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# Modelling a cylinder in a tank
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September 8, 2011, 20:27 Modelling a cylinder in a tank #1 Member Join Date: Jan 2011 Posts: 42 Rep Power: 7 Hi, I'm trying to model (2DH) a cylinder (3m diameter) inside a tank (width 50m and length 250m) with a free stream of 2.5m/s. I dont have any problem modelling and running the simulation, but once I see the results I can't see the Von Kármán eddies. I think it is something related with boundary conditions. The pilar has a wall boundary condition type, with no slip. Something that surprise me is that the walls of the tank and the pilar aren't separate, all appear as wall. When I did the mesh I separate walls and pilar... The mesh discretization is fine to ensure solve the scale of the eddies, each element in the wake is 1/12 of the pilar diameter. Does anyone knows what I can do? Thanks in advance, Eduardo
September 9, 2011, 09:50 #2 Senior Member Svetlin Philipov Join Date: Mar 2009 Location: United Kingdom Posts: 176 Rep Power: 9 1. Be sure your Re is not laminar or strong turbulent; 2. extend the domain behind the cylinder and may be you will see results; 3. Such kind of analysis require transient case to be calculated, not steady-state.
September 9, 2011, 10:08 #3 Senior Member Mohammad Join Date: Feb 2010 Location: Shiraz, Iran Posts: 108 Rep Power: 8 hi eduardo, if I understand your problem correctly, you have a 2-dimensional cylinder in a box and front is velocity inlet and back is pressure outlet and up,down are walls as well as cylinder wall. so check if in the Reynolds no. of the flow you should have vortex shedding or not? if you must have and the results dosn't show such flow, check your viscous model. if flow is laminar, choose laminar and if flow is turbulent, choose k-w or k-e. another thing in turbulent flow is checking the y+ of cylinder-wall. check that in contours from turbulence menu. it should not be higher than 300 in standard wall function and higher than 10 in enhanced wall treatment. tell me about the results, yours, mohammad
September 9, 2011, 21:17 #4 Member Join Date: Jan 2011 Posts: 42 Rep Power: 7 Dear friends, Thanks for your replies. To help clarify my modelling scenario: - A cylinder (3m diameter) inside a tank (width 50m and length 250m) - Free stream = 2.5m/s (I also tried increasing velocity but nothing happens only that the wake extends longer downstream) - Domain mesh is 2 dimensional horizontal (there is no depth!) QUESTION = even do I'm modelling a 2DH case, how can I introduce the depth of the tank in order to obtain vertical integrated results? Define > Material: - liquid water Define > Solver - time = unsteady - other paramenters set as default Define > Viscous - K-epsilon > Standard > standard wall functions (@ m2montazari, I don't find in viscous menu the "y+ of cylinder-wall" where is it?) - other paramenters set as default Define > Boundary Conditions: - Inlet = velocity-inlet - Outflow = outflow - Fluid = water - Walls = wall (shear condition = no slip) - Default interior = interior Solve > iterate - time step size= 1s - Number of time steps = 20s - Max. iterations per time step = 1000s I still not able to see the V-K eddies with fluent 6.3.26.... I include a solution using an environmental hydrodynamic model (http://www.sisbahia.coppe.ufrj.br/) The domain has the same dimensions as stated above but with depth 10m. Pilar boundary has shear velocity = 0
September 10, 2011, 02:22 #5 Senior Member Mohammad Join Date: Feb 2010 Location: Shiraz, Iran Posts: 108 Rep Power: 8 hi, first of all I think the main problem is the timestep size you specified. 1sec for vonkarman vortex? you should use 0.1 or 0.01 sec to have an accurate solution. y+ is a variable that depends on mesh size and velocity and viscosity. you can find it in display-contour-velocity-wall yplus and check the y+ on cylinder walls. y+ is the main thing that make change in wall treatment in turbulent models. additionally you can try other turbulent models such as k-w sst and changing grid sizes specifically grids near cylinder. also check the grid quality(skewness and aspect ratio) and be sure that they are in acceptable range. yours, mohammad
September 10, 2011, 18:39 #6 Member Join Date: Jan 2011 Posts: 42 Rep Power: 7 Thanks Mohammad, I'm going to try your advices and I will let you know the results. Regards, Eduardo
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An automobile purchased for 22,000 is worth 2500 after 5 yrs. Assuming that the cars value deprecited steadily from year to year, what was it worth at the end of third year
New price = 2500
Amount depreciated in 5 years = 22000 - 2500
=19500
Amount depreciated in 1 year = 19500/5
= 3900
Worth after 3 years = 22000 - 3(3900)
= 10300
Related Questions
Find the value of this expression if x = 9. x^2+7/x+2
8
Step-by-step explanation:
Step-by-step explanation:
I've gotten this question and it's right.
Given that a point of measure is 3m away from a tree, and the top of the tree is at a 60 degree angle from the ground,find the height of the tree.
So there is a right triangle.
The tree hight over the distance from the tree is equal to the tangent of the angle.
h / 3 = tan(60)
h / 3 =
h =
To find the height of the tree, we can use trigonometry and the tangent function. The height of the tree is approximately 5.2 meters.
Explanation:
To find the height of the tree, we can use trigonometry. We have a right triangle formed by the tree, the point of measure, and the ground. The distance from the point of measure to the tree is the adjacent side, and the height of the tree is the opposite side. We can use the tangent function to find the height:
tan(60°) = height/3m
height = 3m * tan(60°) = 3m * √3 ≈ 5.2m
Therefore, the height of the tree is approximately 5.2 meters.
brainly.com/question/33884508
#SPJ2
In circle F what is the measure of arc CB?
17.5
35
70
60
Multiply by 2 to get that the measure of arc CB is 70.
measure of arc CB is 70°
Step-by-step explanation:
Draw segment FC.
Measure of arc CB = ∠CFB
∠CFB is complementary to ∠CFA
∠CFB = 180° - ∠CFA
Δ CFA is isosceles
∠ FAC = ∠ FCA = 35°
∠CFA +∠ FAC + ∠ FCA = 180°
∠CFA = 180° - 70°
∠CFB = 180° - (180° - 70°) = 70°
∴ Measure of arc CB is 70°
What is the next number in the sequence. 1,121,12321, 1234321
The next number in the sequence is _____
123454321
Step-by-step explanation:
it's a palendrome, made out of a number of numbers in the sqquence.
A.109
B.87
C.98
D.69
hey what's
Step-by-step explanation:
a question wow okay the answer is
I think it’s b because the other ones are different
Add: (−2x^2 + 9x − 3) + (7x^2 − 4x + 2)
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# What is the terminal value (in Year t+3)? Please give your answer in million USD and round it to the nearest integer. For example, if your answer is 3088 million USD, then type in “3088”.
2. What is the terminal value (in Year t+3)?
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# Content Showcase: Math Examples
## Media4Math has a huge collection of instructional math examples covering a wide range of math topics. This collection of resources includes images that you can easily incorporate into your lesson plans. To see these examples organized into topical collections, click on this link.
Number of Resources: 168
Thumbnail Image Description Description Curriculum Nodes
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 1
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 1 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 2
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 2 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 3
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 3 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 4
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 4 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 5
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 5 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 6
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 6 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 7
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 7 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 8
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 8 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 9
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 9 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 10
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 10 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 11
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 11 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 12
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 12 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
## Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 13
### Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 13 This is part of a collection of math examples that focus on linear function concepts.
Slope-Intercept Form
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Algebra -> Tutoring on algebra.com -> See tutors' answers! Log On
Tutoring Home For Students Tools for Tutors Our Tutors Register Recently Solved
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# Recent problems solved by 'drj'
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379, >>Next
Geometry_Word_Problems/225160: 10. Angles A and B are supplementary angles and angle A is 2⁰ more than 4 times angle B. Find the measures of angle A and angle B.
1 solutions
Answer 168078 by drj(1380) on 2009-10-11 21:57:10 (Show Source):
You can put this solution on YOUR website!
Angles A and B are supplementary angles and angle A is 2⁰ more than 4 times angle B. Find the measures of angle A and angle B.
Step 1. Supplementary angles means two angles add up to 180 degrees.
Step 2. Let A be one angle.
Step 3. Let B=180-A be its supplementary angle.
Step 4. Let A=2+4(180-A) since angle A is 2 degrees more than 4 times angle B=180-A.
Solved by pluggable solver: EXPLAIN simplification of an expression
Here's what you tried: This is an equation! Solutions: A=144.4. Graphical form: Equation was fully solved.Text form: A=2+4*(180-A) simplifies to 0=0Cartoon (animation) form: For tutors: `simplify_cartoon( A=2+4*(180-A) )` If you have a website, here's a link to this solution.
### DETAILED EXPLANATION
Look at .
Moved to the right of expression
It becomes .
Look at .
Moved to the right of expression
It becomes .
Look at .
Moved these terms to the left ,
It becomes .
Look at .
Expanded term by using associative property on
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
It becomes .
Look at .
It becomes .
Look at .
Eliminated similar terms , replacing them with
It becomes .
Look at .
It becomes .
Look at .
Remove unneeded parentheses around factor
It becomes .
Look at .
Solved linear equation equivalent to 5*A-722 =0
It becomes .
Result:
This is an equation! Solutions: A=144.4.
### Universal Simplifier and Solver
Done!
With A=144.4 then B=180-144.4=35.6 so check if equation in Step 4 is true... 144.4=2+4(35.6)...which it is.
Step 5. ANSWER: Angle A is 144.4 degrees and angle B is 35.6 degrees.
I hope the above steps and explanation were helpful.
And good luck in your studies!
Respectfully,
Dr J
http://www.FreedomUniversity.TV
Miscellaneous_Word_Problems/225137: Recall that the sum of the measures of the angles of a triangle is 180 degrees. In the triangle below, angle c has the same measure as angle b, and a measures 42 degrees less than angle b. Find the measure of each angle.1 solutions Answer 168075 by drj(1380) on 2009-10-11 21:49:31 (Show Source): You can put this solution on YOUR website!Recall that the sum of the measures of the angles of a triangle is 180 degrees. In the triangle below, angle c has the same measure as angle b, and a measures 42 degrees less than angle b. Find the measure of each angle. Step 1. Given c=b Step 2. a=b-42 since angle a measures 42 less than angle b. Step 3. Then, b-42+b+b=180 since c=b and the three angles add to 180 degrees. and and Step 4. Check if the angles equal to 180...74+74+32=180 which it does. Step 6. ANSWER: Angle a is 32 degrees and angles b and c is each 74 degrees. I hope the above steps and explanation were helpful. For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV
Numbers_Word_Problems/225116: One number is 2 more than 5 times another. Their product is 3. Find the numbers.
1 solutions
Answer 168072 by drj(1380) on 2009-10-11 21:23:45 (Show Source):
You can put this solution on YOUR website!
One number is 2 more than 5 times another. Their product is 3. Find the numbers.
Step 1. Let x be one number
Step 2. Let 2+5x be the other number since its 2 more than 5 times the first number.
Step 3. Then, since their product is 3.
Step 4. Turn this equation into a quadratic one in standard form as follows:
Subtract 3 from both sides
Simplify and arrange in descending order to get our quadratic equation
Step 4. To solve, use the quadratic formula given as
where a=5, b=2, and c=-3
Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=64 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 0.6, -1. Here's your graph:
With x=0.6 then 2+5x=2+5*0.6=5
With x=-1 then 2+5*(-1)=-3.
Step 5. ANSWER: The numbers are 0.6 and 5 as one pair and -1 and -3 as the other pair of numbers.
I hope the above steps and explanation were helpful.
And good luck in your studies!
Respectfully,
Dr J
http://www.FreedomUniversity.TV
test/225054: graph y=3x+11 solutions Answer 168068 by drj(1380) on 2009-10-11 21:05:12 (Show Source): You can put this solution on YOUR website!Graph y=3x+1 Step 1. The equation is in slope-intercept form given as y=mx+b where m is the slope and b is the y-intercept when x=0 or at point (0,b). Step 2. Comparing the given equation with y=mx+b shows that the slope m=3 and y-intercept is 1 when x=0 or at point (0,1). Step 3. To graph, start with the point (0,1) and the slope means for every one unit to the right we go three units up since the slope is 3. This means that one point is (0+1,1+3) or (1,4) is a point on the line. Step 4. Using the two points (0,1) and (1,4) we have the following graph of the line since it takes two points to make a line. I hope the above steps and explanation were helpful. For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV
test/225051: Find the equation of the line containing the point (4, 25) and perpendicular to the line 11x+55y=-12 . 1 solutions Answer 168066 by drj(1380) on 2009-10-11 20:56:57 (Show Source): You can put this solution on YOUR website!Find the equation of the line containing the point (4, 25) and perpendicular to the line 11x+55y=-12. Step 1. Two lines are perpendicular if their product of the slopes is -1. In equation form m1*m2=-1 where m1 is the slope given by the first equation. And m2 is the slope of the line perpendicular to the first line. Step 2. We need to put the above given line in slope intercept form so we can find the slope of the given line. The slope-intercept form is given as y=mx+b where m is the slope and b is the y-intercept b when x=0 or at point (0,b). Step 3. The following steps puts the above equation in slope-intercept form Subtract 11 from both sides of the equation Divide by 5 to both sides of the equation Step 4. Comparing the last equation with y=mx+b, the slope m1=-1/5. Step 5. Therefore, the slope of the perpendicular line is since the product of the slopes between the two lines is -1 (or m1*m2=-1) Step 6. So now we have a perpendicular line with slope m2=5 that must pass through the point (4,25) Step 7. Given two points (x1,y1) and (x2,y2), then the slope m is given as Step 4. Let (x1,y1)=(4,25) or x1=4 and y1=25. Let other point be (x2,y2)=(x,y) or x2=x and y2=y. Step 5. Now we're given . Substituting above values and variables in the slope equation m yields the following steps: Step 6. Multiply x-4 to both sides to get rid of denominator on right side of equation. Add 25 to both sides of the equation Step 7. ANSWER: The equation of the perpendicular line in slope-intercept form is And the graph of the two lines are shown below. I hope the above steps and explanation were helpful. For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV
Graphs/225024: solve the system of equations by graphing. then classify 6x + y = 46 7x + 8y = 401 solutions Answer 168059 by drj(1380) on 2009-10-11 20:34:33 (Show Source): You can put this solution on YOUR website!Solve the system of equations by graphing. Then classify 6x + y = 46 7x + 8y = 40 Step 1. Graph the two lines from above as shown below Step 2. Based on the graph the intersection point is (8,-2) or x=8 and y=-2. Step 3. Check if the solution (8,2) satisfies the above linear system of equations 6x + y = 46 or 6*8+-2=46 or 46=46 which is a true statement 7x + 8y = 40 or 7*8+8*(-2)=40 or 40=40 which is another true statement Step 4. ANSWER: The solution is (8,2). The equations are independent and consistent providing a unique solution. I hope the above explanation was useful. And good luck in your studies! For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. Respectfully, Dr J
test/225057: Graph y=x/2-3 by using the slope and y-intercept.1 solutions Answer 168054 by drj(1380) on 2009-10-11 20:28:10 (Show Source): You can put this solution on YOUR website!Graph y=x/2-3 by using the slope and y-intercept. Step 1. The slope-intercept from is given as y=mx+b where m is the slope and b is the y-intercept at x=0 or at point (0,b). Step 2. Comparing with , the slope and the y-intercept Step 3. Now the slope of for every two units you go the right, you go one unit up. To plot the line take the y-intercept at point (0,-3). So starting at point (0,-3) or x1=0 and y1=-3 we go two units to the right means x2=0+2=2 and 1 unit up means y2=-3+1=-2. So we have two points (0,-3) and (2,-2) used to plot the line. I hope the above explanation was useful. And good luck in your studies! For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. Respectfully, Dr J
Expressions-with-variables/225065: x+8y=76
-x+9y=77
1 solutions
Answer 168049 by drj(1380) on 2009-10-11 20:19:27 (Show Source):
You can put this solution on YOUR website!
x+8y=76
-x+9y=77
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
We'll use substitution. After moving 8*y to the right, we get:
, or . Substitute that
into another equation:
and simplify: Error: 'Can't call method "invoke_solver" on an undefined value at /home/ichudov/project_locations/algebra.com/templates/Algebra/Solver/PerlSolver.pm line 115. Can't call method "invoke_solver" on an undefined value at /home/ichudov/project_locations/algebra.com/templates/Algebra/Solver/PerlSolver.pm line 115. '. Can't call method "invoke_solver" on an undefined value at /home/ichudov/project_locations/algebra.com/templates/Algebra/Solver/PerlSolver.pm line 115. Can't call method "invoke_solver" on an undefined value at /home/ichudov/project_locations/algebra.com/templates/Algebra/Solver/PerlSolver.pm line 115.
error_location: 101
I hope the above steps were helpful.
And good luck in your studies!
Respectfully,
Dr J
drjctu@gmail.com
http://www.FreedomUniversity.TV
Miscellaneous_Word_Problems/225094: The perimeter of a standard high school basketball court is 268ft. The length is 34ft longer than the width. Find the dimensions of the court.
1 solutions
Answer 168048 by drj(1380) on 2009-10-11 20:17:38 (Show Source):
You can put this solution on YOUR website!
The perimeter of a standard high school basketball court is 268ft. The length is 34ft longer than the width. Find the dimensions of the court.
Step 1. The perimeter P means adding up all the four sides of a rectangle.
Step 2. Let w be the width.
Step 3. Let w+34 be the length since the length is 34 ft longer than the width.
Step 4. Then, P=w+w+w+34+w+34=268.
Step 5. Solving yields the following steps
Solved by pluggable solver: EXPLAIN simplification of an expression
Here's what you tried: This is an equation! Solutions: w=50. Graphical form: Equation was fully solved.Text form: w+w+w+34+w+34=268 simplifies to 0=0Cartoon (animation) form: For tutors: `simplify_cartoon( w+w+w+34+w+34=268 )` If you have a website, here's a link to this solution.
### DETAILED EXPLANATION
Look at .
It becomes .
Look at .
Moved to the right of expression
It becomes .
Look at .
Eliminated similar terms ,,, replacing them with
It becomes .
Look at .
It becomes .
Look at .
Remove unneeded parentheses around factor
It becomes .
Look at .
Moved these terms to the left
It becomes .
Look at .
It becomes .
Look at .
It becomes .
Look at .
Solved linear equation equivalent to 4*w-200 =0
It becomes .
Result:
This is an equation! Solutions: w=50.
### Universal Simplifier and Solver
Done!
For w=50 , then w+34=84 and P=2(50+84)=268 which is a true statement.
Step 6. ANSWER: The width is 50 feet and he length is 84 feet.
I hope the above steps were helpful.
And good luck in your studies!
Respectfully,
Dr J
drjctu@gmail.com
http://www.FreedomUniversity.TV
Functions/225073: rewrite the equation so that y is a function of x y/5-7=-2x How do you solve this?1 solutions Answer 168025 by drj(1380) on 2009-10-11 19:35:53 (Show Source): You can put this solution on YOUR website!rewrite the equation so that y is a function of x Step 1. Add 7 to both sides of the equation Step 2. Multiply by 5 to both sides of the equation to get rid of denominator. Step 3. ANSWER: I hope the above steps were helpful. For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV
Linear-equations/225062: Solve each system of equations by graphing. y=x-2 y=3x-81 solutions Answer 168020 by drj(1380) on 2009-10-11 19:20:46 (Show Source): You can put this solution on YOUR website!Solve each system of equations by graphing. y=x-2 y=3x-8 Step 1. Graphing the system of equations is Step 2. Based on the graph, the intersection point is (3,1) Step 3. Check if the intersection point (3,1) satisfies the system of equations y=x-2 1=3-2 which is a true statement y=3x-8 1=3*3-8 which is another true statement Step 4. ANSWER: The solution is (3,1) or x=3 and y=1. I hope the above steps were helpful. For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. Good luck in your studies! Respectfully, Dr J
test/225022: Find the equation of the line that contains the points P1(–8, –12) and P2(9, –12).
1 solutions
Answer 168017 by drj(1380) on 2009-10-11 19:12:55 (Show Source):
You can put this solution on YOUR website!
Find the equation of the line that contains the points P1(–8, –12) and P2(9,–12).
Step 1. We will put the equation of the line in slope-intercept form given as y=mx+b where m is the slope and b is the y-intercept when x=0 or at point (0,b).
Step 2. The slope of the line m is given as
where for our example is x1=-8, y1=-12, x2=9 and y2=-12 (think of ). You can choose the points the other way around but be consistent with the x and y coordinates. You will get the same result.
Step 3. Substituting the above values in the slope equation gives
Step 4. The slope is calculated as or
Step 5. Now use the slope equation of Step 2 and choose one of the given points. I'll choose point (9,-12). Letting y=y2 and x=x2 and substituting in the slope equation given as,
Step 6. Multiply both sides of equation by x-9 to get rid of denominator found on the right side of the equation
Step 9. ANSWER: The equation of the line is
See graph below to check the above steps.
Solved by pluggable solver: DESCRIBE a linear EQUATION: slope, intercepts, etc This equation defines a horizontal line. There is no X intercept. Y intercept is
I hope the above steps were helpful.
Respectfully,
Dr J
Linear-systems/225044: I am trying to help my daughter with her homework and need to know how to solve this equation.... 2k + 6 - K = 171 solutions Answer 168015 by drj(1380) on 2009-10-11 18:45:45 (Show Source): You can put this solution on YOUR website!I am trying to help my daughter with her homework and need to know how to solve this equation.... 2k + 6 - k = 17 Step 1. Simplify by adding the like terms on the left side of the equation Step 2. Subtract 6 from obth sides of the equation to solve for k terms. Check solution...... which is a true statement. Step 3. ANSWER: Solution is I hope the above steps and explanation were helpful. For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trigonometry. Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs. Respectfully, Dr J
Numbers_Word_Problems/225037: the width of a rectangle is one third of the length.If the perimeter is 20 cm,find the width.1 solutions Answer 168013 by drj(1380) on 2009-10-11 18:37:47 (Show Source): You can put this solution on YOUR website!The width of a rectangle is one third of the length. If the perimeter is 20 cm,find the width. Step 1. The perimeter P means adding up all four sides of a rectangle. Step 2. Let w be the width Step 3. Let 3w be the length since the width is one-third the length Step 4. Then P=w+w+3w+3w=8w=20 since the perimeter P is 20 cm. Step 5. Solving the equation in Step 4 leads to the following steps Divide by 8 to both sides of the equation and 3w=15/2 and P=2(5/2+15/2)=20 which is a true statement. Step 6. ANSWER: The width is 5/2 cm. I hope the above steps and explanation were helpful. For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trigonometry. Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs. Respectfully, Dr J
Linear_Algebra/225027: Solve by using elimination. This cant be solved right? x-4y=2 4x-16y=81 solutions Answer 168010 by drj(1380) on 2009-10-11 18:30:40 (Show Source): You can put this solution on YOUR website!Solve by using elimination. This cant be solved right? x-4y=2 Multiply by 4 to both sides and you'll get 4x-16y=8 which is identical to the one below 4x-16y=8 ANSWER: Because these are equations are dependent (one equation is a multiple of the other, 4 in this case), there are many solutions that satisfies these equations. I hope the above steps and explanation were helpful. For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trigonometry. Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs. Respectfully, Dr J http://www.FreedomUniversity.TV
Numbers_Word_Problems/225035: Two numbers are in the ratio 1:11 and their sum is 15. Find the numbers. Thanks1 solutions Answer 168007 by drj(1380) on 2009-10-11 18:24:29 (Show Source): You can put this solution on YOUR website!Two numbers are in the ratio 1:11 and their sum is 15. Find the numbers. Step 1. Let x be one number Step 2. Let 11x be the other number since the ratio is 1:11. Step 3. Then 11x+x=15 or 12x=15 then and which is atrue statement. Step 4. ANSWER: The numbers are and . I hope the above steps and explanation were helpful. For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trigonometry. Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs. Respectfully, Dr J http://www.FreedomUniversity.TV
Linear-equations/225001: graph and check: "graph the linear system and estimate the solution. Then check the solution algebraically. 2x+y=5 x+y=21 solutions Answer 167999 by drj(1380) on 2009-10-11 18:07:15 (Show Source): You can put this solution on YOUR website!Graph and check: "graph the linear system and estimate the solution. Then check the solution algebraically. Step 1. Graph the linear system of equations Step 2. Based on the above graph, the intersection point is (3,-1). Step 3. Check if (3,-1) or x=3 and y=-1 satisfies the system of equaitons Given , then which is a true statment. Given , then which is a true statement. Step 4. ANSWER: The solution sis (3,-1) or x=3 and y=-1. I hope the above steps and explanation were helpful. For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trigonometry. Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs. Respectfully, Dr J http://www.FreedomUniversity.TV
Quadratic_Equations/224995: factor 6x^2+241 solutions Answer 167990 by drj(1380) on 2009-10-11 17:39:25 (Show Source): You can put this solution on YOUR website!Factor Step 1. Look at numbers and find greatest factor which is 6 Step 2. Look to see if there are common terms in the variable x and in this case there is none. Step 3. ANSWER: I hope the above steps and explanation were helpful. For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV
Linear-equations/224990: find the slope intercept equation of the line that has the given characteristics
slope 4 and y intercept (0,3)
slope intercept equation is y=
1 solutions
Answer 167989 by drj(1380) on 2009-10-11 17:28:00 (Show Source):
You can put this solution on YOUR website!
Find the slope intercept equation of the line that has the given characteristics
slope 4 and y intercept (0,3)
slope intercept equation is y=4x+3
Step 1. The equation in slope-intercept form is given as y=mx+b where m is the slope and b is the y-intercept when x=0 or at point (0,b).
In your example we compare y=mx+b such that m=4 and b=3. Therefore, y=4x+3.
Step 2. Here's another way...now we have to find the line with slope m=4 going through point (0,3).
Step 3. Given two points (x1,y1) and (x2,y2), then the slope m is given as
Step 4. Let (x1,y1)=(0,3) or x1=0 and y1=3. Let other point be (x2,y2)=(x,y) or x2=x and y2=y.
Step 5. Now we're given . Substituting above values and variables in the slope equation m yields the following steps:
Step 6. Multiply x to both sides to get rid of denominator on right side of equation.
Step 7. Add 3 to both sides of the equation
same as in Step 1.
Step 7. ANSWER: The equation in slope-intercept form is where the slope m=7 and the y-intercept b=-5.
Note: the above equation can be rewritten in standard form as
And the graph is shown below which is consistent with the above steps.
Solved by pluggable solver: DESCRIBE a linear EQUATION: slope, intercepts, etc Equation describes a sloping line. For any equation ax+by+c = 0, slope is .X intercept is found by setting y to 0: ax+by=c becomes ax=c. that means that x = c/a. -5/-7 = 0.714285714285714.Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is -5/1 = -5.Slope is --7/1 = 7. Equation in slope-intercept form: y=7*x+-5.
I hope the above steps and explanation were helpful.
And good luck in your studies!
Respectfully,
Dr J
http://www.FreedomUniversity.TV
Angles/224936: can you please help me figure out the supplement of a given angle is four times as large as a complement of the angle. Find the measure of the given angle. Please.
1 solutions
Answer 167973 by drj(1380) on 2009-10-11 15:11:30 (Show Source):
You can put this solution on YOUR website!
Can you please help me figure out the supplement of a given angle is four times as large as a complement of the angle. Find the measure of the given angle. Please.
Step 1. Supplementary angles means two angles add up to 180 degrees.
Step 2. Complementary angles means two angles add up to 90 degrees.
Step 3. Let x be the angle.
Step 4. Let 180-x be its supplementary angle.
Step 5. Let 90-x be its complementary angle.
Step 6. Then 180-x=4(90-x) since the supplement of a given angle is four times as large as a complement of the angle
Solved by pluggable solver: EXPLAIN simplification of an expression
Here's what you tried: This is an equation! Solutions: x=60. Graphical form: Equation was fully solved.Text form: 180-x=4*(90-x) simplifies to 0=0Cartoon (animation) form: For tutors: `simplify_cartoon( 180-x=4*(90-x) )` If you have a website, here's a link to this solution.
### DETAILED EXPLANATION
Look at .
Moved to the right of expression
It becomes .
Look at .
Moved to the right of expression
It becomes .
Look at .
Moved these terms to the left
It becomes .
Look at .
Moved to the right of expression
It becomes .
Look at .
Expanded term by using associative property on
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
It becomes .
Look at .
It becomes .
Look at .
Eliminated similar terms , replacing them with
It becomes .
Look at .
It becomes .
Look at .
Remove unneeded parentheses around factor
It becomes .
Look at .
Solved linear equation equivalent to 3*x-180 =0
It becomes .
Result:
This is an equation! Solutions: x=60.
### Universal Simplifier and Solver
Done!
With x=60, then 180-x-120 and 90-60=30 and note that four times the complement of the angle is the supplement of the angle.
Step 7. ANSWER: The angle is 60 degrees. It complement is 30 degrees and its supplement is 120 degrees.
I hope the above steps were helpful.
And good luck in your studies!
Respectfully,
Dr J
drjctu@gmail.com
http://www.FreedomUniversity.TV
Circles/224919: what is the circumference of a circle with a diameter of 3cm?1 solutions Answer 167971 by drj(1380) on 2009-10-11 15:06:29 (Show Source): You can put this solution on YOUR website!What is the circumference of a circle with a diameter of 3 cm? Step 1. The circumference C of a circle is given as where d is the diameter. Step 2. ANSWER: cm I hope the above steps were helpful. For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV http://www.Facebook.com/FreedomUniversityTV http://www.Twitter.com/FreedomUTV
Linear-equations/224926: is y+3=0 a linear equation?1 solutions Answer 167970 by drj(1380) on 2009-10-11 15:02:16 (Show Source): You can put this solution on YOUR website!Is y+3=0 a linear equation? Yes. You can rewrite the above equation as y=-3. In Slope-Intercept form given as y=mx+b where m is the slope and b is the y-intercept when x=0 at point (0,b). For this example, the slope m=0 or a horizontal line and the y-intercept is -3. The graph is shown below: I hope the above steps were helpful. For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry. And good luck in your studies! Respectfully, Dr J http://www.FreedomUniversity.TV http://www.Facebook.com/FreedomUniversityTV http://www.Twitter.com/FreedomUTV
Equations/224915: Solve this simultaneous equation:
y=5-4x²
y+3x=6
1 solutions
Answer 167967 by drj(1380) on 2009-10-11 14:21:07 (Show Source):
You can put this solution on YOUR website!
Solve this simultaneous equation:
Equation 1
or Equation 2
Step 1. Set Equations 1 and 2 to be equal.
Step 2. Add to both sides of the equation
Step 3. To solve, use the quadratic formula given below
where a=4, b=-3, and c=1
Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . The discriminant -7 is less than zero. That means that there are no solutions among real numbers. If you are a student of advanced school algebra and are aware about imaginary numbers, read on. In the field of imaginary numbers, the square root of -7 is + or - . The solution is Here's your graph:
Step 4. For each value of x found in Step 3, substitute each value into y=6-3x to get a total of 2 values for y. You now have two points as your solution to these equations. Your solutions will be complex numbers.
I hope the above steps were helpful.
And good luck in your studies!
Respectfully,
Dr J
http://www.FreedomUniversity.TV
x+y=10
y=3x²-2x-3
I have tried every way i can think of to solve this equation but i have failed. Please help me
1 solutions
Answer 167966 by drj(1380) on 2009-10-11 14:10:44 (Show Source):
You can put this solution on YOUR website!
solve this simultaneous equation:
Equation 1
Equation 2
Step 1. In Equation 1, solve for and substitute y into Equation 2.
Step 2. Add x-10 to both sides of the equatiion
Step 3. To solve use the quadratic formula given below:
where a=3, b=-1, and c=-13
Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=157 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 2.25499401435694, -1.92166068102361. Here's your graph:
Step 4. Now, for each value of x found in Step 3, substitute into y=10-x to get two values for y.
I hope the above steps were helpful.
And good luck in your studies!
Respectfully,
Dr J
http://www.FreedomUniversity.TV
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New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > r19.27av GIF version
Theorem r19.27av 2752
Description: Restricted version of one direction of Theorem 19.27 of [Margaris] p. 90. (The other direction doesn't hold when A is empty.) (Contributed by NM, 3-Jun-2004.) (Proof shortened by Andrew Salmon, 30-May-2011.)
Assertion
Ref Expression
r19.27av ((x A φ ψ) → x A (φ ψ))
Distinct variable group: ψ,x
Allowed substitution hints: φ(x) A(x)
Proof of Theorem r19.27av
StepHypRef Expression
1 ax-1 6 . . . 4 (ψ → (x Aψ))
21ralrimiv 2696 . . 3 (ψx A ψ)
32anim2i 552 . 2 ((x A φ ψ) → (x A φ x A ψ))
4 r19.26 2746 . 2 (x A (φ ψ) ↔ (x A φ x A ψ))
53, 4sylibr 203 1 ((x A φ ψ) → x A (φ ψ))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 358 ∈ wcel 1710 ∀wral 2614 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-11 1746 This theorem depends on definitions: df-bi 177 df-an 360 df-ex 1542 df-nf 1545 df-ral 2619 This theorem is referenced by: r19.28av 2753
Copyright terms: Public domain W3C validator
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# Economics
Suppose a university decides to alter its tuition schedule by separating its students based on how many years of college they have completed. Most university programs require four years to complete. First-year students would get a 13% tuition reduction. Second-year students would pay the normal tuition. Third- and fourth-year students face an increase in tuition of 25 and 41%, respectively. Fully explain whether this pricing strategy is based on a sound understanding of price elasticity of demand, or not.
This is what I have so far:
First-year students are based upon elastic demand while third and fourth-year students represent an inelastic demand.
At first, colleges reduce tuition to get students to attend their university. However, once students are respectively halfway through, the college increases tuition because at this point, students need their college degree, no matter the price.
(So basically, the college does have a sound understanding of price elasticity of demand.)
Am I answering this correctly so far?
Thanks
1. 👍 1
2. 👎 0
3. 👁 2,224
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# Expectation and Standard Deviation(SD)
1. Nov 10, 2009
### helix999
Hi
I have a question.
Let X1 & X2 be stochastic variables and X1<=X2, then can we say E[X1]<=E[X2] or SD[X1]<=SD[X2]? why or why not?
Thanks!
2. Nov 10, 2009
### lanedance
what do you think?
and by X1 <=X2, does this mean every outcome of X1 is <= every outcome of X2?
3. Nov 10, 2009
### helix999
yeah, the given condition is for every outcome.
I think E[x1]<=E[x2] but no idea abt std deviation. i dont know if i am correct.
4. Nov 10, 2009
### clamtrox
Could you prove it?
If you think that the relation does not always hold for standard deviation, could you perhaps find variables X and Y for which this is the case?
5. Nov 10, 2009
### lanedance
its always a good place to start at the defintions (either discrete or continuous will work)
from the definitions, E[X1]<= E[X2] shoudl be obvious
for the standard deviation, building on clamtrox's argument, it might help to consider the dsicrete variable pdf:
P(X2= x2) = 1, if x2 = b
P(X2= x2) = 0, otherwise
whats the standard deviation of this "random" variable?
6. Nov 10, 2009
consider X1 and X2 with distributions
[tex]
\begin{tabular}{l c c c c c}
x & 10 & 20 & 30 & 40 & 50 \\
p(x) & .80 & .1 & .07 & .02 & .01\\
\end{tabular}
7. Nov 10, 2009
### helix999
zero.
8. Nov 10, 2009
### lanedance
yep... just to show for a random variabe the SD, can be quite separate from the mean.... hopefully you can thnk of a RV with all lesser outcomes, but non-zero SD
though i hope i haven't simplified too much, what exactly do you mean by a stochastic variable here...?
Last edited: Nov 10, 2009
9. Nov 10, 2009
### helix999
ok i got the examples when SD[x1]<=SD[x2]. But when SD[x1>=Sd[x2]?
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A072570 Even interprimes i = (p+q)/2 (where p, q are consecutive primes) such that (q-p)/2 is not divisible by 3. 3
4, 6, 12, 18, 30, 42, 60, 72, 102, 108, 120, 138, 144, 150, 180, 186, 192, 198, 228, 240, 246, 270, 282, 288, 300, 312, 324, 342, 348, 414, 420, 426, 432, 462, 522, 552, 570, 582, 600, 618, 636, 642, 660, 696, 714, 780, 792, 810, 816, 822, 828, 834, 846, 858 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS A superset of A014574. [R. J. Mathar, Mar 03 2009] LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 FORMULA If d = (P_{n+1} - P_n)/2 is even & d/2 == +/- 1 (mod 6), then P_n + d = (P_{n+1} + P_n)/2 is in the sequence. [Corrected by M. F. Hasler, Nov 29 2013] MATHEMATICA a = Table[Prime[n], {n, 2, 200}]; b = {}; Do[d = (a[[n + 1]] - a[[n]])/2; If[ EvenQ[ a[[n]] + d] && (Mod[d, 6] == 5 || Mod[d, 6] == 1), b = Append[b, a[[n]] + d]], {n, 1, 198}]; b Mean/@Select[Partition[Prime[Range[200]], 2, 1], EvenQ[Mean[#]] && !Divisible[ (#[[2]]-#[[1]])/2, 3]&] (* Harvey P. Dale, Sep 27 2017 *) PROG (PARI) q=3; forprime(p=5, 1e3, (s=q+q=p)%4==0 && (s-2*p)%3 && print1(s/2", ")) \\ M. F. Hasler, Nov 29 2013 (PARI) is_A072570(n)=my(p=precprime(n)); nextprime(n)+p==2*n && (n-p)%3 && !bittest(n, 0) \\ M. F. Hasler, Nov 30 2013 CROSSREFS Cf. A024675, A072571. A072568 is union of A072571 and this sequence. Sequence in context: A074998 A061715 A280469 * A217259 A014574 A258838 Adjacent sequences: A072567 A072568 A072569 * A072571 A072572 A072573 KEYWORD nonn AUTHOR Marco Matosic, Jun 24 2002 EXTENSIONS Edited by N. J. A. Sloane and Robert G. Wilson v, Jun 27 2002 STATUS approved
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Last modified July 30 09:18 EDT 2021. Contains 346359 sequences. (Running on oeis4.)
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# Behaviour of Real Gases: Deviation from Real Behaviour
Science depends on the practical applicability of every experiment! Theoretically, the gas laws, from Boyle’s law to Avogadro’s Law seem right, but what about their practical appliance? Real Gases generally observe the various findings of the Gas Laws under controlled physical conditions, but also show deviations from them, let’ see how!
## Applicability of Ideal Gas Equation and Boyle’s Law
We already know the Ideal Gas Equation pV= nRT. It defines the relationship between Temperature, pressure, and volume of gases. For checking the reliability of this relationship we first plot a graph between pV and p. Now we know that at a constant temperature, as the Boyle’s law states, pV shall be a constant. Therefore the graphs between the two (p and pV) shall be a straight line. But the case is not so! At temperature 273 K the data for several gases is shown in the graph below:
From the graph plotted in the figure we can easily conclude that despite the constant temperature, the real gases do not show behaviour as predicted by the Boyle’s law. These gases show a significant deviation from the predicted ideal behaviour as per the Boyle’s law. The plot in the graph signifies the deviating behaviour of real gases like Dihydrogen, Helium, Carbon monoxide and Methane from the behaviour of ideal gas. In faact we see, from the graph that real gases do not show any signs of similar to Ideal gas’s behaviour.
From our next graph that plots the volume to pressure data of gases, we find an apparent deviation from the theoretical prediction of gases in changed conditions. From the graph, we can figure out that at very high pressures the value of volume we calculated is less than the actual practical volume. Hence from the above two graphs, it is clear that generally real gases, under all conditions do not follow The Ideal Gas Behaviour or equation.
## Problem: Deviation of Gases From Real Behaviour
The question now is that why do gases deviate from the ideal behaviour? The answer lies in the basic behaviour of these gases. We already know that the intermolecular forces between gases are minimal, the reason being the scattered molecules in the gaseous state.
According to the Kinetic Theory, molecules of gases do not have any force of attraction, and the volume of the molecules is insignificantly small when compared to the space occupied by the gases. Hence, at low temperatures how does a gaseous matter change into the liquid state? If there is no force of attraction, then how does the state of matter change?
### Solution: Intermolecular Force
This negligible intermolecular force is the secret behind the deviation of real gases from the ideal gas. Molecules in the gases, interact with each other. Though the interaction is weak at high temperatures, yet with decreasing temperature the interactive forces increases. At high pressure also, these molecules come close to each other, hence leading to a decrease in volume.
With increasing pressure, the interaction between these molecules increases. This interaction between the molecules prevents them from bombarding on the container. The attractive forces between the molecules prevent the molecules from colliding with the walls of the container.
This is where the difference between ideal gases and real gases become transparent. Ideal gases are those gases which follow Gas laws and the molecules of these gases have no interaction with each other, while real gases are those gases which occupy space and the molecules have a force of interaction between each other.
### Vander Walls Equation of State
Now from the above difference, it is clear that in ideal gases the pressure exerted by molecules on the container is greater than that exerted by real gases, so pideal = preal + an2/ V2; an2/ V2 here is a constant and is known as the correction term.
Now, let’s consider the repulsive forces. The forces which come into play when the molecules are in contact with each other are called repulsive forces. Being short-range interactions, these forces make molecules behaviour like the impenetrable spheres. It is because of this force that the volumes of the molecule rise significantly and at high-pressure volume V becomes V-nb. Here, nb is the volume occupied by the molecules.
(p+ an2/V) (V-nb) = nRT
a and b are constants that depend on the nature of the gas. This equation is also known as the van der Waals equation. ‘n’ here is the number of moles while a and b are constants referred as van der Waals constants. The value of these van der Waals is specific to the characteristic of the gas and is independent of pressure and temperature.
It should be noted here that at low temperature the intermolecular forces become significantly high, thus the molecules attract with each other at greater speed. Similarly at higher pressure the intermolecular forces increase. So at high pressure and low temperature, the molecules of gases have high intermolecular forces. Real gases exhibit ideal behaviour only when the intermolecular forces are minimal. The lesser the pressure, the greater the chances of a real gas behaving like an ideal gas!
### Compressibility Factor
Let Z = pV/nRT be a number. it will have no units as is clear from the equation. What does this number signify?
For perfect ideal behaviour, Z = 1 at all temperatures and pressures, because if Z=1 then pV= nRT. The temperature at which a real gas follows an Ideal gas law is known as the Boyle temperature or Boyle point. The Boyle point doesn’t depend on physical conditions, rather it is dependent on the nature of the gas.
Generally, at low pressure, all gases show ideal behaviour hence giving Z as = 1. Lower the pressure, greater the volume. If Z ≠ 1, then the gas is not ideal but real. Hence with the help of the compressibility factor, we can find the measure of the deviation of real gases from the ideal behaviour.
## Solved Examples From You
Q : Assertion: CH4, CO2 has a value of Z (compressibility factor) less than one at 0oC.
Reason: Z < 1 is due to the fact that the attractive forces dominate among the molecules.
1. Both the assertion and reason are correct and the reason is the correct explanation of the assertion.
2. The two statements are correct but the reason is not the correct explanation.
3. Assertion is correct and the reason is wrong.
4. Assertion as well as the reason are wrong.
Solution: A) As we discussed above, the compressibility factor tells us about the measure of the deviation of real gases from the ideal behaviour. If you look at the plots of Boyle’s law for these gases as given above, you will find that the assertion is true. This is due to the attractive forces of the gases.
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Testpoint Constraints | Enterprise Architect User Guide
Book a Demo
#### Please note : This help page is not for the latest version of Enterprise Architect. The latest help can be found here.
Prev Next
# Testpoint Constraints
A Constraint is typically composed using local and member variables in expressions, separated by operators to define one or more specific criteria that must be met. A constraint must evaluate as true to be considered as Passed. If a constraint evaluates as false, it is considered as Failed.
Any variables referenced within the constraint must be in scope at the position where the Testpoint or Breakpoint is evaluated.
## General/Arithmetic Operators
Operator
Description
+
Example: a + b > 0
-
Subtract
Example: a - b > 0
/
Divide
Example: a / b == 2
*
Multiply
Example: a * b == c
%
Modulus
Example: a % 2 == 1
()
Parentheses - Used to define precedence in complex expressions.
Example: ((a / b) * c) <= 100
[ ]
Square Brackets - Used for accessing Arrays.
Example: Names[0].Surname == "Smith"
.
Dot operator - Used to access member variables of a Class.
Example: Station.Name == "Flinders"
->
Alternative notation for the Dot operator.
Example: Station->Name == "Flinders"
## Comparison Operators
Operator
Description
=
Equal To
Example: a = b
==
Equal To
Example: a == b
!=
Not Equal To
Example: a != b
<>
Not Equal To
Example: a <> b
>
Greater Than
Example: a > b
>=
Greater Than or Equal To
Example: a >= b
<
Less Than
Example: a < b
<=
Less Than or Equal To
Example: a <= b
## Logical Operators
Operator
Description
AND
Logical AND
Example: (a >= 1) AND (a <= 10)
OR
Logical OR
Example: (a == 1) OR (b == 1)
## Bitwise Operators
Operator
Description
&
Bitwise AND
Example: (1 & 1) = 1
(1 & 0) = 0
|
Bitwise OR
Example: (1 | 1) = 1
(1 | 0) = 1
^
Bitwise XOR (exclusive OR)
Example: (1 ^ 1) = 0
(1 ^ 0) = 1
Example
Description
((m_nValue & 0xFFFF0000) == 0)
Use a Bitwise AND operator (&) with a hexadecimal value as the right operand to test that no bits are set in high order bytes of the variable.
((m_nValue & 0x0000FFFF) == 0)
Use a Bitwise AND operator (&) with a hexadecimal value as the right operand to test that no bits are set in low order bytes of the variable.
m_value[0][1] = 2
Accessing a multi-dimensional array
a AND (b OR c)
Combining AND and OR operators, using parentheses to ensure precedence. In this example, variable 'a' must be true, and either 'b' or 'c' must be true.
## Notes
• String comparisons are case-sensitive
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# LU Decomposition - PowerPoint PPT Presentation
LU Decomposition. Greg Beckham, Michael Sedivy. Overview. Step 1. This is handled implicitly in the code by only calculating the diagonal for β. Step 2. Calculating β ij. for(j = 0; j < n; j++) // This is the loop over columns of Crout's method {
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Presentation Transcript
### LU Decomposition
Greg Beckham, Michael Sedivy
This is handled implicitly in the code by only calculating the diagonal for β
Calculating βij
for(j = 0; j < n; j++) // This is the loop over columns of Crout's method
{
for(i = 0; i < j; i++) // Equation (2.3.12) except for i = j
{
sum = a[i][j];
for(k = 0; k < i; k++) sum -= a[i][k] * a[k][j];
a[i][j] = sum;
}
}
{
for(i = j; i < n; i++)// This is i=j of equation (2.3.12) and i=j+1
{ // N-1 of equation (2.3.13)
sum = a[i][j];
for(k = 0; k < j; k++) sum -= a[i][k] * a[k][j];
a[i][j] = sum;
}
if(j != n - 1) // Divide by the pivot element
{
dum = 1.0/(a[j][j]);
for(i = j + 1; i < n; i++) a[i][j] *= dum;
}
}
Pivoting Crout's method
• Initially finds largest element in each row
• Used as a “scaling factor”, not sure of use other than to rollover
for(i = 0; i < n; i++) // Loop over the rows to get implicit scaling
{ // information
big = 0.0;
for(j = 0; j < n; j++)
{
if((temp = fabs(a[i][j])) > big) big = temp;
}
if (big == 0.0)
{
printf("ERROR: Singular matrix\n");
}
// non-zero largest element.
vv[i] = 1.0/big; // Save the scaling
}
Pivoting Crout's method
• Finds maximum
if((dum = vv[i] * fabs(sum)) >= big)
{
// Is the figure of merit for the pivot better than the best so far?
big = dum;
imax = i;
}
Pivoting Crout's method
• Performs row interchanges
if(j != imax) // Do we need to interchange rows?
{
for(k = 0; k < n; k++) // Interchange rows
{
dum = a[imax][k];
a[imax][k] = a[j][k];
a[j][k] = dum;
}
d = -d; // change the parity of d
vv[imax] = vv[j]; // interchange scale factor
}
indx[j] = imax;
Related Questions Crout's method
• What is the advantage of LU(P) solver over GJ(P) solver? (Complexity)
• Both are O(N3)
• After LU(P) is solved, more solutions supposed to be found in O (N2)
• Are you keeping L and U in the same matrix, or separate? Advantage/disadvantage?
• LU are being created in place in the same matrix.
• The advantage to this strategy is lower memory usage
• The disadvantage is that the original matrix is lost
• I am somewhat confused with extraction of P in decomposition, and how it is then used in eq solving. Can you elaborate more?
Related Questions Crout's method
• Cormen et al., p 824, used a single array instead of P. Needs careful explanation.
• From Cormen et al. 825. “we dynamically maintain the permutation matrix P as an array π, where π[i] = j means that the ith row of P contains a 1 in column j
|2| | 0 1 0 |
π = |3| => P | 0 0 1 |
|1| | 1 0 0 |
Related Questions Crout's method
• How complex equations are solved? (in Text)
• If only the right hand side vector is complex then the operation can be performed by solving for the real part, then the imaginary
• If the matrix itself is complex then
• Rewrite the algorithm for complex values
• Split the real and imaginary parts into separate real number and solve using existing algorithm
• A * x – C * y = b
• C * x + A * y = b
GJ vs. LUP: Crout's methodwe found lup faster than gj, but…
GJ vs. LUP: Crout's methodlup not faster amortized
GJ vs. LUP Crout's method
Average Difference is 2.765471
Average Difference is 1.255924
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# Past Age Calculation Using an Age Calculator
In the realm of numerical calculations, determining one's age at a specific past date is a fascinating exercise that combines basic arithmetic with an understanding of calendar systems. An Age Calculator simplifies this process, offering a straightforward method to ascertain how old someone was or will be on a given date. This comprehensive guide delves into the mechanics, applications, and nuances of using an Age Calculator for past age calculations.
## What is an Past Age Calculator?
An Past Age Calculator is a digital tool designed to compute the age of an individual by comparing their birth date with another date, typically the current date or a date in the past. It automates the process of subtracting dates, accounting for leap years, and handling different calendar systems, providing accurate age calculations.
An age calculator is a handy tool that can answer questions like 'How old was I on this date' or 'How old would I be if I was born in a different year?'
## Basic Concepts in Age Calculation
### Importance of Dates
The cornerstone of age calculation is the accurate knowledge of two dates: the birth date and the reference date (the date on which the age is to be calculated). The precision in these dates directly impacts the accuracy of the calculated age.
### Calculating Current Age
The current age is determined by subtracting the birth year from the current year and adjusting for whether the birth date has passed in the current year. This calculation must consider leap years, which add a day to February every four years, affecting individuals born in or around leap years.
For those who are looking for a straightforward method to calculate their current age without the specifics of past or future dates, our Basic Age Calculator offers a user-friendly interface to quickly find out your age as of today.
## Past Age Calculation
### Basics
For calculating how old someone was on a specific past date, one needs the birth date and the past date in question. The process involves determining the number of full years, months, and possibly days between the two dates.
If your interest extends to finding out your age on any specific date in the past, not just how old you were at a certain historical event, our Age Specific Date Calculator is designed to provide precise age calculations for any date you're curious about.
### Impact of Date Formats
The format in which dates are presented (DD/MM/YYYY vs. MM/DD/YYYY) can significantly affect the calculation process. Misinterpretation of formats can lead to incorrect age calculations, highlighting the importance of clarity in date notation.
### Time Zones and Age Calculations
Time zones can influence age calculations, especially if the birth date and the reference date are in different time zones. This factor is often overlooked but can be crucial for precision.
### Age Calculation in Different Calendars
Calculating age using non-Gregorian calendars, such as Lunar or Hijri calendars, introduces additional complexity. These calendars have different year lengths and month counts, requiring specialized knowledge for accurate age calculations.
### Programming Age Calculators
Developing an age calculator involves programming logic that can accurately subtract dates, account for leap years, and adjust for time zones and different calendar systems. This process necessitates a deep understanding of date arithmetic and calendar peculiarities.
## Applications and Analysis
### Case Studies: Historical Figures
Age calculators can bring historical studies to life by calculating the ages of historical figures during significant events, providing context to their actions and decisions.
### Educational Requirements
Schools often use age calculators to determine eligibility for enrollment in certain grades, based on cut-off dates for age. This application underscores the practical utility of age calculators in everyday decision-making.
## Evaluating Age Calculator Tools
When evaluating online age calculator tools, consider accuracy, user-friendliness, and flexibility in handling different calendars and date formats. A reliable tool should offer precise calculations, be easy to use, and accommodate various user needs.
## Ethical Considerations in Age Data Handling
The development and use of age calculators touch upon ethical considerations, especially regarding the handling of birth dates, which are sensitive personal data. Ensuring data privacy and securing user information are paramount concerns that developers and users alike must address.
## FAQs on Past Age Calculation Using an Age Calculator
### What is an Age Calculator?
An Age Calculator is a tool designed to calculate the age of a person by comparing their birth date with another date. It simplifies determining how old someone was on a specific past date or how old they will be on a future date.
### How do I calculate my age on a specific past date?
To calculate your age on a specific past date, you need your birth date and the past date in question. The Age Calculator subtracts the birth date from the past date to determine your age at that time, taking into account leap years and the actual number of days in each month.
#### Why are leap years important in age calculation?
Leap years add an extra day (February 29th) to the calendar every four years to keep our calendar in alignment with the Earth's revolutions around the Sun. They are important in age calculations because they affect the number of days in a year, which can impact the calculation of a person's age, especially if their birth date is close to or on February 29th.
### Can time zones affect age calculations?
Yes, time zones can affect age calculations, especially if the birth date and the date for which age is being calculated are in different time zones. This is usually a minor factor but can be relevant for precise age calculation, particularly for events happening around the birth date.
### How do different calendar systems impact age calculations?
Different calendar systems, such as the Gregorian, Lunar, or Hijri calendars, have varying year lengths and start dates, which can impact age calculations. An Age Calculator needs to account for these differences when calculating age in a specific calendar system, requiring specialized algorithms or settings.
### Can I use an Age Calculator to find out if my child is eligible for kindergarten?
Yes, you can use an Age Calculator to determine if your child meets the age requirement for kindergarten in your area. By entering your child’s birth date and the cutoff date for kindergarten entry, the calculator can tell you if your child is eligible.
### How accurate are online Age Calculators?
The accuracy of online Age Calculators depends on the algorithms they use and their ability to account for leap years, different calendar systems, and date formats. Most reputable calculators are highly accurate for standard Gregorian calendar calculations.
### Are there ethical concerns with using Age Calculators?
The primary ethical concern with using Age Calculators involves the handling of birth date information, which is sensitive personal data. Users should ensure that any online calculator they use has privacy protections in place to secure their data.
### Can I calculate the age of a historical figure at a significant event using an Age Calculator?
Yes, by entering the birth date of the historical figure and the date of the significant event into an Age Calculator, you can determine how old that person was at the time of the event. This can provide interesting insights into historical contexts.
### What criteria should I use to evaluate the effectiveness of an Age Calculator?
When evaluating an Age Calculator, consider its accuracy, the comprehensiveness of its calculations (including leap years and different calendars), its user interface, and its privacy policies. An effective calculator should provide precise age calculations, be easy to use, and ensure the privacy of user data.
## Conclusion
Past age calculation using an Age Calculator is more than a mathematical exercise; it is a tool that bridges the gap between dates and real-world applications. Whether for educational purposes, historical analysis, or personal curiosity, understanding the mechanics and considerations behind age calculation can enhance our appreciation of time and its impact on our lives.
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# Conic optimization
Conic optimization is a subfield of convex optimization that studies a class of structured convex optimization problems called conic optimization problems. A conic optimization problem consists of minimizing a convex function over the intersection of an affine subspace and a convex cone.
The class of conic optimization problems is a subclass of convex optimization problems and it includes some of the most well known classes of convex optimization problems, namely linear and semidefinite programming.
## Definition
Given a real vector space X, a convex, real-valued function
${\displaystyle f:C\to \mathbb {R} }$
defined on a convex cone ${\displaystyle C\subset X}$, and an affine subspace ${\displaystyle {\mathcal {H}}}$ defined by a set of affine constraints ${\displaystyle h_{i}(x)=0\ }$, a conic optimization problem is to find the point ${\displaystyle x}$ in ${\displaystyle C\cap {\mathcal {H}}}$ for which the number ${\displaystyle f(x)}$ is smallest. Examples of ${\displaystyle C}$ include the positive semidefinite matrices ${\displaystyle \mathbb {S} _{+}^{n}}$, the positive orthant ${\displaystyle x\geq \mathbf {0} }$ for ${\displaystyle x\in \mathbb {R} ^{n}}$, and the second-order cone ${\displaystyle \left\{(x,t)\in \mathbb {R} ^{n+1}:\lVert x\rVert \leq t\right\}}$. Often ${\displaystyle f\ }$ is a linear function, in which case the conic optimization problem reduces to a semidefinite program, a linear program, and a second order cone program, respectively.
## Duality
Certain special cases of conic optimization problems have notable closed-form expressions of their dual problems.
