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# 2.2 notation and algebra of functions
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### 2.2 notation and algebra of functions
1. 1. Notation and Algebra of Functions http://www.lahc.edu/math/precalculus/math_260a.html
2. 2. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). Notation and Algebra of Functions
3. 3. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Notation and Algebra of Functions
4. 4. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Notation and Algebra of Functions
5. 5. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions
6. 6. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered),
7. 7. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost).
8. 8. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost). So this procedure is a function.
9. 9. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost). So this procedure is a function. Its domain is D = {1, 2, 3, …}
10. 10. Recall that the pizzas at Pizza Grande cost \$8 each and there is \$10 delivery charge so the cost is \$50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost). So this procedure is a function. Its domain is D = {1, 2, 3, …} and its range is R = {various \$-cost}.
11. 11. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range.
12. 12. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.
13. 13. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver.
14. 14. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = the range R =
15. 15. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R =
16. 16. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}.
17. 17. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}. b. A procedure that takes any driver–license number as an input and produces the name(s) of all of the cousins of that license holder.
18. 18. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}. b. A procedure that takes any driver–license number as an input and produces the name(s) of all of the cousins of that license holder. This is not a function because the license holder may have many cousins so there might be several outputs.
19. 19. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions.
20. 20. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y,
21. 21. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
22. 22. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. the domain D the range R Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
23. 23. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range R Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
24. 24. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
25. 25. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name”. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
26. 26. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
27. 27. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing: the input x
28. 28. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = “Joe Blow”. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing: the input x the output y
29. 29. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = “Joe Blow”. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing: the input x the output y 123456 “Joe Blow” D ={Lic–numbers} R={Names} Licen#-to-name
30. 30. Notation and Algebra of Functions The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}.
31. 31. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}.
32. 32. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function.
33. 33. We may define functions with tables, for example: x y=f(x) –1 4 2 3 5 –3 6 4 7 2 Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function.
34. 34. We may define functions with tables, for example: x y=f(x) –1 4 2 3 5 –3 6 4 7 2 Notation and Algebra of Functions We search the table for outputs so f(2) = 3, f(6) = 4, etc.. There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function.
35. 35. x y=f(x) –1 4 2 3 5 –3 6 4 7 2 The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function. We may define functions with tables, for example: We search the table for outputs so f(2) = 3, f(6) = 4, etc..
36. 36. x y=f(x) –1 4 2 3 5 –3 6 4 7 2 The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}. Note that f(–1) = f(6) = 4, Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function. We may define functions with tables, for example: We search the table for outputs so f(2) = 3, f(6) = 4, etc..
37. 37. x y=f(x) –1 4 2 3 5 –3 6 4 7 2 The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}. Note that f(–1) = f(6) = 4, so that a function may assign multiple inputs to the same output. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function. We may define functions with tables, for example: We search the table for outputs so f(2) = 3, f(6) = 4, etc..
38. 38. Notation and Algebra of Functions Functions may be given graphically as the ones here:
39. 39. For instance, Nominal Price(1975) \$0.50 Notation and Algebra of Functions (1975, \$0.50) Nominal–price is the price that’s posted at the gas stations. Functions may be given graphically as the ones here:
40. 40. Domain (Nominal Price) = {year 1918 2005} For instance, Nominal Price(1975) \$0.50 Notation and Algebra of Functions (1975, \$0.50) 1918 2005 Nominal–price is the price that’s posted at the gas stations. Functions may be given graphically as the ones here:
41. 41. Domain (Nominal Price) = {year 1918 2005} Range (Nominal Price) = {\$0.20\$2.51} For instance, Nominal Price(1975) \$0.50 Notation and Algebra of Functions (1975, \$0.50) \$0.20 \$2.51 1918 2005 Nominal–price is the price that’s posted at the gas stations. Functions may be given graphically as the ones here:
42. 42. Notation and Algebra of Functions (1975, \$0.50) (1975, \$1.85) Adjusted Price(1975) \$1.85 Adjusted–price is the inflation adjusted price in 2007–dollar. Functions may be given graphically as the ones here:
43. 43. Notation and Algebra of Functions Domain (Adjusted Price) = {year 1918 2005} Adjusted Price(1975) \$1.85 (1975, \$0.50) (1975, \$1.85) 1918 2005 Adjusted–price is the inflation adjusted price in 2007–dollar. Functions may be given graphically as the ones here:
44. 44. Notation and Algebra of Functions Domain (Adjusted Price) = {year 1918 2005} Range (Adjusted Price) = {\$1.25\$3.50} Adjusted Price(1975) \$1.85 (1975, \$0.50) (1975, \$1.85) 1918 2005 \$1.25 \$3.50 Adjusted–price is the inflation adjusted price in 2007–dollar. Functions may be given graphically as the ones here:
45. 45. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f.
46. 46. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically.
47. 47. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out.
48. 48. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out. x y x y x y x yA. B. C. D.
49. 49. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out. Hence A and B below are graphs of functions x y x y x y x y A and B are functions because for each input x there is one output y. A. B. C. D. (x, y) (x, y)
50. 50. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out. Hence A and B below are graphs of functions but C and D are not graphs of functions. x y x y (x, y2) y x y A and B are functions because for each input x there is one output y. A. B. C. D. C and D are not since there are multiple points lining up vertically. (x, y) (x, y) x (x, y1) (x, y3)
51. 51. Notation and Algebra of Functions Most functions are given by mathematical formulas.
52. 52. For example, f(X) = X2 – 2X + 3 = y Notation and Algebra of Functions Most functions are given by mathematical formulas.
53. 53. For example, f(X) = X2 – 2X + 3 = y name of the function Notation and Algebra of Functions Most functions are given by mathematical formulas.
54. 54. For example, f(X) = X2 – 2X + 3 = y name of the function Notation and Algebra of Functions Most functions are given by mathematical formulas. input box
55. 55. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function Notation and Algebra of Functions Most functions are given by mathematical formulas. input box
56. 56. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box
57. 57. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula.
58. 58. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) =
59. 59. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3
60. 60. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
61. 61. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y. Real functions are functions whose domain and range are subsets of real numbers (no complex numbers).
62. 62. Domain of Functions
63. 63. Domain of Functions There are two main restrictions to consider when determining the domains of real functions.
64. 64. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.
65. 65. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6)
66. 66. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0
67. 67. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3
68. 68. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3 So the domain = {all numbers except x = –3}.
69. 69. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3 So the domain = {all numbers except x = –3}. b. f(x) = 2x + 6
70. 70. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3 So the domain = {all numbers except x = –3}. b. f(x) = 2x + 6 We can only extract square roots of nonnegative numbers
71. 71. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3 So the domain = {all numbers except x = –3}. b. f(x) = 2x + 6 We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0 x ≥ –3.
72. 72. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3 So the domain = {all numbers except x = –3}. b. f(x) = 2x + 6 We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0 x ≥ –3. So the domain = {all numbers x ≥ –3}.
73. 73. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3 So the domain = {all numbers except x = –3}. b. f(x) = 2x + 6 We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0 x ≥ –3. So the domain = {all numbers x ≥ –3}. The requirement of having “the radicands ≥ 0” applies to the 4th root, or the 6th root, or any even-order root.
74. 74. Domain of Functions c. Find the domain of f(x) = 1 x + 6 – 14
75. 75. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 14
76. 76. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 4
77. 77. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. 4
78. 78. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 4
79. 79. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 Draw the real line and sample points, – 5– 6 UDF 4
80. 80. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 Draw the real line and sample points, we get : – 5– 6 UDF – –+– – 4
81. 81. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1 x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 Draw the real line and sample points, we get : – 5– 6 UDF – –+– – The domain consists of the non–negative portion i.e. the domain is the interval (–6 ,–5]. 4
82. 82. Algebra of Functions We may form expressions using functions or use expressions as inputs for functions.
83. 83. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) b. f(a+b)
84. 84. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, b. f(a+b)
85. 85. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, b. f(a+b)
86. 86. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b)
87. 87. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3
88. 88. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
89. 89. c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] Algebra of Functions insert [ ] for subtraction We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
90. 90. c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] = 4a2 – 4a + 3 – 3a – 3b + 4 Algebra of Functions insert [ ] for subtraction We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
91. 91. c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] = 4a2 – 4a + 3 – 3a – 3b + 4 = 4a2 – 7a – 3b + 7 Algebra of Functions insert [ ] for subtraction We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
92. 92. Notation and Algebra of Functions One important function-expression for any f(x) is its "difference quotient": h f(x+h) – f(x) where h is another variable.
93. 93. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
94. 94. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
95. 95. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
96. 96. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
97. 97. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
98. 98. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] = h 4xh + 2h2 – 3h Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
99. 99. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] = h 4xh + 2h2 – 3h = h h(4x + 2h – 3) Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
100. 100. Example E. Let f(x) = 2x2 – 3x + 4, simplify Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] = h 4xh + 2h2 – 3h = h h(4x + 2h – 3) = 4x + 2h – 3 One important function-expression for any f(x) is its "difference quotient":
101. 101. Composition of Functions When one function is used as the input of another function, the outcome is called their composition.
102. 102. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).
103. 103. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f.
104. 104. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2)
105. 105. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2))
106. 106. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5
107. 107. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5)
108. 108. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20
109. 109. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20
110. 110. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2)
111. 111. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2))
112. 112. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1
113. 113. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1
114. 114. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x)
115. 115. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3
116. 116. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3 = –12x + 23
117. 117. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3 = –12x + 23 In general, (f○g)(x) = (g○f)(x).
118. 118. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1)
119. 119. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) b. (f○g○h)(1)
120. 120. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) b. (f○g○h)(1)
121. 121. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) b. (f○g○h)(1)
122. 122. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 b. (f○g○h)(1)
123. 123. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1))
124. 124. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1)) (1, ½ )
125. 125. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1)) ≈ f(g(1/2)), (1, ½ )
126. 126. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1)) ≈ f(g(1/2)), this is UDF because x = ½ in not in the domain of g. (1, ½ )
127. 127. Composition of Functions Equations may be posed in the language of functions.
128. 128. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x)
129. 129. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2
130. 130. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 x2 – 4x – 5 = 0Setting one side 0:
131. 131. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 x2 – 4x – 5 = 0Setting one side 0:
132. 132. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0Setting one side 0:
133. 133. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0 f(x) = g(x) means that x2 – 4x – 3 = –3x – 1 Setting one side 0:
134. 134. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0 f(x) = g(x) means that x2 – 4x – 3 = –3x – 1 (x + 1)(x – 2) = 0 so x = –1, or 2, x2 – x – 2 = 0 Setting one side 0: Setting one side 0:
135. 135. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0 f(x) = g(x) means that x2 – 4x – 3 = –3x – 1 (x + 1)(x – 2) = 0 so x = –1, or 2, x2 – x – 2 = 0 i.e. f(–1) = g(–1) and f(2) = g(2) Setting one side 0: Setting one side 0:
136. 136. Composition of Functions c. Solve for x where f(x) ≥ g(x).
137. 137. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1.
138. 138. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0
139. 139. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0
140. 140. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 2–1 (x + 1)(x – 2)
141. 141. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 0 2–1 + + + – – – – + + + + (x + 1)(x – 2)
142. 142. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 0 2–1 + + + – – – – + + + + (x + 1)(x – 2)
143. 143. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 So f(x) ≥ g(x) for x’s from: (–∞, –1] U [2, ∞). Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. 0 2–1 + + + – – – – + + + + (x + 1)(x – 2)
144. 144. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 So f(x) ≥ g(x) for x’s from: (–∞, –1] U [2, ∞). Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. 0 2–1 + + + – – – – + + + + (x + 1)(x – 2) The step of “setting one side 0” allows us to transform all equations and inequalities into problems of finding roots (zeros) or sign–charting.
145. 145. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 So f(x) ≥ g(x) for x’s from: (–∞, –1] U [2, ∞). Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. 0 2–1 + + + – – – – + + + + (x + 1)(x – 2) The step of “setting one side 0” allows us to transform all equations and inequalities into problems of finding roots (zeros) or sign–charting. This step will be done as a routine from now on. | 16,778 | 49,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-09 | latest | en | 0.84368 |
https://link.springer.com/article/10.1186/s13660-016-1072-6 | 1,726,042,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00768.warc.gz | 336,690,312 | 62,121 | ## 1 Introduction
Let H be a real Hilbert space with the inner product $$\langle\cdot ,\cdot\rangle$$, which induces the norm $$\|\cdot\|$$.
Let C be a nonempty, closed, and convex subset of H. Let T be a nonlinear mapping of C into itself; we denote with $$\operatorname{Fix}(T)$$ the set of fixed points of T, that is, $$\operatorname{Fix}(T)=\{z\in C: Tz=z\}$$.
We recall that a mapping $$T:C\rightarrow C$$ is said to be k-strict pseudo-contractive (in the sense of Browder-Petryshyn) if there exists $$k\in[0,1)$$ such that
$$\Vert Tx-Ty\Vert ^{2}\leq \Vert x-y\Vert ^{2}+k\bigl\Vert (I-T)x-(I-T)y\bigr\Vert ^{2}, \quad \forall x,y \in C.$$
(1.1)
Note that the class of strict pseudo-contractions includes the class of nonexpansive mappings, which are mappings T on C such that
$$\|Tx-Ty\|\leq\|x-y\|, \quad \forall x,y \in C.$$
The problem of finding fixed points of nonexpansive mappings via Mann’s algorithm [2] has been widely investigated in the literature (see e.g. [3]).
Mann’s algorithm generates, on initializing with an arbitrary $$x_{1}\in C$$, a sequence according to the recursive formula
$$x_{1}\in C, \quad x_{n+1}=\alpha_{n} x_{n}+(1-\alpha_{n})Tx_{n},\quad \forall n\geq1,$$
(1.2)
where $$(\alpha_{n})_{n\in\mathbb{N}}\subset(0,1)$$.
If $$T:C\rightarrow C$$ is a nonexpansive mapping with a fixed point in a closed and convex subset of a uniformly convex Banach space with a Frechét differentiable norm, and if the control sequence $$(\alpha _{n})_{n\in\mathbb{N}}$$ is chosen so that $$\sum_{n=1}^{\infty}\alpha _{n}(1-\alpha _{n})=\infty$$, then the sequence $$(x_{n})$$ generated by Mann’s algorithm converges weakly to a fixed point of T [3]. However, this convergence is in general not strong (see the counterexample in [4]).
On the other hand, iterative algorithms for strict pseudo-contractions are still less developed than those for nonexpansive mappings, despite the pioneering work of Browder and Petryshyn [5] dating from 1967. However, strict pseudo-contractions have many applications, due to their ties with inverse strongly monotone operators. Indeed, if A is a strongly monotone operator, then $$T=I-A$$ is a strict pseudo-contraction, and so we can redraft a problem of zeros for A in a fixed point problem for T, and vice versa (see e.g. [6, 7]).
The Mann algorithm has weak convergence also in the broader setting of strict pseudo-contractions mapping, containing the nonexpansive mappings.
### Theorem 1.1
(Marino and Xu [8], 2007, Mann’s method)
Let C be a closed and convex subset of a Hilbert space H. Let $$T:C\rightarrow C$$ be a k-strict pseudo-contraction for some $$0\leq k<1$$. Assume that T admits a fixed point in C. Let $$(x_{n})$$ be the sequence generated by $$x_{0}\in C$$ and the Mann algorithm
$$x_{n+1}=\alpha_{n}x_{n}+ (1-\alpha_{n} )Tx_{n}.$$
Assume that the control sequence $$(\alpha_{n})$$ is chosen so that $$k<\alpha_{n}<1$$ for all n and
$$\sum_{n=0}^{+\infty} (\alpha_{n}-k ) (1-\alpha _{n} )=+\infty.$$
Then $$(x_{n})$$ converges weakly to a fixed point of T.
It is not possible, in general, to obtain strong convergence, in view of the celebrated counterexample of Genel and Lindenstrauss [4].
So, to obtain strong convergence, one can try to modify the Mann algorithm and strengthen the hypotheses on the mapping.
We recall here some obtained results.
### Theorem 1.2
(Li et al. [9], 2013, modified Halpern’s method)
Let C be a closed and convex subset of a real Hilbert space H, $$T:C\rightarrow C$$ be a k-strict pseudo-contraction such that $$\operatorname{Fix}(T)\neq\emptyset$$. For an arbitrary initial value $$x_{0}\in C$$ and fixed anchor $$u\in C$$, define iteratively a sequence $$(x_{n})$$ as follows:
$$x_{n+1}=\alpha_{n}u+\beta_{n}x_{n}+ \gamma_{n}Tx_{n},$$
where $$(\alpha_{n})$$, $$(\beta_{n})$$, $$(\gamma_{n})$$ are three real sequences in $$(0,1)$$ satisfying $$\alpha_{n}+\beta_{n}+\gamma_{n}=1$$ and $$0< k<\frac {\beta_{n}}{\beta_{n}+\gamma_{n}}$$. Suppose that $$(\alpha_{n})$$ satisfies the conditions:
$$\lim_{n\to\infty}\alpha_{n}=0 , \qquad \sum _{n=1}^{+\infty}\alpha _{n}=+\infty.$$
Then $$(x_{n})$$ converges strongly to $$x^{*}=P_{\operatorname{Fix}(T)}u$$, where $$P_{\operatorname{Fix}(T)}$$ is the metric projection from H onto $$\operatorname{Fix}(T)$$.
### Theorem 1.3
(Marino and Xu [8], 2007, CQ method)
Let C be a closed convex subset of a Hilbert space H. Let $$T:C\rightarrow C$$ be a k-strict pseudo-contraction for some $$0\leq k<1$$ and assume that $$\operatorname{Fix}(T)\neq\emptyset$$. Let $$(x_{n})$$ be the sequence generated by the following (CQ) algorithm:
$$\left \{ \textstyle\begin{array}{l} x_{0}\in C, \\ y_{n}=\alpha_{n}x_{n}+ (1-\alpha_{n} )Tx_{n}, \\ C_{n}= \{z\in C: \Vert y_{n}-z\Vert ^{2}\leq \Vert x_{n}-z\Vert ^{2}+ (1-\alpha_{n} ) (k-\alpha_{n} )\Vert x_{n}-Tx_{n}\Vert ^{2} \}, \\ Q_{n}= \{z\in C: \langle x_{n}-z, x_{0}-x_{n} \rangle \geq 0 \}, \\ x_{n+1}=P_{C_{n}\cap D_{n}}x_{0}. \end{array}\displaystyle \right .$$
Assume that the control sequence $$(\alpha_{n})$$ is chosen so that $$\alpha_{n}<1$$ for all n. Then $$(x_{n})$$ converges strongly to $$P_{\operatorname{Fix}(T)}x_{0}$$.
### Theorem 1.4
(Shang [10], 2007, viscosity method)
Let C be a closed convex subset of a Hilbert space H and let $$T:C\rightarrow C$$ be a k-strict pseudo-contraction with $$\operatorname{Fix}(T)\neq \emptyset$$. Let $$f:C\rightarrow C$$ be a contraction. The initial value $$x_{0}\in C$$ is chosen arbitrarily, and we have sequences $$(\alpha_{n})$$ and $$(\beta_{n})$$ satisfying the following conditions:
1. (1)
$$\lim_{n\to\infty}\alpha_{n}=0$$, $$\sum_{n=1}^{+\infty }\alpha _{n}=+\infty$$;
2. (2)
$$0< a<\beta_{n}<\gamma$$ for some $$a\in(0,\gamma]$$ and $$\gamma =\min \{1,2k \}$$;
3. (3)
$$\sum_{n=1}^{+\infty} \vert \alpha_{n+1}-\alpha_{n} \vert <+\infty$$ and $$\sum_{n=1}^{+\infty} \vert \beta_{n+1}-\beta_{n} \vert <+\infty$$.
Let $$(x_{n})$$ be the composite process defined by
$$\left \{ \textstyle\begin{array}{l} y_{n}= (1-\beta_{n} )x_{n}+\beta_{n}Tx_{n}, \\ x_{n+1}=\alpha_{n}f(x_{n})+ (1-\alpha_{n} )y_{n}. \end{array}\displaystyle \right .$$
Then $$(x_{n})$$ converges strongly to a fixed point $$p\in \operatorname{Fix}(T)$$.
### Theorem 1.5
(Osilike and Udomene [11], 2001, Ishikawa type method)
Let H be a Hilbert space. Let C be a nonempty, closed, and convex subset of H, $$T:C\rightarrow C$$ a demicompact k-strict pseudo-contraction with $$\operatorname{Fix}(T)\neq\emptyset$$. Let $$(\alpha_{n})$$ and $$(\beta_{n})$$ be real sequences in $$[0,1]$$ satisfying the following conditions:
1. (1)
$$0< a<\alpha_{n}\leq b< (1-k ) (1-\beta _{n} )$$, $$\forall n\geq1$$ and for some constants $$a,b\in(0,1)$$;
2. (2)
$$\sum_{n=1}^{+\infty}\beta_{n}<+\infty$$.
Then the sequence $$(x_{n})$$ generated from an arbitrary $$x_{1}\in K$$ by the Ishikawa iteration method
$$\left \{ \textstyle\begin{array}{l} y_{n}= (1-\beta_{n} )x_{n}+\beta_{n}Tx_{n}, \\ x_{n+1}= (1-\alpha_{n} )x_{n}+\alpha_{n}Tx_{n}, \quad n\geq1, \end{array}\displaystyle \right .$$
converges strongly to a fixed point of T.
The mentioned results are probably neither the most general, nor the more recent, but certainly they represent very well some of the different modifications of the original Mann approximation method, made to get strong convergence.
We would like to point out that the differences with the original method are remarkable. So it is quite surprising that recently, in [1] there was obtained a strong convergence method for nonexpansive mappings that is ‘almost’ the Mann method (the difference is given only by a smaller and smaller amount). In [1] was proved the convergence of this method only for nonexpansive mappings.
### Theorem 1.6
(Hussain, Marino et al. [1], 2015)
Let H be a Hilbert space and $$T:H\rightarrow H$$ a nonexpansive mapping. Let $$(\alpha_{n})$$, $$(\mu_{n})$$ be sequences in $$(0,1]$$ such that
• $$\lim_{n\to\infty}\alpha_{n}=0$$;
• $$\sum_{n=1}^{+\infty}\alpha_{n}\mu_{n}=+\infty$$;
• $$\vert \mu_{n+1}-\mu_{n}\vert =o(\mu_{n})$$;
• $$\vert \alpha_{n+1}-\alpha_{n}\vert =o(\alpha_{n}\mu_{n})$$.
Then the sequence $$(x_{n})$$ generated by
$$x_{n+1}=\alpha_{n}x_{n}+ (1-\alpha_{n} )Tx_{n}-\alpha _{n}\mu_{n}x_{n}$$
strongly converges to a point $$x^{*}\in \operatorname{Fix}(T)$$ with minimum norm
$$\bigl\Vert x^{*}\bigr\Vert =\min_{x\in \operatorname{Fix}(T)}\Vert x \Vert .$$
We would like to emphasize that:
1. (1)
In general, the mapping T cannot be defined on a closed convex subset C of H, since $$x_{n+1}$$ is not a convex combination of two elements in C. However, since we can write
$$x_{n+1}=\alpha_{n} (1-\mu_{n} )x_{n}+ (1-\alpha _{n} )Tx_{n},$$
then $$x_{n+1}$$ is meaning full if $$T:C\rightarrow C$$ is a self-mapping defined on a cone C, that is, a particular convex set, closed with respect to linear combinations with positive coefficients.
