 12.2: Conditions for Static Equilibrium
Learning Objectives
• Identify the physical conditions of static equilibrium.
• Draw a free-body diagram for a rigid body acted on by forces.
• Explain how the conditions for equilibrium allow us to solve statics problems.
We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. We say that a rigid body is in static equilibrium when it is at rest in our selected frame of reference. Notice that the distinction between the state of rest and a state of uniform motion is artificial—that is, an object may be at rest in our selected frame of reference, yet to an observer moving at constant velocity relative to our frame, the same object appears to be in uniform motion with constant velocity. Because the motion is relative, what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Since the laws of physics are identical for all inertial reference frames, in an inertial frame of reference, there is no distinction between static equilibrium and equilibrium.
According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or
$$\sum_{k} \vec{F}_{k} = m \vec{a}_{CM} \ldotp \label{12.1}$$
Here, the sum is of all external forces acting on the body, where m is its mass and $$\vec{a}_{CM}$$ is the linear acceleration of its center of mass (a concept we discussed in Linear Momentum and Collisions on linear momentum and collisions). In equilibrium, the linear acceleration is zero. If we set the acceleration to zero in Equation \ref{12.1}, we obtain the following equation:
First Equilibrium Condition
The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium:
$$\sum_{k} \vec{F}_{k} = \vec{0} \ldotp \label{12.2}$$
The first equilibrium condition, Equation \ref{12.2}, is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws.
This vector equation is equivalent to the following three scalar equations for the components of the net force:
$$\sum_{k} F_{kx} = 0,\; \sum_{k} F_{ky} = 0,\; \sum_{k} F_{kz} = 0 \ldotp \label{12.3}$$
Analogously to Equation \ref{12.1}, we can state that the rotational acceleration $$\vec{\alpha}$$ of a rigid body about a fixed axis of rotation is caused by the net torque acting on the body, or
$$\sum_{k} \vec{\tau}_{k} = I \vec{\alpha} \ldotp \label{12.4}$$
Here I is the rotational inertia of the body in rotation about this axis and the summation is over all torques $$\vec{\tau}_{k}$$ of external forces in Equation \ref{12.2}. In equilibrium, the rotational acceleration is zero. By setting to zero the right-hand side of Equation \ref{12.4}, we obtain the second equilibrium condition:
Second Equilibrium Condition
The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium:
$$\sum_{k} \vec{\tau}_{k} = \vec{0} \ldotp \label{12.5}$$
The second equilibrium condition, Equation \ref{12.5}, is the equilibrium condition for torques that we encountered when we studied rotational dynamics. It is worth noting that this equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). Again, this vector equation is equivalent to three scalar equations for the vector components of the net torque:
$$\sum_{k} \tau_{kx} = 0,\; \sum_{k} \tau_{ky} = 0,\; \sum_{k} \tau_{kz} = 0 \ldotp \label{12.6}$$
The second equilibrium condition means that in equilibrium, there is no net external torque to cause rotation about any axis. The first and second equilibrium conditions are stated in a particular reference frame. The first condition involves only forces and is therefore independent of the origin of the reference frame. However, the second condition involves torque, which is defined as a cross product, $$\vec{\tau}_{k} = \vec{r}_{k} \times \vec{F}_{k}$$, where the position vector $$\vec{r}_{k}$$ with respect to the axis of rotation of the point where the force is applied enters the equation. Therefore, torque depends on the location of the axis in the reference frame. However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. The explanation for this is fairly straightforward.
Suppose vector $$\vec{R}$$ is the position of the origin of a new inertial frame of reference S′ in the old inertial frame of reference S. From our study of relative motion, we know that in the new frame of reference S′, the position vector $$\vec{r}′_{k}$$ of the point where the force $$\vec{F}_{k}$$ is applied is related to $$\vec{r}_{k}$$ via the equation
$$\vec{r}'_{k} = \vec{r}_{k} - \vec{R} \ldotp$$
Now, we can sum all torques $$\vec{\tau}'_{k} = \vec{r}'_{k} \times \vec{F}_{k}$$ of all external forces in a new reference frame, S':
$$\sum_{k} \vec{\tau}'_{k} = \sum_{k} \vec{r}'_{k} \times \vec{F}_{k} = \sum_{k} (\vec{r}_{k} - \vec{R}) \times \vec{F}_{k} = \sum_{k} \vec{r}_{k} \times \vec{F}_{k} - \sum_{k} \vec{R} \times \vec{F}_{k} = \sum \vec{\tau}_{k} - \vec{R} \times \sum_{k} \vec{F}_{k} = \vec{0} \ldotp$$
In the final step in this chain of reasoning, we used the fact that in equilibrium in the old frame of reference, S, the first term vanishes because of Equation \ref{12.5} and the second term vanishes because of Equation \ref{12.2}. Hence, we see that the net torque in any inertial frame of reference S′ is zero, provided that both conditions for equilibrium hold in an inertial frame of reference \ref{S}.
The practical implication of this is that when applying equilibrium conditions for a rigid body, we are free to choose any point as the origin of the reference frame. Our choice of reference frame is dictated by the physical specifics of the problem we are solving. In one frame of reference, the mathematical form of the equilibrium conditions may be quite complicated, whereas in another frame, the same conditions may have a simpler mathematical form that is easy to solve. The origin of a selected frame of reference is called the pivot point.
In the most general case, equilibrium conditions are expressed by the six scalar equations (Equations \ref{12.3} and \ref{12.6}). For planar equilibrium problems with rotation about a fixed axis, which we consider in this chapter, we can reduce the number of equations to three. The standard procedure is to adopt a frame of reference where the z-axis is the axis of rotation. With this choice of axis, the net torque has only a z-component, all forces that have non-zero torques lie in the xy-plane, and therefore contributions to the net torque come from only the x- and y-components of external forces. Thus, for planar problems with the axis of rotation perpendicular to the xy-plane, we have the following three equilibrium conditions for forces and torques:
$$F_{1x} + F_{2x} + \cdots + F_{Nx} = 0 \label{12.7}$$
$$F_{1y} + F_{2y} + \cdots + F_{Ny} = 0 \label{12.8}$$
$$\tau_{1} + \tau_{2} + \cdots + \tau_{N} = 0 \label{12.9}$$
where the summation is over all N external forces acting on the body and over their torques. In Equation \ref{12.9}, we simplified the notation by dropping the subscript z, but we understand here that the summation is over all contributions along the z-axis, which is the axis of rotation. In Equation \ref{12.9}, the z-component of torque $$\vec{\tau}_{k}$$ from the force $$\vec{F}_{k}$$ is
$$\tau_{k} = r_{k} F_{k} \sin \theta \label{12.10}$$
where rk is the length of the lever arm of the force and Fk is the magnitude of the force (as you saw in Fixed-Axis Rotation). The angle $$\theta$$ is the angle between vectors $$\vec{r}_{k}$$ and $$\vec{F}_{k}$$, measuring from vector $$\vec{r}_{k}$$ to vector $$\vec{F}_{k}$$ in the counterclockwise direction (Figure $$\PageIndex{1}$$). When using Equation \ref{12.10}, we often compute the magnitude of torque and assign its sense as either positive (+) or negative (−), depending on the direction of rotation caused by this torque alone. In Equation \ref{12.9}, net torque is the sum of terms, with each term computed from Equation \ref{12.10}, and each term must have the correct sense. Similarly, in Equation \ref{12.7}, we assign the + sign to force components in the + x-direction and the − sign to components in the − x-direction. The same rule must be consistently followed in Equation \ref{12.8}, when computing force components along the y-axis.
Note
View this demonstration to see two forces act on a rigid square in two dimensions. At all times, the static equilibrium conditions given by Equation \ref{12.7} through Equation \ref{12.9} are satisfied. You can vary magnitudes of the forces and their lever arms and observe the effect these changes have on the square.
In many equilibrium situations, one of the forces acting on the body is its weight. In free-body diagrams, the weight vector is attached to the center of gravity of the body. For all practical purposes, the center of gravity is identical to the center of mass, as you learned in Linear Momentum and Collisions on linear momentum and collisions. Only in situations where a body has a large spatial extension so that the gravitational field is nonuniform throughout its volume, are the center of gravity and the center of mass located at different points. In practical situations, however, even objects as large as buildings or cruise ships are located in a uniform gravitational field on Earth’s surface, where the acceleration due to gravity has a constant magnitude of g = 9.8 m/s2. In these situations, the center of gravity is identical to the center of mass. Therefore, throughout this chapter, we use the center of mass (CM) as the point where the weight vector is attached. Recall that the CM has a special physical meaning: When an external force is applied to a body at exactly its CM, the body as a whole undergoes translational motion and such a force does not cause rotation.
When the CM is located off the axis of rotation, a net gravitational torque occurs on an object. Gravitational torque is the torque caused by weight. This gravitational torque may rotate the object if there is no support present to balance it. The magnitude of the gravitational torque depends on how far away from the pivot the CM is located. For example, in the case of a tipping truck (Figure $$\PageIndex{2}$$), the pivot is located on the line where the tires make contact with the road’s surface. If the CM is located high above the road’s surface, the gravitational torque may be large enough to turn the truck over. Passenger cars with a low-lying CM, close to the pavement, are more resistant to tipping over than are trucks.
Note
If you tilt a box so that one edge remains in contact with the table beneath it, then one edge of the base of support becomes a pivot. As long as the center of gravity of the box remains over the base of support, gravitational torque rotates the box back toward its original position of stable equilibrium. When the center of gravity moves outside of the base of support, gravitational torque rotates the box in the opposite direction, and the box rolls over. View this demonstration to experiment with stable and unstable positions of a box.
Example 12.1: Center of Gravity of a Car
A passenger car with a 2.5-m wheelbase has 52% of its weight on the front wheels on level ground, as illustrated in Figure 12.4. Where is the CM of this car located with respect to the rear axle?
Strategy
We do not know the weight w of the car. All we know is that when the car rests on a level surface, 0.52w pushes down on the surface at contact points of the front wheels and 0.48w pushes down on the surface at contact points of the rear wheels. Also, the contact points are separated from each other by the distance d = 2.5 m. At these contact points, the car experiences normal reaction forces with magnitudes FF = 0.52w and FR = 0.48w on the front and rear axles, respectively. We also know that the car is an example of a rigid body in equilibrium whose entire weight w acts at its CM. The CM is located somewhere between the points where the normal reaction forces act, somewhere at a distance x from the point where FR acts. Our task is to find x. Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in Figure $$\PageIndex{4}$$.
We are almost ready to write down equilibrium conditions Equation \ref{12.7} through Equation \ref{12.9} for the car, but first we must decide on the reference frame. Suppose we choose the x-axis along the length of the car, the yaxis vertical, and the z-axis perpendicular to this xy-plane. With this choice we only need to write Equation \ref{12.7} and Equation \ref{12.9} because all the y-components are identically zero. Now we need to decide on the location of the pivot point. We can choose any point as the location of the axis of rotation (z-axis). Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. At this point, we are ready to write the equilibrium conditions for the car.
Solution
Each equilibrium condition contains only three terms because there are N = 3 forces acting on the car. The first equilibrium condition, Equation \ref{12.7}, reads
$$+F_{F} - w + F_{R} = 0 \ldotp \label{12.11}$$
This condition is trivially satisfied because when we substitute the data, Equation \ref{12.11} becomes +0.52w − w + 0.48w = 0. The second equilibrium condition, Equation \ref{12.9}, reads
$$\tau_{F} + \tau_{w} + \tau_{R} = 0 \label{12.12}$$
where $$\tau_{F}$$ is the torque of force FF, $$\tau_{w}$$ is the gravitational torque of force w, and $$\tau_{R}$$ is the torque of force FR. When the pivot is located at CM, the gravitational torque is identically zero because the lever arm of the weight with respect to an axis that passes through CM is zero. The lines of action of both normal reaction forces are perpendicular to their lever arms, so in Equation \ref{12.10}, we have |sin $$\theta$$| = 1 for both forces. From the free-body diagram, we read that torque $$\tau_{F}$$ causes clockwise rotation about the pivot at CM, so its sense is negative; and torque $$\tau_{R}$$ causes counterclockwise rotation about the pivot at CM, so its sense is positive. With this information, we write the second equilibrium condition as
$$-r_{F} F_{F} + r_{R} F_{R} = 0 \ldotp \label{12.13}$$
With the help of the free-body diagram, we identify the force magnitudes FR = 0.48w and FF = 0.52w, and their corresponding lever arms rR = x and rF = d − x. We can now write the second equilibrium condition, Equation \ref{12.13}, explicitly in terms of the unknown distance x:
$$-0.52(d - x)w + 0.48xw = 0 \ldotp \label{12.14}$$
Here the weight w cancels and we can solve the equation for the unknown position x of the CM. The answer is x = 0.52d = 0.52(2.5 m) = 1.3 m. Solution Choosing the pivot at the position of the front axle does not change the result. The free-body diagram for this pivot location is presented in Figure 12.6. For this choice of pivot point, the second equilibrium condition is
$$-r_{w} w + r_{R} F_{R} = 0 \ldotp \label{12.15}$$
When we substitute the quantities indicated in the diagram, we obtain
$$-(d - x)w + 0.48dw = 0 \ldotp \label{12.16}$$
The answer obtained by solving Equation \ref{12.13} is, again, x = 0.52d = 1.3 m.
## Significance
This example shows that when solving static equilibrium problems, we are free to choose the pivot location. For different choices of the pivot point we have different sets of equilibrium conditions to solve. However, all choices lead to the same solution to the problem.
Exercise 12.1
Solve Example 12.1 by choosing the pivot at the location of the rear axle.
Exercise 12.2
Explain which one of the following situations satisfies both equilibrium conditions: (a) a tennis ball that does not spin as it travels in the air; (b) a pelican that is gliding in the air at a constant velocity at one altitude; or (c) a crankshaft in the engine of a parked car.
A special case of static equilibrium occurs when all external forces on an object act at or along the axis of rotation or when the spatial extension of the object can be disregarded. In such a case, the object can be effectively treated like a point mass. In this special case, we need not worry about the second equilibrium condition, Equation \ref{12.9}, because all torques are identically zero and the first equilibrium condition (for forces) is the only condition to be satisfied. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion and Applications of Newton’s Laws. You will see a typical equilibrium situation involving only the first equilibrium condition in the next example.
View this demonstration to see three weights that are connected by strings over pulleys and tied together in a knot. You can experiment with the weights to see how they affect the equilibrium position of the knot and, at the same time, see the vector-diagram representation of the first equilibrium condition at work.
Example 12.2: A Breaking Tension
A small pan of mass 42.0 g is supported by two strings, as shown in Figure 12.7. The maximum tension that the string can support is 2.80 N. Mass is added gradually to the pan until one of the strings snaps. Which string is it? How much mass must be added for this to occur?
Strategy
This mechanical system consisting of strings, masses, and the pan is in static equilibrium. Specifically, the knot that ties the strings to the pan is in static equilibrium. The knot can be treated as a point; therefore, we need only the first equilibrium condition. The three forces pulling at the knot are the tension $$\vec{T}_{1}$$ in the 5.0-cm string, the tension $$\vec{T}_{2}$$ in the 10.0-cm string, and the weight $$\vec{w}$$ of the pan holding the masses. We adopt a rectangular coordinate system with the y-axis pointing opposite to the direction of gravity and draw the free-body diagram for the knot (see Figure 12.8). To find the tension components, we must identify the direction angles $$\alpha_{1}$$ and $$\alpha_{2}$$ that the strings make with the horizontal direction that is the x-axis. As you can see in Figure 12.7, the strings make two sides of a right triangle. We can use the Pythagorean theorem to solve this triangle, shown in Figure 12.8, and find the sine and cosine of the angles $$\alpha_{1}$$ and $$\alpha_{2}$$. Then we can resolve the tensions into their rectangular components, substitute in the first condition for equilibrium (Equation \ref{12.7} and Equation \ref{12.8}), and solve for the tensions in the strings. The string with a greater tension will break first.
Solution
The weight w pulling on the knot is due to the mass M of the pan and mass m added to the pan, or w = (M + m)g. With the help of the free-body diagram in Figure 12.8, we can set up the equilibrium conditions for the knot:
in the x-direction, $$-T_{1x} + T_{2x} = 0$$in the y-direction, $$+T_{1y} + T_{2y} - w = 0 \ldotp$$
From the free-body diagram, the magnitudes of components in these equations are
$$\begin{split} T_{1x} & = T_{1} \cos \alpha_{1} = \frac{T_{1}}{\sqrt{5}},\quad T_{1y} = T_{1} \sin \alpha_{1} = \frac{2T_{1}}{\sqrt{5}} \\ T_{2x} & = T_{2} \cos \alpha_{2} = \frac{2T_{2}}{\sqrt{5}},\quad T_{2y} = T_{2} \sin \alpha_{2} = \frac{T_{2}}{\sqrt{5}} \ldotp \end{split}$$
We substitute these components into the equilibrium conditions and simplify. We then obtain two equilibrium equations for the tensions:
in x-direction, $$T_{1} = 2T_{2}$$in y-direction, $$\frac{2T_{1}}{\sqrt{5}} + \frac{T_{2}}{\sqrt{5}} = (M + m)g \ldotp$$
The equilibrium equation for the x-direction tells us that the tension T1 in the 5.0-cm string is twice the tension T2 in the 10.0-cm string. Therefore, the shorter string will snap. When we use the first equation to eliminate T2 from the second equation, we obtain the relation between the mass m on the pan and the tension T1 in the shorter string:
$$\frac{2.5T_{1}}{\sqrt{5}} = (M + m)g \ldotp$$
The string breaks when the tension reaches the critical value of T1 = 2.80 N. The preceding equation can be solved for the critical mass m that breaks the string:
$$m = \frac{2.5}{\sqrt{5}} \frac{T_{1}}{g} - M = \frac{2.5}{\sqrt{5}} \frac{2.80\; N}{9.8\; m/s^{2}} - 0.042\; kg = 0.277\; kg = 277.0\; g \ldotp$$
Significance
Suppose that the mechanical system considered in this example is attached to a ceiling inside an elevator going up. As long as the elevator moves up at a constant speed, the result stays the same because the weight w does not change. If the elevator moves up with acceleration, the critical mass is smaller because the weight of M + m becomes larger by an apparent weight due to the acceleration of the elevator. Still, in all cases the shorter string breaks first.
## Contributors
• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). | 5,454 | 21,590 | {"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": 2, "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} | 4.15625 | 4 | CC-MAIN-2020-10 | latest | en | 0.875765 |
https://forums.unrealengine.com/t/electrical-circuit-simulation/111415 | 1,675,110,921,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00280.warc.gz | 300,206,603 | 7,592 | # Electrical Circuit Simulation
I’ve been trying for months to create a way to simulate electrical circuits. I’ve tried so many different ways, but this is what I currently have.
Circuit Board actor has components attached to it that are the elements (ex: sources, resistors, ground, etc). Circuit Board actor has an array of connection structures which describe element connections (which component’s terminal to which component terminal). Every circuit element has a terminal map, integers as keys, with an energy struct that holds voltage and amperage as values.
Here is an organized way to explain this:
• CircuitBoard Actor
• Components (scene components)
• Terminals (Int to S_Energy Map
• Key of Int
• Value of S_Energy
• Voltage
• Amperage
• Data
• Resistance
• Voltage Drop (ex: diode)
• Other variables (EX: Integrity)
• Connections (struct array)
• Each connection (struct)
• Object 1 Terminal (int)
• Object 2 Terminal (int)
• Object 1 (component ref)
• Object 2 (component ref)
All energy flow and actions should take place on the circuit board blueprint, the component is just a data holder. I’m having trouble simulating the flow of energy from component to component. I usually use Set Members in S_Energy on each terminal for each connection. Am I going about this all wrong? If anybody can give some suggestions or ideas it would be greatly appreciated.
Here are some sites I use for refrences
I have a general understanding (about 2-3 years) of electronics, C++/C# and blueprints experience.
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Oh man wish you had an answer for this, only thing i have been able to create is a generator that when on allows you to turn on/off anything in a boxtrace radius that has a power tag.
Im more versed in dc electronics than blue prints. And im not sure how youve structured your code But if i were thinking like i do.
Complicated Electrical circuits are lots of simple math problems like divide voltage and resistance to get current. And formulas to keep up with capacities, wattage, ext. and how all these affect the circuit can be summed up per circuit and saved as variables per circuit based on all components in the circuit.
I might make a macro for basic math equations. All that you would use, like circuit voltage(input) X component resistance(input) = circuit current(output) and then you could probably just place your macro and hook up a component and the circuit, then it would automatically grab the values of that component and run them with the circuits equation one by one to get the final values and use them for simulating.
In one blueprint that might be easy but i haven’t mastered casting so i dont know on that part. Ive just got the math theory.
Sounds like a fun experiment. If its what you’re talking about i might try to write my own test to see
I would have no experience on visualizing it but i got a few ideas. Like spawning and animated somethe between each component in the direction of flow.
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we’re on the same page with inputs & outputs…
i think an electrical circuit simulator could be designed along the principles of “Flow-Based Programming”, which is what Blueprints is.
i’m not sure how this would be setup, but each electrical component would be a “node” and you connect them with wires, but the power source node doesn’t switch on unless it gets an input.
I mean, necroing a thread is always fun
Circuit simulation is a solved problem.
Each component connects each of its terminals to a “network” which is like a bit of wire with zero resistance/inductance/capacitance.
The CPU then solves the equation for the component using very small time steps (1-10 megaherts step rate is not uncommon) by iterating over all components current state, solving the differential equation for its transfer/response function, and updating the next step version of the component.
The only thing to look out for is that all components should READ the input from the current step, but WRITE the output to the next step, so you’ll typically need to double buffer this state.
Or you can just download an open source circuit simulator like NGSpice and drop that in. Seems easier
nice, i was looking for some kind of packaged solution, but as i couldn’t find one for UE i ended up here making a suggestion, which sounds like what you described. so, I’ll check out NGSpice, and search github for C++ libraries to implement a simulator.
this got me thinking about generally implementing “flow-based programming” in-game, maybe as a plugin. i recently saw gameplay of a game about building a factory, weaving a web of conveyor belts and machines, and it struck me as an implementation of FBP as playable game logic. there are several types of (mini-)games that could be built as in-game objects representing custom nodes and connectors… electricity, plumbing, conveyors, train tracks, etc.
There is indeed a whole “simulation” sub-genre of games based on that paradigm, yes!
Satisfactory is a nice Unreal Engine based factory simulator; Factorio is the classic 2D based game in the same vein.
Most of the “sim city” / “cities skylines” type games use a similar system for water, electricity, and roads.
On Roblox, there are many “miner tycoon” type games that do something similar, one way or another.
And then there are Zachtronics games like SpaceChem and Opus Magnum and Shenzhen I/O and …
Anyway, NGSpice isn’t “integrated with Unreal Engine” but it is open source, and you can probably break into the library in an appropriate place, to configure it with the components/models you want, and run some number of simulation iterations per game frame. Most third party libraries can be made to work by starting them in another thread, and pushing a little message queue of some sort into the middle, where you can post a message from the game to read state out, or control how they otherwise run.