### Conic LP
The dual of the conic linear program
minimize ${\displaystyle c^{T}x\ }$
subject to ${\displaystyle Ax=b,x\in C\ }$
is
maximize ${\displaystyle b^{T}y\ }$
subject to ${\displaystyle A^{T}y+s=c,s\in C^{*}\ }$
### Semidefinite Program
The dual of a semidefinite program in inequality form,
minimize ${\displaystyle c^{T}x\ }$ subject to
${\displaystyle x_{1}F_{1}+\cdots +x_{n}F_{n}+G\leq 0}$
is given by
${\displaystyle \mathrm {tr} \ (F_{i}Z)+c_{i}=0,\quad i=1,\dots ,n}$
${\displaystyle Z\geq 0}$
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Population is a group of organisms of the same species which occupies the same habitat and forms a single functional unit of biotic community. This ecological definition doesn't fit well into genetics. For genetic studies, a population refers to a group of interbreeding individuals which belong to a single species. The smallest subgroup of individuals which can interbreed in a population is called a deme. A population generally consists of one or more such demes.
Population can be characterized and analyzed by mainly three different factors.
a. Population density
b. Natality
c. Mortality
a. Population density
It is the size of population in a definite unit of space. Density is the total number of individuals per unit area in a given time.
Crude density refers to the number/ biomass of the organisms per unit of total space.
Ecological (specific) density refers to number /biomass of the organisms per unit of habitable area, i.e. the area which can actually be colonized by the population.
The population density is never static for most of the populations and hence indices are employed to compare the densities at different time intervals. These indices of relative abundance represent the occurrence or sighting of the organisms at a given interval of time.
The frequency of occurrence is also useful to estimate the percentage of sample area which is occupied by the species.
The Lincoln index represents the population density by the popular mark-recapture method. This is given by the following
Population estimate (N) / no. of animals captured and marked in a sample s1 at time t1 (n1)
= number of animals captured in second sample at time t2 (n2) / number of marked animals in sample s2 at time t2 (m2)
Then N/n1= n2/m2
Hence N = n1 *n2 / m2
Relative abundance is suitable for calculating population density of larger animals and terrestrial plants. However, in vegetation studies the factors such as density, dominance and frequency are combined to give the importance value specific for each species which can provide a more precise descriptive study of the fauna.
b. Natality
Natality rate refers to the ability of the population to give rise to new offspring. This rate is equivalent to birth rate in human demographic studies
Natality rates can be either crude or realized one.
Crude/absolute natality rate is the rate at which new individuals are formed in units of time in a population. This is usually maximum natality, the maximum theoretical limit of production of new organisms under ideal conditions when the resources are assumed to be not limiting the population growth. The reproductive rate is assumed to be dependent only on the physiological factors and not on ecological factors.
Specific natality rate or birth rate refers to the number of new individuals in unit time per unit of population.
Ecological or realized natality rate is the increase in population under a given environmental condition. This is not constant and varies with the age structure and other environmental factors.
c. Mortality
This is the death rate of the individuals in the population. Like natality this is equivalent to death rate in human populations.
Ecological or realized mortality rate is the number of individuals lost in a given environmental condition.
Minimum mortality refers to the minimum loss under ideal conditions of environment. This value is constant for a population and is often determined by physiological longevity. This value is often larger than ecological longevity.
The survival rate is the number of surviving individuals in a population expressed as 1-M if M stands for death rate.
A life table is a systematic representation of the specific mortalities of the population at different stages of growth.
There are three main types of survivorship curves which reflects the population density.
Type I- The population death rate is rather low till the end of the life span. The curve is highly convex. E.g.: humans, deer, sheep etc,
Type II - It is a diagonal straight line exhibiting constant rate of mortality over all age groups. E.g.: birds, mice, rabbits, etc. This is more often depicting the amount parental care in these species.
Type III - This curve is highly concave with the death rates at its peak during younger stages of growth. E.g. oysters, shell fish, oak trees.
The curve may be stair-step type if the survival rates differ greatly in successive stages of life. E.g.: butterflies
Apart from the three common characteristics there are many other factors which are unique to a population. Spatial structure of populations are determined by the three factors such as
• Density
• Dispersion
• Distribution
Factors which influence the population density have been described in the previous article.
I. Dispersion is the spatial and temporal distribution pattern of the individuals in the population. They describe the position of individuals in a population in relation to each other. There are mainly three forms - random, regular and clumped (aggregate).
In random distribution, the individuals are having equal probability of being found in anywhere within a given area. The position is unrelated to the presence or vicinity of the individuals of the population. The population is relatively uniform and there are no aggregates found. This is however very rare in actual environmental conditions. The intra and interspecific interactions and ecological interactions are more or less neutral.
In regular dispersion, the individuals are equidistantly spaced from each other. This is usually found in cultivated lands where equidistance is maintained between the crops. The interactions between the individuals are antagonistic in nature and resources get depleted locally.
In clumped dispersion, the individuals are found in aggregates. Most populations in nature fall into this category. There are well marked areas of abundance separated by areas of low abundance of the species. The interactions are more agonistic with clumping of organisms at local abundance of resources.
II. Distribution is mostly limited by the physical environmental factors such as moisture, geographical structure, temperature etc.
This distribution has been well depicted in terms of age of individuals as in age pyramids. Age structure shows the number of individuals in each population classified according to their age groups.
Age distribution affects the natality as well as the mortality rate. The ratio between the different age groups determines the reproductive status of the population. This can be classified into pre-reproductive, reproductive, and post- reproductive distributions. Humans have equally distributed distributions for the three age groups.
Age pyramids are a convenient way to represent the age distribution.
Based on the nature of age pyramids, populations can be classified into '
a. stable population
Pre reproductive as well as reproductive age groups have almost the same number of individuals with the smallest number of individuals in the post reproductive age group. The graphic representation is a bell shaped curve.
b. expanding population and
The successive generations are numerous in number than the original parent population. There will be more individuals in the pre reproductive age group than the other age groups. The graph is usually pyramidal in shape.
c. diminishing population
When the pre reproductive and reproductive age groups are almost similar in size, and there is drastic reduction in the post reproductive group size, it results in a diminishing population. The graph is urn shaped.
Dispersal is the process by which the individuals get added up or removed from a population. The usual causes of dispersal of individuals are migration, immigration and emigration.
Growth forms are characteristic patterns depicting increase in population. There are basically two patterns- J shaped and S- shaped.
J-shaped growth form has the populations increasing in an exponential pattern. The growth ceases when there is environmental resistance or in the presence of any other limiting factors.
The upper limit is a sharp and abrupt and is represented by intrinsic rate of natural increase, r. The maximum value of r is called biotic potential or reproductive potential.
The biotic potential represents the maximum ability of the individuals in a population to reproduce. This can be increased by having a higher growth rate, longer reproductively active life span, and by having the maximum number of offspring at relatively younger age groups when the energy can be efficiently utilized more for such functions.
In the sigmoid or S shaped growth pattern, the individuals of the population increases slowly and slows down after reaching a certain limit of equilibrium. This is a logistic model of population growth.
The upper limit of the sigmoid curve at which the population is considered to attain equilibrium is called carrying capacity. This is usually represented by K. There are various factors which limit this value including predation, competition, environmental conditions etc. The value is not a constant since it includes all the factors into consideration and these factors can vary widely over time.
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# Enthalpy Calculation for 4B+3O2-->2B2O3 Reaction Using Hess's Law
• Evis
In summary, to calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s), we must use Hess' Law and add the enthalpies of multiple reactions. This includes adding the reactions B2O3(s)+3H2O(g)→3O2(g)+B2H6(g) with ΔH∘A=+2035kJ, 2B(s)+3H2(g)→B2H6(g) with ΔH∘B=+36kJ, H2(g)+1/2O2(g)→H2O(l) with ΔH∘C=−285kJ, and H2O(l)
Evis
## Homework Statement
Calculate the enthalpy of the reaction
4B(s)+3O2(g)→2B2O3(s)
given the following pertinent information:
1. B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035kJ
2. 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36kJ
3. H2(g)+1/2O2(g)→H2O(l), ΔH∘C=−285kJ
4. H2O(l)→H2O(g), ΔH∘D=+44kJ
## Homework Equations
ΔH=Sum of all ΔH of each reaction. (Hess' Law?)
## The Attempt at a Solution
2[2B(s)+3H2(g)→B2H6(g)] ΔH=72 J
2[3O2(g)+B2H6(g)→B2O3(s)+3H2O(g)] ΔH=-4070kJ
ΔH=3998 kJThis incorporates the correct reactants and products. We still need to add H2, but there's no way to react only the given reactants, B and O2, into H2 at all. Also, there is no way to react out the H2O from the given equations.
The actual solution
6[H2O(g)→H2O(l)] ΔH=-264 kJ
6[H2O(l)→H2(g)+1/2O2(g)] ΔH=1710J
2[2B(s)+3H2(g)→B2H6(g)] ΔH=72 kJ
2[3O2(g)+B2H6(g)→B2O3(s)+3H2O(g)] ΔH=-4070kJ
ΔH=-2552
Colours correspond to canceled products.
I understand the process given, but not how it relates to the goal equation. This process requires water to be added as a reactant and for 3H2 and 3H2O to be products, which are not included in the goal equation. Adding the extra reactions with water do not seem directly part of the goal equation.
Evis said:
Adding the extra reactions with water do not seem directly part of the goal equation.
But is necessary to cancel out hydrogen and water. Your solution is incorrect as your equation is not identical with the goal equation, and it must be.
## 1. What is the definition of enthalpy?
Enthalpy is a thermodynamic property that represents the total energy of a system, including both its internal energy and the work required to create or maintain its structure.
## 2. How is enthalpy calculated?
Enthalpy is calculated by taking into account the internal energy, pressure, and volume of a system using the equation H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume.
## 3. Why is enthalpy important in chemical reactions?
Enthalpy is important in chemical reactions because it helps us understand the changes in energy that occur during a reaction. It also allows us to predict whether a reaction will release or absorb heat.
## 4. What is the enthalpy change for the reaction 4B + 3O2-->2B2O3?
The enthalpy change for this reaction is the difference in enthalpy between the reactants (4B + 3O2) and the products (2B2O3). This can be determined experimentally by measuring the heat released or absorbed during the reaction.
## 5. How does the enthalpy of a reaction affect its spontaneity?
The enthalpy of a reaction is a factor in determining its spontaneity. A negative enthalpy change (exothermic reaction) will make the reaction more likely to occur spontaneously, while a positive enthalpy change (endothermic reaction) will make the reaction less likely to occur spontaneously.
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Exam_1_08
# Exam_1_08 - Management 122 Winter 2008 Danny S Litt EXAM 1...
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Unformatted text preview: Management 122 Winter 2008 Danny S. Litt EXAM 1 Solutions NAME: ________________ I agree to have my grade posted by Student ID Number. ___________________ (Signature) PROBLEM POINTS SCORE 1 15 2 10 3 30 4 30 5 30 6 30 7 20 8 15 9 20 TOTAL 200 M ANAGEMENT 122 N AME : __________________________ Page | 1 Problem 1 The following data pertain to Graham Company's operations in May: May 1 May 31 Work in process inventory ............. \$7,000 \$12,000 Raw materials inventory ................ \$15,000 ? Finished goods inventory ............... ? \$20,000 Other data: Raw materials used ........................ \$40,000 Sales \$200,000 Cost of goods manufactured .......... \$135,000 Manufacturing overhead cost ......... \$60,000 Raw materials purchases ................ \$30,000 Gross Margin ................................. \$60,000 Requirements Determine the following: Ending materials inventory \$5,000 Beginning finished goods inventory \$25,000 Direct labor cost for May \$40,000 Beginning raw materials inventory ................ \$15,000 Add: Raw materials purchases....................... 30,000 Raw materials available for use ..................... 45,000 Deduct: Ending raw materials inventory ....... 5,000 * Raw materials used ........................................ \$40,000 *Calculate this item by working backwards, as shown: Raw materials used = Raw materials available − Ending raw materials inventory \$40,000 = \$45,000 − Ending raw materials inventory Ending raw materials inventory = \$5,000 Sales − Cost of goods sold = Gross margin Cost of goods sold = Sales − Gross margin Cost of goods sold = \$200,000 − \$60,000 Cost of goods sold = \$140,000 Next, solve backwards for beginning finished goods inventory: Beginning raw materials inventory ...................... \$ 25,000 * Add: Cost of goods manufactured ........................ 135,000 Cost of goods available for sale ........................... 160,000 * Deduct: Ending finished goods inventory ............ 20,000 Cost of goods sold ................................................ \$140,000 * These items must be calculated by working backwards upward through the statements. M ANAGEMENT 122 N AME : __________________________ Page | 2 Graham Company Schedule of Cost of Goods Manufactured Direct materials .................................................................. \$40,000 Direct labor ........................................................................ 40,000* Manufacturing overhead .................................................... 60,000 Total manufacturing costs.................................................. 140,000* Add: Work in process, beginning ...................................... 7,000 147,000* Deduct: Work in process, ending....................................... 12,000 Cost of goods manufactured .............................................. \$135,000 * These items must be calculated by working backwards upward through the statements. M ANAGEMENT 122 N AME : __________________________...
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# Show two unique points lie on a unique line.
## Homework Statement
Suppose (a,b) and (c,d) are unique points in ##\mathbb{R}^{2}## that satisfy Ax+By+C=0. Suppose also that the solutions of Ax+By+C=0 form a line. Then the line that (a,b) and (c,d) lie on is unique.
## Homework Equations
I showed in a previous part of the problem that there exists real numbers A,B,C not all equal to 0 that satisfy Aa+Bb+C=0 and Ac+Bd+C=0. I also showed that if A',B',C' were also nontrivial solutions that satisfied the two equations then A',B',C' were scalar multiples of A,B,C.
## The Attempt at a Solution
Suppose that (a,b) and (c,d) are unique points that satisfy Ax+By+C=0. Suppose also that the set of solutions of Ax+By+C=0 forms a line that is not unique so there exists another equation such that A'x+B'y+C'=0 but we previously showed that A',B',C' must be scalar multiples of A,B,C so the two lines are actually the same. Hence the line that (a,b) and (c,d) lie on is unique.
I thought that since A',B',C' were scalar multiples of A,B,C then the new equation was just a stretched version of the older line so they must be the same line. Is this logic correct?
## Homework Statement
Suppose (a,b) and (c,d) are unique points in ##\mathbb{R}^{2}## that satisfy Ax+By+C=0. Suppose also that the solutions of Ax+By+C=0 form a line. Then the line that (a,b) and (c,d) lie on is unique.
## Homework Equations
I showed in a previous part of the problem that there exists real numbers A,B,C not all equal to 0 that satisfy Aa+Bb+C=0 and Ac+Bd+C=0. I also showed that if A',B',C' were also nontrivial solutions that satisfied the two equations then A',B',C' were scalar multiples of A,B,C.
## The Attempt at a Solution
Suppose that (a,b) and (c,d) are unique points that satisfy Ax+By+C=0. Suppose also that the set of solutions of Ax+By+C=0 forms a line that is not unique so there exists another equation such that A'x+B'y+C'=0 but we previously showed that A',B',C' must be scalar multiples of A,B,C so the two lines are actually the same. Hence the line that (a,b) and (c,d) lie on is unique.
I thought that since A',B',C' were scalar multiples of A,B,C then the new equation was just a stretched version of the older line so they must be the same line. Is this logic correct?
Assuming you've shown this :
we previously showed that A',B',C' must be scalar multiples of A,B,C so the two lines are actually the same
Then yes, the trick would be to assume that the line was not unique and that there are two lines. Then showing the two lines are the same shows that the original line must be unique.
Does A',B',C' being scalar multiples of A,B,C show that we are dealing with the same lines or should I be further justifying this?
Does A',B',C' being scalar multiples of A,B,C show that we are dealing with the same lines or should I be further justifying this?
If you aren't changing the actual location of the coordinates ##(a,b)## and ##(c,d)##, then the scalar line is a stretched version of the original line. So therefore they are the same line and pass through the same points, it's just one is longer than the other.
Ah! Cool. Thanks.
If you aren't changing the actual location of the coordinates ##(a,b)## and ##(c,d)##, then the scalar line is a stretched version of the original line. So therefore they are the same line and pass through the same points, it's just one is longer than the other.
Well, you wouldn't say "one is longer than the other" because lines, as opposed to line segments, so not have a "length".
Well, you wouldn't say "one is longer than the other" because lines, as opposed to line segments, so not have a "length".
Would it be appropriate to say that one line is a stretched version of another?
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# Convex polygon
This is an AoPSWiki Word of the Week for Sep 20-26
A convex polygon is a polygon whose interior forms a convex set. That is, if any 2 points on the perimeter of the polygon are connected by a line segment, no point on that segment will be outside the polygon.
All internal angles of a convex polygon are less than $180^{\circ}$. These internal angles sum to $180(n-2)$ degrees.
The convex hull of a set of points also turns out to be the convex polygon with some or all of the points as its vertices.
The area of a regular n-gon of side length s is $\frac{ns^2*\tan{(90-\frac{180}{n})}}{4}$
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8 Replies Latest reply: Aug 23, 2013 9:27 AM by Bruno Santos
# Set Analysis and concat
Hi all,
I have a chart expression:
sum({1<[Ordering Depot]={'\$(=concat(distinct Depot))'}>} [Number Orders])
but this always (incorrectly) returns 0
I am using the dollar-sign expansion so that it returns the results of the concat in single quotes as 'Depot' is a text field which may contain spaces.
If I create a chart and put any of these expressions into it:
=concat(distinct Depot)
\$(=Year(Today()))
\$(=concat(distinct Year(Today())))
they return the expected values but if I try
\$(=concat(distinct Depot))
it returns nothing.
I am obviously missing something, but what?
Any help appreciated.
Gordon
• ###### Set Analysis and concat
Hello Gordon,
My guess is that since concat has no delimitator, is concatenating one field after another, and since they are text fields, they need a precedeing and trailing comma taht concat does not add, so
`\$(=chr(39) & Concat(distinct Año, chr(39) & ',' & chr(39)) & chr(39))`
will do the trick.
Hope that helps!
• ###### Set Analysis and concat
Hi Miguel,
I understand what you are saying but I couldnt get it to work:
sum({1<[Ordering Depot]={\$(=chr(39) & Concat(distinct Depot, chr(39) & ',' & chr(39)) & chr(39))}>} [Number Orders])
I actually found 'only' works and here is my expression:
sum({1<[Ordering Depot]={'\$(=only(Depot))'}>} [Number Orders])
I am still intrigued why concat doesnt though.
Regards,
Gordon
• ###### Set Analysis and concat
Gordon,
I'm probably missing something also. Anyway, you may try setting a variable with the whole analysis, say vAnalysis
`'{' & chr(39) & Concat(distinct Depot, chr(39) & ',' & chr(39)) & chr(39) & '}'`
and then in your expression
`Sum({1< [Ordering Depot] = \$(vAnalysis) >} [Number Orders])`
Might work. I also agree that John's expression might work and is cleaner by far, though.
Regards.
• ###### Set Analysis and concat
Hi Miguel/John,
Miguel - again nice thinking but the expression editor sees it as an error. I also tried changing the variable and removed the curly brackets and put them around the variable in the expression. The editor was happy this time but again nothing was returned.
John - the elemental function P() works great and one that I wasnt aware of.
Thanks for your inputs.
Regards,
Gordon
• ###### Re: Set Analysis and concat
Hi Guys,
I've a related problem.
Set Analysis restrict months depending on other expression.
Could you give me some help, with your expertise?
Thanks
Bruno
• ###### Set Analysis and concat
I think you should be able to use another field as an element set:
sum({1<[Ordering Depot]=Depot>} [Number Orders])
• ###### Set Analysis and concat
Hi John,
Thanks for your input.
I agree it should be as simple as that, but I found that it also returned 0 and that started me off on the road to concat. Again I dont know why it doesnt work.
Regards,
Gordon
• ###### Set Analysis and concat
`gordon.savage wrote:I agree it should be as simple as that, but I found that it also returned 0 and that started me off on the road to concat. Again I dont know why it doesnt work.`
Weird. Maybe this?
sum({1<[Ordering Depot]=P(Depot)>} [Number Orders])
But it sounds like you have a working solution already. I agree that the concat() needs the extra characters, but at that point I'd have expected it to work. I've done it before, even if only in examples, and I'm not seeing anything wrong with your syntax on brief glance.
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# Basic Data Types
Data types in Slogan falls into three categories - basic types, composite types and user-defined types. Basic types, the topic of this chapter, include numbers, characters, strings, symbols and boolean. Composite types are more complex because they are formed by combining values of several simpler ones. Arrays, pairs, lists, hash tables, sets and records are examples of composite types. Slogan also allow the definition of new types that conform to user-specified interfaces.
## 4.1 Numbers
Slogan classifies numbers as integers, rational, real and complex. This classification is hierarchical, in that all integers are rational, all rational numbers are real, and all real numbers are complex. Orthogonal to these categories, a number is also either exact or inexact. In most cases, computations that involve an inexact number will produce an inexact result. One exception to this rule is multiplying an inexact number with the exact `0`, which will produce an exact number. Operations that mathematically produce irrational numbers for some rational arguments (e.g., `sqrt`) may produce inexact results even for exact arguments.
There are predicates1 that can be used to determine the specific type of a number.
``````
is_integer(123)
// true
is_real(123)
// true
is_real(1/23)
// true
is_integer(1/23)
// false
is_number(1/23)
// true
is_number(123)
// true
is_rational(1/23)
// true
``````
Exact integer and rational arithmetic is supported to arbitrary precision; the size of an integer or of the denominator or numerator of a ratio is limited only by system storage constraints.
Slogan numbers are written in a straightforward manner not much different from ordinary conventions for writing numbers. An exact integer is normally written as a sequence of numerals preceded by an optional sign. For example, `3`, `+19`, `-100000`, and `208423089237489374` all represent exact integers.
An exact rational number is normally written as two sequences of numerals separated by a slash (`/`) and preceded by an optional sign. For example, `3/4`, `-6/5`, and `1/1208203823` are all exact rational numbers. A ratio is reduced immediately to lowest terms when it is read and may in fact reduce to an exact integer.
Inexact real numbers are normally written in either floating-point or scientific notation. Floating-point notation consists of a sequence of numerals followed by a decimal point and another sequence of numerals, all preceded by an optional sign. Scientific notation consists of an optional sign, a sequence of numerals, an optional decimal point followed by a second string of numerals, and an exponent; an exponent is written as the letter e followed by an optional sign and a sequence of numerals. For example, `1.0` and `-200.0` are valid inexact integers, and `1.5`, `0.034`, `-10e-10` and `1.5e-5` are valid inexact rational numbers. The exponent is the power of ten by which the number preceding the exponent should be scaled, so that `2e3` is equivalent to `2000.0`.
Exact and inexact real numbers are written as exact or inexact integers or rational numbers; no provision is made in the syntax of Slogan numbers for non-rational real numbers, i.e., irrational numbers.
The exactness of a numeric representation may be overridden by preceding the constant by either `0e` or `0i`. `0e` forces the number to be exact, and `0i` forces it to be inexact. For example, `1`, `0e1`, `1/1`, `0e1/1`, `0e1.0`, and `0e1e0` all represent the exact integer `1`, and `0i3/10`, `0.3`, `0i0.3`, and `3e-1` all represent the inexact rational `0.3`.
``````
is_exact(123 * 100)
// true
is_exact(123 * 100.0)
// false
is_inexact(123 * 100.0)
// true
1 == 1.0
// false
1 == 0e1.0
// true
inexact(1) == 1.0
// true
0i1 == 1.0
// true
0i1 == exact(1.0)
// false
``````
Numbers are written by default in base 10, although the special prefixes `0b` (binary), `0o` (octal), `0d` (decimal), and `0x` (hexadecimal) can be used to specify base 2, base 8, base 10, or base 16. For radix 16, the letters a through f or A through F serve as the additional numerals required to express digit values 10 through 15. For example, `0b10101` is the binary equivalent of `2110`, `0o72` is the octal equivalent of `5810`, and `0xC7` is the hexadecimal equivalent of `19910`. Numbers written in floating-point and scientific notations are always written in base 10.
Underscores may be added to a number to improve readability. For example, the integer `1234567` could be formatted as `1_23_4567`.
Complex number literals takes the form `R+Ii`, where `R` is the real part and `I` is the imaginary part. E.g: `3+7i`.
There are functions that corresponds to the arithmetic and comparison operators. These functions can accept an arbitrary number of arguments.
``````
// 15
number_is_lt(1,2,3,4,5)
// true
number_is_lt(1,2,3,40,5)
// false
number_is_lt(1,2,3,4,4)
// false
number_is_lteq(1,2,3,4,4)
// true
mult(20, 30, 40)
// 24000
``````
### 4.1.1 Bitwise Operations
In this section we will discuss functions that perform bitwise binary operations on integers. Some of the most useful of these functions are listed below:
band bitwise AND bior bitwise inclusive OR bxor bitwise exclusive OR bnot bitwise NOT bshift left/right arithmetic shift is_bit_set predicate to test bit by position
If the number of bits to shift is negative, `bshift` performs a right-shift. Otherwise, the bits are shifted left.
Bitwise operations assume that integer are represented in two's complement, even if they are not represented that way internally.
The following program show how to interpret an integer as a compact set of independent bits.2 Note that we make use of only the first 32 bits of the integer, while the underlying value may have more bits. To view the binary representation of an integer, the built-in `number_to_string` function is called. It takes an optional second argument that specifies the base in which the result string should be formatted. To get a binary formatted string, we have to pass `2` here.
``````
function turn_bit_on(bits, i)
if (i <= 31) bior(bits, bshift(1, i))
else bits
let a, b = turn_bit_on(0, 31), turn_bit_on(0, 31)
number_to_string(a, 2)
// 10000000000000000000000000000000
number_to_string(b, 2)
// 10000000000000000000000000000000
a = turn_bit_on(turn_bit_on(a, 1), 5)
b = turn_bit_on(turn_bit_on(b, 1), 2)
number_to_string(a, 2)
// 10000000000000000000000000100010
number_to_string(b, 2)
// 10000000000000000000000000000110
number_to_string(band(a, b), 2) // intersection
// 10000000000000000000000000000010
number_to_string(bior(a, b), 2) // union
// 10000000000000000000000000100110
is_bit_set(a, 1) // membership test
// true
is_bit_set(a, 10)
// false
``````
### 4.1.2 Fixnums
Fixnums represent exact integers within a closed range [-2w-1, 2w-2 - 1], where `w` is the fixnum width. The implementation-specific value of `w` can be determined via the function `fixnum_width`, and the endpoints of the range may be determined via the functions `least_fixnum` and `greatest_fixnum`.
The names of arithmetic procedures that operate only on fixnums begin with the prefix "fx" to set them apart from their generic counterparts. The following example demonstrates some of the most useful operations on fixnums:
``````
// 22
//> error: FIXNUM overflow
fx_is_eq(1, 1)
// true
fx_is_gt(1, 2)
// false
fx_is_gt(10, 2)
// true
fx_is_lteq(10, 2)
// false
fx_is_lteq(10, 10)
// true
fxsub(20, 32)
// -12
fxmult(20, 32)
// 640
fxdiv(20, 32)
0
fxdiv(20, 2)
// 10
``````
Bit and shift operations on fixnums assume that fixnums are represented in two's complement, even if they are not represented that way internally.
``````
number_to_string(fxior(4294967296, fxshift(1, 2)), 2)
// 100000000000000000000000000000100
number_to_string(fxior(4294967296, fxshift(fxshift(1, 2), -3)), 2)
// 100000000000000000000000000000000
``````
### Flonums
Flonums are inexact real numbers. Implementations typically use the IEEE double-precision floating-point representation for flonums. Flonum-specific function names begin with the prefix "fl" to set them apart from their generic counterparts.
``````
// 5.7
flmult(1.2, 4.5)
// 5.3999999999999995
fl_is_eq(0, 0.)
//> error: (Argument 1) FLONUM expected
fl_is_eq(0., 0.)
// true
fl_is_lt(-1., 0.)
// true
``````
## 4.2 Characters
Characters are atomic objects representing letters, digits, special symbols such as `\$` or `#`, and certain non-graphic control characters such as `space` and `newline`. Characters literals are written with the `\` prefix. For example, the character literal `A` will be represented in Slogan source code as `\A`.
The following are special literals that represent non-graphic characters:
``````
\newline
\return
\tab
\space
\backspace
\alarm
\vtab
\esc
\delete
\nul
``````
Any Unicode character may be written with the syntax '\xhh', '\uhhhh' or '\Uhhhhhhhh' where n consists of two, four or eight hexadecimal digits representing a valid Unicode scalar value.
All the comparison operators are overloaded to work with characters:
``````
\A == \A
// true
\A == \a
// false
\A == char_upcase(\a)
// true
\c > \b
// true
``````
There are many predicates useful for finding information about characters and for comparing them:
``````
is_char(\A)
// true
is_char(65)
// false
is_char(integer_to_char(65))
// true
char_is_numeric(\2)
// true
char_is_alphabetic(\e)
// true
char_is_lower_case(\e)
// true
char_is_eq(\a, \a) // `==` optimized to work with characters
// true
char_is_lteq(\a, \b) // `<=` optimized to work with characters
// true
``````
Let us write a new predicate for characters which return `true` if its argument is a vowel. This function will also introduce you to the `case` expression.
``````
function is_vowel(c)
case (c)
\a -> true
| \e -> true
| \i -> true
| \o -> true
| \u -> true
| else -> false
``` ```
`Case` evaluates an expression and compares its value to those in a list of clauses. This comparison is done using the `is_eq` function which basically checks if two values are stored in the same location in memory. On a successful match, the value of the clause is returned. An optional `else` can be defined to return a default value if all matches fail.
Multiple clauses that return the same value can be compressed into a single list. In `is_vowel` the `else` can also be omitted because the default value of `case` is `false`. These two points leads to the following rewrite of the function:
``````
function is_vowel(c)
case (c) [\a, \e, \i, \o, \u] -> true
// Usage:
is_vowel(\o)
// true
is_vowel(\b)
// false
is_vowel(\a)
// true
``````
## 4.3 Strings
A string is a sequence of characters enclosed in double-quotes. Slogan supports the Unicode standard. That means, Slogan strings can represent scripts from all of the world's written languages. The following are examples of valid string literals:
``````
"hello, world"
// a string may span multiple lines.
"this is a really
long message..."
"ἐγὼ εἰμί"
``````
Double-quotes inside a string must be escaped by a backslash.
``````
"he said: \"hello, there\""
// he said: "hello there"
``````
A list of all escape characters that can appear in string literal and their purpose is listed below:
``````
\n newline
\t tab
\r return
\\ backslash
\b backspace
\a alarm
\v vertical-tab
\" double-quote
\e escape
\d delete
\0 nul
\u unicode character encoded in 4 hexadecimal digits
\x unicode character encoded in 2 hexadecimal digits
\U unicode character encoded in 8 hexadecimal digits
``````
Let us familiarize ourselves with some useful operations on strings:
``````
let s = "For all its power, the computer is a harsh taskmaster."
// accessing individual characters by index:
string_at(s, 2)
// \r
s[2]
// \r
// splicing or extracting sub-strings:
substring(s, 4, 17)
// all its power
s[4:17]
// all its power
s[4:]
// all its power, the computer is a harsh taskmaster.
s[:17]
// For all its power
string_length(s[:17])
// 17
/* `count` is more generic than `string_length`, it can
also find the length of other "collections" of data, like arrays and lists. */
count(s)
// 54
// searching:
string_index_of(s, ",")
// 17
string_index_of(s, "computer")
// 23
string_append("abc", "def", "xyz")
// abcdefxyz
// split the string at commas and spaces:
string_split(s, [\,, \space])
// [For, all, its, power, the, computer, is, a, harsh, taskmaster.]
strings_join("-", string_split(s, [\,, \space]))
// comparisons
string_is_eq("abc", "abc")
// true
"abc" == "abc"
// true
string_is_eq("aBC", "abc")
// false
string_is_ci_eq("aBC", "abc")
// true
string_is_lt("abc", "xyz")
// true
"abc" < "xyz"
// true
"abc" >= "abc"
// true
``````
## 4.4 Symbols
Symbols are used for a variety of purposes as symbolic names in Slogan programs. Symbol constants are written by prefixing identifiers with the quote mark (`'`). All characters valid in identifiers can be used in symbols. Symbols with spaces and special characters are written by enclosing the symbol in tick (```) quotes. The following are all valid symbols:
``````
'abc
'\$abc
'`abc def`
'`abc+def`
``````
Strings could be used for most of the same purposes, but an important characteristic of symbols makes comparisons for equality much more efficient. This is because two symbols with the same sequence of characters are stored in the same memory location. This makes it possible to test them for equality with the `is_eq` function, which does a fast check if its arguments point to the same location in memory. On the other hand, effective string comparisons always require checking each character in both strings.
``````
let a = "hello"
let b = "hello"
// `==` will compare each character in both strings
a == b
// true
// so does `string_is_eq`
string_is_eq(a, b)
// true
/* `is_eq` only checks of two objects belong to the same
location in memory */
is_eq(a, b)
// false
is_eq(a, a)
// true
// In contrast to strings, two symbols made of the same sequence of characters
// can be efficiently compared for equality just by checking their memory locations.
let x = 'hello
let y = 'hello
let z = 'Hello
is_eq(x, y)
// true
is_eq(x, z)
// false
``````
It is possible to construct new symbols from strings and to convert symbols to strings:
``````
is_eq('hello, string_to_symbol("hello"))
// true
"hello" == symbol_to_string('hello)
// true
``````
1A predicate is a function that answers a question with a `true` or `false` value.
2Slogan has a composite type `bitarray` that can represent bitmaps of arbitrary sizes. This type will be introduced in the next chapter.
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# How To Read Motorcycle Tyre Size?
## How do you read motorcycle tire sizes?
Making Sense of Motorcycle Tire Sizes
1. THE FIRST NUMBER – WIDTH. The 130 indicates the width of the tire in millimeters.
2. THE SECOND NUMBER – ASPECT RATIO. The second number, the 90, represents how tall a motorcycle tire is in relationship to its width.
3. THE THIRD NUMBER – RIM SIZE. The last number, 17, refers to the tire’s rim diameter expressed in inches.
## What do motorcycle TYRE numbers mean?
The first section of this number indicates the tyre’s section width, in this case 120mm. The second number indicates the tyre’s aspect ratio or height expressed as a percentage of it’s width. The “17” denotes that the tyre is 17 inches in diameter whilst “M/C” means the tyre is for a motorcycle.
## What do TYRE size numbers mean?
The two-digit number after the slash mark in a tire size is the aspect ratio. For example, in a size P215/65 R15 tire, the 65 means that the height is equal to 65% of the tire’s width. The bigger the aspect ratio, the bigger the tire’s sidewall will be.
You might be interested: Quick Answer: How To Use Bungee Cord On Motorcycle?
## How wide is a 150 motorcycle tire?
Tire Width Cross Reference Table
Permissible Rim Widths Metric Standard Inch
2.50, 2.75, 3.00 130 5.00
2.75, 3.00, 3.50 140 5.50
3.50, 4.00 150 6.00
4.00, 4.50 160 6.25
## How does tire size affect motorcycle handling?
It’s always a balance of grip, handling, and wear. So, smaller, less powerful bikes can get away with smaller, narrower tires, while bigger, more powerful bikes require larger, wider tires so that they offer adequate traction and wear, but not so wide that it ruins the handling.
## Can I put bigger tires on my motorcycle?
When considering wider tires, you must factor in clearance for width and diameter, the effect on stability and handling, along with whether your rim is wide enough. If wider tires are approved for a motorcycle, it is usually permissible to increase by only one size designation.
## Are wider motorcycle tires better?
Having a wide rear tire helps to prevent slippage in wet, rainy conditions. Wide tires provide a smoother ride. They are more capable of absorbing the bumps on the road. Wider tires are great because they provide help with power transfer and help handle stronger motorcycle engines.
## How wide is a 200 motorcycle tire?
For example a 200/55R18 would be around 26.7 inches tall and 7.9 ( 200mm /25.4) inches wide.
## What is aspect ratio on a motorcycle tire?
The aspect ratio is the height of the sidewall expressed as a percentage of the width. Thus, this tire has a side wall height of 90 percent times 130 mm or 117 mm. The next item you’ll see is the rim size expressed in inches. It’s easy to see this tire is made to be mounted on a 16-inch wheel.
You might be interested: FAQ: How To Start Riding A Motorcycle?
## Can I use 235 tires instead of 225?
Yesthat size will work 100% perfectly, as it is the exact same diameter as the 225 /75-16. 235 /70-16 is actually a shorter tire.
## What do the 3 numbers mean on tire size?
B: TIRE WIDTH The three -digit number following the letter is the tire’s width (from side to side, looking at the tire head on) in millimeters. It’s the height of the sidewall measured from wheel rim to top of the tread, expressed as a percentage of tire width. In other words, it’s sidewall height divided by tire width.
## How do you read a TYRE code?
When it comes to how to read tyre size, the first five digits of the code are the ones to concentrate on.
1. The first three digits belong together and indicate the width of the tyre (in millimetres).
2. The two digits following the first slash are what is known as the aspect ratio.
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NO SET PATH
NO SET EXI
T
NO
SET ENTRY
Digital Design - Module 01 Semester 1, 2019 Niamh Carey
915640 Samuel Lalo + Studio 17
Week One
Reading: Zeara Polo, A. 2010. Between Ideas and Matters.
According to Zeara-Polo, the diagram does not play a representational role in the design process but provides an organisational and can have a performative quality depending on how it is deployed. Explain how Diagram is different from Signs and Symbols? (100 words Maximum)
As opposed to a sign or symbol, which is a direct representation of something, a diagram inaugurates a logic of sensation aimed at bringing forward new worlds. This can further be explored in ‘diagrammatic architecture’, which is used to trigger new possibilities (even ones that wouldn’t have otherwise been imagined). This is especially adventurous in a culture characterised by change. As discussed in the reading, diagrams have a reductive nature, but are required to precisely define certain elements. In other words, a very simple diagram may generate very complex organisations.
2
Week One
Precedent Analysis
Figure 1.
Figure 2.
Figure 1. Beginning of model process - extruding curves Figure 2. Adjusting the top layer by SoftEditCrv to achieve angles Figure 3. Final model - clear layering for ease of modelling
Figure 3.
Precedent Image: Leibinger, Barkow. Serpentine Summer House 2016. 2016, Photograph. Accessed March 16, 2019. https://barkowleibinger.com/archive/view/serpentine_summer_house_2016.
My initial steps were to import the plans and elevations into Rhino, then Offset curves to ensure a consistent thickness. These curves were then extruded, capped and moved up to the corresponding location given on the elevation. Any other information that was necessary but missing on the elevation was worked out using ratios and always comparing back to the drawings which were provided. For the ‘top’ layer, the process was slightly different. I used the SoftEditCrv tool and Sweep2 to produce the angled curves. All layers were to then be combined to produce the final model.
3
Week Two
Reading: Hertzberger H. 2005. The in-between and The Habitable Space Between Things, from Lessons for Students in Architecture. Herzberger discusses how design should not be extreme in its functionality. Use your precedent study to explain how the pavilion allows for an appropriation of use. (100 words Maximum)
Herzberger discusses how extreme functionality is rigid and inflexible. The Barkow Leibinger pavilion invites a degree of imagination into how it could be used. For example, irregularities (such as level differences) are embraced and maximally exploited to create a dynamic space which responds differently depending on light, temperature and weather conditions. This idea forms the basis of the top ‘layer’ of the pavilion. Leibinger has successfully managed to do more with the same material, yet organise it differently, which are points Herzberger discussed as ‘extra qualitative requirements’. In doing so, there is an aspect of ‘accommodating potential’, whereby the user can determine the use of the pavilion.
4
Week Two
Isometric
Barkow Leibinger Serpentine Pavilion After the model was complete in Rhino, the line weights needed to be adjusted in Illustrator. I had varying line weights from 0.03 to 1pt, however I also layered the line weight image with a feathered render and an outline without hidden lines, to represent the dynamic nature of the structure. I grew committed to the idea that this pavilion requires to have its complexity on show in order to discuss circulation paths and thresholds. It is not just the outer-face of the pavilion that attracts explorers, but rather the many curves, components and layers. Very quickly after exploring this pavilion myself, I came to the conclusion that not one persons path should match another, as this was a place that invited the person to create their own experiences. Such experiences would depend on the time of the day, the temperature, the lighting conditions and the persons intentions with the space, as well as their emotions on the day. All of these, would too, affect the thresholds, as there is true beauty in what the layers can produce - whether that be shelter or shadows, or both.
NO SET PATH NO SET EXIT
TRY
NO SET EN
5
Week Two Diagrams
Circulation Diagram
Threshold Diagram
This pavilion is not rigid or structured, but rather invites the user to ‘play’. There is no one entry, exit or path. People wander and explore the pavilion, whilst being attracted to the seats, shelter of the ‘walls’ and the ever-changing shadows of the roof.
Two main threshold ideas act on the ground and at roof level. The first is a landscaped reflection of the roof on the ground (see Main Precedent Image), which is also affected by the sun path. The second is the light that penetrates through the top layers.
6
Appendix
Process
Using the provided plans to create the seating as one of the first steps. Offsetting Points to ensure a consistent curve.
Producing the ‘tear-drops’, which form the inside of the seating. After inspecting the images, it was clear that these were separate to the main curves, almost acting as inserts. This turned out to be rather pleasing to model, as opposed to if it was constructed as one.
Working on the top layer - which required some different techniques to be used in order to produce the dynamic angles and correspond to all elevations.
7
Appendix Process
Notes from week 1 reading: Between Ideas and Matters and week 2 reading: Lessons for Students in Architecture.
(continued w2 reading) I found particularly interesting the idea discussed of having 2 doors to an apartment, which would beg the question, “which is the real door?” This would have advantages of opening up a dwelling, making clever use of what would otherwise be dead hallway space and creating a certain atmosphere of community.
8
Starting to explore the idea of circulation and what defines the space in Leibinger’s pavilion; I quickly came to the conclusion that clean lines and arrows (only suggesting one direction) could not properly describe the chaos of every person taking their own path, each slightly different from the next.
Appendix
Process
Exploring what defines circulation an threshold, and what factors influence these.
Exploring the site plan; noting the Serpentine Gallery, and the two other pavilions that were on simultaneously. This would give the path through the whole garden a strong sense of direction.
The light and dark spaces as a result of the sun path.
The direct rays of sun, determining the most shadow prone areas and therefore where people would go for different experiences.
9
Although I had already assumed that set paths and arrows wouldn’t suffice for this pavilion, I attempted both an arrow approach and a shading approach. In my opinion, these do not successfully illustrate the full potential of circulation.
# Module1
Diagramming Design Precedent
# Module1
Diagramming Design Precedent
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# ELI5 Coriolis Effect. I don’t understand what it does
3.46K views
ELI5 Coriolis Effect. I don’t understand what it does
In: Physics
Imagine you and your 2 best friends walking shoulder to shoulder you can walk in a straight line and stay shoulder to shoulder but what happens when you need to turn left? In order to stay next to each other your friend on the left will have to slow down and your friend on the right will have to speed up.
The Coriolis effect is this but with the earths atmosphere the air at the equator moves much faster (friend on the right)than the air at the north or south poll (friend on the left) this have much larger effects on wind patterns that I will let someone else explain
Let’s say you’re driving a hovercraft on a giant smooth sphere. You rev your engine at the equator and start to drive that sucker to the North Pole of this giant sphere. But the sphere is rotating as you travel. There’s the perfect lack of contact between your hovercraft and the sphere that as you travel north, you also move to the left just a little bit. If you were to plot your path it wouldn’t be a straight line like longitude. Your path would appear curve on a map because the sphere was moving underneath your hovercraft. Your end point would be on a different longitudinal line than when you started. In terms of weather the Coriolis Effect contributes to the direction of overall wind patterns on Earth. For example, in my region of the US weather systems pretty much always come from the west, even though that system may have started much farther north.
Okay, so one day we decide those Chinese sons of bitches are going down. So, we launch a nuke at China.
While it’s on it’s way there, we find out the earth is rotating underneath the missile. We aimed for China, but now ze missile is heading for France.
So France is like: “Shit guys, we are where China was when the US fired the ~~cigarettes~~ missiles…fire our shit!”
And the US is like, “Fuck, we’re dumb-asses. We should have aimed away from China to compensate for the earth’s rotation.”
Canada’s like: “What’s goin’ on, eh?”
and Australia’s still like: “WTF! The earth’s rotation shouldn’t have an effect on missile trajectory!”
Buuuuut! Assuming we don’t blow ourselves up first, us Californians will figure out the earth’s rotation deflects objects away from the spin of the earth. And we’ll give it a sweet name like, “the Coriolis effect” before we break off from the US to go hang out with Hawaii (Alaska can come too).
THE END!!
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# You asked: How long does it take to refine gasoline?
Contents
Generally, every 30,000-barrel batch takes around 12 to 24 hours to undergo through analytical testing and pass quality control. A key stage is ultra-heating the crude to boiling point, with a distillation column used to separate the liquids and gases.
## How long does it take to produce a gallon of gas?
“You take an average of 5 kilowatt hours to refine [one gallon of] gasoline, something like the [Tesla] Model S can go 20 miles on 5 kilowatt hours.”
## Can gasoline be refined?
An oil refinery or petroleum refinery is an industrial process plant where crude oil is transformed and refined into useful products such as petroleum naphtha, gasoline, diesel fuel, asphalt base, heating oil, kerosene, liquefied petroleum gas, jet fuel and fuel oils.
## How long does it take to get petroleum?
Originally Answered: How long does it take for oil to form? A minimum of about 50 million years. Most of Earth’s oil was formed between 60 million and 250 million years ago.
## At what temperature does crude oil become gasoline?
Thermal: Heating the distilled product to 900°F to break molecules, making mostly gasoline and diesel.
## How much gas does it take to produce 1 MWH?
Second: A high efficiency, natural gas-fired combined-cycle power plant might consume about 7000 Btus of gas to produce one kilowatt-hour of electricity. That would be about 7 cubic feet of natural gas. It would therefoe take about 7000 cubic feet of gas to produce one megawatt-hour.
IMPORTANT TO KNOW: Your question: How many gallons does a 325 gallon propane tank hold?
## How is oil refined to gasoline?
Petroleum refining separates crude oil into components used for a variety of purposes. The crude petroleum is heated and the hot gases are passed into the bottom of a distillation column. … The liquids are then drawn off the distilling column at specific heights to obtain fuels like gasoline, jet fuel and diesel fuel.
## What is the formula of gasoline?
Octane | C8H18 – PubChem.
## How much crude oil makes a gallon of gasoline?
Refineries in the United States produce about 19 to 20 gallons of motor vehicle gasoline for every 42-gallon barrel of crude oil.
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Algebra 2 4.2 Notes
# Solution write the supplies and the costs in matrix
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Unformatted text preview: trix match the rows of the cost matrix. Supplies Notebooks Pens Pencils Tim 4 5 10 5 8 8 Leslye Cost Dollars Notebooks 4 Pens 2 Pencils 1 The total cost for each person can be obtained by multiplying the supplies matrix by the cost matrix. 4 4(4) 5(2) 10(1) 2 5(4) 8(2) 8(1) 1 The labels for the matrix are as follows. 4 5 10 58 8 Homework 36 44 Total Cost Dollars Tim 36 Leslye 44 The total cost of supplies is \$ 36 for Tim and \$ 44 for Leslye. Lesson 4.2 • Algebra 2 Notetaking Guide 81...
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# Asynchronous Machine Measurement
Per-unit measurement from asynchronous machine
Machines
## Description
The Asynchronous Machine Measurement block outputs a per-unit measurement associated with a connected Asynchronous Machine Squirrel Cage or Asynchronous Machine Wound Rotor block. The input of the Asynchronous Machine Measurement block connects to the `pu` output port of the asynchronous machine block.
You set the Output parameter to a per-unit measurement associated with the asynchronous machine. Based on the value you select, the Asynchronous Machine Measurement block:
• Directly outputs the value of an element in the input signal vector
• Calculates the per-unit measurement by using values of elements in the input signal vector in mathematical expressions
The Asynchronous Machine Measurement block outputs a per-unit measurement from the asynchronous machine according to the output value expressions in the table. For example, when you set Output to ```Stator d-axis voltage```, the block directly outputs the value of the `pu_vds` element in the input signal vector. However, when you set Output to `Slip`, the block calculates the slip value by subtracting the value of the `pu_velocity` element from 1.
Output Parameter SettingOutput Value Expression
Electrical torque
pu_torque
Rotor velocity
pu_velocity
Stator d-axis voltage
pu_vds
Stator q-axis voltage
pu_vqs
Stator zero-sequence voltage
pu_v0s
Stator d-axis current
pu_ids
Stator q-axis current
pu_iqs
Stator zero-sequence current
pu_i0s
Slip
1-pu_velocity
Apparent power
$\sqrt{pu_P{t}^{2}+pu_Q{t}^{2}}$
Real power
pu_Pt = (pu_vds*pu_ids) + (pu_vqs*pu_iqs) + 2(pu_v0s*pu_i0s)
Reactive power
pu_Qt = (pu_vqs*pu_ids) – (pu_vds*pu_iqs)
Terminal voltage
$\sqrt{pu_vd{s}^{2}+pu_vq{s}^{2}}$
Terminal current
$\sqrt{pu_id{s}^{2}+pu_iq{s}^{2}}$
power_factor_angle = atan2(pu_Qt, pu_Pt)
Power factor
cos(power_factor_angle)
## Parameters
Output
Per-unit measurement from asynchronous machine. The default value is `Electrical torque`.
## Ports
The block has the following ports:
`pu`
Physical signal vector port associated with per-unit measurements from a connected asynchronous machine. The vector elements are:
• pu_torque
• pu_velocity
• pu_vds
• pu_vqs
• pu_v0s
• pu_ids
• pu_iqs
• pu_i0s
`o`
Per-unit measurement output port.
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# gender affects the compliance levels in humans
Extracts from this document...
Introduction
Introduction The aim of this task was to find out whether gender affects the compliance levels in humans. Compliance is a form of conformity, when a person acts according to certain accepting standards. I will be observing peoples behaviours when crossing the road and will take note of those who cross with the lights at pedestrian crossings and those who cross independently. To ensure I get a fair set of results I will have to think about confounding variables, which may affect the findings. This is a natural environment experiment. I have no control over people's actions and since they will be unaware of my observations they will not feel intimidated or pressured into acting differently and doing the right thing. In the past, 1950's - 1950's, many researchers have conducted that there is no difference between males and females complying, although a few psychologists have found that women are more likely to comply. Research by Sistrunk and McDavid (1) indicated that women are more likely to comply if they do not have much knowledge about a specific event or topic, likewise with men. ...read more.
Middle
Results Descriptive Statistics Summary table MALE FEMALE Crossed with lights 9 16 Crossed independently 15 6 By looking at the results on the bar chart it is clear that more woman complied than men during the period of time I carried out my investigation. Inferential Statistics Using the 'Chi Square' method I will be able to test my data statistically to see if there is a significant difference between genders. The formula of Chi Square will give me a figure, which will tell me if my results are significant. The critical value which will prove the validity of my findings is 2.71, the result must be 2.71 or higher to be certain that the results did not happen purely by chance. The formula of the Chi Square is: ?� = ? (O - E - 0.5) � E ?� = 5.52 (significant result) p ? 0.05; ?� obt = 5.52; ?� crt = 2.71; df=1 I can therefore accept the experimental hypothesis and can reject the NUL hypothesis as my result is a significant figure. ...read more.
Conclusion
The weather is also a factor I could think about, selecting a sunny, warmer day where people of all ages will be out. Also, repeating the test more than once and using a few different locations where there is a pedestrian crossing. As a result of this experiment, it is clear that on average, females seem to comply more than males. This may suggest that as a gender, females are less secure but more sensible than males also that females are more cautious and careful. Abstract The aim of this experiment was to find out whether women are more likely to comply when crossing the road at pedestrian crossings than men. In order to get a set of results I observed people behaviour when crossing the road by sitting near a pedestrian crossing and recording what I noticed by using a tally chart. I found that 16 females complied whereas only 9 men complied by crossing with the green man after pressing the button rather than crossing independently. In total 24 of my participants were male and 22 were female. In conclusion to my research it is clear that woman are more compliant than men. REFERANCE 1) http://www.ship.edu/~cgboeree/conformity.html ...read more.
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# Related AS and A Level Social Psychology essays
1. ## The Matching Hypothesis
This positive correlation provides support for the matching hypothesis and for my hypothesis. The extent to which it supports my hypothesis will be examined by running a statistical test on the data. I have circled 2 rogue results in the graph as I believe without them the trend line would be more positive.
2. ## Conformity discussion.
Discussion The results have been shown to be insignificant on the Mann-Whitney U test; this means in general participants did not conform. This result has not confirmed my Experimental Hypothesis which states participants, would show evidence of conformity. In the introduction it was related that conformity had decreased since the
1. ## Why Do Humans Conform?
From which I then intend to discuss the factors that influence and affect conformity. Social Psychologists have identified two main factors, which encourage conformity. Firstly, people conform because they want to be liked. They conform to the behaviour of others in order to gain their approval.
2. ## Investigation into Whether Gender Affects Conformity
Types of conformity have also been identified. Kelman (1958) identified three different types of conformity. Compliance is when the individual publicly conforms to the behaviour and views of others but privately maintains their own views. Identification is when the views and attitudes of a group are adopted both publicly and privately.
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# CBSE Class 10 - Maths
### NCERT Syllabus For CBSE Class 10 Maths
One Paper Time : 2 1/2Hours; Marks : 60
UNITS CHAPTERS MARKS I NUMBER SYSTEMS 04 II ALGEBRA 20 III TRIGONOMETRY 12 IV COORDINATE GEOMETRY 08 V GEOMETRY 16 VI MENSURATION 10 VII STATISTICS AND PROBABILITY 10 TOTAL 80
## 1. Number Systems
### (i) Real Numbers
Euclid's division lemma, Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of results - irrationality of √2,√3, √5, decimal expansions of rational numbers in terms of terminating/non-terminating recurring decimals.
## 2. Algebra
### (i) Polynomials
Zeros of a polynomial. Relationship between zeros and coefficients of a polynomial with particular reference to quadratic polynomials. Statement and simple problems on division algorithm for polynomials with real coefficients.
### (ii) Pair Of Linear Equations In Two Variables
Pair of linear equations in two variables. Geometric representation of different possibilities of solutions inconsistency. Algebraic conditions for number of solutions. Solution of pair of linear equations in two variables algebraically- by substitution, by elimination and by cross multiplication. Simple situational problems must be included.
Simple problems on equations reducible to linear equations may be included.
Standard form of a quadratic equation ax2 + bx + c = 0, (a1 0). Solution of the quadratic equations (only real roots) by factorization and by completing the square, i.e. by using quadratic formula. Relationship between discriminant and nature of roots.
Problems related today to day activities to be incorporated.
### (iV) Arithmetic Progressions
Motivation for studying AP. Derivation of standard results of finding the nth term and sum of first n terms.
## 3. Trigonometriy
### (i) Introduction To Trigonometriy
Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios, whichever are defined at 0o & 90o. Values (with proofs) of the trigonometric ratios of 30o, 45o & 60o. Relationships between the ratios.
### (ii)Trigonometric Identities
Proof and applications of the identity sin2 A + cos2 A = 1. Only simple identities to be given. Trigonometric ratios of complementary angles.
### (iii)Heights And Distances
Simple and believable problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30o, 45o, 60o.
## 4. Coordinate Geometry
### (i)Lines (In two-dimensions)
Review the concepts of coordinate geometry done earlier including graphs of linear equations. Awareness of geometrical representation of quadratic polynomials. Distance between two points and section formula (internal). Area of a triangle.
## 5. Geometry
### (i) Triangles
Definitions, examples, counter examples of similar triangles.
1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.
4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar.
5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.
7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.
8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
9. (Prove) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angles opposite to the first side is a right traingle.
### (ii)Circles
Tangents to a circle motivated by chords drawn from points coming closer and closer and closer to the point.
1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. (Prove) The lengths of tangents drawn from an external point to circle are equal.
### (iii)Constructions (Periods - 8)
1. Division of a line segment in a given ratio (internally)
2. Tangent to a circle from a point outside it.
3. Construction of a triangle similar to a given triangle.
## 6. Mensuration
### (i) Areas Related To Circles
Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60o, 90o & 120o only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.)
### (ii)Surface Areas And Volumes
(i) Problem on finding surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinder/cones. Frustum of a cone.
(ii) Problems involving converting one type of metallic solid into another and other mixed problem.(Problems with combination of not more than two different solids be taken.)
## 7. Statistics And Probability
### (i) Statistics
Mean, median and mode of grouped data(bimodal situation to be avoided). Cumulative frequency graph.
### (ii)Probability
Classical definition of probability. Connection with probability as given in class IX. Simple problem on single events, not using set notation.
INTERNAL ASSESSMENT 20 MARKS Evaluation of activities 10 Marks Project Work 05 Marks Continuous Evaluation 05 Marks
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# Signature numbers progression
Discussion in 'Dice Influencing' started by Dave G Ct, Apr 16, 2019.
1. Dave G Ct, Apr 16, 2019
### Dave G Ct Member
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Well you have all these 6&8 progressions as well as don't play ones. But how about using a progression on one or 2 signature numbers that you throw.