2. (2)
The proof of Theorem 1.6 is easy using the properties of nonexpansive mappings and cannot be adjusted to the strict pseudo-contractive mappings. The purpose of the present paper is to show that the result is true also for strict pseudo-contractions. The proof uses completely different techniques, as well as the assumptions on coefficients. For all we know, this is the algorithm most similar to the original iterative Mann’s method (and the one most easy to implement), providing strong convergence.
3. (3)
Our techniques can also be used to clarify the proofs of main results in [12] and [13].
## 2 Preliminaries
We need some tools in a real Hilbert space H, and some facts about k-strict pseudo-contractive mappings which are listed in the following auxiliary lemmas.
The first result is very well known and easy to prove.
### Lemma 2.1
Let H be a Hilbert space, then:
1. (i)
$$\|tx+(1-t)y\|^{2}=t\|x\|^{2}+(1-t)\|y\|^{2}-t(1-t)\|x-y\|^{2}$$, for all $$x,y\in H$$ and for all $$t\in[0,1]$$;
2. (ii)
$$\|x+y\|^{2}\leq\|x\|^{2}+2\langle y, x+y \rangle$$, for all $$x,y\in H$$.
A pertinent tool for us is the following well-known lemma of Xu.
### Lemma 2.2
[14]
Let $$(a_{n})_{n\in\mathbb{N}}$$ be a sequence of nonnegative real numbers satisfying the following relation:
$$a_{n+1}\leq(1-\alpha_{n})a_{n}+ \alpha_{n}\sigma_{n}+\gamma_{n}, \quad n\geq0,$$
where:
• $$(\alpha_{n})_{n\in\mathbb{N}}\subset[0,1]$$, $$\sum_{n=1}^{\infty}\alpha_{n}=\infty$$;
• $$\limsup_{n\rightarrow\infty}\sigma_{n}\leq0$$;
• $$\gamma_{n}\geq0$$, $$\sum_{n=1}^{\infty}\gamma _{n}<\infty$$.
Then we have
$$\lim_{n\rightarrow\infty}a_{n}=0.$$
### Lemma 2.3
Let C a nonempty, closed, and convex subspace of H, T a mapping from C into itself such that $$I-T$$ is demiclosed at 0, let $$(y_{n})\subset C$$ be a bounded sequence.
If $$\Vert y_{n}-Ty_{n}\Vert \rightarrow0$$, then
$$\limsup_{n} \langle-\bar{p},y_{n}-\bar{p} \rangle \leq0,$$
where $$\bar{p}=P_{\operatorname{Fix}(T)}(0)$$ is the unique point in $$\operatorname{Fix}(T)$$ that satisfies the variational inequality
$$\langle-\bar{p},x-\bar{p} \rangle\leq0,\quad \forall x\in \operatorname{Fix}(T).$$
(2.1)
### Proof
Let satisfy (2.1). Let $$(y_{n_{k}})$$ be a subsequence of $$(y_{n})$$ for which
$$\limsup_{n} \langle-\bar{p}, y_{n}-\bar{p} \rangle=\lim_{k} \langle-\bar{p}, y_{n_{k}}-\bar{p} \rangle.$$
Select a subsequence $$(y_{n_{k_{j}}})$$ of $$(y_{n_{k}})$$ such that $$y_{n_{k_{j}}}\rightharpoonup v$$ (this is possible by the boundedness of $$(y_{n})$$). By the hypothesis $$\Vert y_{n}-Ty_{n}\Vert \rightarrow 0$$, and by demiclosedness of $$I-T$$, we have $$v\in \operatorname{Fix}(T)$$, and
$$\limsup_{n} \langle-\bar{p}, y_{n}-\bar{p} \rangle=\lim_{j} \langle-\bar{p}, y_{n_{k_{j}}}-\bar{p} \rangle= \langle-\bar{p},v-\bar{p} \rangle.$$
So the claim follows by (2.1). □
Finally, a crucial tool for our results is the following lemma, proved by Maingé.
### Lemma 2.4
[15]
Let $$(\gamma_{n})_{n\in\mathbb{N}}$$ be a sequence of real numbers such that there exists a subsequence $$(\gamma_{n_{j}})_{j\in\mathbb{N}}$$ of $$(\gamma _{n})_{n\in\mathbb{N}}$$ such that $$\gamma_{n_{j}}<\gamma_{n_{j}+1}$$, for all $$j\in\mathbb{N}$$. Then there exists a nondecreasing sequence $$(m_{k})_{k\in\mathbb{N}}$$ of $$\mathbb {N}$$ such that $$\lim_{k\to\infty}m_{k}=\infty$$ and the following properties are satisfied by all (sufficiently large) numbers $$k\in \mathbb{N}$$:
$$\gamma_{m_{k}}\leq\gamma_{m_{k}+1} \quad \textit{and} \quad \gamma _{k}\leq \gamma_{m_{k}+1}.$$
In fact, $$m_{k}$$ is the largest number n in the set $$\{1,\ldots,k\}$$ such that the condition $$\gamma_{n}<\gamma_{n+1}$$ holds.
Before proving our convergence result for strict pseudo-contractions, we recall some properties of these mappings.
### Lemma 2.5
[8]
Assume C is a closed convex subset of a Hilbert space H and $$T:C\rightarrow C$$ be a self-mapping of C. If T is a k-strict pseudo-contraction, then:
1. (1)
T satisfies the Lipschitz condition:
$$\Vert Tx-Ty\Vert \leq\frac{1+k}{1-k}\Vert x-y\Vert ;$$
2. (2)
the mapping $$I-T$$ is demiclosed at 0; that is, if $$(x_{n})$$ is a sequence in C such that $$x_{n}\rightharpoonup\hat{x}$$ and $$(I-T)x_{n}\rightarrow0$$, then $$T\hat{x}=\hat{x}$$;
3. (3)
the set $$\operatorname{Fix}(T)=\{x\in C: Tx=x\}$$ is closed and convex, so that the projection $$P_{\operatorname{Fix}(T)}$$ is well defined.
Moreover, we have the following auxiliary result.
### Lemma 2.6
Let be $$T:C\rightarrow C$$ a k-strict pseudo-contractive self-mapping of a closed and convex subset of a Hilbert space H, and suppose that $$\operatorname{Fix}(T)\neq\emptyset$$; then
$$(1-k )\Vert Tx-x\Vert ^{2}\leq2 \langle x-p,x-Tx \rangle,\quad \forall p\in \operatorname{Fix}(T), \forall x\in C.$$
(2.2)
### Proof
Let $$p\in \operatorname{Fix}(T)$$. Putting $$y=p$$ in the definition of T, we get
$$\Vert Tx-p\Vert ^{2}\leq \Vert x-p\Vert ^{2}+k\Vert x-Tx\Vert ^{2}$$
so
\begin{aligned}& \langle Tx-p,Tx-p \rangle\leq \langle x-p,x-Tx \rangle+ \langle x-p,Tx-p \rangle+k\Vert x-Tx\Vert ^{2} \\& \quad \Rightarrow\quad \langle Tx-p,Tx-x \rangle\leq \langle x-p,x-Tx \rangle+k \Vert x-Tx\Vert ^{2} \\& \quad \Rightarrow\quad \langle Tx-x,Tx-x \rangle+ \langle x-p,Tx-x \rangle\leq \langle x-p,x-Tx \rangle +k\Vert x-Tx\Vert ^{2}, \end{aligned}
from which we get (2.2). □
## 3 The main result
Now we can prove our theorem. We use the notation $$\omega_{l}(x_{n})$$ to denote the set of weak limit points of $$(x_{n})$$.
### Theorem 3.1
Let H be a Hilbert space and let C be a nonempty closed cone of H. Let $$T:C \to C$$ be a k-strict pseudo-contractive mapping such that $$\operatorname{Fix}(T)\neq\emptyset$$. Suppose that $$(\alpha_{n})_{n\in\mathbb {N}}$$ and $$(\mu _{n})_{n\in\mathbb{N}}$$ are real sequences, respectively, in $$(k,1)$$ and in $$(0,1)$$ satisfying the conditions:
1. (1)
$$k<\liminf_{n\rightarrow\infty}\alpha_{n}\leq \limsup_{n\rightarrow\infty}\alpha_{n}<1$$;
2. (2)
$$\lim_{n\rightarrow\infty}\mu_{n}=0$$;
3. (3)
$$\sum_{n=1}^{\infty}\mu_{n}=\infty$$.
Let us define a sequence $$(x_{n})_{n\in\mathbb{N}}$$ as follows:
$$x_{1} \in C, \quad x_{n+1}=\alpha_{n} (1- \mu_{n} ) x_{n}+ (1-\alpha_{n} )Tx_{n},\quad n \in\mathbb{N}.$$
(3.1)
Then $$(x_{n})_{n\in\mathbb{N}}$$ converges strongly to $$\bar{x}\in \operatorname{Fix}(T)$$, that is, the unique solution of the variational inequality
$$\langle-\bar{x},y-\bar{x}\rangle\leq0,\quad \forall y\in \operatorname{Fix}(T).$$
### Proof
We begin by proving that $$(x_{n})_{n\in\mathbb{N}}$$ is bounded.
First of all, observe that from the conditions $$\mu_{n}\rightarrow0$$ and $$k<\liminf\alpha_{n}\leq\limsup\alpha_{n}<1$$, it follows that there exists an integer $$n_{0}\in\mathbb{N}$$ such that
$$\mu_{n}\leq1-\frac{k}{\alpha_{n}},\quad \forall n\geq n_{0},$$
i.e.
$$k-\alpha_{n} (1-\mu_{n} )\leq0.$$
(3.2)
Let be $$p\in \operatorname{Fix}(T)$$ and put $$r=\max \{\Vert x_{n_{0}}-p\Vert , \Vert p\Vert \}$$. We have
\begin{aligned} x_{n+1}-p&=\alpha_{n} \bigl[ (1-\mu_{n} )x_{n}-p \bigr]+ (1-\alpha _{n} ) [Tx_{n}-p ] \\ &= \alpha_{n} \bigl[ (1-\mu_{n} ) (x_{n}-p )+ \mu _{n} (-p ) \bigr]+ (1-\alpha_{n} ) [Tx_{n}-p ]. \end{aligned}
Regarding Lemma 2.1(ii), we derive that
\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} =& \alpha_{n}\bigl\Vert (1-\mu_{n} ) (x_{n}-p )+\mu_{n} (-p ) \bigr\Vert ^{2}+ (1-\alpha _{n} )\Vert Tx_{n}-p\Vert ^{2} \\ &{}-\alpha_{n} (1-\alpha_{n} )\bigl\Vert (1- \mu_{n} )x_{n}-Tx_{n}\bigr\Vert ^{2} \\ \leq&\alpha_{n} \bigl[ (1-\mu_{n} )\Vert x_{n}-p\Vert ^{2}+\mu_{n}\Vert p\Vert ^{2}-\mu_{n} (1-\mu_{n} )\Vert x_{n} \Vert ^{2} \bigr] \\ &{}+ (1-\alpha_{n} ) \bigl[\Vert x_{n}-p\Vert ^{2}+k\Vert x_{n}-Tx_{n}\Vert ^{2} \bigr] \\ &{}-\alpha_{n} (1-\alpha_{n} )\bigl\Vert (1- \mu_{n} ) (x_{n}-Tx_{n} )+\mu_{n} (-Tx_{n} )\bigr\Vert ^{2} \\ =&\alpha_{n} \bigl[ (1-\mu_{n} )\Vert x_{n}-p\Vert ^{2}+\mu _{n}\Vert p\Vert ^{2}-\mu_{n} (1-\mu_{n} )\Vert x_{n} \Vert ^{2} \bigr] \\ &{}+ (1-\alpha_{n} ) \bigl[\Vert x_{n}-p\Vert ^{2}+k\Vert x_{n}-Tx_{n}\Vert ^{2} \bigr] \\ &{}-\alpha_{n} (1-\alpha_{n} ) \bigl[ (1- \mu_{n} )\Vert x_{n}-Tx_{n}\Vert ^{2}+\mu_{n}\Vert Tx_{n}\Vert ^{2}- \mu_{n} (1-\mu _{n} )\Vert x_{n}\Vert ^{2} \bigr] \\ \leq&\alpha_{n} (1-\mu_{n} )\Vert x_{n}-p \Vert ^{2}+\alpha _{n}\mu _{n}\Vert p\Vert ^{2}+ (1-\alpha_{n} )\Vert x_{n}-p\Vert ^{2} \\ &{}+ (1-\alpha_{n} )k\Vert x_{n}-Tx_{n} \Vert ^{2}-\alpha_{n} (1-\alpha_{n} ) (1- \mu_{n} )\Vert x_{n}-Tx_{n}\Vert ^{2} \\ =& (1-\alpha_{n}\mu_{n} )\Vert x_{n}-p \Vert ^{2}+\alpha_{n}\mu _{n}\Vert p\Vert ^{2}+ (1-\alpha_{n} ) \bigl[k-\alpha_{n} (1-\mu _{n} ) \bigr]\Vert x_{n}-Tx_{n}\Vert ^{2} \\ \mbox{(from (3.2))} \leq& (1-\alpha_{n} \mu_{n} )\Vert x_{n}-p\Vert ^{2}+ \alpha_{n}\mu_{n}\Vert p\Vert ^{2} \\ \leq&\max \bigl\{ \Vert x_{n}-p\Vert ^{2},\Vert p \Vert ^{2} \bigr\} \leq \max \bigl\{ \Vert x_{n_{0}}-p\Vert ^{2},\Vert p\Vert ^{2} \bigr\} = r^{2}. \end{aligned}
Thus, we conclude that the sequence $$(x_{n})$$ is bounded.
Now we shall prove that, for $$p\in \operatorname{Fix}(T)$$,
\begin{aligned} (1-\alpha_{n} ) (\alpha_{n}-k )\Vert x_{n}-Tx_{n}\Vert ^{2} \leq& \bigl(\Vert x_{n}-p\Vert ^{2}-\Vert x_{n+1}-p\Vert ^{2} \bigr) \\ &{}-2\alpha_{n}\mu_{n} \langle x_{n},x_{n+1}-p \rangle. \end{aligned}
(3.3)
Regarding (3.1), we easily observe that
\begin{aligned} x_{n+1}-p&=\alpha_{n} (1-\mu_{n} )x_{n}+ (1-\alpha_{n} )Tx_{n}-p \\ &= \bigl[1- \bigl(1-\alpha_{n} (1-\mu_{n} ) \bigr) \bigr]x_{n}+ (1-\alpha_{n} )Tx_{n}-p \\ &= (x_{n}-p )- (1-\alpha_{n} ) (x_{n}-Tx_{n} )-\alpha _{n}\mu_{n} x_{n}, \end{aligned}
and so
\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} \leq&\bigl\Vert (x_{n}-p )- (1-\alpha _{n} ) (x_{n}-Tx_{n} )\bigr\Vert ^{2}-2\alpha_{n}\mu_{n} \langle x_{n},x_{n+1}-p \rangle \\ =&\Vert x_{n}-p\Vert ^{2}-2 (1-\alpha_{n} ) \langle x_{n}-Tx_{n},x_{n}-p \rangle \\ &{}+ (1-\alpha_{n} )^{2}\Vert x_{n}-Tx_{n} \Vert ^{2}-2\alpha_{n}\mu _{n} \langle x_{n},x_{n+1}-p \rangle \\ \mbox{(from (2.2))} \leq&\Vert x_{n}-p\Vert ^{2}+ (1-\alpha _{n} ) (k-\alpha_{n} )\Vert x_{n}-Tx_{n}\Vert ^{2}-2\alpha_{n} \mu _{n} \langle x_{n},x_{n+1}-p \rangle, \end{aligned}
so (3.3) is proved. Moreover, since $$\alpha_{n}\in(k,1)$$,
$$\Vert x_{n+1}-p\Vert ^{2}\leq \Vert x_{n}-p \Vert ^{2}-2\alpha_{n}\mu _{n}\langle x_{n},x_{n+1}-p\rangle.$$
Now we prove the strong convergence of $$(x_{n})$$ concerning two cases:
Case 1. Suppose that $$\Vert x_{n}-p\Vert$$ is monotone nonincreasing. Then $$\Vert x_{n}-p\Vert$$ converges and hence
$$\lim_{n\to\infty} \Vert x_{n+1}-p\Vert ^{2}- \Vert x_{n}-p\Vert ^{2}=0.$$
From this and from the assumptions $$\lim_{n}\mu_{n}=0$$, and $$k<\liminf_{n}\alpha_{n}\leq\limsup_{n}\alpha_{n}<1$$, by (3.3) we get
$$\lim_{n\to\infty} \Vert x_{n}-Tx_{n}\Vert =0;$$
from this and boundedness of $$(x_{n})$$, thanks to demiclosedness of $$I-T$$ we deduce $$\omega_{l}(x_{n})\subseteq \operatorname{Fix}(T)$$.
Now we put
$$z_{n}=\alpha_{n} x_{n}+ (1- \alpha_{n} )Tx_{n}= \bigl(1- (1-\alpha _{n} ) \bigr)x_{n}+ (1-\alpha_{n} )Tx_{n},$$
from which we have
$$z_{n}-x_{n}= (1-\alpha_{n} ) (Tx_{n}-x_{n} ).$$
(3.4)
Hence, we find that
\begin{aligned} x_{n+1}&=z_{n}-\alpha_{n} \mu_{n} x_{n} \\ &= (1-\alpha_{n}\mu_{n} )z_{n}+ \alpha_{n}\mu_{n} (z_{n}-x_{n} ) \\ \mbox{(from (3.4))}&= (1-\alpha_{n} \mu_{n} )z_{n}+\alpha _{n}\mu _{n} (1- \alpha_{n} ) (Tx_{n}-x_{n} ). \end{aligned}
(3.5)
Let $$\bar{x}=P_{\operatorname{Fix}(T)}(0)\in \operatorname{Fix}(T)$$ the unique solution of the variational inequality
$$\langle-\bar{x},y-\bar{x} \rangle\leq0,\quad \forall y\in \operatorname{Fix}(T).$$
(3.6)
From the definition of $$z_{n}$$,
\begin{aligned} \Vert z_{n}-\bar{x}\Vert ^{2}&=\bigl\Vert x_{n}-\bar{x}- (1-\alpha_{n} ) (x_{n}-Tx_{n} )\bigr\Vert ^{2} \\ &=\Vert x_{n}-\bar{x}\Vert ^{2}-2 (1- \alpha_{n} ) \langle x_{n}-Tx_{n},x_{n}- \bar{x} \rangle+ (1-\alpha_{n} )\Vert x_{n}-Tx_{n} \Vert ^{2} \\ \mbox{(from(2.2))} &\leq \Vert x_{n}- \bar{x}\Vert ^{2}- (1-\alpha_{n} ) \bigl[ (1-k )- (1- \alpha_{n} ) \bigr]\Vert x_{n}-Tx_{n}\Vert ^{2} \\ &\leq \Vert x_{n}-\bar{x}\Vert ^{2}. \end{aligned}
(3.7)
So,
\begin{aligned} \Vert x_{n+1}-\bar{x}\Vert ^{2} =&\mbox{(from (3.5))}=\bigl\Vert (1-\alpha_{n} \mu_{n} )z_{n}+\alpha_{n}\mu _{n} (1- \alpha_{n} ) (Tx_{n}-x_{n} )-\bar{x}\bigr\Vert ^{2} \\ =&\bigl\Vert (1-\alpha_{n}\mu_{n} ) (z_{n}-\bar {x} )+\alpha_{n}\mu_{n} \bigl[ (1- \alpha_{n} ) (Tx_{n}-x_{n} )-\bar {x} \bigr]\bigr\Vert ^{2} \\ \mbox{(from Lemma 2.1)} \leq& (1-\alpha_{n}\mu _{n} )^{2}\Vert z_{n}-\bar{x}\Vert ^{2}+2\alpha_{n}\mu_{n} \bigl\langle (1-\alpha _{n} ) (Tx_{n}-x_{n} ),x_{n+1}-\bar{x} \bigr\rangle \\ &{}+2\alpha_{n}\mu_{n} \langle- \bar{x},x_{n+1}-\bar {x} \rangle \\ \mbox{(from (3.7))} \leq& (1- \alpha_{n}\mu _{n} )\Vert x_{n}-\bar{x}\Vert ^{2} \\ &{}+2\alpha_{n}\mu_{n} \bigl( (1-\alpha_{n} ) \langle Tx_{n}-x_{n},x_{n+1}-\bar{x} \rangle+ \langle-\bar {x},x_{n+1}-\bar {x} \rangle \bigr). \end{aligned}
(3.8)
Now, since $$(x_{n})$$ is bounded and $$\omega_{l}(x_{n})\subseteq \operatorname{Fix}(T)$$, there exists an appropriate subsequence $$x_{n_{k}}\rightharpoonup p_{0}\in \operatorname{Fix}(T)$$ such that
$$\limsup_{n} \langle-\bar{x},x_{n+1}-\bar{x} \rangle =\lim_{k} \langle-\bar{x},x_{n_{k}}-\bar{x} \rangle= \langle -\bar{x},p_{0}-\bar{x} \rangle\leq0.$$
(3.9)
From this, it follows that all the hypotheses of Lemma 2.2 are satisfied and finally by (3.8) we can conclude
$$\lim_{n\to\infty} \Vert x_{n}-\bar{x}\Vert =0.$$
Let now $$\bar{x}\in \operatorname{Fix}(T)$$ be defined by the variational inequality (3.6).
Case 2. If $$\Vert x_{n}-\bar{x}\Vert$$ does not be monotone nonincreasing, there exists a subsequence $$(x_{n_{k}})$$ such that $$\Vert x_{n_{k}}-\bar{x}\Vert <\Vert x_{n_{k}+1}-\bar{x}\Vert$$, $$\forall k\in\mathbb{N}$$. So by Lemma 2.4, $$\exists\tau(n)\uparrow +\infty$$ such that
1. (1)
$$\Vert x_{\tau(n)}-\bar{x}\Vert <\Vert x_{\tau (n)+1}-\bar {x}\Vert$$;
2. (2)
$$\Vert x_{n}-\bar{x}\Vert <\Vert x_{\tau(n)+1}-\bar {x}\Vert$$.