1 Like | 1,269 | 5,886 | {"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-2023-06 | latest | en | 0.900898 |
https://theconfidencemag.com/how-to-evaluate-a-deep-learning-model/ | 1,718,237,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00459.warc.gz | 526,643,573 | 56,060 | There are several ways to evaluate the performance of a deep learning model. Some of the most common metrics include accuracy, precision, recall, F1 score, and MSE (mean squared error). Accuracy is simply determined by whether the algorithm correctly classifies all instances in the testing set.
Precision measures how many samples the algorithm correctly identifies as being positive for the outcome variable. Recall refers to how well the algorithm performs at identifying negatives or false positives. The simple math formula for precision and recall is:
Precision = true Positives / (true Positives + false Negatives)
Recall = true Positives / (true Positives + false Negatives)
The F-score metric combines both precision and recall into one value using the equation below:
2 * Precision * Recall/(Precision+Recall)
MSE is an average measure that takes into account the difference between each instance’s predicted label and its actual label. It can be thought of as the mean square deviation of the prediction from the truth.
## Look at the source code
It is important to look into the source code of your model to determine how well it performed. If you are able to easily manipulate the parameters or run the trained algorithm through different test samples, then it proves that the model was capable of performing its task.
Interpreting the results of the source code can also help identify whether the model under-performed or over-performed. For instance, if the model’s performance decreases when there were no changes to the input materials, this may indicate that the model does not generalize very well.
By looking at both the accuracy and complexity of the models, we are able to evaluate their effectiveness. Although more complex models usually take longer to train, they will always perform better than simpler ones!
Product recommendations are one of the most essential functions in any online platform. By using ML algorithms to mine for potential products, sites like Amazon and Netflix have been able to provide high quality content to their users.
## Look at the output
A very important part of evaluating a deep learning model is looking at what the model outputs. Does it have clear, unambiguous answers? If not, then you will need to evaluate whether this makes sense for the task.
The most common way to do this is by comparing the results with an external standard – typically the current state-of-the-art method.
If there’s no such standard, you can instead compare against data in how well the model performs on similar tasks and datasets. This can be done either within the same domain or across different domains (for example, if your model was designed to identify cats, you could look to see how well it works on other animals).
By doing both of these comparisons, you can get a good picture of whether the model worked for the given task and set parameters. You may also want to try altering those settings to see if the model responds in a useful way.
## Run the model on a test set
Now that you have your accuracy numbers, it is time to evaluate how well these models work! While there are several ways to do this, one of the most common is to run the model on an independent test dataset. This way, you can assess how well the model works overall without being influenced by either over- or underfitting.
By running the model on our validation set, we were able to determine whether or not the model was fitting in when it predicted “Yes” for questions about Donald Trump and Vladimir Putin (which it did not). We also determined whether or not the model was correctly identifying people as political allies or enemies (it seems to be working quite well on that front!).
However, evaluating the performance of a deep learning model on another set of data raises other issues.
## Use an evaluation metric to determine how good the model is
An evaluation metricis a way to measure how well your model performed. There are many different metrics that can be used, with no one perfect solution!
The most common ones are accuracy, precision, recall, F1 score, and MSE (mean squared error). Accuracy measures whether the model’s predictions match the actual label for each example in the test set.
Precisionmeasures how often the model predicts the correct classfor examples that it does not correctly predict as belonging to another category. Recall means how oftenthe model predicts the right classes for instances of the target concept.
And finally, the F1score adds both accuracy and recallto produce a more balanced number compared to either individual metric alone.
The final metric – MSE– tries to find the average mean square deviation between the true labelsand the predicted labels. Because this value will always be higher than accuracy or other similar metrics, thisone is sometimes considered less important than the others.
However, we should use MSE only if there is enough variabilityin the data-set.
## Use the model’s accuracy
Accuracy is one of the most important metrics in determining how well a neural network performed its job. Accuracy is the ratio of items that are correctly identified by the model to the total number of examples the model was trained with.
The more instances of the item being classified, the higher the accuracy. A common way to measure accuracy is using the precision-recall curve. The recall is calculated as the percentage of times the model identifies an instance of the item (true positive). Precision is the proportion of time it correctly does not identify something that is not part of the category (false negative) over all cases when predicting which categories do not belong to the given example (negative predictive value or NPV).
By taking both these values together, you get the precision-recall curve, where the area under the curve (AUC) represents the accuracy of the model. An ideal score for this metric is 1, meaning every item is always predicted as either belonging to the class or not, and the proportions are perfect. This is impossible due to limited data, but we can make sure the models perform reasonably well.
## Look at the confusion matrix
The term “confusion matrix” comes from machine learning, where it is used to evaluate how well an algorithm performs. A confusion matrix is like a Venn diagram for numbers. It has two sets of circles that are connected by rows and columns. The top row and column contain either 0 or 1 values, while the other rows and columns have a value in them.
The number in each cell represents whether the model gave that result often (1) or rarely (0). For example, if there was one case when the model predicted something wrong, then the proportion of times it got that prediction right is zero percent! More typically, however, the model gets the prediction correct most of the time, which results in a higher percentage than zero.
A confusion matrix with only ones and zeros would not tell you much because it does not give you any information about how frequently the model makes false predictions. Therefore, we need some proportions too so that we can determine how likely it is for the model to make a mistake. These proportions are called accuracy measures.
There are many different ways to calculate these accuracies, but they all depend on the same thing: what the model is being tested against. When doing research, you will find several variations of accuracy metrics depending on what type of test set the models were evaluated on.
This article will go into more detail about these tests and accuracy metrics, as well as talk about why some are better than others.
## Look at the output of each class
A common beginner mistake is thinking that if a model produces a high score for one category, then it must be better than another model which does not produce as strong a score for that same category.
This isn’t always the case!
There are several reasons why this might occur. It could be because the models were trained on different datasets with differing numbers in each category. Or maybe there was no such thing as a perfect match for that category during training, so both models struggled to make a good prediction.
Another reason could be due to computational precision. The model you evaluated may have used very specific mathematical techniques or features which made it difficult for it to apply those concepts to your test cases.
By looking into the details of individual classes instead of just the overall accuracy, we can avoid making assumptions like these that may not hold up under scrutiny.
Interpretation of individual categories is also much more straightforward than trying to figure out what makes an overall accurate classification.
## Use an online judge to get peer feedback
An increasingly popular way to evaluate machine learning models is using what’s called an open-source, publicly available model. This allows for comparisons across different models because anyone can inspect the code and determine whether or not it works well.
There are several sites that offer this feature. Some of the most common ones include Kaggle, GCP, and GitHub. By submitting your own project or testing another person’s project, you can quickly gain insights into how effective their model is and if it works better than yours.
These websites also typically have rules set up to ensure fairness as far as data goes. For example, people may be allowed to use only locally available data sets or may be required to add extra labels to make sure the model can identify those new categories.
Overall, these types of evaluations are very helpful in the field since they create an unbiased comparison.
Caroline Shaw is a blogger and social media manager. She enjoys blogging about current events, lifehacks, and her experiences as a millennial working in New York. | 1,898 | 9,871 | {"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-2024-26 | latest | en | 0.92578 |
https://testbook.com/question-answer/the-area-of-an-equilateral-triangle-is-3-w--5cee9a7afdb8bb0f432c4364 | 1,642,954,252,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00333.warc.gz | 598,561,562 | 32,003 | # The area of an equilateral triangle is √3. What is the perimeter of the triangle?
This question was previously asked in
GATE IN 2018 Official Paper
View all GATE IN Papers >
1. 2
2. 4
3. 6
4. 8
Option 3 : 6
Free
CT 1: Ratio and Proportion
4929
10 Questions 16 Marks 30 Mins
## Detailed Solution
Let the side of an equilateral triangle is ‘a’.
The area of an equilateral triangle of side ‘a’ is given by
$$A = \frac{{\sqrt 3 }}{4}{a^2}$$
In the question, it is given that the area is √3
$$\Rightarrow \sqrt 3 = \frac{{\sqrt 3 }}{4}{a^2}$$
⇒ a = 2
The perimeter of an equilateral triangle = 3a = 6 | 207 | 607 | {"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} | 4.1875 | 4 | CC-MAIN-2022-05 | latest | en | 0.855745 |
https://www.askiitians.com/forums/Algebra/22/23521/sequence-and-series.htm | 1,709,537,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00812.warc.gz | 661,063,885 | 44,362 | # ABCD is a square. The midpoints of the sides are joined to form a square A1,B1,C1,D1. The same way we form another square A2,B2,C2,D2 from A1,B1,C1,D1. This process of forming squares is continued. Question: The ratio of the areas of the circum circle and area of the square is ? A. Depends on the radius of the circle. B. Independent of the radius of circle. If A why ? If B why?
420 Points
13 years ago
Dear Vivek
Question is not clear.i.e. ratio of which circumcircle and which square beacause as mentioned in the question there are many squares ABCD,A1B1C1D1, A2B2C2D2..etc.
All the best.
AKASH GOYAL
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
509 Points
13 years ago
let the side of outermost square is a....
after drawing the diagram it is clear that radius of circumcircle is 1/21/2 times the side of square ....
side of inner square is 1/21/2 times side of outer square.....
now if the side of outermost square is a then we have
a1,a2,a3,a4....... = a , a/21/2 , a/2 , a/2.21/2 and so on...
r1,r2,r3,r4........ = a/21/2 , a/2 , a/2.21/2 , a/4 and so on
now
area's of circumcircle is ,
Ac = pi(r12 + r22 + r32 ............infinity)
=(pi)a2 (1/2 +1/4 +1/8 +1/16 .....infinity) (sum of infinite GP is a/a-r)
=(pi)a2 ............1
area's of square is ,
As =a2 (1 + 1/2 + 1/4 + 1/8 ...........infinity)
As = 2a2 ...............2
dividing 1 and 2
eq1/eq2=pi/2 independent of radius of circle...
hence option B is correct | 544 | 1,805 | {"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 | 4 | CC-MAIN-2024-10 | latest | en | 0.827831 |
http://mathhelpforum.com/calculus/67238-help-needed-finding-second-derivative-print.html | 1,519,022,602,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812405.3/warc/CC-MAIN-20180219052241-20180219072241-00656.warc.gz | 231,531,784 | 3,260 | # Help needed finding the second derivative.
• Jan 7th 2009, 05:44 PM
Fallen186
Help needed finding the second derivative.
Find the second derivative
$
f(x) = ((x-2)^3)/(x^2)
$
Attempt:
f'(x)/g(x) = (f'*g-g'*f)/(g^2)
$
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)
$
f(x)/g(x) = (f'*g-g'*f)/(g^2)
f(x)*d(x) = f'd + d'f
f''(x) = ((6(x-2)*x^2+2x*(3*(x-2)^2) -(2*((x-2)^3)+2x*(3(x-2)^2))*(x^4)-(4x^3)*((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^8)
*Sorry the math thing can't fit the size of this
Simplifying doesnt give me the right answer. The correct answer is $24(x-2)/(x^4)$.
How do i do this?
• Jan 7th 2009, 05:51 PM
Mathstud28
Quote:
Originally Posted by Fallen186
Find the second derivative
$
f(x) = ((x-2)^3)/(x^2)
$
Attempt:
f'(x)/g(x) = (f'*g-g'*f)/(g^2)
$
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)
$
I do it again
f(x)/g(x) = (f'*g-g'*f)/(g^2)
f(x)*d(x) = f'd + d'f
f''(x) = ((6(x-2)*x^2+2x*(3*(x-2)^2) -(2*((x-2)^3)+2x*(3(x-2)^2))*(x^4)-(4x^3)*((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^8)
*Sorry the math thing can't fit the size of this
Simplifying doesnt give me the right answer. The correct answer is $24(x-2)/(x^4)$.
How do i do this?
This looksl like a prime candidate for logarthimic differenatiation.
Since $f(x)=\frac{(x-2)^3}{x^2}\implies \ln f(x)=\ln\left(\frac{(x-2)^3}{x^2}\right)=3\ln(x-2)-2\ln(x)$
Differentiating both sides gives
$\frac{f'(x)}{f(x)}=\frac{3}{x-2}-\frac{2}{x}\implies f'(x)=f(x)\left\{\frac{3}{x-2}-\frac{2}{x}\right\}$
But remember that $f(x)=\frac{(x-2)^3}{x^2}$
So $f'(x)=\frac{(x-2)^3}{x^2}\left\{\frac{3}{x-2}-\frac{2}{x}\right\}$
• Jan 8th 2009, 12:40 AM
Opalg
Quote:
Originally Posted by Fallen186
Find the second derivative
$
f(x) = ((x-2)^3)/(x^2)
$
Attempt:
f'(x)/g(x) = (f'*g-g'*f)/(g^2)
$
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)
$
That is correct, but you should simplify it before differentiating it again.
$f'(x) = \frac{3(x-2)^2x^2-2x(x-2)^3}{x^4} = \ldots = \frac{(x-2)^2(x+4)}{x^3}$
Now differentiate, either by using the quotient rule again, or (preferably) by logarithmic differentiation as in Mathstud28's suggestion. | 941 | 2,114 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-09 | longest | en | 0.658601 |
https://eurekamathanswers.com/factorization-of-expressions-of-the-form-a-cube-plus-b-cube/ | 1,638,557,273,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00409.warc.gz | 319,207,063 | 10,029 | Learn how to factorize Expressions of the Form a3 + b3 by referring to the entire article. Here we will use the concept of algebraic identity to factor the given expressions. Factorizing is the process of writing an expression into a product of two expressions which results in the given expression.
Know how to factorize the expressions in the form or can be put in the form a3 + b3 with the step-by-step explanations provided. Check out the Problems on Factorization of Expressions of the form a3 + b3 and learn the problem-solving approach used.
Also, Check:
## Problems on Factorization of Expressions a^3 + b^3
Example 1.
Factorize x3 + 64y3?
Solution:
Given Expression = x3 + 64y3
=(x)3+(4y)3
=(x+4y){(x)2 – (x)(4y) + (4y)2}
=(x+4y)(x2-4xy+16y2)
Example 2.
Factorize a6 + b6?
Solution:
Given Expression = a6 + b6
=(a2)3 + (b2)3
= (a2 + b2){(a2)2 – a2 ∙ b2 + (b2)2}
= (a2 + b2)(a4 – a2b2 + b4)
Example 3.
Factorize 1 + 216z3?
Solution:
Given Expression = 1 + 216z3
=13 + (6z)3
= (1 + 6z{12 – 1 ∙ 6z + (6z)2 }
=(1 + 6z)(1 – 6z +36z2)
Example 4.
Factorize 27x3 + 1/x3?
Solution:
Given Expression = 27x3 + 1/x3
=(3x)3+(1/x)3
=(3x+1/x)((3x)2-3x.1/x+(1/x)2)
=(3x+1/x)(9x2-3+1/x2)
Example 5.
Factorize a6 + 1?
Solution:
Given Expression = a6 + 1
=(a2)3+(1)3
=(a2+1)((a2)2-a2.1+(1)2)
=(a2+1)(a4-a2.1+1)
Example 6.
Factorize the Expression 125u3 + 64v3?
Solution:
Given Expression = 125u3 + 64v3
= (5u)3+(4v)3
=(5u+4v)((5u)2-5u.4v+(4v)2)
=(5u+4v)(25u2-20uv+16v2) | 607 | 1,468 | {"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} | 4.875 | 5 | CC-MAIN-2021-49 | latest | en | 0.698907 |
https://www.teacherspayteachers.com/Product/Addition-and-Subtraction-envision-Ten-Frame-Common-Core-and-MAFS-1459842 | 1,490,504,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189092.35/warc/CC-MAIN-20170322212949-00502-ip-10-233-31-227.ec2.internal.warc.gz | 962,636,092 | 26,385 | Total:
\$0.00
# Addition and Subtraction envision Ten Frame Common Core and MAFS
Subjects
Resource Types
Common Core Standards
Product Rating
3.9
File Type
PDF (Acrobat) Document File
55.04 MB | 30 pages
### PRODUCT DESCRIPTION
This math pack includes Tens frame activities worksheets and Work Mats with counters for exploring the number ten.
Print and laminate mats for longer use. Students can use dry markers to write down the missing part or use ocean counters I have provided. Use it as a fun center game. Students can use counters to complete equation. Mats have an addition and subtraction equation blanks. Can be uses whole group teaching or small group. Teacher can use it for intervention. It can also be used as an assessment.
This file goes along with Envision Math series Topic 1, 2, and 3.
This pack consists of one Mat design in both full color and black and white. There are six different work sheets with missing addend or whole. There is a blank worksheet to create your own problem. Pack comes in bright color border with white background. Cute monkey and banana clip art are on each card. Two options of counters: bananas or palm trees.
Remember this is intended for one classroom only. If you enjoy my product please consider following me on Teacher Pay Teachers store and rate my down load.
Please email me if you have any question. I can adjust any part of the product to meet the needs of your students. Thanks to Scrapadottles for their wonderful clip art.
Packet Includes:
2 sheets of ocean theme counters (four designs)
2 Standards poster ( common core and another MAFS)
12 MATS with addition and subtraction equation fill ins
2 blank- Mat full color no graphics and one in black and white
3 Worksheet each with an answer key – color and black and white
Standards Aligned:
MAFS. 1.OA.2.4
Understand subtraction as an unknown addend problem. For example, subtract 10 -8, by finding the number that makes 10 when added to 8. Cognitive Complexity: Level 2: Basic Application of Skills & Concepts
MAFS. 1.OA.4.8
Determine the unknown whole number in an addition or subtraction equation relating to three whole numbers.
For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 10, , 5 + 5 = 10
Cognitive Complexity: Level 2: Basic Application of Skills & Concepts
Common Core Standards:
Common Core: 1.O.A.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10
CCSS.Math.Content.K.CC.A.3 Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects).
CCSS.Math.Content.K.CC.B.5 Count to answer “how many?” questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1–20, count out that many objects.
Be sure to check out the preview file to see what's included!
Total Pages
30
Included
Teaching Duration
45 Minutes
### Average Ratings
3.9
Overall Quality:
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3.9
Total:
20 ratings
\$4.00
User Rating: 4.0/4.0
(616 Followers)
\$4.00 | 778 | 3,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-13 | longest | en | 0.88332 |
http://www.loisterms.com/qckquiz2.htm | 1,490,420,879,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188824.36/warc/CC-MAIN-20170322212948-00304-ip-10-233-31-227.ec2.internal.warc.gz | 587,321,770 | 2,196 | ## HEY STUDENTS! CHECK OUT YOUR KNOWLEDGE OF THE RULES FOR INTEGERS.
In each set of equations, two of the equations are false and one of the equations is true. Click on the number of the equation that is true.
Question 1:
`#1 +2 + -8 = -6#2 -2 + +8 = -6#3 -2 + -8 = -6`
Question 2:
`#1 -4 - +5 = -1#2 -4 - -5 = -1#3 +4 - +5 = -1`
Question 3:
`#1 +4 x +4 = -16#2 -4 x -4 = -16#3 -4 x +4 = -16`
Question 4:
`#1 -24 / +6 = +4#2 -24 / -6 = +4#3 +24 / -6 = +4`
Question 5:
`#1 +11 + -4 = +7#2 -11 + -4 = +7#3 -11 - -4 = +7`
Question 6:
`#1 +15 - +6 = -9#2 -15 + -6 = -9#3 -15 - -6 = -9`
SCORING
Score one point for each question that you got correct.
1 - 2 points - You're are just beginning. Go to Issue 21 to read about the rules for integers.
3 - 5 points - Check out Issue 21and keep working. You'll get there!
6 points - You've mastered the rules for integers. Congratulations!
I hope that you have enjoyed this Quickie Quiz. Keep looking for more quizzes in the future! | 452 | 1,103 | {"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} | 3.5 | 4 | CC-MAIN-2017-13 | longest | en | 0.891371 |
https://www.cise.ufl.edu/research/sparse/matrices/Meszaros/p0201.html | 1,716,521,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00318.warc.gz | 607,729,535 | 2,303 | Matrix: Meszaros/p0201
Description: linear programming problem, C. Meszaros test set
(bipartite graph drawing)
• Home page of the UF Sparse Matrix Collection
• Matrix group: Meszaros
• Click here for a description of the Meszaros group.
• Click here for a list of all matrices
• Click here for a list of all matrix groups
• download as a MATLAB mat-file, file size: 3 KB. Use UFget(1744) or UFget('Meszaros/p0201') in MATLAB.
• download in Matrix Market format, file size: 6 KB.
• download in Rutherford/Boeing format, file size: 4 KB.
Matrix properties number of rows 133 number of columns 334 nonzeros 2,056 structural full rank? yes structural rank 133 # of blocks from dmperm 1 # strongly connected comp. 1 explicit zero entries 0 nonzero pattern symmetry 0% numeric value symmetry 0% type integer structure rectangular Cholesky candidate? no positive definite? no
author editor C. Meszaros date 2004 kind linear programming problem 2D/3D problem? no
Additional fields size and type b full 133-by-1 c full 334-by-1 lo full 334-by-1 hi full 334-by-1 z0 full 1-by-1
Notes:
```http://www.sztaki.hu/~meszaros/public_ftp/lptestset
Converted to standard form via Resende and Veiga's mpsrd:
minimize c'*x, subject to A*x=b and lo <= x <= hi
```
Ordering statistics: result nnz(V) for QR, upper bound nnz(L) for LU, with COLAMD 12,446 nnz(R) for QR, upper bound nnz(U) for LU, with COLAMD 2,888
SVD-based statistics: norm(A) 287.419 min(svd(A)) 1 cond(A) 287.419 rank(A) 133 sprank(A)-rank(A) 0 null space dimension 0 full numerical rank? yes
singular values (MAT file): click here SVD method used: s = svd (full (A)) ; status: ok
For a description of the statistics displayed above, click here.
Maintained by Tim Davis, last updated 12-Mar-2014.
Matrix pictures by cspy, a MATLAB function in the CSparse package.
Matrix graphs by Yifan Hu, AT&T Labs Visualization Group. | 533 | 1,887 | {"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-2024-22 | latest | en | 0.632041 |
https://elsner.edu.pl/en/swimming-pace-chart.html | 1,701,907,939,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00755.warc.gz | 258,803,861 | 5,599 | # Swimming Pace Chart
Swimming Pace Chart - Male 1 mile swimming times. Web the swim pace calculator simplifies the process of determining the required pace for various swimming distances. Web speed charts are a great tool to motivate athletes when they are doing race pace training. Web a good 1 mile time is 32:24. Web fast = zone 5 or 6 (example: Web this calculator can help you determine your pace, time, and distance based on your goals and performances. Web swim pace chart do you need to know the pace you need to maintain to finish a distance in a certain amount of time? A good 1 mile time for a man is. Listed below are speed charts. Web the pace calculator quickly works out the swimming pace, time and distance depending on what is required.
### Web So If You Swim 1,000M In 21 Minutes And 58 Seconds, Your Pace Would Be 2:12 Per 100M (Which Is 2.7Km/H, Or.
Web a swimming pace calculator is a tool used to determine the pace (speed) of a swimmer based on the distance and time it takes to. Web enter your time for 400 (meters or yards) step 1. Web swim pace chart do you need to know the pace you need to maintain to finish a distance in a certain amount of time? Web this calculator can help you determine your pace, time, and distance based on your goals and performances.