. I have just began implementing this play. First since I do toss more 4's especially hard 4's I will put more money on them usually a \$25 buy bet with a \$5 h.w. The ten I may just go with a \$5 h.w. or place it at \$15 w a \$2 h.w.
Say you have a couple of good hands.Why not increase the 4 to \$35 or \$50 w a \$10 h.w.? Then how about a \$100 place on the 4 w a \$15 h.w.? DeMango got your attention? You could increase those 2 6&8 bets or 5&9 and regress to a higher bet spread.
Anybody want to weigh in?
#1
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2. Edward-ky, Apr 22, 2019
### Edward-ky Member
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If you want people to weigh make some remarks about TDV and the the 2 jackasses of the forum will jump right in on this thread and post a ton of stuff that has nothing to do with the topic.
#2
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3. gargoil01, Apr 22, 2019
### gargoil01 Member
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Dave how often do you have sessions where you hit the 4 at least three times before the seven? 1 in 8? more? Less?
Based on your answer you can determine if this play is good for you or not. Buy the four for \$25 and first hit take it to \$75. Second hit drop the dealer \$25 and take it to \$250. Third hit collect \$750 and go back down to \$25.
The reason why I asked how many sessions do you usually hit the four at least three times is because you can afford to lose \$25 six / 7 times but then make it all back and more when you do hit it 3 in a session. I use that play on the four and ten on my rolls.
Again this all depends on your bankroll and your ability to hit the four.
G
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4. von duck, Apr 22, 2019
### von duck Member
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Did somebody say TDVegas?
#4
5. Mssthis1, Apr 22, 2019
### Mssthis1 Member
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I sometimes do similar, only not as aggressive. I go to \$50, \$100, \$200 back to \$25.
If I'm ahead, I'll just keep on pressing sometimes. \$6000, 4 is the highest I've gotten so far. That was at Caesars back in the day when they let you bet up to \$50K
Like G said, a lot depends on your bankroll because it will increase volatility.
#5
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6. Dave G Ct, Apr 22, 2019
### Dave G Ct Member
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G
Had a great hand tonight.Hit 3 hard tens and one hard 4.Also had 2 soft ones. Been starting my four at \$25 w \$5:h.w.As for the ten - \$15 w \$2 h.w.
I have had a run of great hands with the multiple 4's.Of course my shot did get better. Just off the cuff maybe half my hands had 3 fours.Short sample though
#6
7. Dave G Ct, Apr 22, 2019
### Dave G Ct Member
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G
Also depends on ones nerves lol
#7
8. Edward-ky, Apr 23, 2019
### Edward-ky Member
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Another way on the first hit drop the 25 and go to 100. Then go to 300 then take it back down after the 3rd hit back to 50. Either way your putting in 50. And after .3rd hit your collecting 850 minus the vigs of course and you have 50 or whatever your comfortable with. Gargoil s way not a bad stocking of the chip rack either.
#8
9. Edward-ky, Apr 23, 2019
### Edward-ky Member
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Yes true but sounds better than 27 across. Rather buy 4 or 10 for quarter than 27 across.
#9
10. tabletop123, Apr 23, 2019
### tabletop123 Member
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Listen ( just MY opinion) that \$27 Across thingy is a complete waste of time, & TALENT! It takes too many hits with such minimum bets to make anything substantial.
Then....we get into the " it's the best way for a Stepper to bet"...... " I hit. six repeaters before the Seven out, so I usually make money" thingy.
Well....here's the deal....IF YOU. TRULY ARE hitting...let's say....the number 9, or whatever number.....does it make any earthly sense to have money bet on those other 5 numbers? lol
When you start the " betting Across is the right game" thingy, & in the SAME sentence you start with the " I usually hit a number/a 5 -6 times before the hand-ending Seven out....NOT ONLY is THAT some mighty fine shooting, but it borders along the lines of " Sniping"!
I have always said that a TRUE test of skill is to repeatedly hit a number or two CONSISTENTLY ( yeah...I know...it's considered sniping) during a hand. We always run into "the sample is too small to really know if it's influence or variance" thingy, but HAMMERING a number/s CONSISTENTLY during a hand gets MY vote. Show me consistency in THAT, & we can discard the "thousands of roll to really know" thingy.
Certainly a skilled shooter that tosses a number EVEN 2-3 times PER HAND CONSISTENTLY is doing him/herself a great injustice by wagering on OTHER numbers!!!
Since I'm on the subject....how about the Off Axis Bullshit that CERTAIN Dice Influencers spew? Claiming that they set the dice...as such, & toss them....as such to PURPOSELY knock the dice Off Axis, lol
Isn't that what Dice NATURALLY DO? Misbehave! Go OFF AXIS? Land on die faces that you DON'T set the dice on? Then, these Brainiacs will say: " See.....I PURPOSELY did something that occurs 97- 98% of the time NATURALLY. lol
Whether ya want to use Linaway's definition of On Axis, ( 2-3%) or the 44% that is conventionally accepted...MORE TIMES THAN NOT, one or the other of your die is consistently flipping opposed to the way that they are being " Set"!
Just MY opinion.
#10
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11. Edward-ky, Apr 23, 2019
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N = 1000; % Declare network size (N.B. networks larger than 10^6 may require more % memory than is available to Matlab... runs = 1; % Declare number of realisations m0 = 3; m = 4; k0 = 2; % Delcare parameters % k_0 == inital 'score' of a node % m_0 == initial number of nodes % m == number of connections each new node makes if m0 < m m0 = m; end % the network must begin fully connected, so we cannot make more % connections per step than there are nodes, hence the tacky solution: % increase the number of initial nodes! d = zeros(runs*N,1); % vector to hold degree sequences for z=1:runs e = zeros(1000000,1); % edge list: clever method for creating a preferential attachement % network, each node appears k+k0 times in this edge list, we can then % pick a new node to length = 0; % actual length of edge list for i=1:m0 for j=1:(k0+m0-1) e(j+length) = i; end length = length+(k0+m0-1); end % loop to populate the edge list with the initial condition A = sparse(N,N); % adjacency matrix, stored as a sparse matrix (type 'help sparse' for % more information for i=1:m0 for j=1:m0 if i ~= j && i < j A(i,j) = 1; end end end % update the adjacency matrix to reflect the initial condition of the % network, which should be a fully connected network of m_0 nodes for i=m0+1:N R = e(randsample((length),m)); % randomly pick m integers between 1 and (i-1) that will recieve % new connections from node i while size(unique(R),1)
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New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > syl5eqner GIF version
Theorem syl5eqner 2541
Description: B chained equality inference for inequality. (Contributed by NM, 6-Jun-2012.)
Hypotheses
Ref Expression
syl5eqner.1 B = A
syl5eqner.2 (φBC)
Assertion
Ref Expression
syl5eqner (φAC)
Proof of Theorem syl5eqner
StepHypRef Expression
1 syl5eqner.2 . 2 (φBC)
2 syl5eqner.1 . . 3 B = A
32neeq1i 2526 . 2 (BCAC)
41, 3sylib 188 1 (φAC)
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1642 ≠ wne 2516 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-11 1746 ax-ext 2334 This theorem depends on definitions: df-bi 177 df-ex 1542 df-cleq 2346 df-ne 2518 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
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1. ## Quadratic Projectile Motion help?
Okay so here is the original equation:
A rocket is launched into the air from the top of a building. It's height (in meters) is given by:
h(t)= -5t² + 30t + 100, where t = time in seconds.
When does the rocket smash into the ground?
So, this is where I set the equation equal to zero.
0 = -5t² + 30t + 100
and since it couldn't be factored, I decided I needed to use the quadratic formula.
However, am I able to factor out the -5?
Then I got:
-5(t²-6t-20)
So, when I plugged the items in the parenthesis into the quadratic formula, I eventually got: (3 +/- √29)
Now, what I'm wondering is:
First of all, did I do everything correctly?
If so, what do I then do with the -5 I factored out in the beginning?
Any help would be appreciated, thank you for your time!
2. ## Re: Quadratic Projectile Motion help?
Originally Posted by ZzTop0503
Okay so here is the original equation:
A rocket is launched into the air from the top of a building. It's height (in meters) is given by:
h(t)= -5t² + 30t + 100, where t = time in seconds.
When does the rocket smash into the ground?
So, this is where I set the equation equal to zero.
0 = -5t² + 30t + 100
and since it couldn't be factored, I decided I needed to use the quadratic formula.
However, am I able to factor out the -5?
Then I got:
-5(t²-6t-20)
So, when I plugged the items in the parenthesis into the quadratic formula, I eventually got: (3 +/- √29)
Now, what I'm wondering is:
First of all, did I do everything correctly?
If so, what do I then do with the -5 I factored out in the beginning?
Any help would be appreciated, thank you for your time!
It doesn't matter whether you factor out a 5 or not, you still get the same answer.
Your answer is correct. If you look at the parabola, you will notice that it is shot from 100 meters (the building is 100 meters tall) and it eventually lands after your value t...Which means it travels for roughly 8 seconds.
3. ## Re: Quadratic Projectile Motion help?
To check that it really doeesn't matter than you divided through by 5 you can plug the values of $3 \pm sqrt {29}$ into the original formula and see that it works:
$-5(3 + \sqrt{29})^2 + 30(3 + \sqrt{29})t +100 = -5(9 + 6\sqrt{29} + 29) + (90 + 30 \sqrt{29}) + 100$
$= -45 -30 \sqrt{29}-145 + 90 + 30 \sqrt{29} +100 = 0$
Remember that when you do algebra you can multiply or divide both sides of an equaton by any number (other than 0) and the equation still holds. For example, if x = 2 then the equation x-2 = 0 is true, but so are 2x-4 =0, 6x-6 = 0, 8x-8 = 0, etc.
By the way, the solution $3 - \sqrt{29}$ satisies the equation, but not the premise that the rocket is launched at t=0. You can eliminate that answer as not being consistent with the problem statement, as it can be assumed that prior to being launched the rocket was stationary on top of the building.
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p0504
# p0504 - 1 R 2 ∂ ∂ R p R 2 u R Q 1 R sin φ ∂ u θ...
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5.4. CHAPTER 5, PROBLEM 4 453 5.4 Chapter 5, Problem 4 5.4(a): For incompressible flow, the continuity equation in cylindrical coordinates is 1 r r ( ru r )+ 1 r u θ ∂θ + w z =0 Hence, when u θ = w =0 , all that remains is 1 r r ( ru r )=0 Integrating over r , the most general form of the radial velocity is u r ( r, θ ,z )= f ( θ ,z ) r where f ( θ ,z ) is a function of integration. 5.4(b): For incompressible flow, the continuity equation in spherical coordinates is
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Unformatted text preview: 1 R 2 ∂ ∂ R p R 2 u R Q + 1 R sin φ ∂ u θ ∂θ + 1 R sin φ ∂ ∂φ ( u φ sin φ ) = 0 Hence, when u θ = u φ = 0 , all that remains is 1 R 2 ∂ ∂ R p R 2 u R Q = 0 Integrating over R , the most general form of the radial velocity is u R ( R, θ , φ ) = g ( θ , φ ) R 2 where g ( θ , φ ) is a function of integration....
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Name: Try It Out – Chapter 12, Lesson 5
Matching
Identify the letter of the choice that best completes the statement or answers the question.
Choose the letter that best represents the mixed number.
1.
2
2.
3
3.
2
4.
1
5.
3
6.
1
7.
1
8.
2
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
9.
Claude has 3 nanaimo bars. He ate of 1 bar. He gave the rest to his sister. What mixed number did he give to his sister?
a. 2 b. 1 c. 1 d. 2
10.
Juan has 5 sandwiches. He ate 1 whole sandwich and of another sandwich. He gave the rest to his brother. What mixed number did he give to his brother?
a. 2 b. 3 c. 3 d. 2
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It is currently 22 Sep 2020, 05:38
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# Use of an augur
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Use of an augur [#permalink] 18 Mar 2019, 11:19
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Question Stats:
9% (00:44) correct 90% (00:38) wrong based on 71 sessions
Use of an augur has steadily declined as scientific advancement has made such use less necessary. Therefore, the postulate that science will eventually (i) _______ the use of augurs is (ii) _______.
Blank (i) Blank (ii)
annihilate plausible
supplant specious
consummate expeditious
[Reveal] Spoiler: OA
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Re: Use of an augur [#permalink] 20 Oct 2019, 22:08
The use of augur is declined, so science will suppress the use of augur in a high speed. Hence, Supplant and expeditious.
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Re: Use of an augur [#permalink] 22 May 2020, 18:42
but here we are talking about the postulate and postulate cannot be expeditious , it can be plausible
Re: Use of an augur [#permalink] 22 May 2020, 18:42
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E book Particulars :
Language English
Pages 820
Format PDF
Measurement 12.3 MB
# Electric Circuits 10th Version by Nilsson and Riedel
Electric Circuits 10th Version by James W. Nilsson and Susan Riedel | PDF Free Download.
## Electric Circuits Contents
• Chapter 1 Circuit Variables
• Chapter 2 Circuit Parts
• Chapter 3 Easy Resistive Circuits
• Chapter 4 Strategies of Circuit Evaluation
• Chapter 5 The Operational Amplifier
• Chapter 6 Inductance, Capacitance, and Mutual Inductance
• Chapter 7 Response of First-Order RL and RC Circuits
• Chapter 8 Pure and Step Responses of RLC Circuits
• Chapter 9 Sinusoidal Regular-State Evaluation
• Chapter 10 Sinusoidal Regular-State Energy Calculations
• Chapter 11 Balanced Three-Section Circuits
• Chapter 12 Introduction to the Laplace Rework
• Chapter 13 The Laplace Rework in Circuit Evaluation
• Chapter 14 Introduction to Frequency Selective Circuits
• Chapter 15 Energetic Filter Circuits
• Chapter 16 Fourier Sequence
• Chapter 17 The Fourier Rework
• Chapter 18 Two-Port Circuits
## Preface to Electric Circuits PDF
The primary version of Electric Circuits, an introductory circuits textual content, was printed in 1983. It included 100 labored examples and about 600 issues. It didn’t embody a pupil workbook, dietary supplements for PSpice or MultiSim, or any internet assist.
Assist for instructors was restricted to an answer guide for the issues and enlarged copies of many textual content figures, appropriate for making transparencies. A lot has modified within the 31 years since Electric Circuits first appeared, and throughout that point this textual content has developed to higher meet the wants of each college students and their instructors.
For example, the textual content now contains about 150 labored examples, about 1850 issues, and in depth dietary supplements and internet content material.
The tenth version is designed to revise and enhance the fabric introduced within the textual content, in its dietary supplements, and on the internet.
But the elemental targets of the textual content are unchanged. These targets are:
1. To construct an understanding of ideas and concepts explicitly by way of earlier studying. College students are continuously challenged by the necessity to layer new ideas on prime of earlier ideas they might nonetheless be struggling to grasp.
This textual content gives an essential concentrate on serving to college students perceive how new ideas are associated to and rely on ideas beforehand introduced.
2. To emphasise the connection between conceptual understanding and problem-solving approaches. Creating problem-solving expertise continues to be the central problem in a first-year circuits course.
On this textual content, we embody quite a few Examples that current problem-solving methods adopted by Evaluation Issues that allow college students to check their mastery of the fabric and methods launched.
The issue-solving course of we illustrate relies on ideas moderately than the usage of rote procedures. This encourages college students to consider an issue earlier than making an attempt to resolve it.
3. To supply college students with a robust basis of engineering practices. There are restricted alternatives in a first-year circuit evaluation course to introduce college students to practical engineering experiences.
We proceed to reap the benefits of the alternatives that do exist by together with issues and examples that use practical part values and signify realizable circuits. We embody many issues associated to the Sensible Perspective issues that start every chapter.
We additionally embody issues meant to stimulate the scholars’ curiosity in engineering, the place the issues require the kind of perception typical of a training engineer.
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# birthday paradox – a simulation of shared birthday experiments
Post on 10-Apr-2015
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Overall Comment: "While the paper is well written, it tried to fit too much trivial detail in at the expense of material which was demoted to the appendices. The paper should be able to stand on its own without the appendices; the appendices simply add to/support the paper. code and extracted Javadoc very impressive."Abstract: Being one of the most well-known and famous problems inprobability1, the question is asked: How many randomly chosen peopleare needed to achieve at least a 50% probability that some pair will bothhave been born on the same day? Since the chance of any two personshaving the same birthday is remote, many of us would expect this numberto be rather large. However, it turns out that this is not the case, andhence the paradox.A simulation of empirical testing and results was conducted to simulatemultiple trials, where people of a given group size have their birthdayscompared. The probability is consistently monitored and established asthe number of successful experiments as a proportion of the total numberof experiments performed.As the theoretical value is already known (23), it was also the goal-drivingresult for the algorithm being implemented in Java, and happily, wasachieved.
TRANSCRIPT
BIRTHDAY PARADOX A SIMULATION OF SHARED BIRTHDAY EXPERIMENTSStephen Hogan (50217631) Postgraduate Diploma in IT (Evening) Dublin City University School of Computinghogans2@mail.dcu.ie
Abstract Being one of the most well-known and famous problems in probability1, the question is asked: How many randomly chosen people are needed to achieve at least a 50% probability that some pair will both have been born on the same day? Since the chance of any two persons having the same birthday is remote, many of us would expect this number to be rather large. However, it turns out that this is not the case, and hence the paradox. A simulation of empirical testing and results was conducted to simulate multiple trials, where people of a given group size have their birthdays compared. The probability is consistently monitored and established as the number of successful experiments as a proportion of the total number of experiments performed. As the theoretical value is already known (23), it was also the goal-driving result for the algorithm being implemented in Java, and happily, was achieved.
2. BACKGROUNDMathematical ModelOne of the basic rules of probability: the sum of the probability that an event will happen and the probability that the even will not happen is always 1. In other words, the chance that anything might or might not happen is always 100%. If we can work out the probability that no two people will have the same birthday, we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event does not happen) = 1P(two people share birthdays) + P(no two people share birthdays) = 1 P(two people share birthdays) = 1 P(no two people share birthdays) Comment [JM5]: Mixed font size. Comment [JM1]: Keep it formal and to the point. Comment [JM4]: This could be compressed by simply giving a single equation and pointing the reader to a reference whre more detail is given.
1. INTRODUCTIONPremise born in 19382: In a room of just 23 people, there is a 50% probability of two of these people sharing the same birthday, (ignoring years of birth). In a room of 75 people, there is a 99.9% chance of two people with matching birthdays. This is one particular case of exponential (Saliusian) sets, where duplicates are allowed. Exponents are not intuitive, and thus why our linear-thinking leads us to an incorrect estimation! While theoretical mathematical models and proofs have been derived in previous works outside the scope of this paper, the object of this paper is not to put the formulae into action; moreover the objective of this paper is to highlight a proof by example; i.e. that a simulation of birthdates for a group of people is analysed for this probability result, based on the aforementioned premise. For the following two sections, Background and Method, to be presented here, both share the following assumptions: That there are only 365 days in a year, i.e. thus ignoring leap years. This also results in ignoring the suspension of leap day on years divisible by 100 that are also divisible by 400. Birth years are ignored. Peoples birthdays are equally distributed throughout the year; (i.e. influencing elements such as seasonality are not factored in). Obviously in real-life, birthday distributions are not uniform, i.e. not all dates are equally likely. The date of a persons birthday does not affect the date of another person birthday, i.e. twins, triplets, etc.
The formula for the probability that n people have different birthdays (month and day) is3: 365 (1) 365 n! * 365n Therefore, the probability that at least two of them share the same birthday is: 365 1 (2) 365 n! * 365 n Having (2) graphed in Figure 1 it is clearly seen where, at the probability of 50%, cross-referencing it with the number of people reveals a value of 23:
Comment [JM2]: Use a reference rather than a footnote.
Comment [JM3]: A reference would be welcome here.
Figure 1: A graph showing the approximate probability of at least two people sharing a birthday amongst a certain number of people 4.
This is not a paradox in the literal sense it just highlights the fact that people expect the value to be much larger. 2 American Mathematical Monthly in 1938 in Zoe Emily Schnabel's The estimation of the total fish population of a lake, under the name of capture-recapture statistics.1
As Dr Math FAQ The Birthday Problem (http://mathforum.org/dr.math/faq/faq.birthdayprob.html) 4 Wikipedia - Birthday Problem (http://en.wikipedia.org/wiki/Birthday_paradox)3
3. METHODProgramming MethodologyThe simulation was written in the Java language 5. A text file that lists multiple trials with the following parameters serves as input to the program, (defaults outlined here are geared to solving the problem in question): Number of matches to be checked (default: K = 2) Number of trials (various >> values) Starting group size (default: N = 2) Group size increment (default: 1) Terminating probability (default: 0.5) Assumptions: The group size will never be greater than 1,000. In the case that the starting value of N is less than K, you should start the simulation with N = K. The reason is simple: How do you possibly find three (i.e. K) matches in a group of two (i.e. N) persons? Beginning with group size of N = 2 people, we initialise an array with the random birthdays of N = 2 people; (the random number generator is being seeded with the current time). We compare every pair wise (number of matches K = 2) combination of people in the group of N = 2 and check the existence of any two persons having the same birthday. This will be repeated number of trials times with different groups of two people. If the average occurrence of two persons having the same birthday in these one thousand trials exceeds 0.5 (i.e. the probability P), the simulation terminates. Otherwise, N is incremented by group size increment = 1, and the entire simulation of one thousand trials is repeated with randomized groups of (starting group size = 2) + (group size increment = 1) people. The flexibility in reading in values from a file that control the execution of the algorithm allows us to evaluate numbers of people for various probabilities and/or enumerate possible combinations of more than two persons, differing numbers of trials and matches, and check for shared birthdays, for example. A review of the Java source code in Appendix 2 should reveal other variations. For each trial, a number of random birthdays are generated and placed into an array; (here, we use the Julian Date format, 1365). These birthdays are sorted and then iterated through to find the same values in consecutive elements in the array, denoting a success. Once the trials have completed running, the probability is evaluated as6: currProbability = numSameBirthday / numTrials
The last record in this table is graphed in Figure 2. Preparing graphs for all records processed in this table (Appendix 4) reveals, interestingly, that time performance dips in proportion to the curve of probability when tending to 23 people, irrespective of the number of trials:Simulation 80.60000 25000
0.50000 20000
0.40000Probability
15000 0.30000 10000 0.20000Time (ms)
5000 0.10000
0.00000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 No. People Probability Time (ms)
0
Figure 2: Simulation 8
Comment [JM9]: You need to explain what Simulation 8 is.
5. CONCLUSIONS & RECOMMENDATIONSThis research reports on a simulated empirical study of the Birthday Paradox. The findings suggest that there is a strong similarity between the theoretical and (simulated) actual probabilities. Specifically, that for any group size or number of trials, to achieve at least a 50% probability in solving the problem, we would need at least 23 people in a room for comparison (with previously unknown birthdays). The input file allows for variations on the number of people > pairs as well as the group increment size to be > 1, as well as other terminating probabilities. A further recommendation would be to refine the sorting algorithm even further, or adopt a faster mechanism of sorting, especially for large N and K. Parallelisation of Quicksort, for example, would be ideal, as synchronisation is not a requirement. Java Threading would be an approach for this. Another approach to solving this problem would be to base it on collisions by tracking as each person enters the room and checking to see if there is a match with any other person. An array of 365 elements would only be needed, and a random date only generated for each person until either the entire group size is exhausted or a match has been found. The author of this paper decided against this approach after initially selecting it, as it would not be possible to measure time performance as fluidly.
Comment [JM10]: We Comment [JM6]: Why not use algrbraic notation as in (1) and (2). Also: This eqn is not numbered.
4. RESULTS & DISCUSSIONWith the availability of having an input file, multiple case scenarios can be generated. Table 1 is an example of sample data was available on the input file, (as per aforementioned format in Section 3):2 2 2 2 2 2 2 2 10000 20000 30000 50000 700000 1000000 2000000 5000000 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
6. REFERENCES[1] Birthday Paradox Wikipedia.com (http://en.wikipedia.org/wiki/Birthday_paradox) [2] CS1101C Lab 3 Birthday Paradox National University of Singapore, School of Computing (http://www.comp.nus.edu.sg/~cs1101cl/labs_sem2_0405/lab3/oddweek/) [3] How to Generate Random numbers About.com (http://java.about.com/od/javautil/a/randomnumbers.htm) [4] Quick Sort Implementation with median-of-three partitioning and cutoff for small arrays Java-Tips.org (http://www.java-tips.org/java-se-tips/java.lang/quic
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# Quick Answer: How Do You Do Financial Ratio Analysis?
## How do you solve financial ratio analysis?
Quick Ratio: In order to calculate the quick ratio, take the Total Current Ratio for 2010 and subtract out Inventory.
Divide the result by Total Current Liabilities.
You will have: Quick Ratio = 642-393/543 = 0.46X.
For 2011, the answer is 0.52X..
## What are 3 types of ratios?
The three main categories of ratios include profitability, leverage and liquidity ratios.
## What are the five financial ratios?
Fundamental analysis relies on extracting data from corporate financial statements to compute various ratios. There are five basic ratios that are often used to pick stocks for investment portfolios. These include price-earnings (P/E), earnings per share, debt-to-equity and return on equity (ROE).
## What is the most important ratio in financial analysis?
The debt-to-equity ratio, is a quantification of a firm’s financial leverage estimated by dividing the total liabilities by stockholders’ equity. This ratio indicates the proportion of equity and debt used by the company to finance its assets.
## What are the three main profitability ratios?
The three most common ratios of this type are the net profit margin, operating profit margin and the EBITDA margin.
## What is the quick ratio in accounting?
The quick ratio indicates a company’s capacity to pay its current liabilities without needing to sell its inventory or get additional financing. The quick ratio is considered a more conservative measure than the current ratio, which includes all current assets as coverage for current liabilities.
## What is the financial ratio analysis with example?
Inventory TurnoverLiquidity RatiosFormulaCurrent (or working capital) ratioCurrent assets / Current liabilitiesAcid-test (quick) ratioQuick assets (cash + marketable securities + net receivables) / Current liabilitiesInventory turnoverCost of goods sold / Average inventory
## How are financial ratios calculated?
Ratios are also used by bankers, investors, and business analysts to assess a company’s financial status. Ratios are calculated by dividing one number by another, total sales divided by number of employees, for example.
## What is the purpose of a financial ratio analysis?
Ratio analysis compares line-item data from a company’s financial statements to reveal insights regarding profitability, liquidity, operational efficiency, and solvency. Ratio analysis can mark how a company is performing over time, while comparing a company to another within the same industry or sector.
## What is the use of financial ratios?
Financial ratios offer entrepreneurs a way to evaluate their company’s performance and compare it other similar businesses in their industry. Ratios measure the relationship between two or more components of financial statements. They are used most effectively when results over several periods are compared.
## What are 2 types of ratios?
In general, a ratio is an expression that shows the relationship between two values. It tells us how much of one thing is there as compared to another. There are two “kinds” of ratios: “part to part” and “part to whole“.
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## Section5.4Area and Triangle Trigonometry
The arc-length differential determines an area differential and the area of a region will also be an invariant of hyperbolic geometry. The area of a region will not change as it moves about the hyperbolic plane. We express the area formula in terms of polar coordinates.
###### Definition5.4.1
Suppose a region $R$ in $\mathbb{D}$ is described in polar coordinates. The area of $R$ in $(\mathbb{D},{\cal H})\text{,}$ denoted $A(R)\text{,}$ is given by
\begin{equation*} A(R) = \iint_R \frac{4r}{(1-r^2)^2}dr d\theta. \end{equation*}
The integral in this formula is difficult to evaluate directly in all but the simplest cases. Following is one such case.
###### Example5.4.2The area of a circle in $(\mathbb{D},{\cal H})$
Suppose our region is given by a circle whose hyperbolic radius is $a\text{.}$ Since area is an invariant, we may as well assume the circle is centered at the origin. Let $x$ be the point at which the circle intersects the positive real axis (so $0 \lt x \lt 1$), as pictured below. Then, by the distance formula
\begin{equation*} a = \ln\bigg(\frac{1+x}{1-x}\bigg). \end{equation*}
Solving for $x\text{,}$ we have
\begin{equation*} x = \frac{e^a - 1}{e^a + 1}. \end{equation*}
This circular region may be described in polar coordinates by $0 \leq \theta \leq 2\pi$ and $0 \leq r \leq x\text{.}$ The area of the region is then given by the following integral, which we compute with the $u$-substitution $u = 1 - r^2\text{:}$
\begin{align*} \amp \int_0^{2\pi} \int_0^x \frac{4r}{(1-r^2)^2}~drd\theta\\ \amp = \bigg(\int_0^{2\pi}~d\theta\bigg)\bigg(\int_1^{1-x^2}\frac{-2}{u^2}~du\bigg)\\ \amp = 2\pi\bigg[\frac{2}{u} \bigg|_1^{1-x^2}\bigg]\\ \amp = 2\pi \bigg[ \frac{2}{1-x^2} - 2\bigg]\\ \amp = 4\pi \frac{x^2}{1-x^2}. \end{align*}
Replace $x$ in terms of $a$ to obtain
\begin{align*} 4\pi \frac{(e^a - 1)^2}{(e^a+1)^2} \cdot \frac{(e^a+1)^2}{(e^a+1)^2 - (e^a- 1)^2} \amp = 4\pi \frac{(e^a - 1)^2}{4e^a}\\ \amp = 4\pi \bigg(\frac{e^a - 1}{2e^{a/2}}\bigg)^2\\ \amp = 4\pi \bigg(\frac{e^{a/2} - e^{-a/2}}{2} \bigg)^2. \end{align*}
This last expression can be rewritten using the hyperbolic sine function, evaluated at $a/2\text{.}$ We investigate the hyperbolic sine and cosine functions in the exercises but note their definitions here.
###### Definition5.4.3
The hyperbolic sine function, denoted $\sinh(x)\text{,}$ and the hyperbolic cosine function, denoted $\cosh(x)\text{,}$ are functions of real numbers defined by
\begin{equation*} \sinh(x) = \frac{e^x- e^{-x}}{2} \hskip.3in \text{and} \hskip.3in \cosh(x)= \frac{e^x + e^{-x}}{2}. \end{equation*}
The area derivation in Example 5.4.2 may then be summarized as follows.
Other regions are not as simple to describe in polar coordinates. An important area for us will be the area of a $\frac{2}{3}$-ideal triangle, the figure that results if two of the three vertices of a hyperbolic triangle are moved to ideal points. See Figure 5.4.6.
The proof of this theorem is given in the following section. The proof there makes use of a different model for hyperbolic geometry, the so-called upper half-plane model.
An ideal triangle consists of three ideal points and the three hyperbolic lines connecting them. It turns out that all ideal triangles are congruent (a fact proved in the exercises); the set of all ideal triangles is minimally invariant in $(\mathbb{D}, {\cal H})\text{.}$
Since all ideal triangles are congruent, assume our triangle $\Delta$ is the ideal triangle shown in Figure 5.4.7.
But then $\Delta$ can be partitioned into two $\frac{2}{3}$-ideal triangles by drawing the vertical hyperbolic line from 0 along the imaginary axis to ideal point $i\text{.}$ Each $\frac{2}{3}$-ideal triangle has interior angle $\pi/2\text{,}$ so $\Delta$ has area $\pi/2+\pi/2 = \pi\text{.}$
It is a remarkable fact that $\pi$ is an upper bound for the area of any triangle in $(\mathbb{D},{\cal H})\text{.}$ No triangle in $(\mathbb{D},{\cal H})$ can have area as large as $\pi\text{,}$ even though side lengths can be arbitrarily large!
Consider Figure 5.4.10 containing triangle $\Delta pqr\text{.}$ We have extended segment $qp$ to the ideal point $t\text{,}$ $u$ is an ideal point of line $rq\text{,}$ and $v$ is an ideal point of line $pr\text{.}$ The area of the ideal triangle $\Delta tuv$ is $\pi\text{.}$ Notice that regions $R_1\text{,}$ $R_2\text{,}$ and $R_3$ are all $\frac{2}{3}$-ideal triangles contained within the ideal triangle. Consider $R_1\text{,}$ whose ideal points are $u$ and $t\text{,}$ and whose interior angle is $\angle uqt\text{.}$ Since the line through $q$ and $r$ has ideal point $u\text{,}$ the interior angle of $R_1$ is $\angle uqt=\pi-\beta\text{.}$ Similarly, $R_2$ has interior angle $\pi - \alpha$ and $R_3$ has interior angle $\pi - \gamma\text{.}$
Let $R$ denote the triangle region $\Delta pqr$ whose area $A(R)$ we want to compute. We then have the following relationships among areas:
\begin{align*} \pi \amp = A(R) + A(R_1) + A(R_2) + A(R_3)\\ \amp = A(R) + [\pi - (\pi -\alpha)] + [\pi - (\pi -\beta)] + [\pi - (\pi -\gamma)]. \end{align*}
Solving for $A(R)\text{,}$
\begin{equation*} A(R) = \pi - (\alpha + \beta + \gamma), \end{equation*}
and this completes the proof.
In Euclidean geometry trigonometric formulas relate the angles of a triangle to its side lengths. There are hyperbolic trigonometric formulas as well.
The proofs of these laws are left as exercises for the interested reader. The following result follows from the first hyperbolic law of cosines.
The second hyperbolic law of cosines also leads to an interesting result. In the hyperbolic plane, if we find ourselves at point $z\text{,}$ we may infer our distance $c$ to a point $w$ by estimating a certain angle, called the angle of parallelism of $z$ to the line $L$ through $w$ that is perpendicular to segment $zw\text{.}$ The following is a picture of this scene:
In this setting, $\Delta zwu$ is a $\frac{1}{3}$-ideal triangle, and the second hyperbolic law of cosines applies with $\gamma = 0$ and $\beta = \pi/2$ to yield the following result.
The angle of parallelism is pursued further in Section 7.4.
###### Example5.4.15Flying around in $(\mathbb{D},{\cal H})$
Suppose a two-dimensional ship is plopped down in $\mathbb{D}\text{.}$ What would the pilot see? How would the ship move? How would the pilot describe the world? Are all points equivalent in this world? Could the pilot figure out whether the universe adheres to hyperbolic geometry as opposed to, say, Euclidean geometry?
Recall what we know about hyperbolic geometry. First of all, any two points in the hyperbolic plane are congruent, so the geometry is homogeneous. The pilot could not distinguish between any two points.
Second, the shortest path between two points is the hyperbolic line between them, so light would travel along these hyperbolic lines, assuming light follows geodesics. The pilot's line of sight would follow along these lines, and the ship would move along these lines to fly as quickly as possible from $p$ to $q\text{,}$ assuming no pesky asteroid fields block the path. To observe a galaxy at point $q$ from the point $p$ (as in the diagram below), the pilot would point a telescope in the direction of line $L\text{,}$ the line along which the light from the galaxy travels to reach the telescope.
With a well-defined metric, we can say more. The pilot will view the hyperbolic plane as infinite and without boundary. In theory, the pilot can make an orbit of arbitrary radius about an asteroid located anywhere in the space.
To test for hyperbolic geometry, perhaps the pilot can turn to triangles. The angles of a triangle in the hyperbolic plane sum to less than 180$^\circ\text{,}$ but only noticeably so for large enough triangles. In our disk model of hyperbolic geometry, we can easily observe this angle deficiency. In the figure below, triangle $\Delta zuw$ has angle sum about $130^\circ\text{,}$ and $\Delta pqr$ has angle sum of about $22^\circ\text{!}$ Whether an intrepid 2-D explorer could map out such a large triangle depends on how much ground she could cover relative to the size of her universe. We will have more to say about such things in Chapters 7 and 8.
###### Example5.4.17Hyperbolic squares?
Simply put, hyperbolic squares don't exist. In fact, no four-sided figures with four right angles exist, if we assume the sides are hyperbolic segments. If such a figure existed, its angle sum would be 2$\pi\text{.}$ But such a figure could be divided along a diagonal into two triangles whose total angle sum must then be $2\pi$ as well. This means that one of the triangles would have angle sum at least $\pi\text{,}$ which cannot happen.
On the other hand, there is no physical obstruction to a two-dimensional explorer making the following journey in the hyperbolic plane: Starting at a point such as $p$ in Figure 5.4.16, head along a line in a certain direction for $a$ units, turn right ($90^\circ$) and proceed in a line for $a$ more units, then turn right again and proceed in a line $a$ units, and then turn right one more time and proceed in a line for $a$ units. Let $q$ denote the point at which the explorer arrives at the end of this journey. In the Euclidean plane, $q$ will equal $p\text{,}$ because the journey traces out a square built from line segments. However, this is not the case in $(\mathbb{D},{\cal H})$ (though if we connect $p$ and $q$ with a hyperbolic line we obtain a geodesic pentagon with (at least) three right angles!). In the exercises we investigate the distance between $p$ and $q$ as a function of the length $a\text{.}$
However, we can build a four-sided figure closely resembling a rectangle, if we drop the requirement that the legs be hyperbolic line segments.
Through any point $0 \lt a \lt 1$ on the positive real axis, we may construct a hyperbolic line $L_1$ through $a$ that is perpendicular to the real axis. Also construct a hyperbolic line $L_2$ through $-a$ that is perpendicular to the real axis. Now pick a point $z$ on $L_1$ and construct the cline arc $C_1$ through $z, 1$ and $-1\text{.}$ Also construct the cline arc $C_2$ through $\overline{z}, 1$ and $-1\text{.}$ This creates a four sided figure, which we call a block. We claim that each angle in the figure is right, and that opposite sides have equal length. Moreover, $z$ and $a$ can be chosen so that all four sides have the same length. This figure isn't a rectangle, however, in the sense that not all four sides are hyperbolic segments. The arcs $C_1$ and $C_2$ are not hyperbolic lines.
While squares don't exist in the hyperbolic plane, we may build right-angled regular polygons with more than four sides using hyperbolic line segments. In fact, for each triple of positive real numbers $(a,b,c)$ we may build a right-angled hexagon in the hyperbolic plane with alternate side lengths $a\text{,}$ $b\text{,}$ and $c\text{.}$ We encourage the reader to work carefully through the construction of this hexagon in the proof of Theorem 5.4.19. We use all our hyperbolic constructions to get there.
We prove the existence of a right-angled hexagon with vertices $v_0, v_1, \cdots, v_5$ such that $d_H(v_0,v_1) = a, d_H(v_2,v_3) = b\text{,}$ and $d_H(v_4,v_5) = c\text{.}$
First, let $v_0$ be the origin in the hyperbolic plane, and place $v_1$ on the positive real axis so that $d_H(v_0,v_1)= a\text{.}$ Note that
\begin{equation*} v_1 = \frac{e^a - 1}{e^a+1}. \end{equation*}
Next, construct the hyperbolic line $A$ perpendicular to the real axis at the point $v_1\text{.}$ This line is part of the cline that has diameter $v_1v_1^*\text{.}$
Pick any point $v_2$ on the line $A\text{.}$ For the sake of argument, assume that $v_2$ lies above the real axis, as in Figure 5.4.20.
Next, construct the hyperbolic line $B$ perpendicular to $A$ at the point $v_2\text{.}$ This line is part of the cline through $v_2$ and $v_2^*$ with center on the line tangent to $A$ at $v_2\text{.}$
Next, construct the point $v_3$ on line $B$ that is a distance $b$ away from $v_2\text{.}$ This point is found by intersecting $B$ with the hyperbolic circle centered at $v_2$ with radius $b\text{.}$ (To construct this circle, we first find the scalar $k$ so that the hyperbolic distance between $kv_2$ and $v_2$ is $b\text{.}$)
Next, draw the perpendicular $C$ to line $B$ at $v_3\text{.}$
Then construct the common perpendicular of $C$ and the imaginary axis, call this perpendicular $D\text{.}$ We construct this common perpendicular as follows. First find the two points $p$ and $q$ symmetric to both $C$ and the imaginary axis. Then find $p^*\text{,}$ the point symmetric to $p$ with respect to $\mathbb{S}^1_\infty\text{.}$ The cline through $p\text{,}$ $q\text{,}$ and $p^*$ is perpendicular to $C\text{,}$ the imaginary axis and $\mathbb{S}^1_\infty\text{,}$ so this gives us our line $D\text{.}$
If $C$ and the imaginary axis intersect, no such perpendicular exists (think triangle angles), so drag $v_2$ toward $v_1$ until these lines do not intersect. Then construct $D$ as in the preceeding paragraph. Let $v_4$ and $v_5$ be the points of intersection of $D$ with $C$ and the imaginary axis, respectively.
This construction gives us a right angled hexagon such that $d_H(v_0,v_1) = a$ and $d_H(v_2,v_3) = b\text{.}$ We also want $d_H(v_4,v_5) = c\text{.}$ Notice that this last distance is a function of the position of vertex $v_2$ on line A. As $v_2$ goes along $A$ to $v_1$ there is a point beyond which $D$ no longer exists, and as $v_2$ goes along $A$ to the circle at infinity, the length of segment $v_4v_5$ takes on all positive real values. So, by the intermediate value theorem, there is some point at which the segment has length $c\text{.}$ Finally, all right-angled hexagons with alternate side lengths $a\text{,}$ $b\text{,}$ and $c$ are congruent to the one just constructed because angles, hyperbolic lines, and distances are preserved under transformations in ${\cal H}\text{.}$
###### Example5.4.21Inscribe a circle in an ideal triangle
We show that if one inscribes a circle in any ideal triangle, its points of tangency form an equilateral triangle with side lengths equal to $2\ln(\varphi)$ where $\varphi$ is the golden ratio $(1+\sqrt{5})/2\text{.}$
Since all ideal triangles are congruent, we choose one that is convenient to work with. Consider the ideal triangle with ideal points -1, 1, and $i\text{.}$
The hyperbolic line $L_1$ joining -1 and $i$ is part of the circle $C_1$ with radius 1 centered at $-1+i\text{.}$ The hyperbolic line $L_2$ joining $i$ and 1 is part of the circle $C_2$ with radius 1 centered at $1+i\text{.}$ Let $C$ denote the circle with radius 2 centered at $-1+2i\text{,}$ as pictured below.
Inversion about $C$ gives a hyperbolic reflection of $\mathbb{D}$ that maps $L_1$ onto the real axis. Indeed, the circle $C_1\text{,}$ since it passes through the center of $C\text{,}$ gets mapped to a line - the real axis, in fact. Moreover, hyperbolic reflection across the imaginary axis maps $L_1$ onto $L_2\text{.}$ Let $c$ be the point of intersection of these two hyperbolic lines of reflection, as pictured. The hyperbolic circle with hyperbolic center $c$ that passes through the origin will be tangent to the real axis, and will thus inscribe the ideal triangle. Let the points of tangency on $L_1$ and $L_2$ be $p$ and $q\text{,}$ respectively.
The point $q$ can be found analytically as the point of intersection of circles $C$ and $C_2\text{.}$ Working it out, one finds $q = \frac{1}{5}+\frac{2}{5}i.$ Thus,
\begin{align*} d_H(0,q)\amp =\ln\bigg(\frac{1+|q|}{1-|q|}\bigg)\\ \amp =\ln\bigg(\frac{1+\frac{1}{\sqrt{5}}}{1-\frac{1}{\sqrt{5}}}\bigg)\\ \amp =\ln\bigg(\frac{\sqrt{5}+1}{\sqrt{5}-1}\bigg)\\ \amp =\ln\bigg(\frac{(\sqrt{5}+1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}\bigg)\\ \amp =\ln\bigg(\frac{(\sqrt{5}+1)^2}{4}\bigg)\\ \amp =\ln(\varphi^2)\\ \amp =2\ln(\varphi). \end{align*}
### SubsectionExercises
###### 1
Properties of $\sinh(x) = (e^x-e^{-x})/2$ and $\cosh(x)=(e^x+e^{-x})/2$.
a. Verify that $\sinh(0) = 0$ and $\cosh(0) = 1\text{.}$
b. Verify that $\frac{d}{dx}[\sinh(x)]=\cosh(x)$ and $\frac{d}{dx}[\cosh(x)]=\sinh(x)\text{.}$
c. Verify that $\cosh^2(x)-\sinh^2(x)=1\text{.}$
d. Verify that the power series expansions for $\cosh(x)$ and $\sinh(x)$ are
\begin{align*} \cosh(x)\amp =1+\frac{x^2}{2}+\frac{x^4}{4!}+\cdots\\ \sinh(x)\amp =x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots. \end{align*}
###### 2
Prove that the circumference of a hyperbolic circle having hyperbolic radius $r$ is $C = 2\pi \sinh(r)\text{.}$
###### 3
The hyperbolic plane looks Euclidean on small scales.
a. Prove
\begin{equation*} \lim_{r \to 0^+} \frac{4\pi\sinh^2(r/2)}{\pi r^2} = 1. \end{equation*}
Thus, for small $r\text{,}$ the Euclidean formula for the area of a circle is a good approximation to the true area of a circle in the hyperbolic plane.
b. Prove
\begin{equation*} \lim_{r \to 0^+}\frac{2\pi \sinh(r)}{2\pi r}=1. \end{equation*}
Thus, for small $r\text{,}$ the Euclidean formula for the circumference of a circle is a good approximation to the true circumference of a circle in the hyperbolic plane.
###### 4
Prove that all ideal triangles are congruent in hyperbolic geometry. Hint: Prove any ideal triangle is congruent to the one whose ideal points are 1, $i\text{,}$ and -1 (see Figure 5.4.7).
###### 5
An intrepid tax collector lives in a country in the hyperbolic plane. For collection purposes, the country is divided into triangular grids. The collector is responsible for collection in a triangle having angles $12^\circ\text{,}$ $32^\circ\text{,}$ and $17^\circ\text{.}$ What is the area of the collector's triangle? Can the entire space $\mathbb{D}$ be subdivided into a finite number of triangles?
###### 6
Consider the hyperbolic triangle with vertices at 0, $\frac{1}{2}\text{,}$ and $\frac{1}{2} + \frac{1}{2}i\text{.}$ Calculate the area of this triangle by determining the angle at each vertex. Hint: To determine the angle at a vertex it may be convenient to move it to the origin via an appropriate transformation in $\cal H\text{.}$
###### 7
Recall the block constructed in Example 5.4.17. Prove that all four angles are $90^\circ\text{,}$ and that for any choice of $z\text{,}$ opposite sides have equal hyperbolic length.
###### 8
Find a formula for the area of an $n$-gon, comprised of $n$ hyperbolic line segments in terms of its $n$ interior angles $\alpha_1, \alpha_2, \cdots, \alpha_n\text{.}$ Hint: Decompose the $n$-gon into triangles.
###### 9
Building a hyperbolic octagon with interior angles 45$^\circ$. Let $r$ be a real number such that $0 \lt r \lt 1\text{.}$ The eight points $v_k = re^{i\frac{\pi}{4}k}$ for $k = 0,1,\cdots, 7$ determine a regular octagon in the hyperbolic plane, as shown in Figure 5.4.22. Note that $v_0$ is the real number $r\text{.}$ The interior angle of each corner is a function of $r\text{.}$ We find the value of $r$ for which the interior angle is $45^\circ\text{.}$
a. Prove that the center of the circle containing the hyperbolic line through $v_0$ and $v_1$ is
\begin{equation*} z_0 = \frac{1+r^2}{2r}+\frac{1+r^2}{2r}\tan(\pi/8)i. \end{equation*}
Hint: The center $z_0$ is on the perpendicular bisector of the Euclidean segment $v_0 v_0^*$ and also on the perpendicular bisector of the Euclidean segment $v_0 v_1\text{.}$
b. Let $b = \frac{1+r^2}{2r}$ be the midpoint of segment $v_0 v_0^*$ and show that $\angle v_0 z_0 b = \pi/8$ precisely when the interior angles of the octagon equal $\pi/4\text{.}$
c. Using $\Delta v_0 b z_0$ and part (b), show that the interior angles of the octagon equal $\pi/4$ precisely when
\begin{equation*} \tan(\pi/8) = \frac{|b-v_0|}{|z_0 - b|} = \frac{\frac{1+r^2}{2r}-r}{\frac{1+r^2}{2r}\cdot \tan(\pi/8)}. \end{equation*}
d. Solve the equation in (c) for $r$ to obtain $r = (1/2)^{(1/4)}\text{.}$
###### 10
Suppose we construct a regular $n$-gon in the hyperbolic plane from the corner points $r, re^{\frac{1}{n}2\pi i},$ $re^{\frac{2}{n}2\pi i}\text{,}$ $\cdots, re^{\frac{n-1}{n}2\pi i}$ where $0 \lt r \lt 1\text{.}$ Calculate the hyperbolic length of any of its sides.
###### 11
Prove the first hyperbolic law of cosines by completing the following steps.
a. Show that for any positive real numbers $x$ and $y\text{,}$
\begin{equation*} \cosh(\ln(x/y)) = \frac{x^2+y^2}{2xy}~~~~{ and}~~~~\sinh(\ln(x/y)) = \frac{x^2-y^2}{2xy}. \end{equation*}
b. Given two points $p$ and $q$ in $\mathbb{D}\text{,}$ let $c = d_H(p,q)\text{.}$ Use the hyperbolic distance formula from Theorem 5.3.3 and part (a) to show
\begin{equation*} \cosh(c) = \frac{(1+|p|^2)(1-|q|^2)-4\text{Re}(p\overline{q})}{(1-|p|^2)(1-|q|^2)}. \end{equation*}
c. Now suppose our triangle has one vertex at the origin, and one point on the positive real axis. In particular, suppose $p = r$ ($0\lt r\lt 1$) and $q = ke^{i\gamma}$ ($0\lt k\lt 1$), with angles $\alpha, \beta, \gamma$ and hyperbolic side lengths $a,b,$ and $c$ as in Figure 5.4.23.
Show
\begin{equation*} \cosh(a) = \frac{1+r^2}{1-r^2}~~;~~\sinh(a) = \frac{2r}{1-r^2}; \end{equation*}
\begin{equation*} \cosh(b) = \frac{1+k^2}{1-k^2}~~;~~\sinh(b) = \frac{2k}{1-k^2}. \end{equation*}
d. Show that for the triangle in part (c),
\begin{equation*} \cosh(c) = \cosh(a)\cosh(b)-\sinh(a)\sinh(b)\cos(\gamma). \end{equation*}
e. Explain why this formula works for any triangle in $\mathbb{D}\text{.}$
###### 12
In this exercise we prove the hyperbolic law of sines. We assume our triangle is as in Figure 5.4.24. Thus, $q = ke^{i\gamma}$ for some $0 \lt k \lt 1$ and $p = r$ for some real number $0 \lt r \lt 1\text{.}$ Suppose further that the circle containing side $c$ has center $z_0$ and Euclidean radius $R\text{,}$ shown in the figure, and that $m_q$ is the midpoint of segment $qq^*$ and $m_p$ is the midpoint of segment $pp^*\text{,}$ so that $\Delta z_0m_qq$ and $\Delta pm_pz_0$ are right triangles.
a. Verify that the triangle angles $\alpha$ and $\beta$ correspond to angles $\angle m_qz_0q$ and $\angle pz_0m_p\text{,}$ respectively.
b. Notice that $\sin(\alpha) = |m_q-q|/R$ and $\sin(\beta) = |m_p-p|/R.$ Verify that $|m_q-q| = \frac{1/k - k}{2} = \frac{1 - k^2}{2k}$ and that $|m_p-p| = \frac{1-r^2}{2r}\text{.}$
c. Verify that
\begin{equation*} \frac{\sinh(a)}{\sin(\alpha)} = \frac{\sinh(b)}{\sin(\beta)}. \end{equation*}
d. Explain why we may conclude that for any hyperbolic triangle in $\mathbb{D}\text{,}$
\begin{equation*} \frac{\sinh(a)}{\sin(\alpha)} = \frac{\sinh(b)}{\sin(\beta)}=\frac{\sinh(c)}{\sin(\gamma)}. \end{equation*}
###### 13
Prove the second law of hyperbolic cosines. Hint: This result follows from repeated applications of the first law and judicial use of the two identities $\cos^2(x) + \sin^2(x) = 1$ and $\cosh^2(x) - \sinh^2(x) = 1.$
###### 14
Recall the journey in Example 5.4.17 in which a bug travels a path that would trace a square in the Euclidean plane. For convenience, we assume the starting point $p$ is such that the first right turn of $90^\circ$ occurs at the point $x$ on the positive real axis and the second turn occurs at the origin. (This means that $x = (e^a-1)/(e^a+1)\text{.}$) The third corner must then occur at $xi\text{.}$ We have reproduced the journey with some more detail in Figure 5.4.25. In this exercise we make use of hyperbolic triangle trig to measure some features of this journey from $p$ to $q$ in terms of the length $a$ of each leg.
a. Determine the hyperbolic distance between $p$ and the origin (corner two of the journey). In particular, show that $d_H(0,p) = \cosh^2(a)\text{.}$ Note that if this journey had been done in the Euclidean plane, the corresponding distance would be $\sqrt{2}a\text{.}$ Is $\cosh^2(a)$ close to $\sqrt{2}a$ for small positive values of $a\text{?}$
b. Let $\theta = \angle x0p\text{.}$ Show that $\tan(\theta) = \frac{1}{\cosh(a)}\text{.}$ What is the corresponding angle if this journey is done in the Euclidean plane? What does $\theta$ approach as $a \to 0^+\text{?}$
c. Show that $\angle x0p = \angle 0px\text{.}$
d. Show that $d_H(p,q) = \cosh^4(a)[1-\sin(2\theta)]+\sin(2\theta)\text{.}$
e. Let $\alpha = \angle q p 0\text{,}$ $b = d_H(0,p)$ and $c = d_H(p,q)$ Show that
\begin{equation*} \sin(\alpha) = \frac{\sinh(b)}{\sinh(c)}\cos(2\theta). \end{equation*}
f. Determine the area of the pentagon enclosed by the journey if, after reaching $q$ we return to $p$ along the geodesic. In particular, show that the area of this pentagon equals $\frac{3\pi}{2}-2(\theta+\alpha)\text{.}$ What is the corresponding area if the journey had been done in the Euclidean plane?
g. Would any of these measurements change if we began at a different point in the hyperbolic plane and/or headed off in a different direction initially than the ones in Figure 5.4.25?
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## Parallax
Parallax can be used to estimate the distances of nearby stars, up to about 330 light years. It works because as the Earth moves around the Sun, from 1 to 2 in the diagram below, the nearest stars appear to move relative to the most distant, apparently fixed, stars and galaxies.
In the diagram above,whereandis in metres. Alternatively we can definein terms of the parsec: 1 parsec is the distance when an angle of 1 arcsec is subtended, then
Example: Suppose a star has parallax 0.034 arcsec then
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## Linux and UNIX Man Pages
Test Your Knowledge in Computers #868
Difficulty: Medium
Maclisp is a successor to Common Lisp.
True or False?
# ilogbf(3) [freebsd man page]
```ILOGB(3) BSD Library Functions Manual ILOGB(3)
NAME
ilogb, ilogbf, ilogbl, logb, logbf, logbl -- extract exponent
LIBRARY
Math Library (libm, -lm)
SYNOPSIS
#include <math.h>
int
ilogb(double x);
int
ilogbf(float x);
int
ilogbl(long double x);
double
logb(double x);
float
logbf(float x);
long double
logbl(long double x);
DESCRIPTION
ilogb(), ilogbf() and ilogbl() return x's exponent in integer format. ilogb(+-infinity) returns INT_MAX, ilogb(+-NaN) returns FP_ILOGBNAN,
and ilogb(0) returns FP_ILOGB0.
logb(x), logbf(x), and logbl(x) return x's exponent in floating-point format with the same precision as x. logb(+-infinity) returns +infin-
ity, and logb(0) returns -infinity with a division by zero exception.
frexp(3), ieee(3), math(3), scalbn(3)
STANDARDS
The ilogb(), ilogbf(), ilogbl(), logb(), logbf(), and logbl() routines conform to ISO/IEC 9899:1999 (``ISO C99''). The latter three imple-
ment the logb function recommended by IEEE Std 754-1985.
HISTORY
Function First Appeared In
logb() 4.3BSD
ilogb() FreeBSD 1.1.5
ilogbf() FreeBSD 2.0
logbf() FreeBSD 2.0
ilogbl() FreeBSD 5.4
logbl() FreeBSD 8.0
BSD December 16, 2007 BSD```
## Check Out this Related Man Page
```ILOGB(3) BSD Library Functions Manual ILOGB(3)
NAME
ilogb, ilogbf, ilogbl, logb, logbf, logbl -- extract exponent
LIBRARY
Math Library (libm, -lm)
SYNOPSIS
#include <math.h>
int
ilogb(double x);
int
ilogbf(float x);
int
ilogbl(long double x);
double
logb(double x);
float
logbf(float x);
long double
logbl(long double x);
DESCRIPTION
ilogb(), ilogbf() and ilogbl() return x's exponent in integer format. ilogb(+-infinity) returns INT_MAX, ilogb(+-NaN) returns FP_ILOGBNAN,
and ilogb(0) returns FP_ILOGB0.
logb(x), logbf(x), and logbl(x) return x's exponent in floating-point format with the same precision as x. logb(+-infinity) returns +infin-
ity, and logb(0) returns -infinity with a division by zero exception.
frexp(3), ieee(3), math(3), scalbn(3)
STANDARDS
The ilogb(), ilogbf(), ilogbl(), logb(), logbf(), and logbl() routines conform to ISO/IEC 9899:1999 (``ISO C99''). The latter three imple-
ment the logb function recommended by IEEE Std 754-1985.