Now, we have
\begin{aligned} 0&\leq\liminf_{n} \bigl(\Vert x_{\tau(n)+1}-\bar{x} \Vert - \Vert x_{\tau (n)}-\bar{x}\Vert \bigr) \\ &\leq\limsup_{n} \bigl(\Vert x_{\tau(n)+1}-\bar{x} \Vert -\Vert x_{\tau (n)}-\bar{x}\Vert \bigr) \\ &\leq\limsup_{n} \bigl(\Vert x_{n+1}-\bar{x} \Vert -\Vert x_{n}-\bar {x}\Vert \bigr) \\ &\leq\limsup_{n} \bigl(\Vert x_{n}-\bar{x} \Vert +\sqrt{\mu _{n}}M-\Vert x_{n}-\bar{x}\Vert \bigr)=0. \end{aligned}
Thus, we derive that
$$\Vert x_{\tau(n)+1}-\bar{x}\Vert ^{2}-\Vert x_{\tau(n)}- \bar {x}\Vert ^{2}\longrightarrow0,$$
from which
$$\Vert x_{\tau(n)}-Tx_{\tau(n)}\Vert \longrightarrow0.$$
(3.10)
Now, from (3.8), we get
\begin{aligned} \Vert x_{\tau(n)+1}-\bar{x}\Vert ^{2} \leq& (1-\alpha _{\tau(n)}\mu_{\tau(n)} )\Vert x_{\tau(n)}-\bar{x}\Vert ^{2} \\ &{}+2\alpha_{\tau(n)}\mu_{\tau(n)} (1-\alpha_{\tau (n)} ) \langle Tx_{\tau(n)}-x_{\tau(n)},x_{\tau(n)+1}-\bar{x} \rangle \\ &{}+2\alpha_{\tau(n)}\mu_{\tau(n)} \langle-\bar {x},x_{\tau (n)+1}-\bar{x} \rangle \\ =&\Vert x_{\tau(n)}-\bar{x}\Vert ^{2}+2 \alpha_{\tau (n)}\mu _{\tau(n)} (1-\alpha_{\tau(n)} ) \langle Tx_{\tau (n)}-x_{\tau(n)},x_{\tau(n)+1}-\bar{x} \rangle \\ &{}+2\alpha_{\tau(n)}\mu_{\tau(n)} \langle-\bar {x},x_{\tau (n)+1}-\bar{x} \rangle \\ &{}-2\alpha_{\tau(n)}\mu_{\tau(n)} \biggl(\frac{\Vert x_{\tau (n)}-\bar {x}\Vert ^{2}}{2} \biggr). \end{aligned}
(3.11)
Putting in (3.11)
\begin{aligned} A_{\tau(n)} =& (1-\alpha_{\tau(n)} ) \langle Tx_{\tau (n)}-x_{\tau(n)},x_{\tau(n)+1}- \bar{x} \rangle \\ &{}+ \langle-\bar{x},x_{\tau(n)+1}-\bar{x} \rangle-\frac {\Vert x_{\tau(n)}-\bar{x}\Vert ^{2}}{2}, \end{aligned}
we have
$$\Vert x_{\tau(n)+1}-\bar{x}\Vert ^{2}\leq \Vert x_{\tau (n)}-\bar {x}\Vert ^{2}+2\alpha_{\tau(n)} \mu_{\tau(n)}A_{\tau(n)}.$$
(3.12)
Notice that we cannot use Lemma 2.2 as in Case 1 (or in [12, 13]) since we could not guarantee that $$\sum_{n=1}^{+\infty}\mu _{\tau(n)}=+\infty$$. So, we proceed as follows. Assume by contradiction that $$\Vert x_{\tau(n)}-\bar{x}\Vert$$ does not converge to 0. Then there exist $$(n_{j})$$ and an $$\epsilon>0$$ such that
$$\Vert x_{\tau(n_{j})}-\bar{x}\Vert \geq2\epsilon.$$
(3.13)
By (3.9) and (3.10) we know that there exist $$n_{0},n_{1}\in \mathbb{N}$$ such that
$$(1-\alpha_{\tau(n)} ) \langle Tx_{\tau(n)}-x_{\tau (n)},x_{\tau(n)+1}- \bar{x} \rangle< \frac{\epsilon}{3},\quad \forall n\geq n_{0}$$
(3.14)
and
$$\langle-\bar{x},x_{\tau(n)+1}-\bar{x} \rangle< \frac {\epsilon }{3},\quad \forall n\geq n_{1} .$$
(3.15)
Hence, if we take $$n_{j_{0}}\geq\max \{n_{0},n_{1} \}$$ one obtains by the definition of $$A_{\tau(n)}$$,
$$A_{\tau(n)}< \frac{\epsilon}{3}+\frac{\epsilon}{3}-\epsilon=- \frac {\epsilon}{3}< 0,\quad \forall n\geq n_{j_{0}}.$$
So, by (3.12) we have $$\Vert x_{\tau(n)+1}-\bar{x}\Vert ^{2}\leq \Vert x_{\tau(n)}-\bar{x}\Vert ^{2}$$, which contradicts $$\Vert x_{\tau(n)}-\bar{x}\Vert <\Vert x_{\tau(n)+1}-\bar{x}\Vert$$, ∀n. This implies that
$$\Vert x_{\tau(n)}-\bar{x}\Vert \longrightarrow0,$$
and so, using $$\Vert x_{n}-\bar{x}\Vert <\Vert x_{\tau (n)+1}-\bar {x}\Vert$$, we finally obtain
$$\Vert x_{n}-\bar{x}\Vert \longrightarrow0.$$
□
### Example 3.2
The mapping $$T:\mathbb{R}\rightarrow\mathbb{R}$$ defined by $$Tx=-2x$$ is $$\frac {1}{3}$$-strict pseudo-contractive. Taking $$\alpha_{n}=\frac{1}{2}$$, $$\mu _{n}=\frac{1}{n}$$, our algorithm becomes
$$x_{n+1}=-\frac{1}{2}\frac{n+1}{n}x_{n} ,$$
which goes to $$0=\operatorname{Fix}(T)$$ swinging around it.
### Open questions
1. (1)
Does the result hold in Banach spaces?
2. (2)
Does the result hold for families of strict pseudo-contractive mappings?
3. (3)
Does the result hold for Lipschitzian pseudo-contractive mappings? | 10,468 | 26,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.814952 |
https://forum.philosophynow.org/viewtopic.php?f=26&t=26198&start=15 | 1,600,866,536,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210996.32/warc/CC-MAIN-20200923113029-20200923143029-00401.warc.gz | 367,134,088 | 12,793 | ## What the theorems of Incompleteness or Undecidability assert...
What is the basis for reason? And mathematics?
Moderators: AMod, iMod
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm
### Re: What the theorems of Incompleteness or Undecidability assert...
Scott Mayers wrote: Wed Feb 27, 2019 1:10 am I have to say I like your background even if we might not be precisely in agreement.
This actually proves something about reality itself and similar to what I used to demonstrate HOW you can begin with absolutely nothing in totality to derive something. That is, you think that the computer program is presenting a contradiction when it is actually the underlying logic of reality itself (as a whole).
information It's everywhere. 1s and 0s. Trues and Falses.
Scott Mayers wrote: Wed Feb 27, 2019 1:10 am 1,0
0 == (0 and 1)
but
1 != (0 and 1)
...which implies that 1 can only equal itself.... an Identity.
I can't decide if Aristotle made an error of if he was a genius.
Scott Mayers wrote: Wed Feb 27, 2019 1:10 am Assigning 0 to (A == A) is equivalent to assigning 0 == 1 in a way that one might say, "From nothing, we can get something."
See! Inversion!
Eventually you start questioning those damn "laws" and you begin to contradict THEM out of our own mind.
And you emerge on the other side with intuitionistic logic.
The laws are like training wheels until you can learn to think
And then you can take off the training wheels yourself.
I am going to finish responding to the rest of your post in the morning. It's 2am. My brain scanned over it and said "cannot compute".
Last edited by Logik on Wed Feb 27, 2019 1:27 am, edited 1 time in total.
surreptitious57
Posts: 4217
Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Logic wrote:
You cannot prove ANYTHING about reality with logic . NOTHING . NADA . ZILCH . ZIP
The concept of proof is valid only within the context of logic NOT in the context of using logic FOR THE PURPOSES OF reasoning about reality
Logic is fundamentally the systematic study of argument form and inference and paradox / contradiction
It is an attempt to make sense of assumptions about the world that are not based upon direct observation
By contrast reason is the application of logic with reference to understanding the world as it appears to be
Reason is therefore a sub set of logic within the hierarchy of epistemology
Proof is not only valid within the context of logic but also empiricism
For example Einsteins Theory Of General Relativity disproved / falsified Newtons Theory Of Universal Gravitation
This was demonstrated by the orbit of Mercury which had a 43 arc second discrepancy that only GR could explain
Disproof / falsification of a hypothesis is the best way to obtain knowledge about the observable world
Knowledge that would be unreliable if it could not be empirically demonstrated since it could be false
The word science is derived from the Latin root scientia which literally means knowledge / to know
You cannot know anything about natural phenomena just by assuming it since that is not knowledge
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm
### Re: What the theorems of Incompleteness or Undecidability assert...
surreptitious57 wrote: Wed Feb 27, 2019 1:27 am
Logic wrote:
You cannot prove ANYTHING about reality with logic . NOTHING . NADA . ZILCH . ZIP
The concept of proof is valid only within the context of logic NOT in the context of using logic FOR THE PURPOSES OF reasoning about reality
Logic is fundamentally the systematic study of argument form and inference and paradox / contradiction
It is an attempt to make sense of assumptions about the world that are not based upon direct observation
By contrast reason is the application of logic with reference to understanding the world as it appears to be
Reason is therefore a sub set of logic within the hierarchy of epistemology
Proof is not only valid within the context of logic but also empiricism
For example Einsteins Theory Of General Relativity disproved / falsified Newtons Theory Of Universal Gravitation
This was demonstrated by the orbit of Mercury which had a 43 arc second discrepancy that only GR could explain
Disproof / falsification of a hypothesis is the best way to obtain knowledge about the observable world
Knowledge that would be unreliable if it could not be empirically demonstrated since it could be false
The word science is derived from the Latin root scientia which literally means knowledge / to know
You cannot know anything about natural phenomena just by assuming it since that is not knowledge
Logic is the study of abstract structure/patterns. Logic is metaphysics. Logic/Mathematics/Computer science/Physics are all merged into one.
As of a few days ago I am 100% convinced that Lambda calculus is the ultimate metaphysic ( https://en.wikipedia.org/wiki/Metalanguage )
Once you have acquired knowledge you express it in language/logic/mathematics
surreptitious57
Posts: 4217
Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Logic wrote:
Logic / Mathematics / Computer science / Physics are all merged into one
Logic is the foundation of both philosophy and mathematics but they are also separate disciplines
Computer science is a branch of both mathematics and science which are also separate disciplines
Disciplines are also separated by time : physics came first [ Big Bang ] then chemistry then biology
Also logic like science is not a simple discipline but many variations within the general definition :
Classical / Temporal / Fuzzy / Modal / Categorical / Constructive / Ordinal / Doxastic / Infinitary
surreptitious57
Posts: 4217
Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Logic wrote:
Once you have acquired knowledge you express it in language / logic / mathematics
Yes you can but you still need a reliable methodology that can demonstrate that what you actually have is knowledge
If your observations about the world are not rigorous enough then language / logic / mathematics will also be no use
This is why from a scientific perspective empiricism is the most important thing of all
The map is very important but not as much as the thing being studied : the territory
Scott Mayers
Posts: 1692
Joined: Wed Jul 08, 2015 1:53 am
### Re: What the theorems of Incompleteness or Undecidability assert...
surreptitious57 wrote: Wed Feb 27, 2019 2:08 am
Logic wrote:
Logic / Mathematics / Computer science / Physics are all merged into one
Logic is the foundation of both philosophy and mathematics but they are also separate disciplines
Computer science is a branch of both mathematics and science which are also separate disciplines
Disciplines are also separated by time : physics came first [ Big Bang ] then chemistry then biology
Also logic like science is not a simple discipline but many variations within the general definition :
Classical / Temporal / Fuzzy / Modal / Categorical / Constructive / Ordinal / Doxastic / Infinitary
Welcome to the discussion surreptitious57. I gather that logik shares much the same views but we all come from distinct approaches. Does your own background reach the classics of Plato and Aristotle?
And do you also agree to the motivational problems from the first post? (or agree to these as what you might have asked yourself on these independent of reading by other philosophers?
Scott Mayers
Posts: 1692
Joined: Wed Jul 08, 2015 1:53 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Logik wrote: Wed Feb 27, 2019 1:26 am
Scott Mayers wrote: Wed Feb 27, 2019 1:10 am I have to say I like your background even if we might not be precisely in agreement.
This actually proves something about reality itself and similar to what I used to demonstrate HOW you can begin with absolutely nothing in totality to derive something. That is, you think that the computer program is presenting a contradiction when it is actually the underlying logic of reality itself (as a whole).
information It's everywhere. 1s and 0s. Trues and Falses.
Scott Mayers wrote: Wed Feb 27, 2019 1:10 am 1,0
0 == (0 and 1)
but
1 != (0 and 1)
...which implies that 1 can only equal itself.... an Identity.
I can't decide if Aristotle made an error of if he was a genius.
Although these questions were likely long before Aristotle, his own contribution was more specifically about a mix of quantification, static propositions or nouns and a predicate of existence. So it was not as expansive as using 'classics' of the 1800s with George Boole's take. His take on it was incredible for its day though and it covers enough of most everyday conversations.
I believe from the context of the Judaeo-Christian Bibles and related books that they probably though of all of this but labelled the unknown variable sources as having an extension of human-like qualities, like gods. This is too much of a digression but I think that the ancient ancients had already thought of these questions but we get them now as 'myths' about literal 'gods' rather than variables unknown. [we don't give them much the credit they likely deserved. I'm guessing that the dark ages were about the fear of this very problem's impact upon the general population.
Scott Mayers wrote: Wed Feb 27, 2019 1:10 am Assigning 0 to (A == A) is equivalent to assigning 0 == 1 in a way that one might say, "From nothing, we can get something."
See! Inversion!
Eventually you start questioning those damn "laws" and you begin to contradict THEM out of our own mind.
And you emerge on the other side with intuitionistic logic.
The laws are like training wheels until you can learn to think
And then you can take off the training wheels yourself.
I am going to finish responding to the rest of your post in the morning. It's 2am. My brain scanned over it and said "cannot compute".
So you are returning to the nothing? We'll have to wait until you become one again.
surreptitious57
Posts: 4217
Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Scott Mayers wrote:
( I ) Can we find some universal logic that can cover the full range of all specific logical systems ?
( 2 ) Are we permitted to use any logic especially some possible universal one with initial simple premises
( including possibly none ) to prove all of reality beyond mere abstraction such as scientific truths ?
The notion of a universal logic is flawed because universal knowledge applied to any discipline is simply not possible
Knowledge by definition has to be a posteriori but science is inductive so its knowledge base can never be complete
No logical system outside of science could absolutely prove it for this very reason but could do so if it was deductive
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm
### Re: What the theorems of Incompleteness or Undecidability assert...
surreptitious57 wrote: Wed Feb 27, 2019 2:08 am Logic is the foundation of both philosophy and mathematics but they are also separate disciplines
Computer science is a branch of both mathematics and science which are also separate disciplines
I will accept your premises as valid for now, simply so that I can help you unify these concepts into one.
What is the foundation of logic?
Grammar and semantics. The rules for forming coherent, consistent and meaningful thoughts.
These coherent thoughts are then expressed through language.
Every unique set of "rules for forming coherent, consistent and meaningful thoughts" a new language/logic makes.
Since you can have a near-infinite set of rules. And a near-infinite set of rule permutations - you can have near-infinite languages (this is what Postmodern philosophers have been on about. This is what Derida's means by "il n'y a pas de hors-texte" .
It alludes to the fact that text, symbols, squiggles and characters contain absolutely no meaning without knowing the grammar, semantics and most importantly - the INTENTION of the person who expressed any particular thought into language/logic/mathematics.
In the simplest term possible: in a universe where you are free to invent an near-infinite number of languages - what are the odds that any two people speak the EXACT same language? But I am getting a little off-topic.
Back to near-infinite number of languages. How do we enumerate ALL of them?
https://en.wikipedia.org/wiki/Chomsky_h ... 0_grammars
Type-0 grammars include all formal grammars. They generate exactly all languages that can be recognized by a Turing machine.
Turing machines are the foundational concept of Computer Science.
Now that we have unpacked all the things to the bottom. We have established these two truths:
Turing machines are computers.
Grammar is foundational of language.
Now follow my finger up to the surface.
Turing machines can enumerate ALL Type-0 formal grammars: logic, mathematics, programming languages!
You brain is a Turing machine.
Understanding how your mind works is computer science. Once you understand your mind. All languages can be unified.
surreptitious57
Posts: 4217
Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
I agree that your brain is a Turing machine however it is not exclusively so only sometimes
We think emotionally as well as logically unlike actual machines which only think logically
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm
### Re: What the theorems of Incompleteness or Undecidability assert...
surreptitious57 wrote: Wed Feb 27, 2019 9:38 am I agree that your brain is a Turing machine however it is not exclusively so only sometimes
We think emotionally as well as logically unlike actual machines which only think logically
What is emotion? Is it not information?
It's a feedback loop from your body to CPU that something requires attention.
https://en.wikipedia.org/wiki/Interrupt
Our brains are a bunch of systems working together as a team. Sometimes even a janitors requires the full, undivided attention of the CEO.
Feed back loops
https://en.wikipedia.org/wiki/Feedback
surreptitious57
Posts: 4217
Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Emotion is information but information is not always knowledge so will not always make sense
There is in actual fact no requirement for ANY emotion to be rigorous / consistent like logic is
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm
### Re: What the theorems of Incompleteness or Undecidability assert...
surreptitious57 wrote: Wed Feb 27, 2019 10:58 am Emotion is information but information is not always knowledge so will not always make sense
There is in actual fact no requirement for ANY emotion to be rigorous / consistent like logic is
So? Know thyself
The colloquial term for this is EQ - emotional intelligence. It's a scientific process like any other - it starts with observation first.
We, men are somewhat disadvantaged in this department unlike women. It comes much more naturally and instinctive to them than it comes to us. Because their emotions ring louder bells in their brains than in ours. So you know - they are forced to listen. Men just go "shut up! I am busy now."
Introspection goes a long way to addressing this, but if you want to set yourself up for success then nothing brings about compound interest in returns like having a loving and emotionally intelligent woman in your life who can set you straight whey you fall astray.
(DON'T tell my fiance I said that - I am STILL pretending she is not in control)
surreptitious57
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Joined: Fri Oct 25, 2013 6:09 am
### Re: What the theorems of Incompleteness or Undecidability assert...
Peace of mind for me is being as detached as possible both physically and psychologically
Having a woman around me would disturb this balance and so I prefer to be alone instead
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm
### Re: What the theorems of Incompleteness or Undecidability assert...
surreptitious57 wrote: Wed Feb 27, 2019 12:12 pm Peace of mind for me is being as detached as possible both physically and psychologically
Having a woman around me would disturb this balance and so I prefer to be alone instead
Who are you trying to convince? Me or yourself
"physical and psychological detachment" is an overly-sophisticated way of describing death. | 3,818 | 16,494 | {"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} | 3.296875 | 3 | CC-MAIN-2020-40 | latest | en | 0.931082 |
http://www.drumtom.com/q/find-a-cubic-function-f-x-ax-3-bx-2-cx-d-that-has-a-local-maximum-value-of-60-at-x-1-and-a-local-minimum-value-of-4-at-x-3 | 1,481,222,322,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00123-ip-10-31-129-80.ec2.internal.warc.gz | 442,569,042 | 7,942 | # Find a cubic function f(x)=ax^3+bx^2+cx+d that has a local maximum value of 60 at x=1 and a local minimum value of -4 at x=3?
• Find a cubic function f(x)=ax^3+bx^2+cx+d that has a local maximum value of 60 at x=1 and a local minimum value of -4 at x=3?
Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 4 at x = 2 and a local minimum value of 0 at x = 1
Positive: 53 %
Answer to Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 4 at x = -3 and a ... x = -3 and a local minimum value of 0 ...
Positive: 50 %
### More resources
Answer to Find a cubic function f(x)=ax^3-bx^2+cx-d that has a local max value of 60 ... =ax^3-bx^2+cx-d that has a local max value of 60 at 1 and local ...
Positive: 53 %
Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at x = −3 and a local ... 3 at x = −3 and a local minimum value ...
Positive: 48 %
F X Ax 3 Bx 2 Cx D Docs. ... and Models A quadratic function in general form is f(x) ... Cubic Functions A cubic..... By Activity 2.3, if f (x) ax3 bx2 ... | 367 | 1,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-50 | longest | en | 0.725603 |
https://www.formulaconversion.com/formulaconversioncalculator.php?convert=imperialgallon_to_cupsmetric | 1,621,297,518,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00518.warc.gz | 809,007,590 | 16,366 | # Imperial gallon to cups (metric) (imperial gallon to c) Metric conversion calculator
Welcome to our imperial gallon to cups (metric) (imperial gallon to c) conversion calculator. You can enter a value in either the imperial gallon or cups (metric) input fields. For an understanding of the conversion process, we include step by step and direct conversion formulas. If you'd like to perform a different conversion, just select between the listed Volume units in the 'Select between other Volume units' tab below or use the search bar above. Tip: Use the swap button to switch from converting imperial gallon to cups (metric) to cups (metric) to imperial gallon.
## cups (metric) (c)
(not bookmarks)
Swap
< == >
1 imperial gallon = 18.18436 c 1 c = 0.05499231 imperial gallon
Algebraic Steps / Dimensional Analysis Formula
imperialgallon * 1 c0.055 imperialgallon = c
cubic feet 0 cubic meters 0 cubic millimeters 0 cubic yards 0 cups 0 fluid ounces 0 gallons (liquid) 0 liters 0 milliliters 0
If you would like to switch between Volume units, select from the tables below
bushels centiliters cubic centimeters cubic decimeters cubic dekameters cubic feet cubic gigameters cubic hectometers cubic inches cubic kilometers cubic megameters cubic meters cubic micrometers cubic miles cubic millimeters cubic nanometers cubic yards cups cups (metric) deciliters dekaliters fluid ounces gallons (dry) gallons (liquid) gigaliters gills hectoliters imperial gallon kiloliters liters megaliters microliters milliliters nanoliters pecks pints (dry) pints (liquid) quarts (dry) quarts (liquid) tablespoons teaspoons < == > bushels centiliters cubic centimeters cubic decimeters cubic dekameters cubic feet cubic gigameters cubic hectometers cubic inches cubic kilometers cubic megameters cubic meters cubic micrometers cubic miles cubic millimeters cubic nanometers cubic yards cups cups (metric) deciliters dekaliters fluid ounces gallons (dry) gallons (liquid) gigaliters gills hectoliters imperial gallon kiloliters liters megaliters microliters milliliters nanoliters pecks pints (dry) pints (liquid) quarts (dry) quarts (liquid) tablespoons teaspoons
Active Users | 509 | 2,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-21 | longest | en | 0.573377 |
http://www.mathworks.com/help/images/remove-noise-from-color-image-using-pretrained-neural-network.html?requestedDomain=www.mathworks.com&nocookie=true | 1,508,359,488,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823114.39/warc/CC-MAIN-20171018195607-20171018215607-00795.warc.gz | 503,656,247 | 13,358 | # Documentation
### This is machine translation
Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.
## Remove Noise from Color Image Using Pretrained Neural Network
This example shows how to remove Gaussian noise from an RGB image. Convert the noisy image to the L*a*b* color space, and remove noise on the luminance channel L* by using a pretrained denoising neural network.
In practice, image color channels frequently have correlated noise. You will have better denoising results if you train a denoising network on color images. For more information, see Train and Apply Denoising Neural Networks.
Read a color image into the workspace and convert the data to type `double`. Display the image.