### The Chart Provides General Paces In Minutes And Seconds For Set Distances From 100 Yards.
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Related Post: | 531 | 2,309 | {"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-2023-50 | latest | en | 0.893544 |
https://stats.stackexchange.com/questions/377816/limiting-distribution-of-a-ratio-using-basus-theorem | 1,713,312,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00482.warc.gz | 496,119,078 | 40,952 | # Limiting distribution of a ratio using Basu's theorem
Edit: there's seems to be a typo in original question.
This is a past exam question that I'm trying to solve. Suppose that $$X_1,\ldots, X_n$$ are i.i.d. Uniform (0, $$\theta$$) random variables. Let $$X_{n:n} = \max_{1 \leq i \leq n} X_i.$$
Find the limiting distribution of: $$\frac{2\sum_{i=1}^n X_i - \theta}{\sqrt{n}X_{n:n}},$$ as $$n \rightarrow \infty$$. Here's what I've been able to do:
• 1) Can show that the limiting distribution of $$X_{n:n}$$ is $$\theta$$ by working on the CDF of $$X_{n:n}$$ as $$n$$ tends to infinity;
• 2) Using Basu's Theorem, I was able to show that $$\sum_{i=1}^n X_i/\theta$$ is independent of $$X_{n:n}$$ since the distribution of $$\sum_{i=1}^n X_i/\theta$$ is independent of $$\theta$$ and is thus an ancillary statistic. I was also able to show that $$X_{n:n}$$ was a boundedly complete sufficient statistic.
I can thus work on the initial term to get the following: $$\frac{2\sum_{i=1}^n X_i - \theta}{\sqrt{n}X_{n:n}} = \frac{\theta}{X_{n:n}}\frac{\left(2 \sum_{i=1}^n X_i/\theta - 1\right)}{\sqrt{n}}.$$ From 1), I know that $$\theta/X_{n:n}$$ converges in distribution to 1. But, for the second term $$\frac{\left(2 \sum_{i=1}^n X_i/\theta - 1\right)}{\sqrt{n}}$$, I'm not able to find a limiting distribution. I tried to use the CLT, but without any success. Is my work so far wrong? Thanks for your help.
• the scaling of the sum is incorrect... and the call to Basu's theorem does not seem right, since $\sum_{i=1}^{n}X_{i}/\theta$ is not a statistic, being also a function of $\theta$. Nov 19, 2018 at 19:10
• What do you mean the scaling of the sum is incorrect? I don't see my mistake, I'm sorry. You have a good point as for the call of the Basu's theorem, $\sum_{i=1}^{n} X_i/\theta$ can't be a statistic since it depends on $\theta$. I'm quite lost on what I should do. How would go about computing the following expectation: $E(\sum_{i=1}^{n}X_i/X_{n:n}).$ I used a similar method than in my question in order to be able to use Basu's theorem. Nov 19, 2018 at 19:28
If one applies the CLT to $$\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$$it implies that $$\sqrt{n}\left(\bar{X}_n-\frac{\theta}{2}\right)\longrightarrow \mathcal{N}(0,\theta^2/12)$$ Further,$$\frac{X_{n:n}}{\theta}\stackrel{\text{a.s.}}{\longrightarrow}1$$This should be enough to conclude. I believe the question as expressed $$\frac{2\sum_{i=1}^n X_i - \theta}{\sqrt{n}\,X_{n:n}}$$contains a typo and it should instead be $$\frac{\sum_{i=1}^n (2X_i - \theta)}{\sqrt{n}\,X_{n:n}}$$
• Thank you for your response. I'm sorry but I don't understand how this solves the problem. I understand that the CLT implies the convergence in distribution of $\sqrt{n}\left(\bar{X}_{n} - \theta/2\right)$ to a gaussian of zero mean and variance of $\theta^{2}/12$. What I don't understand is how to use the CLT in my problem. For example: $\frac{\left(2 \sum_{i=1}^{n}X_i - \theta \right)}{\sqrt{n}X_{n:n}} = ... = \frac{\sqrt{n}2\left( \bar{X}_{n} - \theta/2n\right)}{X_{n:n}}$. My issue is with the term $\theta/2n$, I'm not able to get only $\theta/2$. Nov 19, 2018 at 22:54 | 1,056 | 3,143 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.922166 |
https://www.clutchprep.com/chemistry/practice-problems/73861/how-many-moles-are-contained-in-each-number-of-grams-of-fructose-c6h12o6-molar-m | 1,643,465,063,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00082.warc.gz | 735,237,313 | 21,214 | # Problem: How many moles are contained in each number of grams of fructose (C 6H12O6 molar mass 180.2 g/mol), a carbohydrate that is about twice as sweet as table sugar? "Lite" food products use half as much fructose as table sugar to achieve the same sweet taste, but with fewer calories. Enter your answer in scientific notation. a. 19.0 g b. 0.0820 g a. x 10 molb. x 10 mol
###### FREE Expert Solution
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###### FREE Expert Solution
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###### Problem Details
How many moles are contained in each number of grams of fructose (C 6H12O6 molar mass 180.2 g/mol), a carbohydrate that is about twice as sweet as table sugar? "Lite" food products use half as much fructose as table sugar to achieve the same sweet taste, but with fewer calories. Enter your answer in scientific notation.
a. 19.0 g b. 0.0820 g
a. x 10 mol
b. x 10 mol
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Our tutors have indicated that to solve this problem you will need to apply the Mole Concept concept. You can view video lessons to learn Mole Concept Or if you need more Mole Concept practice, you can also practice Mole Concept practice problems .
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Based on our data, we think this problem is relevant for Professor Hanson's class at WMU. | 430 | 1,854 | {"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-2022-05 | latest | en | 0.87519 |
https://goprep.co/ex-5.a-q28-in-the-given-figure-abc-is-an-equilateral-i-1njpu0 | 1,606,933,948,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141715252.96/warc/CC-MAIN-20201202175113-20201202205113-00176.warc.gz | 314,750,124 | 43,034 | # In the given figure, ABC is an equilateral triangle, PQ ‖AC and AC is produced to R such that CR=BP. Prove that QR bisects PC.
Given: ABC is an equilateral triangle, PQ AC and CR=BP
To prove: QR bisects PC or PM = MC
Proof:
Since, ∆ABC is equilateral triangle,
A = ACB = 60°
Since, PQ AC and corresponding angles are equal,
BPQ = ACB = 60°
In ∆BPQ,
B= ACB = 60°
BPQ = 60°
Hence, ∆BPQ is an equilateral triangle.
PQ = BP = BQ
Since we have BP = CR,
We say that PQ = CR …(1)
Consider the triangles ∆PMQ and ∆CMR,
PQM = CRM …alternate angles
PMQ = CMR … vertically opposite angles
PQ = CR … from 1
Thus by AAS property of congruence,
∆PMQ CMR
Hence, we know that, corresponding parts of the congruent triangles are equal
PM = MC
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https://answerofmath.com/solved-how-to-test-if-some-data-points-is-drawn-from-a-distribution-with-linear-pdf/ | 1,696,255,847,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00843.warc.gz | 109,494,734 | 21,000 | # Solved – How to test if some data points is drawn from a distribution with linear PDF
I have some data in the range [0, 1], and from the histogram below, it seems that they might be drawn from a distribution with linear probability density function (what's the name of that kinds of distribution?). How do I estimate the parameters of that distribution and how do I test how likely the data were drawn from it in R?
Update:
Now I know (thanks to Glen_b) if I can define the PDF (e.g., \$ P(x) = ax + b\$), I can test it with Kolmogorov-Smirnov test, but how do I estimate \$a\$ and \$b\$ from the sample?
Contents
At least as a rough approximation, one might regard that as having a pdf that increases linearly from 0:
(However, you should check whether there's actually a spike at exactly 1 – and perhaps a small one at exactly 0. Are the values near 1 all less than 1, or are there a number at exactly 1? Similarly with 0.)
Such a linear-increasing pdf might be regarded as a special case of the triangular distribution or as a particular beta distribution (a beta(2,1)).
You can do a hypothesis test for a fully-specified distribution using a Kolmogorov-Smirnov test (for example; there are other choices). In R, that's `ks.test`.
`` f <- function(x) pbeta(x,2,1) ks.test(x,f) One-sample Kolmogorov-Smirnov test data: x D = 0.1309, p-value = 0.3291 alternative hypothesis: two-sided ``
However, two caveats:
1) If you chose the shape to test for based on the same data you run the test on, the p-values are pretty much meaningless.
2) In any case, even if that wasn't at issue, a formal hypothesis test is usually not what you want if your question is "is it reasonable to use this as a model?" — it mostly answers the wrong question.
Alternatively, perhaps you might have regarded the pdf as starting not from 0 but from something higher than 0:
That's not a triangle, but a trapezoid (often called a trapezium if you're not in the US).
Both the triangle I initially drew and this are linear of course.
if I can define the PDF (e.g., P(x)=ax+b), I can test it with Kolmogorov-Smirnov test, but how do I estimate a and b from the sample?
1) if you intended a triangular pdf (with peak at the right) you have a=2 and b=0
2) If you instead assume that it's a trapezoid(/trapezium) as drawn above there aren't two free parameters, but 1 (think instead of the height of the density at 0.5 and the slope – the height at 0.5 is restricted to be exactly 1, leaving you only with the slope)
3) The K-S test is for fully specified distributions (the triangle I had thought you mean was fully specified, now if we're dealing with the trapezoid above, you're estimating a parameter). If you use the same test statistic as the Kolmogorov-Smirnov, you are now doing what's called a Lilliefors test and you'll need to simulate to obtain the distribution of the test statistic.
4) you said you had spikes at exactly 0 and 1; this trapezoid-shaped model we're now discussing doesn't. If you have spikes in your data, a model without them won't fit. But perhaps you mean to have spikes as well as some linear density in between. The density in between doesn't look to me to be particularly linear though (which is why I suggested a beta model), but it might do well enough.
Let's imagine that we condition only on the data between 0 and 1 and we're fitting that trapzoidal-shaped (linear pdf) model, \$f(x) = 1+beta(x-frac{_1}{^2}),,quad -2leqbetaleq 2\$.
(The term "trapezoidal distribution" usually means a shape with the parallel sides parallel to the x-axis, not the y-axis as here. So perhaps I should stick with your 'linear' characterization after all.)
This parameter may be estimated in a number of ways (the most common approaches would be method of moments and maximum likelihood estimation).
Let's look at MLE.
\$\$cal{L}(beta) = prod_{i=1}^n 1+beta(x_i-frac{_1}{^2})\$\$
\$\$log(cal{L}(beta))=cal{l}(beta) = sum_{i=1}^n log(1+beta(x_i-frac{_1}{^2}))\$\$
Now this is a nice smooth function and it's quite possible to take the derivative, but we can't solve it for \$hat{beta}\$ in closed form (or at least I don't think so) by setting that derivative to zero. However, we can take our sample, evaluate the log-likelihood at any value of \$beta\$, and so we can use optimization routines to locate the maximum.
The maximum for this sample occurs at about 1.2875. (The actual population value I generated the data from was \$frac{4}{3}approx 1.3333\$.)
Method of moments:
The mean of a random variable with the linear pdf is \$int_0^1 x [ 1+beta(x-frac{1}{2})] dx = [frac{x^2}{2} + beta (frac{x^3}{3} -frac{x^2}{4})]_0^1 = frac{1}{2} + frac{beta}{12}\$
Equating sample and population mean, \$hat{beta}=12(bar{x}-frac{1}{2})\$
This has the very nice advantage of great simplicity, but of course if \$bar{x}\$ is not between \$frac{1}{3}\$ and \$frac{2}{3}\$, this yields an impossible estimate for \$beta\$ (in that the resulting density isn't actually a density).
In the data I used in the MLE example, the method of moments estimate was 1.324 .
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https://www.physicsclassroom.com/Physics-Video-Tutorial/Momentum-and-Collisions/Impulse-Momentum-Change-Theorem | 1,721,368,899,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00653.warc.gz | 788,428,401 | 27,564 | ## Impulse-Momentum Change Theorem Video Tutorial
The Impulse-Momentum Change Theorem Video Tutorial explains how an impulse changes an object's momentum. But what exactly is impulse? And how can it be mathematically related to momentum change in order to understand collisions? Mr. H explains the important connection between impulse and momentum and explains how to use the idea..The video lesson answers the following questions:
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https://openstax.org/books/physics/pages/10-introduction | 1,723,439,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00081.warc.gz | 361,942,156 | 82,405 | Physics
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Figure 10.1 Special relativity explains why travel to other star systems, such as these in the Orion Nebula, is unlikely using our current level of technology. (s58y, Flickr)
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Start a discussion that taps into the longing of humans to explore worlds beyond our planet. Is this basic human nature? Perhaps it is; humans have now been almost everywhere there is to go on this planet. Ask students why we have not traveled to other star systems yet. Is it just a matter waiting a few years for technological advances, or is there a more daunting problem? If no one knows, tell them it all has to do with achievable speeds, and use this as a lead-in to Einstein’s postulate related to the speed of light.
Have you ever dreamed of traveling to other planets in faraway star systems? The trip might seem possible by traveling fast enough, but you will read in this chapter why it is not. In 1905, Albert Einstein developed the theory of special relativity. Einstein developed the theory to help explain inconsistencies between the equations describing electromagnetism and Newtonian mechanics, and to explain why the ether did not exist. This theory explains the limit on an object’s speed among other implications.
Relativity is the study of how different observers moving with respect to one another measure the same events. Galileo and Newton developed the first correct version of classical relativity. Einstein developed the modern theory of relativity. Modern relativity is divided into two parts. Special relativity deals with observers moving at constant velocity. General relativity deals with observers moving at constant acceleration. Einstein’s theories of relativity made revolutionary predictions. Most importantly, his predictions have been verified by experiments.
In this chapter, you learn how experiments and puzzling contradictions in existing theories led to the development of the theory of special relativity. You will also learn the simple postulates on which the theory was based; a postulate is a statement that is assumed to be true for the purposes of reasoning in a scientific or mathematic argument.
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Before students begin this chapter, it is useful to review the following concepts:
• Using significant figures in calculations—Demonstrate how to use the proper number of significant figures when adding and multiplying.
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To reinforce this description, and to open the door for a discussion of frame of reference, take an object, place it in front of the class, and ask someone to describe its motion. Students will likely respond that the object is at rest. Explain that this correct, but it is not the only correct answer. Help students to understand that the object is sitting still but also moving at a high rate of speed as the earth rotates, orbits the sun, etc. It all depends on how you define the frame of reference.
Initiate a discussion aimed at making relativity theory less intimidating. Dispel the misconception that “Only three people in the world understand Einstein’s theories.” Stories like this come about because Einstein’s second relativity theory, called general relativity, was more difficult to understand. In this chapter, we will only learn about special relativity. | 780 | 4,037 | {"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} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.93608 |
https://www.experts-exchange.com/questions/10342753/Loop-That-Pa.html | 1,495,738,296,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608120.92/warc/CC-MAIN-20170525180025-20170525200025-00433.warc.gz | 873,708,482 | 32,081 | Solved
# Loop That Pa
Posted on 2000-05-10
157 Views
Have input for \$fields[11] like so:
ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL
Have the following loop reading input:
while(<IN>){
@anomalies=split(/;/,\$fields[11]);
\$number=0;
\$i=1;
#LIST OF ANOMALIES
\$1="NO"; #ACFT EQUIPMENT PROBLEM/CRITICAL
\$2="NO"; #ACFT EQUIPMENT PROBLEM/LESS SEVERE
\$3="NO"; #ALT DEV/EXCURSION FROM ASSIGNED
\$4="NO"; #ALT DEV/OVERSHOOT ON CLB OR DES
for(\$i;\$i< 4;\$i++){
\$number++;
if(@anomalies[i]="ACFT EQUIPMENT PROBLEM/CRITICAL"){
\$1="YES";
@count[\$number];
}
elsif(@anomalies[i]="ACFT EQUIPMENT PROBLEM/CRITICAL"){
\$2="YES";
@count[\$number];
}
elseif(@anomalies[i]="ACFT EQUIPMENT PROBLEM/LESS SEVERE"){
\$3="YES";
@count[\$number];
}
else(@anomalies[i]="ALT DEV/EXCURSION FROM ASSIGNED"){
\$4="YES";
\$count++;
}
}
print OUT "\$1|\$count[\$number]|\$2|\$count[\$number]|\$3|\$count[\$number]|\$4|\$count[\$number]";
}
The line of input in the example should print
NO|0|YES|2|NO|0|YES|1
How would u correct the code to properly produce the output.
0
Question by:tomatocans
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ID: 2797973
Have input for \$fields[11] like so:
ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL
Have the following loop reading input:
while(<IN>){
@anomalies=split(/;/,\$fields[11]);
\$number=0;
\$i=1;
#LIST OF ANOMALIES
\$1="NO"; #ACFT EQUIPMENT PROBLEM/CRITICAL
\$2="NO"; #ACFT EQUIPMENT PROBLEM/LESS SEVERE
\$3="NO"; #ALT DEV/EXCURSION FROM ASSIGNED
\$4="NO"; #ALT DEV/OVERSHOOT ON CLB OR DES
for(\$i;\$i< 4;\$i++){
\$number++;
if(@anomalies[i]="ACFT EQUIPMENT PROBLEM/CRITICAL"){
\$1="YES";
@count[\$number];
}
elsif(@anomalies[i]="ACFT EQUIPMENT PROBLEM/CRITICAL"){
\$2="YES";
@count[\$number];
}
elseif(@anomalies[i]="ACFT EQUIPMENT PROBLEM/LESS SEVERE"){
\$3="YES";
@count[\$number];
}
else(@anomalies[i]="ALT DEV/EXCURSION FROM ASSIGNED"){
\$4="YES";
@count[\$number];
}
}
print OUT "\$1|\$count[\$number]|\$2|\$count[\$number]|\$3|\$count[\$number]|\$4|\$count[\$number]";
}
The line of input in the example should print
NO|0|YES|2|NO|0|YES|1
How would u correct the code to properly produce the output.
0
LVL 16
Expert Comment
ID: 2797991
given that
\$field[11] contains ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL
& also give that
\$1 looks for ACFT EQUIPMENT PROBLEM/CRITICAL
\$2 looks for ACFT EQUIPMENT PROBLEM/LESS SEVERE
\$3 looks for ALT DEV/EXCURSION FROM ASSIGNED
\$4 looks for ALT DEV/OVERSHOOT ON CLB OR DES
shouldn't the output be....
NO|1|YES|0|NO|1|YES|0
??
i am assuming that the numbers are the count of times that each sub-string has been found in the mail string (\$field[11]).
0
LVL 16
Accepted Solution
maneshr earned 50 total points
ID: 2798020
based on my above assumption, here is a working solution...
NOTE: you cannot assign values yourself to \$1,\$2 etc.. since these are PERL read-only variables.
================tcans.pl
#!/usr/local/bin/perl
\$fields[11]="ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL";
@anomalies=split(/\;/,\$fields[11]);
\$number=0;
\$i=1;
#LIST OF ANOMALIES
\$one="NO"; #ACFT EQUIPMENT PROBLEM/CRITICAL
\$two="NO"; #ACFT EQUIPMENT PROBLEM/LESS SEVERE
\$three="NO"; #ALT DEV/EXCURSION FROM ASSIGNED
\$four="NO"; #ALT DEV/OVERSHOOT ON CLB OR DES
\$count{'one'}=0;
\$count{'two'}=0;
\$count{'three'}=0;
\$count{'four'}=0;
foreach(@anomalies){
if (\$_ eq "ACFT EQUIPMENT PROBLEM/CRITICAL"){
\$one="YES";
\$count{'one'}++;
}elsif (\$_ eq "ACFT EQUIPMENT PROBLEM/LESS SEVERE"){
\$two="YES";
\$count{'two'}++;
}elsif (\$_ eq "ALT DEV/EXCURSION FROM ASSIGNED"){
\$three="YES";
\$count{'three'}++;
}elsif (\$_ eq "ALT DEV/OVERSHOOT ON CLB OR DES"){
\$four="YES";
\$count{'four'}++;
}
}
print "\$one|\$count{'one'}|\$two|\$count{'two'}|\$three|\$count{'three'}|\$four|\$count{'four'}\n";
0
Author Comment
ID: 2798032
No it is the postion in the string not the count.
I apologize.
0
Author Comment
ID: 2798037
The output should return the what position in the string the anomaly appears.
0
Author Comment
ID: 2798051
In the example
ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL
ALT DEV/EXCURSION FROM ASSIGNED is first
ACFT EQUIPMENT PROBLEM/CRITICAL is second
so the ouput should be
YES|2|NO|0|YES|1|NO|0
Sorry about the confusion
0
LVL 12
Expert Comment
ID: 2798107
If I understand your question correctly, this may be what you are looking for:
#!/usr/local/bin/perl
\$lst = "ACFT EQUIPMENT PROBLEM/CRITICAL";
\$lst .= ":ACFT EQUIPMENT PROBLEM/CRITICAL";
\$lst .= ":ACFT EQUIPMENT PROBLEM/LESS SEVERE";
\$lst .= ":ALT DEV/EXCURSION FROM ASSIGNED";
@a = split /:/, \$lst;
\$c[0] = "ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL";
\$c[1] = "ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL";
\$c[1] .= "; ;ALT DEV/OVERSHOOT ON CLB OR DES";
\$c[3] = " ; ; ;ALT DEV/EXCURSION FROM ASSIGNED";
\$rst = "";
%b = ();
# while(<IN>){
for \$j (0..\$#c) {
# @anomalies=split(/;/,\$fields[11]);
@anomalies=split(/;/,\$c[\$j]);
for \$i (0..\$#anomalies) {
if ("\$anomalies[\$i]" eq "\$a[\$i]") { ++\$b{'YES'}[\$i];
} else { ++\$b{'NO'}[\$i]; }
}
}
# }
for \$i (0..3) {
if (\$b{'YES'}[\$i] > 0) {
if (\$rst) { \$rst .= "|YES|" . \$b{'YES'}[\$i];
} else { \$rst = "YES|" . \$b{'YES'}[\$i]; }
} else {
if (\$rst) { \$rst .= "|NO|0"; } else { \$rst = "NO|0"; }
}
}
# print OUT "\$rst\n";
print "\$rst\n";
# The line of input in the example should print
# NO|0|YES|2|NO|0|YES|1
The output is
NO|0|YES|2|NO|0|YES|1
0
Author Comment
ID: 2798130
Sorry this is too hard coded the input
i have has 66,000 lines with
30 types of anomalies.
I agree this is a solution for the
specific example but
I am look for something like mansher
was doing but with position instead
of count.
In your example geotiger I would
need to much hard coding for the \$c[N]
area.
Is there a proper solution within the framework
that Maneshr was exploring. Just substituting position for count.
0
LVL 16
Expert Comment
ID: 2800493
tomatocans,
i have a Q. can the string contain the same substring twice ??
Eg.
ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL;ALT DEV/EXCURSION FROM ASSIGNED
if yes, what should the result be in this case since the sub-string ALT DEV/EXCURSION FROM ASSIGNED appears at position 1 and 3!!
if the answer is no, i have a solution ready for you.
0
Author Comment
ID: 2800528
The answer is not it will not be contained twice
in the substring
0
LVL 16
Expert Comment
ID: 2800531
here is your solution....
i have had to make some very small changes to the original script to make it work.
========tcans.pl
#!/usr/local/bin/perl
\$fields[11]="ALT DEV/EXCURSION FROM ASSIGNED;ACFT EQUIPMENT PROBLEM/CRITICAL";
@anomalies=split(/\;/,\$fields[11]);
\$number=0;
\$i=1;
#LIST OF ANOMALIES
\$one="NO"; #ACFT EQUIPMENT PROBLEM/CRITICAL
\$two="NO"; #ACFT EQUIPMENT PROBLEM/LESS SEVERE
\$three="NO"; #ALT DEV/EXCURSION FROM ASSIGNED
\$four="NO"; #ALT DEV/OVERSHOOT ON CLB OR DES
\$count{'one'}=0;
\$count{'two'}=0;
\$count{'three'}=0;
\$count{'four'}=0;
#foreach(@anomalies){
foreach(\$[ .. \$#anomalies){
if (\$anomalies[\$_] eq "ACFT EQUIPMENT PROBLEM/CRITICAL"){
\$one="YES";
\$count{'one'}=(\$_+1);
}elsif (\$anomalies[\$_] eq "ACFT EQUIPMENT PROBLEM/LESS SEVERE"){
\$two="YES";
\$count{'two'}=(\$_+1);
}elsif (\$anomalies[\$_] eq "ALT DEV/EXCURSION FROM ASSIGNED"){
\$three="YES";
\$count{'three'}=(\$_+1);
}elsif (\$anomalies[\$_] eq "ALT DEV/OVERSHOOT ON CLB OR DES"){
\$four="YES";
\$count{'four'}=(\$_+1);
}
}
print "\$one|\$count{'one'}|\$two|\$count{'two'}|\$three|\$count{'three'}|\$four|\$count{'four'}\n";
0
Author Comment
ID: 2802546
Adjusted points from 38 to 50
0
Author Comment
ID: 2802547
Thanks
0
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# Analyzing the Experiment (Part VI)
Design of Experiments
In the last Blog, we explored the use of contour plots and other tools (such as a response optimizer) to help us quickly find solutions to our models. In this blog, we will look at the uncertainty in these predictions. We will also discuss model validation to ensure that technical assumptions that are inherent in the modeling process is satisfied.
We again start by revisiting the battery life DOE example that was discussed in the previous blog. Recall that previously we used the optimizer to find a solution for the wall thickness that would produce a target battery life of 45. As a reminder, we constrained the solution to use a Lithium battery since this made the response (battery life) insensitive to changes in ambient temperature. In the table below, we see that the wall thickness (uncoded) should be set at 1.44 (mm).
An important question with any model is how well does it predict? Suppose we actually produced a bunch of Lithium batteries with the wall thickness = 1.44. Would we always expect to get a battery life of exactly 45? The answer is NO!. Our model was not perfect (not all variation is explained) and even if it was, the model effects and parameters will based on average responses. The solution we found predicts that the average battery life will be 45.0 if we set wall thickness at 1.44.
At the bottom of the output above, we see both a 95% Confidence Interval (CI) and a 95% Prediction Interval (PI). The confidence interval tells us how much uncertainty exists in the average response. Thus, if we produce a lithium battery with a wall thickness of 1.44 mm, then we could expect that the range 43.475 to 46.525 has a 95% probability of containing the true average response. The prediction interval is always wider because it provided the range over which we may expect individual response values to fall. Thus, if we produce a lithium battery with a wall thickness of 1.44 mm, then we could expect that the range 41.339 to 48.661 has a 95% probability of containing the individual values. This is how much variation we could expect, given the uncertainty in our model. Note that the model uncertainty is both a function of the amount of data used to build the model as well as the experimental error observed in the study.
In the next blog, we will look a bit more at the uncertainty in these predictions. We will also, talk about model validation to ensure that technical assumptions that are inherent in this process are satisfied.
It is important to recognize the degree of uncertainty when using predictive models for making predictions, helping to set specifications, etc.
Next, let’s look at Model Validation. The basic reason for validating the model are summarized in the graphic below. To perform the validation, we calculate the residuals associated with each treatment.
The residuals (errors) are calculated by any DOE software program, but they are not difficult to compute. We’ll look at a simple example.
The experiment above is an experiment with 3 factors (2 levels). The right-most column contains the observed response and the model that was developed is shown at the bottom. (X3 and X1X3 were significant). For each row, we can plug in the actual coded values for X1 and X3 and compute the result, using the model. For the first row, both X1 and X3 are “low” (-1). So, we have:
y-hat = 20 + 4.5(-1) – 2.5(-1)(-1)
= 20 – 4.5 -2.5 = 13
Thus, the predicted value is 13 and the observed value is 13, so the residual is 13-13 = 0. The residuals for each row are calculated similarly.
Once we compute all the residuals, we can determine how large they are and also graph them on various plots to determine if there are any significant violations of the model assumptions. The graphic below summarizes how residuals should behave if our modeling assumptions are satisfied.
Let’s consider some examples of residual plots.
The upper left plot shows a normal probability plot which is used to check if residuals are reasonably described by a normal distribution. Since the points are close to linear on this plot, the normality assumption is satisfied. The upper right shows residuals vs. the model predicted values (fitted value). We are just looking for random values around zero and this one looks fine.
Let’s look at a few examples of model violations. In the plot below (residuals vs. predicted values) the module tends to over-predicts for smaller predicted values and under-predicts for larger predicted values. A valid model should not exhibit this pattern as it should predict similarly across the range of predicted values.
In the same kind of plot below, we see that the variability in residuals is not constant across the range of predicted values. This condition is called heterskedasticity, and means that the size of the model errors changes significantly across the range of predicted values. Constant variance of residuals is a requirement of the modeling method.
Below is a plot of residuals vs. the factor levels for a given factor. Just like across response levels, we shouldn’t see a pattern across factor levels.
Finally, we should look at the residuals vs. the run order of the experiment. Non-random patterns may be indicative of a change occurring during the conduct of the experiment (assuming that the experiment was randomized). For example, in the plot below the first half of the runs look very different than the second half with regard to the predictive ability.
Violations of these rules may indicate non-linear responses, missing important factors, or other issues. Non-constant variances across the range or lack of normality can often be corrected by transforming the response values before developing the predictive model.
In the next blog, we will look at using Monte Carlo simulation with our predictive model to do some further optimization of target responses while considering variation in the factor levels. | 1,249 | 5,990 | {"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} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.925219 |
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In this activity, students examine how different balls react when colliding with different surfaces. Also, they will have plenty of opportunity to learn how to calculate momentum and understand the principle of conservation of momentum.
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Students act as Mars exploratory rover engineers, designing, building and displaying their edible rovers to a design review. To begin, they evaluate rover equipment and material options to determine which parts might fit in their given NASA budget. With provided parts and material lists, teams analyze their design options and use their findings to design their rovers.
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In this hands-on activity rolling a ball down an incline and having it collide into a cup the concepts of mechanical energy, work and power, momentum, and friction are all demonstrated. During the activity, students take measurements and use equations that describe these energy of motion concepts to calculate unknown variables and review the relationships between these concepts.
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In this hands-on activity, students learn about two types of friction static and kinetic and the equation that governs them. They also measure the coefficient of static friction and the coefficient of kinetic friction experimentally.
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# G2ELAB
Model Order Reduction
Partial Report
Mateus Antunes Oliveira Leite
24/04/2015
Partial report containing the theory of model order reduction
1. Once the columns of 𝐁 are linearly independent. However. the matrix is not singular and its inverse exists. Equation (2) can be regarded as change of basis. In Equation (1). 𝛂 = 𝐁−𝟏 𝐯 (3) Projections In this section. 𝐯 = 𝐁𝛂 (2) It must be pointed out that vectors are geometric objects and thus are independent of the basis one chooses the represent them. a simple 3-dimensional example of projection is presented. The columns of matrix 𝐁 are the base vectors and the vector 𝛂 contains the coefficients of each of the base vectors. v e p b1 b2 Ω Figure 1 . some important topics in linear algebra are presented. this relation can be expressed as in Equation (2). Vector basis Any vector in a ℝ𝑛 space can be represented as a linear combination of n linearly independent base vectors. the derived results are general and can be applied directly to an arbitrary number of dimensions.Projection into a 2-dimensional space . The vector 𝐯 is said to be in the column space of B. Therefore. This permits to express the inverse mapping as represented in Equation (3). 𝐛i represents a base vector of ℝ𝑛 with index i and 𝐯 is the vector being composed. The vectors 𝐯 and 𝛂 can be regarded as the same geometric vector but in different basis. 𝐯 = 𝛼1 𝐛𝟏 + 𝛼2 𝐛𝟐 + ⋯ + 𝛼𝑛 𝐛𝐧 (1) In matrix notation. Topics in Linear Algebra In order to introduce the concepts of model order reduction in a self-contained manner.
one should choose a subspace 𝚿 orthogonal to e. the projected vector can be obtained by Equation (7). Thus. This kind of projection has a very important relation to the following optimization problem: find a vector 𝐩 lying in the subspace 𝛀 that best approximates the vector v. the best approximation is obtained when the square of the length of the error is minimized.The Figure 1 represents a projection of the vector 𝐯 into the subspace . . Using the law of cosines. 𝐯 = 𝛀𝛂 + 𝐞 (5) To determine the vector . In the current context. the above relation can be written as in Equation (10).Ψ = 𝛀(𝚿T 𝛀) 𝚿T (8) If both the subspaces 𝛀 and 𝚿 are the same. the error vector represented in Figure 1 is given by Equation (9). obtaining an explicit expression for . This process is shown in (11). p and e is given by Equation (4). as shown in Equation (6). See the appendix for the used math property. The subspace represented by the matrix 𝛀 can be constructed by arranging the base vectors 𝐛𝟏 and 𝐛𝟐 as its columns. the relation among the vectors v. any vector lying in this subspace can be represented by a relation similar to Equation (2). 𝐯=𝐩+𝐞 (4) The fact that 𝐩 is in the column span of 𝛀 allows writing equation (4) as in Equation (5). |𝐞|2 = 𝐯 T 𝐯 − 2𝐯 T 𝐩 + 𝐩T 𝐩 (9) Using the fact that 𝐩 is in the column space of . this projection is called an orthogonal projection. −1 𝚷Ω. one can isolate the vector . Therefore. The resulting or projected vector is 𝐩 and 𝐞 is the different among these two quantities. −1 𝐩 = 𝛀(𝚿T 𝛀) 𝚿T 𝐯 (7) The linear operator representing the oblique projection of the components of 𝐯 that are orthogonal to 𝚿 into 𝛀 can be extracted directly from this relation as expressed in Equation (8). |𝐞|2 = 𝐯 T 𝐯 − 2𝐯 T 𝛀𝛂 + (𝛀𝛂)T 𝛀𝜶 (10) Taking the gradient of the right hand side in relation to α and imposing the condition of it being zero. −1 𝛂 = (𝚿T 𝛀) 𝚿T 𝐯 (6) Therefore.
once that the number of base vector to span the totality of ℝ𝑛 is equal to n but only m base vectors are available. . one can conclude that the optimal approximation vector is just an orthogonal projection into the subspace . 𝐱 ∈ ℝ𝑚 and 𝐛 ∈ ℝ𝑛 with n > m. the available basis represents a subspace of dimension m embedded on a higher n-dimensional space. 𝐀T 𝐀𝐱 = 𝐀T 𝐛 (15) Therefore. it is only necessary to multiply the system of equations by the transpose of the coefficient matrix. imagine that one wants to solve the linear system of Equation (13) with 𝐀 ∈ ℝ𝑛. 𝐀𝐱 = 𝐛 (13) This problem is overdetermined and can be interpreted geometrically as trying to find the representation of the vector 𝐛 using the columns of 𝐀 as the base vectors. that represents the application of a transformation A to a vector x. it is possible to conclude that to obtain the closest approximation to Equation (13). it is natural to wonder how to transform the operator between coordinate systems. It is very unlikely that this task is possible. Therefore.𝛁(𝐯 T 𝐯 − 2𝐯 T 𝛀𝛂 + (𝛀𝛂)T 𝛀𝜶) = 𝟎 𝛁(−2𝐯 T 𝛀𝛂) + 𝛁[(𝛀𝛂)T 𝛀𝜶] = 𝟎 −2[𝛁𝐯 T 𝛀𝛂 + 𝛁(𝛀𝛂)T 𝐯] + 2𝛁(𝛀𝛂)T 𝛀𝜶 = 𝟎 (11) 𝛀T 𝐯 = 𝛀T 𝛀𝜶 −1 𝜶 = (𝛀T 𝛀) 𝛀T 𝐯 Therefore the vector that minimizes the square of the modulus of the error is given by Equation (12). −1 𝐀𝐱 = 𝐀(𝐀T 𝐀) 𝐀T 𝐛 (14) One can multiply both sides of the above equation by 𝐀T to obtain Equation (15). This results in shown in Equation (14). −1 𝐩 = 𝛀(𝛀T 𝛀) 𝛀T 𝐯 (12) Comparing Equations (8) and (12). A very important application of this fact that will be largely used in model order reduction is that it can be used to approximately solve overdetermined systems. an example of this is the projection operator. in the sense of least squares. As an example. Similarity Transformation Matrices may be used to represent linear operators.𝑚 . If two bases for the same linear space are related by a change of coordinates as shown by Equation (16). One approach to solve this problem in an approximated fashion is to project the vector 𝐛 into the column span of A. One may start with Equation (17).
A similar argument shows that the matrix V is composed by the eigenvectors of the matrix A*A. ̅ 𝐁𝐲 = 𝐓 −1 𝐀𝐓𝐲 = 𝒚 (18) This type of relation has a very important property: the eigenvalues of A are the same as B’s. This is shown in Equation (18). |𝐓 −1 𝐀𝐓 − 𝜆𝐈| = |𝐓 −1 𝐀𝐓 − 𝜆𝐓 −1 𝐓| = |𝐓 −1 ||𝐀 − λ𝐈||𝐓| = |𝐀 − λ𝐈| (19) In a similar fashion. This is very important because. 𝐀𝐫 = 𝐏 T 𝐀𝐏 (20) For this analysis the focus is not to show that the eigenvalues are the same. 𝐀 = 𝐔𝚺𝐕 ∗ (23) To be able to determine these matrices. as will be explained latter in this document. 𝐀𝐀∗ = 𝐔𝚺𝐕 ∗ 𝐕𝚺 ∗ 𝐔 ∗ = 𝐔𝚺𝚺 ∗ 𝐔 ∗ (24) This leads to conclude that the matrix U is build using the eigenvectors of the matrix AA*.𝐱 = 𝐓𝐲 (16) 𝐀𝐱 = 𝐱̅ (17) Using Equation (16) into (17). but to demonstrate that their signs are preserved.n may be decomposed as a product of two orthonormal and a diagonal matrix as shown in Equation (23). one may wonder of what are the relations of the eigenvalues of a transformation of the kind indicated in Equation (20). the relation for y and the transformed operator B may be obtained. one may write the product of Equation (24). If A is symmetric. there is no guarantee that the resulting reduced system will be stable. and each of the diagonal entries of Σ is the square root of the not null eigenvalues of AA*. which they are not. and the system is stable. This can be easy shown by the development in (19). The first n columns of matrix U are an orthonormal basis for the range of A. the sign of the eigenvalues determines if the system is stable or not. For a general A matrix. 𝑨 = −𝑩𝑻 𝑩 (21) This allows to prove that the reduced matrix is negative definite as shown in (22). then it can be decomposed as written in Equation (21). . 𝐱 𝐓 𝐀𝐫 𝐱 = 𝐱 𝐓 𝐏 T 𝐀𝐏𝐱 = −𝐱 𝐓 𝐏𝐓 𝑩𝑻 𝑩𝑷𝐱 = −(𝑩𝑷𝒙)𝑻 (𝑩𝑷𝒙) = −‖𝑩𝑷𝒙‖ ≤ 𝟎 (22) Singular Value Decomposition Any matrix 𝐀 ∈ ℝm.
𝐱 ∈ ℝ𝑛 . 𝐃 ∈ ℝ𝑠. If this is not the case in a particular application. 𝐱̇ = 𝐄−1 𝐀𝐱 + 𝐄−1 𝐁𝐮 = 𝐀′ 𝐱 + 𝐁′ 𝐮 (28) Many textbooks present this formulation as the standard form for the state space system.𝑟 .𝑛 . 𝐱 = 𝐏𝐲 (30) Accordingly.This decomposition allows writing any matrix as a sum of products of rank 1. To achieve this. 𝐀 = σ1 𝐮𝟏 𝐯𝟏∗ + σ1 𝐮𝟏 𝐯𝟏∗ + ⋯ + σr 𝐮𝐫 𝐯𝐫∗ (25) 2.𝑟 . 𝐁 ∈ ℝ𝑛. y is the output vector and u is the input vector. 𝐂 ∈ ℝ𝑠.𝑛 . 𝐄 [ 11 𝟎 𝐀 𝟎 𝐲̇1 ] [ ] = [ 11 𝐀 𝐲̇ 𝟎 2 21 𝐀12 𝐲1 𝐁 ] [ ] + [ 1] 𝐮 𝐀22 𝐲2 𝐁2 (29) To obtain this form. problems arise when E is singular. This is written in Equation (25). one may use a permutation matrix P to create the transformation of Equation (30). the transformed system is given by Equation (31). The first step is to write Equation (26) as in Equation (29). 𝐄𝐱̇ = 𝐀𝐱 + 𝐁𝐮 (26) It must be pointed out that Equations (26) and (27) are not coupled. permutations of the lines and columns of the E matrix should be followed by equivalent permutations of A and B. This indicates the presence of algebraic states that have to be threated explicitly. the matrix D is considered of all zeros. 𝐮 ∈ ℝ𝑟 .𝑛 . 𝐏 T 𝐄𝐏𝐲 = 𝐏 T 𝐀𝐏𝐲 + 𝐏 T 𝐁𝐮 (31) In this stage it is possible to apply a Kron reduction to Equation (29) to obtain (32). The vector x is the state vector. It is possible to show that truncating this series at position r leads to the best rank r approximation of matrix A. the reasoning presented below can be easily adapted. −1 𝐄11 𝐲̇ 1 = (𝐀11 − 𝐀12 𝐀−1 22 𝐀 21 )𝐲𝟏 + (𝐁1 − 𝐀12 𝐀22 𝐁2 )𝐮 (32) . Model Order Reduction by Projection State Space Representation The representation of time invariant linear systems used in this document is given by Equations (26) and (27). 𝐀 ∈ ℝ𝑛. If the matrix E is not singular it is possible to write the system in a simpler form as shown in (28). However. In these equations: 𝐄 ∈ ℝ𝑛. 𝐲 = 𝐂𝐱 + 𝐃𝐮 (27) For now. 𝐲 ∈ ℝs .
In the framework of model order reduction. Thus. 𝑡 x(t) = ∫ 𝑒 𝐀(t−𝜏) 𝐁𝐮(τ)dτ (34) 0 If one chooses the input function as in Equation (35). 𝑡 𝐱(𝑡) = 𝑒 𝐀(𝑡−𝑡0 ) 𝐱(𝑡0 ) + ∫ 𝑒 𝐀(𝑡−𝜏) 𝐁𝐮(τ)dτ (33) 𝑡0 If in the beginning of the simulations the system is found with zero initial conditions. For simplicity. The system response to this scenario is given by Equation (38). ∗ 𝐮(𝜏) = 𝐁∗ 𝑒 𝐀 (𝑡−𝜏) 𝓟−1 (𝑡)𝐱(𝑡) 𝑡 ∗ 𝑡 (35) ∗ 𝓟 = ∫ 𝑒 𝐀(t−τ) 𝐁𝐁∗ 𝑒 𝐀 (𝑡−𝜏) 𝑑𝜏 = ∫ 𝑒 𝐀𝜉 𝐁𝐁∗ 𝑒 𝐀 𝜉 𝑑𝜉 0 (36) 0 The energy in this signal can by written as in (37). The equation that describes the state evolution of this system is given by (34). the response to an arbitrary input signal is given by Equation (33). The matrix 𝒫 which is called the controllability Gramian is defined by Equation (36). Its physical interpretation leads to the conclusion that the controllability Gramian measures how hard is to achieve a certain state.This representation is a reduced state space equation that allows direct reduction to standard form. 𝐱0 = 𝐂𝑒 𝐀𝑡 𝐱𝟎 (38) . Observability and Reachability There are two quantities that are very important to characterize the system. the initial time is set to zero. There may be an infinite number of signals that can accomplish this task. it is a natural question to ask which is the input signal u(t) that may be used to drive the system to a given state. Some authors use t = ∞ to define this quantity and they call it the infinite Gramian. but without any loss of generality. one may want that this signal is optimal in the sense that it is capable to drive the system to the desired state using the minimum amount of energy. Suppose a system written as in Equation (28). 𝑡 Ec = ∫ 𝐮(𝜏)∗ 𝐮(𝜏)𝑑𝜏 = 𝐱 0∗ 𝓟−1 𝐱0 (37) 0 The above discussion involved solely the states of the system and ignored its output. A dual concept called observability can be derived if one analyses the amount of energy that a certain state delivers to the output if no input signal is present. To reduce the number of choices. writing the system as in Equation (26) or as in Equation (28) is interchangeable and both formulations may be used when they are more convenient. states that are difficult to reach are good candidates for truncation. the above identity is satisfied in the least amount of energy [1].
the last two terms of the right hand side can be interpreted as an error factor. 𝑡 ∗ Eo = 𝐱0∗ ∫ 𝑒 𝐀 𝝉 𝐂 ∗ 𝐂𝑒 𝐀𝜏 𝑑𝜏 𝐱 0 = 𝐱𝟎∗ 𝓠𝐱0 (39) 0 Controllability and Observability are dual concepts and their relation is shown in Table 1. Equation (26) may be written as in (41). 𝐄𝐕𝐱̇ 𝐫 = 𝐀𝐕𝐱𝐫 + 𝐁𝐮 + 𝝐 (42) The above system in xr is overdetermined. V contains q base vector and U the last n – q. any two bases for ℝn can be related by a transformation matrix. Therefore.The energy of this signal can be calculates with aid of Equation (37) and the result is given by Equation (39) which is also the definition of the observability Gramian. the relation is exact and does not lead to any computational improvement. As presented in Equation (2). Table 1 – Duality relation among controllability and observability Controllability Observability A A* B C* Reduction by Projection If it is assumed that n is a large number. the burden of solving Equation (26) is very high or even prohibitive. if the last term of the right rand side of (43) is neglected. allowing one to obtain Equation (40). a reduction of the system may be used to allow a faster solution with an acceptable loss of accuracy. The transformation matrix can be partitioned into two smaller matrices. The same process used in Equations (13). However. This is shown in Equations (44) and (45). This is shown in (43). one can write the equations for the reduced system. 𝐱 = 𝐕𝐱𝐫 + 𝐔𝐱𝐬 (40) Then. 𝐖 T 𝐄𝐕𝐱̇ 𝐫 = 𝐖 T 𝐀𝐕𝐱𝐫 + 𝐖 T 𝐁𝐮 + 𝐖 T 𝝐 (43) Until this point. (14) and (15) can be used to allow the solution of this equation. 𝐖 T 𝐄𝐕𝐱̇ 𝐫 = 𝐖 T 𝐀𝐕𝐱𝐫 + 𝐖 T 𝐁𝐮 (44) 𝐲 = 𝐂𝐕𝐱𝐫 (45) . Thus. 𝐄𝐕𝐱̇ 𝐫 = 𝐀𝐕𝐱𝐫 + 𝐁𝐮 + 𝐀𝐔𝐱 𝐬 − 𝐄𝐔𝐱̇ 𝐬 (41) One may choose the basis in a way that the vectors of V are more important in representing the dynamics of the system than the vectors in U.