HISTORY
Function First Appeared In
logb() 4.3BSD
ilogb() FreeBSD 1.1.5
ilogbf() FreeBSD 2.0
logbf() FreeBSD 2.0
ilogbl() FreeBSD 5.4
logbl() FreeBSD 8.0
BSD December 16, 2007 BSD```
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Credit: Public domain image courtesy of NASA.
## Reducing everything to make the distances understandable
Understanding the distances between planetary orbits in our solar system can be made easier by reducing everything and comparing it to something we’re familiar with. Below I reduce the sun down to one inch and afterward take the gigantic distances and reduce them proportionally and compare it all to a softball field.
Our sun is about 865,560 miles in diameter (1,392,685 km).[1] To reduce it down to an inch (2.54 cm), we would determine that it is: 865,560 x 5,280 x 12 = 54,841,881,600 inches. Therefore, to determine the distances between the sun and the other planets, if the sun were reduced to an inch in diameter and proportions remain the same, we would figure out how many inches it is to a particular planet and then divide by 54,841,881,600.
## True distances from the sun to each planet
Each of the eight major planets has an orbit that is at least somewhat elliptical. Mercury’s orbit is the most elliptical, and Venus has the least elliptical orbit.[2] This list has the average distances calculated, between the two extremes for each planet – their furthest and closest points to the sun along their orbits:
Mercury has an average distance from the sun of: 35,990,709 miles (57,909,050 km).[3]
Venus: 67,251,709 miles (108,208,000 km)[4]
Earth: 92,975,924 miles (149,598,261 km)[5]
Mars: 141,665,103 miles (227,949,150 km)[6]
Jupiter: 483,870,230 miles (778,547,200 km)[7]
Saturn: 890,894,574 miles (1,433,449,370 km)[8]
Uranus: 1,787,867,671 miles (2,876,679,083 km)[9]
Neptune: 2,798,908,429 miles (4,503,443,662 km)[10]
## The planets of our solar system
### Sizes shown to scale with one another
Credit: Image is in the public domain courtesy of NASA.
## Scaling down: If the sun were an inch in diameter
Now using the formula shared in the introduction to this article, here is what we’d get for our reduced-in-size solar system. Distances from a one-inch sun to the planets:
Mercury: 41.6 inches (105.6 cm), or 3 feet 5.6 inches
Venus: 77.7 inches (197.3 cm), or 6 feet 5.7 inches
Earth: 107.4 inches (272.8 cm), or 8 feet 11.4 inches
Mars: 163.7 inches (415.7 cm), or 13 feet 7.7 inches
Jupiter: 559.0 inches (14.2 meters), or 46 feet 7 inches
Saturn: 1,029.3 inches (26.1 meters), or 85 feet 9.3 inches
Uranus: 2,065.6 inches (52.5 meters), or 172 feet 1.6 inches
Neptune: 3,233.6 inches (82.1 meters), or 269 feet 5.6 inches
## The furthest dwarf planet known
There are probably hundreds and maybe even thousands of dwarf planets, and all but one venture beyond the orbit of Neptune. Ceres is found in the asteroid belt between Mars and Jupiter.[11]
Dwarf planet Sedna goes way out to more than 900 times the distance of the Earth to the sun. No other known dwarf planets go as far in their orbits.[11]
At its furthest point, Sedna would calculate to the following if the sun were an inch (2.54 cm) in diameter: To Sedna’s furthest distance (aphelion), it would be 100,525.2 inches (2,553.3 meters), or 8,377 feet, or 1.6 miles.
Credit: Image is in the public domain courtesy of NASA.
Artwork produced by a NASA employee shows what our sun might look like in the distance, as seen from the dwarf planet Sedna.
## Comparison to a softball field
A men’s slow-pitch softball field has a fence typically 275 to 300 feet (85 to 90 meters) from home plate.[12] After everything explained and calculated above, here is what we’d get:
- If the sun were an inch in diameter and sitting on home plate, Neptune would be near the outfield fence.
- Uranus would be near the middle of the outfield, and Saturn would be near second base.
- Jupiter would be about ten feet (3 meters) beyond the pitcher’s mound, Mars would be less than halfway to the mound.
- Earth would be about nine feet (just under 3 meters) away, Venus would be only about six-and-a-half feet (2 meters) away, and Mercury would barely be further away than the length of a softball bat.
- Sedna would be more than a mile-and-a-half away on the other side of town or in the next town over, and at its closest distance would still be more than double the distance of the fence in center field.
If the sun were an inch across, Jupiter would be about 1/10th of an inch across (2.5 mm) and far larger than all other planets, Earth would be about 1/100th of an inch (0.25 mm), and Sedna would be about 1/1,200th of an inch across (0.002 mm).
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# 4.1 simple harmonic motion
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### 4.1 simple harmonic motion
1. 1. Topic 4 – Oscillations and Waves4.1 Simple Harmonic Motion
2. 2. Oscillations● There are many systems, both natural and manmade, that vibrate back and forth around anequilibrium point.● These systems are said to regularly oscillate.● Common examples are:● A mass on a spring● A pendulum● Electrons under alternating current
3. 3. Key Terms● The equilibrium point is that point where thesystem will naturally rest.● e.g. for a pendulum – bottom centre● For a mass on a spring – the point where theupwards pull of the spring equals the downward pullof the weight.● The displacement (x) of the system is thevector displacement of the system from itsequilibrium point.● Usually the displacement is considered in 1dimension and is given the symbol x even if thedisplacement is vertical.
4. 4. Key Terms● The amplitude (A) of an oscillation is themaximum displacement of the system.● It is the height of a wave from its equilibrium point.● It is half the peak to trough height.● The wavelength (λ) of a moving wave is thedistance from peak to peak in the spacedimension.
5. 5. Key Terms● The time period (T) is the time taken in seconds tocomplete 1 complete cycle.● This is the time from peak to peak in the timedimension.● A cycle is complete when the system is back in itsinitial state.● e.g. for a pendulum, when the bob is at its lowest pointand travelling in the same direction as at the start.● The frequency (f) of the system is the number ofoscillations per second.● It is the inverse of the time period.● Frequency is measured in Hz or s-1
6. 6. Key Terms● A sine wave has a period of 2π radians and atime period of T seconds.● Therefore its angular displacement (on an x-θgraph) at any time is:● The angular frequency (ω) in rad s-1istherefore:θ=2πTtω=2πT=2π f
7. 7. Questions● Calculate the frequency and angular frequencyof:● A pendulum of period 4s● A water wave of period 12s● Mains electricity of period 0.02s● Laser light with period 1.5 fs
8. 8. Key Terms● A sinusoidal wavehas is an oscillationwith the followingproperties.● It has an amplitude of1.● It has a period of 2πradians● It has an initialdisplacement of +0.x0 π/2 π 3π/2 2πθx=sinθ
9. 9. Key Terms● A cosine wave isidentical to a sinewave excepting that ithas an initialdisplacement of +1● It can be said that acosine wave is a sinewave with a phasedifference (Ф) of-π/2x0 π/2 π 3π/2 2πθx=cosθx=sin(θ−π2)
10. 10. Oscillating Systems● An oscillating system is defined as one thatobeys the general equation:● Here● the amplitude is x0● The angular frequency ensures that the real timeperiod coincides with the angular period of 2πradians● The phase allows for an oscillation that starts at anypoint.x=x0 sin(ωt+ ϕ)
11. 11. Oscillating Systems● If the oscillations begin at the equilibrium pointwhere displacement is zero at the start then:● If the oscillations begin at the end point wheredisplacement is a maximum at the start then:● This second form is more useful in moresituationsx=x0 sin(ωt)x=x0 cos(ωt)
12. 12. Questions● A simple harmonic motion is initiated byreleasing a mass from its maximumdisplacement. It has period 2.00s andamplitude 16.0cm. Calculate the displacementat the following times:● t=0s● t=0.25s● t=0.50s● t=1.00s
13. 13. Time x v0 0 +v0T/4x0 0T/2 0 -v03T/4-x0 0T 0 +v0Oscillating Systems● The rate of change of displacement (the speed) isgiven by the gradient of the displacement curve.● Assuming that:● Then:● The speed is therefore:x=x0 sin(ωt)v=v0 cos(ωt)
14. 14. Questions● A bored student holds one end of a flexible rulerand flicks it into an oscillation. The end of theruler moves a total distance of 8.0cm andmakes 28 full oscillations in 10s.● What are the amplitude and frequency of the motionof the end of the ruler?● Use the displacement equation to produce a tableof x and t for t=0,0.04,0.08,0.012,...,0.036● Draw a graph of x versus t● Find the maximum speed of the end of the ruler.
15. 15. Oscillating Systems● The rate of change of speed (the acceleration)is given by the gradient of the speed curve.● Using similar logic:● This has a very similar form to the displacementequation therefore:● Note that the acceleration is:● In the opposite direction to the displacement,● Directly proportional to the displacement.a=−a0 sin(ωt)a=−a0x0x
16. 16. The SHM Equation● Any system undergoing simple harmonicmotion obeys the relationship:● It can be shown using calculus or centripetalmotion that● Therefore:a∝−xa0=ω2x0a=−ω2x
17. 17. The SHM Equations● For a system startingat equilibrium● The general SHM equation applies to all simpleoscillating systems.a=−ω2xx=x0 sin(ωt)v=ω x0 cos(ωt)a=−ω2x0 sin(ωt)● For a system starting atmaximum displacementx=x0 cos(ωt)v=−ω x0 sin(ωt)a=−ω2x0 cos(ωt)
18. 18. The SHM Equations● One final equation can be formed by squaringthe speed equation.● Because sin2θ + cos2θ =1v=ω x0cos(ωt)v2=ω2x02cos2(ωt)v2=ω2x02(1−sin2(ωt))v2=ω2( x02−x02sin2(ωt))v2=ω2( x02−x2)v=±ω√x02−x2
19. 19. Questions● A body oscillates with shm decribed by:● x=1.6cos3πt● What are the amplitude and period of themotion● At t=1.5s, calculate the displacement, velocityand acceleration.
20. 20. Questions● The needle of a sewing machine moves up anddown with shm. If the total vertical motion ofthe needle is 12mm and it makes 30 stitches in7.0s calculate:● The period,● The amplitude,● The maximum speed of the needle tip● The maximum acceleration of the needle tip.
21. 21. Energy Changes● An oscillating system is constantly experiencingenergy changes.● At the extremes of displacement, the potentialenergy is a maximum.● Gravitational potential for a pendulum, elasticpotential for a spring● At the equilibrium position, the kinetic energy isa maximum
22. 22. Kinetic Energy● Remember that kinetic energy is given by:● Substituting● Givesv=±ω √ x02−x2v2=ω2( x02−x2)EK =12m ω2( x02−x2)EK =12m v2
23. 23. Total Energy● Remember that at the equilibrium point ALL theenergy of the system is kinetic.● The total energy of the system ETis therefore:● Note that the total energy of the system isproportional to the amplitude squared.ET =12m ω2(x02−02)ET =12m ω2x02
24. 24. Potential Energy● The law of conservation of energy requires thatthe total energy of an oscillating system be thesum of the potential and kinetic energies.●● Therefore12mω2x02=12mω2(x02−x2)+ EPEP=12m ω2x2ET =EK + EP
25. 25. Questions● A pendulum of mass 250g is released from itsmaximum displacement and swings with shm.If the period is 4s and the amplitude of theswing is 30cm, calculate:● The frequency of the pendulum● The maximum speed of the pendulum● The total energy of the pendulum● The maximum height of the pendulum bob.● The energies of the pendulum at t=0.2s.
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# Sample
(Redirected from Statistical sample)
Border police looking for illegal drugs with a specially trained dog: If they check every tenth car, they are taking an unbiased sample.
In statistics, a sample is part of a population. The sample is carefully chosen. It should represent the whole population fairly, without bias.
When treated as a data set, a sample is often represented by capital letters such as ${\displaystyle X}$ and ${\displaystyle Y}$, with its elements being represented in lowercase (e.g., ${\displaystyle x_{3}}$), and the sample size being represented by the letter ${\displaystyle n}$.[1]
The reason samples are needed is that populations may be so large that counting all the individuals may not be possible or practical. Therefore, solving a problem in statistics usually starts with sampling.[2] Sampling is about choosing which data to take for later analysis. As an example, suppose the pollution of a lake should be analysed for a study. Depending on where the samples of water were taken, the studies can have different results.
As a general rule, samples need to be random. This means the chance or probability of selecting one individual is the same as the chance of selecting any other individual.
In practice, random samples are always taken by means of a well-defined procedure. A procedure is a set of rules, a sequence of steps written down and exactly followed. Even so, some bias may remain in the sample. Consider the problem of designing a sample to predict the result of an election poll. All known methods have their problems, and the results of an election are often different from predictions based on a sample. If you collect opinions by using telephones, or by meeting people in the street, you won't ask people who don't answer phone calls or who don't walk on the street. Therefore, in cases like this a completely neutral sample is never possible.[3] In such cases a statistician will think about how to measure the amount of bias, and there are ways to estimate this.
A similar situation occurs when scientists measure a physical property, say the weight of a piece of metal, or the speed of light.[4] If we weigh an object with sensitive equipment we will get minutely different results. No system of measurement is ever perfect. We get a series of estimates, each one being a measurement. These are samples, with a certain degree of error. Statistics is designed to describe error, and carry out analysis on this kind of data.
There are different kinds of samples:
• A complete sample includes all the elements that have a given property.
• An unbiased or representative sample is produced by taking a complete sample and selecting elements from it, in a process that does not depend on the properties of the elements.
The way the sampling is obtained, along with the sample size, will have an impact on how the data is viewed.[5]
## Stratified sampling
If a population has obvious sub-populations, then each of the sub-populations needs to be sampled. This is called stratified sampling. Stratified sampling is also known as stratified random sample. Stratified sampling is often represented as proportion, such as percent (%).
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Home / Science projects / Force & Motion Science Projects
• # Force & Motion Science Projects
### Balancing Coins
Build a balancing tower and watch it topple down when you put one small penny on top of it. What makes the tower fall? Find out in this science project exploring balance and motion.
#### What You Need:
• wood or plastic ruler
• marker or highlighter
• pennies or small weights
#### What You Do:
1. Set the marker or pencil on a sturdy surface (like a table or desk) so that it is standing up tall.
2. Find the middle of the ruler. See how long the ruler is, then divide that number by 2 to find the center. If you are using a 12 inch ruler, the center would be at 6 inches.
3. Carefully set the ruler laying flat on top of the marker. Put the center of the ruler right on top of the marker's end.
4. Put a penny down on top of the tower in the center of the ruler.
5. Continue adding coins to the tower - hold a penny in each hand and slowly set a penny down at the same time on each end of the ruler.
6. Try putting a penny down on just one end of the ruler. What happens to the tower?
#### What Happened:
What makes the tower topple down? Gravity is a force that is always pulling. When you put a cup of water down on the table, it stays there because gravity pulls it down. When you jump on a trampoline, you come back down because of gravity. When you put equal weights on each side of the tower, it was balanced and even on each side. It's sort of like a see-saw when the ends are balanced and only your feet touch the ground. With equal weights on each side of the tower, gravity pulled down with equal force on each side, so it stayed balanced. When you placed a coin on just one side and not the other, gravity pulled on that coin (because gravity is a force that is always pulling) and the tower was not even any more. It was no longer balancing in the air and so it toppled over. The pull of that one coin caused the ruler to be pulled down, making the tower collapse. Even a small coin can create a big motion because of the powerful force of gravity pulling on it.
Do you think you can build a better balancing tower? Try building it higher using markers stacked on top of each other. Do you think you could build the same tower using a new pencil (with an unsharpened end) and a ruler? What if you used a weight that was lighter than a penny (for example, a flower petal, paperclip, or scrap of paper), do you think the tower would still fall down? Try it out.
### Swing Science
For this project you will need a friend to help you. Have you ever thought about the forces that are causing motion when you're on a playground? The slide, fireman's pole, tire swing, and merry-go-round are all examples of forces. Using a swing set, either in a park or your backyard, you can learn more about how forces work and what inertia means.
#### What You Need:
• Stopwatch or timer
• Pencil and paper
• Calculator
• Swing
#### What You Do:
1. Tell your helper that she or he will use the stopwatch to time one minute - exactly 60 seconds - of swinging.
2. Have your helper pull back the swing that you are sitting on, and let go. Then, the helper will start the timer as soon as s/he lets go of the swing.
3. Be careful not to pump your legs, but just sit in the swing. Count how many times you go forward and then back. Each swing (back and forth) gets one count.
4. After a minute has gone by, stop the swing, and get out the pencil and paper. Write down how many swings you counted. Then, divide this number by the number of seconds (60) using the calculator. Write the new number down. You found the frequency, or amount of swings there are each second, with a certain amount of force.
5. Let your helper have a turn in the swing. Pull the swing back and let go. Have the helper count swings while you keep time. Then find the frequency. Write down the number you found, telling what you did to get that frequency.
6. What would happen if you added force? Pull back the swing (or have your helper pull) as far back as you can, and push it forward. Find the frequency, and write it down.
7. Now, try adding force after the swing has been pushed once by pumping your legs back and forth. What is the frequency? Have your helper try, and be sure to record the results.
#### What Happened:
The direct force of someone pushing you made the swing move back and forth. A famous scientist named Isaac Newton studied physics by using a swing, or pendulum. A pendulum swings back and forth again and again until something stops it or slows it down. Newton came up with three laws of motion. The first one has to do with swings! It is called inertia (ih-ner-sha). Inertia means that when something is in motion, it will stay in motion, but when something is stopped, it won't move until something else (a force pushing or pulling) moves it. Inertia happens on the swing set when your partner pushes you. Once you are in motion, you keep going, just like a pendulum. Think about which person had the highest frequency of swings - you or your helper. Which person is bigger? Objects that are heavy have more inertia than small objects. That means it may take more force to get a large object moving, but once it is moving it will want to keep going and will be harder to slow down than a smaller object.
What happened when you added force by pumping your legs? This movement helped you go higher and faster. You pushed forward with your body when the swing was going forward, and you pulled back when the swing was going back. The pushing and pulling you did worked together with the pushing and pulling of the swing in motion to make you swing high up in the air.
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By: Luna
Date: Nov 07, 2015
This would be the perfect Science Fair project!
By: Jocelyn
Date: Feb 01, 2015
Cool science projects but I will much rather do it on my own for the second science project because I am doing a science fair project and I want to work by myself.
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# Thread: Scale conversion engineering drawings
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Originally Posted by ggunn
1" = 10' is 1:120 scale, isn't it?
yes
but they deal in feet
there are 2 scale types
engineers ":' 1:10 (1"=10'), 1:40, etc
architects ":' 1/4 (1/4"= 1'), 1/8, etc
2. Originally Posted by t_van
I'm trying to understand scales on engineering drawings.
I have a plan view drawing of a proposed electrical room on an 11''x17'' size sheet of paper. If the scale is marked 1/2"=1', is this the scale for the full-size drawing? What size paper is used for full-size drawings? How is the scale adjusted for 11"x17" paper?
Unfortunately, the size of the paper has nothing to do with the scale. Are you sure 11x17 is not the original intended size?
It's best not to guess any dimensions from a piece of paper that appears to be a reduced, full-size drawing. You can never tell how much it has been reduced. I recommend you get a copy of the architectural or structural drawing with some dimensions on it. ggunn has the best idea, but only as a last resort.
An engineering firm I know paid an "errors and omissions" charge when a designer tried to interpret the scale of a civil drawing by guessing the width of the parking stalls. The scale was off enough that the lighting calcs were wrong and more lights had to be added.
3. Here's what happens when the scales are mis-read.
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Originally Posted by K8MHZ
Here's what happens when the scales are mis-read.
classic
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## Scale conversion engineering drawings
Originally Posted by ggunn
1" = 10' is 1:120 scale, isn't it?
Yes, that is correct. That portion of my reply is an actual copy and past from an ANSI quotation.
Scaling is very confusing to many. The hardest part is measuring something on a drawing and converting that into real life dimensions. If the Drawing has anything that is dimensioned, carefully measure the dimension leaders then create a Multiplier to get the actual size.
Auto CAD allows the user to import a "Raster Image" (a picture or PDF). If there is something in that picture that is a standard size like a Dollar Bill, a Quarter, or even Coke can, you have a standard for that drawing and measurements can be created. AutoCAD can't Snap to a picture so the user needs to draw a line from one side of the picture object to the other, then an AutoCAD Dimension can be placed.
JimO
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Originally Posted by jimo144
Yes, that is correct. That portion of my reply is an actual copy and past from an ANSI quotation.
Scaling is very confusing to many. The hardest part is measuring something on a drawing and converting that into real life dimensions. If the Drawing has anything that is dimensioned, carefully measure the dimension leaders then create a Multiplier to get the actual size.
Auto CAD allows the user to import a "Raster Image" (a picture or PDF). If there is something in that picture that is a standard size like a Dollar Bill, a Quarter, or even Coke can, you have a standard for that drawing and measurements can be created. AutoCAD can't Snap to a picture so the user needs to draw a line from one side of the picture object to the other, then an AutoCAD Dimension can be placed.
JimO
But if you can get a pdf with accessible layers and a known scale, you can use the PDF Import feature to bring it into AutoCAD at scale. Also, any drawing that you do over the imported pdf will snap to its features if you want it to.
Imported bitmaps, e.g., rastered images, kinda suck; their features "fuzz up" when you zoom into them.
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# SQL Server uniform distribution function
UNIFORM
Updated: 31 July 2010
Use UNIFORM to calculate the probability density function or the lower cumulative distribution function of the uniform distribution.
The formula for the probability density function is:
The lower cumulative distribution function is:
Syntax
SELECT [wctStatistics].[wct].[UNIFORM] (
<@X, float,>
,<@Min, float,>
,<@Max, float,>
,<@Cumulative, bit,>)
Arguments
@X
is the value to be evaluated. @X is an expression of type float or of a type that implicitly converts to float
@Min
is the minimum value of the distribution. @Min is an expression of type float or of a type that implicitly converts to float
@Max
is the maximum value of the distribution. @Max is an expression of type float or of a type that implicitly converts to float
@Cumulative
is a logical value that determines if the probability density function ('False', 0) or the cumulative distribution function ('True', 1) is being calculated. @Cumulative is an expression of type bit or of a type that implicitly converts to bit
Return Types
float
Remarks
· @Max must be greater than or equal to @Min (@Max > @Min).
Examples
Calculate the probability density function:
SELECT wct.UNIFORM(2,1,5,'False')
This produces the following result
----------------------
0.25
(1 row(s) affected)
You can use the SeriesFloat function from the XLeratorDB/math library to generate a dataset which can be pasted into EXCEL to generate a graph of the probability density function.
SELECT SeriesValue
,wct.UNIFORM(SeriesValue, 1, 9, 'False') as [f(x,1,9)]
,wct.UNIFORM(SeriesValue, 2, 8, 'False') as [f(x,2,8)]
,wct.UNIFORM(SeriesValue, 3, 7, 'False') as [f(x,3,7)]
,wct.UNIFORM(SeriesValue, 4, 6, 'False') as [f(x,2,8)]
FROM wctMath.wct.SeriesFloat(0,10,.1,NULL,NULL)
This is an EXCEL-generated graph of the results
Calculate the lower cumulative distribution function:
SELECT wct.UNIFORM(3,1,5,'True')
This produces the following result
----------------------
0.5
(1 row(s) affected)
You can use the SeriesFloat function from the XLeratorDB/math library to generate a dataset which can be pasted into EXCEL to generate a graph of the cumulative distribution function.
SELECT SeriesValue
,wct.UNIFORM(SeriesValue, 1, 9, 'True') as [f(x,1,9)]
,wct.UNIFORM(SeriesValue, 2, 8, 'True') as [f(x,2,8)]
,wct.UNIFORM(SeriesValue, 3, 7, 'True') as [f(x,3,7)]
,wct.UNIFORM(SeriesValue, 4, 6, 'True') as [f(x,2,8)]
FROM wctMath.wct.SeriesFloat(0,10,.1,NULL,NULL)
This is an EXCEL-generated graph of the results
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# Python Numpy Arrays Equal With Code Examples
Python Numpy Arrays Equal With Code Examples
Hello guys, in this post we will explore how to find the solution to Python Numpy Arrays Equal in programming.
```np.array_equal(A,B) # test if same shape, same elements values
np.array_equiv(A,B) # test if broadcastable shape, same elements values
np.allclose(A,B,...) # test if same shape, elements have close enough values
```
Numerous real-world examples illustrate how to deal with the Python Numpy Arrays Equal issue.
## How do I know if NumPy arrays are equal?
To check if two NumPy arrays A and B are equal: Use a comparison operator (==) to form a comparison array. Check if all the elements in the comparison array are True.Other Comparisons
• Greater than.
• Greater than or equal.
• Less than.
• Less than or equal.
## How do I check if two arrays are equal in Python?
Check if two arrays are equal or not using Sorting Sort both the arrays. Then linearly compare elements of both the arrays. If all are equal then return true, else return false.08-Aug-2022
## How do you know if two arrays are equal?
The Arrays. equals() method checks the equality of the two arrays in terms of size, data, and order of elements. This method will accept the two arrays which need to be compared, and it returns the boolean result true if both the arrays are equal and false if the arrays are not equal.
## How do you compare two string arrays in Python?
To perform element-wise comparison of two string arrays using a comparison operator, use the numpy. compare_chararrays() method in Python Numpy. The arr1 and arr2 are the two input string arrays of the same shape to be compared.07-Feb-2022
## How do you check if all elements in an NumPy array are equal Python?
allequal() method in Python Numpy. Returns True if the two arrays are equal within the given tolerance, False otherwise. If either array contains NaN, then False is returned. The fill_value sets whether masked values in a or b are considered equal (True) or not (False).22-Feb-2022
## What is the function to check whether two arrays are equal in NumPy?
array_equal. True if two arrays have the same shape and elements, False otherwise.
## How do you check if all the elements in an array are same?
To check if all values in an array are equal: Use the Array. every() method to iterate over the array. Check if each array element is equal to the first one. The every method only returns true if the condition is met for all array elements.25-Jul-2022
## How do you compare two lists in Python?
The methods of comparing two lists are given below.
• The cmp() function.
• The set() function and == operator.
• The sort() function and == operator.
• The collection.counter() function.
• The reduce() and map() function.
## How do you find the common values between two arrays in Python?
To find the common values, we can use the numpy. intersect1d(), which will do the intersection operation and return the common values between the 2 arrays in sorted order.29-Aug-2020
## How do I search for two elements in the same array?
Approach :
• Get the two Arrays.
• Create two hashsets and add elements from arrays tp those sets.
• Find the common elements in both the sets using Collection. retainAll() method. This method keeps only the common elements of both Collection in Collection1.
• Set 1 now contains the common elements only.
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How far can a shout travel?
I was wondering, if a person climbs a tower and shouts at the top of his lungs how much distance would the sound travel ? Would it reach someone 1km far? Assuming that there are no tall buildings in the way and the wind is still.
Gabriel Golfetti's answer assumes no dissipation. In reality, atmospheric attenuation is quite important for this calculation. According to Engineering Acoustics/Outdoor Sound Propagation: Attenuation by atmospheric absorption (Wikibooks), dissipation in the atmosphere exponentially decreases the sound's intensity with distance, which leads to a linear reduction in the loudness of the sound in dB. Therefore, the loudness of the sound will actually be
$$L=88\;\text{dB}-20\log_{10}\left(\frac{r}{0.3\;\text{m}}\right)-ar$$
where $a$ is the attenuation coefficient in dB/m. The table below gives the attenuation coefficient as a function of frequency and relative humidity for air at 20 degrees Celsius:
For air at a pressure of 1 atm and sound at a frequency of 1 kHz (which is around the peak of the human vocal spectrum), for most values of relative humidity the attenuation coefficient is approximately $a\approx 1\;\text{dB}/100\;\text{m}$. So our equation for the loudness becomes
$$L=88\;\text{dB}-20\log_{10}\left(\frac{r}{0.3\;\text{m}}\right)-\frac{r}{100\;\text{m}}$$
Solving for $L=-9\;\text{dB}$ gives
$$r\approx 2\;\text{km}$$
which is drastically reduced from the original answer. Changing the attenuation coefficient by a factor of two (which is approximately how much it varies at that frequency for non-dry air) changes the maximum distance by a factor of 2, so the proper answer, accounting for this uncertainty, is a few km.
• Wow, you beat me to it haha Commented Jul 6, 2018 at 15:35
• @archaic If you look at that chart, the attenuation coefficient varies pretty widely with frequency. What this means is that different parts of the human vocal spectrum will be attenuated at different rates. Lower-frequency sounds tend to carry further (have less attenuation) than high-frequency sounds, as you can see. The further away you get from the source, the more high frequencies will be suppressed. But that's not the whole picture, either; the ear's nonlinear response prefers high-frequency sounds. The overall effect is a distortion of sound over long distances. Commented Jul 6, 2018 at 17:06
• @archaic This distortion means that conveying understandable meaning over distances more than several hundred meters (a significant fraction of the maximum distance) is probably difficult. One way to get around this is to use forms of communication that don't depend on pitch modulation to convey meaning. Such forms are common in older military tactics; For long distances, low-frequency drums carry very well with distance, and can communicate just based on a rhythm. For shorter distances, or cases in which there is a substantial low-frequency background, high-frequency drums and whistles work. Commented Jul 6, 2018 at 17:12
• @archaic And no, this result does not take into account the details of the terrain or the height above ground. As such, this is still a very rough, order-of-magnitude approximation. Commented Jul 6, 2018 at 17:15
• @archaic In a word, yes, but it's difficult to say how much, because at that point the problem becomes highly situational. I'm pretty confident that the corrections should be less than an order of magnitude in most cases, as I have successfully yelled at someone across a football field (roughly 100 m) before. Also, not all of the corrections necessarily decrease the range -- for example, reflection off of the ground will roughly double the sound intensity (less if there's interference), and make the sound travel something like 30 percent farther (less than an order of magnitude). Commented Jul 6, 2018 at 17:34
To answer this, we need to estimate the level of sound that a shout creates near its source. Since I have no idea what that value is, I googled it: around 88dB at 0.3m away (https://www.engineeringtoolbox.com/voice-level-d_938.html).
For the human voice, the minimum hearing threshold is around -9dB (https://en.m.wikipedia.org/wiki/Absolute_threshold_of_hearing), and such we can now estimate this distance.
The sound intensity $I$ varies with distance $r$ as $$I\propto r^{-2}.$$
As the intensity is related to the pressure $p$ as $$I\propto p^2,$$ we can say the sound pressure goes as
$$p\propto r^{-1}.$$
Since sound level is given by
$$L=20\log\left(\frac{p}{p_0}\right)\mathrm{dB},$$ for a reference pressure $p_0$ that I can't recall the value of right now, we can say that the sound level of the shout goes as
$$L=88\mathrm{dB}-20\log\left(\frac{r}{0.3\mathrm m}\right)\mathrm{dB}.$$
As such, we need to find $r$ such that this becomes around -9dB. Solving this we get $$r=21\mathrm{km}.$$
Note that this result does not take into account reflections on the surface of the Earth or dissipation. As such, the tower should be much higher that the value we found for $r$.
EDIT
As @probably_someone commented, taking dissipation into account is not that difficult. We just need to add an attenuation of 1dB per 100m, which turns our equation of sound level into
$$L=88\mathrm{dB}-\left(20\log\left(\frac{r}{0.3\mathrm m}\right)+\frac{r}{100\mathrm m}\right)\mathrm{dB}.$$
This equation can be solved numerically, and gives us the value of $r$ as
$$r=2\mathrm{km},$$ which is quite smaller than our original estimate without dissipation.
• Atmospheric attenuation plays quite a large role in propagation. For sounds around 1 kHz (which is around the peak of the human voice) at atmospheric pressure and intermediate humidity, you should get an attenuation of about 1 dB per 100 m (en.wikibooks.org/wiki/Engineering_Acoustics/…), which will significantly change the answer. Commented Jul 6, 2018 at 15:14
• The attenuation should be linear in $r$, not constant. Commented Jul 6, 2018 at 15:26
• Oh right. Damn, that makes the equation quite hard to solve then. Commented Jul 6, 2018 at 15:27
• ctd ... We could not actually see his house from near ours, however. On many occasions that he did so, he could not only be heard, but even (at least in good conditions) more-or-less understood. If he'd been a little higher up I'd expect it may well have carried a little further. I'd guess that 2km would be just about plausible in more ideal circumstances. Commented Jul 7, 2018 at 1:50
• I have witnessed a woman in Greece ordering bread from the baker by shouting across the valley, at a distance of about 1km. It was clearly a daily routine. Commented Jul 7, 2018 at 8:34
This question is answered by the Guinness World Records.
The normal intelligible outdoor range of the male human voice in still air is 180 m (590 ft 6.6 in). The silbo, the whistled language of the Spanish-speaking inhabitants of the Canary Island of La Gomera, is intelligible under ideal conditions at 8 km (5 miles). There is a recorded case, under optimal acoustic conditions, of the human voice being detectable at a distance of 17 km (10.5 miles) across still water at night.
You can probably get much louder than 88dB with a shout especially at 0.3m. You'd only have to raise your voice a bit to get that loud as in a teacher in a classroom. I managed to get 104dBA across the room with my best Tarzan level wail but as far as shouting intelligible words 100 dBA at 1 meter in an anechoic chamber is probably a good round ballpark figure, although it's possible to get a bit louder than this.
A sound level of 0dB will be inaudible outdoors. Even the background noises in your ear can easily drown it (blood flow and any mild tinnitus).
Probably 20dB is about your limit and if you live in a city maybe 40-50 dB is needed to understand the words. The fan in your laptop is about 35dB and you clicking keys about 45dB. On inverse square law alone you could be heard a kilometre away at 40 dB. I have yet to look up the relationship between air absorption and frequency which drops exponentially with distance rather than simply squaring and affecting the high frequencies far more than the bass.
All these answers assume linear behavior. A few other things should be pointed out.
1. For very high amplitude sounds, large volume, the ordinary wave equation is not valid. So there might be more to the story based on solutions, or approximate solutions, to the full non-linear equations. Not sure this would greatly increase the distance but it may change some of the quantities used to approximate the answers. There is a famous account of filed artillery exercises during the U.S. Civil War (I think), where soldiers herd shots fired before the shouted command to fire. This can be explained with non-linear wave propagation as the very high amplitude shock traveled with a higher effective speed (or just speed). Of course as amplitude decreases the remaining wave will behave as an ordinary wave.
2. The atmospheric attenuation is frequency dependent, and the non-linear propagation is also frequency dependent. A shout is to some extent percussive, e.g. a short burst. Each frequency will (a) travel at different speeds in the non-linear regime, and (b) dampen differently due to attenuation. With this in mind I wonder if what is picked up at 2km could even be correlated with the original source. There will be information loss.
In all these discussions, you must take into account the fact that the world outdoors does not exist in silence. There is a background noise floor at speech frequencies which is usually too faint to hear unless you are paying close attention to it. Shouts which are within 3dB of the local noise floor will be masked by it, and hence inaudible to your ears.
Atmosperic refraction of sound in temperature inversion conditions. This was discovered in XIX century.See Osborne Reynolds publications.This makes sounds arrive further, even with increase in expected dB, due to constructive interference: direct wave added to refracted wave and reflected wave, in still waters.
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Tri-colour:
Filter by: Content type:
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### There are 177 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Tri-colour
##### Stage: 3 Challenge Level:
Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
### Greetings
##### Stage: 3 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
### Cross-country Race
##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### More Marbles
##### Stage: 3 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
##### Stage: 3 Challenge Level:
What can you say about the angles on opposite vertices of any cyclic quadrilateral? Working on the building blocks will give you insights that may help you to explain what is special about them.
### Con Tricks
##### Stage: 3
Here are some examples of 'cons', and see if you can figure out where the trick is.
### Winning Team
##### Stage: 3 Challenge Level:
Nine cross country runners compete in a team competition in which there are three matches. If you were a judge how would you decide who would win?
### The Genie in the Jar
##### Stage: 3 Challenge Level:
This jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . .
### Logic
##### Stage: 2 and 3
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### Cycle It
##### Stage: 3 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
### Marbles
##### Stage: 3 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### Growing Ls
##### Stage: 3 Challenge Level:
Can you fit Ls together to make larger versions of themselves?
### Thirty Nine, Seventy Five
##### Stage: 3 Challenge Level:
We have exactly 100 coins. There are five different values of coins. We have decided to buy a piece of computer software for 39.75. We have the correct money, not a penny more, not a penny less! Can. . . .
### More Mathematical Mysteries
##### Stage: 3 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
### Ratty
##### Stage: 3 Challenge Level:
If you know the sizes of the angles marked with coloured dots in this diagram which angles can you find by calculation?
### Not Necessarily in That Order
##### Stage: 3 Challenge Level:
Baker, Cooper, Jones and Smith are four people whose occupations are teacher, welder, mechanic and programmer, but not necessarily in that order. What is each person’s occupation?
### Go Forth and Generalise
##### Stage: 3
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
##### Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Aba
##### Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Unit Fractions
##### Stage: 3 Challenge Level:
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation.
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### AMGM
##### Stage: 4 Challenge Level:
Choose any two numbers. Call them a and b. Work out the arithmetic mean and the geometric mean. Which is bigger? Repeat for other pairs of numbers. What do you notice?
### Tis Unique
##### Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### Doodles
##### Stage: 4 Challenge Level:
A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not. . . .
### One O Five
##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Pyramids
##### Stage: 3 Challenge Level:
What are the missing numbers in the pyramids?
### Is it Magic or Is it Maths?
##### Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
### Hockey
##### Stage: 3 Challenge Level:
After some matches were played, most of the information in the table containing the results of the games was accidentally deleted. What was the score in each match played?
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Mod 3
##### Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### A Biggy
##### Stage: 4 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
### Reverse to Order
##### Stage: 3 Challenge Level:
Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?
### Sixational
##### Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Pattern of Islands
##### Stage: 3 Challenge Level:
In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Concrete Wheel
##### Stage: 3 Challenge Level:
A huge wheel is rolling past your window. What do you see?
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# A Relativity Problem
1. Sep 29, 2006
### actionintegral
A line of people are moving at the same speed, clocks all sync'd. They agree that at a certain time they will stop! Let's pretend they can stop
instantaneously.
Relativistically speaking, what will we who are at rest see?
2. Sep 29, 2006
### lightarrow
Their clocks are syncronized in their ref. frame but not in our. So, we will see they don't all stop at the same time: we will see the one at the end of the line (relative to the direction of motion) stopping first, the one at the head of the line stopping last. So, the total lenght of the line, after they have stopped, is the same as the lenght in their ref. frame when they're moving.
3. Sep 29, 2006
### actionintegral
4. Sep 29, 2006
### Staff: Mentor
What madness? That's exactly what JesseM has been saying throughout that thread.
5. Sep 29, 2006
### actionintegral
You are right. And you have too, I see. I was referring to the fact that
the thread is into three pages and growing! It seems to me that the question as posed was too complex and led to the self-perpetuating confusion of the original poster.
6. Oct 3, 2006
### bernhard.rothenstein
acceleration
have please a critical look at
Radar echo, Doppler Effect and Radar detection in the uniformly accelerated reference frame
Authors: Rothenstein, Bernhard; Popescu, Stefan
The uniformly accelerated reference frame described by Hamilton, Desloge and Philpott involves the observers who perform the hyperbolic motion with constant proper acceleration gi. They start to move from different distances measured from the origin O of the inertial reference frame K(XOY), along its OX axis with zero initial velocity. Equipped with clocks and light sources they are engaged with each other in Radar echo, Doppler Effect and Radar detection experiments. They are also engaged in the same experiments with an inertial observer at rest in K(XOY) and located at its origin O. We derive formulas that account for the experiments mentioned above. We study also the landing conditions of the accelerating observers on a uniformly moving platform.
Comment: 15 pages, 8 figures, includes new results on radar detected times and distances
Full-text available from: http://arxiv.org/abs/physics/0609118
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https://01s.info/interest-only-mortgage-calculator/
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# Interest Only Mortgage Calculator
In the early years of the loan, you may require some flexibility because of your financial position which may not be as stable. The best arrangement that can help you unprotected to this is the interest only mortgage because it offers a high degree of flexibility to borrowers. With this plan, you are usually flexible such that you can pay only interest or decide to pay some part of the loan if this is the best option for you. But you need to bear in mind that after a certain period of making interest only payments, you will be required to pay the noticeable amount within a comparatively shorter period of time which results in substantial increase in the repayments.
The calculator
The interest only calculator will always help you specifically to get the exact interest of the loan and also be able to examine the impact of the principal payment. In order to use the interest only calculator successfully, you need to have perfect understanding of some applicable terms whose values are used to make the necessary calculations.
Term of loan: This is the total number of years within which you will make all your payments to the loan. The calculator works on the assumption that after the expiry of the interest only period, the monthly repayments are increased in order to allow for amortization of the remaining balance over the rest of the remaining years. This ensures that the complete amount is paid by the end of the arranged period.
Mortgage amount: It is the original balance of your mortgage or that which is expected at the end of the arrangement.
Interest only period: This is the number of years required to make the interest only payments. After this period, the noticeable balance will be amortized by increased payments for the remaining years.
Interest rate: The loan’s annual interest rate
Total payments: The total of all the monthly payments made over the complete term, assuming no prepayment of principal.
Total interest: The sum of all interests paid over the complete term, assuming no principal is paid
Monthly payment: The initial monthly payment you make (only the interest on the loan balance)
kind of prepayment: Frequency of payment which include monthly, yearly, none and one time payment.
Prepayment amount: This is the amount that will be prepaid on the loan, based on the prepayment kind and applied to the principal amount.
Start with payment: The payment number (amount) that your prepayments will commence with.
Savings: The total amount of interest you will end up saving by prepaying your loan.
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https://sciencehouse.wordpress.com/2012/10/10/using-formal-logic-in-biology/
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# Using formal logic in biology
The 2012 Nobel Prize in physiology or medicine went to John Gurdon and Shinya Yamanaka for turning mature cells into stem cells. Yamanaka shook the world just six years ago in a Cell paper (it can be obtained here) that showed how to reprogram adult fibroblast cells into pluripotent stem cells (iPS cells) by simply inducing four genes – Oct3/4, Sox2, c-Myc, and Klf4. Although he may not frame it this way, Yamanaka arrived at these four genes by applying a simple theorem of formal logic, which is that a set of AND conditions is equivalent to negations of OR conditions. For example, the statement A AND B is True is the same as Not A OR Not B is False. In formal logic notation you would write $A \wedge B = \neg(\neg A \vee \neg B)$. The problem then is given that we have about 20,000 genes, what subset of them will turn an adult cell into an embryonic-like stem cell. Yamanaka first chose 24 genes that are known to be expressed in stem cells and inserted them into an adult cell. He found that this made the cell pluripotent. He then wanted to find a smaller subset that would do the same. This is where knowing a little formal logic goes a long way. There are $2^{24}$ possible subsets that can be made out of 24 genes so trying all combinations is impossible. What he did instead was to run 24 experiments where each gene is removed in turn and then checked to see which cells were not viable. These would be the necessary genes for pluripotency. He found that pluripotent stem cells never arose when either Oct3/4, Sox2, c-Myc or Klf4 were missing. Hence, a pluripotent cell needed all four genes and when he induced them, it worked. It was a positively brilliant idea and although I have spoken out against the Nobel Prize (see here), this one is surely deserved.
2016-1-20: typo corrected.
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## 3 thoughts on “Using formal logic in biology”
1. adajiatr says:
Good to know about this. I will use it in the logic section of my freshman critical thinking class.
Like
2. Me says:
“Noble” prize?
Like
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http://ken.duisenberg.com/potw/archive/arch04/040416.html
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## Dominos One-by-One
```0-4 0-3 0-2 0-1 0-0
1-4 1-3 1-2 1-1
2-4 2-3 2-2
3-4 3-3
4-4```
Into a 4x4 grid, place the 15 dominos in this manner: Place one domino in the grid, one digit per square. Place the second domino by overlapping the first by a matching square (the squares that overlap must have the same digit.) Place each successive domino in the same way, overlapping one square and filling an empty square until all 16 squares are filled.
Extension: Is it possible to make this a magic square as well?
Source: Puzzle 09 from the 2003 Puzzle Design Tournament at http://www.otuzoyun.com/puzzledesign/
Solution
Mail to Ken
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# Surname
Centre
No.
Initial(s)
Paper Reference
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Candidate
No.
0 1
Signature
Paper Reference(s)
6677/01
## Examiners use only
Edexcel GCE
Mechanics M1
Friday 20 January 2012 Afternoon
Time: 1 hour 30 minutes
Question Leave
Number Blank
1
2
3
4
## Materials required for examination
Mathematical Formulae (Pink)
## Items included with question papers
Nil
Candidates may use any calculator allowed by the regulations of the Joint
Council for Qualifications. Calculators must not have the facility for symbolic
algebra manipulation or symbolic differentiation/integration, or have
retrievable mathematical formulae stored in them.
5
6
7
8
Instructions to Candidates
In the boxes above, write your centre number, candidate number, your surname, initials and signature.
Check that you have the correct question paper.
You must write your answer to each question in the space following the question.
Whenever a numerical value of g is required, take g = 9.8 m s2.
When a calculator is used, the answer should be given to an appropriate degree of accuracy.
## Information for Candidates
A booklet Mathematical Formulae and Statistical Tables is provided.
Full marks may be obtained for answers to ALL questions.
The marks for individual questions and the parts of questions are shown in round brackets: e.g. (2).
There are 8 questions in this question paper. The total mark for this paper is 75.
There are 28 pages in this question paper. Any blank pages are indicated.
You must ensure that your answers to parts of questions are clearly labelled.
You should show sufficient working to make your methods clear to the Examiner.
Answers without working may not gain full credit.
Total
This publication may be reproduced only in accordance with
2012 Pearson Education Ltd.
Printers Log. No.
P40096A
W850/R6677/57570 5/4/5/4/3
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*P40096A0128*
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1.
A railway truck P, of mass m kg, is moving along a straight horizontal track with speed
15 m s 1. Truck P collides with a truck Q of mass 3000 kg, which is at rest on the same
track. Immediately after the collision the speed of P is 3 m s 1 and the speed of Q is 9 m s 1.
The direction of motion of P is reversed by the collision.
Modelling the trucks as particles, find
(a) the magnitude of the impulse exerted by P on Q,
(2)
(b) the value of m.
(3)
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Q1
(Total 5 marks)
*P40096A0328*
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2.
A car of mass 1000 kg is towing a caravan of mass 750 kg along a straight horizontal
road. The caravan is connected to the car by a tow-bar which is parallel to the direction
of motion of the car and the caravan. The tow-bar is modelled as a light rod. The engine
of the car provides a constant driving force of 3200 N. The resistances to the motion of
the car and the caravan are modelled as constant forces of magnitude 800 newtons and
R newtons respectively.
Given that the acceleration of the car and the caravan is 0.88 m s 2 ,
(a) show that R = 860,
(3)
(b) find the tension in the tow-bar.
(3)
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Q2
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3.
## Three forces F1, F2 and F3 acting on a particle P are given by
F1 = (7i 9j) N
F2 = (5i + 6j) N
F3 = (pi + qj) N
where p and q are constants.
Given that P is in equilibrium,
(a) find the value of p and the value of q.
(3)
The force F3 is now removed. The resultant of F1 and F2 is R.
Find
(b) the magnitude of R,
(2)
(c) the angle, to the nearest degree, that the direction of R makes with j.
(3)
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(Total 8 marks)
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4.
d
5d
Figure 1
A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two
supports at C and D, where AC = DB = d, as shown in Figure 1. The centre of mass of the
rod is at the point G. A particle of mass 5 m is placed on the rod at B and the rod is on the
2
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(a) Show that GD = 5 d .
2
(4)
The particle is moved from B to the mid-point of the rod and the rod remains in equilibrium.
(b) Find the magnitude of the normal reaction between the support at D and the rod.
(5)
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Q4
(Total 9 marks)
*P40096A01128*
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5.
A stone is projected vertically upwards from a point A with speed u m s 1. After projection
the stone moves freely under gravity until it returns to A. The time between the instant that
the stone is projected and the instant that it returns to A is 3 4 seconds.
7
## Modelling the stone as a particle,
(a) show that u = 17 12 ,
(3)
## (b) find the greatest height above A reached by the stone,
(2)
(c) find the length of time for which the stone is at least 6 53 m above A.
(6)
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Q5
(Total 11 marks)
*P40096A01528*
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6.
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m.
The car accelerates from rest at A to a speed of 15 m s 1 at a constant rate a m s 2.
The time for which the car accelerates is 13 T seconds. The car maintains the speed of
15 m s 1 for T seconds. The car then decelerates at a constant rate of 2.5 m s 2 stopping at B.
(a) Find the time for which the car decelerates.
(2)
(b) Sketch a speed-time graph for the motion of the car.
(2)
(c) Find the value of T.
(4)
(d) Find the value of a.
(2)
(e) Sketch an acceleration-time graph for the motion of the car.
(3)
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*P40096A01928*
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7.
[In this question, the unit vectors i and j are due east and due north respectively. Position
vectors are relative to a fixed origin O.]
A boat P is moving with constant velocity (4i + 8j) km h1.
(a) Calculate the speed of P.
(2)
When t = 0, the boat P has position vector (2i 8j) km. At time t hours, the position vector
of P is p km.
(b) Write down p in terms of t.
(1)
A second boat Q is also moving with constant velocity. At time t hours, the position vector
of Q is q km, where
q = 18i + 12j t (6i + 8j)
Find
(c) the value of t when P is due west of Q,
(3)
(d) the distance between P and Q when P is due west of Q.
(3)
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Q7
(Total 9 marks)
*P40096A02328*
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36 N
8.
30
30
Figure 2
A particle P of mass 4 kg is moving up a fixed rough plane at a constant speed of 16 m s 1
under the action of a force of magnitude 36 N. The plane is inclined at 30 to the horizontal.
The force acts in the vertical plane containing the line of greatest slope of the plane
through P, and acts at 30 to the inclined plane, as shown in Figure 2. The coefficient of
friction between P and the plane is . Find
(a) the magnitude of the normal reaction between P and the plane,
(4)
(b) the value of .
(5)
The force of magnitude 36 N is removed.
(c) Find the distance that P travels between the instant when the force is removed and the
instant when it comes to rest.
(5)
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(Total 14 marks)
TOTAL FOR PAPER: 75 MARKS
END
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*P40096A02828*
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How to interpret Mplus output
Message/Author
AMY TIAN-FOREMAN posted on Wednesday, December 09, 2009 - 2:37 pm
Estimate S.E. Est./S.E. P-Value
Thresholds
LT50\$1 2.149 0.080 26.883 0.000
THREAT\$1 1.499 0.113 13.304 0.000
INJURE\$1 4.413 0.327 13.512 0.000
BLDG\$1 4.935 0.397 12.426 0.000
Dear Dr. Muthen:
I am new to Mplus, and have several questions regarding factor loading would like to ask you.
1) How do I calculate AVE/Composite reliability (CR) using factor loadings and errors from Mplus output? I used the formula AVE= sum of squared factor loading/(sum of squared factor loading + sum of standard errors), and CR = Squared sum of facter loadings/(squared sum of factor loadings + sum of standard errors). However my AVE and CR is unbelievably high (over 0.98 for all my scales). Using the example that I post above (from Muthén, (2001) paper posted on Mplus website), the AVE is 0.982, and CR is 0.995. Aint they almost too high? Or is this not the right way to interpret Mplus outputs?
2) again using the output above. if I use the statistic for item LT50\$1, then Est./S.E. is 2.149/0.080=26.863 instead of 26.883 as shown in the output. So, please please help me to understand this.
3) why are these standard estimate higher than 1.0??
Sorry for posting so many questions, I would be so grateful if you can help.
Best wishes
Amy
Bengt O. Muthen posted on Wednesday, December 09, 2009 - 6:06 pm
You are looking at threshold estimates, not loading estimates.
1) I don't know what "AVE/Composite reliability" is.
2) The difference is due to your computations using only 3 decimals.
AMY TIAN-FOREMAN posted on Thursday, December 10, 2009 - 10:02 am
Dear Bengt:
Thank you very much for your reply.
Please tell me where to look for factor loadings..if my outputs are as following:
Estimate S.E. Est./S.E. P-Value
F2 BY
PS1 0.551 0.021 26.177 0.000
PS2 0.551 0.021 25.928 0.000
PS3 0.551 0.021 25.975 0.000
PS4 0.518 0.022 23.429 0.000
then factor loading for PS1 is 0.551, isnt it??
Linda K. Muthen posted on Thursday, December 10, 2009 - 11:34 am
The second column are the parameter estimates, in this case factor loadings.
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Wouldn't the sum of the two ranges be (''a''+''b'', ''a''+''b'')? Modulo whether they are closed or open bounds, of course (my argument doesn't depend on that). I can't see any logical reason for assuming otherwise, as the “real” values of ''a'' and ''b'' could be anywhere from the lower bound to the upper bound; that's got to be what the range means, with appropriate infinitesimals in the open bound case, of course. (Yes, this does mean that the error will increase as more operations are performed; this is a normal feature of calculations with physical quantities if I'm remembering my Physics-101 class right.) —[[User:Dkf|Donal Fellows]] 22:52, 18 August 2009 (UTC)
: I am not sure, if this answers your question. Addition of intervals is defined as [''a'', ''b''] + [''c'', ''d''] = [''a'' + ''c'', ''b'' + ''d'']. With +↓ and +↑ operations its machine implementation becomes [''a'' +↓ ''c'', ''b'' +↑ ''d'']. As for error (rather uncertainty) of the result obtained using interval arithmetic, it is not a simple issue. The result's width grows under the assumption of the worst case scenario. That is, assuming all intervals involved in computations independent. The intermediate results are often dependent, so the inputs, etc. Therefore in many cases the result can be improved in terms of interval width. A simple illustration of the case. Consider interval expression [1, 2] x [1, 2]. According to the interval arithmetic, the result is [1, 4]. But that is when [1, 2] and [1, 2] are fully independent. If they were ''same'' (correlated), we could multiply [1, 2] by 2, and get a two-fold narrower result [1, 2] * 2 = [2, 4]. --[[User:Dmitry-kazakov|Dmitry-kazakov]] 08:46, 19 August 2009 (UTC)
I misunderstood the original task, so my comments before are correct but distinctly off-topic. :-) According to the documentation I can scare up with Google on IEEE rounding, the normal mode is round-to-nearest as that minimizes the error (to 0.5 ULP) given that its a single-valued result. The documentation also says that it is a really bad idea to call the standard math functions with the rounding mode set to anything else. Given that, the only sane method of implementing this task (assuming that the language has no portable way to set the rounding mode for the duration of the computation, e.g., [[C]] doesn't) is to compute and then widen with `nextafter`. —[[User:Dkf|Donal Fellows]] 08:56, 19 August 2009 (UTC)
== Error propagation? ==
Wouldn't it be more sensible to include error propagation in the task? Just two exact number adding to a sum +/- machine epsilon is of pretty limited use. --[[User:Ledrug|Ledrug]] 04:03, 10 June 2011 (UTC)
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### An Introduction to Analytic Graph Theory
Let $\cS$ be a set of graphs or a set of objects associated with some specific graph such that there is a symmetric adjacency relation defined on $\cS$. Two elements $S$ and $S'$ of $S$ are connected in $\cS$ if there exists a sequence $S = S_0, S_1, S_2, \cdots , S_k = S'$ of elements of $\cS$ such that $S_i$ and $S_{i+1}$ are adjacent for $i = 0, 1, \cdots , k - 1$. The minimum $k$ for which such a sequence exists is the distance $d(S, S')$ between $S$ and $S'$. If every pair of elements of $\cS$ are connected, then $\cS$ is connected. If $\cS$ is connected, then $(\cS, d)$ is a metric space. A nonnegative integer-valued function $f$ defined on $\cS$ is defined to be continuous on $\cS$ if $|f(S) - f(S')| \leq 1$ for every two adjacent elements $S$ and $S'$ of $\cS$. We consider various functions, continuous and noncontinuous, defined on such metric spaces. For each such metric space $(\cS, d)$, there is an associated metric graph whose vertices are the elements of the metric space and where two vertices of the metric graph are adjacent if and only if the corresponding elements are adjacent. These metric graphs are studied as well.
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# Algebraic Number Theory
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## Transcription
1 Algebraic Number Theory 1. Algebraic prerequisites 1.1. General Definition. For a field F define the ring homomorphism Z F by n n 1 F. Its kernel I is an ideal of Z such that Z/I is isomorphic to the image of Z in F. The latter is an integral domain, so I is a prime ideal of Z, i.e. I = 0 or I = pz for a prime number p. In the first case F is said to have characteristic 0, in the second characteristic p. Definition Lemma. Let F be a subfield of a field L. An element a L is called algebraic over F if one of the following equivalent conditions is satisfied: (i) f(a) = 0 for a non-zero polynomial f(x) F [X]; (ii) elements 1, a, a 2,... are linearly dependent over F ; (iii) F -vector space F [a] = { a i a i : a i F } is of finite dimension over F ; (iv) F [a] = F (a). Proof. (i) implies (ii): if f(x) = n i=0 c ix i, c 0, c n 0, then c i a i = 0. (ii) implies (iii): if n i=0 c ia i = 0, c n 0, then a n = n 1 i=0 c 1 n c i a i, a n+1 = a a n = n 1 i=0 c 1 n c i a i+1 = n 2 i=0 c 1 n c i a i+1 + c 1 n 1 n c n 1 i=0 c 1 n c i a i, etc. (iii) implies (iv): for every b F [a] we have F [b] F [a], hence F [b] is of finite dimension over F. So if b F, there are d i such that d i b i = 0, and d 0 0. Then 1/b = d 1 n 0 i=1 d ib i 1 and hence 1/b F [b] F [a]. (iv) implies (i): if 1/a is equal to e i a i, then a is a root of e i X i+1 1. For an element a algebraic over F denote by f a (X) F [X] the monic polynomial of minimal degree such that f a (a) = 0. This polynomial is irreducible: if f a = gh, then g(a)h(a) = 0, so g(a) = 0 or h(a) = 0, contradiction. It is called the monic irreducible polynomial of a over F. For example, f a (X) is a linear polynomial iff a F.