```RGB = imread('lighthouse.png'); RGB = im2double(RGB); figure imshow(RGB) title('Pristine Image')```
Add Gaussian noise with a variance of 0.01 to the image. `imnoise` adds noise to each color channel independently. Display the noisy image.
```noisyRGB = imnoise(RGB,'gaussian',0,0.01); figure imshow(noisyRGB) title('Noisy Image')```
Load the pretrained denoising neural network.
`net = denoisingNetwork('dncnn');`
Convert the image to the L*a*b* color space.
`LAB = rgb2lab(RGB);`
Noise is primarily in the luminance channel. Remove the noise from this channel only, by using the pretrained denoising neural network.
`LAB(:,:,1) = denoiseImage(LAB(:,:,1),net);`
Convert the image back to the RGB color space.
`denoisedRGB = lab2rgb(LAB);`
Display the denoised image.
```figure imshow(denoisedRGB) title('Denoised Image')```
Calculate the peak signal-to-noise ratio (PSNR) for the noisy and denoised images. A larger PSNR indicates that noise has a smaller relative signal, and is associated with higher image quality.
```noisyPSNR = psnr(noisyRGB,RGB); fprintf('\n The PSNR value of the noisy image is %0.4f.',noisyPSNR);```
``` The PSNR value of the noisy image is 20.6395. ```
```denoisedPSNR = psnr(denoisedRGB,RGB); fprintf('\n The PSNR value of the denoised image is %0.4f.',denoisedPSNR);```
``` The PSNR value of the denoised image is 53.7825. ```
Calculate the structural similarity (SSIM) index for the noisy and denoised images. An SSIM index close to 1 indicates good agreement with the reference image, and higher image quality.
```noisySSIM = ssim(noisyRGB,RGB); fprintf('\n The SSIM value of the noisy image is %0.4f.',noisySSIM);```
``` The SSIM value of the noisy image is 0.7393. ```
```denoisedSSIM = ssim(denoisedRGB,RGB); fprintf('\n The SSIM value of the denoised image is %0.4f.',denoisedSSIM);```
``` The SSIM value of the denoised image is 0.9999. ``` | 709 | 2,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-43 | latest | en | 0.744276 |
http://www.slideserve.com/imani-powers/week-4 | 1,495,970,876,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609613.73/warc/CC-MAIN-20170528101305-20170528121305-00337.warc.gz | 817,722,120 | 18,499 | 1 / 19
# Week 4 - PowerPoint PPT Presentation
Week 4. Vocabulary Review. cent( i )/ hect (o)(a ) – one hundred. centimeter centipede hectograph. dec ( im )/ dec (a) - ten. decimal decade decagon. du(o)/bi - two. bicycle bifocal duet. mill(e)/kilo. one thousand millipede millimeter kilogram. non(a) - nine. nonagon
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Week 4
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
## Week 4
Vocabulary Review
• centimeter
• centipede
• hectograph
• decimal
• decagon
• bicycle
• bifocal
• duet
• one thousand
• millipede
• millimeter
• kilogram
• nonagon
• nonagenarian
### octo/octa - eight
• octagon
• octane
• octopus
• quartet
• tetrachloride
• quintuplets
• pentagon
• pentameter
• semicircle
• demigod
• hemisphere
• September
• heptagon
• septuplets
• tricycle
• trilogy
• triangle
### uni/mono - one
• unicycle
• mononucleosis
• unicorn
## Vocab Words
Explanations & Examples
• The teacher’s students were a little more candid about the fact that his sense of humor was lame than he would have wished.
• Other forms
candidness (n)
• Similar to (synonyms)
blunt
forthright
• Not like (antonyms)
deceitful
tricky
### discern (v.)-To distinguish one thing from another
• Discerning minds can tell the difference between Star Wars and Star Trek.
• Other forms
discerner(n)
• Similar to (synonyms)
differentiate
distinguish
• Not like (antonyms)
neglect
overlook
### disdain (n.) - contempt, intense dislike
• The mouse’s unusual disdain for cheddar undoubtably saved his life.
• Other forms
• Similar to (synonyms)
antipathy
aversion
• Not like (antonyms)
respect
### eccentric (adj.) -odd, unusual, quirky
• The man was a little eccentric and only wore polka dot clothing.
• Other forms
eccentricity (n)
• Similar to (synonyms)
anomalous
unconventional
• Not like (antonyms)
standard
ordinary
### subtle (adj.) -not obvious, hard to spot
• The employee was not subtle about the fact that he was bored with his job.
• Other forms
subtlety (n)
subtleness (n)
• Similar to (synonyms)
inconspicuous
faint
• Not like (antonyms)
harsh
noisy | 745 | 2,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-22 | longest | en | 0.777349 |
https://www.statistics-lab.com/category/math6370-algebraic-number-theory/ | 1,716,812,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00153.warc.gz | 885,318,809 | 30,726 | ## MATH6370 Algebraic number theory课程简介
This course is a basic introduction to algebraic number theory. A core topic deals with the ideal theory of Dedekind domains as applied to the rings of integers of number fields (finite extensions of $Q$ ). $A$ major purpose of the theory is to overcome the lack of unique factorisation into primes in these rings.
The course will also cover the fundamental finiteness theorems: the finiteness of the ideal class group (via Minkowski’s geometric theory of numbers), and the structure (finite generation, determination of the rank etc.) of the unit group. Additional topics which will be discussed if time permits: law of quadratic reciprocity, elementary Diophantine equations, completions (p-adic numbers), zeta-functions, distribution of primes in arithmetic progressions.
## PREREQUISITES
The lack of unique factorization into primes in rings of integers of number fields is indeed a significant obstacle, and the development of the ideal theory of Dedekind domains provides a powerful tool for overcoming it. The fact that the ideal class group is finite is a crucial result in this context, and Minkowski’s geometric theory of numbers provides a beautiful geometric interpretation of this fact.
The structure of the unit group is another important topic, and it is closely related to the ideal class group. The determination of the rank of the unit group is a difficult problem in general, but there are many interesting results in special cases.
The law of quadratic reciprocity is one of the most famous results in number theory, and its proof involves many of the techniques and ideas from algebraic number theory. The distribution of primes in arithmetic progressions is another fascinating topic, and it has important applications in cryptography and other areas.
## MATH6370 Algebraic number theory HELP（EXAM HELP， ONLINE TUTOR）
(1) Prove that $\sin \left(1^{\circ}\right)$ is algebraic. (One approach is to begin by showing that the coefficient of $x^j$ in $g_m(x)$, as given in the lemma, is 0 if $j \not \equiv m(\bmod 2)$. There are easier approaches, but I won’t give hints for them.)
To prove that $\sin \left(1^{\circ}\right)$ is algebraic, we will use the following lemma:
Lemma: Let $m$ be a positive integer, and let $g_m(x)=\sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{x^{m-2k}}{(m-2k)!}$. Then, the coefficient of $x^j$ in $g_m(x)$ is 0 if $j \not \equiv m(\bmod 2)$.
Proof of lemma: We can write $g_m(x)$ as $\sum_{k=0}^{\infty} (-1)^k \frac{x^{m-2k}}{(m-2k)!} – \sum_{k=\lfloor (m-1)/2 \rfloor + 1}^{\infty} (-1)^k \frac{x^{m-2k}}{(m-2k)!}$. The first sum is the power series for $\cos x$, and the second sum is the power series for $\sin x$ multiplied by $x$. Therefore, the coefficient of $x^j$ in $g_m(x)$ is given by $\frac{1}{j!} \left(\frac{d^j}{dx^j} g_m(x)\right){x=0} = \frac{1}{j!} \left(\frac{d^j}{dx^j} \cos x – x\frac{d^j}{dx^j} \sin x\right){x=0}$. This expression is 0 if $j \not \equiv m(\bmod 2)$, since both $\cos x$ and $\sin x$ are periodic with period $2\pi$.
Now, let’s return to the main problem of proving that $\sin \left(1^{\circ}\right)$ is algebraic. We can write $\sin \left(1^{\circ}\right) = \sin \left(\frac{\pi}{180}\right)$. Using the Taylor series expansion for $\sin x$, we have:
$$\sin x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$
Substituting $x=\frac{\pi}{180}$, we get:
$$\sin \left(\frac{\pi}{180}\right) = \sum_{k=0}^{\infty} (-1)^k \frac{\left(\frac{\pi}{180}\right)^{2k+1}}{(2k+1)!}$$
Let $m=2k+1$. Then, the coefficient of $\left(\frac{\pi}{180}\right)^m$ in the above expression is $(-1)^k \frac{1}{m!}$. By the lemma, this coefficient is 0 if $m \not \equiv 1(\bmod 2)$, i.e., if $k$ is even. Therefore, the only non-zero coefficients are for odd values of $m$, which correspond to even values of $k$. In particular, the coefficient of $\left(\frac{\pi}{180}\right)^{2n+1}$ is algebraic for any non-negative integer $n$.
Therefore, $\sin \left(1^{\circ}\right)$ is a sum of algebraic numbers, and hence algebraic itself. This completes the proof.
(2) (a) Using Theorem 1 , determine the values of $\theta \in \mathbb{Q} \cap[0,2)$ for which $\sin ^2(\pi \theta) \in \mathbb{Q}$.
(b) Using Theorem 1 , determine the values of $\theta \in \mathbb{Q} \cap[0,2)$ for which $\sin (\pi \theta) \in \mathbb{Q}$.
(a) According to Theorem 1, if $\theta \in \mathbb{Q} \cap [0,2)$ and $\sin^2(\pi \theta) \in \mathbb{Q}$, then $\sin(\pi \theta) \in {0,\pm 1/2, \pm 1}$.
First, suppose $\sin(\pi \theta) = 0$. Then $\theta$ is an integer. Next, suppose $\sin(\pi \theta) = \pm 1/2$. Then $\pi \theta = n\pi \pm \frac{\pi}{6}$ for some integer $n$, and so $\theta = \frac{n}{2} \pm \frac{1}{12}$. Finally, suppose $\sin(\pi \theta) = \pm 1$. Then $\pi \theta = n\pi \pm \frac{\pi}{2}$ for some integer $n$, and so $\theta = \frac{n}{2} \pm \frac{1}{4}$.
Therefore, the values of $\theta \in \mathbb{Q} \cap [0,2)$ for which $\sin^2(\pi \theta) \in \mathbb{Q}$ are of the form $n, \frac{n}{2} \pm \frac{1}{12}, \frac{n}{2} \pm \frac{1}{4}$ for some integer $n$.
(b) According to Theorem 1, if $\theta \in \mathbb{Q} \cap [0,2)$ and $\sin(\pi \theta) \in \mathbb{Q}$, then $\sin(\pi \theta) \in {0, \pm 1/2, \pm 1}$.
First, suppose $\sin(\pi \theta) = 0$. Then $\theta$ is an integer. Next, suppose $\sin(\pi \theta) = \pm 1/2$. Then $\pi \theta = n\pi \pm \frac{\pi}{6}$ for some integer $n$, and so $\theta = \frac{n}{2} \pm \frac{1}{12}$. Finally, suppose $\sin(\pi \theta) = \pm 1$. Then $\pi \theta = n\pi \pm \frac{\pi}{2}$ for some integer $n$, and so $\theta = \frac{n}{2}$.
Therefore, the values of $\theta \in \mathbb{Q} \cap [0,2)$ for which $\sin(\pi \theta) \in \mathbb{Q}$ are of the form $n, \frac{n}{2} \pm \frac{1}{12}$ for some integer $n$.
## Textbooks
• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.
Statistics-lab™可以为您提供cornell.eduMATH6370 Algebraic number theory代数数论的代写代考和辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。 | 2,056 | 6,502 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.894983 |
https://jeopardylabs.com/play/khs-mcdowell-online-physical-science-chapter-11 | 1,540,109,139,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513804.32/warc/CC-MAIN-20181021073314-20181021094814-00165.warc.gz | 735,992,040 | 8,476 | Machines1
Machines2
Machines3
Machines4
Machines5
### 100
A simple machine that consists of an inclined plane wrapped around a cylinder or cone is this.
What is a screw?
### 100
Efficiency and a comparison of mechanical advantage and ideal mechanical advantage gives you an idea of this.
What is how much energy is lost due to friction in a machine?
### 100
A simple machine consisting of a bar that rotates around a fixed point called a fulcrum is this.
What is a lever?
### 100
The number of times a machine multiplies input force is known as this.
### 100
This is the rate at which work is done.
What is power?
### 200
A type of lever there the fulcrum is located on the far side of the input force area, has an ideal mechanical advantage less than 1, and does not result in a change in the direction of force is this.
What is a third class lever?
### 200
This is the SI unit for power.
What is Watt?
### 200
A simple machine that consists of two connected rings or cylinders, one inside the other, which both turn in the same direction around a single center point is this.
What is a wheel and axle?
### 200
The SI unit for work is this.
What is Joule?
### 200
When the force is applied to the wheel of a wheel and axle machine, the axle has this amount of output force relative to the input force.
What is more?
### 300
A type of lever where the fulcrum is between the input and output force areas, and results in a change in the direction of the force is this.
What is a first class lever?
### 300
Compound machines tend to have a lower efficiency than simple machines due to this.
What is friction?
### 300
This is a device that makes work easier by changing a force.
What is a machine?
### 300
This value associated with a compound machine is the product of all the simple machines incorporated into the compound machine.
### 300
When the force is applied to the axle of a wheel and axle machine, the wheel has this amount of output force relative to the input force.
What is less?
### 400
This is the use of force to move an object.
What is work?
### 400
A compound pulley is made up of these.
What are fixed and movable pulleys?
### 400
A machine made of more than one simple machine is this.
What is a compound machine?
### 400
A simple machine that consists of two inclined plane and only is functional when it moves is this.
What is a wedge?
### 500
A simple machine that consists of a rope and grooved wheel is known as this.
What is a pulley?
### 500
This represents the multiplication of input force that would be achieved in the absence of friction.
### 500
A simple machine consisting of a sloped surface that connects lower and higher elevations is this.
What is an inclined plane?
### 500
The percent of input work that becomes output work is known as this.
What is efficiency?
### 500
A type of lever where the fulcrum is on the far side of the output force area, has an ideal mechanical advantage greater than 1, and does not result in a change in the direction of force is this.
What is a second class lever? | 718 | 3,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-43 | latest | en | 0.952065 |
https://communities.sas.com/t5/General-SAS-Programming/Sum-digits-of-a-string/td-p/263050?nobounce | 1,531,926,030,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590199.42/warc/CC-MAIN-20180718135047-20180718155047-00494.warc.gz | 625,524,810 | 27,286 | Solved
Contributor
Posts: 21
# Sum digits of a string
Hello
My question is if I have a string such as '6423BS24F', how do I write a step that sums only the numbers of the string?
THANK YOU!!!!!
Accepted Solutions
Solution
04-11-2016 08:53 PM
Contributor
Posts: 21
## Re: Sum digits of a string
Posted in reply to Astounding
Hello Astounding,
The code actually works!! You are the best
Thank you!!!
All Replies
Super User
Posts: 6,764
## Re: Sum digits of a string
data want;
string = '6423BS24F';
total = 0;
do i=1 to length(string);
total + input(substr(string, i, 1), ?? 1.);
end;
run;
Solution
04-11-2016 08:53 PM
Contributor
Posts: 21
## Re: Sum digits of a string
Posted in reply to Astounding
Hello Astounding,
The code actually works!! You are the best
Thank you!!!
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https://techcommunity.microsoft.com/t5/excel/looking-to-make-a-forecast-formula/m-p/3256877 | 1,656,529,843,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00654.warc.gz | 634,529,450 | 53,976 | New Contributor
# Looking to make a forecast formula
Hello, I'm looking to make a basic change in value formula for interest on a credit card.
ex.
Jan = \$136
Feb = 138
Mar = 124
I want the column to auto fill that my change in interest was \$2 Jan. > Feb., and -\$14 Feb. > Mar.
3 Replies
# Re: Looking to make a forecast formula
For example:
The formula in C3 is =IF(B3="","",B3-B2)
This can be filled down as far as you want.
# Re: Looking to make a forecast formula
Hi, @Hans Vogelaar, I may need to be linked to a video or something, because it does not seem to be working for me.
# Re: Looking to make a forecast formula
It should be
=IF(K4="","",K4-K3) | 190 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.926761 |
https://www.cfd-online.com/Forums/fluent/136970-grid-convergence-index.html | 1,511,206,764,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806124.52/warc/CC-MAIN-20171120183849-20171120203849-00743.warc.gz | 788,954,242 | 15,860 | Grid Convergence Index
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June 7, 2014, 16:36 Grid Convergence Index #1 New Member Mahdi Join Date: Nov 2012 Location: Malaysia Posts: 27 Rep Power: 6 Hello dear friend I'm studying on the grid convergence index, presented by Celik, et al whose their article in this case can be found in the link below: http://www.taoxing.net/web_documents/p4enew.pdf Now I have got some question that how can I calculate area of each cell mentioned in step 1 of this article and then do I need to calculate area of all cells? How come it is possible? I'm So confused
June 20, 2014, 11:23 #2 Member Join Date: Nov 2010 Posts: 95 Rep Power: 9 Check out these papers: Roache 1997 "Quantification of numerical uncertainty in computational fluid dynamics". Freitas, 2002 "The issue of numerical uncertainty" There is a clear explanation on how to compute GCI. I don't think you need to compute the area of each cell. Cheers!
March 19, 2017, 06:58
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Hazem
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Originally Posted by metmet Hello dear friend I'm studying on the grid convergence index, presented by Celik, et al whose their article in this case can be found in the link below: http://www.taoxing.net/web_documents/p4enew.pdf Now I have got some question that how can I calculate area of each cell mentioned in step 1 of this article and then do I need to calculate area of all cells? How come it is possible? I'm So confused
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https://crypto.stackexchange.com/revisions/60254/8 | 1,582,148,029,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00123.warc.gz | 329,389,381 | 18,103 | 8 of 15 Clarify
The birthday problem can't help to find "vanity" public/private key pairs; where "vanity" is some arbitrary characteristic/metric of the public key, like (in the question) having as low as possible a $$X$$ in Cartesian coordinates, or (more often) a long recognizable string in the text representation of the public key.
The only known methods essentially try many key pairs, and keep the "best" according to the vanity metric. We do not know how to make the tries better than random, hence if the vanity of a random public key has probability $$p$$ to be satisfactory, we'll need to test about $$1/p$$ public keys. The cost to compute each key tested can be reduced to one elliptic curve group operation, but not much less (in particular, with the customary format of public keys in Bitcoin, it seems we need at least one inversion in the base field).
A simple method can be:
• draw a random $$u$$ in $$[1,n)$$ where $$n$$ is the order of $$G$$, with $$n=2^{256}-\mathtt{14551231950b75fc4402da1732fc9bebf_h}$$ for secp256k1
• compute associated point $$P_0=u\times G$$
• keep adding $$G$$ to that point, computing $$P_i=P_{i-1}+G$$ until that's a public key with suitable vanity
• find the corresponding private key as $$u+i\bmod n$$ (the modulus step is unnecessary in practice).
Usually point addition is more costly than point doubling, and it might be better to use:
• draw a random $$u$$ in $$[1,n)$$
• compute associated point $$P_0=u\times G$$
• keep doubling that point, computing $$P_i=P_{i-1}+P_{i-1}$$ until that's a public key with suitable vanity
• find the corresponding private key as $$2^i u\bmod n$$.
1KATWALSHTHECuz5DFHoNqJax6hnDbL6KN (I can sign with it if you want)
Any vanity pubkeys category? If so, I win:
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
The following concludes that the public key with the hex form of the third line does not match the text string[*] of the first line. My reading is thus that it is not even alleged that it is known how to make a signature that verifies against a public key with an $$x$$ having 90 leading zero bits (as the one in the third line does). And, for the reasons in the first part of this answer, I doubt that it is.
[*] an address, that is the hash of a matching public key among about $$2^{257-320}=2^{97}$$.
That helpful schematic shows how the public key and its text form relate.
The first string is the Base58Check encoding of a "Bitcoin pubkey hash", that is the Base58 encoding of 1+20+4 bytes (obtained using this tool) with the last 4 bytes equal to the first 4 bytes of the SHA-256 hash of the SHA-256 hash of the first 1+20 bytes.
00 c73c219421726619b3d680cbcf07d70783a91d40 b2f57ab5
The last string is a point in compressed coordinate (as recognizable from the leading 02) with $$x$$ (in hex)
00000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
We compute $$y$$ as the solutions to $$y^2=x^3+7\bmod p$$ with $$p=2^{256}-2^{32}-977$$, and that is (hex)
3f3979bf72ae8202983dc989aec7f2ff2ed91bdd69ce02fc0700ca100e59ddf3
c0c686408d517dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
The leading 02 in the compressed form means to select the even (second) value of $$y$$, therefore the uncompressed public key complete with 04 prefix is
04 00000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63 c0c686408d517dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
which, hashed by SHA-256 then RIPEMD-160, gives (according to this tool) 2a8cfefd81c810f789e2f75c4fce37922c939310 which does not match c73c219421726619b3d680cbcf07d70783a91d40 from the text representation. I tried with the other $$y$$ coordinate and the compressed form, no match. | 1,123 | 3,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-10 | latest | en | 0.907278 |
http://www.scribd.com/doc/126581152/Inelastic-Design-Spectra-Newmark-Hall-Design-Spectra-Yoppy-Soleman-2006 | 1,441,358,323,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645339704.86/warc/CC-MAIN-20150827031539-00067-ip-10-171-96-226.ec2.internal.warc.gz | 693,540,152 | 52,940 | Welcome to Scribd. Sign in or start your free trial to enjoy unlimited e-books, audiobooks & documents.Find out more
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Inelastic Design Spectra _Newmark Hall Design Spectra_[Yoppy Soleman, 2006]
# Inelastic Design Spectra _Newmark Hall Design Spectra_[Yoppy Soleman, 2006]
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Excel spreadsheet for counting Inelastic Design Spectra (IDS) to implement A.K. Chopra Method. This is one of several methods that are included in Displacement-Based Method Procedures for Earthquake Resistant Structure Design.
Excel spreadsheet for counting Inelastic Design Spectra (IDS) to implement A.K. Chopra Method. This is one of several methods that are included in Displacement-Based Method Procedures for Earthquake Resistant Structure Design.
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Inelastic Design Spectra (IDS)_Prof. A.K.