The above set of equations allows the solution of an approximation to the original system but with a reduced order. Therefore. 𝐄𝐫 𝐱̇ 𝐫 = 𝐀𝐫 𝐱𝐫 + 𝐁𝐫 𝐮 (46) 𝐲 = 𝐂𝐫 𝐱𝐫 (47) To deal with initial conditions. the transfer function of the system is given by Equation (48). If the matrices K and L are nonsingular. This relation is presented in Table 2. the above equations represent the projection of the system into a subspace spanned by V and orthogonal to W. This allows writing the reduced system in standard form. a relation between the original system and reduced system matrices can be made. one could use the projection of the initial condition vector into the span of the reduced basis. respectively. defined in Equations (49) and (50). the matrices V’ and W’. −1 𝐇 = 𝐂𝐕 ′ 𝐊 −1 (s(𝐖 ′ 𝐋−1 )T 𝐄𝐕 ′ 𝐊 −1 − (𝐖 ′ 𝐋−1 )T 𝐀𝐕 ′ 𝐊 −1 ) (𝐖𝐋−1 )T 𝐁 (51) Developing the above expression.Reduced System Coefficient Matrices Original System Reduced System 𝐄 𝐖 T 𝐄𝐕 𝐀 𝐖 T 𝐀𝐕 𝐁 𝐖T𝐁 𝐂 𝐂𝐕 The reduced system can be written as in Equations (46) and (47). This is valid for the full and reduced system as well if one keeps in mind the relations of Table 2.Under the light of projection matrices. For simplicity. Invariance of the Transfer Function In the frequency domain. it may be concluded that the transfer function is invariant under a change of base. have the same column span of V and W. The important aspects of matrices V and W are their column span. the resulting transfer function is unchanged. 𝐇 = 𝐂(s𝐄 − 𝐀)−1 𝐁 (48) If instead of using matrices V and W for the projection. . Table 2 . 𝐕 ′ = 𝐕𝐊 (49) 𝐖 ′ = 𝐖𝐋 (50) Substituting these matrices into Equation (48). Equation (51) is obtained. the expression to the transformed reduced system is found to be identical to the original reduced system. one utilizes different matrices but with the same column span.
̃ = 𝑻−𝟏 𝓟𝑻−∗ 𝓟 ̃ = 𝑻∗ 𝓠𝑻 𝓠 (52) It is possible to choose the matrix T such that both Gramians are equal and diagonal. The problem of this method is the burden to calculate the Gramians that is of the order of n3 [2]. the states that have small associated diagonal values. The proper orthogonal decomposition aims at finding a projection operator of fixed rank (Π) that minimizes the quadratic error created by the simulation of a lower order system. This is shown in Equation (56). The verification of this fact can be made by direct substitution of the definition of the Gramians. However. As any vector can be written as the sum of the projection into a subspace and the projection to the orthogonal subspace. the development in (57) is possible. The interested reader may look at this work for further discussion. the transformed Gramians are given by (52). The standard way to do this is to solve the Lyapunov equations given by (54). This can be achieved by a singular value decomposition of the product (53). the following development raises a lot of understanding of the problem structure. minimizing the error is equal to maximize the projection into the orthogonal space. In this equation. 𝑻 𝜖 = ∫ ‖𝐱(𝑡) − 𝚷𝐱(𝑡)‖2 𝑑𝑡 (55) 𝟎 The integrand is the projection of the vector x into the space that is orthogonal to the projected one.Balanced Truncation The observability and reachability of each state is dependent of the system realization. 𝑻 𝑻 2 𝑻 𝑀 = ∫ ‖𝚷𝐱(𝑡)‖2 𝑑𝑡 = ∫ ‖𝛀T 𝐱(𝑡)‖ 𝑑𝑡 = 𝑡𝑟 (𝛀𝑻 ∫ 𝐱(𝑡)𝐱(𝑡)𝑇 𝑑𝑡 𝛀) 𝟎 𝟎 𝟎 (57) . If one applies a base change represented by the matrix T. 𝓟𝓠 = 𝑻𝚺 𝟐 𝑻−𝟏 (53) In order to reduce the system. 𝑻 𝑀 = ∫ ‖𝚷𝐱(𝑡)‖2 𝑑𝑡 (56) 𝟎 Using Equation (12) and imposing the orthonormality of the base. For the case of continuous time. the quadratic error is given by Equation (55). 𝑨𝓟 + 𝓟𝑨∗ + 𝑩𝑩∗ = 𝟎 𝑨∗ 𝓠 + 𝓠𝑨 + 𝑪∗ 𝑪 = 𝟎 (54) Proper Orthogonal Decomposition The development of this section is largely inspired in [3]. tr represents the trace. We already know that one may truncate Equation (25) to solve directly this problem.
to obtain Equation (62). in relation to Ω. Mathematically this is equivalent to compute V by the eigenvalue decomposition of the product STS. To calculate K in a digital computer.This motivates the definition of the POD Kernel given by Equation (58). 𝐊𝐕 = 𝐕𝐀 (62) This condition implies that V must span a subspace that is invariant under the application of the linear operator K. 𝜶) = 𝒕𝒓(𝛀𝑻 𝑲𝛀) − ∑ 𝜶𝒊. To compute U in an efficient way. 𝜶) = 𝑡𝑟(𝛀𝑻 𝑲𝛀) − 𝑡𝑟(𝑨𝑻 𝑽𝑻 𝑽) + 𝑡𝑟(𝑨) 𝑨𝒊𝒊 = 𝜶𝒊𝒊 . one may calculate the first order optimality condition. the eigenvectors that we are looking for are the columns of the U matrix of the singular value decomposition of the kernel. The first order optimality condition for the α values leads to the obvious restriction given by Equation (63). 𝓛(𝑽.𝒋 It is possible to cast the orthogonality restriction as in Equation (60) if one defines the matrix A as in (61). the burden to compute the eigenvectors may be very high. that has a size much smaller than the Kernel. Equation shows a way to approximate the kernel. The S matrix is defined in (65). a discretization is necessary. 𝑻 𝐊 = ∫ 𝑥(𝑡) 𝑥(𝑡)𝑇 𝑑𝑡 = ∑ 𝒙(𝒕𝒊 )𝒙(𝒕𝒊 )𝑻 𝚫𝒕𝒊 = 𝑺𝑺𝑻 𝟎 (64) 𝒊 𝐒 = [𝒙𝟎 √𝚫𝒕𝟎 𝒙𝟏 √𝚫𝒕𝟏 ⋯ 𝒙𝒏𝒔 √𝚫𝒕𝒏𝒔 ] (65) As the kernel is an n-by-n matrix. 𝑆𝑆 𝑇 = (𝑈Σ𝑉 𝑇 )(𝑈Σ𝑉 𝑇 )𝑇 = 𝑈ΣΣ𝑇 𝑈 𝑇 (66) Therefore. 𝓛(𝑽. 𝑻 𝐊 = ∫ 𝐱(𝑡) 𝐱(𝑡)𝑇 𝑑𝑡 (58) 𝟎 This development can be cast into an optimization problem whose Lagrangian is given by (59).𝒋 (𝒗𝑻𝒊 𝒗𝒋 − 𝜹𝒊𝒋 ) (59) 𝒊. 𝑨𝒊𝒋 = 𝟏 𝜶 𝟐 𝒊𝒋 (60) (61) Using the properties in the appendix. However it is possible to avoid this by the application of the singular value decomposition of S. the ‘economic’ version of the SVD may be used. 𝐕T𝐕 = ℑ (63) Both of these conditions are satisfied if V is built with the eigenvectors of K. and then using Equation (67) to get . This is shown in Equation (66).
These moments are centered at the zero frequency.Equivalence for decentered moments Centered at Zero Centered at σ 𝑠 𝑠−𝜎 𝐀 𝐀 − 𝜎𝐄 Using these relations of can directly write the moments for any frequency. Moment Matching The transfer function in Equation (48) can be rewritten as in Equation (68). This does not alter the problem once that the column span of the product is unchanged by the multiplication by Σ. The first moment of the reduced model is given by Equation (72). . This relation can be used to extend other results obtained to the zero centered expression to an arbitrary placed frequency. In this method. the transfer function can be written as in Equation (85).only some eigenvectors. only the system states are taken into account. This is written in Equation (71). ∞ 𝐇 = − ∑ 𝐂(𝐀−1 𝐄)𝑗 𝐀−1 𝐁s𝑗 (69) 𝑗=0 The negative of these terms in s are called the moments of the transfer function. To obtain the moments for other frequencies. Using information from these two systems allows to approximate the balanced truncation method in a less expensive way. 𝐇 = −𝐂(s𝐀−1 𝐄 − 𝐈)−1 𝐀−1 𝐁 (68) The central term can be expanded to obtain a series representation of the transfer function. This can be easily solved by using a second system that satisfies the relations in Table 1. Table 3 . −1 𝐇 = 𝐂((𝑠 − σ)𝐄 − (𝐀 − σ𝐄)) 𝐁 (70) Direct comparison of Equations (48) and (70) allows determining the equivalences pointed out in Table 3. This is shown in (69). 𝑈Σ = 𝑆𝑆 𝑇 𝑉 (67) It must be pointed out that this is only the mathematical description of the solution. ∞ 𝐇 = − ∑ 𝐂[(𝐀 − 𝜎𝐄)−1 𝐄]𝑗 (𝐀 − 𝜎𝐄)−1 𝐁(s − σ)j (71) 𝑗=0 The moment matching technique aims at choosing the columns of V and W in a way that some of the moments of the transfer function are exactly matched. In a computer implementation this procedure may be replaced by more efficient ones. called the dual system. No information is obtained from the system output.
q moments of the transfer function of the full system are matched if the matrix V is given by Equation (80). If A and E are symmetric and CT = B. 𝐯. 𝒦(𝐆. To express this result in a simple way. −𝟏 𝐦1 = 𝐂𝐕(𝐖 𝐓 𝐀𝐕) 𝐖 𝐓 𝐄𝐀−1 𝐛 (76) If A-1EA-1B is in the column span of V. −𝟏 𝐦1 = 𝐂𝐕(𝐖 𝐓 𝐀𝐕) −𝟏 𝐖 𝐓 𝐄𝐕(𝐖 𝐓 𝐀𝐕) 𝐖𝐓𝐁 (75) Using (73). one may introduce the Krylov subspaces. impedance probing. 𝐕𝐫𝟏 = 𝐀−1 𝐄𝐀−1 𝐁 (77) This relation can be used to write (76) as in Equation (78). the zeroth and the first moments match. 𝐀−𝟏 𝐁. −𝟏 𝐦1 = 𝐂𝐕(𝐖 𝐓 𝐀𝐕) 𝐖 𝐓 𝐀𝐕𝐫1 = 𝐂𝐀−1 𝐄𝐀−1 𝐁 (78) Therefore.g. if A-1B and A-1EA-1B are in the column span of V. Equation (75) can be reduced to (76). Therefore. both of the subspaces are the same. 𝑞) = [𝐯 𝐆𝐯 𝐆𝟐 𝐯 ⋯ 𝐆 𝐪−𝟏 𝐯 𝐆 𝐪 𝐯] (79) Using this definition. A simple choice for this matrix is to make it equal to V. 𝑞) (80) It must be pointed out that nothing was imposed over the matrix W. 𝐀−𝐓 𝑪𝑻 . . one may use a very similar argument to show that W may be chosen in order to match even more moments. −𝟏 𝐦0 = 𝐂𝐕(𝐖 T 𝐀𝐕) 𝐖 T 𝐀𝐕𝐫0 = 𝐂𝐀−1 𝐁 (74) The first moment for the reduced system is given by Equation (75). there is no need to calculate both subspaces. there exists a vector r1 as shown in Equation (77). there exists a vector r0 as shown in Equation (88). 𝐖 = 𝒦(𝐀−𝐓 𝐄𝐓 . A very similar argument shows that if W is chosen in accordance to Equation (81). then other q moments of the full system are matched. e. 𝐕𝐫0 = 𝐀−1 𝐁 (73) Equation (72) can be written as in (74).−𝟏 𝐇 = 𝐂𝐕(𝐖 T 𝐀𝐕) 𝐖T𝐁 (72) If A-1B is in the column span of V. This process can be continued to match the successive moments. the zeroth moment of the reduced system is equal of the zeroth moment of the full system. 𝑞) (81) In some special systems. defined in Equation (79). 𝐕 = 𝒦(𝐀−𝟏 𝐄. However.
A graph. The vectors 𝐢r .) and the nodes are the connection points. current source.2 zero entries. In order to translate this structure into matrix notation.|E|) with its elements satisfying 𝐀ij ∈ {−1. Before introducing the linear equations for circuit simulation. From this matrix. 𝐀r 𝐑−1 𝐯𝑟 + 𝐀c 𝐂 d𝐯𝑐 + 𝐀l 𝐢l = 𝐀s 𝐢s dx (84) . inductors and current sources will be considered. If the edges node of known potential. Circuit Simulation This section is largely related to the work in [4]. 𝐢c . denoted by G(V. In this particular analysis. The nonzero elements must be placed accordingly with the incidence of the edges into the nodes. capacitor. ̃ = [𝐀s 𝐀 𝐀r 𝐀c 𝐀l ] (82) Using these submatrices is possible to write the Kirchhoff’s circuit laws for the circuit as shown in Equation (83). Each column of this matrix is directly associated with an edge of the underlying graph and each row with a node. These components and its orientations are illustrated in Figure 2. one entry -1 and |V| . each column must contain exactly one entry 1. containing tuples of elements of V. 0.Circuit elements with voltage and current orientations Suppose the existence of an oriented graph representing the circuit. some notation and important concepts are introduced. To build the matrix. This new matrix is denoted by 𝐀 of this matrix are placed in a way that components of the same kind are side by side. one utilizes an edge-node incidence matrix (𝐀 ∈ ℝ|V|. 𝐀r 𝐢r + 𝐀c 𝐢c + 𝐀l 𝐢l = 𝐀s 𝐢s (83) Direct application of resistors and capacitors terminal relations leads to Equation (84). is a group of two sets: the set of vertices (V) and the set of edges (E). only resistors. etc. 𝐢l e 𝐢s contains the currents of each of the different types of circuit elements. 1}.3. some submatrices can be identified as shown in (82). it is possible to derive a reduced matrix excluding the line representing the ̃ . an edge can be tough as a circuit component (resistor. i V i i V V i V Figure 2 . One way to model a circuit into the computer is by an abstraction called graph. The graph orientation is used to determine the sign of the entry. For the case of circuit representation. usually the ground node. E). capacitors.
This allows calculating only one subspace and having the precision achieved by two subspaces in the general case.RLC ladder For this experiment a pattern as indicated in Table 4 was chosen. 𝐀Ti 𝐯 = 𝐯i (85) Using this relation. Figure 3 . 𝐀r 𝐑−1 𝐀Tr 𝐯 + 𝐀c 𝐂𝐀Tc d𝐯 + 𝐀l 𝐢l = 𝐀s 𝐢s dx (86) Writing the inductors terminal voltage and using Equation (85). This allows obtaining a unified equation with the node voltages and currents in the inductors as unknowns. The Bode diagram for the impedance measured at the terminals of the current source is plotted in Figure 4. both right and left Krylov subspaces are the equal. Equation (84) can be rewritten as in Equation (86). This circuit consists basically in a current source in parallel with a capacitor that feeds a ladder of adjustable size that contains a pattern of inductance.The relation between the voltage in each of the elements and the potential of each of the nodes of the circuit is given by Equation (85). Application Example The formulation developed in the last section is used to build an electric circuit as shown in Figure 3. it is possible to write Equation (87). . resistance and capacitance. 4. l or s. The output is an ordinary resistor whose choice is independent of the other. The index i is a placeholder and can be replaced by r. The same phenomenon happens with the original and dual systems in the Proper Orthogonal Decomposition. 𝐀Ti 𝐯 = 𝐯i T [𝐀c 𝐂𝐀𝑐 𝟎 −1 T 𝟎 ] 𝑑 [𝐯n ] = − [𝐀r 𝐑 𝐀r 𝐀Ti −𝐋 𝑑𝑥 𝐢l (87) 𝐀i 𝐯n 𝐀 ] [ 𝐢 ] + [ 𝑠 ] 𝐢𝑠 𝟎 l 𝟎 (88) If formulated in this fashion. c. This is presented in Equation (88).
Finally.Value patterns for the RLC ladder Parameter Value Input capacitor 1 pF Output resistor 1Ω Resistor pattern 1Ω Inductor pattern 1 nH. 10 nH.Table 4 . . 10nF Figure 4 – Bode diagram of the system The resulting reduced order models for an order equal to 20 is shown for the POD method in Figure 5. 1μH Capacitor pattern 1pF. Figure 6 shows the result for the Krylov method with 12 logarithmically distributed expansion points for the same system. 1nF. Figure 7 shows the result for a Balanced Truncation method. The sampling time was one nanosecond.
Figure 5 – POD Approximation Figure 6 – Krylov Approximation .
Figure 7 – Balanced Truncation 5. Appendix – Mathematical development Gradient property 𝑛 = ∑ 𝑢𝑘 𝑣𝑘 (89) 𝑘 𝑑𝑛 𝑑𝑢𝑘 𝑑𝑣𝑘 =∑ 𝑣𝑘 + 𝑢𝑘 𝑑𝑥𝑖 𝑑𝑥𝑖 𝑑𝑥𝑖 (90) ∇𝑥 (𝑢𝑇 𝑣) = (∇𝑥 𝑢𝑇 )𝑣 + (∇𝑥 𝑣 𝑇 )𝑢 (91) 𝑘 .
First gradient of the trace 𝑡𝑟(𝑉 𝑇 𝐾𝑉) = ∑ ∑ 𝑉𝑙𝑢 ∑ 𝐾𝑙𝑘 𝑉𝑘𝑢 𝑢 𝑙 (92) 𝑘 𝑑 𝑡𝑟(𝑉 𝑇 𝐾𝑉) 𝑑𝑉𝑙𝑢 𝑑𝑉𝑘𝑢 = ∑∑( ∑ 𝐾𝑙𝑘 𝑉𝑘𝑢 + 𝑉𝑙𝑢 ∑ 𝐾𝑙𝑘 ) 𝑑𝑉𝑖𝑗 𝑑𝑉𝑖𝑗 𝑑𝑉𝑖𝑗 (93) 𝑑 𝑡𝑟(𝑉 𝑇 𝐾𝑉) = ∑ 𝐾𝑖𝑘 𝑉𝑘𝑗 + ∑(𝐾 𝑇 )𝑖𝑙 𝑉𝑙𝑗 𝑑𝑉𝑖𝑗 (94) ∇𝑉 𝑡𝑟(𝑉 𝑇 𝐾𝑉) = 𝐾𝑉 + 𝐾 𝑇 𝑉 (95) 𝑢 𝑙 𝑘 𝑘 𝑘 𝑙 Second gradient of the trace 𝑡𝑟(𝐴𝑇 𝑉 𝑇 𝑉) = ∑ ∑ 𝐴𝑙𝑢 ∑ 𝑉𝑘𝑙 𝑉𝑘𝑢 𝑢 𝑙 (96) 𝑘 𝑑 𝑡𝑟(𝐴𝑇 𝑉 𝑇 𝑉) 𝑑𝑉𝑘𝑙 𝑑𝑉𝑘𝑢 = ∑ ∑ 𝐴𝑙𝑢 ∑ ( 𝑉𝑘𝑢 + 𝑉𝑘𝑙 ) 𝑑𝑉𝑖𝑗 𝑑𝑉𝑖𝑗 𝑑𝑉𝑖𝑗 (97) 𝑑 𝑡𝑟(𝐴𝑇 𝑉 𝑇 𝑉) = ∑ 𝑉𝑖𝑢 (𝐴𝑇 )𝑢𝑗 + ∑ 𝑉𝑖𝑙 𝐴𝑙𝑗 𝑑𝑉𝑖𝑗 (98) ∇𝑉 𝑡𝑟(𝐴𝑇 𝑉 𝑇 𝑉) = 𝑉𝐴𝑇 + 𝑉𝐴 (99) 𝑢 𝑙 𝑘 𝑢 𝑙 . | 10,194 | 27,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-47 | latest | en | 0.932203 |
https://www.physicsforums.com/threads/direct-substitution-problem.221562/ | 1,575,839,692,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00036.warc.gz | 826,020,192 | 18,040 | # Direct substitution problem
Gold Member
1. Homework Statement
verify by direct substitution that the wave function for a standing wave given in equation below is a solution to the general linear wave equatin.
$$y= (2A sin kx)cos \omega t$$
$$\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2}$$
2. Homework Equations (above)
How do I do this by direct substittution ?
I originally went and just proved it through partial differentiation but I've never heard of direct substitution before.
Thank you
## Answers and Replies
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Hootenanny
Staff Emeritus
Gold Member
I have a sneaking suspicion that you've verified the solution by direct substitution without knowing it. The method of direct substitution simply means substituting the proposed solution into the equation and seeing if it works. In other words, taking the second partial derivatives with respect to position and time and seeing if they fit the equation.
tiny-tim
Homework Helper
Yeah, sometimes examiners are really sneaky, and use words which you think are technical, but actually they're just obscure ordinary English.
Gold Member
I have a sneaking suspicion that you've verified the solution by direct substitution without knowing it. The method of direct substitution simply means substituting the proposed solution into the equation and seeing if it works. In other words, taking the second partial derivatives with respect to position and time and seeing if they fit the equation.
well not exactly. I couldn't get them to equal since I wasnt sure what to do after I got:
$$\frac{\delta^2 y} {\delta x^2}|_t= -k^2(2A sin kx)cos \omega t)$$
and
$$\frac{\delta^2 y} {\delta t^2}|_x= -\omega^2 2A cos\omega t(sin kx)$$
and if that's right...I'm not sure how to make them equal Is there a equation I can use to do that?
$$(-k^22A sin kx)(cos \omega t)= \frac{1} {v^2} (-\omega^2 cos\omega t)(2A sin kx)$$
Thanks
Last edited:
tiny-tim
Homework Helper
well not exactly. I couldn't get them to equal since I wasnt sure what to do after I got:
$$\frac{\delta^2 y} {\delta x^2}|_t= -k^2(2A sin kx)cos \omega t)$$
and
$$\frac{\delta^2 y} {\delta t^2}= -\omega^2 2A cos\omega t(sin kx)$$
Hi christina!
erm … they're the same! … except for a factor, which you can call v.