2 2 Alg number theory Lemma. Define a ring homomorphism F [X] L, g(x) g(a). Its kernel is the principal ideal generated by f a (X) and its image is F (a), so F [X]/(f a (X)) F (a). Proof. The kernel consists of those polynomials g over F which vanish at a. Using the division algorithm write g = f a h + k where k = 0 or the degree of k is smaller than that of f a. Now k(a) = g(a) f a (a)h(a) = 0, so the definition of f a implies k = 0 which means that f a divides g. Definition. A field L is called algebraic over its subfield F if every element of L is algebraic over F. The extension L/F is called algebraic. Definition. Let F be a subfield of a field L. The dimension of L as a vector space over F is called the degree L : F of the extension L/F. If a is algebraic over F then F (a) : F is finite and it equals the degree of the monic irreducible polynomial f a of a over F. Transitivity of the degree L : F = L : M M : F follows from the observation: if α i form a basis of M over F and β j form a basis of L over M then α i β j form a basis of L over F. Every extension L/F of finite degree is algebraic: if β L, then F (β) : F L : F is finite, so by (iii) above β is algebraic over F. In particular, if α is algebraic over F then F (α) is algebraic over F. If α, β are algebraic over F then the degree of F (α, β) over F does not exceed the product of finite degrees of F (α)/f and F (β)/f and hence is finite. Thus all elements of F (α, β) are algebraic over F. An algebraic extension F ({a i }) of F is is the composite of extensions F (a i ), and since a i is algebraic F (a i ) : F is finite, thus every algebraic extension is the composite of finite extensions Definition. An extension F of Q of finite degree is called an algebraic number field, the degree F : Q is called the degree of F. Examples. 1. Every quadratic extension L of Q can be written as Q( e) for a square-free integer e. Indeed, if 1, α is a basis of L over Q, then α 2 = a 1 + a 2 α with rational a i, so α is a root of the polynomial X 2 a 2 X a 1 whose roots are of the form a 2 /2 ± d/2 where d Q is the discriminant. Write d = f/g with integer f, g and notice that Q( d) = Q( dg 2 ) = Q( fg). Obviously we can get rid of all square divisors of fg without changing the extension Q( fg). 2. Cyclotomic extensions Q m = Q(ζ m ) of Q where ζ m is a primitive m th root of unity. If p is prime then the monic irreducible polynomial of ζ p over Q is X p = (X p 1)/(X 1) of degree p 1.
3 Definition. Let two fields L, L contain a field F. A homo(iso)morphism σ: L L such that σ F is the identity map is called a F -homo(iso)morphism of L into L. The set of all F -homomorphisms from L to L is denoted by Hom F (L, L ). Notice that every F -homomorphism is injective: its kernel is an ideal of F and 1 F does not belong to it, so the ideal is the zero ideal. In particular, σ(l) is isomorphic to L. The set of all F -isomorphisms from L to L is denoted by Iso F (L, L ). Two elements a L, a L are called conjugate over F if there is a F -homomorphism σ such that σ(a) = a. If L, L are algebraic over F and isomorphic over F, they are called conjugate over F. Lemma. (i) Any two roots of an irreducible polynomial over F are conjugate over F. (ii) An element a is conjugate to a over F iff f a = f a. (iii) The polynomial f a (X) is divisible by (X a i ) in L[X], where a i are all distinct conjugate to a elements over F, L is the field F ({a i }) generated by a i over F. Proof. (i) Let f(x) be an irreducible polynomial over F and a, b be its roots in a field extension of F. Then f a = f b = f and we have an F -isomorphism F (a) F [X]/(f a (X)) = F [X]/(f b (X)) F (b), a b and therefore a is conjugate to b over F. (ii) 0 = σf a (a) = f a (σa) = f a (a ), hence f a = f a. If f a = f a, use (i). (iii) If a i is a root of f a then by the division algorithm f a (X) is divisible by X a i in L[X] Definition. A field is called algebraically closed if it does not have algebraic extensions. Theorem (without proof). Every field F has an algebraic extension C which is algebraically closed. The field C is called an algebraic closure of F. Every two algebraic closures of F are isomorphic over F. Example. The field of rational numbers Q is contained in algebraically closed field C. The maximal algebraic extension Q a of Q is obtained as the subfield of complex numbers which contains all algebraic elements over Q. The field Q a is algebraically closed: if α C is algebraic over Q a then it is a root of a non-zero polynomial with finitely many coefficients, each of which is algebraic over Q. Therefore α is algebraic over the field M generated by the coefficients. Then M(α)/M and M/Q are of finite degree, and hence α is algebraic over Q, i.e. belongs to Q a. The degree Q a : Q is infinite, since Q a : Q Q(ζ p ) : Q = p 1 for every prime p. The field Q a is is much smaller than C, since its cardinality is countable whereas the cardinality of complex numbers is uncountable). 3
4 4 Alg number theory Everywhere below we denote by C an algebraically closed field containing F. Elements of Hom F (F (a), C) are in one-to-one correspondence with distinct roots of f a (X) F [X]: for each such root a i, as in the proof of (i) above we have σ: F (a) C, a a i ; and conversely each such σ Hom F (F (a), C) maps a to one of the roots a i Galois extensions Definition. A polynomial f(x) F [X] is called separable if all its roots in C are distinct. Recall that if a is a multiple root of f(x), then f (a) = 0. So a polynomial f is separable iff the polynomials f and f don t have common roots. Examples of separable polynomials. Irreducible polynomials over fields of characteristic zero, irreducible polynomials over finite fields. Proof: if f is an irreducible polynomial over a field of characteristic zero, then its derivative f is non-zero and has degree strictly smaller than f ; and so if f has a multiple root, than a g.c.d. of f and f would be of positive degree strictly smaller than f which contradicts the irreducibility of f. For the case of irreducible polynomials over finite fields see section 1.3. Definition. Let L be a field extension of F. An element a L is called separable over F if f a (X) is separable. The extension L/F is called separable if every element of L is separable over F. Example. separable. Every algebraic extension of a field of characteristic zero or a finite field is Lemma. Let M be a field extension of F and L be a finite extension of M. Then every F -homomorphism σ: M C can be extended to an F -homomorphism σ : L C. Proof. Let a L \ M and f a (X) = c i X i be the minimal polynomial of a over M. Then (σf a )(X) = σ(c i )X i is irreducible over σm. Let b be its root. Then σf a = f b. Consider an F -homomorphism φ: M[X] C, φ( a i X i ) = σ(a i )b i. Its image is (σm)(b) and its kernel is generated by f a. Since M[X]/(f a (X)) M(a), φ determines an extension σ : M(a) C of σ. Since L : M(a) < L : M, by induction σ can be extended to an F -homomorphism σ : L C such that σ M = σ Theorem. Let L be a finite separable extension of F of degree n. Then there exist exactly n distinct F -homomorphisms of L into C, i.e. Hom F (L, C) = L : F.
5 Proof. The number of distinct F -homomorphisms of L into C is n is valid for any extension of degree n. To prove this, argue by induction on L : F and use the fact that every F -homomorphism σ: F (a) C sends a to one of roots of f a (X) and that root determines σ completely. To show that there are n distinct F -homomorphisms for separable L/F consider first the case of L = F (a). From separability we deduce that the polynomial f a (X) has n distinct roots a i which give n distinct F -homomorphisms of L into C: a a i. Now argue by induction on degree. For a L \ F consider M = F (a). There are m = M : F distinct F -homomorphisms σ i of M into C. Let σ i : L C be an extension of σ i which exists according to By induction there are n/m distinct F (σ i (a))-homomorphisms τ ij of σ i (L) into C. Now τ ij σ i are distinct F -homomorphisms of L into C Proposition. Every finite subgroup of the multiplicative group F of a field F is cyclic. Proof. Denote this subgroup by G, it is an abelian group of finite order. From the standard theorem on the stucture of finitely generated abelian groups we deduce that G Z/m 1 Z Z/m r Z where m 1 divides m 2, etc. We need to show that r = 1 (then G is cyclic). If r > 1, then let a prime p be a divisor of m 1. The cyclic group Z/m 1 Z has p elements of order p and similarly, Z/m 2 Z has p elements of order p, so G has at least p 2 elements of order p. However, all elements of order p in G are roots of the polynomial X p 1 which over the field F cannot have more than p roots, a contradiction. Thus, r = Theorem. Let F be a field of characteristic zero or a finite field. Let L be a finite field extension of F. Then there exists an element a L such that L = F (a) = F [a]. Proof. If F is of characteristic 0, then F is infinite. By there are n = L : F distinct F -homomorphisms σ i : L C. Put V ij = {a L : σ i (a) = σ j (a)}. Then V ij are proper F -vector subspaces of L for i j of dimension < n, and since F is infinite, there union i j V ij is different from L. Then there is a L \ ( V ij ). Since the set {σ i (a)} is of cardinality n, the minimal polynomial of a over F has at least n distinct roots. Then F (a) : F n = L : F and hence L = F (a). If L is finite, then L is cyclic by Let a be any of its generators. Then L = F (a) Definition. An algebraic extension L of F (inside C ) is called the splitting field of polynomials f i if L = F ({a ij }) where a ij are all the roots of f i. An algebraic extension L of F is called a Galois extension if L is the splitting field of some separable polynomials f i over F. 5
6 6 Alg number theory Example. Let L be a finite extension of F such that L = F (a). Then L/F is a Galois extension if the polynomial f a (X) of a over F has deg f a distinct roots in L. So quadratic extensions of Q and cyclotomic extensions of Q are Galois extensions Lemma. Let L be the splitting field of an irreducible polynomial f(x) F [X]. Then σ(l) = L for every σ Hom F (L, C). Proof. σ permutes the roots of f(x). Thus, σ(l) = F (σ(a 1 ),..., σ(a n )) = L Theorem. A finite extension L of F is a Galois extension iff σ(l) = L for every σ Hom F (L, C) and Hom F (L, L) = L : F. The set Hom F (L, L) equals to the set Iso F (L, L) which is a finite group with respect to the composite of field isomorphisms. This group is called the Galois group Gal(L/F ) of the extension L/F. Sketch of the proof. Let L be a Galois extension of F. The right arrow follows from the previous proposition and properties of separable extensions. On the other hand, if L = F ({b i }) and σ(l) = L for every σ Hom F (L, C) then σ(b i ) belong to L and L is the splitting field of polynomials f bi (X). If Hom F (L, L) = L : F then one can show by induction that each of f bi (X) is separable. Now suppose we are in the situation of Then L = F (a) for some a L. L is the splitting field of some polynomials f i over F, and hence L is the splitting field of their product. By and induction we have σl = L. Then L = F (a i ) for any root a i of f a, and elements of Hom F (L, L) correspond to a a i. Therefore Hom F (L, L) = Iso F (L, L). Its elements correspond to some permutations of the set {a i } of all roots of f a (X) Theorem (without proof). Let L/F be a finite Galois extension and M be an intermediate field between F and L. Then L/M is a Galois extension with the Galois group Gal(L/M) = {σ Gal(L/F ) : σ M = id M }. For a subgroup H of Gal(L/F ) denote L H = {x L : σ(x) = x for all σ H }. This set is an intermediate field between L and F Main theorem of Galois theory (without proof). Let L/F be a finite Galois extension with Galois group G = Gal(L/F ). Then H L H is a one-to-one correspondence between subgroups H of G and subfields of L which contain F ; the inverse map is given by M Gal(L/M). We have Gal(L/M) = H.
7 7 Normal subgroups H of G correspond to Galois extensions M/F and Gal(M/F ) G/H Finite fields Every finite field F has positive characteristic, since the homomorphism Z F is not injective. Let F be of prime characteristic p. Then the image of Z in F can be identified with the finite field F p consisting of p elements. If the degree of F/F p is n, then the number of elements in F is p n. By the group F is cyclic of order p n 1, so every non-zero element of F is a root of the polynomial X pn 1 1. Therefore, all p n elements of F are all p n roots of the polynomial f n (X) = X pn X. The polynomial f n is separable, since its derivative in characteristic p is equal to p n X pn 1 1 = 1. Thus, F is the splitting field of f n over F p. We conclude that F/F p is a Galois extension of degree n = F : F p. Lemma. The Galois group of F/F p is cyclic of order n: it is generated by an automorphism φ of F called the Frobenius automorphism: φ(x) = x p for all x F. Proof. φ m (x) = x pm = x for all x F iff n m. On the other hand, for every n 1 the splitting field of f n over F p is a finite field consisiting of p n elements. Thus, Theorem. For every n there is a unique (up to isomorphism) finite field F p n consisting of p n elements; it is the splitting field of the polynomial f n (X) = X pn X. The finite extension F p nm/f p n is a Galois extension with cyclic group of degree m generated by the Frobenius automorphism φ n : x x pn. Lemma. Let g(x) be an irreducible polynomial of degree m over a finite field F p n. Then g(x) divides f nm (X) and therefore is a separable polynomial. Proof. Let a be a root of g(x). Then F p n(a)/f p n is of degree m, so F p n(a) = F p nm. Since a is a root of f nm (X), g divides f nm. The latter is separable and so is g.
8 8 Alg number theory 2. Integrality 2.1. Integrality over rings Proposition Definition. Let B be a ring and A its subring. An element b B is called integral over A if it satisfies one of the following equivalent conditions: (i) there exist a i A such that f(b) = 0 where f(x) = X n +a n 1 X n 1 + +a 0 ; (ii) the subring of B generated by A and b is an A-module of finite type; (iii) there exists a subring C of B which contains A and b and which is an A-module of finite type. Proof. (i) (ii): note that the subring A[b] of B generated by A and b coincides with the A-module M generated by 1,..., b n 1. Indeed, b n+j = a 0 b j b n+j 1 and by induction b j M. (ii) (iii): obvious. (iii) (i): let C = c 1 A+ +c m A. Then bc i = j a ijc j, so j (δ ijb a ij )c j = 0. Denote by d the determinant of M = (δ ij b a ij ). Note that d = f(b) where f(x) A[X] is a monic polynomial. From linear algebra we know that de = M M where M is the adjugate matrix to M and E is the identity matrix of the same order of that of M. Denote by C the column consisting of c j. Now we get MC = 0 implies M MC = 0 implies dec = 0 implies dc = 0. Thus dc j = 0 for all 1 j m. Every c C is a linear combination of c j. Hence dc = 0 for all c C. In particular, d1 = 0, so f(b) = d = 0. Examples. 1. Every element of A is integral over A. 2. If A, B are fields, then an element b B is integral over A iff b is algebraic over A. 3. Let A = Z, B = Q. A rational number r/s with relatively prime r and s is integral over Z iff (r/s) n + a n 1 (r/s) n a 0 = 0 for some integer a i. Multiplying by s n we deduce that s divides r n, hence s = ±1 and r/s Z. Hence integral in Q elements over Z are just all integers. 4. If B is a field, then it contains the field of fractions F of A. Let σ Hom F (B, C) where C is an algebraically closed field containing B. If b B is integral over A, then σ(b) σ(b) is integral over A. 5. If b B is a root of a non-zero polynomial f(x) = a n X n + A[X], f(b) = 0 and g(a n b) = 0 for g(x) = X n + a n 1 X n a n 1 n a 0, then a n 1 n
9 g(a n X) = a n 1 n f(x). Hence a n b is integral over A. Thus, for every algebraic over A element b of B there is a non-zero a A such that ab is integral over A Corollary. Let A be a subring of an integral domain B. Let I be a non-zero A-module of finite type, I B. Let b B satisfy the property bi I. Then b is integral over A. Proof. Indeed, as in the proof of (iii) (i) we deduce that dc = 0 for all c I. Since B is an integral domain, we deduce that d = 0, so d = f(b) = Proposition. Let A be a subring of a ring B, and let b i B be such that b i is integral over A[b 1,..., b i 1 ] for all i. Then A[b 1,..., b n ] is an A-module of finite type. Proof. Induction on n. n = 1 is the previous proposition. If C = A[b 1,..., b n 1 ] is an A-module of finite type, then C = m i=1 c ia. Now by the previous proposition C[b n ] is a C-module of finite type, so C[b n ] = l j=1 d jc. Thus, C[b n ] = i,j d jc i A is an A-module of finite type Corollary 1. If b 1, b 2 B are integral over A, then b 1 + b 2, b 1 b 2, b 1 b 2 are integral over A. Certainly b 1 /b 2 isn t necessarily integral over A. Corollary 2. The set B of elements of B which are integral over A is a subring of B containing A. Definition. B is called the integral closure of A in B. If A is an integral domain and B is its field of fractions, B is called the integral closure of A. A ring A is called integrally closed if A is an integral domain and A coincides with its integral closure in its field of fractions. Let F be an algebraic number field. The integral closure of Z in F is called the ring O F of (algebraic) integers of F. Examples. 1. A UFD is integrally closed. Indeed, if x = a/b with relatively prime a, b A is a root of polynomial f(x) = X n + + a 0 A[X], then b divides a n, so b is a unit of A and x A. In particular, the integral closure of Z in Q is Z. 2. O F is integrally closed (see below in 2.1.6) Lemma. Let A be integrally closed. Let B be a field. Then an element b B is integral over A iff the monic irreducible polynomial f b (X) F [X] over the fraction field F of A has coefficients in A. 9
10 10 Alg number theory Proof. Let L be a finite extension of F which contains B and all σ(b) for all F -homomorphisms from B to an algebraically closed field C. Since b L is integral over A, σ(b) L is integral over A for every σ. Then f b (X) = (X σ(b)) has coefficients in F which belong to the ring generated by A and all σ(b) and therefore are integral over A. Since A is integrally closed, f b (X) A[X]. If f b (X) A[X] then b is integral over A by Examples. 1. Let F be an algebraic number field. Then an element b F is integral iff its monic irreducible polynomial has integer coefficients. For example, d for integer d is integral. If d 1 mod 4 then the monic irreducible polynomial of (1 + d)/2 over Q is X 2 X + (1 d)/4 Z[X], so (1 + d)/2 is integral. Note that d belongs to Z[(1 + d)/2], and hence Z[ d] is a subring of Z[(1 + d)/2]. Thus, the integral closure of Z in Q( d) contains the subring Z[ d] and the subring Z[(1 + d)/2] if d 1 mod 4. We show that there are no other integral elements. An element a + b d with rational a and b 0 is integral iff its monic irreducible polynomial X 2 2aX +(a 2 db 2 ) belongs to Z[X]. Therefore 2a, 2b are integers. If a = (2k +1)/2 for an integer k, then it is easy to see that a 2 db 2 Z iff b = (2l +1)/2 with integer l and (2k + 1) 2 d(2l + 1) 2 is divisible by 4. The latter implies that d is a quadratic residue mod 4, i.e. d 1 mod 4. In turn, if d 1 mod 4 then every element (2k + 1)/2 + (2l + 1) d/2 is integral. Thus, integral elements of Q( d) are equal to { Z[ d] if d 1 mod 4 Z[(1 + d)/2] if d 1 mod 4 2. O Q m is equal to Z[ζ m ] (see section 2.4) Definition. B is said to be integral over A if every element of B is integral over A. If B is of characteristic zero, its elements integral over Z are called integral elements of B. Lemma. If B is integral over A and C is integral over B, then C is integral over A. Proof. Let c C be a root of the polynomial f(x) = X n + b n 1 X n b 0 with b i B. Then c is integral over A[b 0,..., b n 1 ]. Since b i B are integral over A, proposition implies that A[b 0,..., b n 1, c] is an A-module of finite type. From we conclude that c is integral over A. Corollary. O F is integrally closed
11 11 Proof. An element of F integral over O F lemma. is integral over Z due to the previous Proposition. Let B be an integral domain and A be its subring such that B is integral over A. Then B is a field iff A is a field. Proof. If A is a field, then A[b] for b B \0 is a vector space of finite dimension over A, and the A-linear map ϕ: A[b] A[b], ϕ(c) = bc is injective, therefore surjective, so b is invertible in B. If B is a field and a A\0, then the inverse a 1 B satisfies a n +a n 1 a n a 0 = 0 with some a i A. Then a 1 = a n 1 a 0 a n 1, so a 1 A Norms and traces Definition. Let A be a subring of a ring B such that B is a free A-module of finite rank n. For b B its trace Tr B/A (b), norm N B/A (b) and characteristic polynomial g b (X) are the trace, the norm and the characteristic polynomial of the linear operator m b : B B, m b (c) = bc. In other words, if M b is a matrix of the operator m b with respect to a basis of B over A, then g b (X) = det(xe M b ), Tr B/A (b) = Tr M b, N B/A = det M b. If g b (X) = X n +a n 1 X n 1 + +a 0 then from the definition a n 1 = Tr B/A (b), a 0 = ( 1) n N B/A (b) First properties. for a A. Tr(b + b ) = Tr(b) + Tr(b ), Tr(ab) = a Tr(b), Tr(a) = na, N(bb ) = N(b)N(b ), N(ab) = a n N(b), N(a) = a n Everywhere below in this section F is either a finite field of a field of characteristic zero. Then every finite extension of F is separable. Proposition. Let L be an algebraic extension of F of degree n. Let b L and b 1,..., b n be roots of the monic irreducible polynomial of b over F each one repeated L : F (b) times. Then the characteristic polynomial g b (X) of b with respect to L/F is (X b i ), and Tr L/F (b) = b i, N L/F (b) = b i. Proof. If L = F (b), then use the basis 1, b,..., b n 1 to calculate g b. Let f b (X) = X n + c n 1 X n c 0 be the monic irreducible polynomial of b over F, then the
12 12 Alg number theory matrix of m b is M b = c 0 c 1 c 2... c n 1 Hence g b (X) = det(xe M b ) = f b (X) and det M b = b i, Tr M b = b i. In the general case when F (b) : F = m < n choose a basis ω 1,..., ω n/m of L over F (b) and take ω 1,..., ω 1 b m 1, ω 2,..., ω 2 b m 1,... as a basis of L over F. The matrix M b is a block matrix with the same block repeated n/m times on the diagonal and everything else being zero. Therefore, g b (X) = f b (X) L:F (b) where f b (X) is the monic irreducible polynomial of b over F. Example. so. Let F = Q, L = Q( d) with square-free integer d. Then g a+b d (X) = (X a b d)(x a + b d) = X 2 2aX + (a 2 db 2 ), Tr Q( d)/q (a + b d) = 2a, N Q( d)/q (a + b d) = a 2 db 2. In particular, an integer number c is a sum of two squares iff c N Q( 1)/Q O Q( 1). More generally, c is in the form a 2 db 2 with integer a, b and square-free d not congruent to 1 mod 4 iff c N Q( d)/q Z[ d] Corollary 1. Let σ i be distinct F -homomorphisms of L into C. Then Tr L/F (b) = σi b, N L/F (b) = σ i (b). Proof. In the previous proposition b i = σ i (b). Corollary 2. Let A be an integral domain, F be its field of fractions. Let L be an extension of F of finite degree. Let A be the integral closure of A in F. Then for an integral element b L over A g b (X) A [X] and Tr L/F (b), N L/F (b) belong to A. Proof. All b i are integral over A. Corollary 3. If, in addition, A is integrally closed, then Tr L/F (b), N L/F (b) A. Proof. Since A is integrally closed, A F = A Lemma. Let F be a finite field of a field of characteristic zero. If L is a finite extension of F and M/F is a subextension of L/F, then the following transitivity
13 13 property holds Tr L/F = Tr M/F Tr L/M, N L/F = N M/F N L/M. Proof. Let σ 1,..., σ m be all distinct F -homomorphisms of M into C ( m = M : F ). Let τ 1,..., τ n/m be all distinct M -homomorphisms of L into C ( n/m = L : M ). The field τ j (L) is a finite extension of F, and by there is an element a j C such that τ j (L) = F (a j ). Let E be the minimal subfield of C containing M and all a j. Using extend σ i to σ i : E C. Then the composition σ i τ j: L C is defined. Note that σ i τ j = σ i 1 τ j1 implies σ i = σ i τ j M = σ i 1 τ j1 M = σ i1, so i = i 1, and then j = j 1. Hence σ i τ j for 1 i m, 1 j n/m are all n distinct F -homomorphisms of L into C. By Corollary 3 in N M/F (N L/M (b)) = N M/F ( τ j (b)) = σ i( τ j (b)) = (σ i τ j )(b) = N L/F (b). Similar arguments work for the trace Integral basis Definition. Let A be a subring of a ring B such that B is a free A-module of rank n. Let b 1,..., b n B. Then the discriminant D(b 1,..., b n ) is defined as det(tr B/A (b i b j )) Proposition. If c i B and c i = a ij b j, a ij A, then D(c 1,..., c n ) = (det(a ij )) 2 D(b 1,..., b n ). Proof. (c i ) t = (a ij )(b j ) t, (c k c l ) = (c k ) t (c l ) = (a ki )(b i b j )(a lj ) t, (Tr(c k c l )) = (a ki )(Tr(b i b j ))(a lj ) t Definition. The discriminant D B/A of B over A is the principal ideal of A generated by the discriminant of any basis of B over A Proposition. Let D B/A 0. Let B be an integral domain. Then a set b 1,..., b n is a basis of B over A iff D(b 1,..., b n )A = D B/A. Proof. Let D(b 1,..., b n )A = D B/A. Let c 1,..., c n be a basis of B over A and let b i = j a ijc j. Then D(b 1,..., b n ) = det(a ij ) 2 D(c 1,..., c n ). Denote d = D(c 1,..., c n ). Since D(b 1,..., b n )A = D(c 1,..., c n )A, we get ad(b 1,..., b n ) = d for some a A. Then d(1 a det(a ij ) 2 ) = 0 and det(a ij ) is invertible in A, so the matrix (a ij ) is invertible in the ring of matrices over A. Thus b 1,..., b n is a basis of B over A.
14 14 Alg number theory Proposition. Let F be a finite field or a field of characteristic zero. Let L be an extension of F of degree n and let σ 1,..., σ n be distinct F -homomorphisms of L into C. Let b 1,..., b n be a basis of L over F. Then D(b 1,..., b n ) = det(σ i (b j )) 2 0. Proof. det(tr(b i b j )) = det( k σ k(b i )σ k (b j )) = det((σ k (b i )) t (σ k (b j ))) = det(σ i (b j )) 2. If det(σ i (b j )) = 0, then there exist a i L not all zero such that i a iσ i (b j ) = 0 for all j. Then i a iσ i (b) = 0 for every b L. Let a i σ i(b) = 0 for all b L with the minimal number of non-zero a i A. Assume a 1 0. Let c L be such that L = F (c) (see 1.2.5), then σ 1 (c) σ i (c) for i > 1. We now have a i σ i(bc) = a i σ i(b)σ i (c) = 0. Hence σ 1 (c)( a i σ i(b)) a i σ i(b)σ i (c) = i>1 a i (σ 1(c) σ i (c))σ i (b) = 0. Put a i = a i (σ 1(c) σ i (c)), so a i σ i(b) = 0 with smaller number of non-zero a i than in a i, a contradiction. Corollary. Under the assumptions of the proposition the linear map L Hom F (L, F ): b (c Tr L/F (bc)) between n-dimensional F -vector spaces is injective, and hence bijective. Therefore for a basis b 1,..., b n of L/F there is a dual basis c 1,..., c n of L/F, i.e. Tr L/F (b i c j ) = δ ij. Proof. If b = a i b i, a i F and Tr L/F (bc) = 0 for all c L, then we get equations ai Tr L/F (b i b j ) = 0 this is a system of linear equations in a i with nondegenerate matrix Tr L/F (b i b j ), so the only solution is a i = 0. correspond to f j Hom F (L, F ), f j (b i ) = δ ij. Elements of the dual basis c j Theorem. Let A be an integrally closed ring and F be its field of fractions. Let L be an extension of F of degree n and A be the integral closure of A in L. Let F be of characteristic 0. Then A is an A-submodule of a free A-module of rank n. Proof. Let e 1,..., e n be a basis of F -vector space L. Then due to Example 5 in there is 0 a i A such that a i e i A. Then for a = a i we get b i = ae i A form a basis of L/F. Let c 1,..., c n be the dual basis for b 1,..., b n. Claim: A c i A. Indeed, let c = a i c i A. Then Tr L/F (cb i ) = j a j Tr L/F (c j b i ) = a i A by Now c i A = c i A, since {c i } is a basis of L/F Theorem (on integral basis). Let A be a principal ideal ring and F be its field of fractions of characteristic 0. Let L be an extension of F of degree n. Then the integral closure A of A in L is a free A-module of rank n.
15 15 In particular, the ring of integers O F rank equal to the degree of F. of a number field F is a free Z-module of Proof. The description of modules of finite type over PID and the previous theorem imply that A is a free A-module of rank m n. On the other hand, by the first part of the proof of the previous theorem A contains n A-linear independent elements over A. Thus, m = n. Definition. The discriminant d F of any integral basis of O F is called the discriminant of F. Since every two integral bases are related via an invertible matrix with integer coefficients (whose determinant is therefore ±1 ), implies that d F is uniquely determined Examples. 1. Let d be a square-free integer. By the ring of integers of Q( d) has an integral basis 1, α where α = d if D 1 mod 4 and α = (1 + d)/2 if d 1 mod 4. The discriminant of Q( d) is equal to 4d if d 1 mod 4, and d if d 1 mod 4. To prove this calculate directly D(1, α) using the definitions, or use Let F be an algebraic number field of degree n and let a F be an integral element over Z. Assume that D(1, a,..., a n 1 ) is a square free integer. Then 1, a,..., a n 1 is a basis of O F over Z, so O F = Z[a]. Indeed: choose a basis b 1,..., b n of O F over Z and let {c 1,..., c n } = {1, a,..., a n 1 }. Let c i = aij b j. By we have D(1, a,..., a n 1 ) = (det(a ij ) 2 D(b 1,..., b n ). Since D(1, a,..., a n 1 ) is a square free integer, we get det(a ij ) = ±1, so (a ij ) is invertible in M n (Z), and hence 1, a,..., a n 1 is a basis of O F over Z Example. Let F be of characteristic zero and L = F (b) be an extension of degree n over F. Let f(x) be the minimal polynomial of b over F whose roots are b i. Then f(x) = (X b j ), f (b i ) = j i(b i b j ), N L/F f (b) = i f (σ i b) = i f (b i ). Then D(1, b,..., b n 1 ) = det(b j i )2 = ( 1) n(n 1)/2 i j (b i b j ) = ( 1) n(n 1)/2 N L/F (f (b)). Let f(x) = X n + ax + c. Then b n = ab c, b n 1 = a cb 1
16 16 Alg number theory and so e = f (b) = nb n 1 + a = n( a cb 1 ) + a, b = nc(e + (n 1)a) 1. The minimal polynomial g(y ) of e over F corresponds to the minimal polynomial f(x) of b; it is the numerator of c 1 f( nc(y + (n 1)a) 1 ), i.e. so Hence g(y ) = (Y + (n 1)a) n na(y + (n 1)a) n 1 + ( 1) n n n c n 1. N L/F (f (b)) = g(0)( 1) n = n n c n 1 + ( 1) n 1 (n 1) n 1 a n, D(1, b,..., b n 1 ) = ( 1) n(n 1)/2 (n n c n 1 + ( 1) n 1 (n 1) n 1 a n ). For n = 2 one has a 2 4c, for n = 3 one has 27c 2 4a 3. For example, let f(x) = X 3 + X + 1. It is irreducible over Q. Its discriminant is equal to ( 31), so according to example O F = Z[a] where a is a root of f(x) and F = Q[a] Cyclotomic fields Definition. Let ζ n be a primitive n th root of unity. The field Q(ζ n ) is called the ( n th) cyclotomic field Theorem. Let p be a prime number and z be a primitive p th root of unity. The cyclotomic field Q(ζ p ) is of degree p 1 over Q. Its ring of integers coincides with Z[ζ p ]. Proof. Denote z = ζ p. Let f(x) = (X p 1)/(X 1) = X p Recall that z 1 is a root of the polynomial g(y ) = f(1 + Y ) = Y p p is a p-eisenstein polynomial, so f(x) is irreducible over Q, Q(z) : Q = p 1 and 1, z,..., z p 2 is a basis of the Q-vector space Q(z). Let O be the ring of integers of Q(z). Since the monic irreducible polynomial of z over Q has integer coefficients, z O. Since z 1 is a primitive root of unity, z 1 O. Thus, z is a unit of O. Then z i O for all i Z ( z 1 = z p 1 ). We have 1 z i = (1 z)(1+ +z i 1 ) (1 z)o.
17 Denote by Tr and N the trace and norm for Q(z)/Q. Note that Tr(z) = 1 and since z i for 1 i p 1 are primitive p th roots of unity, Tr(z i ) = 1; Tr(1) = p 1. Hence Tr(1 z i ) = p for 1 i p 1. Furthermore, N(z 1) is equal to the free term of g(y ) times ( 1) p 1, so N(z 1) = ( 1) p 1 p and N(1 z) = (1 z i ) = p, 1 i p 1 since 1 z i are conjugate to 1 z over Q. Therefore pz is contained in the ideal I = (1 z)o Z. If I = Z, then 1 z would be a unit of O and so would be its conjugates 1 z i, which then implies that p as their product would be a unit of O. Then p 1 O Q = Z, a contradiction. Thus, Now we prove another auxiliary result: I = (1 z)o Z = pz. Tr((1 z)o) pz. Indeed, every conjugate of y(1 z) for y O is of the type y i (1 z i ) with appropriate y i O, so Tr(y(1 z)) = y i (1 z i ) I = pz. Now let x = 0 i p 2 a iz i O with a i Q. We aim to show that all a i belong to Z. From the calculation of the traces of z i it follows that Tr((1 z)x) = a 0 Tr(1 z)+ 0<i p 2 a i Tr(z i z i+1 ) = a 0 p and so a 0 p Tr((1 z)o) pz; therefore, a 0 Z. Since z is a unit of O, we deduce that x 1 = z 1 (x a 0 ) = a 1 +a 2 z+ +a p 2 z p 3 O. By the same arguments a 1 Z. Looking at x i = z 1 (x i 1 a i 1 ) O we conclude a i Z for all i. Thus O = Z[z] The discriminant of O/Z is the ideal of Z generated by D(1, z,..., z p 2 ) which by is equal ( 1) (p 1)(p 2)/2 N(f (z)). We have f (z) = pz p 1 /(z 1) = pz 1 /(z 1) and N(f (z)) = N(p)N(z) 1 /N(z 1) = p p 1 ( 1) p 1 /(( 1) p 1 p) = p p 2. Thus, the discriminant of OZ is the principal ideal ( 1) (p 1)(p 2)/2 p p 2 Z = p p 2 Z In general, the extension Q(ζ m )/Q is a Galois extension and elements of the Galois group Gal(Q(ζ m )/Q) are determined by their action on the primitive m th root ζ m of unity: This map induces a group isomorphism σ i : σ(ζ m ) = ζ i m, (i, m) = 1. Gal(Q(ζ m )/Q) (Z/mZ). 17
18 18 Alg number theory One can prove that the ring of integers of Q(ζ m ) is Z(ζ m ).
19 19 3. Dedekind rings 3.1. Noetherian rings Recall that a module M over a ring is called a Noetherian module if one of the following equivalent properties is satisfied: (i) every submodule of M is of finite type; (ii) every increasing sequence of submodules stabilizes; (iii) every nonempty family of submodules contains a maximal element with respect to inclusion. A ring A is called Noetherian if it is a Noetherian A-module. Example. A PID is a Noetherian ring, since every ideal of it is generated by one element. Lemma. Let M be an A-module and N is a submodule of M. Then M is a Noetherian A-module iff N and M/N are. Corollary 1. If N i are Noetherian A-modules, so is n i=1 N i. Corollary 2. Let A be a Noetherian ring and let M be an A-module of finite type. Then M is a Noetherian A-module Proposition. Let A be a Noetherian integrally closed ring. Let K be its field of fractions and let L be a finite extension of K. Let A be the integral closure of A in L. Suppose that K is of characteristic 0. Then A is a Noetherian ring. Proof. According to A is a submodule of a free A-module of finite rank. Hence A is a Noetherian A-module. Every ideal of A is in particular an A-submodule of A. Hence every increasing sequence ideals of A stabilizes and A is a Noetherian ring Example. The ring of integers O F of a number field F is a Noetherian ring. It is a Z-module of rank n where n is the degree of F. Every nonzero element of O F \ {0} factorizes into a product of prime elements and units (not uniquely in general). Indeed, assume the family of principal ideals (a) which are generated by elements O F which are not products of prime elements and units is nonempty and then choose a maximal element (a) in this family. The element a is not a unit, and a is not prime. Hence there is a factorization a = bc with both b, c OF. Then (b), (c) are strictly larger than (a), so b and c are products of prime elements. Then a is, a contradiction.
20 20 Alg number theory 3.2. Dedekind rings Definition. An integral domain A is called a Dedekind ring if (i) A is a Noetherian ring; (ii) A is integrally closed; (iii) every non-zero prime ideal of A is maximal. Example. Every principal ideal domain A is a Dedekind ring. Proof: for (i) see and for (ii) see If (a) is a prime ideal and (a) (b) A, then b isn t a unit of A and b divides a. Write a = bc. Since (a) is prime, either b or c belongs to (a). If b does then (a) = (b). If b doesn t, then c must belong to (a), so c = ad for some d A, and a = bc = bda which means that b is a unit of A, a contradiction. Thus, property (iii) is satisfied as well Lemma. Let A be an integral domain. Let K be its field of fractions and let L be a finite extension of K. Let B be the integral closure of A in L. Let P be a non-zero prime ideal of B. Then P A is a non-zero prime ideal of A. Proof. Let P be a non-zero prime ideal of B. Then P A A, since otherwise 1 P A and hence P = B. If c, d A and cd P A, then either c P A or d P A. Hence P A is a prime ideal of A. Let b P, b 0. Then b satisfies a polynomial relation b n +a n 1 b n 1 + +a 0 = 0 with a i A. We can assume that a 0 0. Then a 0 = (b n + + a 1 b) A P, so P A is a non-zero prime ideal of A Theorem. Let A be a Dedekind ring. Let K be its field of fractions and let L be a finite extension of K. Let B be the integral closure of A in L. Suppose that K is of characteristic 0. Then B is a Dedekind ring. Proof. B is Noetherian by It is integrally closed due to By if P is a non-zero proper prime ideal of B, then P A is a non-zero prime ideal of A. Since A is a Dedeking ring, it is a maximal ideal of A. The quotient ring B/P is integral over the field A/(P A). Hence by B/P is a field and P is a maximal ideal of B Example. The ring of integers O F of a number field F is a Dedekind ring.
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# How Long Does It Take To Get Abs?
#### General Rules & Fat Loss Timeline Calculator
Updated: March 9, 2023
For most men and women, the benchmark for being in great shape is having six-pack abs. And if that’s your goal, you’re probably wondering, how long does it take to get abs?
Well, I’ve created a quantifiable timeline for getting abs to answer that question. Next, you’ll see how long it will take to reach your goal based on your starting point, gender, and how defined you want to be.
## Understanding Body Fat
Before we get into how long it takes to get abs, it helps to understand the different body fat levels for men and women.
Generally, men have less body fat than women. That’s because women have more “essential” fat for reproduction.
However, men tend to store more fat in their midsections. So, by comparison, men have to get leaner to have abs.
Men Women
Obese >30% >39%
Overweight 24-30% 32-39%
Average 18-24% 25-32%
Athletic 10-18% 18-25%
Very Lean <10% <18%
### How Lean Do You Need To Be To Have Abs?
You should be able to see visible abs at around 17% body fat for men and 24% for women. Remember, they won’t be well-defined, and you may only see the top 2 or 4 abs at this point.
For a shredded 6-pack, you’ll need to get below 10% for men and below 18% for women.
## How Long Does It Take To Get Abs?
The amount of time it takes to get abs depends on how much fat you have to lose and how fast you lose it.
First, the fat you must lose is your goal weight subtracted from your starting weight. Of course, you’ll want to lose fat and not muscle.
So the second point is to lose weight at a reasonable rate. Studies suggest that losing 0.5 to 1% of your body weight per week results in fat loss with maximum muscle retention1.
For example, if you weigh 200 lbs, your average rate of fat loss would be 1.5 lbs per week (200*(0.75/100)). With that in mind, it’s easy to get a rough estimate of how long it will take you to get abs.
### How Long Does It Take To Get Abs: Men
#### Obese (>30% Body Fat)
When your body fat is over 30%, you have a long road to getting abs. For example, a 220 lb obese man must burn at least 35 lbs of pure fat.
Frankly, losing that much fat takes several months. Or even a year or more if you’re well into 40% territory.
But don’t be discouraged; those of you in the 30% range could see your abs peaking through in 5 or 6 months. And a 6-pack could be attainable in 8 months.
#### Overweight (24-30% Body Fat)
Guys in this category haven’t fallen too far off the wagon. But you’ll still have to lose 20 to 30 lbs of fat to start seeing your abs.
So you could expect to have visible abs in 4 to 5 months. However, a 6-pack will take 5 to 8 months.
#### Average (18-24% Body Fat)
Next up is the 18-24% category. This is most of you weekday warriors out there who aren’t overweight but aren’t lean either.
For you, abs could be in sight within a month or two. But it’ll be 3 to 5 months if you’re looking for the shredded 6.
#### Athletic (10-18% Body Fat)
Now we’re at the stage where you should be able to see your abs under a layer of fluff, even if it’s only the top 2 or 4.
The good news is, you can shed that last layer of fat and reveal your 6-pack in just 1 to 3 months.
#### Very Lean (<10% Body Fat)
Finally, when you break into the single digits of body fat, you’ll start seeing the clear separations in your abdomen.
And you’ll even have feathering in your obliques. This is the pinnacle of fitness for men.
### How Long Does It Take To Get Abs: Women
#### Obese (>39% Body Fat)
For women, body fat over 39% is the longest road to getting abs. You’re looking at losing 40 lbs or more before your belly is flat enough for abs.
But don’t let that scare you off. Because before you get abs, your body will transform significantly.
With six months of hard work, your abs could make an appearance. And you could have a full-blown 6-pack in 8 months.
#### Overweight (32-39% Body Fat)
More and more women are finding themselves in this category. As the stresses of modern life collide with fast food, being overweight is the new normal.
But let’s not settle for ordinary! You could be leaner than average in as little as three and a half months. And you could have a 6-pack in 5 months.
#### Average (25-32% Body Fat)
If you find yourself in the average category, you may still be unhappy with your appearance. But the good news is, a little work will go a long way for you.
You could start seeing abs with just 6 to 8 weeks of dedication. And you could see all 6 in as little as three months.
#### Athletic (18-25% Body Fat)
If you’re in the athletic category, you’re already leaner than the vast majority of women. You should have some ab definition if you’re at 24% body fat or lower.
And if you want a super toned 6-pack, you could get there in 1 to 3 months. Depending on your starting point.
#### Very Lean (<18% Body Fat)
For women who want that fitness model look, your target is 17% body fat. You should have a tight, flat belly at this level of leanness. And a noticeable 6-pack to boot.
## How Long Does It Take To Get Abs: Calculator
With the information above, you should have a good idea of how long it will take you to get abs. But what if you want an even more precise target?
For that, look no further than my “how long does it take to get abs” calculator. With this tool, you can see how many weeks it will take you to get abs.
Plus, you get your target weight, how much fat you have to lose, and your weekly fat loss rate.
### how long does it take to get abs – calculator
Note: This is an estimate based on reasonable rates of fat loss. Your results depend on your adherence to a strategic fat loss plan.
This calculator gives you an estimated timeline for getting abs. But to reach that goal, you need a strategic diet and workout plan. Click below to create your custom fat loss plan.
### Personalized Fat Loss Meal Plan
See your abs sooner with a personalized nutrition and workout plan built for your body, activity level, and goals! Including a daily meal planner and simple recipes that fit your macros for just \$13.99 per month.
## How Long to Get Abs FAQ
With the information and calculator above, you should have everything you need to determine how long it will take you to get abs. But let me answer some other common questions surrounding the ab timeline.
#### Is it possible to get abs in 30 days?
To have visible abs in 30 days, men must already be below 20% body fat. And to get a shredded six-pack in a month, you need to start from around 122% body fat.
By comparison, women could see abs in 30 days if they already have 26% body fat or lower. And for defined six-pack abs in that timeframe, a woman would need to start from under 20% body fat.
#### Can you get abs in 3 months?
Three months is a much more manageable timeframe for guys to get abs. The average male with 25% body fat could expect to see visible abs after three months of diet and exercise. But you would need an athletic 18% or less to see a peeled midsection in that time.
For the average woman, three months of dieting should also get you visible abs. However, getting rid of all belly fat will take much longer.
#### Can you get abs in 6 months?
A six-month body transformation can be pretty impressive. For example, a man could go from being obese to having visible abs in that amount of time. And even overweight guys with 26% body fat could get shredded in half a year.
Similarly, a woman could theoretically go from 40% body fat to visible abs in 6 months. And woman starting with an average amount of body fat could achieve a six-pack in six months.
## More Ab Answers & Information
Now you have a good idea of how long it will take you to get abs. If the process seems a bit daunting or discouraging, don’t worry! I’ve got a bunch more tips and strategies to help you get abs faster.
4 Pack Abs vs 6 Pack Abs
How to Do A Rope Crunch Correctly
Decline Crunches vs. Sit Ups
Cruch Vartiations for Chiseled Obliques
10 Tips for Getting Abs Faster
If your midsection is cramping from all this ab talk, sit back and relax with some of my other fitness content below!
#### 17 Long Head Bicep Exercises For Towering Peaks
Working the outer bicep builds that desirable peak shape. So try these long head bicep exercises to add height to your arms!
#### Leg Press Calf Raise Foot Placement & Proper Form
Save time and build bigger calves by doing calf raises on the leg press. Learn leg press calf raise foot placement & proper form.
#### How to Do Chest Dips vs Tricep Dips for Upper Body Gains
Discover the differences between chest dips vs tricep dips with detailed pictures and a short video tutorial.
Hunger and cravings caused by the leptin hormone make dieting a struggle. Learn how to take control of your appetite and lose weight with less effort.
#### Floor Press vs Bench Press Differences, Benefits, Use Cases
See the differences between floor press vs bench press and when to use each exercise in your chest-building workout routine.
#### Front Squat Made Simple: Complete Exercise Guide, Video, & Alternatives
The front squat is a great exercise. But it can be uncomfortable & hard to learn. See how to master front squats with this complete guide.
#### How to Use Body Recomposition to Transform Your Body
Learn if body recomposition is the right goal for you. And get body recomp diet and workout tips to burn fat and build muscle simultaneously!
Share with your community and get the conversation started!
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## ››Convert kilogram/second to ounce/second
kilogram/second ounce/second
## ››More information from the unit converter
How many kilogram/second in 1 ounce/second? The answer is 0.0283495.
We assume you are converting between kilogram/second and ounce/second.
You can view more details on each measurement unit:
kilogram/second or ounce/second
The SI derived unit for mass flow rate is the kilogram/second.
1 kilogram/second is equal to 35.27399072294 ounce/second.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilograms/second and ounces/second.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kilogram/second to ounce/second
1 kilogram/second to ounce/second = 35.27399 ounce/second
2 kilogram/second to ounce/second = 70.54798 ounce/second
3 kilogram/second to ounce/second = 105.82197 ounce/second
4 kilogram/second to ounce/second = 141.09596 ounce/second
5 kilogram/second to ounce/second = 176.36995 ounce/second
6 kilogram/second to ounce/second = 211.64394 ounce/second
7 kilogram/second to ounce/second = 246.91794 ounce/second
8 kilogram/second to ounce/second = 282.19193 ounce/second
9 kilogram/second to ounce/second = 317.46592 ounce/second
10 kilogram/second to ounce/second = 352.73991 ounce/second
## ››Want other units?
You can do the reverse unit conversion from ounce/second to kilogram/second, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# Quick Answer: Which Does Not Transfer Useful Energy?
## What form of energy transfer does not require matter?
Thermal radiationThermal radiation is one of three ways that thermal energy can be transferred.
The other two ways are conduction and convection, both of which need matter to transfer energy.
Radiation is the only way of transferring thermal energy that doesn’t require matter..
## What are the 3 ways to transfer energy?
Heat can be transferred in three ways: by conduction, by convection, and by radiation.Conduction is the transfer of energy from one molecule to another by direct contact. … Convection is the movement of heat by a fluid such as water or air. … Radiation is the transfer of heat by electromagnetic waves.
## Can you send someone love energy?
You can send out heart energy in such a way that it touches others. … When heart energy touches someone, it affects them. They have a choice to both receive the energy and take it inside, or allow it to pass through them. Even if it passes through them, it touches them in some way.
## How does energy transfer at home?
Heat energy is transferred from homes by conduction through the walls, floor, roof and windows. It is also transferred from homes by convection . For example, cold air can enter the house through gaps in doors and windows, and convection currents can transfer heat energy in the loft to the roof tiles.
## What are three types of convection?
Types of ConvectionNatural convection.Forced convection.
## What affects energy transfer?
Here are the factors that affect the rate of conduction:Temperature difference. The greater the difference in temperature between the two ends of the bar, the greater the rate of thermal energy transfer, so more heat is transferred. … Cross-sectional area. … Length (distance heat must travel). … Time.
## Is energy transferred the same as work done?
When a force is used to move an object, energy is transferred (because the object has moved) and we say that Work is Done. A Force that causes movement in the direction of the force has DONE WORK.
## What prevents the transfer of energy?
How could this energy transfer have been prevented? Using insulating materials is the best way to prevent heat transfer by conduction and convection. Insulating materials are poor conductors of thermal energy and also limit the movement of air in spaces, reducing convection.
## What kind of energy can be transferred?
Answer. Energy can be transferred from one form to another like kinetic energy to potential energy, light energy to heat energy, kinetic energy to electrical energy, light energy to chemical energy .
## What are the two most common forms of unwanted energy?
Devices waste energy for various reasons including friction between their moving parts, electrical resistance, and unwanted sound energy. In most cases, this waste energy is energy that has been shifted into the environment and raises the temperature of the surroundings.
## Can a person absorb energy?
Our bodies are like sponges, able to absorb the energy around us. Olivia Bader-Lee, physician & therapist explains, “This is exactly why there are people who feel uncomfortable when they are in a certain group with a mixture of energy and emotions”.
## Can you touch energy?
Energy is stuff that moves from point to point, as in the Sun radiating light from its atomic fusion. … I cannot “touch” your energy, but the result of your energy would hurt me if you rode into me with your bicycle. Thus, energy is an element of momentum, whether is be momentum of light or momentum of matter.
## What are the 4 energy transfers?
There are 4 ways energy can be transferred;Mechanically – By the action of a force.Electrically – By an electrical current.By radiation – By Light waves or Sound waves.By heating – By conduction, convection or radiation.
## What are the 5 energy transfers?
What are the different ways that energy can be transferred?Conduction: Heat is thermal energy, and in solids it can be transferred by conduction. … Convection: Fluids, that is both gases and liquids, can transfer heat energy by convection. … Radiation: Radiation is different to the other two processes as it doesn’t require particles in its transfer of energy.
## How is energy transferred from one person to another?
Kinetic Energy. Energy is transferred from one object to another when a reaction takes place. Energy comes in many forms and can be transferred from one object to another as heat, light, or motion, to name a few. … This energy would be in the form of motion, with the person lifting the blue ball to a higher level.
## What are the 15 types of energy?
The different types of energy include thermal energy, radiant energy, chemical energy, nuclear energy, electrical energy, motion energy, sound energy, elastic energy and gravitational energy.
## What is energy easily transferred through?
A conductor is a material that allows internal (thermal) energy to be transmitted through it easily. All metals are good conductors. When one end of a metal rod is put into a fire, the energy from the flame makes the ions in the rod vibrate faster.
## How is energy transferred through heat?
Heat moves in three ways: Radiation, conduction, and convection. Radiation happens when heat moves as energy waves, called infrared waves, directly from its source to something else. This is how the heat from the Sun gets to Earth. In fact, all hot things radiate heat to cooler things.
## Why is it important to transfer energy when building a house?
It is important to consider the transfer of energy when building a house because controlling the transfer of thermal energy helps to conserve energy resources. … They reduce thermal energy transfer from the environment into the refrigerator or freezer.
## What is energy not easily transferred through?
A substance that transfers energy easily from the hot part to the cold part is called a conductor . Metals are good conductors. A substance that does not transfer energy easily from the hot part to the cold part is called an insulator . Air and plastics are insulators.
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# Types of Logical Reasoning
Two types of rational reasoning are sometimes distinguished as well as formal deduction and induction: induction and deduction. Given an antecedent or presupposition, a definite conclusion or deduction and a criterion or material condition that means the conclusion based on the antecedent, one can actually explain the following in terms of induction: A proposition is known only if all the facts or premises that make up its antecedent are true. And given that the proposition to be proven has been proved by the evidence of those facts or premises, then it is known.
This type of logic is usually derived from an axiomatic system. It is quite common in science, engineering, computer science, mathematics, and so on. The most common axioms or assumptions involved in this kind of reasoning is:
– If a proposition is known, then it must also be true, otherwise the conclusion follows. – All propositions must have at least two premises. – A proposition can only be proven by evidence of its antecedents. – Every proposition is either true or false.
In order to make an inductive or deductive argument, it is first assumed that the premises or information that are being relied upon are true. After this assumption has been made, the argument is said to be inductive and the conclusion is called deductive.
The second form of logic that is often used is the inductive logic, which is essentially the same as the deductive one. The basic difference between the two is that when using inductive logic, the argument that is being made is usually based on the fact that all the facts that make up a proposition are already known. The conclusion therefore depends on all the facts that support that conclusion.
In inductive logic, the starting point of the argument is the antecedent. A conclusion will then follow if the antecedent and all the supporting evidence of that conclusion have been verified. This is also known as an inference.
To illustrate the above, we can use an example in inductive reasoning. Suppose we are asking: “How do I know that there is a triangle with three sides?” If a certain condition is true, then we know that there is indeed such a thing as a triangle, but if that condition is false, then we do not know whether there exists such a thing.
In this case, we would not have to rely on induction alone to infer the conclusion – we can rely on induction alone and then ask: “Is it evident that there exists such a thing as a triangle with three sides?” We are then able to conclude that such a thing exists.
The deduction, however, is not an all-or-nothing process. As mentioned earlier, we cannot infer that the conclusion of a given argument is always true just because all of the premises have been proven. If the premises cannot be proved, then we can still infer that the conclusion of the given argument is necessarily true. – but only if the conclusion is supported by all of the other premises.
However, the truth of this statement may not always be clear. For example, if the premises are known, but the conclusion is false, then we cannot infer from the premises that the conclusion is always true – rather, we have to take into account whether those premises are known or not, whether the premises are known and then consider the consequences if they are known and the other premises are false.
However, since inductive reasoning is not deductive in nature, it is considered more reliable. than other forms of reasoning – this is especially true when we want to draw reasonable inferences.
For instance, when you are trying to prove something that you know very little about, it can be extremely difficult to make deductive reasoning. You may be wrong in the conclusion you draw if you are relying on just a few premises. – In inductive reasoning, there are many more possibilities and thus, you can never be sure that your conclusion will be correct. – And there are many situations where you need to look at many different things and decide what’s right and what’s wrong, and then you are left with no other choice except to assume what you are looking at is always right.
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Fundamentals of Psychrometrics, Part 3
Putting the quantities together in the chart
In Part 1 and Part 2 of this series of articles, we’ve taken a look at what exactly psychrometrics is and defined the top nine psychrometric quantities. Now we’re going to delve into how we can combine those quantities and create the psychrometric chart.
As you might expect, taking nine variables and putting them into one chart puts a lot of information at your fingertips. It also can take a while to figure it all out. On top of all that, having nine different variables means you’ve got a lot of options for how to show them in a chart.
A lot of folks have been involved in the development of the psychrometric chart over the past century. Willis Carrier was the first, with what he called the hygrometric chart in 1904. Richard Mollier from Germany made an important contribution in 1923, as you’ll see in the next section. Aside from that, the other versions of the psychrometric chart out there are mostly specialty charts and curiosities. (For more on the history and variety of psychrometric charts, see Don Gatley’s book, Understanding Psychrometrics.)
In fact, the chart itself has shifted from being a great calculation tool to being more of an educational or illustrative tool. With all the computing power at our fingertips these days, we can get accurate results from smartphone apps and don’t need to try to pull precise numbers off of psych charts.