1. Equivalent Linear System (Rosenblueth, Herrera, Chopra)2. Elastic Design Spectrum (Newmark, Hall)3. Inelastic Design Spectrum (Newmark, Hall)4. Metoda IDS (Inelastic Design Spectra)
A. Variasi perioda dan redaman viskos sistem linear ekivalen
1. Hubungan perioda untuk sistem linear ekivalen dengan duktilitas
F a k t o r k e k a k u a n p a s c a - l u l u h ,
a
0 0.5 1 1.5 2 2.5 3 3.5
a =
0.00
1.00 1.00 1.00 1.22 1.41 1.58 1.73 1.87
a =
0.05
1.00 1.00 1.00 1.21 1.38 1.52 1.65 1.76
a =
0.10
1.00 1.00 1.00 1.20 1.35 1.47 1.58 1.67
a =
0.20
1.00 1.00 1.00 1.17 1.29 1.39 1.46 1.53
a
0.000.501.001.502.002.503.003.500 1 2 3 4 5 6 7 8
T
e q
: T
n
Faktor Duktilitas,
m
2. Hubungan redaman viskos ekivalen pada sistem linear ekivalen denga
F a k t o r k e k a k u a n p a s c a - l u l u h ,
a
0 0.5 1 1.5 2 2.5 3 3.5
a =
0.00
0.00 0.00 0.00 0.21 0.32 0.38 0.42 0.45
a =
0.05
0.00 0.00 0.00 0.19 0.27 0.32 0.35 0.37
a =
0.10
0.00 0.00 0.00 0.17 0.25 0.29 0.31 0.31
a =
0.20
0.00 0.00 0.00 0.14 0.20 0.22 0.23 0.23
B. Elastic Design Spectrum
0.000.100.200.300.400.500.600.700 1 2 3 4 5 6 7 8
R e d a m a n v i s k o s e k i v a l e n ,
x
e q
Faktor Duktilitas,
m
1. Faktor Amplifikasi Spektrum Percepatan,
Nilai-nilai
Kecepatan dan Displasemen vs Redaman
Redama
x
x (%)a
A
a
V
a
D
A
2.00
3.66 2.92 2.42
1.0g3.00
3.24 2.64 2.24 981 cm/s2
5.00
2.71
2.30
2.0110.00
1.99 1.84 1.69
23.34
1.1038 1.27 1.310.40789 4.772309
2. Elastic Pseudo-Accelleration Design Spectrum (Newmark-Hall Proced
Tabel Respons Spektra Percepatan Elastik,
x
=
x
0
= 5%
PBA 1.0g
(
m
=1)
1 0.010 1.00002 0.020 1.00003 0.030 1.00004
0.033
1.00005 0.0334 1.06886 0.0335 1.07117 0.0336 1.07338 0.0337 1.07569 0.0338 1.077810
0.125
1.103911 0.130 1.103912 0.140 1.103913 0.150 1.103914 0.160 1.103915 0.170 1.103916 0.180 1.103917 0.190 1.103918 0.200 1.103919 0.210 1.103920 0.220 1.103921 0.230 1.103922 0.240 1.103923 0.250 1.103924 0.260 1.103925 0.270 1.103926 0.280 1.103927 0.290 1.103928 0.300 1.103929 0.310 1.103930 0.320 1.1039
NoPeriod(s)
0.00.20.40.60.81.01.20.0 0.5 1.0 1.5 2.0 2.5
P s e u d o - A c c e l l e r a t i o n S p e c t r a ( g )
Perioda
Pseudo Respons Percep(
x
=5%) Profil Tanah Kera | 1,625 | 3,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-35 | latest | en | 0.351247 |
http://research.stlouisfed.org/fred2/series/LRHUTTFECLA156S | 1,419,264,376,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802775394.157/warc/CC-MAIN-20141217075255-00116-ip-10-231-17-201.ec2.internal.warc.gz | 224,184,785 | 19,570 | # Harmonized Unemployment: Total: Females for Chile©
2013: 6.88263 Percent (+ see more)
Annual, Not Seasonally Adjusted, LRHUTTFECLA156S, Updated: 2014-02-05 8:38 AM CST
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
OECD descriptor ID: LRHUTTFE
OECD unit ID: STSA
OECD country ID: CHL
All OECD data should be cited as follows: OECD, "Main Economic Indicators - complete database", Main Economic Indicators (database),http://dx.doi.org/10.1787/data-00052-en (Accessed on date)
Copyright, 2014, OECD. Reprinted with permission.
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(a) Harmonized Unemployment: Total: Females for Chile©, Percent, Not Seasonally Adjusted (LRHUTTFECLA156S)
OECD descriptor ID: LRHUTTFE
OECD unit ID: STSA
OECD country ID: CHL
All OECD data should be cited as follows: OECD, "Main Economic Indicators - complete database", Main Economic Indicators (database),http://dx.doi.org/10.1787/data-00052-en (Accessed on date)
Copyright, 2014, OECD. Reprinted with permission.
Harmonized Unemployment: Total: Females for Chile©
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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Unformatted text preview: nd fission), and to convert energy to matter (observed in creation of elementary particles). • It is important to note here the law of conservation of mass, which embodies in it the principles that both matter and energy are conserved. In essence, it says that the mass to be conserved includes both the mass of the matter in the system (portion of the universe that is singled out for study) and the mass of energy in the system. Sample Problem 1.5: Einstein’s relation between mass and energy: nuclear reaction. When 1 kg of uranium 235 (235U) undergoes nuclear fission, as in the detonation of an atomic bomb, 0.823 × 1014J of energy is liberated. How much has the mass of sample decreased? Let’s calculate the mass of uranium lost by the use of Einstein’s equation E 0.823 × 1014 J m= 2 = = 0.0916 × 10− 2 kg = 0.000916 kg c (2.998 × 108 ) 2 m 2 s − 2 Thus, the mass of 235U has decreased by 0.0916%. In the above example, a small amount of matter was converted to a large amount of energy. But that mass lost is relatively large compared to mass lost in non-nuclear reactions – reactions that we’re concerned with in general chemistry. Let’s take a very explosive reaction and see how much mass is lost in the process. Sample Problem 1.6: Einstein’s relation between mass and energy: non-nuclear reaction. When 1 kg of nitroglycerine explodes, 8.0 × 106 J of energy is liberated. How much has the mass of sample decreased? Let’s calculate the mass of nitroglycerine lost by the use of Einstein’s equation 0.823 × 106 J E m= 2 = = 0.89 × 10−10 kg = 0.000000000089kg (2.998 × 108 ) 2 m 2 s − 2 c Thus, the mass of nitroglycerine has decreased by 0.0000000089%. From the above example we see that the difference in mass of the product and reactants in non-nuclear reactions is negligible. In fact, it is so slight that you would never detect it with regular weight balances. For all practical purposes, we say that mass of matter is conserved in ordinary chemical reactions. Effectively, for ordinary chemical reactions we can ignore the energy part, and simply say that the total mass of the products of chemical reaction is the same as the total mass of reactants. Mass of matter is conserved in ordinary chemical reactions. 19 Copyright © 2007 by Concise Books Publishing LLC. Visit us at www.concisechem.com to download other free chapters from "The Concise Guide to Chemistry." CHAPTER 1: INTRODUCTION TO CHEMISTRY Section Test Questions 1. Loss of mass is an important factor in: (a) nuclear reactions, (b) ordinary reactions, (c) both? Answers 1. A - see sample problems 1.5 and 1.6. Antimatter, Dark Matter & Dark Energy 1.5 You’ve probably heard of, if not even read, Dan Brown’s bestselling fiction novel Angels and Demons in which he talks about a plot to destroy the Vatican using 1 gram of antimatter stolen from Switzerland’s Conseil Europeen pour la Recherche Nucleaire (CERN). Luckily, antimatter is not present in the universe...
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## This document was uploaded on 09/19/2013.
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Question 29275: At very long hotel in Arnold, MD, there are n rooms located along a very long corridor and numbered consecutively from 1 to n. One night after a party, n people, who have been likewise numbered 1 to n, arrived at this hotel and did as follows: guest 1 opens all the doors. Then guest 2 closed every second door beginning with door 2. Afterwards, guest 3 changed the position of every third door (that is the guest opened the doors that were closed and closed the doors that were open). In a similar way, guest 4 changed the position of doors 4,8,12,.... This process continued until each person had walked the length of the corridor. Of course, the last person, guest n, merely strolled to the end of the corridor, where the guest changed the position of door n. The question is, “Which doors were left open and which ones were left closed at the end of the process?" Click here to see answer by 303795(595)
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Question 25561: 1- Find the mesure of an angle if its measure is 3 degrees less than twice the measure of its supplement Click here to see answer by 303795(595)
Question 25097: In a mixture of concrete there are 3lbs of cement mix for every 1lb of gravel. If the mixture contains a total if 140lbs of these two ingredients how many pound of gravel are there? Click here to see answer by 303795(595)
Question 20123: I really need help on this problem quickly!!! I'm a AP AB Calculus student, but this extra credit problem is giving me serious issues. I can find one of the two equations and can continue no further! O.k. here it is: "A Yankee mixes a certain number of wooden nutmegs, which cost him 1/4 cent apiece, with a quantity of real nutmegs, worth 4 cents apiece, and sells the whole assortment for \$44; and gains \$3.75 by the fruad. How many wooden nutmegs were there?" I've managed this so far: x= number of real nutmegs y= number of fake nutmegs The cost of the total number of nutmegs is 44 - 3.75, which equals 40.25 0.0025x + 0.04y = 40.25 Since I have two variables I need two equations. I know I need the total number of nutmegs but I don't know how to find it! Please help! It would be greatly appreciated! :) Click here to see answer by 303795(595)
Question 65598: A corn mean feed is 10% protein and a feed supplement is 40% protein. How much of each must be conbined to get 2400 pounds of feed that is 19% protein? Click here to see answer by ptaylor(2048)
Question 65729: A barrel contains 30 gallons of 40% antifreeze. How many gallons of an 80% antifreeze solution should be added to make the barrel contain 50% antifreeze? Click here to see answer by stanbon(57323)
Question 65810: The volume of a quantity of an ideal gas was held constant. The inital pressure and temperature were 400 torr and 80 K. What would be the final temperature if the final pressure was 300 torr? Click here to see answer by Edwin McCravy(8908)
Question 65848: A farmer wants to mix milk containing 6% butterfat with 2 quarts of cream that is 15% butterfat to obtain a mixture that is 12% butterfat. How much milk containing 6% butterfat must he use? Click here to see answer by stanbon(57323)
Question 65890: The Queen Mary is a large cruise ship that has 8,12,700 cubic feet of cargo space. Express this nunmber in scientific notation. Click here to see answer by Nate(3500)
Question 65890: The Queen Mary is a large cruise ship that has 8,12,700 cubic feet of cargo space. Express this nunmber in scientific notation. Click here to see answer by uma(370)
Question 66077: Elmer and Rosy were playing a game.in each game the loser givesthe winner a marble from the marble jar.When they finished, elmer had won fifteen times and rosy had twenty two more marbles than when she started.How many games did they play? Click here to see answer by stanbon(57323)
Question 66303: Mix a solution that is 50% alcohol with a soution that is 20%to make 450 mL of a aolution that is 40% alcohol. How much of each solutions should you use ? Click here to see answer by checkley71(8403)
Question 66306: A 6-quart cooling system is checked and found to be filled with a solution that is 20% antifreeze. The desired strength of the solution is 60 % antifreeze. How many quarts of solution need to be drained and replaced with pure antifreeze to reach the desired strength? Click here to see answer by funmath(2925)
Question 66313: Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 800mL of a soultion that is 60% alcoho. How much of each solution should you use? Click here to see answer by ankor@dixie-net.com(15649)
Question 66317: mix a solution that is 40% acid with a solution that is 100% water to make 16L of a solution that is 10% acid. How much of each solution should you use? Click here to see answer by ptaylor(2048)
Question 66332: The weight of carbon (C) in a quantity of c3h3cl5 was 432 grams. What was the weight of the chlorine (cl) in the compound? (C,12;H,1;Cl,35) Click here to see answer by venugopalramana(3286)
Question 66436: A 12 quart cooling system is filled with a solution that is 10% antifreeze. The desired strength of the solution is 40% antifreeze. How many quarts of solution need to be drained and replaced with pure antifreeze to reach the desired strength? Click here to see answer by ptaylor(2048)
Question 66436: A 12 quart cooling system is filled with a solution that is 10% antifreeze. The desired strength of the solution is 40% antifreeze. How many quarts of solution need to be drained and replaced with pure antifreeze to reach the desired strength? Click here to see answer by josmiceli(9671)
Question 66435: Mix a solution that is 50%alcohol with a solution that is 10% alcohol to make 300 L of a solution that is 30% alcohol. How much of each should you use? Click here to see answer by ptaylor(2048)
Question 66467: How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze. How do I go about setting up this equation? Please help! Click here to see answer by ptaylor(2048)
Question 66574: a lab technician has one solution that is 60% chlorinated and another that is 40% chlorinated. how much of each solution is needed to make a 100 L solution that is 50% clorine? Click here to see answer by stanbon(57323)
Question 66582: A jeweler is using 10 lbs. of an alloy that is 2% robium. He has another alloy which is 5% robium. He wishes to combine these mixtures to form a mixture that is 3% robium. How many pounds of the second alloy should be added to the first to get a mixture that is 3% robium. I really need help, so I would ask for step by step instructions please. Thank you. Click here to see answer by stanbon(57323)
Question 66644: 34. Sleep Deprivation In a recent study, volunteers who had 8 hours of sleep were three times more likely to answer correctly on a math test than were sleep-deprived participants. (a) Identify the sample used in the study. (b) What is the sample's population? (c) Which part of the study represents the descriptive branch of statitics? (d) Make an inference based on the results of the study? Click here to see answer by stanbon(57323)
Question 66884: How many pounds of a 35% salt solution and how many pounds of a 14% salt solution should be combined so that 50 pounds of a 20% salt solution are obtained? Click here to see answer by ptaylor(2048)
Question 66883: A candy store sells boxes of candy containing caramels and chocolates. Each box sells for \$12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost \$0.25 to produce and the chocolates cost \$0.45 to produce, how many of each should be in a box to make a profit of \$3.00? Click here to see answer by ptaylor(2048)
Question 66951: Florence Flask has 500 grams of a 72% salt solution. How much water must she add to obtain a 25 % salt solution? Click here to see answer by ptaylor(2048)
Question 67103: I'm having to do an algebra review to start off the semester. I am able to recall doing these types of problems, but it's been a while. The question is as follows: Suppose a fuel mixture is 6% ethanol and 94% gasoline. How much ethanol (in gallons) must you add to one gallon of the fuel so that the new fuel mixture is 10% ethanol? Click here to see answer by Nate(3500)
Question 67414: Michelle had two containers, each containing a mixture of alcohol and disinfectant. One was 40% disinfectant and the other was 80% disinfectant. How much of each should be used to get 600 milliliters of a solution that is 50% disinfectant? Click here to see answer by ptaylor(2048)
Question 67408: A recipe calls for 12 ounces of cottage chees and 6 ounces of cream cheese. If there are 216 calories in 8 ounces of cottage cheese and 106 calories in 1 ounce of cream chees, how many calories are there in 36 ounces of the mixture? Click here to see answer by ptaylor(2048)
Question 66481: This is for a work seminar I'm involed in. I have solved half the problem....but can't get the last question. My brain is tired....help:) The Facts: There are five houses.Each house is a different color.Each house is inhabited by people of different nationalities.Each house has a different pet, and the people in each house drink different drinks and smoke different cigarettes. The clues: 1. The english man lives in the red house. 2. The spaniard owns the dog. 3. Coffee is drunk in the green house. 4. The ukrainia drinks tea. 5. The green house is immediately to the right of the ivory house. 6. The camel smoker owns snails. 7. Kools are smoked in the yellow house. 8. Milk is drunk in the middle house. 9. The norwegian lives in the first house on the left. 10. The man who smokes marlboroughs lives in the house next to the man with the fox. 11. The lucky strike smoker drinks orange juice. 12. Kools are smoked in the house next to the house where the horse is kept. 13. The japanese smokes parliaments. 14. The noewegian lives next to the blue house. The two questions are: Who owns the zebra? Who drinks water? The answer is the Norwegian So my question is who of these upstanding, heavy smoking neighbors, keeps a zebra in his yard.......cause I mean...WHO wouldn't want a zebra?:) Click here to see answer by 303795(595)
Question 67578: A chemist has 300 milliliters (mL) of solution that is 18% acid. How many milliliters of acid are in the solution? Click here to see answer by tutorcecilia(2152)
Question 67717: Here is the problem. A parts person has to spend exactly \$100 and must buy 100 parts. Good parts cost 5\$, regular parts cost 1\$, and cheap parts cost 25 cents. He has to buy 100 parts for \$100 and have some of each part. I tried the trial and error method and got an answer, but wanted to get an anwer algebraically, (sp?) I tried: 5x + y + .25z = \$100 x + y + z = 100 and substituting variables but could not get it to work. I appreciate your help. George Click here to see answer by ankor@dixie-net.com(15649)
Question 67768: At a local community college, five statistics classes are randomly selected and all of the students from each class are interviewed. What sampling tecnique is used? a. stratified b. cluster c. convenience d. random e. systematic Click here to see answer by stanbon(57323)
Question 67819: A liter of cream has 9.2% butterfat. How much skim milk containing 2% butterfat should be added to the cream to obtain a maxture of 6.4% butterfat? Click here to see answer by checkley71(8403)
Question 67819: A liter of cream has 9.2% butterfat. How much skim milk containing 2% butterfat should be added to the cream to obtain a maxture of 6.4% butterfat? Click here to see answer by stanbon(57323)
Question 67898: How many gallons of a 5% acid solution must be mixed with 5 gal of a 10% solution to obtain a 7% solution? Click here to see answer by Earlsdon(6287)
Question 68002: Dimitruis skipped lunch today to get some sun in the school courtyard. Lying on the grass and looking up at an angle of 22 he could see the top of the flagpole which was about 100 ft away. How tall is the flagpole? Click here to see answer by Nate(3500)
Question 68112: Bob needs to buy 5 pounds of candy. one kind costs 1.75 per pound and the second kind costs 2.35 per pound. How much of each will he need if he spends 10.00? Click here to see answer by ankor@dixie-net.com(15649)
Question 68196: An 800 g solution of acid and water contains 128 g of acid. What percent of the solution is acid? Click here to see answer by ptaylor(2048)
Question 68320: An oil company wants to mix two grades of rocket fuel costing \$3 and \$4 per gallon,respectively with purified ethanol costing \$8 per gallon to obtain 140 gallons of hybrid fuel costing NASA \$6 per gallon. If the oil company also wants the amount of lower grade fuel to be twice that of higher-grade rocket fuel,how many gallons of each variety of fueling product should be mixed? - define your unknowns in terms of x,y etc. -write down a linear function of equations representing the information given Click here to see answer by stanbon(57323)
Question 68521: How many pounds of peanuts costing \$2.00 a pound should be mixed with a 4 pounds of cashews costing \$4.50 a pound to obtain a mixture costing \$3.00 a pound? Click here to see answer by checkley71(8403)
Question 68592: Please help to solve One solution was 10% alcohol and the other was 30% alcohol. HOw much of each should be used to get 200 gallons of a solution that is 17% alcohol? Click here to see answer by ptaylor(2048)
Question 68704: one grade of oil has 0.50% of an additive, and a higher grade has .75% of the additive. How many liters of each must be used to have 1000L of a mixture with 0.65% of the additive? I tried to set up a chart but I'm having problems figuring out how to set up this problem. Click here to see answer by stanbon(57323)
Question 68762: 1. How many liters of a 50% solution should be added to a 80% solution to obtain a 60% solution? Click here to see answer by josmiceli(9671)
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http://www.vfxalll.com/104-news.html | 1,539,724,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510867.6/warc/CC-MAIN-20181016201314-20181016222814-00125.warc.gz | 578,416,497 | 7,938 | Using Forex linear regression to see the big picture - Forex Trading: Vfxall
# Using Forex linear regression to see the big picture
Traders are always looking for methods to apply to the financial markets to provide some element of a trading edge.
A trading edge is any strategy which you expect to make a profitable return if used repeatedly in the long term.
Such strategies seek to trade when the odds are skewed in your favour.
One area upon which analysts have focused is statistical methods proven in other fields...
...and whether they can beneficially inform trading decisions.
As we know, some methods of trading apply only to certain instruments.
But here is one of the benefits of regression trading:
...it favours no single market over another.
This is precisely because it is based on general statistical concepts.
So what is the linear regression?
Linear regression attempts to model the relationship between two variables, given a collection of data values.
The technique tries to do so by finding a line of best fit between the two.
With Forex linear regression trading, the two variables we are interested in are time and price.
Existing data values between the two are plentiful, of course.
By observing the data in a given period:
...we theoretically gain insight into the future performance, given that we can find a satisfactory line of best fit.
This is because the line of best fit is effectively what traders normally call the trend.
Most trend regression indicators don't stop there, though.
They usually also provide channels that can help indicate support and resistance.
They do this by tying in probability theory:
...and assuming that price values will fall in a normal distribution around this median line.
If the prices move a sufficiently significant distance away from the median line, they can be thought of as statistical outliers.
At these levels, we might expect to find some form of support or resistance.
So, how do we work out where these price extremes occur?
One way is to utilise the statistical concept of a normal distribution and the accompanying measure of standard deviation.
To understand better this standard deviation Forex strategy, let's quickly have a run-through of what we mean by these terms.
A normal distribution is a probability distribution that follows a bell-shaped curve.
The highest probability density is centred around the mean.
This is also the median and is represented by the dotted line μ in the diagram above.
An important point to note is that all normal distributions are symmetrical.
This places both the mean and the median at the exact centre of the bell curve.
Standard deviation is another statistical measure and quantifies how scattered the values are in a data set.
The larger the standard deviation, the wider the bell curve.
Now, the mathematics that govern this curve are relatively complex.
But here's the good news:
...the concept it represents is actually fairly simple.
The further we get away from the middle of the bell, the smaller the chances are of those values of X occurring.
This means the majority of values for X occur one standard deviation either side of the mean.
In fact, in a normal distribution, we would expect around 68% of the data values to occur in this range.
Two standard deviations either side of the mean cover roughly 95% of all data values.
At the tails of the curve, we get the outliers.
These are extremely rare occurrences.
Now, why does this matter?
Well, if we see a data value that is an outlier…
...it seems a fair assumption that future values may regress toward the mean.
Trend channel trading takes these concepts and applies them to market prices.
It plots parallel regression lines either side of the line of best fit.
These lines give a rule of thumb as to where we might expect to find outlying prices.
Let's take a look at using a regression channel indicator in practice with MetaTrader 4.
### Linear regression channel indicator MT4
Linear regression channel indicator comes as a standard tool with MetaTrader 4.
The image above shows how to select the indicator, via the MT4 Insert tab.
Note that Standard Deviation is also available as an option, of which more later.
To add to your chart once you have selected MT4 Linear Regression channel:
click on the chart where you wish to begin your analysis drag to the right until you reach the point you wish to end your analysis the linear regression channel automatically appears on the chart between the two points.
The image belows shows the regression trend channel, plotted as blue dotted lines, on a EUR/USD four-hour chart.
The median line is the line of best fit for the closing prices contained within the selected period.
The trend lines above and below are at equal distances from the median line.
They are parallel and represent one standard deviation from the median line.
The rules for trading the regression channel are fairly simple.
The strategy revolves around the expectation that once the price moves out to an outlying level…
...there is a good chance that it will revert to the median line.
Therefore:
the upper line represents resistance the lower line represents support.
The best fit line suggests the trend.
So, a suggested method of using this indicator is to assume trend continuation and trade in the direction of the trend.
While that trend persists, we can think of the median line as being a kind of equilibrium point.
Obviously this kind of median line trading is susceptible to breakouts that result in a new trend forming.
Anytime the price breaks out beyond the channel should be treated with caution therefore.
An extended period beyond the channels suggests a new trend may be forming.
This is why you must be careful with trend analysis regression and ensure you are disciplined with your risk management.