Gold Member
Hi christina!
erm … they're the same! … except for a factor, which you can call v.
Of course they are but I can't say that they are unless I show it first.
The question is to verify by direct substitution...is this right ? (below).
I'm thinking of dividing over the cos and sin and then canceling to just get
$$\frac{\omega^2} {k^2} = \frac{1} {v^2}$$
but I'm trying to find a relationship in physics that allows that. If I'm not incorrect.
Gold Member
I think I found the relationship after looking at one of my old posts.
It's interesting though how I can't find this equation anywhere in my book but my lab teacher gave it to the class and it works here. ($$v= \omega /k$$)
$$(-k^22A)(sin kx)(cos \omega t)= \frac{1} {v^2} (-\omega^22A) (cos\omega t)(sin kx)$$
I'm thinking of dividing over the cos and sin to the left and then canceling to just get
$$\frac{\omega^2} {k^2} = \frac{1} {v^2}$$
But I'm confused to as to when I use $$v= \omega /k$$
I end up with
$$\frac{1} {v^2} = (\frac{k} {\omega})^2$$
but is there another way to verify?
Thank you
Gold Member
Actually I think I found what I was supposed to get( above equation)......right?
Last edited:
HallsofIvy
Homework Helper
1. Homework Statement
verify by direct substitution that the wave function for a standing wave given in equation below is a solution to the general linear wave equatin.
$$y= (2A sin kx)cos \omega t$$
$$\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2}$$
2. Homework Equations (above)
How do I do this by direct substittution ?
I originally went and just proved it through partial differentiation but I've never heard of direct substitution before.
Thank you
Suppose you were asked to show that x= 1 is a solution to x5- 3x2+ 5x- 2= 0. Would you solve the equation? Of course not- you would just substitute 1 for x and show that the equation is true. That's "direct substitution".
jtbell
Mentor
It's interesting though how I can't find this equation anywhere in my book but my lab teacher gave it to the class and it works here. ($$v= \omega /k$$)
Look up the definitions of $k$ and $\omega$ in terms of $\lambda$ and $f$ and substitute them into the equation above. You should get something that looks familiar.
Gold Member
Suppose you were asked to show that x= 1 is a solution to x5- 3x2+ 5x- 2= 0. Would you solve the equation? Of course not- you would just substitute 1 for x and show that the equation is true. That's "direct substitution".
I understand. So this case however, it would be pluging an equation into the linear wave equation for direct substitution.
Thank You HallsofIvy | 1,375 | 4,975 | {"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} | 3.765625 | 4 | CC-MAIN-2019-51 | latest | en | 0.913205 |
http://www.finderchem.com/how-do-you-explain-the-law-of-large-numbers-in-simple-terms.html | 1,500,555,905,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423183.57/warc/CC-MAIN-20170720121902-20170720141902-00561.warc.gz | 439,501,061 | 8,426 | # How do you explain the Law of Large Numbers in simple terms?
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Related Questions
Recent Questions | 1,570 | 7,128 | {"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-2017-30 | latest | en | 0.900814 |
http://www.mogenk.com/placing-fractions-on-a-number-line-worksheet/ | 1,566,074,718,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312025.20/warc/CC-MAIN-20190817203056-20190817225056-00404.warc.gz | 278,414,413 | 6,957 | # Placing Fractions On A Number Line Worksheet
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Trending Now | 596 | 2,561 | {"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} | 2.71875 | 3 | CC-MAIN-2019-35 | longest | en | 0.851951 |
https://singaporemathsource.com/category/word-problem-wednesday/?author=Cassyt | 1,601,097,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00330.warc.gz | 606,840,176 | 22,116 | Word Problem Wednesday – Marbles
This month’s Word Problem Wednesday problem comes from an article in the Daily Express, a UK newpaper.
The headline teases: This is the maths puzzle that is baffling everyone – but could you solve it?
Yes, we can! So, don’t lose your marbles over this one.
This appears to come from a Maths No Problem! workbook, probably 2A as the article states it is a problem for 7-year-olds. The author interviews a math professor:
Math expert, Dr. James Hind, of Nottingham Trent University, said the confusing question is above the level it was set for and to reach a conclusion it is best to try a number of equations.
Dr. Hind then proceeds to use a guess and check method to solve the problem. Maybe they asked the wrong expert.
Programs based on a Singapore Math approach start bar model drawing in either 2nd or 3rd grade, making this a challenging problem for many 2nd graders, but not a guess and check problem. Visualization of problem-solving actually starts in kindergarten!
See if you can solve this one like a 7-year- old. Submit your solutions by the end of the month!
Our last Word Problem Wednesday problem was from the chapter on the “Model Method and Algebra” from The Singapore Model Method for Learning Mathematics.
We had several correct answers submitted. Here’s a worked example from Shirley Davis:
How did you do?
Word Problem Wednesday – Ava, Ben, and Carlos
This month’s Word Problem Wednesday problem comes from the chapter on the “Model Method and Algebra” from The Singapore Model Method for Learning Mathematics.
The book models the Unitary Method as well as 3 variations on an algebraic solution. The author’s end this problem with the following commentary
The Model Method is a means, not and end in itself. It helps students formulate an algebraic equation to solve the problem. While more able students can proceed quickly to the absract algebraic method to solve problems without drawing a model, others may still need to rely on drawing the model as a problem-solving heuristic.
-The Singapore Model Method for Learning Mathematics, p. 58
Wise words, indeed! On to the problem:
\$120 is shared among 3 friends, Ava, Ben, and Carlos. If Ava receives \$20 less than Ben, and Ben receives 3 times as much money as Carlos, how much does Carlos receive?
Submit your solutions by the end of the month!
Last month’s Word Problem Wednesday problem was from the chapter on Real-World Problems Math in Focus 2A:
We had several correct answers submitted. Here’s a worked example from Minnesota math teacher and coach Kris Simonsen:
How did you do?
Word Problem Wednesday – Comic Books
This month’s Word Problem Wednesday problem comes from the chapter on Real-World Problems Math in Focus 2A:
Tom has 275 comic books in his collection. Chris sells 82 comic books to Tom. Then Chris has 148 comic books left. How many more comic books does Tom have than Chris now?
Submit your solutions by the end of the month!
Last month’s Word Problem Wednesday problem was from Dimensions Math 4A (available spring of 2019).
How did you do?
Word Problem Wednesday – Screws and Nails
This month’s Word Problem Wednesday problem comes from Dimensions Math 4A (available spring of 2019).
There are twice as many nails as screws in a bin. If 510 of the nails and 75 of the screws are used, there will be the same number of nails as screws. How many nails were in the bin to start with?
Submit your solutions by the end of the month!
Last month’s Word Problem Wednesday problem was from Math in Focus Grade 3.
How did you do?
Word Problem Wednesday – Farmer Fred
This month’s Word Problem Wednesday problem comes from Math in Focus Grade 3.
Farmer Fred makes 4 quarts of orange juice on Monday. He makes 2 quarts more orange juice on Tuesday than on Monday. He makes 2 more quarts on Wednesday than on Tuesday. He carries on making 2 more quarts of orange juice every day than the day before. In how many days will he make a total of 80 pints of orange juice?
Submit your solutions by the end of the month!
How did you do? | 896 | 4,089 | {"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} | 3.890625 | 4 | CC-MAIN-2020-40 | latest | en | 0.944816 |
https://www.canterbury.ac.nz/courseinfo/GetCourseDetails.aspx?course=ENME302&occurrence=20S2(C)&year=2020 | 1,603,349,041,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00020.warc.gz | 671,459,437 | 30,939 | ENME302-20S2 (C) Semester Two 2020
# Computational and Applied Mechanical Analysis
15 points
Details:
Start Date: Monday, 13 July 2020 End Date: Sunday, 8 November 2020
Withdrawal Dates
Last Day to withdraw from this course:
• Without financial penalty (full fee refund): Friday, 24 July 2020
• Without academic penalty (including no fee refund): Friday, 25 September 2020
## Description
Partial differential equations and their classification; boundary and initial conditions; analytical solution methods. Introduction to computational solution techniques and packages in solid mechanics (FEM), fluid dynamics (CFD) and heat/mass transfer.
To extend students’ exposure to, and understanding of the significance and solution of differential equations by adding partial differential equations (PDEs) to the already-familiar ordinary differential equations. Based on this mathematical understanding of PDEs, students will then become familiar with the underlying principles of the numerical solution techniques of these same equations that are utilised in commonly-employed computational packages such as COMSOL, used not in a “black box” manner but, rather, with an appreciation of the underlying mathematics and numerical techniques that are embedded within them. This understanding of computational methods will be further augmented by the students’ own development and implementation of standard algorithms for numerical solution of PDEs.
## Learning Outcomes
• On successful completion of this course students will be able to:
• Understand and apply the basic FEA elements (bar, beam and frame elements) and formulations that are readily extended from 2D to 3D analysis, including the major limitations of the methods.
• Be able to code, in Matlab or similar, the basic FEA assembly and analysis methods, and apply them to solve problems.
• Recognise and classify the different types of partial differential equations (elliptic, parabolic and hyperbolic)
• Recognise and apply, as appropriate, Dirichlet and Neumann boundary conditions (and, for unsteady state, initial conditions)
• Use separation of variables solution method where applicable
• Understand and apply D’Alembert solution and characteristics
• Understand and appreciate the essential components of the PDE models for classical mechanical systems: steady and transient heat transfer; potential and transient flow; elastic bending and waves.
• Confidently apply standard analytic solution methods to the classical PDEs used in mechanical analysis.
• Appreciate properties and limitations of any numerical solution method: accuracy, consistency, convergence
• Recognise and apply different numerical solution terminology and techniques: Spatial discretization; finite differences; weighted residual methods; polynomial interpolating/weighting functions; finite element methods; finite volumes
• Understand the strategies used in coding computational methods to maximize efficiency and minimize processing time.
• Productively and confidently use generic computational packages (e.g. COMSOL) in the solution of “real world” problems in solid mechanics, fluid flow, and heat or mass transfer
• Appreciate both the benefits and the limitations of such packages by comparison of numerical solutions with analytical solutions in situations where this is possible.
This course will provide students with an opportunity to develop the Graduate Attributes specified below:
Critically competent in a core academic discipline of their award Students know and can critically evaluate and, where applicable, apply this knowledge to topics/issues within their majoring subject.
## Timetable 2020
Students must attend one activity from each section.
Activity Day Time Location Weeks Lecture A 01 Thursday 09:00 - 10:00 E8 Lecture Theatre 13 Jul - 23 Aug 7 Sep - 18 Oct Lecture B 01 Monday 11:00 - 12:00 C2 Lecture Theatre 13 Jul - 23 Aug 7 Sep - 18 Oct Lecture C 01 Friday 10:00 - 11:00 E8 Lecture Theatre 13 Jul - 23 Aug 7 Sep - 18 Oct Lecture D 01 Tuesday 09:00 - 10:00 E8 Lecture Theatre 13 Jul - 23 Aug 7 Sep - 18 Oct Computer Lab A 01 Tuesday 16:00 - 17:30 Eng Core 342 CAD Lab 13 Jul - 23 Aug 7 Sep - 18 Oct 02 Wednesday 16:30 - 18:00 Civil - Mech E201 Mech Computer Lab 13 Jul - 23 Aug 7 Sep - 18 Oct 03 Wednesday 15:00 - 16:30 Civil - Mech E201 Mech Computer Lab 13 Jul - 23 Aug 7 Sep - 18 Oct
## Examination and Formal Tests
Activity Day Time Location Weeks Test A 01 Friday 12:30 - 14:00 Civil - Mech E201 Mech Computer Lab 10 Aug - 16 Aug 02 Friday 12:30 - 14:00 Eng Core 216 CAD Lab 10 Aug - 16 Aug 03 Friday 12:30 - 14:00 Eng Core 342 CAD Lab 10 Aug - 16 Aug 04 Friday 11:00 - 12:30 Civil - Mech E201 Mech Computer Lab 10 Aug - 16 Aug 05 Friday 11:00 - 12:30 Eng Core 216 CAD Lab 10 Aug - 16 Aug 06 Friday 11:00 - 12:30 Eng Core 342 CAD Lab 10 Aug - 16 Aug
## Assessment
Assessment Due Date Percentage
Test 1 14 Aug 2020 25%
Assignment 1 14 Sep 2020 14%
Assignment 2 12 Oct 2020 20%
Homeworks & Quizzes 6%
Final exam 35%
## Textbooks / Resources
#### Required Texts
James, Glyn; Advanced modern engineering mathematics; 4th ed; Prentice Hall, 2011.
Patankar, Suhas V; Numerical heat transfer and fluid flow; Hemisphere Pub. Corp ; McGraw-Hill, 1980.
Zienkiewicz & Taylor; The finite element method for solid and structural mechanics; 7th edition; Butterworth-Heinemann, 2014.
Harassment
* Harassment of any sort will not be tolerated. Each UC student is here to learn and to experience a friendly and supportive community.
* It is every student's right to expect: respect and courtesy from staff and other students, including freedom from harassment of any sort; fair treatment; the ability to speak out about any issues that concern them, without fear of consequences for their safety and well-being.
* Furthermore, each student has the responsibility to: respect the rights and property of others; attend to their own health and safety, and that of others; and behave in a manner towards each other that does not reflect badly on the student body or the University.
* If you, or someone you know, has experienced harassment, please talk to your lecturers, directors of study, or head of department.
Dishonest Practice
* Plagiarism, collusion, copying, and ghost writing are unacceptable and dishonest practices.
* Plagiarism is the presentation of any material (test, data, figures or drawings, on any medium including computer files) from any other source without clear and adequate acknowledgment of the source.
* Collusion is the presentation of work performed in conjunction with another person or persons, but submitted as if it has been completed only by the named author(s).
* Copying is the use of material (in any medium, including computer files) produced by another person(s) with or without their knowledge and approval.
* Ghost writing is the use of another person(s) (with or without payment) to prepare all or part of an item submitted for assessment.
Do not engage in dishonest practices. The Department reserves the right to refer dishonest practices to the University Proctor and where appropriate to not mark the work.
The University regulations on academic integrity and dishonest practice can be found here.
## Indicative Fees
Domestic fee \$975.00
International fee \$5,500.00
* Fees include New Zealand GST and do not include any programme level discount or additional course related expenses.
For further information see Mechanical Engineering.
## All ENME302 Occurrences
• ENME302-20S2 (C) Semester Two 2020 | 1,711 | 7,537 | {"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} | 2.6875 | 3 | CC-MAIN-2020-45 | latest | en | 0.908961 |
http://www.cfd-online.com/Forums/main/7816-backward-facing-step.html | 1,444,083,605,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736677946.19/warc/CC-MAIN-20151001215757-00117-ip-10-137-6-227.ec2.internal.warc.gz | 176,928,992 | 17,088 | # Backward facing step.
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July 12, 2004, 09:43 Backward facing step. #1 Fluids Guest Posts: n/a Hi All, I have been looking at some papers about Backward facing step problems. In all of these papers they refer to the step height as h and the free stream (inlet) velocity as Uo. They then give a Reynolds number of, say, 25,000 based on these quantities. However, it is rare that someone says "we used a step height of x m" . Why is this? I assume it is simply because a choice of h defines a choice of Uo through the Reynolds number and this is non-dimensional? They often gien h in terms of the boundary layer thickness at the end of the step, why is this? Also, in the spanwise direction people tend to use 32 nodes. Why is this? The axial and vertical directions are always well refined. I asusme it is because the flow is quasi-2D? Also, does anyone have any guide lines for defining a mesh spacing size (and expansion ratio) along the shear layer (and perpendicular to it) at the end of the step in an LES? Thanks for any help, J.
July 13, 2004, 11:13 Re: Backward facing step. #2 Fluids Guest Posts: n/a Surely someone knows the answer to some of these?! Please!
July 14, 2004, 20:43 Re: Backward facing step. #3 CFDtoy Guest Posts: n/a Hi there How did you generate 2D structured grids for this problem (Backward Facing step) I am looking for some grid generators. I request your kind help in this regard. Thanks CFDtoy
July 15, 2004, 15:13 Re: Backward facing step. #4 Fluids Guest Posts: n/a I did it myself. I simply created the grid points in an array. I am however, looking for a poisson grid smoother if anyone knows where I can get one. Thanks.
July 22, 2004, 05:40 Re: Backward facing step. #5 Morazza Guest Posts: n/a I'm not an expert, but maybe i can help you a little bit why don't you copy the spacing from le, kim & moin (stanford) or some other work? they are all using les bfs is not 2d from Re=1000 to 7000, but it's always simetric (barkley,gomes,henderson) the boundary layer thickness tells you if the flux is turbulent before the step, it influences the entrainment of fluid in the shear layer and if you use adimensionalization with the boundary layer, you can generalize the results with other configurations ciao
July 29, 2004, 18:20 Re: Backward facing step. #6 CFDtoy Guest Posts: n/a Hello Fluids Can you send me a structured grid for backward facing step ? I wanna check my code on a structured grid. my email is CFDtoy@hotmail.com Thanks CFDtoy
December 2, 2012, 19:41 #7 New Member Mahfuz Sarwar Join Date: Nov 2011 Posts: 13 Rep Power: 5 hi! I am recently working on simulation of backward-facing step. For validation, I will need the DNS data of Le & Moin - 'Direct numerical simulation of turbulent flow over a backward-facing step'. Please let me know if anyone can help me. mahfuzsarwar@yahoo.com
December 4, 2012, 12:28 #8 New Member RZA Join Date: Nov 2012 Posts: 25 Rep Power: 4 Why u not using the mesh independence study? Start with the initial mesh size mentioned in the literature , make it course and fine. Check the results for the mesh independence and use it for your detailed study.
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# Improving Q44 to Q49/Q50 Range?
Author Message
Manager
Joined: 12 Dec 2009
Posts: 116
Location: United States (AL)
Concentration: Finance
Schools: Harvard Business School (HBS) - Class of 2014
GMAT 1: 770 Q50 V46
GPA: 3.91
WE: Corporate Finance (Venture Capital)
Followers: 16
Kudos [?]: 105 [0], given: 6
Improving Q44 to Q49/Q50 Range? [#permalink]
### Show Tags
13 Mar 2010, 14:57
I have no issues with verbal but always struggled with math, which blows my mind since I'm an Asian engineer. But anyway, my first GMAT I got Q41. I went through GMAT Hack's Math Bible, about 3/4 of Knewton, MGMAT Number Properties, kept an error log and tried to tackle certain perceived weaknesses, and have improved my Quant to about Q44/Q45 on a consistent basis.
What should be my next step if I want to score in the Q49/Q50 range? I know the GMATClub tests (which I have access to but haven't done yet) are a good resource - but are they enough? I really liked the MGMAT Number Properties guide but given that I probably don't need very basic review, is it worth purchasing the rest of the Math guides? Should I study additional "prep materials" or just do more problems?
Thank you!
Manager
Joined: 12 Dec 2009
Posts: 116
Location: United States (AL)
Concentration: Finance
Schools: Harvard Business School (HBS) - Class of 2014
GMAT 1: 770 Q50 V46
GPA: 3.91
WE: Corporate Finance (Venture Capital)
Followers: 16
Kudos [?]: 105 [0], given: 6
Re: Improving Q44 to Q49/Q50 Range? [#permalink]
### Show Tags
13 Mar 2010, 15:17
I should add that I already plan to do the OG problems + 2 full length tests from the GMAT but will save those till the end.
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 13982
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3453
Kudos [?]: 20572 [0], given: 4340
Re: Improving Q44 to Q49/Q50 Range? [#permalink]
### Show Tags
13 Mar 2010, 16:41
I would encourage you to focus more on theory than practice questions - you will achieve your goal much faster.
From MGMAT collection, would also recommend MGMAT Word Translations Guide.
Potentially would also suggest targeted practice for the question types you are weak with and potentially going through the MGMAT Number Properties again (through the questions you got wrong or had issues with).
Any specific areas that you feel is lacking or based on your error log, what is the greatest weakness you have?
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Manager
Joined: 12 Dec 2009
Posts: 116
Location: United States (AL)
Concentration: Finance
Schools: Harvard Business School (HBS) - Class of 2014
GMAT 1: 770 Q50 V46
GPA: 3.91
WE: Corporate Finance (Venture Capital)
Followers: 16
Kudos [?]: 105 [0], given: 6
Re: Improving Q44 to Q49/Q50 Range? [#permalink]
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14 Mar 2010, 06:25
Thanks. My biggest weakness used to be number properties which is why I bought the MGMAT number properties book and it helped tremendously. Basically my improvement came in that area. Now the error log seems to be showing pretty consistent misses across all math topics, which is why I'm debating what resources I should use to gain maximum results.
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 13982
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3453
Kudos [?]: 20572 [0], given: 4340
Re: Improving Q44 to Q49/Q50 Range? [#permalink]
### Show Tags
14 Mar 2010, 15:25
Hm.... MGMAT Word Translations is a good book and potentially a step in a different direction is MGMAT Math Foundations.
Also, Kaplan Math Workbook is decent (I used this one) - here is a list of the best practice questions from it: which-one-to-start-first-manhattan-quans-or-kaplan-math-work-83184.html?view-post=623687#p623687
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Re: Improving Q44 to Q49/Q50 Range? [#permalink]
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26 Mar 2012, 23:53
For all those who wish to improve quant score......
MGMAT books are the best source..... go over and over again and each time you will find new concept
Intern
Joined: 21 Aug 2011
Posts: 27
GMAT Date: 04-02-2012
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Re: Improving Q44 to Q49/Q50 Range? [#permalink]
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27 Mar 2012, 22:48
Charm - Why does your profile say otherwise? I am confused.