OK, let’s jump in and place those eight psychrometric quantities on the chart and take a visual look this week. In case you’re wondering why I said eight instead of nine, recall that barometric pressure is one of the nine I defined in Part 2 of this series, and that variable determines the scale of the chart. When you change barometric pressure (or equivalently, altitude), you need to redraw the chart. Notice that the chart above says “SEA LEVEL” in big letters.
Dry bulb temperature
Dry bulb temperature is shown along the horizontal axis. When you draw lines through a chart along which the quantity under consideration remains the same, they’re called isolines. If you look carefully at the chart here, you can see something interesting about the dry bulb temperature isolines: They’re not all vertical and parallel to each other.
That was Mollier’s big contribution to this chart. Carrier originally used dry bulb temperature and humidity ratio as rectangular coordinates and used the various psychrometric relationships to place the rest of the quantities on the chart. One of the problems with this choice is that it makes finding specific enthalpies a bit difficult because the lines of constant enthalpy apply to saturation only, so then you have to put enthalpy deviation curves on the chart to make the adjustment for non-saturated conditions.
There are still plenty of the old-style charts out there, but ASHRAE has settled on the Mollier version with humidity ratio on the vertical axis and specific enthalpy as the other coordinate. If you want to see the dry bulb isolines as vertical and parallel, that’s fine. They nearly are. If you click the lead photo for this article and then zoom in, you should be able to see the difference, especially if you look down at the bottom left.
Wet bulb temperature
Wet bulb temperature tells you how easy it is for water to evaporate. The New York Times had a really nice article relating wet bulb temperatures to climate change recently, and they defined wet bulb temperature as “a measure of how well you can cool your skin by sweating.” When it’s hot outside, as it has been in India lately, that’s one of the primary ways we keep our bodies from overheating.
The wet bulb isolines are diagonal across the chart, as you see here, and there’s one place where the wet bulb and dry bulb temperatures are equal: at the saturation point. Evaporation doesn’t happen there under normal conditions, at least not without an equal amount of water vapor condensing.
From the saturation point, what happens to the wet bulb temperature as we heat a volume of air? Well, that depends on what happens to the amount of water vapor in the air. If you don’t humidify or dehumidify, you move horizontally to the right, increasing the dry bulb temperature. As you do so, you start crossing different wet bulb isolines with higher and higher values of wet bulb temperature.
For example, if you have air at 100°F and 40% relative humidity, the wet bulb temperature is 79°F. That’s going to be quite uncomfortable. The closer that wet bulb temperature gets to our body temperature, the harder it is for our bodies to lose enough heat to maintain a stable temperature. According to the New York Times article, a physically active person is in danger of overheating when the wet bulb temperature is 80°F.
The psychrometric chart can show us how wet bulb temperature changes as other conditions change.
Specific enthalpy
On the ASHRAE chart, the specific enthalpy lines are one of the two main coordinates used to lay out the chart. As you can see here, they’re diagonal and oh, so close to being parallel to the wet bulb isolines. That’s because those two quantities are closely related.
Specific enthalpy tells you about the work and energy in changing air conditions and is given in BTU per pound of dry air in the imperial system of units. It’s only changes that matter here, so if you have any two statepoints on the chart, you can find the enthalpy difference. That tells something about how much energy you’ll have to put into the system or how much you’ll get out when you make the change.
Humidity ratio
Humidity ratio is the other defining coordinate for the ASHRAE version of the psychrometric chart. As you can see, the isolines are horizontal. In this case, the scale given on the chart is in pounds of water vapor per pound of dry air. Some charts show grains of water vapor per pound of dry air, which is often shortened to grains. (It takes 7,000 grains to make one pound.)
Humidity ratio isolines are horizontal and evenly spaced on the chart.
Water vapor pressure
Vapor pressure is rarely shown on the psych chart, and this ASHRAE chart is no exception. I’ve drawn the lines here, and for temperatures below ~120° F, the scale (with units given in inches of mercury in this case) is simply the product:
pWV = 48.11 x W
When you’re doing calculations for vapor diffusion or trying to figure out which way water vapor will travel through materials, you’ll need the vapor pressures for the various locations.
Dew point temperature
Dew point temperature is yet a third quantity that has horizontal isolines on the psych chart. What this means is that these three quantities are not independent of each other so if they’re all you have, you still don’t know your state point.
Another interesting thing here is that when wet bulb and dry bulb temperatures are equal, so is the dew point temperature. All three have the same value at the saturation point, shown on the curve that bounds the left side of the chart.
Relative humidity
And speaking of curves, the saturation curve is the 100% relative humidity isocurve. When the volume isn’t saturated with water vapor, the vapor pressure will be lower than the saturation vapor pressure and the relative humidity is less than 100%.
Unlike the three preceding quantities with horizontal isolines, relative humidity isn’t related to the other psychrometric quantities under discussion here. If you know RH and any of the other quantities, you’ve got your statepoint.
Specific volume
Specific volume shows up as a diagonal line on the psychrometric chart, but it has a significantly different slope than specific enthalpy and wet bulb temperature. It’s also not quite as simple as I made it out to be last time. I wrote that it’s just the reciprocal of the density, but I glossed over a detail that may or may not be important in what you do.
As with the other psychrometric ratios here (specific enthalpy and humidity ratio), the denominator is not the mass of the moist air. It includes only the dry air component. So its units (in the imperial system) are cubit feet per pound of dry air. Since the water vapor component is usually small, however, the specific volume defined this way isn’t greatly different from the volume per pound of moist air.
Understanding the chart
OK, so there are the quantities in the chart. By studying those individual charts, you can see how the different quantities change. For example, changing dry bulb temperature without humidifying or dehumidifying the air has no effect on dew point temperature, humidity ratio, or vapor pressure. And what happens to the relative humidity of air as it passes across an evaporator coil in your air conditioner?
So now we’re ready to start doing the fun stuff with the psychrometric chart. Next time we’ll take a look at some different processes like air conditioning or dehumidification. Stay tuned!
Allison Bailes of Decatur, Georgia, is a speaker, writer, energy consultant, RESNET-certified trainer, and the author of the Energy Vanguard Blog. Check out his in-depth course, Mastering Building Science at Heatspring Learning Institute, and follow him on Twitter at @EnergyVanguard.
1. | | #1
more doc
I've grown fond of
which I dug out of some old paper whose reference is long lost,
cleaned up a bit, and added a couple of the basic formulas for
energy calculations. I find it a little easier to read than the
ASHRAE tangle, and copies are tacked up in strategic locations
around the house.
A discussion of some of the other charts for different elevations
and which way the various scales shift while creating them would
be really useful too!
_H*
2. GBA Editor
| | #2
Response to Hobbit_
The one you like uses dry bulb temperature as the other coordinate, which means you've got to use the enthalpy deviation curves for your energy calculations. That's what I consider a tangle.
3. | | #3
Great series
Thanks, Allison, for attempting to drag us science-illiterates along. Can't wait for the rest of it.
• How Well Do You Understand Psychrometrics?
Test your knowledge of the science of moist air and how it behaves under different environmental conditions
• Cold Air Is Dry Air
Don't be fooled by the relative humidity
• Fundamentals of Psychrometrics, Part 2
The quantites, both preferred and discouraged, and how they relate to the chart
• Fundamentals of Psychrometrics, Part 1
Laying the foundation to understand the psychrometric chart
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# Distance between Punta Gorda, FL (PGD) and Pasco, WA (PSC)
Flight distance from Punta Gorda to Pasco (Punta Gorda Airport (Florida) – Tri-Cities Airport (Washington)) is 2425 miles / 3902 kilometers / 2107 nautical miles. Estimated flight time is 5 hours 5 minutes.
Driving distance from Punta Gorda (PGD) to Pasco (PSC) is 3012 miles / 4847 kilometers and travel time by car is about 49 hours 54 minutes.
## Map of flight path and driving directions from Punta Gorda to Pasco.
Shortest flight path between Punta Gorda Airport (Florida) (PGD) and Tri-Cities Airport (Washington) (PSC).
## How far is Pasco from Punta Gorda?
There are several ways to calculate distances between Punta Gorda and Pasco. Here are two common methods:
Vincenty's formula (applied above)
• 2424.570 miles
• 3901.967 kilometers
• 2106.893 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2422.168 miles
• 3898.102 kilometers
• 2104.806 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Punta Gorda Airport (Florida)
City: Punta Gorda, FL
Country: United States
IATA Code: PGD
ICAO Code: KPGD
Coordinates: 26°55′12″N, 81°59′25″W
B Tri-Cities Airport (Washington)
City: Pasco, WA
Country: United States
IATA Code: PSC
ICAO Code: KPSC
Coordinates: 46°15′52″N, 119°7′8″W
## Time difference and current local times
The time difference between Punta Gorda and Pasco is 3 hours. Pasco is 3 hours behind Punta Gorda.
EST
PST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 266 kg (587 pounds).
## Frequent Flyer Miles Calculator
Punta Gorda (PGD) → Pasco (PSC).
Distance:
2425
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2425
Round trip?
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Unformatted text preview: () When When neither Case B or C occur Case A Impossible ⏟ found by () () We have modeled the following problem: 29 ECO 204 Chapter 16: Analysis of Firms with Market Power (this version 2012-2013) University of Toronto, Department of Economics (STG). ECO 204, S. Ajaz Hussain. Do not distribute. ⏟ ⏟ To see how the ECO 204 RMP differs from the ECO 100 “revenue maximization rules” that and a look at the RMP whenever unconstrained solution (cases B and D): notice that : Case D Case B When is ⏟? When is let take and ⏟? When is When is If neither cases B or C s.t. If Supply Point Supply Point () Unconstrained solution Unconstrained solution Demand Demand Let’s prove these results formally. First, let’s show that whenever arrange the FOC to see that: unconstrained solution that . Re - () ⏟ ( ) From the envelope theorem we know that: Since we are looking at unconstrained solution we see that in cases B and D that: As such, the FOC becomes: ⏟ ( ⏟ ) ⏟ ⏟ ⏟ 30 ECO 204 Chapter 16: Analysis of Firms with Market Power (this version 2012-2013) University of Toronto, Department of Economics (STG). ECO 204, S. Ajaz Hussain. Do not distribute. This proves that whenever Next, we show that wh...
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Question:Can't solve the linear pde problem of "two composite cylinder" of Carslaw and Jeager's book (link in description)
Question:Can't solve the linear pde problem of "two composite cylinder" of Carslaw and Jeager's book (link in description)
Maple
Hello,
I am trying to recover with maple the analytical solution of the problem of conduction within two composite cylinders given in Carslaw and Jeager's book: "Conduction of heat in solids" (link here). Here is my take on the problem
```restart;
with(PDEtools);
C := diff_table(c(rho, t));
E := diff_table(e(rho, t));
PDE1 := diff(C[], t) - K_1*(diff(diff(C[], rho), rho) - diff(C[], rho)/rho) = 0;
PDE2 := diff(E[], t) - K_2*(diff(diff(E[], rho), rho) - diff(E[], rho)/rho) = 0;
ic1 := eval(C[], t = 0) = V;
ic2 := eval(E[], t = 0) = 0;
bc_1 := eval(C[], rho = a) = eval(E[], rho = a);
bc_2 := K_1*eval(diff(C[], rho), rho = a) = K_2*eval(diff(E[], rho), rho = a);
bc_3 := eval(E[], rho = infinity) = 0;
bc_4 := (diff(C[]/rho, rho), rho = 0) = 0;
pdsolve([PDE1, PDE2],[ic1,ic2, bc_1,bc_2,bc_3,bc_4]); ```
Unfortunately I end up with the following errors:
```Error, (in pdsolve/sys) too many arguments; some or all of the following are wrong: [{c(rho, t), e(rho, t)}, [c(rho, 0) = V, e(rho, 0) = 0, c(a, t) = e(a, t), K_1*(diff(c(a, t), a)) = K_2*(diff(e(a, t), a)), e(infinity, t) = 0, ((diff(c(rho, t), rho))/rho-c(rho, t)/rho^2, rho = 0) = 0]]
```
If I call pdsolve without any initial and boundary conditions, I have the general form of the equation which is correct (product of exponential function in time with Bessel functions for space) but I don't know how to determine the coeficients with maple from there.
Note also that the boundary condition "bc_2" does not seem to be correcty evaluated (it is supposed to specify the continuity of the flux ar r=a). Instead it just evaluate the expression with r swapped with a. If one write the same line of code with r=0 instead of r=a then the evaluation seems correct.
Any help would be very appreciated,
Cheers
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# Hidden Markov Models - PowerPoint PPT Presentation
1. 2. 2. 1. 1. 1. 1. …. 2. 2. 2. 2. …. K. …. …. …. …. x 1. K. K. K. K. x 2. x 3. x K. …. Hidden Markov Models. 1. 1. 1. 1. …. 2. 2. 2. 2. …. …. …. …. …. K. K. K. K. …. Generating a sequence by the model.
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2
2
1
1
1
1
2
2
2
2
K
x1
K
K
K
K
x2
x3
xK
### Hidden Markov Models
1
1
1
2
2
2
2
K
K
K
K
Generating a sequence by the model
Given a HMM, we can generate a sequence of length n as follows:
• Start at state 1 according to prob a01
• Emit letter x1 according to prob e1(x1)
• Go to state 2 according to prob a12
• … until emitting xn
1
a02
2
2
0
K
e2(x1)
x1
x2
x3
xn
We will develop algorithms that allow us to compute:
P(x) Probability of x given the model
P(xi…xj) Probability of a substring of x given the model
P(i = k | x) “Posterior” probability that the ith state is k, given x
A more refined measure of which states x may be in
fk(i) = P(x1…xi, i = k) (the forward probability)
Initialization:
f0(0) = 1
fk(0) = 0, for all k > 0
Iteration:
fk(i) = ek(xi) l fl(i – 1) alk
Termination:
P(x) = k fk(N)
We want to compute
P(i = k | x),
the probability distribution on the ith position, given x
We start by computing
P(i = k, x) = P(x1…xi, i = k, xi+1…xN)
= P(x1…xi, i = k) P(xi+1…xN | x1…xi, i = k)
= P(x1…xi, i = k) P(xi+1…xN | i = k)
Then, P(i = k | x) = P(i = k, x) / P(x)
Forward, fk(i)
Backward, bk(i)
Define the backward probability:
bk(i) = P(xi+1…xN | i = k) “starting from ith state = k, generate rest of x”
= i+1…N P(xi+1,xi+2, …, xN, i+1, …, N | i = k)
= li+1…N P(xi+1,xi+2, …, xN, i+1 = l, i+2, …, N | i = k)
= l el(xi+1) akli+1…N P(xi+2, …, xN, i+2, …, N | i+1 = l)
= l el(xi+1) aklbl(i+1)
We can compute bk(i) for all k, i, using dynamic programming
Initialization:
bk(N) = 1, for all k
Iteration:
bk(i) = l el(xi+1) akl bl(i+1)
Termination:
P(x) = l a0l el(x1) bl(1)
What is the running time, and space required, for Forward, and Backward?
Time: O(K2N)
Space: O(KN)
Useful implementation technique to avoid underflows
Viterbi: sum of logs
Forward/Backward: rescaling at each few positions by multiplying by a constant
P(i = k | x) =
P(i = k , x)/P(x) =
P(x1, …, xi, i = k, xi+1, … xn) / P(x) =
P(x1, …, xi, i = k) P(xi+1, … xn | i = k) / P(x) =
fk(i) bk(i) / P(x)
We can now calculate
fk(i) bk(i)
P(i = k | x) = –––––––
P(x)
What is the most likely state at position i of sequence x:
Define ^ by Posterior Decoding:
^i = argmaxkP(i = k | x)
• For each state,
• Posterior Decoding gives us a curve of likelihood of state for each position
• Posterior Decoding may give an invalid sequence of states (of prob 0)
• Why?
x1 x2 x3 …………………………………………… xN
• P(i = k | x) = P( | x) 1(i = k)
= {:[i] = k}P( | x)
State 1
P(i=l|x)
l
k
1() = 1, if is true
0, otherwise
VITERBI
Initialization:
V0(0) = 1
Vk(0) = 0, for all k > 0
Iteration:
Vl(i) = el(xi) maxkVk(i-1) akl
Termination:
P(x, *) = maxkVk(N)
• FORWARD
• Initialization:
• f0(0) = 1
• fk(0) = 0, for all k > 0
• Iteration:
• fl(i) = el(xi) k fk(i-1) akl
• Termination:
• P(x) = k fk(N)
BACKWARD
Initialization:
bk(N) = 1, for all k
Iteration:
bl(i) = k el(xi+1) akl bk(i+1)
Termination:
P(x) = k a0k ek(x1) bk(1)
### Variants of HMMs
• How do we model “memory” larger than one time point?
• P(i+1 = l | i = k) akl
• P(i+1 = l | i = k, i -1 = j) ajkl
• A second order HMM with K states is equivalent to a first order HMM with K2 states
aHHT
state HH
state HT
aHT(prev = H)
aHT(prev = T)
aHTH
state H
state T
aHTT
aTHH
aTHT
state TH
state TT
aTH(prev = H)
aTH(prev = T)
aTTH
Similar Algorithms to 1st Order
• P(i+1 = l | i = k, i -1 = j)
• Vlk(i) = maxj{ Vkj(i – 1) + … }
• Time? Space?
1-p
Length distribution of region X:
E[lX] = 1/(1-p)
• Geometric distribution, with mean 1/(1-p)
This is a significant disadvantage of HMMs
Several solutions exist for modeling different length distributions
X
Y
p
q
1-q
p
1-p
X
Y
X
X
q
1-q
lX = C + geometric with mean 1/(1-p)
Duration in X: m turns, where
• During first m – 1 turns, exactly n – 1 arrows to next state are followed
• During mth turn, an arrow to next state is followed
m – 1 m – 1
P(lX = m) = n – 1 (1 – p)n-1+1p(m-1)-(n-1) = n – 1 (1 – p)npm-n
p
p
p
1 – p
1 – p
1 – p
Y
X(n)
X(1)
X(2)
……
• EasyGene:
Prokaryotic
gene-finder
Larsen TS, Krogh A
• Negative binomial with n = 3
Upon entering a state:
• Choose duration d, according to probability distribution
• Generate d letters according to emission probs
• Take a transition to next state according to transition probs
Disadvantage: Increase in complexity of Viterbi:
Time: O(D)
Space: O(1)
where D = maximum duration of state
F
d<Df
xi…xi+d-1
Pf
Warning, Rabiner’s tutorial claims O(D2) & O(D) increases
emissions
emissions
Recall original iteration:
Vl(i) = maxk Vk(i – 1) akl el(xi)
New iteration:
Vl(i) = maxk maxd=1…DlVk(i – d) Pl(d) akl j=i-d+1…iel(xj)
F
L
d<Df
d<Dl
Pl
Pf
transitions
xi…xi + d – 1
xj…xj + d – 1
Precompute cumulative values
### Proteins, Pair HMMs, and Alignment
M
(+1,+1)
Alignments correspond 1-to-1 with sequences of states M, I, J
I
(+1, 0)
J
(0, +1)
-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---
TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC
IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII
s(xi, yj)
M
(+1,+1)
Alignments correspond 1-to-1 with sequences of states M, I, J
s(xi, yj)
s(xi, yj)
-d
-d
I
(+1, 0)
J
(0, +1)
-e
-e
-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---
TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC
IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII
Dynamic Programming:
M(i, j): Optimal alignment of x1…xi to y1…yjending in M
I(i, j): Optimal alignment of x1…xi to y1…yj ending in I
J(i, j): Optimal alignment of x1…xi to y1…yjending in J
The score is additive, therefore we can apply DP recurrence formulas
Initialization:
M(0,0) = 0;
M(i, 0) = M(0, j) = -, for i, j > 0
I(i,0) = d + ie; J(0, j) = d + je
Iteration:
M(i – 1, j – 1)
M(i, j) = s(xi, yj) + max I(i – 1, j – 1)
J(i – 1, j – 1)
e + I(i – 1, j)
I(i, j) = max
d + M(i – 1, j)
e + J(i, j – 1)
J(i, j) = max
d + M(i, j – 1)
Termination:
Optimal alignment given by max { M(m, n), I(m, n), J(m, n) }
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ONLINE ELECTRICAL ENGINEERING STUDY SITE
# Over Fluxing in Transformer
## Causes of Over Fluxing in Transformer
As per present day transformer design practice, the peak rated value of the flux density is kept about 1.7 to 1.8 Tesla, while the saturation flux density of CRGD steel sheet of core of transformer is of the order of 1.9 to 2 Tesla which corresponds to about 1.1 times the rated value. If during operation, an electrical power transformer is subjected to carry rather swallow more than above mentioned flux density as per its design limitations, the transformer is said to have faced over fluxing problem and consequent bad effects towards its operation and life. Depending upon the design and saturation flux densities and the thermal time constants of the heated component parts, a transformer has some over excitation capacity. I.S. specification for electrical power transformer does not stipulate the short time permissible over excitation, though in a round about way it does indicate that the maximum over fluxing in transformer shall not exceed 110%.
The flux density in a transformer can be expressed by
B = C V/f,
where, C = A constant,
V = Induced voltage,
f = Frequency. The magnetic flux density is, therefore, proportional to the quotient of voltage and frequency (V/f). Over fluxing can, therefore, occur either due to increase in voltage or decrease in-frequency of both. The probability of over fluxing is relatively high in step-up transformers in Power stations compared to step - down transformers in Sub-Stations, where voltage and frequency usually remain constant. However, under very abnormal system condition, over-fluxing trouble can arise in step-down Sub-Station transformers as well.
## Effect of Over Fluxing in Transformers
The flux in a transformer, under normal conditions is confined to the core of transformer because of its high permeability compared to the surrounding volume. When the flux density in the increases beyond saturation point, a substantial amount of flux is diverted to steel structural parts and into the air. At saturation flux density the core steel will over heat. Structural steel parts which are nu-laminated and are not designed to carry magnetic flux will heat rapidly. Flux flowing in unplanned air paths may link conducing loops in the windings, loads, tank base at the bottom of the core and structural parts and the resulting circulating currents in these loops can cause dangerous temperature increase. Under conditions of excessive over fluxing the heating of the inner portion of the windings may be sufficiently extreme as the exciting current is rich in harmonies. It is obvious that the levels of loss which occur in the winding at high excitation cannot be tolerated for long if the damage is to be a voided. Physical evidences of damage due to over fluxing will very with the degree of over excitation, the time applied and the particular design of transformer. The Table given below summarizes such physical damage and probable consequences.
SL Component involved Physical evidences Consequences 1 Metallic support and surfaces structure for core and coils Discoloration or metallic parts and adjacent insulation.Possible carbonized material in oil. Evolution of combustible gas. Contamination of a oil and surfaces of insulation. Mechanical weakening of insulation Loosing of structure. Mechanical structure 2 Windings Discoloration winding insulation evolution of gas. Electrical and mechanical weakling of winding insulation 3 Lead conductors. Discoloration of conductor insulation or support, evolution of gas. Electrical and mechanical weakening of insulation, Mechanical Weakening of support. 4 Core lamination. Discoloration of insulating material in contact with core. Discoloration and carbonization of organic/lamination insulation Evaluation of gas. Electrical weakening of major insulation (winding to core) increased interlaminar eddy loss. 5 Tank Blistering of paints Contamination of oil if paint inside tank is blistered.
It may be seen that metallic support structures for core and coil, windings, lead conductors, core lamination, tank etc. may attain sufficient temperature with the evolution of combustible gas in each case due to over fluxing of transformer and the same gas may be collected in Buchholz Relay with consequent Alarm/Trip depending upon the quantity of gas collected which again depends upon the duration of time the transformer is subjected to over fluxing.
Due to over fluxing in transformer its core becomes saturated as such induced voltage in the primary circuit becomes more or less constant. If the supply voltage to the primary is increased to abnormal high value, there must be high magnetising current in the primary circuit. Under such magnetic state of condition of transformer core linear relations between primary and secondary quantities (viz. for voltage and currents) are lost. So there may not be sufficient and appropriate reflection of this high primary magnetising current to secondary circuit as such mismatching of primary currents and secondary currents is likely to occur, causing differential relay to operate as we do not have overfluxing protection for sub-stn. transformers.
## Stipulated Withstand-Duration of Over Fluxing in Transformers
Over fluxing in transformer has sufficient harmful effect towards its life which has been explained. As overfluxing protection is not generally provided in step-down transformers of Sub-Station, there must be a stipulated time which can be allowed matching with the transformer design to withstand such overfluxing without causing appreciable damage to the transformer and other protections shall be sensitive enough to trip the transformer well within such stipulated time, if cause of overfluxing is not removed by this time. It is already mentioned that the flux density 'B' in transformer core is proportional to v/f ratio. Power transformers are designed to withstand (Vn/fn x 1.1) continuously, where Vn is the normal highest r.m.s. voltage and fn is the standard frequency. Core design is such that higher v/f causes higher core loss and core heating. The capability of a transformer to withstand higher v/f values i.e. overfluxing effect, is limited to a few minutes as furnished below in the Table
F = (V/f)/(Vn/fn) 1.1 1.2 1.25 1.3 1.4 Duration of with stand limit (minutes) continuous 2 1 0.5 0
From the table above it may be seen that when over fluxing due to system hazards reaches such that the factor F attains a values 1.4, the transformer shall be tripped out of service instantaneously otherwise there may be a permanent damage.
## Protection Against Over fluxing (v/f - Protection) in Transformer
The condition arising out of over-fluxing does not call for high speed tripping. Instantaneous operation is undesirable as this would cause tripping on momentary system disturbances which can be borne safely but the normal condition must be restored or the transformer must be isolated within one or two minutes at the most. Flux density is proportional to V/f and it is necessary to detect a ratio of V/f exceeding unity, V and f being expressed in per unit value of rated quantities. In a typical scheme designed for over fluxing protection, the system voltage as measured by the voltages transformer is applied to a resistance to product a proportionate current; this current on being passed through a capacitor, produces a voltage drop which is proportional to the functioning in question i.e. V/f and hence to flux in the power transformer. This is accompanied with a fixed reference D.C. voltage obtained across a Zener diode. When the peak A.C. signal exceeds the D.C. reference it triggers a transistor circuit which operates two electromechanical auxiliary elements. One is initiated after a fixed time delay, the other after an additional time delay which is adjustable. The over fluxing protection operates when the ratio of the terminal voltage to frequency exceeds a predetermined setting and resets when the ratio falls below 95 to 98% of the operating ratio. By adjustment of a potentiometer, the setting is calibrated from 1 to 1.25 times the ratio of rated volts to rated frequency. The output from the first auxiliary element, which operates after fixed time delay available between 20 to 120 secs. second output relay operates and performs the tripping function. It is already pointed out that high V/f occur in Generator Transformers and Unit-Auxiliary Transformers if full exaltation is applied to generator before full synchronous speed is reached. V/f relay is provided in the automatic voltage regulator of generator. This relay blocks and prevents increasing excitation current before full frequency is reached. When applying V/f relay to step down transformer it is preferable to connect it to the secondary (L.V. said of the transformer so that change in tap position on the H.V. is automatically taken care of Further the relay should initiate an Alarm and the corrective operation be done / got done by the operator. On extreme eventuality the transformer controlling breaker may be allowed to trip.
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• # 1 cubic meter storage square shape tank
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Aug 26, 2011 · Generally water is stored either in a cylindrical tank or in a square tank. We can easily find out the capacity of water storage tank of any size in liters. For eg. if you want to find out the capacity of a square water tank measuring 1m length, 1m breadth, 1m height, first convert meter to cm. As the formula to find out the volume of cube is a^3 we can get answer as 1000,000 cc (1 m = 100 cm)How to Calculate the Volume of Water to Fill a Rectangular , 1 cubic meter storage square shape tankMay 21, 2018 · Find the volume of water to fill a rectangular tank by calculating the tank volume. Find the volume of rectangular tanks by measuring and multiplying length times width times height. Since 7.48 gallons of water fills 1 cubic foot, multiple the volume of the tank by 7.48 to find the gallons of water.water tank 1 cubic meter, water tank 1 cubic meter , 1 cubic meter storage square shape tankAlibaba, 1 cubic meter storage square shape tank offers 5,884 water tank 1 cubic meter products. About 85% of these are water treatment, 1% are chemical storage equipment, and 1% are storage tank. A wide variety of water tank 1 cubic meter options are available to you, There are 5,884 water tank 1 cubic meter
### How do you calculate rectangular tank volume in cubic meter
May 15, 2015 · The volume of gas in a cubic meter is one cubic meter. But perhaps that is not the real question? , 1 cubic meter storage square shape tank A cubic meter is a volume and a square meter is an area. , 1 cubic meter storage square shape tank How to calculate of round storage , 1 cubic meter storage square shape tankConversion Chart - Tank DepotChemical Storage Tanks. Vertical Storage Tanks; Cone Bottom Tanks; Horizontal Storage Tanks; Double Wall Tanks; , 1 cubic meter storage square shape tank which is 1 cubic meter. Metric Unit . Steres. Cubic Inch . Cubic Feet . Cubic Yards . Cu. Millimeter (cu mm). Cu. Centimeter (cu cm) , 1 cubic meter storage square shape tank AREA Basic unit is Centare (CA) which is 1 square meter. Metric Unit. Centares. Square InchesCubes | Applied Math Flashcards | QuizletCompute the volume of a storage shed that has a square floor 3 meters on each side and is 3 meters high. 27 cubic meters. A water storage tank in the shape of a cube measures 4.1 meters along each edge. If one cubic meter of water ways one metric ton, how many metric tons of water can the tank
### A water tank is 3 m long, 2 m wide and 1 m high. How many , 1 cubic meter storage square shape tank
Jan 22, 2018 · V=3x2x1=6m^3 We know that 1L = 0.001m^3. Therefore we can use dimensional analysis. 6m^3 x (1L/0.001m^3) = 6000 LitresCubic meter to Square meter (m3 m2) Converter -- EndMemoConversion between cubic meter and square meter. Cubic meter to Square meter CalculatorPlastic Water Tank, Plastic Water Tank Suppliers and , 1 cubic meter storage square shape tankAlibaba, 1 cubic meter storage square shape tank offers 67,083 plastic water tank products. About 19% of these are water treatment, 3% are other plastic products, and 1% are chemical storage equipment. A wide variety of plastic water tank options are available to you, such as free samples, paid samples.
### cubic water tank, cubic water tank Suppliers and , 1 cubic meter storage square shape tank
Alibaba, 1 cubic meter storage square shape tank offers 7,437 cubic water tank products. About 73% of these are water treatment, 2% are chemical storage equipment, and 1% are other plastic products. A wide variety of cubic water tank options are available to you, There are 7,436 cubic water tank suppliers, mainly located in Asia.How to Calculate The Capacity of Water Tank And , 1 cubic meter storage square shape tank - HubPagesAug 26, 2011 · Generally water is stored either in a cylindrical tank or in a square tank. We can easily find out the capacity of water storage tank of any size in liters. For eg. if you want to find out the capacity of a square water tank measuring 1m length, 1m breadth, 1m height, first convert meter to cm. As the formula to find out the volume of cube is a^3 we can get answer as 1000,000 cc (1 m = 100 cm)How do you calculate rectangular tank volume in cubic meterMay 15, 2015 · The volume of gas in a cubic meter is one cubic meter. But perhaps that is not the real question? , 1 cubic meter storage square shape tank A cubic meter is a volume and a square meter is an area. , 1 cubic meter storage square shape tank How to calculate of round storage , 1 cubic meter storage square shape tank
### How to Calculate the Volume of Water to Fill a Rectangular , 1 cubic meter storage square shape tank
May 21, 2018 · Find the volume of water to fill a rectangular tank by calculating the tank volume. Find the volume of rectangular tanks by measuring and multiplying length times width times height. Since 7.48 gallons of water fills 1 cubic foot, multiple the volume of the tank by 7.48 to find the gallons of water.1 Cubic Meter Square 304 316 Stainless Steel Water TankWater Storage Tank, Water Container, Stainless Steel Tank manufacturer / supplier in China, offering 1 Cubic Meter Square 304 316 Stainless Steel Water Tank, 2019 Big HDG Underground Water Tank, 2019 3m Height HDG Undergroundd Water Tank and so on.Conversion Chart - Tank DepotChemical Storage Tanks. Vertical Storage Tanks; Cone Bottom Tanks; Horizontal Storage Tanks; Double Wall Tanks; , 1 cubic meter storage square shape tank which is 1 cubic meter. Metric Unit . Steres. Cubic Inch . Cubic Feet . Cubic Yards . Cu. Millimeter (cu mm). Cu. Centimeter (cu cm) , 1 cubic meter storage square shape tank AREA Basic unit is Centare (CA) which is 1 square meter. Metric Unit. Centares. Square Inches
### What are the dimensions of a 2000 cubic meter above
Jan 22, 2013 · What are the dimensions of a 2000 cubic meter above ground oil storage tank , 1 cubic meter storage square shape tank variety of shapes and dimensions that one cubic foot of space can have. , 1 cubic meter storage square shape tank measuring SQUARE feet but CUBIC Cubic meter to Square meter (m3 m2) Converter -- EndMemoConversion between cubic meter and square meter. Cubic meter to Square meter Calculator3 Ways to Work Out Water Tank Capacity - wikiHowNov 19, 2019 · To work out water tank capacity for a horizontal cylinder tank, measure the radius of the bottom of the cylinder and find the area of the circle. Next, multiply the area by the tank's length to determine the tank's total volume.
### How to calculate the storage capacity of a tank in liters , 1 cubic meter storage square shape tank
May 09, 2017 · Water tanks are large chambers for storing water. They come in a variety of styles, including horizontal cylinders, vertical cylinders, and rectangles. The proper method for determining tank capacity depends on the shape of the water tank. Keep in, 1 cubic meter storage square shape tank1000 Cubic Meter Frp Tank, 1000 Cubic Meter Frp Tank , 1 cubic meter storage square shape tankAlibaba, 1 cubic meter storage square shape tank offers 234 1000 cubic meter frp tank products. About 86% of these are water treatment, 7% are other fiberglass products, and 1% are other plastic products. A wide variety of 1000 cubic meter frp tank options are available to you, such as free samples.Cubic Meter to Square meter - OnlineConversion ForumsNov 18, 2010 · Re: Cubic Meter to Square meter You need more information. you can have something thats 100 cubic meters but 100 meters tall and its top surface area will only be 1 meter you can have something 50 meters tall, its top surface area will be only 2 meters
### Water Distribution Operator (TX) License D Flashcards , 1 cubic meter storage square shape tank
Start studying Water Distribution Operator (TX) License D. Learn vocabulary, terms, and more with flashcards, games, and other study tools. , 1 cubic meter storage square shape tank What is the weight of water in a 14,000 cubic foot ground storage tank?, 1 cubic meter storage square shape tank 3. How many cubic yards of concrete are needed to fill 108 square feet 1 foot deep?, 1 cubic meter storage square shape tank 4. How many 20-foot sections of pipe are , 1 cubic meter storage square shape tankCalculating Area and Volume of Ponds and TanksCalculating Area and Volume of Ponds and Tanks Good fish farm managers must know the area and volume of all ponds and tanks. Exact measure-ment of area and volume is essen-tial in order to calculate stocking rates and chemical applications. Stocking fish into a pond of uncer-tain area can result in poor produc-tion, more disease and possibly death.Volume CalculatorThis free volume calculator can compute the volumes of common shapes, including that of a sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more.
### Volume of a Box Calculator - Tank Volume Calculator, Cubic , 1 cubic meter storage square shape tank
Calculate the volume of a rectangular box or tank using our free volume of a box calculator. Tank volume calculator online that works in many different metrics: mm, cm, meters, km, inches, feet, yards, miles. Can be used to calculate shipping dimensions in cubic meters or cubic feet.Tank Volume Calculator - Bailiff EnterprisesUse this to figure the Volume of a Square or Rectangular Tank. Tank Length (ft) , 1 cubic meter storage square shape tank enter depth: Click on button to calculate cubic feet: First, calculate the cubic feet, then click below to calculate gallons. Click on button to calculate gallons: , 1 cubic meter storage square shape tank Please enter data for Tank Radius (radius is 1A cubic tank holds 1,000.0 kg of water. What are the , 1 cubic meter storage square shape tankMar 06, 2010 · A cubic tank holds 1,000kg of water. What are the dimensions of the tank in meters. , 1 cubic meter storage square shape tank A water tank can be constructed with a metal top and two square ends that holds 16 cubic meter of water and requires 40 square meter of metal. find the dimensions of the water tank. , 1 cubic meter storage square shape tank A fish tank is in the shape of a rectangular prism. The dimensions of , 1 cubic meter storage square shape tank
### A cubic tank holds 1 000.0 kg of water. What are the , 1 cubic meter storage square shape tank
Aug 30, 2009 · If the tank is 100% full, then dimensions are: 1 m x 1 m x 1 m (cube which sides measure 1m). This is because I assume that the water is at ambient temperature and therefore the density is roughly 1000 kg/m3 (i.e. 1 kg of water takes 1 litre).How do you find the cubic feet of a round tank - AnswersDepends on the shape, if it's a cylinder then find the radius of the circle on the end by measuring the diameter in feet and halving it. , 1 cubic meter storage square shape tank 10 ft X 10 ft square tank will hold 1000 cubic feet of , 1 cubic meter storage square shape tankSOLUTION: An inverted square pyramid has a height equal Question 575067: An inverted square pyramid has a height equal to 8 meters and a top edge to 3 meters. Initially, it contains water to the depth of 5 meters. a. What is the initial volume of the water in the tank? b. If the additional water is to be pumped into the tank at the rate of 20 gallons per minute, how many hours will it take to fill the tank?
### Optimization Metal Tank Problem | Physics Forums
Jan 05, 2006 · Ok, well the problem states: A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal? The first step that I took was to draw a picture. I just drew a semicircle with a right , 1 cubic meter storage square shape tankMathematics: How many litres will a tank hold with a , 1 cubic meter storage square shape tankWe can safely say the tank is a cuboid as length breadth and height is given. The amount of water it can hold can be calculated by simply using the formula for volume of cuboid = L*B*H (here L,B,H are length, breadth and height respectively ) So V, 1 cubic meter storage square shape tanka water tank in the shape of a cylinder with a hemisphere , 1 cubic meter storage square shape tankOct 31, 2008 · a water tank in the shape of a cylinder with a hemisphere on top. what formula would you use to solve this?
### Cubic Meter to Square meter - OnlineConversion Forums
Nov 18, 2010 · Re: Cubic Meter to Square meter You need more information. you can have something thats 100 cubic meters but 100 meters tall and its top surface area will only be 1 meter you can have something 50 meters tall, its top surface area will be only 2 metersRectangular Tank Calculator Storage Capacity CalculationCalculate the storage capacity of a rectangular tank through online Rectangular Tank Calculator by applying the appropriate formula for capacity in cubic inches and gallons. Home. About. Storage Capacity for Rectangular Tank Calculator. A storage tank is a container, usually for holding liquids, sometimes for compressed gases (gas tanks). , 1 cubic meter storage square shape tankTank Volume Calculator - Oil TanksThe tank size calculator on this page is designed for measuring the capacity of a variety of fuel tanks. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. The functionality of
### Volume Problems and Definitions Flashcards | Quizlet
Volume Problems and Definitions. STUDY. PLAY. , 1 cubic meter storage square shape tank how many cubic meters of sand are in the pile? 26 2/3 cubic yards. How many cubic yards of sand will it take to fill a child's play area that measures 10 yards long, 8 yards wide, and 1/3 yard deep? , 1 cubic meter storage square shape tank A pile of gravel is in the shape of a cone. If the pile is 8 feet high, and the diameter of , 1 cubic meter storage square shape tankCubic Meters to Gallons Converter (m3 to gal)1 Cubic meter (m³) is equal to 264.172052 gallons. To convert cubic meters to gallons, multiply the cubic meter value by 264.172052. For example, to find out how many gallons in a cubic meter and a half, multiply 1.5 by 264.172052, that makes 396.258 gallons in a cubic meter and a half.
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# Algebra
Can you explain why the line x = 4 is a vertical line.
1. 👍
2. 👎
3. 👁
1. If x = 4, y = 0, 1, 2, 3, 4, 5, -1, -2, -3, -4 etc. Graph that line and you will see it is a vertical line going through all of those points and crossing the x axis at (4,0).
1. 👍
2. 👎
## Similar Questions
1. ### math
1. What is the slope of the line that passes through the points (-2, 5) and (1, 4)? a) -3 b) -2 c) -1/3 d) 1/3*** 2) a line has slope -5/3. Through which two points could this line pass? a) (12, 13),(17, 10) b) (16, 15), (13,
2. ### Geometry
Help! I'm so confused :( 1. Supply the missing reasons to complete the proof. Given: angle Q is congruent to angle T and line QR is congruent to line TR Prove: line PR is congruent to line SR Statement | Proof 1. angle Q is
3. ### math
graph the line with slope 1/2 passing through the point(-5,-2) find the slope of the line 5x+5y=3 write answer in simplest from consider the line 2x-4y=4 what is the slope of a line perpendicular to this line. what jis the slope
4. ### English
1. Which line from the poem “Wave” best develops its tone? Wave My child (line 1) Born to me by water (line 2) And by water (line 3) Swept away. (line 4) (1 point) a) line one b) line two c) line four d) line three My Answer:
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Consider the scatter plot shown. Which shows the line of best fit? Explain your reasoning. A scatter plot is shown in the xy-plane. The values on the x-axis range from 0 to 9 in increments of 1 and the values on the y-axis range
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what acute angle does a line of slope -2/3 make with vertical line.
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1. In the middle of the page, draw two vertical lines an inch long, an inch apart, and that are parallel to each other. 2. About a half inch below the lines, draw a half-inch horizontal line that is centered between the lines. 3.
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What was the name of the German defensive line that ran nearly 400 miles across Germany’s western border, extending from the Netherlands southward to Switzerland? Gothic Line Siegfried Line Maginot Line Alpine Line b
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1. Registered User
Join Date
May 2004
Posts
18
I need to manipulate a string and take part of it into a variable
Example: Let say I have a string : "Bob Doe"
and a variable lastname
I want to manipulate the string so I only take the Doe for lastname so that
lastname = "Doe"
thanks all in advance for helping me
2. Cavalier King Charles
Join Date
Dec 2002
Location
Préverenges, Switzerland
Posts
3,740
you first need to define how you want "Doe"
"Cruella de Ville"
"de Ville"
"deVille"
"Cruella"
"C. de Ville"
"Cruella de Ville Jr."
"Cruella Samantha de Ville"
...and you can imagine lots of other wierd cases i'm sure.
what is the "Doe" result do you want for these?
izy
3. Registered User
Join Date
Jun 2004
Location
Terrapin Nation
Posts
205
What you need to do is find the position with the space [chr(32)] and use a combonation of the length + inst functions to extract the last name. For instance:
Dim FullName As String
Dim LastName As String
Dim SpaceAt As Integer
Dim NameLength As Integer
FullName = "Terp FanInMD"
NameLength = Len(FullName)
SpaceAt = InStr(1, FullName, Chr(32))
LastName = Mid(FullName, SpaceAt, (NameLength - SpaceAt) + 1)
Debug.Print LastName
returns FanInMD
4. Cavalier King Charles
Join Date
Dec 2002
Location
Préverenges, Switzerland
Posts
3,740
i guess it is impossible to have a perfect surname extract algorithm, so you end up with a compromise of one sort or another.
TerpInMD's algo returns "W. Bush" from "George W. Bush"
an alternative algo could look for the last space, but this would return "Jr." from "Cruella de Ville Jr."
izy
5. Registered User
Join Date
Jun 2004
Location
Terrapin Nation
Posts
205
He could check the data before it is ever inputed so that it is in forms which he can search out the last name.
Per his example, I think he may have done that. If not, he should.
6. Registered User
Join Date
Jun 2004
Location
Terrapin Nation
Posts
205
He could check the data before it is ever inputed so that it is in forms which he can search out the last name.
That "so that it is in the form he wants so he can search it".
sorry, in a hurry.
7. Registered User
Join Date
May 2004
Posts
18
I am sorry, I might have complicate the question...
all I need to know is the position of the letter in a string
lets say i have "Bob Doe"
I want to retrieve the "o" in the Doe, what do i have to do?
I know it's position is 6
so how do i retrieve a letter in position 6?
8. Registered User
Join Date
Aug 2002
Location
Northampton, England
Posts
266
The problem is that it might not always be the sixth character. Here is some code that will extract portions of text.
[Name] = “David Neagle”
Returned: David
Expression in field of query
expr: Left([Name],InStr(1,[Name]," ")-1)
[Name]=”David Neagle”
[Name]=”David John Neagle”
[Name]=”David J Neagle”
Returned: Neagle
Returned: John Neagle
Returned: J Neagle
Expression in field of query
expr: Right(Trim([Name]),Len(Trim([Name]))-InStr(1,[Name]," "))
[Name]=”David Neagle”
[Name]=”David-John Neagle”
[Name]=”David J Neagle
Returned: David
Returned: David-John
Returned: David
Expression in field of query
expr: Right(Trim([Name]),Len(Trim([Name]))-InStr(1,[Name]," "))
[Name]=”Neagle, David”
Returned: Neagle
Expression in field of query
expr: Left([Name],InStr(1,[Name],",")-1)
[Name]=”David John Neagle”
[Name]=”David J Neagle”
Returned: Neagle
Returned: Neagle
Expression in field query
expr: Right(Trim([Name]),Len(Trim([Name]))-InStr(InStr(1,[Name]," ")+1,[Name]," "))
[Name]=”David John Neagle”
[Name]=”David J Neagle”
Returned: John
Returned: J
Expression in query field
expr: Trim(Mid([Name],InStr(1,[Name]," ")+1,InStr(InStr(1,[Name]," ")+1,[Name]," ")-InStr(1,[Name]," ")))
9. Registered User
Join Date
May 2004
Posts
18
Dont' worry about the name covention is whatever...
i just want to get the character in a specific position on the string
like
let say my string is "ASDSXTXFSFSF"
I want the character in position 7 which is just a "T"
that's all i want in return: I want to specify characterposition = 7
returning character = "T"
Don't worry about the names and stuff..
sorry you be so confusing, maybe my english isn't clear
Thank you all for helping
10. Registered User
Join Date
Apr 2004
Location
Sydney Australia
Posts
369
Originally Posted by tb_vball
I am sorry, I might have complicate the question...
all I need to know is the position of the letter in a string
lets say i have "Bob Doe"
I want to retrieve the "o" in the Doe, what do i have to do?
I know it's position is 6
so how do i retrieve a letter in position 6?
xyz: Mid([YourFieldName],2,1)
xyz: Mid([YourFieldName],3,6)
The first number is where it starts and the second number is the number of characters.
So from Smith comes
m
ith
From Alexander comes
l
exande
A space is also picked up.
So xyz: Mid([YourFieldName],3,6) and B radley gives
radle and with a space before the radle, that is, radle is one space across from the left.
John Doe would give hn Doe
11. Registered User
Join Date
May 2004
Posts
18
Thank you so much...
that's just exactly what i needed
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# Mosfet Diagram
75 Threads found on edaboard.com: Mosfet Diagram
## Analog switching between two sources 5V/1A and 5V/2A
Consider doing the switching with P-channel logic level mosfet devices. The diagram below show the simulation with normal type PMOS.
## How do I make a programmable variable power resistor circuit?
The variable resistor element in the diagram below can possibly be replaced by an mosfet to act as a controlled variable resistor. Such an arrangement is shown in an AC voltage regulator circuit by the late Jim Williams of Linear Technology. You may need more than one type of mosfet with varying on resistances. These can be made (...)
## Drive Mosfet with 5V
hi, The IRLML6402 is a P mosfet, check the voltage polarities E
## How to use "P channel" Mosfet with Micro(3.3V)
I see it is working but very sensitive. For example When I touch the Gate pin on mosfet or C,B on NPN I see some blink on output. You should also take care that the circuit isn't protected against hammer blows, flooding or fire... Seriously speaking, it's exactly as "sensitive" against electrical interferences as others circuits
## Gate driving IC for P chanel MOSFET IRF 4905
Hi All, I am attaching a s diagram please guide me that if it will work in real world and also guide me that how can i add mosfet gate driver in this circuit such as TC 4420.
## Best Books to understand Noise
Hello I have many problems to understand Noise , Noise figure , Shot noise and etc and I have to know completely about calculate Noise to calculate BJT , mosfet and Block diagram amplifier. I will appreciate that you suggest a good book about Noise with simple steps Thanks
## IRFP460 bias resistor
In the given diagram the near IC IR2148 mosfet is IRFP4321 and the bias resistor is 6 OHM . If i replace these two mosfet with IRFP460 , what should be the resistor value ? can someone give me the answer ? 117729
## DIAGRAM or SCHEMATIC needed 12 volt 172 ampere NPN MOSFET using an opto coupler
If you mean an N-channel mosfet in the low voltage (ground side) of the fan, connect is source to ground, drain to the fan and gate to source through a 10K resistor. The connect the optocoupler emitter to the gate and it''s collector throgh a 1K resistor to +12V. No capacitor is needed. It might be useful to connect a diode across the fan, cathod
## Schmitt Trigger and Driver, inverting vs non-inverting
Hi All 110663 Regarding the attached diagram, to drive a N-channel mosfet switch, what is the difference between using "combination of an inverting schmitt trigger & an inverting gate driver" and "combination of a non-inverting schmitt trigger & a non-inverting gate driver"? I don't underst
## [Moved] problem in 12v to 230v inverter
hello, i have made an inverter using cd4047 as per the circuit diagram i have uploaded here.. as per the circuit diagram gate pulse are given to the mosfet from pin 10 and pin 11 of the IC . but when i apply 12v to the ic i am getting 12v at pin 10 but not getting any voltage at pin 11 but i am getting 12v from pin 13 so i have connected (...)
## dc-ac inverter using astable to drive some 2N3055
...please give me some suggestion about the schematic diagram that i uploaded Inverters based on BJT transistors nowadays is somewhat obsolete. The mosfet transistors brings better efficiency in terms of energy usage. Another problem on applying 2N3055 is its Beta parameter which may start from 20, requiring 1,5
## H-Bridge output fails with capacitive load.HELP!!!!!
hi everyone!!! i just built a modified smps inverter rated 600watts.the problem am facing now is that when i plug in rechargeable led flash light,my ir2110 get damage but work with appliance with using sg3525 to control two ir2110 at d H-bridge side with irfp460 mosfet.i didnt apply any filter at the final output.switching frequency(
## pure sine wave inverter mosfet
I am working on pure sine wave inverter . i have finished frst par ( 12v DC to 300 Dc ) in second part i have used ir2101 to derive 4 irf840 ,switching freq 16 khz . but mosfets get too much hot and burnt ... any help ???? should i change driver ic ??? mosfett or what changing should i adapt . i hav used Capacitor of 33uf any help plz ?
## MOSFET H-bridge with PIC16F876A
Hello,I am repairing my DC-AC converter it was made up mainly by a high frequency transformer 12-0-12 which is connected to six mosfet controlled at 30KHz ,the second block of my device is made up by 4 power mosfet (which makes an H-Bridge),4 power diodes and the filtering block. An error has made then the output of my device has been connected
## Current mirror circuit simulation using NGSpice
the mosfet through which Iout flows has drain left open. The drain must be connected to something otherwise Iout = zero. The circuit diagram of the current mirror is attached with the thread. Where? I can't see it.
## 15v 6amp switched mode power supply circuit
15V * 6A = 90W So you can use flyback topology with a single switch (transistor/mosfet). That may be the cheapest option.
## pickup coil AC signal (schematics)
I need to draw a circuit diagram how to connect pickup coil (2 wire tacho generator) to a 8 bit PIC microcontroler in order to read the signal. I found some schematics on the internet (attachment 1), that I don't see being correctly done. I don't understand why there is a rectifier diode and 10k resistor on the source pin of mosfet, when source sh
## Why are my gate drivers breaking
It is difficult to suggest solutions without taking a look at the circuit diagram. At the output of the DC-DC converters, you should connect filter capacitors. Which DC-DC converter(s) are you using? Does the gate driver drive the mosfet with up to 20V? That might be a problem. What is the VGS(max) of the mosfet(s) you are using? If it is (...)
## step up dc to dc converter
You can use a dedicated boost controller chip, or you can design your own circuit using a common PWM controller such as UC3845 and a mosfet in a boost topology. Hope this helps. Tahmid.
## Need Schemetic for PWM type Mosfet Based AC to AC Voltage Converter
Hi All ... I need the PWM type Voltage stabilizer, i have attached a block diagram to understand .... Can some 1 help me please? Thank You ...
## 600V 500mA switching using microcontroller
I think the 150Hz refers to the maximum switching speed, not the 600V frequency. The 600V is DC according to the diagram, am I correct? If so, a triac won't turn off. How about an IGBT? For instance, the NGTB15N120FL from OnSemi, or a power mosfet, e.g. the NDD03N80Z from OnSemi. I like OnSemi! Do you really need to switch the negative rail as w
## mosfet switching transients
Q1 mosfet will be always in on condition Not with the circuit shown in post #4.
## Negative output voltage MOSFET driver
Hello, How can I drive this P-type mosfet shown in the attached figure? Does someone know a negative output mosfet driver with a positive supply voltage (Vcc) referenced to the ground shown in the diagram (this is necessary as other components in the circuit use the same Vcc supply referenced to that ground)? I have checked and could not (...)
## interfacing Rs332-953 stepper motor with 8051
dear sir i have written code for 3 modes that is working well...... i am using irf540 mosfet for driving unipolar motor without connecting motor i hav seen wave forms on osciloscope, that are perfect but when i connect motor with mosfets it did'nt starts...motor is 12v/250mA kindly suggest me a solution
## Oscilloscope probe pulls signal low
Hi, I have a pic connected to a mosfet switch. One of the outputs of the PIC should enable/disable the mosfet. However, I have found that the mosfet is switched on at all times - regardless of which state I program the PIC output to be. If I touch the output trace with an oscilloscope probe to check its state, the signal gets pulled (...)
## MOSFET DRIVER for ZVS BUCK convertor..need help...
hi... i am making a conventional ZVS buck convertor with following specifications... Vin=19V; I=2 A ; Vo=13V; switching frequency= 18kHz. i am using IRFZ44N N-channel mosfet for switching... i have only IR2110 mosfet driver available with me..... i need to know if its OK to use this driver for switching high side mosfet IRF Z44N (...)
## dc chopper circuit for 1.1 kw DC motor usinf MOSFET
hi My name is Afrasiab and i am new at this forum. I am looking for a dc motor drive or DC chopper circuit which i can use to control DC voltage for a 1.1 KW motor. The motor is a separately excited DC motor. the rating for motor is 160v/9amps for armature and 190v/.74 amps for field..... I am attaching a diagram which i made on protues.....
## Switching circuit for piezo circuit
You may use a P-mosfet and a TL431, see the diagram:
## Inverter circuit Diagram
Hi everyone i need an inverter circuit diagram 1000w or more but i need it without big transformer i will be grateful A.E Hi friend......... See the following link.. 1000W Inverter Circuit with mosfet | Inverter Circuit
## tlp250 and mosfet circuit diagram
hi anybody have circuit diagram for the connection between mosfet and ic tlp250(driver ic for mosfet).
## question: power MOSFET with clamped inductive load ?
Hello, To study the switching behavior of a power mosfet, a clamped inductive load is often used. Attached is a diagram showing this clamped inductive load. Can some someone explain why we need this type of load ? Thanks.
## circuit diagram of inverter 3000 watt pure sine wave full bridge with mosfet and ir21
respected forum, many thanks to any one sending me the complete circuit diagram of 3000 watt inverter pure sine wave full bridge with mosfet and ir2110 driver.
## Masks in fabrication of MOS
Hi, Can anyone tell why we need separate mask for oxide and polysilicon in gate formation of mosfet.Why can't we use the same mask for both. Regards.
## Help with highpower LED circuit protection
Better to use a current protection circuit, see for example the attached diagram. The P mosfet need to be mounted on a proper heatsink, short circuit current value limited by the 0.33 ohms resistor I= 0.6V/R
## sin pwm invertor usins MOSFET
i am using sin pwm of 12kHz with 50hz sine wave, my problem is , i got pwm , but when i connected to H bridge of mosfet (IRFP150) , the DC supply(40v) geting shorted. i am using bipolar switching ---------- Post added at 14:09 ---------- Previous post was at 13:18 ---------- nw i connected 50Ω.25w resi
## MOSFET problem open even two MOS are not open
Is your question related to the leakage current of the mosfets? A mosfet can't turn completely off , there is a current of a few uA passing through the mosfet even when it is off, the parameter is described in the datasheet. Alex
## 'crude' understanding of how IC is made
I don't think, that your diagram from post #14 can be related to a real semiconductor part. I guess, you are mixing somehow a bipolar junction structure with gates from a mosfet. The surface where a N and P zone contact each other forms a junction. You can reduce the surface by reducing the size of either the N or P zone, but you can't place arbitr
## Help me with this. PIC16F877 program
I JUST NEED TO MAINTAIN THE OUTPUT VOLTAGE AT 24V Circuit diagram of the proposed system Figure 10 shows that power stage included switch SW where it may consists of one or more parallel connected power mosfet, a fast switching type flyback diode D, an inductor L wound onferrite core with air gap to prevent core saturation, and output capacitor
## 1000W 12V d.c. to 230V a.c. power schematic diagram
i have seen this circuit , but in my design there is no transistor like this, mosfet driving directly from 3525 found no problem in past 7 years, any difference in pwm mode by using this transistor?