Now it should also be pointed out that you are not limited to using only a single regression channel.
If we select standard deviation, we can add lines that are a multiple number of standard deviations away from the median.
To do this:
...we first draw our channels in exactly the same way as we did with the linear regression tool.
This plots the channels one standard deviation away, exactly as before.
We then need to edit the parameters of the tool.
We do this by selecting chart, objects and objects list.
In the objects list, select StdDev Channel and then Edit.
In the image above, the value of deviations equals 2.00.
Additionally, Ray is ticked, which causes the lines to extend infinitely to the right on the chart.
It now adds a second regression channel, with lines two standard deviations either side of the median line.
We can also draw smaller channels within a larger one in order to identify smaller trends within the overall larger trend.
If you're interested in exploring regression trading further, there are other, more complex versions with which you can experiment.
Moving average regression and polynomial regression forex analysis are just a couple of examples.
Correspondingly, moving linear regression indicator and polynomial regression channel are analysis tools that would involve the areas mentioned above.
For an even greater selection of cutting-edge tools, try out MetaTrader 4 Supreme Edition?
It's the ultimate plugin for MT4, with its own package of indicators and a wealth of trading aids.
### Final thoughts on linear regression Forex trading
At its heart, linear regression is a method of estimating the undefined relationship between price and time.
If you want to try out linear regression trading without any risk, our demo trading account is a good place to experiment.
Regression channels are just one type of trend channel trading system.
There are many other popular types of trend channel analysis that you can use, such as Keltner channels and Donchian channels..
As a final rule of thumb:
...it's always sensible with technical analysis to try and confirm your thinking with a second-look method.
For example, the image below contains the commodity channel index as a confirming indicator.
Another way of confirming your technical analysis linear regression signals is by looking at multiple time frames.
Looking at the same channels on a longer timeframe may reveal aspects you hadn't noticed before.
We hope you found this article interesting and we wish you success with forex linear regression method of trading. | 1,615 | 8,494 | {"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-2018-43 | longest | en | 0.922833 |
http://mathhelpforum.com/calculators/78978-how-apply-chain-rule-product-rule-ti-89-a-print.html | 1,527,240,206,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00413.warc.gz | 189,664,079 | 3,296 | # how to apply chain rule and product rule on Ti-89
• Mar 16th 2009, 05:02 AM
User Name
how to apply chain rule and product rule on Ti-89
Hi to all great members(Happy)
has anyone know a trick to apply these rules in a ti-89?
I can't really get the right answer for this equation:(x^2+4x+4)^2
which cane be done by the chain rule.
all the best.
• Mar 16th 2009, 07:00 AM
Jameson
Because you are mentioning the chain rule, you must be referring to derivatives. The TI-89 can easily take the derivative of a problem like the one you posted. It's under Calc -> Derivative, or something like that. The syntax I believe is find the derivative function on the 89 under calc, input the expression you want to be differentiated, then put a comma and then x. That tells the calculator that x is the variable of differentiation. If you want to apply it at a certain point you can put another comma and the point I believe. Either way, add a closing parenthesis.
• Mar 16th 2009, 08:03 AM
skeeter
Quote:
Originally Posted by User Name
Hi to all great members(Happy)
has anyone know a trick to apply these rules in a ti-89?
I can't really get the right answer for this equation:(x^2+4x+4)^2
which cane be done by the chain rule.
all the best.
see the screenshot ... calculator syntax is highlighted on the bottom line.
• Mar 16th 2009, 01:33 PM
User Name
Quote:
Originally Posted by Jameson
Because you are mentioning the chain rule, you must be referring to derivatives. The TI-89 can easily take the derivative of a problem like the one you posted. It's under Calc -> Derivative, or something like that. The syntax I believe is find the derivative function on the 89 under calc, input the expression you want to be differentiated, then put a comma and then x. That tells the calculator that x is the variable of differentiation. If you want to apply it at a certain point you can put another comma and the point I believe. Either way, add a closing parenthesis.
Thanks buddy, I know to find X we should add comma and than x on the end of the equation, is there anything else we can put there? like to find tangent can we like put x=2 to find normal and tangent?
Quote:
Originally Posted by skeeter
see the screenshot ... calculator syntax is highlighted on the bottom line.
Thanks a lot!!!
now i know what was my mistake! (Happy)
• Mar 23rd 2009, 06:45 AM
earboth
Quote:
Originally Posted by User Name
... is there anything else we can put there? like to find tangent can we like put x=2 to find normal and tangent?
Thanks a lot!!!
now i know what was my mistake! (Happy)
If you want to calculate the slope of the tangent for a specific x-value then you can add the x-value by using the "|" operator. That's the key directly below the equal sign. See attachment.
The syntax is highlighted in the input line. | 690 | 2,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-22 | latest | en | 0.926995 |
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# Which of the following statements about the cube above must be true?
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Which of the following statements about the cube above must be true? [#permalink]
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08 Jan 2019, 05:04
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Which of the following statements about the cube above must be true?
I. FD is parallel to GA.
II. ΔGCF and ΔAHD have the same area.
III. AF = GD
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
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Re: Which of the following statements about the cube above must be true? [#permalink]
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08 Jan 2019, 05:08
All the sides are equal in a cube.
Hence all the three statements sre true.
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Re: Which of the following statements about the cube above must be true? [#permalink] 08 Jan 2019, 05:08
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# 6900 ft to m6900 Foot to Metres
ft
=
m
## How to convert 6900 foot to metres?
6900 ft * 0.3048 m = 2103.12 m 1 ft
## Convert 6900 ft to common lengths
Units of measurementLengths
Nanometre2.10312e+12 nm
Micrometre2103120000.0 µm
Millimetre2103120.0 mm
Centimetre210312.0 cm
Inch82800.0 in
Foot6900.0 ft
Yard2300.0 yd
Metre2103.12 m
Kilometre2.10312 km
Mile1.3068181818 mi
Nautical mile1.1355939525 nmi
## Alternative spelling
6900 Feet to m, 6900 Feet in m, 6900 Foot to Metres, 6900 Foot in Metres, 6900 ft to m, 6900 Feet in Metres, 6900 Foot to Metre, 6900 Feet to Metre, 6900 Feet in Metre, | 256 | 620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.539798 |
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# 01.10.21
Maths Homework:
WALT order numbers
This week, we have been ordering and comparing numbers.
For your homework, put these numbers in order from smallest to largest:
1) 6, 3, 9
2) 7, 10, 15, 1
3) 8, 18, 20, 14
4) 6, 14, 9, 12, 7
5) 5, 17, 16, 13
If you need more support, use objects to represent each number. This could be anything - use leaves, pencils, plates etc!
If you would like a challenge, try this question afterwards:
What could the missing numbers be?
4, ______, 7, 10, 15, ______, 20
Top | 182 | 544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.900045 |
https://oswalpublishers.com/ncert-solutions/ncert-solutions-class-12-mathematics/chapter-1-relations-and-functions/ex-1-3/ | 1,708,873,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00595.warc.gz | 441,131,562 | 70,727 | # NCERT Solutions for Class 12 Maths Chapter 1 - Relations and Functions - Exercise 1.3
### Access Exercises of Class 12 Maths Chapter 1 - Relations and Functions
Exercise 1.1 Solutions : 16 Questions (14 Short Answers, 2 MCQ)
Exercise 1.2 Solutions : 12 Questions (10 Short Answers, 2 MCQ)
Exercise 1.3 Solutions : 14 Questions (12 Short Answers, 2 MCQ)
Exercise 1.4 Solutions : 13 Questions (12 Short Answers, 1 MCQ)
Miscellaneous Exercise Solutions: 19 Questions (7 Long answers, 9 Short Answer Type, 3 MCQs)
Exercise 1.3 :
1. Let f = {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f : {(1, 2), (3, 5), (4, 1)} and g = {(1, 3),
(2, 3), (5, 1)}. Write down gof.
Sol. The functions f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} are defined as f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.
∴ gof(1) = g(f(1)) = g(2) = 3
[∴ f(1) = 2 and g(2) = 3] gof(3) = g(f(3)) = g(5) = 1
[∴ f(3) = 5 and g(5) = 1] gof(4) = g(f(4)) = g(1) = 3
[∴ f(4) = 1 and g(1) = 3]
∴ gof = {(1, 3), (3, 1), (4, 3)}
2. Let f, g and h be function from R to R. Show that
(f + g)oh = foh + goh;
(f.g)oh = (foh).(goh)
Sol. To prove (f + g)oh = foh + goh Consider
LHS = ((f + g)oh) (x) = (f + g)(h(x))
= f(h(x)) + g(h(x))
= (foh)(x) + (goh)(x)
= {(foh) + (goh)}(x)
∴ ((f + g)oh)(x) = ((foh) + (goh)}(x) ∀ x ∈ R
Hence, (f + g)oh = foh + goh
To prove (f.g)oh = (foh).(goh) Consider LHS = ((f.g)oh)(x)
= (f.g)(h(x)) = f(h(x)).g(h(x))
= (foh)(x).(goh)(x)
= {(foh).(goh)}(x)
∴ ((f.g)oh(x) = {(foh).(goh)}(x) ∀ x ∈ R
Hence, (f.g)oh = (foh).(goh)
3. Find gof and fog, if
(i) f(x) = |x| and g(x) = |5x – 2|
(ii) f(x) = 8x3 and 8(x) = x1/3.
Sol. (i) f(x) = |x| and g(x) = |5x – 2|
∴ (gof)(x) = g(f(x)) = g(|x|)
= |5|x| – 2|
and fog(x) = f(g(x)) = f(|5x – 2|)
= ||5x – 2| = |5x – 2|
(ii) f(x) = 8x3 and g(x) = x1/3
∴ (gof)(x) = g(f(x))
= g(8x3) = (8x3)1/3 = 2x
and fog(x) = f(g(x)) = f(x1/3)
= 8(x1/3)3 = 8x
$$\textbf{4. If f( x)}=\frac{(4x+3)}{(6x-4)},x≠\frac{2}{3},\text{show that fof(x) = x, for}\\\text{all}\space x≠\frac{2}{3}\space \text{What is the inverse of f?}$$
$$\textbf{Sol.}\text{It is given that f(x)}=\frac{(4x+3)}{(6x-4)},x≠\frac{2}{3},$$
$$\text{(fof)(x) = f(f(x)) }=f(\frac{(4x+3)}{(6x-4)})\\=\frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4}\\=\frac{16x+12+18x-12}{24x+18-24x+16}\\=\frac{34x}{34}=x\\\text{Therefore, fof(x) = x, for all}\space x≠\frac{2}{3}\space\text{ fof =I}$$
Hence, the given function f is invertible and the inverse of f is itself.
5. State with reason whether following functions have inverse :
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Sol. (i) Function f : {1, 2, 3, 4} → {10} defined as f = {(1, 10), (2, 10), (3, 10), (4, 10)} From the given definition of f, we can see that f is a many one.
Function as f(1) = f(2) = f(3) = f(4) = 10.
Therefore, f is not one-one.
Hence, function f does not have an inverse.
(ii) Function g : {5, 6, 7, 8} → {1, 2, 3, 4} defined as g = {(5, 4), (6, 3), (7, 4), (8, 2)} From the given definition of g, it is seen that g is a many one.
Function as g(5) = g(7) = 4. Therefore, g is not one-one.
Hence, function g does not have an inverse.
(iii) Function h : {2, 3, 4, 5} → {7, 9, 11, 13} defined as h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
Therefore, function h is one-one.
Also, h is onto, since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5} have distinct images under h. Thus, h is one-one and onto function. Hence, h has an inverse.
Note : Student should remember that any function has inverse exist only when f is one-one and onto.
$$\textbf{6. Show that f : [– 1, 1] → R, given by f(x)}\\=\frac{x}{(x+2)},$$$$\textbf{is one-one. Find the inverse of the function}$$$$\textbf{f : [– 1, 1] → Range f.}$$
$$\textbf{Sol}.\text{ Show that f : [– 1, 1] → R, given by f(x)}$$$$=\frac{x}{(x+2)},$$
Let f(x) = f(y)
$$⇒ \frac{x}{x+2}=\frac{y}{y+2}⇒ xy + 2x = xy + 2y\\⇒ 2x = 2y ⇒ x = y\\\text{Therefore, f is a one-one function.}\\\text{Let}\space y=\frac{x}{x+2}⇒x = xy + 2y\\⇒x=\frac{2y}{1-y}$$
So, for every y except 1 in the range there exists x in the domain such that f(x) = y. Hence, function f is onto.
Therefore, f : [– 1, 1] → Range f is one-one and onto and therefore, the inverse of the function f : [– 1, 1] → Range f exists.
Let y be an arbitrary element of range f.
Since, f : [– 1, 1] → Range f is onto, we have
y = f(x) for some x → [– 1, 1]
$$⇒y= \frac{x}{x+2}⇒xy+2y=x$$
$$⇒ x(1 – y) = 2y ⇒ x =\frac{2y}{1-y},y≠1$$
Now, let us define g : Range f → [– 1, 1] as
$$g(y) =\frac{2y}{1-y},y≠1\\Now, (gof)(x) = g(f(x)) =g(\frac{x}{x+2})$$
$$=\frac{2(\frac{x}{x+2})}{1-\frac{x}{x+2}}=\frac{2x}{x+2-x}=\frac{2x}{2}=2$$
(fog)(y) = f(g(y))
$$=f(\frac{2y}{1-y})=\frac{(\frac{2y}{1-y})}{\frac{2y}{1-y}+2}\\=\frac{2y}{2y+2-2y}=\frac{2y}{y}=y$$
Therefore, gof = fog = IR, Therefore, f –1 = g
$$\text{Therefore,},f^{-1}(y)=\frac{2y}{1-y},y≠1$$
7. Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Sol. Here, f : R → R is given by f(x) = 4x + 3
Let x, y ∈ R, such that
f(x) = f(y) ⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y ⇒ x = y
Therefore, f is a one-one function.
Let y = 4x + 3
$$\text{⇒ There exist,}\space x=(\frac{y-3}{4})\space \varepsilon R,\forall \space y\varepsilon R$$
$$\text{Therefore, for any y ∈ R, there exist x}=\frac{y-3}{4}∈R$$
$$\text{such that}\space f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3=y$$
Therefore, f is onto function.
Thus, f is one-one and onto and therefore, f–1
exists.
$$\text{Let us define g : R → R by g(x) =}\frac{x-3}{4}\\Now, (gof)(x) = g(f(x)) = g(4x + 3)\\=\frac{(4x+3)-3}{4}=x\\(fog)(y) = f(g(y)) =f(\frac{y-3}{4})\\=4(\frac{y-3}{4})+3=y-3+3=y$$
Therefore, gof = fog = IR
Hence, f is invertible and the inverse of f is given
$$\text{by} f^{–1}(y) = g(y) =\frac{y-3}{4}.$$
8. Consider f : R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f–1 of f given by f–1(y) =√( y − 4 ), where R+ is the set of all non-negative real numbers.
Sol. Here, function f : R+ → [4, ∞) is given as f(x) = x2 + 4
Let x, y ∈ R+, such that f(x) = f(y)
⇒ x2 + 4 ⇒ x2 = y2
⇒ x = y [as x = y ∈ R+]
Therefore, f is one-one function.
For y ∈ [4, ∞), let y = x2 + 4
⇒ x2 = y – 4 ≥ 0 [as y ≥ 4]
⇒ x = (y − 4) ≥ 0
Therefore, for any y ∈ R+, There exists
x = √(y − 4) ∈ R+ such that
f(x) = f (√( y − 4)) = ( √(y − 4))2 + 4
= y – 4 + 4 = y
Therefore, f is onto. Thus, f is one-one and onto
and therefore, f–1 exists.
Let us define g : [4, ∞) → R+ by g(y) =√(y-4)
Now, gof(x) = g(f(x)) = g(x2 + 4)
=√(x2+4)-4=√x2=x
and fog(y) = f(g(y)) =f(√(y-4))
=(√(y-4))2+4
= (y – 4) + 4 = y
Therefore, gof = IR+ and I + fog =I [4, ∞)
f–1(y) = g(y) =√(y-4)
9. Consider f : R+ → [–5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with f–1(y)
$$=(\frac{(\sqrt{y+6})-1}{3})$$
Sol. Here, function f : R+ → [– 5, ∞) is given as
f(x) = 9x2 + 6x – 5.
Let y be any arbitrary element of [–5, ∞).
Let y = 9x2 + 6x – 5
⇒ y = (3x + 1)2 – 1 – 5
= (3x + 1)2 – 6
⇒ (3x + 1)2 = y + 6
⇒ 3x + 1 = √(y + 6)
[as y ≥ – 5 ⇒ y + 6 ≥ 0]
$$⇒x=\frac{\sqrt{y+6}-1}{3}$$
Therefore, f is onto, there by range f = [– 5, ∞).
$$\text{Let us define g : [– 5, ∞) → R+ as g(y)}=\frac{\sqrt{y+6}-1}{3}$$
Now, (gof)(x) = g(f(x)) = g(9x2 + 6x – 5)
= g((3x + 1)2 – 6)
$$=\frac{\sqrt{(3x+1)^2-6+6}-1}{3}\\\frac{3x+1-1}{3}=x\\\text{and (fog)(y) = f(g(y))}\\=f(\frac{\sqrt{(y+6)-1}}{3})\\=[3(\frac{\sqrt{(y+6)-1}}{3})+1]^2-6\\=(\sqrt{y+6})^2-6\\= y + 6 – 6 = y$$
Therefore, gof = IR+ and fog = I[– 5, ∞)
Hence, f is invertible and the inverse of f if given
by
$$f^{–1}(y) = g(y) =\frac{\sqrt{y+6}-1}{3}$$
10. Let f : X → Y be an invertible function. Show that f has unique inverse. (Hint : suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, (fog1)(y) = Iy(y) = fog2(y). Use one-one of f).
Sol. Let f : X → Y be an invertible function.
Also, suppose f has two inverses (say g1 and g2).
Then, for all y ∈ Y, we have
⇒ fog1(y) = Iy(y) = fog2(y)
⇒ f(g1(y)) = f(g2(y))
⇒ g1(y) = g2(y)
[f is invertible ⇒ f is one-one]
⇒ g1 = g2 [g is one-one]
Hence, f has a unique inverse.
11. Consider f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f–1 and show that (f–1)–1 = f.
Sol. Here, function f : {1, 2, 3} → {a, b, c
is given by,
f(1) = a, f(2) = b and f(3) = c
If we define g : {a, b, c} → {1, 2, 3} as g(a) = 1, g(b)
= 2, g(c) = 3, then we have
(fog)(a) = f(g(a)) = f(1) = a
(fog)(b) = f(g(b)) = f(2) = b
(fog)(c) = f(g(c)) = f(3) = c
and (gof)(1) = g(f(1)) = f(a) = 1
(gof)(2) = g(f(2)) = f(b) = 2
(gof)(3) = g(f(3)) = f(c) = 3
Therefore, gof = IX and fog = IY, where X =
{1, 2, 3} and Y = {a, b, c}
Thus, the inverse of f exists and f–1 = g.
Therefore, f–1 : {a, b, c} → {1, 2, 3} is given by,
f–1(a) = 1, f–1(b) = 2, f–1(c) = 3
Let us now find the inverse of f–1 i.e., find the
inverse of g.
If we define h : {1, 2, 3} → {a, b, c} as
h(1) = a, h(2) = b, h(3) = c, then we have
(goh)(1) = g(h(1))= g(a) = 1
(goh)(2) = g(h(2))= g(b) = 2
(goh)(3) = g(h(3))= g(c) = 3
and (hog)(a) = h(g(a)) = h(1) = a
(hog)(b) = h(g(b)) = h(2) = b
(hog)(c) = h(g(c)) = h(3) = c
Therefore, goh = IX and hog = IY, where X = {1, 2, 3} and Y = {a, b, c}.
Thus, the inverse of g exists and g–1 = h ⇒ (f–1)–1 = h.
It can be noted that h = f. Hence, (f–1) = f.
12. Let f : X → Y be an invertible function. Show that the inverse of f–1 is f, i.e., (f–1)–1 = f.
Sol. Let f : X → Y be an invertible function.
Then, there exists a function g : Y → X such that gof = IX and fog = IY.
Here, f–1 = g
Now, gof = IX and fog = IY
Therefore, f–1of = IX and fof–1 = IY
Hence, f–1 : Y → X is invertible and f is the inverseof f–1.
i.e., (f–1)–1 = f.
$$\textbf{13.\space If f} \textbf{: R}\xrightarrow{}\textbf{R}\space\textbf{be given by}\\\textbf{f(x) = (3-x}^{\textbf{3}})^{\frac{\textbf{1}}{\textbf{3}}}\textbf{,}\space \textbf{then fof(x) is :}\\\textbf{(A)\space x}^{\frac{\textbf{1}}{\textbf{3}}}\\\textbf{(B)\space x}^{\textbf{3}}\\\textbf{(C)\space} \textbf{x}\\\textbf{(D)\space}\textbf{(3-x}^{\textbf{3}})$$
Sol. (C) x
$$\text{Here, function}\space\text{f}: \text{R}\xrightarrow{}\text{R}\space\\\text{is given as f(x) = (3-x}^{3})^{\frac{1}{3}}\\\therefore\space\text{fof(x) = f(f(x)) = f((3-x}^{3})^{\frac{1}{3}})\\=\lbrack 3 - ((3-x^{3})^{\frac{1}{3}})^{3}\rbrack^{\frac{1}{3}}\\=\lbrack 3 - (3 - x^{3})\rbrack^{\frac{1}{3}}\\=(x^{3})^{\frac{1}{3}} = x$$
∴ fof(x) = x.
$$\textbf{14.\space Let}\space \textbf{f : R} - \begin{Bmatrix}-\frac{\textbf{4}}{\textbf{3}}\end{Bmatrix}\xrightarrow{}\textbf{R}\space$$
be a function defined as
$$\textbf{f(x)} =\frac{\textbf{4x}}{\textbf{3x+4}}\textbf{.}\space\textbf{The inverse of }\space\\\textbf{f is the map g : Range}\\\textbf{f}\xrightarrow{}\textbf{R} -\begin{Bmatrix}-\frac{\textbf{4}}{\textbf{3}}\end{Bmatrix}\space\textbf{is given by}\\\textbf{(A)\space g(y) =}\frac{\textbf{3y}}{\textbf{3-4y}}\\\textbf{(B)\space}\textbf{g(y) =} \frac{\textbf{4y}}{\textbf{4-3y}}\\\textbf{(C)\space}\textbf{g(y) = }\frac{\textbf{4y}}{\textbf{3-4y}}\\\textbf{(D)\space g(y) =}\frac{\textbf{3y}}{\textbf{4-3y}}\\\textbf{Sol.\space}\text{(B)} \space g(y) =\frac{4y}{4-3y}$$
$$\text{Given that}\space \text{f : R-}\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}\xrightarrow{}\text{R}\space$$
is defined as
$$\text{f(x) = }\frac{4x}{3x + 4}.$$
Let y be an arbitrary element of range f.
$$\text{Then, there exists x} \epsilon \text{R −}\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}$$
such that y = f(x)
$$\Rarr\space y =\frac{4x}{3x+4}\space\\\Rarr\space 3xy + 4y = 4x\\\Rarr\space x(4-3y) = 4y\\\Rarr\space x = \frac{4y}{4-3y}$$
Let us define g : Range
$$f\xrightarrow{} \text{R} -\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}\space\text{as}\space \text{g(y)} =\frac{4y}{4-3y}.$$
$$\text{Now, (gof)(x) = g(f(x)) =}\\g\bigg(\frac{4x}{3x+4}\bigg)\\=\frac{4\bigg(\frac{4x}{3x+4}\bigg)}{4-3\bigg(\frac{4x}{3x+4}\bigg)}\\=\frac{16x}{12x + 16 - 12 x} =\frac{16x}{16} = x$$
$$\text{and (fog)(y) = f(g(y))} =\\ f\bigg(\frac{4y}{4-3y}\bigg)\\=\frac{4\bigg(\frac{4y}{4-3y}\bigg)}{3\bigg(\frac{4y}{4-3y}\bigg) + 4}\\=\frac{16y}{12y +16- 12y}\\=\frac{16y}{16} = y$$
$$\text{Therefore,}\space\text{gof = I}_{\text{R -}\begin{Bmatrix}-\frac{4}{3}\end{Bmatrix}}\\\text{and fog = I}_{R}$$
Thus, g is the inverse of f i.e., f–1 = g.