Manager
Joined: 17 Sep 2011
Posts: 182
Concentration: Strategy, Operations
Schools: ISB '15
GMAT 1: 720 Q48 V40
GPA: 3.18
WE: Supply Chain Management (Manufacturing)
Followers: 13
Kudos [?]: 54 [0], given: 34
Re: Improving Q44 to Q49/Q50 Range? [#permalink]
### Show Tags
28 Mar 2012, 01:23
Shko1 wrote:
Charm - Why does your profile say otherwise? I am confused.
because this was posted in 2010 my dear....evidently some of the tips ,given by BB ,worked
Re: Improving Q44 to Q49/Q50 Range? [#permalink] 28 Mar 2012, 01:23
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Display posts from previous: Sort by
# Improving Q44 to Q49/Q50 Range?
Moderators: HiLine, WaterFlowsUp
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®. | 1,983 | 6,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-36 | longest | en | 0.915402 |
https://www.carlboettiger.info/2015/01/13/detrend-example.html | 1,701,645,546,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00561.warc.gz | 775,181,100 | 6,884 | # Detrend Example
library(knitr)
library(nimble)
library(earlywarning)
library(ggplot2)
library(tidyr)
opts_chunk$set(dev='png', fig.width=5, fig.height=5, results='hide') some sample data from earlywarning: set.seed(123) data(ibms) plot(ibm_critical) raw <- as.data.frame(ibm_critical) names(raw) <- "x" Rather than explicitly modeling the trend element predicted by the linearization, let us simply remove it: N <- length(raw$x)
raw$t <- 1:N detrend <- loess(x ~ t, raw) data <- data.frame(x = detrend$residuals/sqrt(var(detrend$residuals))) qplot(raw$t, data$x, geom='line') ## LSN version Modify the LSN model to explicitly model the changing parameter as a hidden, stochastic variable lsn <- nimbleCode({ theta ~ dunif(-100.0, 100.0) sigma_x ~ dunif(1e-10, 100.0) sigma_y ~ dunif(1e-10, 100.0) m ~ dunif(-1e2, 1e2) x[1] ~ dunif(-100, 100) y[1] ~ dunif(-100, 100) for(i in 1:(N-1)){ mu_x[i] <- x[i] + y[i] * (theta - x[i]) x[i+1] ~ dnorm(mu_x[i], sd = sigma_x) mu_y[i] <- y[i] + m * t[i] y[i+1] ~ dnorm(mu_y[i], sd = sigma_y) } }) Constants in the model definition are the length of the dataset, $$N$$ and the time points of the sample. Note we’ve made time explicit, we’ll assume uniform spacing here. constants <- list(N = N, t = raw$t)
Initial values for the parameters
inits <- list(theta = 6, m = 0, sigma_x = 1, sigma_y = 1, y = rep(1,N))
and here we go:
Rmodel <- nimbleModel(code = lsn,
constants = constants,
data = data,
inits = inits)
Cmodel <- compileNimble(Rmodel)
mcmcspec <- configureMCMC(Rmodel, print=TRUE,thin=2e2)
Rmcmc <- buildMCMC(mcmcspec)
Cmcmc <- compileNimble(Rmcmc, project = Cmodel)
Cmcmc$run(1e6) NULL and examine results samples <- as.data.frame(as.matrix(Cmcmc$mvSamples))
dim(samples)
[1] 5000 84
samples <- samples[,1:4]
long <- gather(samples)
apply(samples, 2, mean)
m sigma_x sigma_y theta
0.0003790592 1.0792385676 0.1920288851 -0.0150533955
ggplot(long) +
geom_line(aes(seq_along(value), value)) +
facet_wrap(~key, scale='free')
ggplot(long) +
geom_density(aes(value)) +
facet_wrap(~key, scale='free')
sessionInfo()
R version 3.1.2 (2014-10-31)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] methods stats graphics grDevices utils datasets
[7] base
other attached packages:
[1] tidyr_0.2.0 ggplot2_1.0.0 earlywarning_0.0-1
[4] nimble_0.3 yaml_2.1.13 knitr_1.9
loaded via a namespace (and not attached):
[1] codetools_0.2-10 colorspace_1.2-4 deSolve_1.11
[4] digest_0.6.8 evaluate_0.5.5 formatR_1.0
[7] grid_3.1.2 gtable_0.1.2 igraph_0.7.1
[10] labeling_0.3 MASS_7.3-39 mnormt_1.5-1
[13] munsell_0.4.2 parallel_3.1.2 plyr_1.8.1
[16] proto_0.3-10 psych_1.5.1 Rcpp_0.11.4
[19] reshape2_1.4.1 scales_0.2.4 stringr_0.6.2
[22] tools_3.1.2 | 1,125 | 3,122 | {"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": 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} | 2.75 | 3 | CC-MAIN-2023-50 | longest | en | 0.44809 |
https://www.moo.com/blog/2009/09/24/question-52-in-our-cryptic-quiz-%E2%80%9Cthree-is-the-magic-number%E2%80%9D/comment-page-1/ | 1,487,554,736,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170380.12/warc/CC-MAIN-20170219104610-00300-ip-10-171-10-108.ec2.internal.warc.gz | 871,337,956 | 30,684 | ## Question 52 in our cryptic quiz: “Three is the magic number”
24th September 2009 by Pete
It’s time to get your brain in gear and answer today’s first question. What’s in it for me I hear you say?
Well; Today you have another chance to win one of the coolest little gadgets around. We are giving away more Mix Tape USBs from our friends at Suck UK. You can store your MP3s on the USB and be retro and master of all things digital all at the same time!
If you want to see the Mix Tape in action, take a look.
But you have to earn them! So, let the games begin……
### Hardboiled, The Journey
Payback
Nick drew Sam’s short straw, so once we were on the boat it was his turn to face a puzzle challenge.
Sam said, “At this very moment a woman is 21 years older than her child. In exactly 6 years from now she will be precisely 5 times older than the child.”
I could see the wheels turning in Nick’s mind.
“The question is this,” Sam said, “where is the father right now?”
Nick paused a second, surprised. But suddenly he relaxed and said, “he’s right next to the mother, I suspect.”
### QUESTION: How old is the child right now?
Answer below in your comments please, and keep coming back, as the next question will be up soon.
1. Jenny:
The child has just been conceived. -.75 years old.
2. David:
The child is 5.4 years old right now.
3. Maura:
The child is currently being conceived: -9 months old.
4. Aaron McFann:
The child is 1. The mother is 29. In six years the child will be 7 and the mother will be 35. 35/7 = 5.
5. Aaron McFann:
Nevermind, I completely skipped over the mother being 21 years older than the child. The correct answer is:
The child has just been born (if the child was carried for 9 months).
6. Midknyt:
The baby is -0.75, minus 9 months…or in other words, being conceived as we speak.
Tsk tsk.
7. AG:
Hehe. The child is -9 months old…
8. pauline:
child is 3
9. Chi:
The child is currently 6.75 years old (and the mom is 27.75 years old).
Proof: Let c=child’s age, m=mom’s age
1. m = 21 + c (mom is 21 year’s older than child)
=> 27.75 = 21 + 6.75
=> 27.75 = 27.75 is true
2. m + 6 = 5 * c (mom will be 5 times as old as child’s current age in 6 years)
=> 27.75 + 6 = 5 * 6.75
=> 33.75 = 33.75 is true
10. Sally:
-9 months!
11. Nina:
-9 months…
12. Clare:
I imagine the child is just conceived.
13. Irit:
The child is minus nine months old.
14. Joe Tortuga:
The child has just been conceived!
15. Caro:
Minus nine months, or in other words, just conceived!
Honestly MOO, your minds are in the gutter!
16. Isabelle:
He has just been conceived (he’s still in his mother belly with nearly one year to go before being born). No wonder why dad is near
17. Amanda:
The child is -3/4 years old… or, it is 9 months before the child is born. Thus why the father is right next to the mother right now
18. Andrew Sosnkowski:
the child is being conceived
19. Michael:
It´s nine month before it´s birth. That´s why the father should be right beside the mother
20. Cherry:
The child has not yet been born.
21. frankie:
9 months old!
22. Jessica:
three months away from birth
23. Shereen:
The baby was conceived a month ago…I think.
24. Diana Ramirez:
The child was just conceived, so s/he is -9 months old (or -.75 years old).
25. Jenn:
woman = Y
child = X
woman is 21 years older: Y = X+21
in 6 years: Y+6 = 5(X+6)
solving:
Y+6 = 5X+30
Y = 5X+24
5X+24 = X+21
-4X = 3
X = -3/4
the child is -3/4 years old. 3/4 of a year is 9 months, so the child is -9 months old… (s)he’s being conceived!
26. jgardy:
Newborn
27. jgardy:
Oooh no scratch my last answer. The kid was just concieved.
28. Laurel Fan:
The child is -.75 years, so it will be born in about 9 months.
29. spyra:
How risque!
The child is 9 months before birth (-3/4 of a year), which means that they are doing it right now!
30. istoo:
Six and three quarters.
6 years and 9 months old.
32. Sophie:
The child is minus 3/4 of a year, which is near enough to conception as to make Sam’s answer feasible!
33. Anna:
The child is currently -0.75, or minus 9 months, or currently being conceived
34. Rebekah:
The child is being born.
35. katie:
The child is -.75 of a year, or -9 months. He’s just been conceived. And that’s why the father is right next to the mother! It all makes sense now!
LOL good question!
36. Mike:
-.75yr, or more conventionally, nine months away from birth or right around the time of conception.
37. Gabi:
9 month before birth
38. Cleopatra Browne:
The child has just been conceived and is -9 months old.
39. an0305:
the child is conceived at this moment! (it’s -.75 years old, so -9 months!)
we won't publish your email, or spam you
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If you can do us a favour, we'll do one for you in return. Sign up as an affiliate and you could earn up to £5, \$8, or €6 for every new customer you send our way. | 1,569 | 5,676 | {"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} | 3.1875 | 3 | CC-MAIN-2017-09 | longest | en | 0.955713 |
http://ashutosh7s.blogspot.com/2011/08/what-is-probability-that-outside-of.html | 1,526,962,203,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864624.7/warc/CC-MAIN-20180522034402-20180522054402-00162.warc.gz | 23,967,087 | 17,779 | ## Saturday, August 6, 2011
### What is the probability that the outside of this cube is completely black?
Twenty seven identical white cubes are assembled into a single cube,
the outside of which is painted black. The cube is then disassembled
and the smaller cubes are thoroughly shuffled in a bag.
A blindfolded man (who cannot feel the paint) reassembles the pieces
into a cube. What is the probability that the outside of this cube is
completely black?
#### 1 comment:
Jiten said...
I think we need to know area of the wall and area of individual cubes to calculate this | 131 | 581 | {"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} | 3.15625 | 3 | CC-MAIN-2018-22 | latest | en | 0.970553 |
http://barkerhugh.blogspot.com/2011/11/ | 1,529,535,385,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863939.76/warc/CC-MAIN-20180620221657-20180621001657-00396.warc.gz | 34,251,542 | 18,198 | ## Saturday, 19 November 2011
### Symmetry of the prime patterns
Just a quick note on the implications of the K numbers I talked about in my last post. (Edit - since deleted since there were a few issues with the approach - will try to come back and clarify this post at some point...)
In each primorial 0-30, 0-210, 0-2310 etc, there is a limited symmetrical pattern. Up to P#, K numbers (numbers that are either prime or don't have any factors or P or below) on the "way up" are composite, but K numbers on the "way down" (eg the remainder when a K number is subtracted from P#) are prime if they are below P^2. (If these remainders are over P^2 then they are either prime or they have a different set of composite factors to the K number.)
Not a very satisfying symmetry, granted, but it does explain why the assymmetrical prime patterns have such a symmetrical "feel". What we are actually seeing is a fractal, assymmetrical pattern gradually develop. But for every region P^2 to Pn^2 where Pn is the next prime after P, there is a symmetry with the region (Pn# - Pn^2) to (Pn# - P^2) with primes in the former being mirrored by Kpn numbers (numbers that don't have factors of Pn or below) in the latter.
## Saturday, 5 November 2011
### From Euclid to the infinitude of 6n+1 and 6n-1 primes.
I said I'd look at the logic of assuming that there is an infinite number of primes in each axis of the n-factor grid used in previous posts (eg primes in the form of 6n+1 and in the form 6n-1). I'm not sure how to prove this properly. Here's how far I got so far, if anyone can see how to fill in the last step I'd be pleased.
Euclid: Assume X is the highest prime number. Multiply all primes up to X together then add one to the resulting primorial. The result must be either a prime or the product of primes higher than X.
OK, so this number X#+1 is clearly of the form 6n+1 since the first two primes are 2*3. However we haven't yet proved that there is an infinite number of 6n+1 primes, since, if it is a product of primes higher than X, those primes could be 6n-1 numbers (since an even number of 6n-1 numbers multiplied will give us a 6n+1 number).
So let's do something slightly different to Euclid. Assume X is the highest prime number. Multiply all primes up to X together then subtract one from the resulting primorial. The result must be either a prime or the product of primes higher than X. If it is a prime it is a 6n-1 prime. If it is the product of primes higher than X, then at least one of them must be a 6n-1 number, since any number of 6n+1 numbers multiplied together will produce another 6n+1 number.
So that proves there is an infinite number of 6n-1 numbers, but still not that there is an infinite number of 6n+1 numbers.
Intuitively it seems there must be since each number in the Sieve of Eratosthenes will sieve out a fairly even selection of 6n+1 and 6n-1 numbers, but that's only a guess really.
Of course the proof above would also show that there is an infinite number of 6n+1 numbers (since you need a 6n+1 number and a 6n-1 number to multiply together to get a new 6n-1 number) - but only if we could assume that X#-1 wouldn't always be a new 6n-1 prime and would sometimes be a product of other primes instead. So it's not quite there. (Edit - on returning to this post somewhat later on I realise that this paragraph is gobbledygook since a 6n-1 number could always be the product of an odd number of 6n-1 primes, but it doesn't really matter as either way the inifinitude of 6n+1 primes isn't proven using this approach)
I'm probably being stupid and there is probably a proof out there somewhere. (Edit: I think it is implied by Dirichlet's Theorem, so has been proved - I just want to see a proof I can get my head around a bit more easily).
Edit: For the infinitude of 6n+1 primes I'm tempted to suggest using the fraction Euler used (below) when he proved its inverse was equivalent to the sum of the harmonic series and thus diverged. He did this over all prime numbers, but it is clear that within the 6n+1 axis, the same maths is at work, since 4/5 of the numbers are not multiples of 5, 6/7 aren't multiples of 7, 10/11 aren't multiples of 11 and so on. I don't know how to express that idea more rigorously though.
1*2*4*6*….(p-1)
2*3*5*7*……p
(Edit: What I was getting at here was something like this bit from Wikipedia: "Stronger forms of Dirichlet's theorem state that, for any arithmetic progression, the sum of the reciprocals the prime numbers in the progression diverges, and that different arithmetic progressions with the same modulus have approximately the same proportions of primes." - that seems intuitively correct to me and would answer the question posed above, I'll see if I can find any proofs that are easy enough for me to follow).
### Semi-prime distribution on the n-factor grid
Finally, I want to take a look at what we can know about the distribution of semi-primes on this grid. Here it is again as a reminder, with the primes coloured pink and the semi-primes yellow (other colours indicate numbers with more than 2 prime factors).
So what do we know about the primes and semi-primes on this grid.
1. There is an infinite number of primes in the central axes. I think Euclid's proof of the infinitude of primes might also suggest that there is an infinite number in each of these axes but I'm not 100% sure of that - I'll assume it is true for now and try running through the logic in another post.
2. Either way there is an infinite number of semi-primes on the grid. (This is obvious anyway, since each prime number will generate an infinite chain of semi-primes when multiplied by each of the infinite chain of primes.
3. Each prime on the central axes will generate an infinite, intermittent horizontal or vertical line of semi-primes stretching across the grid. This will hit an infinite number of diagonals.
What we don't know:
1. Is there an infinite number of semi-primes on the twin prime diagonal? If so the twin prime conjecture is true. The observations above suggest why this might be true, but they aren't proof of it.
2. Is there an infinite number of semi-primes on every infinite diagonal on the grid? If so, Polignac's conjecture is true (this refers to the larger grid as seen in the second and third image on this page).
3. Is it impossible to take a diagonal from one axis to the other (or from an axis to the diagonal NW-SE boundary - see previous post for a clearer explanation) without hitting a semi-prime. If so the Goldbach conjecture is true.
### The n-factor grid: twin primes, Polignac primes and the Goldbach conjecture.
So, having introduced the grid of n-factor composite numbers in the previous post, how does this help us visualise prime problems?
First, let's look at twin primes. If we are looking for twin primes, we need to look at the pairs of numbers 6k-1 and 6k+1. In this image, the highlighted boxes contain the products of possible twin prime candidates (5*7, 11*13 etc). This diagonal, continued to infinity would contain all such products.
The reason for looking at the products of these numbers is this: where the numbers along this diagonal are semi-prime (yellow), the two factors on the horizontal axes must be twin primes. This is kind of obvious, since only the prime numbers on the central axes will produce a semi-prime when multiplied together. But it does provide some visual explanation of why you are likely to get a proportion of primes along a diagonal of this grid. Put in basic terms, it is obviously going to be pretty hard (especially at low numbers) to draw a diagonal that doesn't hit one of the vertical or horizontal stripes of semi-primes. It also doesn't suffer from the same problem as the original grid I was using as it does continue to distinguish primes from semi-primes and other composites.
Next, we can clearly see on this grid why, as we go up through the Sieve of Eratosthenes, the prime N will remove a maximum of 2/N of remaining twin prime candidates: Since the diagonal will pass through 1/N multiples of N on the horizonal axis and 1/N on the vertical axis, it will encounter up to 2/N multiples of N altogether*.
The twin prime conjecture is true if there is an infinite number of semi-primes on this diagonal, continued to infinity. (I'll talk about this a bit more on the next post).
Incidentally, the other diagonals on the grid all relate to other prime problems. In this diagram, the diagonals highlighted show the products of prime pairs with larger gaps, cousin primes (N, N+4), sexy primes (N, N+6) and some larger examples. If Polignac's conjecture that "for any positive even number n , there are infinitely many prime gaps of size n" is true, then any diagonal on this grid (continued to infinity) contains an infinite number of semi-primes. (Only top right is coloured).
On the grid below, the highlighted diagonals relate to the Goldbach conjecture. In this case the chains are finite - each chain contains all the pairs of 6n+/-1 numbers that could be added together to get a particular sum, for instance for 78 the potential pairs are 73+5, 67+11, 61+17....65+13, 71+7. If each diagonal that stretches from one axis to the other (or from an axis to the diagonal boundary of square numbers to the bottom left) contains at least one semi-prime then the Goldbach conjecture is true.
Edit: I later realise that while this is true, it would demonstrate a slightly stronger conjecture than Goldbach, since Goldbach would allow pairs of primes where one of the pair is 3 and the grid above ignores products of 3.
* 2/N is a maximum because, as previously noted, the first instance of N is not a composite so may be part of a twin prime pair. Also, for higher gaps between the primes, the maths changes slightly - for instance for primes separated by a gap of 30, the number 5 will only remove 1/5 of candidate pairs, since the diagonal will hit the 5-multiples on the horizontal and vertical axis on the same square.
Incidentally this post is a bit wrong - I should have been clearer that the prime N will remove a maximum of 2/N of remaining twin prime candidates in any infinitely repeating chain of numbers separated by the same interval - so for instance after we have sieved to 11, we know that the pair 101/103 is a twin prime. Within any higher primorial length pattern, there will be repetitions of this "twin prime candidate" gap, for instance in 13#, the gap repeats at 2310+101/3, 4620+101/3, 6930+101/3 and on to infinity. Some of these candidates will be eliminated by higher primes, but the prime P will only eliminate 2/P of all instances. All a bit more complicated then...
### The grid of n-factor numbers
So in the previous post I showed a grid of composite numbers of the form 6n+/-1, and also looked at the problem of using it to look at prime patterns.
What I tried next isn't a complete solution to this problem, but I think it brings a bit more clarity to the grid and makes it more visually useful. Instead of the slightly confusing overlaying of rectangles of n-multiples, I used a different colour scheme, and graded the numbers according to how many factors they have. In this image, prime numbers are pink, semi-primes (numbers with 2 factors such as 25, 35, 49, 55 etc) are yellow, 3-factor numbers are green, 4-factor numbers are blue, 5-factor numbers are red, and the one 6-factor number (21875) is purple. (I've only shown the (6n+1)(6n-1) region, but this pattern would extend over the whole grid.)
We still see a similarly fractal pattern starting to emerge. In the next post I will talk about the distribution of semi-primes on this grid, and what we can know about how it will develop.
### The composite 6n+1/6m-1 grid
To help look at prime patterns, I used a spreadsheet to create a grid of multiples of numbers in the form (6n+1) and (6n-1). The reason for this is that all numbers of this form are either prime numbers or multiples of smaller primes of the same form. So for instance, the first few composite numbers of this form are 25 (5*5), 35 (5*7), 49 (7*7), 55 (5*11) and so on. At it's most basic the grid looks like this.
If you imagine this continuing to the right and upwards, it is a small part of the grid of all numbers generated by multiplying together 6n+1 numbers (on the axis to the South and East) and 6n-1 numbers (on the axis to the North and West). The North-East quadrant contains numbers that are of the form (6n+1)(6m-1), the North-West contains (6n-1)(6m-1), the South-East contains those of the form (6n+1)(6m+1). (The whole pattern would mirror across the bottom diagonal row, which contains all the square numbers of this form, thus I leave it blank for clarity).
Next I coloured in stripes of composites with factors of n. So, for instance, I coloured in multiples of 5 yellow.
Overlaying multiples of 7 (brown), 11 (pink) and 13 (blue), you start to see the rather obvious fractal style pattern, a bit like a Sierpinski carpet (except we are eliminating squares of growing size within an infinite plain), but with the centre of the squares off-centre:
(You'll see that where a later prime is on the same line, I've overlayed the colour, so the 55 line is coloured pink for 11, not yellow for 5).
Here is a view of a larger section, also completed up to the multiples of 13.
And one more long shot, to show how the grid extends.
Note that all numbers that appear in the composite area (eg not the central axes) will also appear later on in those central axes as they are higher numbers in the form (6n+/-1). Also numbers with 3 or more factors will appear several times in the composite area, for instance 245 appears as 7*35 as well as 5*49. An infinitely extended version of the grid would (in theory) contain all composite numbers of the form (6n+/-1) as well as all prime numbers of this form (restricted to the central axes).
The problem with the composite 6n+/-1 grid
This grid gives an interesting visual way to observe the build up of symmetrical primorial patterns - basically you go up through the Sieve of Eratosthenes and gradually colour in the map and the patterns that build up are the same patterns I (and others) have observed in the distribution of composite numbers (and, consequently, the remaining numbers that are still "prime candidates", in other words numbers that are either prime or multiples of higher primes).
However there is a problem, which is that the grid of "composite numbers", as it expands, will also include the prime numbers in the central axes. So, when we go on to colour in multiples of 17, the 17 square will be coloured in as though it is a composite. This is one of the basic problems of sieve theory (as I understand it) as this sieve fails to clearly distinguish primes. Extended to infinity, the entire grid would be coloured in as "composite", even though there are an infinite number of primes.
(This is a serious problem for any attempt to understand twin primes and other related problems using these primorial patterns. It is clear that any set of prime numbers up to Pn will leave an infinitely repeating symmetrical pattern of "prime candidates" in subsequent Pn# length primorials. But we aren't actually looking for "prime candidates". We are looking those candidates that get sieved out as "composite" even though they are actually one of a pair of twin primes).
In the next post I will look at a slightly different grid.
### Mea Culpa
I had a bit of a messianic moment in January when I thought I had proved the twin prime conjecture. Very silly of me, and someone from the Mersenne forum patiently told me where I was wrong. On further searching I also see plenty of other people have noticed the primorial patterns that underlied the attempt.
So now, in a more humble frame of mind I've returned to thinking about this problem and the way it relates to a few other issues, so I will be making a few posts to make observations, without any claim to be doing anything more than keeping a record of my thoughts.