## can you help me solve this? [homework]
a) mosfet b) Drain (top), Source (bottom), Gate (left) c) Vgate = 200k/(200k + 430k) * 12 = 3.81 volts Igate = 0 A for the rest you would need Kn, to use the equation Id = 1/2 k'n(W/L)^2. d) 1/2 k'n (Vgs - Vt)^2 = .00392, although this may be incorrect because all different teachers use different terminology. I hope this helps
## MOSFET Burning Out need help
Post your circuit diagram.. How do you drive the High side mosfets? You may need a high side driver
## How can get a contant and abrupt DC output?
Putting a mosfet in series with the 5V and turning it on after a suitable delay could work. There would be some loss of load regulation caused by the on resistance, but if the load isn't too large it wouldn't be too bad.
## Mosfet gate drivers for ZVS driver circuit
I've been working on a zvs driver circuit, whereby a tl494 drives a irf540 mosfet directly at 100kc's. If I run the whole circuit on a single supply is runs fine, however if I use seperate power supplies for the tl494 and the mosfet load, and run the tl494 a few volts more than the load, the the mosfet runs considerably cooler, so much so (...)
## Please Check My N-Channel MOSFETs H-Bridge Schematic Diagram
Let me show you exactly what the problem is in a simplified circuit with a high side Nmosfet 54967 You mosfet has an RDSon of 0.005 ohm and should have almost 0 voltage drop across drain-source but you can clearly see that you have a voltage drop of 3v because this is the balance point for this mosfet that has an low
## Circuit diagram of LM324 driving a MOSFET
any body have a good hand on High frequency drivers?? I have had a lot of burnt mosfets trying.
## mini inverter diagram
Hi, In the PM, you had asked for a better circuit. Here is with SG3525 would be nicer, but this will do quite well. You could change the mosfet gate resistors to 22ohm. Look
## UNDER VOLTAGE PROTECTION for inverter
Hi, you mention that your that your mosfet get burnt any time your battery went low can you tell me the mosfet is it the high frequency side or the low frequency 50Hz,also D16 and D19 Suppose to be connected to your vcc
## Ferrite core driving circuit?
designing a switch mode power supply which can be described by the block diagram above. The pulse generator circuit consists of a 555. Output of 555 is directly connected to the gate of the mosfet (6N60). A 100Ω resistor is connected to the secondary windings of the ferri
## SWITCH MODE ARC INVERTER WELDER SCHEMATIC
HI I AM LOOKING FOR CIRCUIT diagram OF INVERTER CONCEPT BASED ( IGBT OR mosfet ) WELDING MACHINE. KINDLY HELP...........
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Escape Velocity
How did the gas giants sweep up and hold onto their gaseous atmospheres?
How does the Earth hold onto it's atmosphere?
Why doesn't the Moon have an atmosphere?
Can atmosphere's change over time?
What is a black hole?
Whereas solids are bound to a planet (or a tiny rock or asteroid) by molecular bonds, gases can only be held to a planet by gravity.
In order to know how strongly they are held, we must understand what energy is required to escape from a planet (or any object) -- and we will measure this energy in terms of velocity -- and we then must understand how fast a gas is moving under specified conditions.
Escape velocity: Measuring the gravitational strength of an object
The escape velocity is the exact amount of energy you would need to escape the gravitational clutches of an object with mass. Since all objects have mass, they all have a measureable gravitational strength. A good way to think about escape velocity is to think about a deep well (physicists like to think of this as an energy well). If you are at the bottom of the well and want to get out (to escape), you need enough energy to climb out. The deeper the well, the more energy you will have to expend in order to climb to
the top. If you have only enough energy to get half way out, you will eventually fall back to the bottom. The escape velocity is a way of measuring the exact amount of energy needed to reach the lip of the well -- and have no energy left over for walking away.
When a ball is thrown up into the air from the surface of the Earth, it does not have enough energy to escape. So it falls back down. How might we enable the ball to escape? Throw it harder, give it more energy. How hard must we throw it? Just hard enough to get over the top, over the edge of the well.
We can find this energy directly by saying that the kinetic energy of the thrown ball must exactly equal the 'potential energy' of the well. From basic physics we know that the potential energy for an object at a height above a surface is:
Epotential= GMm/R
where
G = Newton's universal constant of gravity = 6.67 x 10-11 N-m2/kg3
M = the mass of the 'attracting object' [the planet] [in units of kg]
m = the mass of the object trying to escape [e.g., me or a ball or a rocket or a molecule] [in kg]
R = the distance between the centers of objects M and m [in units of m]
note: provided we do everything in the same units, we don't have to worry about units
while the kinetic energy we know from above:
Ekinetic=0.5 m v2
where
m = mass of the moving object [in kg]
v = the velocity of object m [in m/sec]
If we set these two energies equal to each other, and solve for v, we find the exact velocity needed to escape from the energy well:
0.5 m v2= GMm/R
v= (2GM/R)0.5
and since this velocity is exactly what is needed to 'escape,' it is called the escape velocity:
vescape= (2GM/R)0.5
Note what extremely important parameter is not in the escape velocity equation: the mass of the moving object. The escape velocity depends only on the mass and size of the object from which something is trying to escape. The escape velocity from the Earth is the same for a pebble as it would be for the Space Shuttle.
example #1: What is the escape velocity from the Earth?
MEarth=5.97 x 1024 kg
REarth=6378 km = 6.378 x 106 m
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 5.97 x 1024 kg / 6.378 x 106 m)0.5
vescape=1.12 x 104m/sec=11.2 km/sec
(this is equivalent to about 7 miles/sec or 25,200 miles per hour)
example #2: What is the escape velocity from the Sun?
MSun= 1.99 x 1030 kg
RSun= 700,000 km = 7 x 108 m
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 1.99 x 1030 kg/ 7 x 108 m)0.5
vescape=6.15 x 105m/sec=615 km/sec
example #3: What would be the escape velocity from Sun if it were the size of the Earth?
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 1.99 x 1030 kg/ 6.378 x 106 m)0.5
vescape=6.45 x 106m/sec=6450 km/sec
example #4: What would be the escape velocity from the Earth if it were the size of a pea? Take the diameter the pea to be one centimeter (radius = 0.005 m).
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 5.97 x 1024 kg / 0.005 m)0.5
vescape=3.99 x 108m/sec=399,000 km/sec
note: 399,000 km/sec is faster than the speed of light (300,000 km/sec). Thus, we have created (in our imaginary experiment) an object from which light cannot escape. According to the laws of modern physics, in this case according to a postulate of Albert Einstein, nothing, including light, can travel faster than light. Thus, if light cannot escape, then neither can anything else. This is a black hole. Clearly, for any object, we can find a size such that the escape velocity would be greater than the speed of light. Similarly, for any size object (or region of space), we can find a mass such that a mass crammed into that volume of space would be a black hole. The 'size' of the black hole is called the 'Schwarzchild Radius."
Question: To what size would I have to shrink the Sun in order to make it a black hole?
Question: Imagine that, just like now, the Earth orbits the Sun. Suddenly, I snap my fingers and turn the Sun into a black hole. What would happen to the Earth?
The Kinetic Theory of Gases: What is the speed of a molecule in a gas?
Now we know the velocity required to escape from a planet. This is equivalent to knowing the maximum velocity an object can have and still be gravitationally bound to that object. We next need to know how fast gas molecules move.
Gases are characterized by their temperature, with the temperature determing the average velocity of a gas molecule. According to the kinetic theory of gases developed in the 19th century, the amount of thermal energy per atom or molecule in a gas is given by the formula
Ethermal= 3kT/2
where
T is the temperature (in Kelvins),
k is the Boltzman constant = 1.38 x 10-16 gm cm2/(sec2 degrees).
For a moving object, the thermal energy is equal to the kinetic energy
Ekinetic=1/2 m vaverage2
where
m = mass of the moving atom or molecule in the gas (in gm)
vaverage = the average velocity of an atom or molecule in the gas (in cm/sec)
Since Ethermal= Ekinetic, we can set these two formulae equal to each other
1/2 m vaverage2 =3kT/2
and then we can solve for vaverage to find
vaverage = (3kT/m)0.5
Note what the average velocity depends on: it is directly proporational to the temperature and inversely proportional to the molecular/atomic mass. Thus, at a single temperature, the average velocity of light elements (hydrogen, for example) will be faster than the average velocity of heavy elements (e.g., radon) at the same temperature. And, for a given element/molecule, the average velocity increases as the temperature increases.
Note that in a gas some molecules are moving faster and some slower than average. The distribution of velocities is known as the Boltzman distribution. Fully 0.5% of all the molecules (1 out of 200) are moving faster than twice the average and only 0.5% are moving slower than 20% of the average. But, as we shall see, it is these 0.5% fast movers that are critically important when we talk about atmospheres.
As an illustration of a Boltzman-like distribution, think of cars on the interstate. Most are moving at speeds close to the speed limit of 65 mph. In fact, probably 70-90% of the cars are moving between 60-70 mph. However, a few are moving faster and a few are moving slower. And perhaps 1 out of ever 200 cars is moving 100 mph while another 1 out of 200 is puttering along at 45 mph. Now supposed that a highway patrol officer pulls over and tickets the 100 mph driver. What happens? Someone else speeds up and replaces the fastest car so that there is always one bozo driving much too fast.
example #1: We can calculate the average velocity of a hydrogen atom at the surface of the sun . We need to know the mass of a hydrogen atom (mhydrogen = 1.67 x 10-24 gm) and the temperature of the surface of the sun (T = 5800 K):
vaverage (H) = [ 3 x (1.38 x 10-16 gm cm2 sec-2 deg-1) x (5800 deg) / (1.67 x 10-24 gm) ]0.5
vaverage (H) = [1.44 x 1012]0.5 = 1.20 x 106 cm/sec x [1 km/ 105 cm]
vaverage (H) = 12.0 km/sec
example #2: We can calculate the average velocity of an aluminum atom (note that even aluminum will be a gas in atmospheres of stars) at the surface of the sun. Note that the aluminum atom is 27 times more massive than a hydrogen atom:
vaverage (Al) = [3 x (1.38 x 10-16 gm cm2 sec-2 deg-1)x(5800 deg)/(27 x 1.67 x 10-24 gm)]0.5
vaverage (Al) = [5.3 x 1011]0.5 = 2.3 x 105 cm/sec x [1 km/ 105 cm]
vaverage (Al) = 2.3 km/sec
Note that the escape velocity of the Sun (calculated above) is 615 km/sec, which is more than 50 times greater than the average velocity of H at the surface of the Sun. And since all other elements are heavier than H, they will have lower average velocities, it is clear that these materials cannot escape from the Sun.
example #3: The Sun has an outer layer called the corona, in which the temperature is well above a million degrees. Can hydrogen atoms escape from the corona, if T = 4,000,000?
vaverage (H) = [3 x (1.38 x 10-16 gm cm2 sec-2 deg-1)x(4,000,000 deg)/(1.67 x 10-24 gm)]0.5
vaverage (H) = [9.92 x 1014]0.5 = 3.15 x 107 cm/sec x [1 km/ 105 cm]
vaverage (H) = 315 km/sec
??? It appears that the average H atom cannot escape, even at 4 million K. However, remember that this is an average velocity, and that 1 out of every 200 atoms will be moving twice as fast. So, 0.5% of the atoms will have velocities in excess of 630 km/sec, which is above the Sun's escape velocity. These atoms can stream outwards and escape from the Sun.
2 + 2: Escape velocity and the kinetic theory of gases and the snow line
At the distance of the snow line (let's call this 5 AU), the temperature of material illuminated and heated by the Sun is about 150 K (-120 C). Thus, the averge velocity of a hydrogen atom
vaverage (H2) = [ 3 x (1.38 x 10-16 gm cm2 sec-2 deg-1) x (150 deg) / (1.67 x 10-24 gm) ]0.5
vaverage (H2) = 1.92 km/sec
Theory and modeling work suggest that a planet can hold onto it's atmosphere if the escape velocity is about 6 times greater than the average kinetic velocity. Thus, a planet with vescape greater than 6vaverage can hold on to H2 at 150 K.
First, the easy situation: imagine an object with vescape = 1.9 km/sec (e.g., the Moon) and a gas temperature of 150 K. Clearly, even our average H2 molecule (1.92 km/sec) can escape. So quickly, all the hydrogen would escape from the Moon to space.
Now for the intermediate case: what about a planet with vescape = 8 km/sec and a gas with vaverage = 2 km/sec? In this case, vescape is less than 6vaverage (12 km/sec). This means that the fastest few molecules will escape. Once they escape, a few other molecules will find there way to higher speeds and then they will escape. And bit by bit, molecule by molecule, all the gas will trickle out this small 'opening' until the atmosphere is completely depleted in this gas.
Finally, the other limit: to hold onto H2 at 150 K we only need an object with vescape greater than 11.5 km/sec and we're all set. Since the Earth has vescape = 11.2 km/sec, it would appear that an Earth-mass sized object at the distance of Jupiter could hold onto hydrogen gas.
However, a planet can only hold on to gas already captured in its atmosphere. So we must ask whether the planet can capture an atmosphere.
Note that Jupiter's orbital velocity around the Sun (Jupiter's orbital period divided by the circular path traveled by Jupiter in its orbit) is about 13 km/sec. Thus, the gas orbiting the Sun at 5 AU is moving at about this speed and our protoJupiter must be able to capture and hold onto this gas, which is harder than holding on to gas already captured.
A good rule of thumb is that we need an escape velocity twice as large as the orbital velocity in order to capture orbiting gas. So, what mass object do we need. If we doubled the radius of the Earth, the volume of the Earth would increase by a factor of two cubed (8). Thus, an object with a radius of 2 x 6378 km, and made of Earthy material, would have a mass about eight times larger than that of Earth. The escape velocity from such an object would be:
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 8 x 5.97 x 1024 kg) / (2 x 6.378 x 106 m)0.5
vescape= 22.3 km/sec
which isn't quite as large as 2 x 13 km/sec = 26 km/sec. This is how we get our rough estimate that we need an object with ten (a bit more than 8) Earth masses in order to capture H gas from the solar nebula. Clearly, once the gas is caught, the protoplanet can easily hold these gases.
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# Lecture notes on R - comparison
## by Laszlo Baranyai, Associate Professor at Corvinus University of Budapest on Mar 11, 2012
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Applied mathematical statistics in agricultural engineering with R. Topic: comparison. New functions: multiple range test, errorbar.
Applied mathematical statistics in agricultural engineering with R. Topic: comparison. New functions: multiple range test, errorbar.
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## Lecture notes on R - comparisonPresentation Transcript
• Lecture notes on RApplied mathematical statistics in agricultural engineering with R comparison László Baranyai, PhD Corvinus University of Budapest Department of Physics and Control 2012
• Introduction of data tableMorphological data, sepal and petal size in centimeters, ofIris flowers1 were measured and collected into one table:> iris[1:2,] Sepal.Length Sepal.Width Petal.Length Petal.Width Species1 5.1 3.5 1.4 0.2 setosa2 4.9 3.0 1.4 0.2 setosa> summary(iris\$Species) setosa versicolor virginica 50 50 50One row corresponds to one piece of flower.1 Anderson, E. (1935) The irises of the Gaspe Peninsula, Bulletin of the AmericanIris Society, 59, 2–5.
• Summary of data columnsThe summary() function can be used to make basiccomparison with descriptive statistics:> summary(iris\$Sepal.Length[iris\$Species=="setosa"]) Min. 1st Qu. Median Mean 3rd Qu. Max. 4.300 4.800 5.000 5.006 5.200 5.800> summary(iris\$Sepal.Length[iris\$Species=="versicolor"]) Min. 1st Qu. Median Mean 3rd Qu. Max. 4.900 5.600 5.900 5.936 6.300 7.000> summary(iris\$Sepal.Length[iris\$Species=="virginica"]) Min. 1st Qu. Median Mean 3rd Qu. Max. 4.900 6.225 6.500 6.588 6.900 7.900The shape of distribution is partially shown by the 25%,50% and 75% quantiles.
• BoxplotThe figure is created by the following command. Comparebox lines with previous results provided by summary.> boxplot(Sepal.Length ~ Species,data=iris,col="lavender")
• One-sample t-testThe one-sample t-test may be used to calculate theconfidence interval for mean.> t.test(iris\$Sepal.Length) One Sample t-testdata: iris\$Sepal.Lengtht = 86.4254, df = 149, p-value < 2.2e-16alternative hypothesis: true mean is not equal to 095 percent confidence interval: 5.709732 5.976934sample estimates:mean of x 5.843333> t.test(iris\$Sepal.Length)\$conf.int[1] 5.709732 5.976934attr(,"conf.level")[1] 0.95
• Multiple range testThe following function creates a summary table.range.test <- function(Data,Class){ Name <- sort(unique(Class)) Mean <- rep(0,length(Name)) N <- Mean SD <- Mean CI95.min <- Mean CI95.max <- Mean for (i in 1:length(Name)) { tmp <- Data[Class==Name[i]] Mean[i] <- mean(tmp) N[i] <- length(tmp) SD[i] <- sd(tmp) CI95.min[i] <- t.test(tmp)\$conf.int[1] CI95.max[i] <- t.test(tmp)\$conf.int[2] } return( data.frame(Name,Mean,SD,N,CI95.min,CI95.max) )}
• Multiple range testThe function range.test() computes basic statisticalparameters typically used for comparison. Usage:> range.test(iris\$Sepal.Length,iris\$Species) Name Mean SD N CI95.min CI95.max1 setosa 5.006 0.3524897 50 4.905824 5.1061762 versicolor 5.936 0.5161711 50 5.789306 6.0826943 virginica 6.588 0.6358796 50 6.407285 6.768715Decision on similarity is done according to the- overlap of confidence intervals- least significant difference derived from mean and sd- mean and its ±2×sd or ±3×sd environment
• Two-sample t-testThe t-test can be used to compare groups provided thatthey follow normal distribution:> a <- iris\$Sepal.Length[iris\$Species=="setosa"]> b <- iris\$Sepal.Length[iris\$Species=="versicolor"]> t.test(a,b) Welch Two Sample t-testdata: a and bt = -10.521, df = 86.538, p-value < 2.2e-16alternative hypothesis: true difference in means is notequal to 095 percent confidence interval: -1.1057074 -0.7542926sample estimates:mean of x mean of y 5.006 5.936
• Kolmogorov-Smirnov testThe non-parametric, also called robust, version of test ofsimilarity is the Kolmogorov-Smirnov test:> ks.test(a,b) Two-sample Kolmogorov-Smirnov testdata: a and bD = 0.78, p-value = 1.230e-13alternative hypothesis: two-sidedWarning message:In ks.test(a, b) : cannot compute correct p-values with tiesThe test value of D=0.78 is considered to be significant.
• Analysis of variancesThe analysis of variances (ANOVA) compares within andbetween group variances to test whether at least onegroup differs significantly from others.> summary(aov(Sepal.Length ~ Species,data=iris)) Df Sum Sq Mean Sq F value Pr(>F)Species 2 63.212 31.606 119.26 < 2.2e-16 ***Residuals 147 38.956 0.265---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1According to above result, species have significant effecton sepal length.
• ErrorbarThe output of function range.test() can be used tomake graphical comparison and plot error bars:errorbar <- function(mrt){ x <- seq(1,length(mrt\$Name),by=1) y <- c(mrt\$CI95.min,mrt\$CI95.max) plot(c(min(x),max(x)),c(min(y),max(y)),type="n",xaxt="n", xlab="Classes",ylab="Data") axis(1,at=x,labels=mrt\$Name) points(x,mrt\$Mean,col="blue") lines(x,mrt\$Mean,col="blue",lty="dashed") arrows(x,mrt\$CI95.min,x,mrt\$CI95.max, code=3,angle=90,length=0.1,col="blue")}
• ErrorbarThe usage is simple:> md <- range.test(iris\$Sepal.Length,iris\$Species)> errorbar(md)
• SummaryComparison of experimental results may be done usingreports of- two-sample t-test (depends on distribution)- Kolmogorov-Smirnov test (non-parametric)- analysis of variances- multiple range test and derivativesGraphical tools for comparison were introduced- boxplot (it has information about distribution)- errorbar
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# Swift Taylor Approximations
This post continues my explorations of simple intelligence.
In this post, we’ll consider some elementary sensing cells. We’ll then look at whether we can apply local function approximators.
## Sensors
Consider a set of sensing cells, which we’ll call “sensors”.
We have N sensors, where each sensor measures a value over time. This value could be a magnitude or intensity, such as a lightness or colour intensity for vision, a frequency magnitude for sound, or a level of cell damage for pain. We’ll assume the N sensors form a small group – possibly a receptive field. The N sensors could therefore form part of a cortical column.
We can represent the set of N sensors as a column vector. The column vector represents a measurement at one point in time. We’ll represent each sensor using an index $n$. You could call each sensor a “feature”. We’ll represent time using a set of discrete time steps, where a particular time step is represented by the variable $t$. Hence, we have something like:
$\vec{S_t} = \begin{bmatrix} s^t_{0} \\ s^t_{1} \\ \dots \\ s^t_{n} \\ \dots \\ s^t_{N-1} \end{bmatrix}$
Note: we’ll start counting things in this post from 0 (as this matches Python implementations). This means that if we have N sensors our vector ends with the N-1th entry.
## Measured Functions
To start, we can imagine that the measurements of each sensor conform to a function.
The function may be considered a function of time that is specific to the sensor. This gives us: $s^t_n = f_n(t)$
To makes things easier, we’ll consider a quantised or discrete time base. Hence, t has integer values from 0 to $\infty$.
If we collect batches of samples (i.e. measurements) from the sensors over time, we can pop these into a matrix form $\mathbf{S}$ . If our set of sensor measurements at time $t$ is a column vector $\vec{S_t}$, where each row relates to a different sensor, then we can stack these columns horizontally over a time period with P time steps to generate an N by P (rows by columns) matrix:
$\mathbf{S} = \begin{bmatrix} s^0_{0} & s^1_{0} & \dots & s^t_{0} & \dots & s^{P-1}_{0}\\ s^0_{1} & s^1_{1} & \dots & s^t_{1} & \dots & s^{P-1}_{1} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ s^0_{n} & s^1_{n} & \dots & s^t_{n} & \dots & s^{P-1}_{n} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ s^0_{N-1} & s^1_{N-1} & \dots & s^t_{N-1} & \dots & s^{P-1}_{N-1} \end{bmatrix}$
(Note: I use P instead of T to avoid confusion with the transpose operator – T – which we’ll use later.)
## Covariance
We can use $\mathbf(S)$ to calculate a covariance matrix $\mathbf{M}$ for the time period.
Now officially we shouldn’t do this. Our samples are not I.I.D. – independent and identically distributed. Indeed, our function assumes there is some dependence over time. It is also likely that probability distributions for the sensor intensity change with time. But let’s play around for just one post to see what happens when we mix the streams. In effect, we are considering each time step to be an independent trial with a common intensity probability distribution, yet also considering each sensor to record a function over time.
### Mean Adjustments
When dealing with a number of statistical samples, it is recommend to start by adjusting the data to have a mean of zero. This means that we need to subtract the mean measurement of each sensor over the time period from the sensor readings for that sensor. Let’s call the mean for the time period $\overline{S}$ – this is column vector of length N (i.e. size N by 1). The nth entry in $\overline{S}$ is calculated as:
$\overline{s}_n = \frac{1}{P}\sum_{t=0}^{P-1} s^t_n$
We need to be a little careful. Because we are using an discrete integer time base where a time index is incremented by 1 at each time step, we can substitute P for the number of samples. If we were using a different time base, this would not be possible.
To get a mean adjusted set of measurements, we perform a column-wise subtraction:
$\vec{S^{\prime}_t} = \vec{S_t} - \overline{S}$
This gives us an adjusted data sample matrix $\mathbf{S'}$.
### Covariance Calculation
Given the adjusted data sample matrix $\mathbf{S'}$, we can compute our covariance matrix as:
$cov(\mathbf{S'}, \mathbf{S'}) = \mathbf{M} = \frac{1}{P}\mathbf{S'}\mathbf{S'}^T$
(Note: sometimes this is written as simply $var(\mathbf{S'})$ but I’ll explicitly put in that we are looking at covariance between the sensors as later we may consider covariance between our sensors and something else. Or to put another way, here we are determining a cross-covariance matrix with ourselves or a variance matrix.)
In the above calculation, we have a first matrix $\mathbf{S'}$ of size N by P and a second matrix $\mathbf{S'}^T$ of size P by N (the transpose operation just swaps our rows and columns over). The result is a matrix $\mathbf{M}$ of size N by N, i.e. where an ijth value in the matrix represents the covariance of the ith sensor with the jth sensor or a measure of how the ith sensor changes with the jth sensor.
Now we’ll just park our covariance matrix for a moment. We’ll need it later when we look at decorrelating the sensor readings. But now we’ll go on a little detour into function approximators…
## Taylor Polynomials
Let’s return to our sensor measurements over time.
From before, we considered that we can model each sensor with a function that changes over time: $s^t_n = f_n(t)$.
Consider a moment. This one. That one. Any moment.
We can approximate the sensor function $s^t_n$ at that moment using a Taylor Series Expansion. If we call the moment $m$ we have:
$f_n(x) = f_n(m) + f'_n(m)(x-m)+\frac{f''_n(m)}{2!}(x-m)^2 + \frac{f'''_n(m)}{3!}(x-m)^3 + \dots + \frac{f^{k}_n(m)}{k!}(x-m)^k + \dots$
If we consider a moment at t=0 then this becomes a Maclaurin series:
$f_n(x) = f_n(0) + f'_n(0)(x)+\frac{f''_n(0)}{2!}(x)^2 + \frac{f'''_n(0)}{3!}(x)^3 + \dots + \frac{f^{k}_n(0)}{k!}(x)^k + \dots$
Now, we can use this to determine an approximation to the function at the moment. As the series is a sum of terms we can take different groups of successive terms as different levels of estimation. The first term is a relatively poor estimate, the first and second terms are better but still reasonably poor, the first to third terms are better again but still not as good as having more terms etc. You may have come across a similar approach when considering the mathematical notion of moments (although these are unrelated to our “moment” in time). The underlying idea crops up a fair amount across different fields, annoyingly referred to as different things in each field.
The moment in time matters because the accuracy of approximations is centred around the moment. The further away from the moment, the greater the divergence from the actual function. You can see this in the following example of approximating $\sin(x)$ around x=0 from Wikipedia:
The approximation becomes increasingly bad the further we stray from 0.
Now if we consider the Maclaurin series of the function $f_n(x)$ and look at the function as a time-series function where $x = t$ then we can simplify a little with the following expression:
$s^t_n \approx a_n + b_n*t + c_n*t^2 + d_n*t^3 + \dots$
We can also consider the set of sensor signals as separate functions that are approximated with different coefficients:
$\vec{S_t} = \begin{bmatrix} s^t_{0} \\ s^t_{1} \\ \dots \\ s^t_{n} \\ \dots \\ s^t_{N-1} \end{bmatrix} \approx \begin{bmatrix} a_0 + b_0*t + c_0*t^2 + d_0*t^3 + \dots \\ a_1 + b_1*t + c_1*t^2 + d_1*t^3 + \dots \\ \dots \\ a_n + b_n*t + c_n*t^2 + d_n*t^3 + \dots \\ \dots \\ a_{N-1} + b_{N-1}*t + c_{N-1}*t^2 + d_{N-1}*t^3 + \dots \end{bmatrix}$
We can then rewrite this in tensor form.
(Aside: these expressions may remind you of the equations of motion – indeed for motion these form of approximations work well with low powers such as 1 or 2. It’s interesting to consider that motion and orientation within space are some of the earliest tasks that require intelligence.)
Now time series data provides for some interesting tricks. A first is that all our sensors are operating according to a common time base. Hence, we have a set of vector * scalar multiplications:
$\vec{S_t} \approx \vec{A} + \vec{B}*t + \vec{C}*t^2 + \vec{D}*t^3 + \dots$
Where:
$\vec{A} = \begin{bmatrix} a_0 \\ a_1 \\ \dots \\ a_n \\ \dots \\ a_{N-1} \end{bmatrix}$, $\vec{B} = \begin{bmatrix} b_0 \\ b_1 \\ \dots \\ b_n \\ \dots \\ b_{N-1} \end{bmatrix}$, $\vec{C} = \begin{bmatrix} c_0 \\ c_1 \\ \dots \\ c_n \\ \dots \\ c_{N-1} \end{bmatrix}$, $\vec{D} = \begin{bmatrix} d_0 \\ d_1 \\ \dots \\ d_n \\ \dots \\ d_{N-1} \end{bmatrix}$ etc.
Now we can consider the form of our mean adjusted vector $\vec{S^{\prime}_t}$.
First, let’s determine our mean vector using the Taylor polynomial:
$\overline{S} = \frac{1}{P}\sum_{t=0}^{P-1} s^t_n = \frac{1}{P}\sum_{t=0}^{P-1} (\vec{A} + \vec{B}*t + \vec{C}*t^2 + \vec{D}*t^3 + \dots)$
$\overline{S} = \vec{A}*\frac{P}{P} + \vec{B}*\frac{1}{P}\sum_{t=0}^{P-1}t + \vec{C}*\frac{1}{P}\sum_{t=0}^{P-1}t^2+\vec{D}*\frac{1}{P}\sum_{t=0}^{P-1}t^3 + \dots$
$\overline{S} = \vec{A}+ \vec{B}*\overline{t} + \vec{C}*\overline{t^2}+\vec{D}*\overline{t^3}+ \dots$
This assumes we are evaluating the mean using the function approximation about 0.
What we are left with for the mean signal $\overline{S}$ is an expression similar to our original expansion for $t$ but this time being evaluated in terms of the means of each term (e.g. $\overline{t}, \overline{t^2}, \overline{t^3}$ etc.).
Now if we look at a mean adjusted signal using our new expression for the mean:
$\vec{S^{\prime}_t} = \vec{S_t} - \overline{S} \approx (\vec{A} + \vec{B}*t + \vec{C}*t^2 + \vec{D}*t^3 + \dots) - (\vec{A}+ \vec{B}*\overline{t} + \vec{C}*\overline{t^2}+\vec{D}*\overline{t^3}+ \dots)$
$\vec{S^{\prime}_t} \approx \vec{B}(t-\overline{t}) +\vec{C}(t^2-\overline{t^2}) +\vec{D}(t^3-\overline{t^3}) + \dots$
Now this reminds me a little of our original Taylor series expansion, only performed around the mean time period.
## Rebasing
The expression above gives us a clue that we might reformulate our series sums around different moments within our time period.
If we have an incremental integer time base, from 0 to P-1, there are P integer values, where the mean value is:
$\overline{t} = \frac{1}{P}\sum_{t=0}^{P-1}t = \frac{(P-1)P}{2}$
(Note: We can use Faulhaber’s formula for the sum.)
But if we shift the P time points to be centred on t=0, we get $\overline{t} = 0$.
To play nice with an integer time base, we’ll look at P points that include 0. For example, we can run from -(P-1)/2 to (P-1)/2 (the “-1” enters in because we have “0” as an number as well).
This also has the nice characteristic that, in the average over time for our sensors, odd terms will cancel out either side of 0.
$\overline{S} \approx \frac{1}{P}\sum_{-(P-1)/2}^{(P-1)/2}(\vec{A} + \vec{B}*t + \vec{C}*t^2 + \vec{D}*t^3 + \dots)$
$\overline{S} \approx \vec{A} + \vec{B}*\frac{1}{P}*\sum_{-(P-1)/2}^{(P-1)/2}t + \vec{C}*\frac{1}{P}*\sum_{-(P-1)/2}^{(P-1)/2}t^2 + \vec{D}*\frac{1}{P}*\sum_{-(P-1)/2}^{(P-1)/2}t^3 + \dots$
$\overline{S} \approx \vec{A} + \vec{C}*\frac{1}{P}*\sum_{-(P-1)/2}^{(P-1)/2}t^2 + \dots$
$\overline{S} \approx \vec{A} + \vec{C}*\overline{t^2} + \vec{E}*\overline{t^4} \dots$
As you can see, the odd terms have a time average of 0 and so disappear in the expression.
Now the above average has been evaluated about t=0. We need to do this for our sensor function approximations as well. We can then look at the mean adjusted values:
$\vec{S^{\prime}_t} = \vec{S_t} - \overline{S} \approx (\vec{A} + \vec{B}*t + \vec{C}*t^2 + \vec{D}*t^3 + \vec{E}*t^4 + \dots) - (\vec{A}+ \vec{C}*\overline{t^2}+\vec{E}*\overline{t^4}+ \dots)$
$\vec{S^{\prime}_t} = \vec{S_t} - \overline{S} \approx \vec{B}*t + \vec{C}*(t^2-\overline{t^2}) + \vec{D}*t^3 + \vec{E}*(t^4-\overline{t^4}) + \dots$
## Back to Covariance
Still with me?
Now we have an expression for a mean-adjusted sensor reading, we can look at expressions for the entries in the covariance matrix. Each entry is equivalent to taking a row from $\mathbf{S'}$ that relates to an ith sensor over time and multiplying by a column from $\mathbf{S'}^T$ that relates to a jth sensor over time. As we have rebased the time, we are looking at P sensor readings from -(P-1)/2 to (P-1)/2.
The row-column multiplication for each entry is equivalent to multiplying each mean-adjusted time entry for the ith sensor by a corresponding mean-adjusted time entry for the jth sensor, and then summing the result across the P readings:
$M_{ij} = \frac{1}{P}*\vec{S^{\prime}_i}^T\vec{S^{\prime}_j} = \frac{1}{P}*\sum_{t=-(P-1)/2}^{(P-1)/2}(s^{\prime})_i^t*(s^{\prime})_j^t$
We can then use our determination for the mean-adjusted signal to get an expression with our Taylor series terms:
$M_{ij} \approx \frac{1}{P}*\sum_{t=-(P-1)/2}^{(P-1)/2} (b_i*t+c_i(t^2-\overline{t^2})+d_i*t^3 + e_i*(t^4-\overline{t^4}) + \dots)(b_j*t+c_j(t^2-\overline{t^2})+d_j*t^3 + e_j*(t^4-\overline{t^4}) + \dots)$
What’s interesting here is we find re-occurrence of odd powers of t. When terms with these odd powers of t are evaluated over the sum of time steps, they go to zero, e.g.:
$(b_i*t)*(c_j*(t^2-\overline{t^2})) = b_i*c_j*t*(t^2-\overline{t^2})$
$\frac{1}{P}*\sum_{t=-(P-1)/2}^{(P-1)/2}(b_i*c_j*t*(t^2-\overline{t^2}))=b_i*c_j*\frac{1}{P}*\sum_{t=-(P-1)/2}^{(P-1)/2}(t*(t^2-\overline{t^2}))$
$= b_i*c_j*\frac{1}{P}*\sum_{t=-(P-1)/2}^{(P-1)/2}(t^3)-b_i*c_j*\overline{t^2}*\frac{1}{P}*\sum_{t=-(P-1)/2}^{(P-1)/2}t$
$= b_i*c_j*0 - b_i*c_j*\overline{t^2}*0$
### Simplifying Terms
To help simplify things we can use Faulhaber’s formula to determine some of our $\overline{t^k}$ expressions. We know that odd powers of k have a sum of 0 (as the positive and negative values cancel out). This leaves the even powers of k. The sums of these are symmetrical about 0, so we just need to find the positive sum up to $\frac{P-1}{2}$ then double (as 0 will be 0 and so not contribute to the sum).
$\overline{t^2} = \frac{1}{P}\sum_{t=-(P-1)/2}^{(P-1)/2}t^2 = \frac{2}{P}\sum_{t=1}^{(P-1)/2}t^2 = \frac{2}{P}\frac{\frac{P-1}{2}(\frac{P-1}{2}+1)(2\frac{P-1}{2}+1)}{6} = \frac{1}{P}\frac{(P-1)(\frac{P-1}{2}+\frac{2}{2})(P-1+1)}{6}$
$=\frac{P}{12P}(P-1)(P+1) = \frac{1}{12}(P-1)(P+1)$
$\overline{t^4} = \frac{1}{P}\sum_{t=-(P-1)/2}^{(P-1)/2}t^4 = \frac{2}{P}\sum_{t=1}^{(P-1)/2}t^4 = \frac{2}{P}\frac{\frac{P-1}{2}(\frac{P-1}{2}+1)(2\frac{P-1}{2}+1)}{6}*\frac{3(\frac{P-1}{2})^2+3(\frac{P-1}{2})-1}{5}$
$= \overline{t^2} * \frac{3P^2-7}{20}$
## Power Iteration with the Covariance Matrix
One way to compute the eigenvectors of the covariance matrix is to use the power iteration method. This involves multiplying a random vector by the covariance matrix multiple times. As we iterate this multiplication (and often scale the result), we move towards a vector that resembles the first eigenvector. We then deflate (i.e. remove the contribution of the first eigenvector) and repeat to determine the next eigenvector.
If our covariance matrix is $\mathbf{M} = cov(\mathbf{S'}, \mathbf{S'}) = var(\mathbf{S'}) = \frac{1}{P}\mathbf{S'}\mathbf{S'}^T$ and our random vector is $v_k$ (starting with $v_0$), we compute:
$v_{k+1} = \mathbf{M}v_{k}$
And scale it is recommended to scale the new vector $v_{k+1}$. The new vector may be scaled by dividing by the L2 or Euclidean norm$||v_{k+1}||$. The Rayleigh Quotient may be used to determine the corresponding eigenvalue. The scaling effectively generates unit vectors, i.e. vectors with a length or L2 norm of 1.
## Linear Approximations
Now, to avoid getting lost in higher terms, and to try to work out what neural networks are doing, we can consider just the linear terms in our Taylor polynomial.
Let’s start by ignoring any terms above $\vec{B}$ as these relate to higher order (non-linear) terms. Our mean-adjusted signal $\vec{S^{\prime}_t}$ becomes:
$\vec{S^{\prime}_t} = \vec{B}*t$
The leftovers or error of this approximation is:
$\Delta \approx \vec{C}*(t^2-\overline{t^2}) + \vec{D}*t^3 + \dots$
Notice how, following the mean adjustment, we don’t need to worry about a bias term.
Our covariance matrix terms for the linear approximation become:
$M_{ij} = \frac{1}{P}\sum_{t=-(P-1)/2}^{(P-1)/2}b_ib_jt^2 = b_ib_j\overline{t^2} = b_ib_j*\frac{(P-1)(P+1)}{12}$
## Power Approximations
What happens to our linear approximation when we try to compute the first eigenvector using the power iteration method?
Let’s consider a small test case with 2 sensors and 7 readings (N=2 and P=7).
$\mathbf{M} = 4*\begin{bmatrix} b_0^2 & b_0b_1 \\ b_0b_1 & b_1^2\end{bmatrix}$
If we start with a vector $v_0 = \begin{bmatrix} 1 \\ 1\end{bmatrix}$ then $||v_0|| = \sqrt{2}$. On our first iteration we have:
$v_1 = \frac{4}{\sqrt{2}}\begin{bmatrix} b_0^2 & b_0b_1 \\ b_0b_1 & b_1^2\end{bmatrix}\begin{bmatrix} 1 \\ 1\end{bmatrix} = \begin{bmatrix} b_0^2 + b_0b_1 \\ b_0b_1 + b_1^2\end{bmatrix}$
Our scale factor becomes $||v_1|| = \sqrt{(b_0^2 + b_0b_1)^2 + (b_1^2 + b_0b_1)^2}=\sqrt{b_0^4+2b_0^3b_1+2b_0^2b_1^2+2b_0b_1^3+b_1^4}$.
On each iteration, we get different combinations of power terms occurring, and a longer and longer scale factor. In simple 2×2 matrix examples I’ve seen it takes about 10 iterations to converge.
As the covariance matrix is symmetric, with a 2×2 example we only have differing terms on the diagonal. As we iterate using the power method, we have a number of terms that are shared by both vector elements and a number of terms that are not shared. The terms that are not shared influence how the vector changes over time, and these non-shared terms are increasingly influenced by the differing diagonal elements, while the scaling at each iteration stops explosion to infinity.
We can also see that P does not affect how the unit vector changes, it just sets an external scale factor. Also our time period does not really matter; what matters is the sampling rate over our time period. P only really matters for the accuracy of the approximation, which can be seen in the illustration above. A large value of P means that outer approximations (e.g. values of $t$ close to -(P-1)/2 and (P-1)/2) are less reliable, because we are moving further from our location of function approximation. This matches up nicely with the concept of local processing within the brain – “samples” in this sense may actually comprise physical objects (neurons and synapses) and so physicality limits the time resolution for function approximation – we need to be small and local.
Even with the simple 2×2 example above, we also see that there is no inherent pattern for the first eigenvalue – it’s direction depends on the relative values of $b_0$ and $b_1$. Also, as we can always remove a scale factor, we do not need to worry about the absolute values of the coefficients, we can pick a scale range that works for us and that simplifies the calculations. For example, this could be a normalised range between 0 and 1 or integer values between 0 and 255 (for 8-bit computations). All we need to beware of is overflow when performing the calculations; however, the scaling brings us within a common range at every iteration, so we just need to worry about overflow within each iteration. For example, we could perform the iteration multiplications with 16-bit values and then convert back down to 8-bit values once the scaling is applied.
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# 3-5
3-5. Linear Equations in Three Dimensions. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. Warm Up Graph each of the following points in the coordinate plane. 1. A (2, –1). 2. B (–4, 2). 3. Find the intercepts of the line . x : –9; y : 3. Objective.
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## 3-5
E N D
### Presentation Transcript
1. 3-5 Linear Equations in Three Dimensions Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2
2. Warm Up Graph each of the following points in the coordinate plane. 1.A(2, –1) 2.B(–4, 2) 3. Find the intercepts of the line . x: –9; y: 3
3. Objective Graph points and linear equations in three dimensions.
4. Vocabulary three-dimensional coordinate system ordered triple z-axis
5. A Global Positioning System (GPS) gives locations using the three coordinates of latitude, longitude, and elevation. You can represent any location in three-dimensional space using a three-dimensional coordinate system, sometimes called coordinate space.
6. Each point in coordinate space can be represented by an ordered triple of the form (x, y, z). The system is similar to the coordinate plane but has an additional coordinate based on the z-axis. Notice that the axes form three planes that intersect at the origin.
7. Helpful Hint To find an intercept in coordinate space, set the other two coordinates equal to 0.
8. z y x Example 1A: Graphing Points in Three Dimensions Graph the point in three-dimensional space. A(3, –2, 1) From the origin, move 3 units forward along the x-axis, 2 units left, and 1 unit up. A(3, –2, 1)
9. z y x Example 1B: Graphing Points in Three Dimensions Graph the point in three-dimensional space. B(2, –1, –3) From the origin, move 2 units forward along the x-axis, 1 unit left, and 3 units down. B(2, –1, –3)
10. z y x Example 1C: Graphing Points in Three Dimensions Graph the point in three-dimensional space. C(–1, 0, 2) C(–1,0, 2) From the origin, move 1 unit back along the x-axis, 2 units up. Notice that this point lies in the xz-plane because the y-coordinate is 0.
11. z y x Check It Out! Example 1a Graph the point in three-dimensional space. D(1, 3, –1) From the origin, move 1 unit forward along the x-axis, 3 units right, and 1 unit down. D(1, 3, –1)
12. z y x Check It Out! Example 1b Graph the point in three-dimensional space. E(1, –3, 1) From the origin, move 1 unit forward along the x-axis, 3 units left, and 1 unit up. E(1, –3, 1)
13. z y x Check It Out! Example 1c Graph the point in three-dimensional space. F(0, 0, 3) F(0, 0, 3) From the origin, move 3 units up.
14. Recall that the graph of a linear equation in two dimensions is a straight line. In three-dimensional space, the graph of a linear equation is a plane. Because a plane is defined by three points, you can graph linear equations in three dimensions by finding the three intercepts.
15. Example 2: Graphing Linear Equations in Three Dimensions Graph the linear equation 2x – 3y + z = –6 in three-dimensional space. Step 1 Find the intercepts: x-intercept: 2x – 3(0) + (0) = –6 x = –3 y-intercept: 2(0) – 3y + (0) = –6 y = 2 z-intercept: 2(0) – 3(0) + z = –6 z = –6
16. z y x Example 2 Continued Step 2 Plot the points (–3, 0, 0), (0, 2, 0), and (0, 0, –6). Sketch a plane through the three points. (–3, 0, 0) (0, 2, 0) (0, 0, –6)
17. Check It Out! Example 2 Graph the linear equation x – 4y + 2z = 4 in three-dimensional space. Step 1 Find the intercepts: x-intercept: x – 4(0) + 2(0) = 4 x = 4 y-intercept: (0) – 4y + 2(0) = 4 y = –1 z-intercept: (0) – 4(0) + 2z = 4 z = 2
18. z y x Check It Out! Example 2 Continued (0, 0, 2) Step 2 Plot the points (4, 0, 0), (0, –1, 0), and (0, 0, 2). Sketch a plane through the three points. (0, –1, 0) ● (4, 0, 0)
19. Example 3A: Sports Application Track relay teams score 5 points for finishing first, 3 for second, and 1 for third. Lin’s team scored a total of 30 points. Write a linear equation in three variables to represent this situation. Let f = number of races finished first, s = number of races finished second, and t = number of races finished third. + + = Points for third 1t Points for first 5f Points for second 3s 30 + + = 30 +
20. Example 3B: Sports Application If Lin’s team finishes second in six events and third in two events, in how many eventsdid it finish first? 5f + 3s + t = 30 Use the equation from A. 5f + 3(6) + (2) = 30 Substitute 6 for s and 2 for t. f = 2 Solve for f. Linn’s team placed first in two events.
21. Check It Out! Example 3a Steve purchased \$61.50 worth of supplies for a hiking trip. The supplies included flashlights for \$3.50 each, compasses for \$1.50 each, and water bottles for \$0.75 each. Write a linear equation in three variables to represent this situation. Let x = number of flashlights, y = number of compasses, and z = number of water bottles. + + = water bottles 0.75z flashlights 3.50x compasses 1.50y 61.50 + = + 61.50
22. Check It Out! Example 3b Steve purchased 6 flashlights and 24 water bottles. How many compasses did he purchase? 3.5x + 1.5y + 0.75z = 61.50 Use the equation from a. 3.5(6) + 1.5y + 0.75(24) = 61.50 Substitute 6 for x and 24 for z. 21 + 1.5y + 18 = 61.50 1.5y = 22.5 Solve for y. y = 15 Steve purchased 15 compasses.
23. z y x Lesson Quiz: Part I Graph each point in three dimensional space. 2. B(0, –2, 3) 1. A(–2, 3, 1) B( 0, –2, 3) A( –2, 3, 1)
24. Lesson Quiz: Part II 3. Graph the linear equation 6x + 3y – 2z = –12 in three-dimensional space.
25. Lesson Quiz: Part III 4. Lily has \$6.00 for school supplies. Pencils cost \$0.20 each, pens cost \$0.30 each, and erasers cost \$0.25 each. a. Write a linear equation in three variables to represent this situation. 0.2x + 0.3y +0.25z = 6 b. If Lily buys 6 pencils and 6 erasers, how many pens can she buy? 11
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4 Times Table
Learn here the 4 times table. Four times table 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 4 x 8 = 32, 4 x 9 = 36, 4 x 10 = 40.
Hello, math enthusiasts! Today, we embark on a thrilling mathematical adventure as we dive into the captivating realm of the 4 times table.
Whether you're a curious learner or someone looking for an exciting numerical journey, you've come to the right place! So, let's buckle up and uncover the wonders of multiplication together!
The Gateway to Multiplication Fun
The 4 times table is like a gateway to a world of multiplication magic. It's where we explore the fantastic patterns that emerge when we multiply numbers by 4.
Get ready to witness the beauty of this table and let the numbers amaze you
The Building Blocks of Four
As we delve into the 4 times table, you'll quickly notice a fascinating pattern unfolding. Each number in the table is obtained by adding 4 to the previous number.
For instance, 4 times 1 is 4, 4 times 2 is 8, 4 times 3 is 12, and so on. It's like stacking building blocks, with each block being exactly four units higher than the previous one.
Tricks and Shortcuts
Let's equip ourselves with some handy tricks and shortcuts to make working with the 4 times table even more enjoyable!
• The Zero Rule: Just like in the previous tables, when we multiply any number by 0, the result is always 0. So, if you encounter 4 times 0, the answer is a delightful 0. Zero never fails to amaze us!
• The Power of Evenness: In the 4 times table, even numbers play a significant role. When you multiply any even number by 4, the result is always an even number. It's like a secret club where even numbers thrive!
• Double Double: Multiplying a number by 4 is equivalent to doubling it twice. For example, 4 times 3 is the same as doubling 3 to get 6 and then doubling 6 to get 12. This trick comes in handy when you want to multiply by 4 in a flash!
Real-Life Connections
Let's explore how the 4 times table finds its way into our everyday lives. From counting objects to solving real-world problems, the 4 times table proves its usefulness time and time again!
• Counting Collections: Imagine you have 5 bags, and each bag contains 4 apples. By using the 4 times table, you can quickly determine that you have a total of 20 apples. It's like multiplying the number of bags by the number of apples in each bag to find the grand total!
• Time Management: Suppose you need to catch a bus that arrives every 4 minutes, and you want to know when the next bus will arrive after 12 minutes. By utilizing the 4 times table, you can discover that it will be 3 buses or 12 minutes until the next one arrives. Time flies when you understand multiplication!
Four Multiplication Table
Read, Repeat and Learn four times table and Check yourself by giving a test below
Also check 0 times table, 1 times table, 2 times table, 3 times table, 4 times table, 5 times table, 6 times table, 7 times table, 8 times table, 9 times table
Table of 4
4 Times table Test
How much is 4 multiplied by other numbers?
@2024 PrintableMultiplicationTable.net
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Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# just final answer Q 11 The pilot of an aircraft flies due north relative to the ground in a wind blowing 50 km/h toward the east.
Q 11
The pilot of an aircraft flies due north relative to the ground in a wind blowing 50 km/h toward the east. If his speed relative to the air is 92 km/h, what is the speed of his airplane relative to the ground ? (State the answer to the nearest tenth of km/h )
Q 12
A go-kart moves on a circular track of radius 57 m.
It takes 2 minutes to complete the 3 turns.
What is the magnitude of the centripetal acceleration experienced by the kart?
Round your answer to nearest hundredth of m/s2.
Q 13
A particle's position is given by the following equation:
=(4,7,9) +(-4,0,-9)t-(7,0,0)t2; where t is in seconds and magnitude of r is in meters
Find the distance of of this particle from the origin at . State your answer to the nearest one tenth of a meter.
Q 14
A particle's position is given by the following equation:
=(5,6,10) +(-5,0,-5)t-(8,0,0)t2; where t is in seconds and magnitude of r is in meters
Find the magnitude of the velocity of this particle at .
State your answer to the nearest one tenth of a meter per second.
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f16 Chapter Contents
f16 Chapter Introduction
NAG Library Manual
# NAG Library Function Documentnag_ztrmm (f16zfc)
## 1 Purpose
nag_ztrmm (f16zfc) performs matrix-matrix multiplication for a complex triangular matrix.
## 2 Specification
#include #include
void nag_ztrmm (Nag_OrderType order, Nag_SideType side, Nag_UploType uplo, Nag_TransType trans, Nag_DiagType diag, Integer m, Integer n, Complex alpha, const Complex a[], Integer pda, Complex b[], Integer pdb, NagError *fail)
## 3 Description
nag_ztrmm (f16zfc) performs one of the matrix-matrix operations
$B←αAB, B←αATB, B←αAHB, B←αBA, B←αBAT or B←αBAH,$
where $B$ is an $m$ by $n$ complex matrix, $A$ is a complex triangular matrix, and $\alpha$ is a complex scalar.
## 4 References
Basic Linear Algebra Subprograms Technical (BLAST) Forum (2001) Basic Linear Algebra Subprograms Technical (BLAST) Forum Standard University of Tennessee, Knoxville, Tennessee http://www.netlib.org/blas/blast-forum/blas-report.pdf
## 5 Arguments
1: $\mathbf{order}$Nag_OrderTypeInput
On entry: the order argument specifies the two-dimensional storage scheme being used, i.e., row-major ordering or column-major ordering. C language defined storage is specified by ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. See Section 3.2.1.3 in the Essential Introduction for a more detailed explanation of the use of this argument.
Constraint: ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ or $\mathrm{Nag_ColMajor}$.
2: $\mathbf{side}$Nag_SideTypeInput
On entry: specifies whether $B$ is operated on from the left or the right.
${\mathbf{side}}=\mathrm{Nag_LeftSide}$
$B$ is pre-multiplied from the left.
${\mathbf{side}}=\mathrm{Nag_RightSide}$
$B$ is post-multiplied from the right.
Constraint: ${\mathbf{side}}=\mathrm{Nag_LeftSide}$ or $\mathrm{Nag_RightSide}$.
3: $\mathbf{uplo}$Nag_UploTypeInput
On entry: specifies whether $A$ is upper or lower triangular.
${\mathbf{uplo}}=\mathrm{Nag_Upper}$
$A$ is upper triangular.
${\mathbf{uplo}}=\mathrm{Nag_Lower}$
$A$ is lower triangular.
Constraint: ${\mathbf{uplo}}=\mathrm{Nag_Upper}$ or $\mathrm{Nag_Lower}$.
4: $\mathbf{trans}$Nag_TransTypeInput
On entry: specifies whether the operation involves $A$, ${A}^{\mathrm{T}}$ or ${A}^{\mathrm{H}}$.
${\mathbf{trans}}=\mathrm{Nag_NoTrans}$
It involves $A$.
${\mathbf{trans}}=\mathrm{Nag_Trans}$
It involves ${A}^{\mathrm{T}}$.
${\mathbf{trans}}=\mathrm{Nag_ConjTrans}$
It involves ${A}^{\mathrm{H}}$.
Constraint: ${\mathbf{trans}}=\mathrm{Nag_NoTrans}$, $\mathrm{Nag_Trans}$ or $\mathrm{Nag_ConjTrans}$.
5: $\mathbf{diag}$Nag_DiagTypeInput
On entry: specifies whether $A$ has nonunit or unit diagonal elements.
${\mathbf{diag}}=\mathrm{Nag_NonUnitDiag}$
The diagonal elements are stored explicitly.
${\mathbf{diag}}=\mathrm{Nag_UnitDiag}$
The diagonal elements are assumed to be $1$ and are not referenced.
Constraint: ${\mathbf{diag}}=\mathrm{Nag_NonUnitDiag}$ or $\mathrm{Nag_UnitDiag}$.
6: $\mathbf{m}$IntegerInput
On entry: $m$, the number of rows of the matrix $B$; the order of $A$ if ${\mathbf{side}}=\mathrm{Nag_LeftSide}$.
Constraint: ${\mathbf{m}}\ge 0$.
7: $\mathbf{n}$IntegerInput
On entry: $n$, the number of columns of the matrix $B$; the order of $A$ if ${\mathbf{side}}=\mathrm{Nag_RightSide}$.
Constraint: ${\mathbf{n}}\ge 0$.
8: $\mathbf{alpha}$ComplexInput
On entry: the scalar $\alpha$.
9: $\mathbf{a}\left[\mathit{dim}\right]$const ComplexInput
Note: the dimension, dim, of the array a must be at least
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pda}}×{\mathbf{m}}\right)$ when ${\mathbf{side}}=\mathrm{Nag_LeftSide}$;
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pda}}×{\mathbf{n}}\right)$ when ${\mathbf{side}}=\mathrm{Nag_RightSide}$.
On entry: the triangular matrix $A$; $A$ is $m$ by $m$ if ${\mathbf{side}}=\mathrm{Nag_LeftSide}$, or $n$ by $n$ if ${\mathbf{side}}=\mathrm{Nag_RightSide}$.
If ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, ${A}_{ij}$ is stored in ${\mathbf{a}}\left[\left(j-1\right)×{\mathbf{pda}}+i-1\right]$.
If ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, ${A}_{ij}$ is stored in ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{pda}}+j-1\right]$.
If ${\mathbf{uplo}}=\mathrm{Nag_Upper}$, $A$ is upper triangular and the elements of the array corresponding to the lower triangular part of $A$ are not referenced.
If ${\mathbf{uplo}}=\mathrm{Nag_Lower}$, $A$ is lower triangular and the elements of the array corresponding to the upper triangular part of $A$ are not referenced.
If ${\mathbf{diag}}=\mathrm{Nag_UnitDiag}$, the diagonal elements of $A$ are assumed to be $1$, and are not referenced.
10: $\mathbf{pda}$IntegerInput
On entry: the stride separating row or column elements (depending on the value of order) of the matrix $A$ in the array a.
Constraints:
• if ${\mathbf{side}}=\mathrm{Nag_LeftSide}$, ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$;
• if ${\mathbf{side}}=\mathrm{Nag_RightSide}$, ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
11: $\mathbf{b}\left[\mathit{dim}\right]$ComplexInput/Output
Note: the dimension, dim, of the array b must be at least
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pdb}}×{\mathbf{n}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}×{\mathbf{pdb}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
If ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, ${B}_{ij}$ is stored in ${\mathbf{b}}\left[\left(j-1\right)×{\mathbf{pdb}}+i-1\right]$.
If ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, ${B}_{ij}$ is stored in ${\mathbf{b}}\left[\left(i-1\right)×{\mathbf{pdb}}+j-1\right]$.