Hence, the inverse of f is the map g : Range
$$f\xrightarrow{}\text{R} -\bigg(-\frac{4}{3}\bigg),\space\\\text{which is given by g(y)} =\frac{4y}{4-3y}.$$ | 6,032 | 12,684 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-10 | longest | en | 0.700323 |
https://bytexd.com/fp16-vs-fp32-what-do-they-mean-and-whats-the-difference/ | 1,726,753,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00336.warc.gz | 123,921,297 | 98,792 | # FP16 vs FP32 – What Do They Mean and What’s the Difference?
You probably came across the floating-point precision formats FP16 and FP32 in GPU specs or in a deep learning application like when training Stable Diffusion with DreamBooth, but did you ever wonder what they mean? Or why do some applications prefer one over the other?
As we all know, computers understand numbers – and pretty much everything else – expressed in the binary number system. Yet, there is no one formula to encode all types of numbers from one number system to another, especially when it comes to floating point numbers.
If you want to encode integers, there are two things you must worry about: the magnitude (informally, the digits) and the sign of the integer. So, when you hear INT8 – which means 8 bits are used to encode an integer, one bit is used to determine the sign of the integer, and the other seven are used to encode the magnitude of our integer.
Here you can speculate that the only hindrance is the value of the integer; if your integers are big, you need to use a format with a wider range.
When it comes to floating-point numbers, things get a little more interesting; we have three sets of bits to represent a floating-point number: the sign (always takes one bit), the exponent (or the magnitude; this tells how big the number is), and the mantissa (the precision, this tells how precise or how many digits the number has). The difference between floating point number formats is how many bits are devoted to the exponent and how many are devoted to the mantissa.
## FP32
The standard FP32 format is supported by almost any modern Processing Unit, and normally FP32 numbers are referred to as single-precision floating points. This format is used in scientific calculations that don’t require a great emphasis on precision; also, it has been used in AI/DL applications for quite a while.
FP32 precision format bits are divided as follows:
1. 1 bit for the sign of the number.
2. 8 bits for the exponent, or the magnitude.
3. 23 bits for the mantissa or the fraction.
FP32 Range:
Every format has a range of numbers that can be represented with, and with FP32, one can represent numbers of the magnitude of order 10^38, with ~7-9 significant decimal digits.
### When do we use FP32 precision?
1. Any scientific computations which don’t require more than 6 significant decimal digits.
2. In neural networks, this is the default format to represent most network weights, biases, and activation, in short, most parameters.
### Software and Hardware Compatibility
FP32 is supported by any CPU and GPU used nowadays; it is represented in popular programming languages by the float type, such as in C and C++. You can also use it in TensorFlow and PyTorch as `tf.float32` and `torch.float/torch.float32` respectively.
## FP16
In contrast to FP32, and as the number 16 suggests, a number represented by FP16 format is called a half-precision floating point number.
FP16 is mainly used in DL applications as of late because FP16 takes half the memory, and theoretically, it takes less time in calculations than FP32. This comes with a significant loss in the range that FP16 covers and the precision it can actually hold.
FP16 precision format bits are divided as follows:
1. 1 bit for the sign, as always.
2. 5 bits for the exponent or the magnitude.
3. 10 bits for the precision or the fraction.
Range:
The representable range for FP16 is in the order of ~10^-8 to ~65504 with 4 significant decimal digits.
### When to Use FP16?
It is mostly used in deep learning applications, where the needed range of numbers is relatively small, also the is no demand for precision.
### Software and Hardware Compatibility
FP16 is supported by a handful of modern GPUs; because there is a move to use FP16 instead of FP32 in most DL applications, also FP16 is supported by TensorFlow by using the type `tf.float16` and in PyTorch by using the type `torch.float16` or `torch.half`.
In other programming languages, the type short float is often used to encode a half-precision floating point number.
## FP16 vs FP32
Both of these formats are best suited for their usages, but there are some points to take into account when you want to choose one of these:
### Range
The range is essential for choosing which format to use; for example, if you are working with integers of order a few 1000s, it is not reasonable to use INT8; because its range is bounded by 255.
INT32 isn’t reasonable either; because you are not going to use many of these bits, and this is a waste of memory.
### Precision
Naturally, precision increases with the bits increment. This means if you need precise results, you should use formats with more precision bits, but this will increase space and time requirements of the calculations.
Using FP16 instead of FP32 in deep learning proved helpful in decreasing the time and space needed for training the models without much loss in the performance of these models.
This transition prevents overfitting to some extent; if the models’ parameters are highly adjustable, this opens a window for overfitting to your training data.
In contrast, FP16 opens a tiny window for overflow and underflow, where you try to compute numbers out of the representable range. Or with unnoticeable differences with regard to this format.
The caveat in DL networks is that the range matters but not the precision, which lead to the invention of BFLOAT16 – short for Google’s Brain float 16. Just an FP32 with it’s precision truncated to leave it with 16 bits.
BFLOAT16 combines the best of both worlds; it has the range of FP32 by using 8 bits as the exponent and 7 bits as the precision part. This makes it possible to represent the whole range of FP32 with BFLOAT16, but with little precision. i.e., you can compare two numbers with a meaningful difference in magnitude, but the same can’t be said for two close numbers (underflow), which isn’t a big issue in DL applications.
A great example to mentalize the difference which is related to computer vision; consider a robotic hand that helps clean valuable pieces; precision is essential in this case. This is opposed to another hand that helps in cutting metals in a factory, which requires fast production rate.
In the example with the first hand, it is plausible to FP32 instead of FP16, which is more suitable to the nature of the industrial-level metal cutting machine!
## Conclusion
In this article, we discussed FP16 and FP32, and we compared them with each other; we knew that if we favor speed over precision, we should use a format with fewer bits and vice versa.
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# hw46 - 4.6 Pages 269-270 1 3(trapezoidal parametric and...
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4.6 Pages 269-270 # 1, 3 (trapezoidal, parametric, and exact), 11, 13, 15, 27, 28, 29 2 1 3 –1 ( ) ; 0.25 8 f x h x = = = 0 1.00 x = 0 ( ) 1 f x = 1 1.25 x = 1 ( ) 0.64 f x = 2 1.50 x = 2 ( ) 0.4444 f x 3 1.75 x = 3 ( ) 0.3265 f x 4 2.00 x = 4 ( ) 0.25 f x = 5 2.25 x = 5 ( ) 0.1975 f x 6 2.50 x = 6 ( ) 0.16 f x = 7 2.75 x = 7 ( ) 0.1322 f x 8 3.00 x = 8 ( ) 0.1111 f x 1. Trapezoidal Rule: 3 0 1 7 8 2 1 1 0.25 [ ( ) 2 ( ) 2 ( ) ( )] 0.6766 2 dx f x f x f x f x x + +…+ + Parabolic Rule: 3 0 1 2 7 8 2 1 1 0.25 [ ( ) 4 ( ) 2 ( ) 4 ( ) ( )] 0.6671 3 dx f x f x f x f x f x x + + +…+ + Fundamental Theorem of Calculus: 3 3 2 1 1 1 1 1 2 1 0.6667 3 3 dx x x = = + = 2 – 0 ( ) ; 0.25 8 f x x h = = = 0 0.00 x = 0 ( ) 0 f x = 1 0.25 x = 1 ( ) 0.5 f x = 2 0.50 x = 2 ( ) 0.7071 f x 3 0.75 x = 3 ( ) 0.8660 f x 4 1.00 x = 4 ( ) 1 f x = 5 1.25 x = 5 ( ) 1.1180 f x 6 1.50 x = 6 ( ) 1.2247 f x 7 1.75 x = 7 ( ) 1.3229 f x 8 2.00 x = 8 ( ) 1.4142 f x 3. Trapezoidal Rule: 2 0 1 7 8 0 0.25 [ ( ) 2 ( ) 2 ( ) ( )] 1.8615 2 xdx f x f x f x f x + +…+ + Parabolic Rule: 2 1 2 3 7 8 0 0.25 [ ( ) 4 ( ) 2 ( ) 4 ( ) ( )] 1.8755 3 xdx f x f x f x f x f x + + +…+ + Fundamental Theorem of Calculus: 2 2 3/2 0 0 2 4 2 1.8856 3 3 xdx x = = 11. ( ) ( ) 2 3 1 2 ; f x f x x x ′′ = − = The largest that ( ) f c ′′ can be on [ ] 1,3 occurs when 1 c = , and ( ) 1 2 f ′′ = ( ) ( ) 3 2 3 1 400 2 0.01; 3 12 n n Round up: n = 12
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3 0 1 11 12 1 1 0.167 [ ( ) 2 ( ) 2 ( ) ( )] 1.1007 2 dx f x f x f x f x x + + + + " 13. ( ) ( ) 3/ 2 1 1 ; 2 4 f x f x x x ′′ = = − The largest that ( ) f c ′′ can be on [ ] 1,4 occurs when 1 c = , and ( ) 1 1 4 f ′′ = . ( ) 3 2 4 1 1 900 0.01; 4 16 12 n n Round up: n = 8 4 0 1 7 8 1 0.375 [ ( ) 2 ( ) 2 ( ) ( )] 4.6637 2 x dx f x f x f x
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hw46 - 4.6 Pages 269-270 1 3(trapezoidal parametric and...
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Ask a homework question - tutors are online | 1,230 | 2,343 | {"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} | 3.703125 | 4 | CC-MAIN-2018-13 | latest | en | 0.569791 |
http://math.stackexchange.com/questions/179064/good-exercises-to-do-examples-to-illustrate-seifert-van-kampen-theorem?answertab=oldest | 1,394,873,281,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678696864/warc/CC-MAIN-20140313024456-00004-ip-10-183-142-35.ec2.internal.warc.gz | 89,700,821 | 23,046 | # Good exercises to do/examples to illustrate Seifert - Van Kampen Theorem
I have just learned about the Seifert-Van Kampen theorem and I find it hard to get my head around. The version of this theorem that I know is the following (given in Hatcher):
If $X$ is the union of path - connected open sets $A_\alpha$ each containing the basepoint $x_0 \in X$ and if each intersection $A_\alpha \cap A_\beta$ is path connected, then the homomorphism $$\Phi:\ast_\alpha \pi_1(A_\alpha) \to \pi_1(X)$$ is surjective. If in addition each intersection triple intersection $A_\alpha \cap A_\beta \cap A_\gamma$ is path-connected, then $\ker \Phi$ is the normal subgroup $N$ generated by all elements of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega^{-1})$, and so $\Phi$ induces an isomorphism $$\pi_1(X) \cong \ast_\alpha \pi_1(A_\alpha)/N.$$
$i_{\alpha\beta}$ is the homomorphism $\pi_1(A_\alpha \cap A_\beta) \to \pi_1(A_\alpha)$ induced from the inclusion $A_\alpha \cap A_\beta \hookrightarrow A_\alpha$ and $\omega$ is an element of $\pi_1(A_\alpha \cap A_\beta)$.
Now I tried to get my head round this theorem by trying to understand the example in Hatcher on the computation of the fundamental group of a wedge sum. Suppose for the moment we look at $X = X_1 \vee X_2$. I cannot just apply the theorem blindly because $X_i$ is not open in $X$. So we need to look at
$$A_1 = X_1 \vee W_2, \hspace{3mm} A_2 = X_2 \vee W_1$$
where $W_i$ is a neighbourhood about the basepoint $x_1$ of $X_1$ that deformation retracts onto $\{x_1\}$, similarly for $W_2$. I believe each of these is open in $X_1 \vee X_2$ because each $A_i$ is the union of equivalence classes that is open in $X_1 \sqcup X_2$. Now how do I see rigorously that $A_1 \cap A_2$ deformation retracts onto the point $p_0$ (that I got from identifying $x_1 \sim x_2$) in $X$? If I can see that, then I know by Proposition 1.17 (Hatcher) that
$$\pi_1(A_1 \cap A_2) \cong \pi_1(p_0) \cong 0$$
from which it follows that $N= 0$ and the Seifert-Van Kampen Theorem tells me that
$$\pi_1(X_1\vee X_2) \cong \pi_1(X_1) \ast \pi_1(X_2).$$
1) Is my understanding of this correct?
2) What other useful exercises/ examples/applications are there to illustrate the power of the Seifert-Van Kampen Theorem? I have also seen that you can use it to prove that $\pi_1(S^n) = 0$ for $n \geq 2$.
I have had a look at the examples section of Hatcher after the proof the theorem, but unfortunately I don't get much out of it. The only example I sort of got was the computation of $\pi_1(\Bbb{R}^3 - S^1)$.
I would appreciate it very much if I could see some other examples to illustrate this theorem. In particular, I heard that you can use it to compute group presentations for the fundamental group - it would be good if I could see examples like that.
Thanks.
Edit: Is there a way to rigorously prove that $A_1 \cap A_2$ deformation retracts onto the point $\{p_0\}$?
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This could be useful. Have a look at example 13.14 and 13.19. – Matt N. Aug 5 '12 at 11:10
You didn't define what $W_1$ and $W_2$ are. I imagine Hatcher wants you to assume that $W_1$ and $W_2$ are both nice enough for the required deformation retract to exist. For example, if both $X_1$ and $X_2$ are manifolds, then such neighbourhoods exist because a manifold is locally euclidean, and so locally convex. – Zhen Lin Aug 5 '12 at 11:16
@ZhenLin Yes I will put that in. – fpqc Aug 5 '12 at 11:17
@ZhenLin How do I prove that $A_1 \cap A_2$ deformation retracts onto $p_0$? – fpqc Aug 5 '12 at 11:18
@BenjaLim: You should check out Lee's "Introduction to Topological Manifolds" (2nd edition), Chapter 10, where applications of Seifert-van Kampen to the fundamental groups of CW complexes and compact surfaces are discussed in rigorous detail. – Jesse Madnick Aug 6 '12 at 2:28
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A more powerful version of the theorem is in terms of groupoids using the notion of the fundamental groupoid $\pi_1(X,A)$ on a set $A$ of base points. This theory is explained in the book "Topology and Groupoids" advertised here. For example, to capture the fundamental group of the circle from the theorem you need two base points. Since the circle is THE basic example in algebraic topology, it is a bit of an anomaly to give a theorem which does not compute this example. So one gets more powerful theorems with hardly any further complication in the proofs, which I thought was a good idea in the 1960s, when the first edition of this book was published. An MAA review of the book is given here.
The more general version of the theorem is also used in the book to give a proof of the Jordan Curve Theorem, via a nice property called the Phragmen-Brouwer Property: the circle does not have this property. (Added later: Take two distinct points $d,e$ on the circle $C$: then $C \backslash \{d\}, C \backslash \{e\}$ are both connected, but $C \backslash (\{d\} \cup \{e\})$ is not connected. The general property involves disjoint closed subspaces $D,E$.)
The groupoid approach also has advantages in considering covering spaces, and orbit spaces.
@user38268: Edit 2013: I'd just like to mention how the fundamental group(oid) of a graph is calculated from a groupoid point of view. So how do we obtain a graph? Take a disjoint union say $Q$ of directed edges. These have a set $Q_0$ of end points. Then a graph $\Gamma$ is obtained by identifying these end points in some way, i.e. by a function $f: Q_0 \to V$, say. Now the fundamental groupoid of $Q$ is the disjoint union $G$ of copies of the "unit interval groupoid" $\mathcal I$, again with object set $Q_0$. In groupoid theory we can start with $G$ and $f$ and produce a new groupoid which in T&G is written $U_f(G)$ with object set $E$, and a good universal property. On p. 343 of T&G, a deduction from the groupoid van Kampen Theorem implies that $U_f(G)$ is the fundamental groupoid of $\Gamma$ on the set of vertices. But this also the free groupoid on the graph $\Gamma$.
This may not be immediately comprehensible, but it shows how a groupoid viewpoint avoids choosing base points and a maximal tree, and simply says that the free groupoid on a graph is obtained analogously to the way the graph itself is obtained by identification of vertices, but in the category of groupoids rather in the category of graphs. See also the book by Higgins Categories and Groupoids.
The emphasis on fundamental groups suggests one would describe a railway timetable in terms of return journeys and change of starting point of the return journeys. There is a more convenient way! Higgins' book was published in 1971.
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Since it happens that I am quite familiar with the corresponding chapter of Hatcher's monograph, the answers for your two questions are:
1) As I understand, $A_1 \cap A_2$ is just $W_1 \vee W_2$ and the requirement is that of being able to "paste" two deformation retracts; however, these may be incompatible (if they are arbitrary). Yet, Hatcher only cares about wedge sums of spheres (or manifolds) if he applies this example; the fact that compatible pairs of retractions do exist is enough in this case.
2) Hatcher is trying to motivate the theorem by giving a few examples with a geometrical "flavor", in which the actual deformation retracts are, in fact, difficult to write down (or even to visualize, as in "Linking Circles"). This is just one peculiarity of his writing, making it difficult at first sight. You're not the first to complain about his style, yet this work is anything but a "light read" (especially deeper into it).
You could read at first the actual theorems: given as a "proposition", 1.26 is computationally the main result of the section, helping you to do what you asked for (computing a group presentation), and also suggesting how to approach the construction of $K(G, 1)$ spaces.
Again, you may reconsider stating the elementary version (only two path-connected components) in terms of the "amalgamated" free product. At least personally, viewing the construction as a solution to a universal mapping problem (pushouts in Grp are exactly that) helped me clear a few confusions in the preamble of Hatcher's exposition about SVK.
Added: to answer the title of the topic, section 1.2 of the book is full with such examples and applications (following the actual proof), and the exercises in the end are non-trivial ones.
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We use the Seifert Van-Kampen Theorem to calculate the fundamental group of a connected graph. This is Hatcher Problem 1.2.5:
It is a fact in graph theory that any connected graph $X$ contains a maximal tree $M$, namely a contractible graph that contains all the vertices of $X$. Now if the maximal tree $M = X$, then we are done because for any $x_0 \in M$, $\pi_1(M,x_0) = \pi_1(X,x_0) = 0$ that is trivially free. Now suppose $M \neq X$. Then there is an edge $e_i$ of $X$ not in $M$. Observe that for each edge $e_i$ we get a loop going in $M \cup e_i$ about some point $x_0 \in M$. Now fix out basepoint $x_0$ to be in $M$ and suppose that the edges not in $M$ are $e_1,\ldots,e_n$. Then it is clear that $$X = \bigcup_{i=1}^n \left(M \cup e_i\right).$$
The intersection of any two $M \cup e_i$ and $M \cup e_j$ contains at least $M$ and is path connected, so is the triple intersection of any 3 of these guys by the assumption that $X$ is a connected graph. So for any $x_0 \in M$, the Seifert-Van Kampen Theorem now tells us that
$$\pi_1(X,x_0) \cong \pi_1(M \cup e_1,x_0) \ast \ldots \ast \pi_1(M \cup e_n,x_0)/N$$
where $N$ is the subgroup generated by words of the form $l_{ij}(w)l_{ji}(w)^{-1}$, where $l_{ij}$ is the inclusion map from $\pi_1((M\cup e_i) \cap (M \cup e_j),x_0) = \pi_1(M \cup (e_i \cap e_j),x_0)$. Now observe that if $i \neq j$ then $M \cup (e_i \cap e_j) = M$ and since $\pi_1(M,x_0) = 0$ we conclude that any loop $w \in \pi_1(M \cup (e_i \cap e_j),x_0)$ in here is trivial. If $i = j$, $l_{ij}$ is just the identity so that our generators for $N$ are just
$$l_{ij}(w)l_{ji}(w)^{-1} = ww^{-1} = 1$$
completing our claim that $N$ was trivial. Now for each $i$, we have that $\pi_1(M\cup e_i,x_0)$ is generated by a loop that starts at $x_0$ and goes around the bounded complementary region formed by $M$ and $e_i$ and back to $x_0$ through the maximal tree. Such a path back to $M$ does not go through any other edge $e_j$ for $j$ different from $i$. It follows that $\pi_1(X,x_0)$ is a free group with basis elements consisting of loops about $x_0 \in M$ as described in the line before.
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If you downvoted because my answer is wrong, please tell me. How will I be able to correct the proof if you don't tell me what is wrong? – fpqc Aug 19 '12 at 5:48
I like easy examples, the easier the better, so I will post two easy ones(sorry for being somewhat late):
A note in advance: I think what you want to understand in order to be able to apply Seifert-van Kampen is the free product with amalgamation and group presentations.
1) I cannot answer your question about whether you understand it all correctly but the example you are trying to do is a general case of the figure eight space. And that is the possibly simplest example of application of Mr. Seifert and Mr. van Kampen:
Then $$\pi_1(U) = \pi_1 (V) = \mathbb Z$$ and $$\pi_1 (U \cap V) = 0$$ so that we compute the fundamental group of the figure of eight space as $$\pi_1 (8) = \pi_1(U) \ast_{\langle \langle \pi_1(U \cap V) \rangle \rangle} \pi_1(V) = \pi_1(U) \ast \pi_1(V) = \mathbb Z \ast \mathbb Z$$
2) Here is a more interesting example. We can take any space of which we already know the fundamental group and recompute it using Seifert-van Kampen's theorem. For example the torus:
Pick any base point $x_0$ in $U \cap V$. Let $\alpha$ be a loop in $U$ based at $x_0$ and let $\beta$ be a loop in $V$ based at $x_0$ and $\gamma$ a loop in $U \cap V$ based at $x_0$. Then $$\pi_1(U,x_0) = \langle \alpha \rangle \cong \mathbb Z$$ $$\pi_1(V,x_0) = \langle \beta \rangle \cong \mathbb Z$$ $$\pi_1(U \cap V,x_0) = \langle \gamma \rangle$$
Now the tricky bit is to see what the normal subgroup of $\pi_1(U) \ast \pi_1(V)$ is. We have the inclusions $\varphi: \gamma \mapsto \alpha$ and $\psi: \gamma \mapsto \beta$.