I'll leave the earlier posts up as a useful reminder of my own hubris. | 3,868 | 16,136 | {"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} | 3.6875 | 4 | CC-MAIN-2018-26 | latest | en | 0.955733 |
https://search.r-project.org/CRAN/refmans/BayesTools/html/prior_factor.html | 1,723,000,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00128.warc.gz | 399,998,905 | 2,784 | prior_factor {BayesTools} R Documentation
## Creates a prior distribution for factors
### Description
prior_factor creates a prior distribution for fitting models with factor predictors. (Note that results across different operating systems might vary due to differences in JAGS numerical precision.)
### Usage
prior_factor(
distribution,
parameters,
truncation = list(lower = -Inf, upper = Inf),
prior_weights = 1,
contrast = "meandif"
)
### Arguments
distribution name of the prior distribution. The possible options are "point"for a point density characterized by a location parameter. "normal"for a normal distribution characterized by a mean and sd parameters. "lognormal"for a lognormal distribution characterized by a meanlog and sdlog parameters. "cauchy"for a Cauchy distribution characterized by a location and scale parameters. Internally converted into a generalized t-distribution with df = 1. "t"for a generalized t-distribution characterized by a location, scale, and df parameters. "gamma"for a gamma distribution characterized by either shape and rate, or shape and scale parameters. The later is internally converted to the shape and rate parametrization "invgamma"for an inverse-gamma distribution characterized by a shape and scale parameters. The JAGS part uses a 1/gamma distribution with a shape and rate parameter. "beta"for a beta distribution characterized by an alpha and beta parameters. "exp"for an exponential distribution characterized by either rate or scale parameter. The later is internally converted to rate. "uniform"for a uniform distribution defined on a range from a to b parameters list of appropriate parameters for a given distribution. truncation list with two elements, lower and upper, that define the lower and upper truncation of the distribution. Defaults to list(lower = -Inf, upper = Inf). The truncation is automatically set to the bounds of the support. prior_weights prior odds associated with a given distribution. The value is passed into the model fitting function, which creates models corresponding to all combinations of prior distributions for each of the model parameters and sets the model priors odds to the product of its prior distributions. contrast type of contrast for the prior distribution. The possible options are "meandif"for contrast centered around the grand mean with equal marginal distributions, making the prior distribution exchangeable across factor levels. In contrast to "orthonormal", the marginal distributions are identical regardless of the number of factor levels and the specified prior distribution corresponds to the difference from grand mean for each factor level. Only supports distribution = "mnormal" and distribution = "mt" which generates the corresponding multivariate normal/t distributions. "orthonormal"for contrast centered around the grand mean with equal marginal distributions, making the prior distribution exchangeable across factor levels. Only supports distribution = "mnormal" and distribution = "mt" which generates the corresponding multivariate normal/t distributions. "treatment"for contrasts using the first level as a comparison group and setting equal prior distribution on differences between the individual factor levels and the comparison level. "independent"for contrasts specifying dependent prior distribution for each factor level (note that this leads to an overparameterized model if the intercept is included).
### Value
return an object of class 'prior'.
prior()
# create an orthonormal prior distribution | 676 | 3,547 | {"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} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.809111 |
https://www.gamedev.net/articles/programming/general-and-gameplay-programming/algorithmic-forays-part-4-r2106?forceDownload=1&_k=880ea6a14ea49e853634fbdc5015a024 | 1,516,408,285,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888302.37/warc/CC-MAIN-20180119224212-20180120004212-00518.warc.gz | 919,853,713 | 27,630 | • # Algorithmic Forays Part 4
General and Gameplay Programming
This series has since been revised and combined
In the last article, I presented the three main techniques in use for generating recognizers from regular expressions. Just to remind you, the techniques are:
• Build a NFA from the regex. "Simulate" the NFA to recognize input.
• Build a NFA from the regex. Convert the NFA to a DFA. Simulate the DFA to recognize input.
• Build a DFA directly from the regex. Simulate the DFA to recognize input. At first, I was determined to spare you from the whole DFA/NFA discussion and just use the third - direct DFA - technique for recognizer generation. Then, I changed my mind, for two reasons. First, the distinction between NFAs and DFAs in the regex world is important. Different tools use different techniques (for instance, Perl uses NFA while lex and egrep use DFA), and it is valuable to have at least a basic grasp of these topics. Second, and more important, I couldn't help falling to the charms of the NFA-from-regex construction algorithm. It is simple, robust, powerful and complete - in one word, beautiful. So, I decided to go for the second technique. [size="5"]Construction of a NFA from a regular expression Recall the basic building blocks of regular expressions: eps which represents "nothing" or "no input"; characters from the input alphabet (we used a and b most often in these articles); characters may be concatenated, like this: abb; alternation a|b meaning a or b; the star * meaning "zero or more of the previous"; and grouping (). What follows is Thompson's construction - an algorithm that builds a NFA from a regex. The algorithm is syntax directed, in the sense that it uses the syntactic structure of the regex to guide the construction process. The beauty and simplicity of this algorithm is in its modularity. First, construction of trivial building blocks is presented.
1. For eps, construct the NFA: Here i is a new start state and f is a new accepting state. It's easy to see that this NFA recognizes the regex eps
2. For some a from the input alphabet, construct the NFA: Again, it's easy to see that this NFA recognizes the trivial regex a.
Now, the interesting part of the algorithm: an inductive construction of complex NFAs from simple NFAs. More specifically, given that N(s) and N(t) are NFA's for regular expressions s and t, we'll see how to combine the NFAs N(s) and N(t) according to the combination of their regexes.
1. For the regular expression s|t, construct the following composite NFA N(s|t): The eps transitions into and out of the simple NFAs assure that we can be in either of them when the match starts. Any path from the initial to the final state must pass through either N(s) or N(t) exclusively. Thus we see that this composite NFA recognizes s|t
2. For the regular expression st (s and then t), construct the composite NFA NFA(st): The composite NFA will have the start state of N(s) and the end state of N(t). The accepting (final) state of N(s) is merged with the start state of N(t). Therefore, all paths going through the composite NFA must go through N(s) and then through N(t), so it indeed recognizes N(st).
3. For the regular expression s*, construct the composite NFA N(s*): Note how simply the notion of "zero or more" is represented by this NFA. From the initial state, either "nothing" is accepted with the eps transition to the final state or the "more than" is accepted by going into N(s). The eps transition inside N(s) denotes that N(s) can appear again and again.
4. For the sake of completeness: a parenthesized regular expression (s) has the same NFA as s, namely N(s).
As you can see, the algorithm covers all the building blocks of regular expressions, denoting their translations into NFAs. [size="5"]An example If you follow the algorithm closely, the following NFA will result for (our old friend,) the regex (a|b)*abb:
Sure, it is much larger than the NFA we saw in the previous article for recognizing the same regex, but this NFA was automatically generated from a regex description using Thompson's construction, rather than crafted by hand. Let's see how this NFA was constructed: First, it's easy to note that states 2 and 3 are the basic NFA for the regex a. Similarly, states 4 and 5 are the NFA for b. Can you see the a|b ? It's clearly states 1,2,3,4,5,6 (without the eps transition from 6 to 1). Parenthesizing (a|b) doesn't change the NFA The addition of states 0 and 7, plus the eps transition from 6 to 1 is the star on NFA(a|b), namely states 0 - 7 represent. The rest is easy. States 8 - 10 are simply the concatenation of (a|b)* with abb. Try to run a few strings through this NFA until you convince yourself that it indeed recognizes (a|b)*abb. Recall that a NFA recognizes a string when the string's characters can be spelled out on some path from the initial to the final state. [size="5"]Implementation of a simple NFA At last, let's get our hands on some code. Now that we know the theory behind NFA-from-regex construction, it's clear that we will be doing some NFA manipulations. But how will we represent NFAs in code? NFA is not a trivial concept, and there are full-blown implementations for general NFAs that are far too complex for our needs. My plan is to code as simple an implementation as possible - one that will be enough for our needs and nothing more. After all, the regex recognizing engine is not supposed to expose its NFAs to the outer world - for us a NFA is only an intermediate representation of a regular expression, which we want to simulate in order to "accept" or "reject" input strings. My philosophy in such cases is the KISS principle: "Keep It Simple, Stupid". The goal is first to code the simplest implementation that fits my needs. Later, I have no problem refactoring parts of the code and inserting new features, on an as-needed basis. A very simple NFA implementation is now presented. We will build upon it in future articles, and for now it is enough just to demonstrate the concept. Here is the interface: #ifndef NFA_H #define NFA_H #include using namespace std; // Convenience types and constants // typedef unsigned state; typedef char input; enum {EPS = -1, NONE = 0}; class NFA { public: // Constructed with the NFA size (amount of states), the // initial state and the final state // NFA(unsigned size_, state initial_, state final_); // Adds a transition between two states // void add_trans(state from, state to, input in); // Prints out the NFA // void show(void); private: bool is_legal_state(state s); state initial; state final; unsigned size; vector > trans_table; }; #endif // NFA_H As promised, the public interface is kept trivial, for now. All we can do is create a NFA object (specifying the amount of states, the start state and the final state), add transitions to it, and print it out. This NFA will then consist of states [font="Courier New"]0 .. size-1[/font], with the given transitions (which are single characters). Note that we use only one final state for now, for the sake of symplicity. Should we need more than one, it won't be difficult to add. A word about the implementation: I don't want to go deep into graph-theory here (if you're not familiar with the basics, a web search can be very helpful), but basically a NFA is a directed graph. It is most common to implement a graph using either a matrix or an array of linked lists. The first implementation is more speed efficient, the second is better space-wise. For our NFA I picked the matrix ([font="Courier New"]vector[/font] of [font="Courier New"]vector[/font]s), mostly because (in my opinion) it is simpler. The classic matrix implementation of a graph has 1 in cell (i, j) when there is an edge between vertex i and vertex j, and 0 otherwise. A NFA is a special graph, in the sense that we are interested not only in whether there is an edge, but also in the condition for the edge (the input that leads from one state to another in FSM terminology). Thus, our matrix holds [font="Courier New"]input[/font]s (a nickname for [font="Courier New"]char[/font]s, as you can see). So, for instance, 'c' in [font="Courier New"]trans_table[arrayi][/arrayi][j][/font] means that the input 'c' leads from state i to state j in our NFA. Here is the implementation for the NFA class: #include #include #include #include #include "nfa.h" using namespace std; NFA::NFA(unsigned size_, state initial_, state final_) { size = size_; initial = initial_; final = final_; assert(is_legal_state(initial)); assert(is_legal_state(final)); // Initialize trans_table with an "empty graph", no transitions // between its states // for (unsigned i = 0; i < size; ++i) { vector v; for (unsigned j = 0; j < size; ++j) { v.push_back(NONE); } trans_table.push_back(v); } } bool NFA::is_legal_state(state s) { // We have 'size' states, numbered 0 to size-1 // if (s < 0 || s >= size) return false; return true; } void NFA::add_trans(state from, state to, input in) { assert(is_legal_state(from)); assert(is_legal_state(to)); trans_table[from][to] = in; } void NFA::show(void) { cout << "This NFA has " << size << " states: 0 - " << size - 1 << endl; cout << "The initial state is " << initial << endl; cout << "The final state is " << final << endl << endl; for (unsigned from = 0; from < size; ++from) { for (unsigned to = 0; to < size; ++to) { input in = trans_table[from][to]; if (in != NONE) { cout << "Transition from " << from << " to " << to << " on input "; if (in == EPS) { cout << "EPS" << endl; } else { cout << in << endl; } } } } } The code is very simple, so you should have no problem understanding what every part of it does. To demonstrate, lets see how we would use this class to create the NFA for (a|b)*abb - the one we built using Thompson's construction earlier in this article (only the driver code is included): #include "nfa.h" int main() { NFA n(11, 0, 10); n.add_trans(0, 1, EPS); n.add_trans(0, 7, EPS); n.add_trans(1, 2, EPS); n.add_trans(1, 4, EPS); n.add_trans(2, 3, 'a'); n.add_trans(4, 5, 'b'); n.add_trans(3, 6, EPS); n.add_trans(5, 6, EPS); n.add_trans(6, 1, EPS); n.add_trans(6, 7, EPS); n.add_trans(7, 8, 'a'); n.add_trans(8, 9, 'b'); n.add_trans(9, 10, 'b'); n.show(); return 0; } This would (quite expectedly) result in the following output: [font="Courier New"]This NFA has 11 states: 0 - 10 The initial state is 0 The final state is 10 Transition from 0 to 1 on input EPS Transition from 0 to 7 on input EPS Transition from 1 to 2 on input EPS Transition from 1 to 4 on input EPS Transition from 2 to 3 on input a Transition from 3 to 6 on input EPS Transition from 4 to 5 on input b Transition from 5 to 6 on input EPS Transition from 6 to 1 on input EPS Transition from 6 to 7 on input EPS Transition from 7 to 8 on input a Transition from 8 to 9 on input b Transition from 9 to 10 on input b[/font] As I mentioned earlier: as trivial as this implementation may seem at the moment, it is the basis we will build upon in future articles. Presenting it in small pieces will, hopefully, make the learning curve of this difficult subject less steep for you.
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• 1929 views | 3,139 | 12,572 | {"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} | 2.734375 | 3 | CC-MAIN-2018-05 | latest | en | 0.91674 |
http://stackoverflow.com/questions/949448/multivariate-mapping-regression-with-objective-function | 1,467,146,709,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397111.67/warc/CC-MAIN-20160624154957-00058-ip-10-164-35-72.ec2.internal.warc.gz | 286,654,432 | 17,077 | # Multivariate mapping / regression with objective function
Overview
I have a multivariate timeseries of "inputs" of dimension N that I want to map to an output timeseries of dimension M, where M < N. The inputs are bounded in [0,k] and the outputs are in [0,1]. Let's call the input vector for some time slice in the series "I[t]" and the output vector "O[t]".
Now if I knew the optimal mapping of pairs <I[t], O[t]>, I could use one of the standard multivariate regression / training techniques (such as NN, SVM, etc) to discover a mapping function.
Problem
I do not know the relationship between specific <I[t], O[t]> pairs, rather have a view on the overall fitness of the output timeseries, i.e. the fitness is governed by a penalty function on the complete output series.
I want to determine the mapping / regressing function "f", where:
``` O[t] = f (theta, I[t])
```
Such that penalty function P(O) is minimized:
``` minarg P( f(theta, I) )
theta
```
[Note that the penalty function P is being applied the resultant series generated from multiple applications of f to the I[t]'s across time. That is f is a function of I[t] and not the whole timeseries]
The mapping between I and O is complex enough that I do not know what functions should form its basis. Therefore expect to have to experiment with a number of basis functions.
Have a view on one way to approach this, but do not want to bias the proposals.
Ideas?
- | 344 | 1,446 | {"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-2016-26 | latest | en | 0.91167 |
https://www.jiskha.com/display.cgi?id=1337860151 | 1,503,081,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105086.81/warc/CC-MAIN-20170818175604-20170818195604-00184.warc.gz | 948,573,143 | 4,181 | finance
posted by .
Assume you are planning to invest \$5,000 each year for six years and will earn 10 percent per year. Determine the future value of this annuity if your first \$5,000 is invested at the end of the first year.
• finance -
1. \$5000 @ end of first yr.
2. 5000*1.1 + 5000=10500 @ end of 2nd yr.
3. 10500*1.1 + 5000 = 16,550. @ end of 3rd yr.
4. 16,550*1.1 + 5000 = 23,205 @ end of
fourth yr.
5. 23,205*1.1 + 5000 = 30,525.50 @ end
of 5th yr.
6. 30,525.50*1.1 + 5000 = \$38,578.05 @
end of 6th yr.
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The future value of an ordinary annuity of 1,000 each year for ten years, deposited at 3 percent, is A. 11,808 B. 8,530 C. 11,464 D. 10,000 | 635 | 2,133 | {"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-2017-34 | latest | en | 0.932022 |
https://www.enotes.com/homework-help/what-difference-between-average-speed-average-394370 | 1,656,367,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00794.warc.gz | 826,811,430 | 19,724 | # What is the difference between average speed and average velocity?
Speed is speed, or how fast something is going, while velocity is speed or how fast something is going in a specific direction. Average speed is the speeds added up and divided by number of speeds, and you do the same tning to calculate average velocity.
Approved by eNotes Editorial Team
Speed is the rate of change in position in a given amount of time. Thus, speed is measured in units of meters/second (m/s), kilometers per hour (km/hr) and more familiarly miles/hour (mph). Average speed is then the total distance traveled divided by the total time spent traveling.
Velocity has identical units to speed but the direction of motion must also be specified. In addition, velocity looks at the displacement (the shortest distance from the start of the motion to the end of the motion). This can lead to some interesting differences between average speed and average velocity.
Consider, for example, a runner on a 400 meter track. Suppose the runner completes exactly 4 laps in 6 minutes. Their average speed is then 1600 m/360 s or 4.44 m/s. However, there average velocity is zero! This is because their displacement is zero.
Another example would be if someone drives 300 m east, then 400 m north in a total time of 5 minutes. Their average speed is 700 m/300 s or 2.33 m/s. However, their displacement is 500 m from where they started. Therefore, their average velocity is 500 m/300 s or 1.67 m/s @ 53.13 degrees north of east.
Approved by eNotes Editorial Team
The basic difference is that speed is a scalar quantity, and velocity is a vector quantity; basically, direction matters when talking about velocity, and it doesn't when we are talking about speed. Speed is how fast something is moving--the rate at which a distance is covered. Velocity measures how fast something changes position. Think of it this way: if you run 100 meters in 10 seconds then immediately run back to your starting point in 10 seconds, your average speed is 10 meters per second--you have covered 200 meters in 20 seconds. Your average velocity is zero--you haven't changed your position after 20 seconds.
Approved by eNotes Editorial Team | 493 | 2,214 | {"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} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.93985 |
https://mathjokes4mathyfolks.wordpress.com/2015/06/11/7-names-to-string-you-along/ | 1,653,331,140,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00665.warc.gz | 460,194,398 | 30,109 | 7 Names to String You Along…
Allow me to alienate 99% of my readership by starting this post about strings with a computer science joke.
An int, a char, and a string walk into a bar and order some drinks. A short while later, the int and char start hitting on the bartendress, who gets very uncomfortable and walks away.
The string approaches the bartendress and says, “Sorry about my friends. Please forgive them. They’re primitive types.”
When I was in fourth grade, my best friend was named Benjamin Patrick.
I thought it was cool that his last name was my first name. That may have even been the reason that he was my best friend. This caused me to create a mental game in which I’d string together a series of people where the first name of person n was the last name of person n – 1.
With our two names, I’d get the string Benjamin Patrick Vennebush. Sadly, there’s no one whose first name is Vennebush (at least, not according to a Google search).
How long is a string?
Seven characters. Eight, if you count the space.
I could add to the front of the string, though.
Arthur Benjamin is a mathemagician at Harvey Mudd College.
Beatrice Arthur starred in the TV shows Maude and The Golden Girls.
Those four names get us to the string…
Beatrice Arthur Benjamin Patrick Vennebush
Frank Beatrice is a realtor in Boston, an operations manager in Indianapolis, and a guitar shop owner in New York.
To that, we could add Anne Frank to get…
Anne Frank Beatrice Arthur Benjamin Patrick Vennebush
However, Beatrice Frank is a professor in Newfoundland, an HR Exec in Philadelphia, and a student in Australia — to be sure, this game was more fun and challenging before social media — and using her name with Frank Beatrice leads to…
Beatrice Frank Beatrice Frank Beatrice Frank Beatrice Frank Beatrice…
That’s a two-name infinite loop.
• Without using search engines or social media, what’s the length of the longest string of names that you can create?
• Bonus points if you create a string where the first name of the first person is the last name of the last person, i.e., you create an infinite loop of names.
• Scoring: +1 for each name in your string; ×2 if you create an infinite loop.
• Use only names that other folks who read this blog would know. (Assume that some of them have actually been outside recently — which may or may not be a good assumption — and haven’t spent the last seven years in an attic trying to trisect an angle.)
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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.
MJ4MF (offline version)
Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature. | 690 | 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} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.906207 |
https://socratic.org/questions/59b5412b11ef6b384fca0464 | 1,597,522,985,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00131.warc.gz | 486,156,462 | 6,104 | # How many molecules are present in a 0.05*cm^3 volume of LIQUID water under standard conditions?
Sep 10, 2017
Approx. $1.67 \times {10}^{21}$ individual water molecules......pretty vast but not infinite.....
#### Explanation:
We approach all these problems dimensionally. We want the answer as a dimensionless number, and our calculation should give us a dimensionless number if we do the correct product/quotient....
${\rho}_{\text{density"="mass"/"volume}}$, and thus $\text{mass"=rhoxx"volume}$
And number of moles $=$ $\text{mass"/"molar mass}$....
And so we take ....................
$\frac{1.0 \cdot g \cdot c {m}^{-} 3 \times 0.05 \cdot c {m}^{3}}{18.01 \cdot g \cdot m o {l}^{-} 1} = 2.78 \times {10}^{-} 3 \cdot m o l$...
And we multiply this molar quantity by $\text{Avocado number}$ to get....
2.78xx10^-3*molxxN_A-=2.78xx10^-3*molxx6.022xx10^23*mol^-1=??
And note that this approach gives us a dimensionless number as required. | 296 | 951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "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} | 3.921875 | 4 | CC-MAIN-2020-34 | latest | en | 0.58684 |
https://www.lmfdb.org/EllipticCurve/Q/6720bg/ | 1,607,058,255,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733122.72/warc/CC-MAIN-20201204040803-20201204070803-00062.warc.gz | 759,104,626 | 44,953 | # Properties
Label 6720bg Number of curves $6$ Conductor $6720$ CM no Rank $0$ Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("6720.k1")
sage: E.isogeny_class()
## Elliptic curves in class 6720bg
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
6720.k6 6720bg1 [0, -1, 0, 639, 8481] [2] 6144 $$\Gamma_0(N)$$-optimal
6720.k5 6720bg2 [0, -1, 0, -4481, 91425] [2, 2] 12288
6720.k4 6720bg3 [0, -1, 0, -23681, -1317855] [2] 24576
6720.k2 6720bg4 [0, -1, 0, -67201, 6727201] [2, 2] 24576
6720.k1 6720bg5 [0, -1, 0, -1075201, 429482401] [2] 49152
6720.k3 6720bg6 [0, -1, 0, -62721, 7658145] [2] 49152
## Rank
sage: E.rank()
The elliptic curves in class 6720bg have rank $$0$$.
## Modular form6720.2.a.k
sage: E.q_eigenform(10)
$$q - q^{3} - q^{5} - q^{7} + q^{9} + 4q^{11} + 2q^{13} + q^{15} + 2q^{17} - 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the Cremona numbering.
$$\left(\begin{array}{rrrrrr} 1 & 2 & 4 & 4 & 8 & 8 \\ 2 & 1 & 2 & 2 & 4 & 4 \\ 4 & 2 & 1 & 4 & 8 & 8 \\ 4 & 2 & 4 & 1 & 2 & 2 \\ 8 & 4 & 8 & 2 & 1 & 4 \\ 8 & 4 & 8 & 2 & 4 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with Cremona labels. | 650 | 1,454 | {"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} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.4408 |
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