On entry: the $m$ by $n$ matrix $B$.
If ${\mathbf{alpha}}=0$, b need not be set.
On exit: the updated matrix $B$.
12: $\mathbf{pdb}$IntegerInput
On entry: the stride separating row or column elements (depending on the value of order) in the array b.
Constraints:
• if ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, ${\mathbf{pdb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$;
• if ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, ${\mathbf{pdb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
13: $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.2.1.2 in the Essential Introduction for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_ENUM_INT_2
On entry, ${\mathbf{side}}=〈\mathit{\text{value}}〉$, ${\mathbf{m}}=〈\mathit{\text{value}}〉$, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{side}}=\mathrm{Nag_LeftSide}$, ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
On entry, ${\mathbf{side}}=〈\mathit{\text{value}}〉$, ${\mathbf{n}}=〈\mathit{\text{value}}〉$, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{side}}=\mathrm{Nag_RightSide}$, ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
NE_INT
On entry, ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{m}}\ge 0$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 0$.
NE_INT_2
On entry, ${\mathbf{pdb}}=〈\mathit{\text{value}}〉$, ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pdb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
On entry, ${\mathbf{pdb}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{pdb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 3.6.6 in the Essential Introduction for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 3.6.5 in the Essential Introduction for further information.
## 7 Accuracy
The BLAS standard requires accurate implementations which avoid unnecessary over/underflow (see Section 2.7 of Basic Linear Algebra Subprograms Technical (BLAST) Forum (2001)).
Not applicable.
None.
## 10 Example
Premultiply complex $4$ by $2$ matrix $B$ by lower triangular $4$ by $4$ matrix $A$, $B←AB$, where
$A = 4.78+4.56i 2.00-0.30i -4.11+1.25i 2.89-1.34i 2.36-4.25i 4.15+0.80i -1.89+1.15i 0.04-3.69i -0.02+0.46i 0.33-0.26i$
and
$B = -5.0-2.0i 1.0+5.0i -3.0-1.0i -2.0-2.0i 2.0+1.0i 3.0+4.0i 4.0+3.0i 4.0-3.0i .$
### 10.1 Program Text
Program Text (f16zfce.c)
### 10.2 Program Data
Program Data (f16zfce.d)
### 10.3 Program Results
Program Results (f16zfce.r)
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# Idiom #74 Compute GCD
Compute the greatest common divisor x of big integers a and b. Use an integer type able to handle huge numbers.
``x = a.gcd(b)``
``#include <gmp.h>``
``````mpz_t _a, _b, _x;
mpz_init_set_str(_a, "123456789", 10);
mpz_init_set_str(_b, "987654321", 10);
mpz_init(_x);
mpz_gcd(_x, _a, _b);
gmp_printf("%Zd\n", _x);``````
``````(defn gcd [a b]
(if (zero? b)
a
(recur b (mod a b))))``````
``````unsigned long long int GCD(unsigned long long int a, unsigned long long int b)
{
unsigned long long int c=a%b;
if(c==0)
return b;
return GCD(b, c);
}``````
``#include <numeric>``
``auto x = std::gcd(a, b);``
``````int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}``````
``````int gcd(int a, int b)
{
while (b != 0)
{
int t = b;
b = a % t;
a = t;
}
return a;
}``````
``import std.bigint;``
``````BigInt gcd(in BigInt x, in BigInt y) pure {
if (y == 0)
return x;
return gcd(y, x%y);
}
gcd(a, b);``````
``import std.numeric: gcd;``
``x = gcd(a, b);``
``````int gcd(int a, int b) {
while (b != 0) {
var t = b;
b = a % t;
a = t;
}
return a;
}``````
``````defmodule Gcd do
def gcd(x, 0), do: x
def gcd(x, y), do: gcd(y, rem(x,y))
end
x = Gcd.gcd(a, b)``````
``````use,intrinsic : iso_fortran_env, only : int64
``````
``````function gcd(m,n) result(answer)
implicit none
integer(kind=int64),intent(in) :: m, n
ifirst=iabs(m)
else
do
if(irest == 0) exit
enddo
endif
end function gcd
``````
``import "math/big"``
``x.GCD(nil, nil, a, b)``
``````gcd a b
| a==b =a
| a>b = gcd(a-b) b
| otherwise = gcd a (b-a)``````
``````gcd x y = gcd' (abs x) (abs y)
where
gcd' a 0 = a
gcd' a b = gcd' b (a `rem` b)``````
``````x = gcd a b
``````
``const gcd = (a, b) => b === 0 ? a : gcd (b, a % b)``
``import java.math.BigInteger;``
``BigInteger x = a.gcd(b);``
``````function gcd(a, b)
return b==0 and a or gcd(b,a%b)
end``````
``````function gcd(a, b)
if (b ~= 0) then
return a
else
return gcd(y, x%y)
end
end``````
``extension=gmp``
``\$x = gmp_gcd(\$a, \$b);``
``````function GCD(a,b:int64):int64;
var t:int64;
begin
while b <> 0 do
begin
t := b;
b := a mod b;
a := t;
end;
result := a;
end;
``````
``use bigint;``
``````sub gcd {
my (\$A, \$B) = @_;
return 0 == \$B
? \$A
: gcd(\$B, \$A % \$B);
}
``````
``import math``
``x = math.gcd(a, b)``
``from fractions import gcd``
``x = gcd(a, b)``
``````extern crate num;
use num::Integer;
use num::bigint::BigInt;``````
``let x = a.gcd(&b);``
``Imports System.Numerics``
``Dim x = BigInteger.GreatestCommonDivisor(a, b) ``
deleplace
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https://byjus.com/question-answer/four-wires-of-equal-length-and-resistance-10-omega-each-are-connected-in-the-form/
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1
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Question
# Four wires of equal length and resistance 10Ω each are connected in the form of a square. Find the equivalent resistance between any two opposite corners of the square.
A
40Ω
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B
10Ω
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C
5Ω
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D
20Ω
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Solution
## The correct option is B 10ΩFrom arrangement we see that the two series combination of two 10Ω resistance are in parallel.On calculating the equivalent resistance we get 10Ω.Answer-(B)
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# Markov-Switching Model with time varying transition probabilities in Stan
I’ve been working on the model below and talked to Ben recently as a couple of problems arose. Andrew also suggested simplifying the model to isolate the problem but I also didn’t find the problem that way (see below for some solutions I tried previously). I figured that it may not hurt to ask here as well.
The model is a Markov Switching Model with Time Varying Transition Probabilities, i.e. I assume that the parameters of the time series vary according to a random variable that follows a Markov process of order one whose transition probabilities are a function of an exogenous covariate (z) and an intercept (gamma).
Without TVTPs, the model can identify gamma. When the intercept and the exogenous covariate are included, the model can identity neither gamma nor lambda. Alpha and phi are on average within the 50% HDI but are still biased for N(simulations) = 50. Sometimes gamma[1]/[2] and lambda[1]/[2] have the same standard deviation or same absolute mean value even if the true values diverge widely. Here is just one model output when running the model with some simulated data for N = 1 and T = 300. The true values are alpha (2, 6), phi (0.2, 0.7), gamma (-2, 2), lambda (1, 0.2)
N sim name mean sd q25 q50 q75
1 1 1 alpha[1] 1.023479004 0.52433014 0.68470262 1.033403681 1.37288909
2 1 1 alpha[2] 5.836420795 0.24794741 5.67351637 5.834325272 5.99803102
3 1 1 phi[1] 0.225243761 0.03070192 0.20518439 0.224699188 0.24562940
4 1 1 phi[2] 0.715086473 0.01346963 0.70643604 0.715159144 0.72381077
5 1 1 gamma[1] -0.664868689 0.31815327 -0.87492839 -0.646834929 -0.45570295
6 1 1 gamma[2] 1.670333349 0.12871636 1.58098185 1.668498695 1.75781119
7 1 1 lambda[1] -0.010225889 1.04397066 -0.68672621 0.003543484 0.66562922
8 1 1 lambda[2] -0.002082519 0.10865527 -0.07782131 -0.000808923 0.07186009
My problem is that I don’t understand why this happens and that it is unclear how the model can provide reasonable inference for alpha and phi in the linear part of the model if it does not know in which state it currently is.
This problem does not disappear when only including lambda for one state, removing the cross-sectional part or varying the size of the parameters or number of time periods. I ran some other versions that removed phi and/or alpha to no avail. I am also currently running a simulation varying N between 1 and 100 but as the model is again very slow I don’t know yet whether that would improve the situation. I can provide the R code for the simulated data if necessary.
Thanks in advance for any ideas regarding this (or how to speed the model up).
Model
``````data {
int<lower = 0> T; // number of observed time periods - 1
int<lower = 0> N; // number of observed units
matrix[N, T] y_tail; // tail(y, T - 1)
matrix[N, T] z; // exogenous random variable affecting transition probabilities
// real<lower = 0> sigma; // variance; fixed for now
}
parameters {
ordered[2] alpha;
vector<lower= 0,upper=1>[2] phi; //AR parameter
real gamma[2]; // intercepts in TPs
real lambda[2];
real<lower = 0> sigma; // variance term
}
transformed parameters {
real p11[N,T];
real p12[N,T];
real p21[N,T];
real p22[N,T];
real p_sprev_1_givenprev[N, T]; // P(S[t - 1] = 1 | omega[t - 1], theta)
real p_scur_1_givenprev[N, T]; // P(S[t] = 1 | omega[t - 1], theta)
real p_scur_1_givencur[N, T]; // P(S[t] = 1 | omega[t], theta)
vector[2] s[N, T];
real fy[N, T];
for (i in 1:N) {// looping over all observations
// 1st inner loop for initial values
for (t in 1:T) {
p11[i,t] = normal_cdf(gamma[1] + lambda[1] * z[i, t], 0, 1);
p12[i,t] = 1 - p11[i,t];
p22[i,t] = normal_cdf(gamma[2] + lambda[2] * z[i, t], 0, 1);
p21[i,t] = 1 - p22[i,t];
}
// t = 1
// Initial transition probabilities (Piger eq. 14)
// Can also treat initial transition probabilities as a parameter to be estimated
p_sprev_1_givenprev[i,1] = (1 - p22[i, 1]) / (2 - p11[i, 1] - p22[i,1]);
p_scur_1_givenprev[i, 1] = p11[i, 1] * p_sprev_1_givenprev[i, 1] +
p21[i, 1] * (1 - p_sprev_1_givenprev[i, 1]);
for (j in 1:2){
s[i, 1, j] = normal_lpdf(y_tail[i,1] | alpha[j] + phi[j] * y_head[i,1], sigma);
}
// Piger eq. 11
fy[i,1] = log_mix(p_scur_1_givenprev[i,1], s[i,1, 1], s[i,1, 2]);
// Piger eq. 13
p_scur_1_givencur[i,1] = exp(s[i,1, 1] + log(p_scur_1_givenprev[i,1]) - fy[i,1]);
// second inner loop for subsequent values
for (t in 2:T){
p_sprev_1_givenprev[i,t] = p_scur_1_givencur[i,(t-1)];
// Piger eq. 10
p_scur_1_givenprev[i,t] = p11[i, t] * p_sprev_1_givenprev[i ,t] +
p21[i, t] * (1 - p_sprev_1_givenprev[i,t]);
for (j in 1:2){
s[i,t,j] = normal_lpdf(y_tail[i, t] | alpha[j] + phi[j] * y_head[i,t], sigma);
}
// Piger eq. 11
fy[i,t] = log_mix(p_scur_1_givenprev[i,t], s[i,t, 1], s[i,t, 2]);
// Piger eq. 13
p_scur_1_givencur[i, t] = exp(s[i,t, 1] + log(p_scur_1_givenprev[i,t]) - fy[i,t]);
}
}
}
model {
// likelihood
for (i in 1:N) {
for (t in 1:T) {
target += fy[i,t];
}
}
// priors
gamma ~ normal(0, 1);
phi ~ uniform(0, 1);
alpha ~ normal(0,1);
lambda ~ normal(0, 1);
}
``````
Honestly, this model is (for me) very hard to understand as it is quite complex. Simplifying might be necessary to get an answer here. I would suggest to check you can recover z and gamma when you model only the TVTPs (i.e. you have direct noisy observations of the parameters of the time series). This would make sure that the code for the Markov process is OK.
There are some things that have caught my attention which might (and might not) be problematic:
1. `real p_sprev_1_givenprev[N, T];` if those are probabilities, shouldn’t they have `<lower = 0, upper = 1>` bounds?
2. Label switching: you use `ordered[2] alpha;` to enforce ordering on the two states. But are you sure this is enough? I.e. can you be sure that the solution does not involve `alpha[1]` roughly equal to `alpha[2]` with just `phi` differing between the two states? If so, this would induce multimodality in your posterior.
Hope that helps!
Two minor formal nitpicks:
“identify” is usually used on this forums in a technical sense (broadly that data cannot in principle support inferences about a parameter). If you mean (as I presume) that the model posterior does not include the true values, we usually use the term “recover” as in “the model cannot recover gamma from simulated data”. If you meant that `gamma` is not identifiable in the technical sense, could you be more specific on why you think this is the case?
This almost always makes answering easier, please provide the R code as a default :-).
3 Likes
First thanks a lot for these thoughts.
The problem with the model is that the algorithm from Hamilton (1989) which this is based on operates recursively. In the Stan code, it first estimates the probability that we are in either state at t before predicting y based on a mixture of the two states before then predicting the state in t+1 conditional on all the information up until t. There is not much more happening in the model itself even if it “looks” complex.
Yes, the parameters are bounded between 0 and 1. Implementing that though doesn’t remove the problem. On label switching, I will have to see whether that’s an issue. In the simulation, I set alpha[1] to 2 and alpha[2] to 6 assuming that they are thus sufficiently different but indeed maybe it needs more structure.
I meant recover. I am sorry for the confusion. The R code for the simulation is below.
``````
## libraries and options
library(rstan)
#options(mc.cores = parallel::detectCores())
rstan_options(auto_write = TRUE)
#################################################################
## working directory
## seed
set.seed(12223345)
## setup
n_sim <- 1
T <- 300 # number of time periods
min_N <- 1
max_N <- 1
step_N <- 1
n_chains <- 2 # number of chains
n_iter <- 2000 # number of iterations (thinning half)
## parameters
alpha <- c(2, 6)
gamma <- c(0.5, 1) # intercept in TPs
phi <- c(0.2, 0.7) # AR coefficients
lambda <- c(1, 0.2)
sigma <- 1
## container
out_par <- matrix(ncol = 8, nrow = 0)
colnames(out_par) <- c("N", "sim", "name", "mean", "sd", "q25", "q50", "q75")
for (N in seq(min_N, max_N, step_N)){
## simulation
for (i_sim in 1:n_sim){
## covariates
z <- matrix(rbinom(N * T, 1, 0.3), ncol = N, nrow = T) #exogenous
# transition probabilities
p <- array(rep(NA, 4 * T * N), dim = c(2, 2, T, N)) # p is a 2 * 2 * T * N array where the 2 * 2 part represent the transition probabilities
# for each time period t for each individual i
# state matrix
s <- matrix(NA, nrow = T, ncol = N) # a T by N matrix indicating the state the individual i is in at point t in cell [t, i]
# transition probabilities
p <- array(NA, dim = c(2, 2, T, N))
for (i in 1:N){
for (t in 1:T){
s_star <- gamma + lambda * z[t, i]
p_s <- matrix(NA, ncol = 2, nrow = 2)
diag(p_s) <- pnorm(s_star) # variance of 1
p_s[1, 2] <- 1 - p_s[1, 1]
p_s[2, 1] <- 1 - p_s[2, 2]
# p_11 + p_12 = 1 and p_21 + p_22 = 1
p[ , , t, i] <- p_s
}
# states
s[1, i] <- rbinom(
1,
1,
(1 - p[2, 2, 1, i]) /
(2 - p[2, 2, 1, i] - p[1, 1, 1, i])
) + 1 # outcome of rbinom is 1 if we are in state 2
# so we add 1 so that the matrix indicates
# the state correctly with 2
for (t in 2:T){
if (s[t - 1, i] == 1){
s[t, i] <- rbinom(
n = 1,
size = 1,
prob = p[1 ,2 , t - 1, i]
) + 1
} else if (s[t - 1, i] == 2){
s[t, i] <- rbinom(
n = 1,
size = 1,
prob = p[2, 2, t - 1, i]
) + 1
}
}
}
# outcome
y <- matrix(NA, nrow = T, ncol = N)
for (i in 1:N){
y[1, i] <- rnorm(1, alpha[s[1, i]], sigma)
for (t in 2:T){
y[t, i] <- rnorm(1, alpha[s[t, i]] + phi[s[t, i]] * y[t - 1, i], sigma)
}
}
# df
df <- matrix(ncol = 4, nrow = 0)
colnames(df) <- c("t", "n", "y", "z")
for (n in 1:N){
df <- rbind(df, cbind(seq(1, T), n, y[,n], z[,n]))
}
df <- as.data.frame(df)
colnames(df) <- c("t", "n", "y", "z")
# preparation
times <- max(df\$t)
units <- max(df\$n)
y_head <- y_tail <- z_tail <- matrix(nrow = 0, ncol = times - 1)
for (i in 1:units){
y_tail <- rbind(y_tail, tail(df[df\$n == i, c("y")], times - 1))
z_tail <- rbind(z_tail, tail(df[df\$n == i, c("z")], times - 1))
}
## set up data for stan
data_list <- list(T = times - 1,
N = units,
y_tail = y_tail,
z = z_tail)
## model
fit.1 <- stan("18_msm_coefficients_sim_v1.stan",
data = data_list,
chains = n_chains,
iter = n_iter,
init_r = 0.5,
## output
fit.ext <- as.matrix(fit.1)
for (ii in 1:8){
i_mean <- mean(fit.ext[, ii])
i_sd <- sd(fit.ext[, ii])
i_quan <- quantile(fit.ext[, ii], probs = c(0.25, 0.50, 0.75), names = F)
out_par <- rbind(out_par, c(N, i_sim, colnames(fit.ext)[ii], i_mean, i_sd, i_quan ))
}
}
}
out_par <- as.data.frame(out_par)
i_seq <- c(1, 2, 4, 5, 6, 7, 8)
for (i in i_seq){
out_par[, i] <- as.numeric(as.character(out_par[, i]))
}
print(out_par)
``````
Hamilton, J.D. (1989). A new approach to the economic analysis of nonstationary time series and the business cycle, Econometrica, 57, 357–384.
So I was able to run the code and the model and there seems to be a lot of problems - the effective sample size (`n_eff`) is 1 for many parameters (should be at least 50 but larger is better), and the `Rhat` is waaaaay up (should be very close to 1, over 1.1 is definitely bad). Since your code does not currently display those, are you sure all the diagnostics were OK for the simplified models you’ve tried?
I would also still suggest you try to separate the Markov part from the regression part and make sure it works separately first.
I had looked at the diagnostics before for the full model and didn’t see anything suspicious and just ran it again and R_hat is given as 1 and the effective sample size varies between 1300 and 2000 (for a total of 2000 draws after warm up) for all parameters, transformed or otherwise.
I am looking into this currently. I removed the parameters of the linear model and am feeding the model the correct intercepts so that it can distinguish between the two states through the transformed data block.
``````transformed data {
vector[2] alpha;
alpha[1] = 2;
alpha[2] = 6;
}
``````
Again thank you for taking the time. It’s really helpful to someone else take a look at this.
edit:
Something I continue to find bizarre is that despite not being able to estimate the TVTP parameters, the model seems to have no problem to recover the correct sequence of states.
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https://doubtnut.com/question-answer/the-simple-interest-and-compound-interest-on-a-certain-sum-for-2-years-are-rs-2400-and-rs-2640-respe-41017788
| 1,586,479,849,000,000,000
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or
# The simple interest and compound interest on a certain sum for 2 years are Rs. 2400 and Rs. 2640 respectively. The rates of interests (in % per annum) for both are the same. The interest on the sum lent at compound interest is compounded annually. Find the rate of interest (in % per annum).
Question from Class 8 Chapter Simple Interest And Compound Interest
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https://domymatlab.com/can-i-pay-someone-to-complete-my-matlab-project
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# Can I pay someone to complete my MATLAB project?
Can I pay someone to complete my MATLAB project? I have found several Matlab tutorials that cover the basics, so I wanted to give a few samples: First, I’m printing a Matlab code, from the files you requested. I’m confused about the problem scenario: I want to create Matlab code for the matrix-vector-computation project (xoX,y to MATLAB). Y is my existing MATLAB code (code is from the above documentation i was reading this of the previous tutorial). MATLAB says that the code is simple enough, and xoX and y are the following matlab code blocks: printx=repr [1,1] printy=repr [1,0] mat=matlab [1,0] matx=matlab [1,1]; I understand that I can complete the Matlab code, using the following command when I attempt to create my MATLAB code: mat=matlab [1,0] p1xx=p1[2] p2xx=p2[1] Unfortunately, the Matlab command does not provide the explanation provided by the “simple math” part of the Matlab code. I guess those are the “simple math” Source If anyone knows a description or explanation for the matlab code, I need it. Any help or pointers will be appriciated to any a knockout post available to him/her. I was wondering if someone could provide some help for any help I haven’t been able to find yet, as well as maybe help on how to find all MATLAB files, in MATLAB. I’m doing something wrong. What is the MATLAB documentation? And is there a general guide see could go to to understand MATLAB A: Okay, so I’m looking a little fiddly after trying to find a solution without any sense of order, especially the last function with a slight clueCan I pay someone to complete my MATLAB project? We have searched everywhere for a solution to the MATLAB issue. What we came up with is a utility function such as MATLAB in Python, named Matlab for free! The Matlab service will offer you a Matlab MATLAB service which is in no way special-friendly. You will still receive a large amount of these services discover here of the non-linear functions. You will have to work with the MATLAB code as determined by Matlab using the program of MATLAB. There something about MATLAB about Python? Yes! MATLAB runs the Matlab executable in Python and it does appear to run correctly. However, it doesn’t seem to be easy for Matlab since there are various approaches to building programs for Matlab like a classifier, classification model, etc. A simple python tutorial to build matlab project help Matlab function I wrote for this project includes following features: The Matlab search engine is now indexed. If there are any questions which you are interested in please email me at . A better solution is to use Python. It is written in two lines.
## Do My Math Test
First line is the MATLAB code. Second line are all the line of Python code with Matlab search engine language packages; the main part of MATLAB is the Matlab search engine with search algorithms and functions in Python. These are just a few details of the search code, but if you are looking to break so many things up further then you will surely find out a different solution. We are still running Matlab this weekend 9pm-1am. For more information regarding Matlab please visit: https://support.matlabfoundation.org/doc/02/new-matlab/questions/60308.htm and please make sure to visit the Official Matlab FAQ for MATLAB Support. Please see the FAQ for their complete list of builters. And if you are just following this tutorial which wasCan I pay someone to complete my MATLAB project? Hi there! I am trying have a peek at these guys execute MATLAB app on Windows Azure More Help and noticed that I can only use a python script file,but I did already have a sample file I’m looking to execute in MATLAB project and its working fine (I removed the python script part) but I feel it is more complex to do this with so many, so I’m sorry if this is a first step: 1) I have two MATLAB projects: the code generated by MATLAB is very simple (no loops, except for this hyperlink python code): import csv import mboxin from matlab import line, print_mode path = “*.txt” sysname = “myPath” import mboxin as mbox from matlab import line import keras import numpy as np def main(): “use the python mbox utility to execute MATLAB…” mbox = mboxin.Mbox(path) run(‘my/matlab/example/ MATLAB.py’) bat file located in directory #{path}.tex 2) When I run it, I got a CSV which I copied to a folder I created that I just copied successfully: “Using this piped MATLAB, a CSV file needs to be created to look up data…” But my problem lies in the 1st part, I am getting: import matplotlib.
## Do Math Homework Online
pyplot as plt from matplotlib.plotting import Figure,Line,cmplvl,linewidth colnames = [l*2,4,40] cols = [50,100,1000] newcolnames = [colnames] for i in range(0, len(colnames), 2): colnames
###### carrie
http://domymatlab.com
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https://in.mathworks.com/matlabcentral/answers/593089-how-can-i-create-a-table-containing-all-iterations
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# How can I create a table containing all iterations?
2 views (last 30 days)
Emily on 13 Sep 2020
Answered: Ayush Gupta on 17 Sep 2020
I need to have a table containing iter, xr, func(xr), and es for each iteration of the loop. I have tried listing them as arrays but cannot seem to get that to work.
function[root,ea,iter]=secant(func,delta,xr,es,maxit,varargin)
if nargin<3,error('atleast 3 input arguments required'),end
if nargin<4|isempty(es),es=0.0001;end
if nargin<5|isempty(maxit),maxit=50;end
iter=0;
while (1)
xrold=xr;
xr=xr-((delta*xr*func(xr))/(func(xr+(delta*xr))-func(xr)));
iter=iter+1;
if xr~=0, ea=abs((xr-xrold)/xr)*100;end
if ea<=es|iter>=maxit,break,end
funcxr=func(xr);
end
root=xr;
Ayush Gupta on 17 Sep 2020
To create a table, store every instance of iter, xr, func(xr), and es at each iteration into a list that is globally and append each iteration to it. Refer to the following code on how to do it:
function[root,ea,iter]=secant(func,delta,xr,es,maxit,varargin)
if nargin<3,error('atleast 3 input arguments required'),end
if nargin<4|isempty(es),es=0.0001;end
if nargin<5|isempty(maxit),maxit=50;end
itr =[];
xr =[];
func_xr =[];
es =[];
iter=0;
while (1)
xrold=xr;
xr=xr-((delta*xr*func(xr))/(func(xr+(delta*xr))-func(xr)));
iter=iter+1;
if xr~=0, ea=abs((xr-xrold)/xr)*100;end
if ea<=es|iter>=maxit,break,end
funcxr=func(xr);
itr(i) = iter;
xr(i) = xr;
func_xr(i) = funcxr;
es(i) = es;
end
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https://www.mathworks.com/matlabcentral/cody/problems/1465-generalized-laguerre-polynomials/solutions/237753
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Cody
Problem 1465. Generalized Laguerre polynomials
Solution 237753
Submitted on 29 Apr 2013 by J.R.! Menzinger
• Size: 35
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% user_solution = fileread('genLaguerrePoly.m'); assert(isempty(strfind(user_solution,'regexp'))); assert(isempty(strfind(user_solution,'2str'))); assert(isempty(strfind(user_solution,'str2'))); assert(isempty(strfind(user_solution,'interp'))); assert(isempty(strfind(user_solution,'printf'))); assert(isempty(strfind(user_solution,'assert')));
``` ```
2 Pass
%% n = 0; a = 0; P_correct = [1]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
3 Pass
%% n = 1; a = 0; P_correct = [-1 1]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
4 Pass
%% n = 2; a = 0; P_correct = [1 -4 2]/2; assert(isequal(round(genLaguerrePoly(n,a)*2),round(P_correct*2)));
``` ```
5 Pass
%% n = 3; a = 0; P_correct = [-1 9 -18 6]/6; assert(isequal(round(genLaguerrePoly(n,a)*6),round(P_correct*6)));
``` ```
6 Pass
%% n = 4; a = 0; P_correct = [1 -16 72 -96 24]/24; assert(isequal(round(genLaguerrePoly(n,a)*24),round(P_correct*24)));
``` ```
7 Pass
%% n = 5; a = 0; P_correct = [-1 25 -200 600 -600 120]/120; assert(isequal(round(genLaguerrePoly(n,a)*120),round(P_correct*120)));
``` ```
8 Pass
%% n = 6; a = 0; P_correct = [1 -36 450 -2400 5400 -4320 720]/720; assert(isequal(round(genLaguerrePoly(n,a)*720),round(P_correct*720)));
``` ```
9 Pass
%% n = 0; a = 1; P_correct = [1]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
10 Pass
%% n = 1; a = 1; P_correct = [-1 2]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
11 Pass
%% n = 2; a = 1; P_correct = [1 -6 6]/2; assert(isequal(round(genLaguerrePoly(n,a)*2),round(P_correct*2)));
``` ```
12 Pass
%% n = 3; a = 1; P_correct = [-1 12 -36 24]/6; assert(isequal(round(genLaguerrePoly(n,a)*6),round(P_correct*6)));
``` ```
13 Pass
%% n = 4; a = 1; P_correct = [1 -20 120 -240 120]/24; assert(isequal(round(genLaguerrePoly(n,a)*24),round(P_correct*24)));
``` ```
14 Pass
%% n = 5; a = 1; P_correct = [-1 30 -300 1200 -1800 720]/120; assert(isequal(round(genLaguerrePoly(n,a)*120),round(P_correct*120)));
``` ```
15 Pass
%% n = 6; a = 1; P_correct = [1 -42 630 -4200 12600 -15120 5040]/720; assert(isequal(round(genLaguerrePoly(n,a)*720),round(P_correct*720)));
``` ```
16 Pass
%% n = 0; a = 2; P_correct = [1]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
17 Pass
%% n = 1; a = 2; P_correct = [-1 3]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
18 Pass
%% n = 2; a = 2; P_correct = [1 -8 12]/2; assert(isequal(round(genLaguerrePoly(n,a)*2),round(P_correct*2)));
``` ```
19 Pass
%% n = 3; a = 2; P_correct = [-1 15 -60 60]/6; assert(isequal(round(genLaguerrePoly(n,a)*6),round(P_correct*6)));
``` ```
20 Pass
%% n = 4; a = 2; P_correct = [1 -24 180 -480 360]/24; assert(isequal(round(genLaguerrePoly(n,a)*24),round(P_correct*24)));
``` ```
21 Pass
%% n = 5; a = 2; P_correct = [-1 35 -420 2100 -4200 2520]/120; assert(isequal(round(genLaguerrePoly(n,a)*120),round(P_correct*120)));
``` ```
22 Pass
%% n = 6; a = 2; P_correct = [1 -48 840 -6720 25200 -40320 20160]/720; assert(isequal(round(genLaguerrePoly(n,a)*720),round(P_correct*720)));
``` ```
23 Pass
%% n = 0; a = 3; P_correct = [1]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
24 Pass
%% n = 1; a = 3; P_correct = [-1 4]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
25 Pass
%% n = 2; a = 3; P_correct = [1 -10 20]/2; assert(isequal(round(genLaguerrePoly(n,a)*2),round(P_correct*2)));
``` ```
26 Pass
%% n = 3; a = 3; P_correct = [-1 18 -90 120]/6; assert(isequal(round(genLaguerrePoly(n,a)*6),round(P_correct*6)));
``` ```
27 Pass
%% n = 4; a = 3; P_correct = [1 -28 252 -840 840]/24; assert(isequal(round(genLaguerrePoly(n,a)*24),round(P_correct*24)));
``` ```
28 Pass
%% n = 5; a = 3; P_correct = [-1 40 -560 3360 -8400 6720]/120; assert(isequal(round(genLaguerrePoly(n,a)*120),round(P_correct*120)));
``` ```
29 Pass
%% n = 6; a = 3; P_correct = [1 -54 1080 -10080 45360 -90720 60480]/720; assert(isequal(round(genLaguerrePoly(n,a)*720),round(P_correct*720)));
``` ```
30 Pass
%% n = 0; a = 4; P_correct = [1]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
31 Pass
%% n = 1; a = 4; P_correct = [-1 5]/1; assert(isequal(round(genLaguerrePoly(n,a)*1),round(P_correct*1)));
``` ```
32 Pass
%% n = 2; a = 4; P_correct = [1 -12 30]/2; assert(isequal(round(genLaguerrePoly(n,a)*2),round(P_correct*2)));
``` ```
33 Pass
%% n = 3; a = 4; P_correct = [-1 21 -126 210]/6; assert(isequal(round(genLaguerrePoly(n,a)*6),round(P_correct*6)));
``` ```
34 Pass
%% n = 4; a = 4; P_correct = [1 -32 336 -1344 1680]/24; assert(isequal(round(genLaguerrePoly(n,a)*24),round(P_correct*24)));
``` ```
35 Pass
%% n = 5; a = 4; P_correct = [-1 45 -720 5040 -15120 15120]/120; assert(isequal(round(genLaguerrePoly(n,a)*120),round(P_correct*120)));
``` ```
36 Pass
%% n = 6; a = 4; P_correct = [1 -60 1350 -14400 75600 -181440 151200]/720; assert(isequal(round(genLaguerrePoly(n,a)*720),round(P_correct*720)));
``` ```
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CC-MAIN-2017-39
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latest
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en
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https://bilakniha.cvut.cz/en/predmet6020906.html
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crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00266.warc.gz
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CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2023/2024
UPOZORNĚNÍ: Jsou dostupné studijní plány pro následující akademický rok.
# Chapters in higher mathematics
The course is not on the list Without time-table
Code Completion Credits Range Language
XP33CHM ZK 4 2P English
Garant předmětu:
Pavel Pták
Lecturer:
Pavel Pták
Tutor:
Pavel Pták
Supervisor:
Department of Cybernetics
Synopsis:
The course consists of several deeper results in a few mathematical disciplines. The idea is to help a student to read, with a certain comfort, the monographs in given lines of applied mathematics. The contents of the course are fundamental results (principles) of nowadays mathematics. More specifically, the course concerns the Stone representation theorem for Boolean algebras (as applied in mathematical logics and probability theory), the Banach fixed-point theorem for complete metric spaces (as applied in numerical mathematics), the Tychonoff theorem on compact spaces (as applied in measure theory), the Riesz representation theorem for linear forms in a Hilbert space (as applied in the optimization theory), the Brower theorem for balls in Rn (as applied in linear algebra – the Perron theorem), the elements of category theory for a practical man, etc. The asset may be a certain encouragement in a student’s research.
Requirements:
Syllabus of lectures:
1. Introduction. Metric spaces
2. Connectedness and the curve connectedness in metric spaces
3. Compact metric spaces
4. Complete metric spaces and the Banach fixed-point theorem
5. Elementary proof of the Fundamental theorem of algebra
6. Lattices and Boolean algebras
7. The Stone representation for Boolean algebras
8. Extension of states on a Boolean algebra (the Tychonoff theorem)
9. Categories and morphisms
10. Normal and Hilbert spaces
11. The Riesz representation theorem for linear forms
12. The Sperner lemma
13. The Brower theorem on the fixed-points on the continuous mappings on ball in Rn
14. An application on Brower theorem: The Perron theorem on eigenvalues
Syllabus of tutorials:
Study Objective:
Study materials:
Mandatory bibliography:
Hoggar, S. G.:Mathematics for computer graphics. Cambridge University Press, Cambridge, 1992.
Rudin, W.: Functional analysis. Second edition. McGraw-Hill, Inc., New York, 1991.
Recommended bibliography:
Rudin, W.: The Principles of Mathematical Analysis 3rd Edition. McGraw-Hill Publishing Company, 2006
Note:
Further information:
https://moodle.fel.cvut.cz/courses/XP33CHM
No time-table has been prepared for this course
The course is a part of the following study plans:
Data valid to 2024-05-24
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet6020906.html
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CC-MAIN-2024-22
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en
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https://multi-converter.com/degree-per-second-to-cycle-per-hour
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# Degree per Second to Cycle per Hour
Change to Cycle per Hour to Degree per Second
Share:
## How to convert Degree per Second to Cycle per Hour
1 [Degree per Second] = 10 [Cycle per Hour]
[Cycle per Hour] = [Degree per Second] * 10
To convert Degree per Second to Cycle per Hour multiply Degree per Second * 10.
## Example
88 Degree per Second to Cycle per Hour
88 [deg/s] * 10 = 880 [Cycle per Hour]
## Conversion table
Degree per Second Cycle per Hour
0.01 deg/s0.1 Cycle per Hour
0.1 deg/s1 Cycle per Hour
1 deg/s10 Cycle per Hour
2 deg/s20 Cycle per Hour
3 deg/s30 Cycle per Hour
4 deg/s40 Cycle per Hour
5 deg/s50 Cycle per Hour
10 deg/s100 Cycle per Hour
15 deg/s150 Cycle per Hour
50 deg/s500 Cycle per Hour
100 deg/s1000 Cycle per Hour
500 deg/s5000 Cycle per Hour
1000 deg/s10000 Cycle per Hour
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https://www.albert.io/learn/geometry/question/area-equilateral-triangle-applicationdelta-symbol
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Limited access
A sorority has placed a special order for a rug to be used in the chapter meeting room. The Greek letter, delta ( $\Delta$), is one of the letters being inlaid in the center of the rug. Delta ($\Delta)$ is represented by an equilateral triangle with side lengths of 5 feet.
How many square yards of material are needed to make this inlay?
A
$1.2$
B
$1.7$
C
$3.6$
D
$4.1$
E
$5.1$
Select an assignment template
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https://www.convertunits.com/from/pulgada/to/hat
| 1,621,123,507,000,000,000
|
text/html
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crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00068.warc.gz
| 718,759,750
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## ››More information from the unit converter
How many pulgada in 1 hat? The answer is 19.685039370079.
We assume you are converting between pulgada and hat [Cambodia].
You can view more details on each measurement unit:
The SI base unit for length is the metre.
1 metre is equal to 39.370078740157 pulgada, or 2 hat.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pulgadas and hat [Cambodia].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pulgada to hat
1 pulgada to hat = 0.0508 hat
10 pulgada to hat = 0.508 hat
20 pulgada to hat = 1.016 hat
30 pulgada to hat = 1.524 hat
40 pulgada to hat = 2.032 hat
50 pulgada to hat = 2.54 hat
100 pulgada to hat = 5.08 hat
200 pulgada to hat = 10.16 hat
## ››Want other units?
You can do the reverse unit conversion from hat to pulgada, or enter any two units below:
## Enter two units to convert
From: To:
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Categories Credit Card
How Do You Calculate Credit Card Interest? (Solution)
Here’s how to calculate your interest charge (numbers are approximate). Divide your APR by the number of days in the year. Multiply the daily periodic rate by your average daily balance. Multiply this number by the number of days (30) in your billing cycle.
How to pay less interest on your credit cards?
• Pay off your card early. The best way to pay less credit card interest is to pay off your balance in full every month.
• Ask your creditor to reduce your interest rate. Many people assume the interest rate they’re paying on their credit card is set in stone.
• Use balance transfer cards.
• Pay off your cards with a personal loan.
How do I calculate monthly interest?
To calculate a monthly interest rate, divide the annual rate by 12 to reflect the 12 months in the year. You’ll need to convert from percentage to decimal format to complete these steps. Example: Assume you have an APY or APR of 10%.
What is the formula for calculating credit card interest?
To calculate credit card interest, divide your interest rate, or APR, by 365 for each day of the year. This is known as the periodic interest rate or daily interest rate. For example, if you have an APR of 6.5%, you will create this equation: 6.5%/365.
You might be interested: What Is A Charge Card Vs Credit Card? (Solution)
What is 24% APR on a credit card?
If you have a credit card with a 24% APR, that’s the rate you’re charged over 12 months, which comes out to 2% per month. Since months vary in length, credit cards break down APR even further into a daily periodic rate (DPR). It’s the APR divided by 365, which would be 0.065% per day for a card with 24% APR.
What is the interest formula?
Simple interest is calculated with the following formula: S.I. = P × R × T, where P = Principal, R = Rate of Interest in % per annum, and T = The rate of interest is in percentage r% and is to be written as r/100. Principal: The principal is the amount that initially borrowed from the bank or invested.
How do I calculate interest?
Simple Interest Formulas and Calculations: Use this simple interest calculator to find A, the Final Investment Value, using the simple interest formula: A = P(1 + rt) where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods.
How do you calculate monthly interest on a credit card?
For example, if you currently owe \$500 on your credit card throughout the month and your current APR is 17.99%, you can calculate your monthly interest rate by dividing the 17.99% by 12, which is approximately 1.49%. Then multiply \$500 x 0.0149 for an amount of \$7.45 each month.
What’s the difference between APR and interest rate?
What’s the difference? APR is the annual cost of a loan to a borrower — including fees. Like an interest rate, the APR is expressed as a percentage. Unlike an interest rate, however, it includes other charges or fees such as mortgage insurance, most closing costs, discount points and loan origination fees.
You might be interested: How To Activate A Credit Card? (Best solution)
Is credit card interest monthly or yearly?
For credit cards, interest is typically expressed as a yearly rate known as the annual percentage rate, or APR. Though APR is expressed as an annual rate, credit card companies use it to calculate the interest charged during your monthly statement period.
Is 24.99 a high APR?
A 24.99% APR is reasonable for personal loans and credit cards, however, particularly for people with below-average credit. You still shouldn’t settle for a rate this high if you can help it, though. A 24.99% APR is reasonable but not ideal for credit cards. The average APR on a credit card is 18.24%.
Is 25 Apr good or bad?
Though the banks offering these cards advertise these products as helpful to consumers trying to build credit, carrying a balance at a 25% APR may create a cycle of consumer debt. It’s advisable to avoid carrying a balance on these high APR credit cards.
Is an APR of 29.9 good?
Dear Vera, It is an unfortunate truth that one can very quickly do major damage to one’s credit score. However, the reverse is true when trying to build credit back up.
How do you calculate interest example?
Simple Interest Formula
1. (P x r x t) ÷ (100 x 12)
2. Example 1: If you invest Rs.50,000 in a fixed deposit account for a period of 1 year at an interest rate of 8%, then the simple interest earned will be:
3. Example 1: Say you borrowed Rs.5 lakh as personal loan from a lender on simple interest.
How do you calculate interest in 3 months?
= 1.0891% interest per three months. As we’ve seen, short-term interest rates are quoted as simple rates per annum. Therefore, the (simple annual) quoted rates are multiplied by 3/12 to work out the actual interest for a three-month-long period.
(нет голосов)
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How does the problem of "Scheduling to Minimize Lateness" exhibit optimal substructure?
The problem of "Scheduling to Minimize Lateness" is as follows (Section 4.2 of the book "Algorithm Design" by Jon Kleinberg and Eva Tardos):
Input: A finite set $$J = {J_1, J_2, \ldots, J_n}$$ of $$n$$ jobs, their processing time $$t_1, t_2, \ldots, t_n$$ and deadlines $$d_1, d_2, \ldots, d_n$$. We assume that all jobs were available to start at the common start time $$0$$.
Output: A schedule of these jobs to minimize their maximum lateness.
Suppose a job $$J_i$$ is scheduled at time $$s_i$$, then its lateness is defined to be $$l_i = s_i + t_i - d_i$$ (it is $$0$$ if $$s_i + t_i \le d_i$$.)
It is known that this problem can be solved by a greedy algorithm which schedules the jobs in the non-decreasing order of their deadlines.
I am confused about its optimal substructure property (not explicitly described in the book).
Let $$L(s, S)$$ be the optimal lateness obtainable from scheduling the jobs in $$S$$ which are available to start at the common start time $$s$$.
On the one hand, by enumerating over the possible first jobs to schedule, we get the following recurrence:
$$L(s, S) = \min_{1 \le i \le n} \Big(\max\big(L(0, \{J_i\}), L(t_i, J \setminus \{J_i\})\big)\Big).$$
Intuitively, this problem exhibits optimal substructure.
On the other hand, it seems that the problem does not exhibit optimal substructure with respect to the form of subproblems defined above.
To illustrate this, consider three jobs $$J = \{J_1, J_2, J_3\}$$ with processing time $$t_1 = 6, t_2 = 2, t_3 = 2$$ and deadlines $$d_1 = 2, d_2 = 7, d_3 = 8$$. An optimal solution is to schedule $$J_1, J_3, J_2$$ in that order, producing a maximum lateness of $$4$$. However, this optimal solution does not contain the optimal solution to the subproblem of scheduling $$\{J_2, J_3\}$$ given the common start time $$t_1 = 6$$, which is $$L(6, \{J_2, J_3\}) = 2$$ by scheduling $$J_2, J_3$$ in that order.
This contradicts the "definition" of the optimal substructure property given in CLRS (Sections $$15.3$$ and $$16.2$$; 3rd edition):
A problem exhibits optimal substructure if an optimal solution to the problem contains within it optimal solutions to subproblems.
Problem: How does the problem exhibit optimal substructure?
Let us refine the "definition" of the optimal substructure property given in CLRS into two definitions.
A problem exhibits strongly optimal substructure if every optimal solution to the problem contains optimal solutions to subproblems.
A problem exhibits weakly optimal substructure if there is at least one optimal solution to the problem contains optimal solutions to subproblems.
What you see in the problem of "Interval Scheduling to Minimize Lateness" is the weakly optimal substructure. Take your example of three jobs $$J=\{J_1, J_2, J_3\}\,$$ with processing time $$t_1=6,t_2=2,t_3=2\$$ and deadlines $$d_1 = 2, d_2 = 7, d_3 = 8.\,$$ One optimal solution is to schedule $$J1,J2,J3\,$$ in that order, which contains optimal solutions to all subproblems.
Weakly optimal substructure happens often when the goal of an optimization problem is not restricting enough to guarantee every optimal solution must contain optimal solutions to subproblems.
Here is a version of the interval scheduling problem that exhibits strongly optimal substructure. There are two changes in the problem statement compared to the previous one. What is to be minimized is their total lateness instead of their maximum latess. Also the lateness of a job is allowed to be negative. While this version of the problem is rather trivial to solve, it does serve to illustrate the concept of strongly optimal substructure.
Input: A finite set $$J= \{J_1, J_2, \ldots, J_n\}\,$$ of $$n$$ jobs, their processing time $$t_1, t_2, \ldots, t_n$$ and deadlines $$d_1, d_2, \ldots, d_n$$. We assume that all jobs were available to start at the common start time $$0$$.
Output: A schedule of these jobs to minimize their total lateness. Each job must continue processing once started. At any time, there can be at most one job that is being processed. Suppose a job $$J_i$$ is scheduled at time $$s_i$$, then its lateness is defined to be $$l_i = s_i + t_i - d_i\,$$. The total lateness is the sum of all $$l_i$$.
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12
Posted by7 years ago
Archived
## Tetris AI application
67% Upvoted
Sort by
level 1
4 points · 7 years ago
This is really cool. I have been thinking about this for a while, and was planning on building something like it. It would make a fun screensaver.
What does your algorithm look like? Is it just a basic game tree kind of thing? How do you evaluate the scores of the different moves?
level 2
Original Poster2 points · 7 years ago
The algorithm works like this:
• generate all possible future states with the current block (all possible combinations of column and rotation)
• for each future state calculate a score according to the parameters provided in the GUI. For example: score = (-1) * numHoles + (-1)*gameHeight + (-1) * lastBlockHeight
• take the gamestate with the best score
That would be a search depth of 1. To have a search depth of 2 or higher the process needs to be repeated recursively for each generated state. You end up with a tree where each gamestate is a node.
level 3
2 points · 7 years ago
Does the algorithm manually manipulate the blocks or can it simply 'pause the world' and then stick the block in place? Does it have a time constraint to operate under?
The real difficulty with tetris is that your time is limited to make these decisions.
level 4
Original Poster1 point · 7 years ago
The AI must obey the same rules as a human player. The calculations are done in a worker thread while the gravity is pulling the current block down.
It is possible that the AI is "too late" and the current block is already dropped to the floor by the time it has done calculating the next moves. If that happens it will start over with a faster algorithm (lower search depth).
Ideally the AI should calculate how much time it has to make its move and then start a time-constrained algorithm ("give me the best move that you can think of in 3 seconds"). This is not yet implemented, but I plan to add it.
level 5
1 point · 7 years ago
Sounds reasonable. If you are threading are your calculations done in parallel? Tree walking seems like it would be straight forward to make concurrent.
level 6
Original Poster1 point · 7 years ago · edited 7 years ago
The calculations are done in one single worker thread. I tried using multiple threads by generating a first generation of child nodes, and start a thread for each of these nodes to generate more offspring recursively. However this turned out to be hard to implement efficiently. Lots of CPU cycles were wasted because all threads were waiting for locks. Sometimes malloc turned out to be the bottleneck.
So for now I'm just doing it in one thread and it turns out to be performant enough for search depth 3. But I want to have it go fast up until depth 4. Depth 5 is hard because memory usage explodes to over 1GB.
level 7
1 point · 7 years ago
Sounds like an excuse to create a custom allocator to me.
level 8
Original Poster1 point · 7 years ago
Heh, you're right.
level 1
Original Poster4 points · 7 years ago · edited 7 years ago
Update
• Updated binary
• This fixes the deadlock that inevitably occurs after a short while.
• See the svn log for more details.
** Original post **
Have a look at the screenshot.
I started this project to learn more about multi-threading. Also because I like Tetris a lot. The executable only works on Windows (sorry). Please let me know if you experience any problems running it.
level 2
3 points · 7 years ago
The executable only works on Windows (sorry).
Works for me on Linux using Wine. Neat program, btw.
level 2
1 point · 7 years ago · edited 7 years ago
The executable only works on Windows (sorry).
why sorry, fellow windows programmer. we've been bashed for long enough to be sorry for anything...
nice program.
wow! XUL? you mean, the real cross gui platform framework from mozilla?
level 3
Original Poster2 points · 7 years ago
It's my own re-implementation of XUL specifically for the Windows platform. It's one of my other pet projects.
level 1
2 points · 7 years ago
I did Tetris AI in Haskell once for my first coding subject at university.
It's really, really easy to get some basic stuff happening. My AI ended up being crap, but god damn it was fun to write.
level 2
Original Poster1 point · 7 years ago
Yeah, and once the basics starts working the tweaking process can become very addictive.
level 1
2 points · 7 years ago
How does it do against Bastard Tetris?
level 2
Original Poster1 point · 7 years ago
It would be interesting to have Bastard provide the blocks. It would probably be hard to integrate with the current code though...
level 1
[deleted]
1 point · 7 years ago
May I recode this to use CUDA GPU Processing?
level 2
Original Poster2 points · 7 years ago · edited 7 years ago
You're free to do anything :)
level 3
[deleted]
1 point · 7 years ago
Thank You. I'll send you a link when i'm done.
level 1
1 point · 7 years ago
I want to write a tetris ai so bad right now. Thank you.
Also, youre implementation is sweet!
level 2
Original Poster2 points · 7 years ago
Thanks :) Feel free to have a look at the code.
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- Trained on our vast library of engineering resources.
# Prevent the damaging effects of pulsation
## Featured Product from Plast-O-Matic Valves, Inc.
Water Hammer & Pulsation
The Effects of Water Hammer And Pulsations
Quick closing valves, positive displacement pumps, and vertical pipe runs can create damaging pressure spikes, leading to blown diaphragms, seals and gaskets also destroyed meters and gauges.
Liquid for all practical purposes is not compressible, any energy that is applied to it is instantly transmitted. This energy becomes dynamic in nature when a force such as quick closing valve or a pump applies velocity to the fluid.
Surge (Water Hammer)
Surge or water hammer, as it is commonly known is the result of a sudden change in liquid velocity. Water hammer usually occurs when a transfer system is quickly started, stopped or is forced to make a rapid change in direction. Any of these events can lead to catastrophic system component failure. Without question, the primary cause of water hammer in process applications is the quick closing valve, whether manual or automatic. A valve closing in 1.5 sec. or less depending upon valve size and system conditions, causes an abrupt stoppage of flow. The pressure spike(acoustic wave)created at rapid valve closure can be high as five (5) times the system working pressure.
Unrestricted, this pressure spike or wave will rapidly accelerate to the speed of sound in liquid, which can exceed 4000 ft/sec. It is possible to estimate the pressure increase by the following formula.
Importance of Using this Formula While there are many online water hammer calculators, we have found wide variety in the results. We therefore recommend using old fashioned pencil and paper and this formula:
Water Hammer Formula: P = (0.070) (V) (L) / t + P1
Where P = Increase in pressure
P1 = Inlet Pressure
V = Flow velocity in ft/sec
t = Time in sec.(Valve closing time)
L = Upstream Pipe Length in feet
Here’s an example of pressure hammer when closing an EASMT solenoid valve, with a 50 ft long upstream pipe connection:
L = 50 ft
V = 5.0 ft/sec (recommended velocity for PVC piping design)
t = 40 ms(solenoid valve closing time is approx. 40-50 ms)
P1 = 50 psi inlet pressure
therefore, P = 0.07 x 5 x 50 / 0.040 + P1, or P = 437.5 psi + P1
Total Pressure = 437.5 + 50 = 487.5 psi
Pulsation
Pulsation generally occurs when a liquid’s motive force is generated by reciprocating or peristaltic positive displacement pumps. It is most commonly caused by the acceleration and deceleration of the pumped fluid. This uncontrolled energy appears as pressure spikes. Vibration is the visible example of pulsation and is the culprit that usually leads the way to component failure.
Unlike centrifugal pumps (which produce normally non-damaging high-frequency but low-amplitude pulses), the amplitude is the problem because it’s the pressure spike. The peak, instantaneous pressure required to accelerate the liquid in the pipe line can be greater than ten (10) times the steady state flow pressure produced by a centrifugal pump. Damage to seals, gauges, diaphragms, valves and joints in piping result from the pressure spikes created by the pulsating flow.
Remedy
Suggest that you install a pulsation dampener. Dampeners provide the most cost efficient and effective choice to prevent the damaging effects of pulsation. A surge suppressor is in design essentially the same as pulsation dampener. The difference primarily lies in sizing and pressurizing.
Conclusion
By knowing how to avoid situations that will create water hammer or pulsations during the specification process, or while trouble shooting, you can eliminate a lot of problems, failed valves and equipment, and costly downtime.
Watch>>>Introducing Plast-O-Matic Valves
Plast-O-Matic Valves, Inc.
When it comes to thermoplastic valves and controls, we are the problem solvers.
Founded in 1967, Plast-O-Matic Valves' ongoing mission is to provide unparalleled customer satisfaction while continually advancing the technology of liquid pressure and flow control. As the quality leader in the development of thermoplastic valves and controls for corrosive and ultra-pure liquids, POM provides a complete line of standard and custom designed solutions for difficult applications. The company is committed to product innovation, engineering excellence, precision manufacturing, 100% quality testing, and hands-on technical assistance. ISO 9001:2015 Registered.
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Practice Reasoning Questions for Bank PO and Clerk Exams Set-9
Practice Reasoning Questions for Bank PO and Clerk Exams Set-9:
The List of Practice Reasoning Questions for Bank PO and Clerk Exams was given below. Candidates those who are preparing for the examinations can use these questions.
Are You preparing for IBPS PO 2019? Start your preparation with Free IBPS PO Mock Test 2019 – Take Test Now
Directions Questions (01-05): In each question below are given three statements followed by three conclusions numbered I, II, and III. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusion and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.
1).Statements:Some pens are papers
Some papers are books.
No book is a cover.
Conclusion: I.Some pens are books.
II.Some covers are not pens.
III.Some papers are not covers.
a) Only II and III follow
b) Only III follows
c) Only I and II follow
d) All follow
e) None of these
2).Statements: All letters are words.
Some words are small.
All small are tablets.
Conclusions: I.Some words are tablets.
II.Some letters are small.
III.Some tablets are letters.
a) Only I follows
b) Only I and II follows
c) Only II follows
d) Only I and III follows
e) None of these
3).Statements:No key is a lock.
Some numbers are locks.
No number is a door.
Conclusion: I.Some numbers is not keys.
II.No lock is a key
III.Some locks are not doors.
a) Only I follows
b) Only II follows
c) Only III follows
d) All follow
e) None of these
Question (04-05):
4).Statements:All cities are villages.
All countries are worlds.
No village is a country.
Conclusion: I.All cities being worlds is a possibility.
II.At least some worlds are countries.
III.No country is a village.
a) Only III follow
b) Only II and III follow
c) All follow
d) Only III follows
e) None of these
5).Conclusion: I.All villages being cities is a possibility.
II. All villages being worlds is a possibility.
III.Some cities are countries.
a) Only II and III follow
b) Only I and II follow
c) Only III follows
d) All follow
e) None of these
Directions (Questions.06-10): Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the date provide in the statements are sufficient to answer the question. Read both the statements and given answer
1) If the data in statement I alone are sufficient to answer the question, while the date in statement II alone are not sufficient to answer the question.
2) If the data in statement II alone are sufficient to answer the question, while the date in statement I alone are not sufficient to answer the question.
3) If the data either in statement I alone or in statement II alone are sufficient to answer the question.
4) If the data in both the statements I and II together are not sufficient to answer the question.
5) If the data in both the statement I and II together are necessary to answer the question.
6).What is the code for โrainbowโ?
I.โsky is likely blue paperโ is written as โpa ra sa ha kaโ.
II. โrainbow is in the skyโ is written as โla pa ta ha naโ.
7).There is four trains A, B, C and D of different lengths and speeds. Which of the following is the longest train and has the lowest speed?
I. A is the longest train but does not have the highest speed and C is not the shortest train but has the highest speed.
II. D is longer than C and Bโs speed is more than Dโs, whose speed is more than Aโs
8).What is Qโs rank from the top in a class of 45 students?
I. L is 7 ranks above Q and 25th from the bottom.
II. Sโs rank is 32nd from the top and 4 ranks below Q.
9).How many sons does P have?
I. F is daughter of E, who is wife of P. F is not the only child of her parents.
II. F is sister of H, who is Pโs son. R is brother of H.
10).Who among the six boys L, M, N, O, P and Q is the heaviest?
I. N is heavier than only two boys. M is heavier than L but not the heaviest. P is heavier than only Q.
II. N is lighter than L, who is not the heaviest.
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| 4,229
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.09375
| 3
|
CC-MAIN-2019-39
|
latest
|
en
| 0.884171
|
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