Then we want the normal subgroup generated by the set $S = \{ \varphi(\gamma^k) \psi(\gamma^k)^{-1} \mid k \in \mathbb Z\} = \{ \alpha^k \beta^{-k} \mid k \in \mathbb Z\}$. The smallest normal subgroup containing $S$ looks like $$\langle \langle S \rangle \rangle = \langle \{ g_1 (\alpha \beta^{-1})^{\mp 1} g_1^{-1} g_2 (\alpha \beta^{-1})^{\mp 1} g_2^{-1} \dots g_n (\alpha \beta^{-1})^{\mp 1} g_n^{-1} \mid g_i \in \langle \alpha , \beta \mid \rangle \} \rangle$$
Now we need to find the corresponding relations in a presentation. We know that the answer should be $\alpha \beta = \beta \alpha$ or equivalently, $\alpha \beta \alpha^{-1} \beta^{-1} = 1$. so that we get
$$\pi_1(\mathbb T) = \pi_1(U) \ast_{\langle \langle S \rangle \rangle} \pi_1(V) = \{ \alpha, \beta \mid \alpha \beta^{-1} = \alpha^{-1} \beta = \mathbb Z \times \mathbb Z\}$$ where I looked up that last equality in a table here. But I'll have to think about this some more. (see here for a related thread)
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Mmmmh... donuts... – Matt N. Aug 26 '12 at 7:45
Haha. $U \cap V$ on the donut is wrong! I get two rings. – Matt N. Aug 26 '12 at 9:37
Donut fail. $U \cap V$ is not path-connected hence SvK cannot be applied. Ouch ouch ouch. But thanks for the upvotes. – Matt N. Aug 26 '12 at 9:39
Ben: I will think of a different "nice" example. One that actually works, this time. – Matt N. Aug 26 '12 at 9:40 | 4,032 | 13,577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2014-10 | longest | en | 0.853757 |
https://everything2.com/title/Too+Much+Light+Makes+the+Baby+Go+Blind | 1,511,318,181,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806447.28/warc/CC-MAIN-20171122012409-20171122032409-00079.warc.gz | 608,519,780 | 9,697 | Created by the Neo-Futurists in Chicago in 1988. The show is an ever-changing attempt to perform 30 plays in 60 minutes. The experience of the audience is designed to approximate the chaos of the real world. Here's how it works, in their words:
1. You wake up in the morning. At some point in your day, you decided to see Too Much Light Makes The Baby Go Blind. You manage to get to the theater by show time.
2. You roll a single six-sided die, letting fate determine your admission cost. The number you roll on the die times \$1 plus an additional \$4 is what you are asked to pay. You do so, or they won't let you in.
3. Before you can enter the theater, a Walkman-wearing host in dark glasses asks you for your name. You reply and are given a "Hello My Name Is" tag with an alternate moniker. You slap it on, leaving your previous name at the door.
4. You enter the theater and meet other people. Some of them are performers, some of them aren't. You're not sure which are which.
5. Someone gives you a program and a "menu". The menu is a list of tonight's play titles, numbered 1 through 30. You notice a clothesline strung across the stage, with pieces of paper hanging on it, numbered 1 through 30. Some say they correspond.
6. Someone shouts "When we sell out, we order out!", and, in fact, if there is at least one person for every seat, someone appears on stage with a telephone and calls a pizza place. You are asked what you'd like on it. An extra large is ordered. You wonder if it will feed 66 (or 154 in Chicago).
7. Another person on stage encourages you to order plays by number from your menu. What you and others sitting around you select will determine the random order in which the plays will appear. The performers ask for a play to start with. You say "five", but realize it was drowned out by the shouts of others. Next time you will say it louder.
8. In order to start the darkroom timer on stage, which will tick off the 60 minutes, the audience is led through a countdown that has little to do with numbers. The clock is started and the first play commences with the word "go!". At the end of the play you hear the word "curtain!" and you say "five" again, a little louder than before. You realize that timing, as well as volume, is involved.
9. You are barraged by a random display of tragic, comic, personal, political, experimental two minute plays over the course of the hour. At one point the clock is stopped to let the pizza delivery person fulfill their duty. The pizza smells good.
10. The thirtieth play concludes with a "CURTAIN!", and hopefully the darkroom timer has not yet buzzed. Someone in the audience rolls a die on stage to determine how many new plays will be created for next week. You eat some pizza, you talk to the performers. You go home and sleep. The next day things are different.
Source: 100 Neo-Futurist Plays from Too Much Light Makes The Baby Go Blind (Chicago Plays, 1993)
Gimmicky? You betcha. But as theatre, the show is one of the most exciting dramatic spectacles I've ever seen. "Why doesn't live theatre generate the excitement of live sports?" Professor Keith Johnstone (Impro) wondered, and he went and created TheatreSports. But in my fifteen years of watching improv, rarely have TheatreSports shows ever reach the level of audience frenzy of a professional sporting event... although performances of TMLMTBGB that I've seen (Chicago, circa 1990 and 1993, and once in San Francisco circa 1994) in each case electrified the audience in the room. There was shouting, there was raucous laughter, there was an adrenalin rush in the crowd... and in San Francisco, due to the small performing space, there was an intense connection between each of the performers and the audience. At the end of the show, there was nonstop talking about what we'd just witnessed. It was the kind of show you had to tell other people about too.
As to the plays, they can hit or miss. They are rarely longer than two minutes, so the bad ones are quickly forgotten, but it is surprising how many of these small plays can pack an emotional wallop. Some of the "plays" I saw came across as choral poetry rather than any kind of narrative, but I do not claim to understand the implications of the Neo-Futurist manifesto when it comes to generating material.
The show premiered in Chicago December 2, 1988 at the Stage Left Theatre, and is still running at the Neo-Futurarium. Made its Off-Broadway premiere at the Joseph Papp Public Theatre in New York in 1993.
One of the short plays, "Disregard this Play" by Greg Kotis was extended into a five-minute long computer animated featurette by Chris Landreth. In the movie, a clown terrorizes a man into believing he is a clown by the name of Bingo. The video was featured at a Siggraph as a marketing piece for Alias Wavefront.
It is delightfully dark in a Marathon Man and Brazil kind of way.
"Bingo."
"Excuse me?"
"Bingo! Bingo the clown."
"I'm not Bing-"
"BINGO! BINGO THE CLOWN-O!"
"But I'm not-"
"Hiiiiiii BINGO!"
Log in or register to write something here or to contact authors. | 1,193 | 5,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-47 | latest | en | 0.959377 |
https://mail.python.org/pipermail/python-list/2015-April/688740.html | 1,568,840,717,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573331.86/warc/CC-MAIN-20190918193432-20190918215432-00194.warc.gz | 571,593,141 | 3,243 | # Best search algorithm to find condition within a range
jonas.thornvall at gmail.com jonas.thornvall at gmail.com
Tue Apr 7 16:10:49 CEST 2015
```Den tisdag 7 april 2015 kl. 15:30:36 UTC+2 skrev Dave Angel:
> On 04/07/2015 05:44 AM, jonas.thornvall at gmail.com wrote:
> >
> >
> > I want todo faster baseconversion for very big bases like base 1 000 000, so instead of adding up digits i search it.
>
> For this and most of the following statements: I can almost guess what
> you're trying to say. However, I cannot. No idea why you're adding up
> digits, that sounds like casting out nines. And in base-N, that would
> be casting out (N-1)'s.
>
> What's the it you're trying to search?
>
> How do you know the baseconversion is the bottleneck, if you haven't
> written any Python code yet?
>
>
> >
> > I need the fastest algorithm to find the relation to a decimal number.
>
> What relation would that be? Between what and what?
>
> > Digmult is an instance of base at a digitplace (base^x) what i try to find is the digit for the below condition is true and the loop break.
> >
>
> You haven't defined a class "Base" yet. In fact, I don't see any Python
> code in the whole message.
>
> >
> > *********************************
> > for (digit=0;digit<=base;digit++) {
> > if((digit+1)*digmult>decNumber)break;
> > }
> > *********************************
>
>
> >
> > So i am looking for the digit where following condition true.
> >
> > if((digit)*digmult<decNumber) AND if((digit+1)*digmult>decNumber) then BREAK;
>
> You could try integer divide. That's just something like
> digit = decNumber // digmult
> But if you think hard enough you'd realize that
>
>
> >
> > One could start at half base searching, but then i Think i've read that using 1/3 closing in faster?
> >
> > I Think also i remember that if the search space so big that at least 22 or 23 guesses, needed.A random Oracle may even faster?
> >
> > Just pick up a number and get lucky, is it any truth to that?
> >
> > Below the actual algorithm.
> >
> >
> >
> > <SCRIPT LANGUAGE="Javascript">
> > //CONVERT A DECIMAL NUMBER INTO ANYBASE
> > function newbase(decNumber,base){
> > digits=1;
> > digmult=1;
> > while(digmult*base<=decNumber){
> > digmult=digmult*base
> > digits++;
> > }
> > digsave=digmult;
> > while(decNumber>0 || digits>0){
> > loop=1;
> > digit=0;
> > for (digit=0;digit<=base;digit++) {
> > if((digit+1)*digmult>decNumber)break;
> > }
> > out[digits]=digit;
> > digmult=digmult*digit;
> > decNumber=decNumber-digmult;
> > digsave=digsave/base;
> > digmult=digsave;
> > digits--;
> > }
> > return out;
> > }
> >
> > var out= [];
> > base=256;
> > number=854544;
> > out=newbase(number,base);
> > out.reverse();
> > document.write("Number = ",out,"<BR>");
> > </script>
> >
>
> If that code were in Python, I could be more motivated to critique it.
> The whole algorithm could be much simpler. But perhaps there is some
> limitation of javascript that's crippling the code.
>
> How would you do it if you were converting the base by hand? I
> certainly wouldn't be doing any trial and error. For each pass, I'd
> calculate quotient and remainder, where remainder is the digit, and
> quotient is the next value you work on.
>
>
> --
> DaveA
I am doing it just like i would do it by hand finding the biggest digit first. To do that i need to know nearest base^exp that is less than the actual number. Add up the digit (multiply) it to the nearest smaller multiple. Subtract that number (base^exp*multiple).
Divide / Scale down the exponent with base. And record the digit.
And start looking for next digit doing same manipulation until remainder = 0.
And that is what i am doing.
``` | 1,067 | 3,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.846062 |
https://www.difference.wiki/nautical-mile-vs-mile/ | 1,726,046,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00793.warc.gz | 676,994,579 | 23,680 | # Nautical Mile vs. Mile: What's the Difference?
Edited by Aimie Carlson || By Harlon Moss || Published on January 7, 2024
A nautical mile, measuring 1.852 kilometers, is based on the Earth's curvature and used in maritime and aviation navigation, while a mile, measuring 1.609 kilometers, is a unit of length in the imperial system.
## Key Differences
A nautical mile is a unit of measurement used primarily in maritime and aviation contexts, defined as exactly 1,852 meters or about 6,076.1 feet. A mile, also known as a statutory mile, is a unit of distance in the imperial system, equivalent to 5,280 feet or approximately 1.609 kilometers.
The nautical mile is derived from the division of the Earth's circumference and is used for measuring distances at sea and in the air. The mile, on the other hand, is commonly used in the United States and the United Kingdom for road distances.
One key difference is in their origins: the nautical mile is based on the Earth's spherical geometry, representing one minute of latitude, while the mile has its origins in Roman measurement systems and has been used in various forms over centuries.
In practical terms, the nautical mile is longer than the mile. This difference is particularly important in navigation, where precise measurements are crucial. The mile is more commonly encountered in everyday use, especially in countries using the imperial system.
When converting, one nautical mile equals approximately 1.15078 miles. This conversion is important in activities like aviation where both measurement systems are in use, and accurate conversion is essential for safety and navigation.
## Comparison Chart
### Length
1.852 kilometers (about 6,076.1 feet)
1.609 kilometers (5,280 feet)
### Usage
Maritime and aviation navigation
General distance measurement, especially in the US and UK
### Basis of Measurement
One minute of latitude, Earth's curvature
Roman measurement systems
### Geographic Relevance
Essential in global navigation
Predominantly used in countries with the imperial system
### Conversion
1 nautical mile = 1.15078 miles
1 mile = 0.868976 nautical miles
## Nautical Mile and Mile Definitions
#### Nautical Mile
A length used to measure distances at sea, approximately 6,076 feet.
The island was 50 nautical miles off the coast.
#### Mile
A length unit originating from Roman measurement systems.
The old trail stretches for over ten miles.
#### Nautical Mile
A unit of measurement used in maritime and aviation equal to 1.852 kilometers.
The ship traveled 200 nautical miles north.
#### Mile
A unit of distance in the imperial system equal to 5,280 feet or about 1.609 kilometers.
The park is two miles down the road.
#### Nautical Mile
A maritime measurement unit based on the Earth's curvature.
The lighthouse is visible from 10 nautical miles away.
#### Mile
A customary unit of measure for distance in daily life in certain countries.
The nearest gas station is a mile away.
#### Nautical Mile
A distance measure representing one minute of arc on the Earth's surface.
Pilots use nautical miles for air route calculations.
#### Mile
A standard measure of distance often used for road distances in the US and UK.
He runs five miles every morning.
#### Nautical Mile
A navigational distance unit longer than a standard mile.
The naval exercise covered several hundred nautical miles.
#### Mile
A common unit for measuring distance in various sporting events.
The marathon includes a challenging twenty-six-mile course.
#### Mile
Abbr. mi. or mi A unit of length equal to 5,280 feet or 1,760 yards (1,609 meters), used in the United States and other English-speaking countries. Also called land mile, statute mile. See Table at measurement.
A nautical mile.
## FAQs
#### Why use nautical miles instead of regular miles at sea?
Because nautical miles are based on the Earth's curvature, making them more suitable for navigation.
#### What exactly is a nautical mile?
A unit of measurement used in marine and aerial navigation, equal to 1.852 kilometers.
#### How did the nautical mile originate?
From dividing the Earth's circumference into 360 degrees and each degree into 60 minutes.
#### What is a mile?
A unit of length in the imperial system, equivalent to 5,280 feet or about 1.609 kilometers.
#### Are miles still relevant today?
Yes, especially in countries like the USA for road distances and cultural references.
#### Is the nautical mile universally accepted in navigation?
Yes, it's the standard unit for maritime and aerial distances globally.
#### Who uses the mile for measurements?
Mostly used in the United States and the United Kingdom.
#### Are nautical miles used in space travel?
No, space travel measurements are usually in kilometers or astronomical units.
#### Are miles used in all countries?
No, miles are primarily used in the US and UK, while other countries use kilometers.
#### Do all GPS systems use miles?
GPS systems can use either miles or kilometers, depending on user settings and regional preferences.
#### Can nautical miles and miles be easily converted?
Yes, though the conversion factor (1 nautical mile = 1.15078 miles) should be accurately used.
#### Do runners use miles or nautical miles?
Runners typically use miles or kilometers, not nautical miles.
#### Is a mile longer than a kilometer?
Yes, a mile (1.609 kilometers) is longer than a kilometer.
#### Do mariners need to understand both nautical miles and miles?
Yes, it's useful for mariners to understand both, especially when communicating with land-based systems.
#### How are nautical miles useful in aviation?
They provide a standard, consistent unit for measuring distances between airports and along air routes.
#### Are there different types of miles?
Yes, including the statute mile (common mile), nautical mile, and historical variations like the Roman mile.
#### Are nautical miles different in the Northern and Southern Hemispheres?
No, a nautical mile is a standard unit and remains the same regardless of the hemisphere.
#### How is a nautical mile measured?
It’s based on one minute of arc of latitude along any meridian.
#### Why don’t we use kilometers instead of miles in the US?
Due to historical and cultural reasons, the US has maintained the use of the imperial system.
#### Why is the mile popular in athletics?
It's a traditional distance for track and field events and a common benchmark for endurance. | 1,366 | 6,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.925309 |
https://glencoe.mheducation.com/sites/0078600510/instructor_view0/unit2/webquest_projects.html | 1,713,115,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00062.warc.gz | 264,571,131 | 7,864 | # Physical Science
## Unit 2: Electricity and Energy Resources
### WebQuest Projects
MAG LEV TRAINS: FLOATING LOCOMOTIVES
Introduction
In this WebQuest, students do some Internet research on magnetic levitation and design, build, and test models of magnetic levitation trains. They learn about magnetism, permanent magnets, electromagnets, and superconductors, and discover how the property of magnetism can be used to propel vehicles along a track very quickly with little or no friction. Students also explore different types of magnetic levitation. After researching designs of model mag lev trains on the Internet, students choose one type for the design of their model mag lev trains, and then design, build, and test their model trains. Finally, students prepare a report of some kind to communicate the results of their tests.
Top
Once students have completed their Internet research, they will design, build, and test model magnetic levitation trains. Students will prepare reports on their findings and share their results with their classmates. Reports may be in the form of written reports, oral reports, videotapes of a test run of a model, an actual test performed in the classroom, or any other creative form devised by the students and approved by the teacher.
You may decide to have students work in small groups for this activity. Be sure that students in small groups assign specific tasks to each group member.
Top
Resources
Students will use Internet links given to find out how magnetic levitation is used in transportation today. They will find out about magnetism, how permanent magnets and electromagnets differ, and why superconductors are needed in certain kinds of mag lev train systems. They will identify the two main types of mag lev systems in use at present – electromagnetic suspension (EMS) and electrodynamic suspension (EDS). The EDS system is presently in use in Japan. The EMS system is soon to be in use in Germany. Whatever mag lev system is used, the result is an extremely fast, smooth ride because the train literally floats above the track.
All mag lev systems use the principles of magnetism and polarity to push and pull trains along their tracks. Students may need to provide the initial "push" to start their trains during testing. This "push" may be in the form of an electrical push, students physically pushing the model, or students pulling it along the track in some fashion.
Students will also need to collect materials needed to build their models. You may decide to provide materials such as strong magnets, copper wire, batteries, large nails, magnetic strips, polystyrene blocks (the vehicles), duct tape, and glue.
Top
Time
1 class period for Internet research; 1 class period for presentations; additional time at home to build and test models and prepare reports.
Top
Process
As students progress through the list of web sites, you may help them to focus on what they need to know in order to design and build a model mag lev train. Many of the Internet links listed have directions for building such vehicles, or ask questions of the students to help them come up with their own plans. Allow students to explore these sites and any others they find in the course of their research. Most students will decide to build a mag lev model based on permanent magnets. If any students wish to make and use electromagnets, you may provide additional classroom resources for this purpose.
Top
Evaluation
Because students have been given the option of preparing different types of reports, your evaluation will be open-ended as well. You may want to use the following rubric to evaluate students’ reports on the testing of their model mag lev trains.
Mag Lev Trains: Floating Locomotives
Presentation Rubric Possible Points* Self- Assessment TeacherAssessment Topic was covered in depth 10 Presentation was well planned and coherent 10 Explanations and reasons were given for conclusions 10 Communication aids were clear and useful 10 Test shown was successful/clear explanation was given if test was unsuccessful 10 Total Possible Points 50
*Rate each category according to the following scale: Excellent – 9-10 points; Very Good – 7-8 points; Good – 5-6 points; Satisfactory – 3-4 points; Poor – 1-2 points; and Unsatisfactory – 0 points.
Top
Conclusion
Using information gathered from the Internet and materials they collect (or you provide), students should be able to design, build, and test models of mag lev trains. In the testing process, students should be able to identify the deficiencies of their models. Students should make conclusions as to whether magnetic levitation is a viable option for transportation needs in the future. | 956 | 4,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.938415 |
http://www.wisegeek.com/what-is-analysis-of-variance.htm | 1,444,186,097,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736680773.55/warc/CC-MAIN-20151001215800-00071-ip-10-137-6-227.ec2.internal.warc.gz | 1,117,273,528 | 18,680 | Category:
# What Is Analysis of Variance?
The standard deviation is more frequently used than variance because it is more intuitive and shares the same units as the mean.
Article Details
• Written By: A. Reed
• Edited By: E. E. Hubbard
2003-2015
Conjecture Corporation
J.D. Salinger was carrying six chapters of "The Catcher in the Rye" when he landed on Omaha Beach during WWII. more...
October 6 , 2004 : A report was released declaring no evidence of weapons of mass destruction were found in Iraq. more...
wiseGEEK Slideshows
When doing research, sometimes it becomes necessary to analyze data comparing more than two samples or groups. A type of inferential statistics test, analysis of variance (ANOVA), permits examination of several samples at the same time for purposes of determining whether a significant relationship exists between them. Reasoning is identical to t-tests, only analysis of variance includes independent variables of two or more samples. Differences between samples as well as the difference within one sample is determined. ANOVA is based upon four assumptions: the level of measurement, the sampling method, the distribution of the population, and the homogeneity of the variance.
In order to determine whether differences are significant, ANOVA is concerned with differences between and within the samples, which is referred to as the variance. The ANOVA can find out if the variance is larger between samples as compared to that among sample members. If this is found to be true, then the differences are considered to be significant.
Conducting an ANOVA test involves acceptance of certain assumptions. The first is that the independent random sampling method is used and the choice of sample members from a single population does not influence the choice of members from later populations. Dependent variables are measured primarily at the interval-ratio level; however, it is possible to apply the analysis of variance to ordinal-level measurements. One can assume that the population is normally distributed, even though this is not verifiable, and population variances are the same, which means that the populations are homogeneous.
The research hypothesis assumes that at least one mean is different from the others, but the different means are not identified as larger or smaller. Only the fact that a difference exists is predicted. The ANOVA tests for the null hypothesis, which means that there is no difference among all of the mean values, such that A = B = C. This requires setting the alpha, referring to the probability level where the null hypothesis will be rejected.
F-ratio is a test statistic used specifically for analysis of variance, as the F score shows where the area of rejection for the null hypothesis begins. Developed by statistician Ronald Fisher, the formula for F is as follows: F = between group variance estimate (MSB) divided by the within group variance estimate (MSW), such that F = MSB/MSW. Each of the variance estimates consists of two parts — the sum of squares (SSB and SSW) and degrees of freedom (df). Using the Statistical Tables for Biological, Agricultural and Medical Research, the alpha can be set and based upon this, and the null hypothesis of no difference can be rejected. It can be concluded that a significant difference exists between all of the groups, if that is the case. | 681 | 3,377 | {"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} | 3.5625 | 4 | CC-MAIN-2015-40 | longest | en | 0.957411 |
https://fastmat.readthedocs.io/en/latest/algorithms/OMP.html | 1,638,009,611,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358180.42/warc/CC-MAIN-20211127103444-20211127133444-00441.warc.gz | 337,407,295 | 6,951 | # OMP Algorithm¶
class fastmat.algorithms.OMP
Bases: fastmat.algorithms.Algorithm.Algorithm
Orthogonal Matching Pursuit
Definition and Interface: For a given matrix $$A \in \mathbb{C}^{m \times N}$$ with $$m \ll N$$ and a vector $$b \in \mathbb{C}^m$$ we approximately solve
$\min\limits_{ x \in \mathbb{C}^N} \Vert x \Vert_0 \quad\mathrm{s.t.}\quad A \cdot x = x.$
If it holds that $$b = A \cdot x_0$$ for some $$k$$-sparse $$x_0$$ and $$k$$ is low enough, we can recover $$x_0$$ via OMP [1].
This type of problem as the one described above occurs in Compressed Sensing and Sparse Signal Recovery, where signals are approximated by sparse representations.
>>> # import the packages
>>> import numpy.linalg as npl
>>> import numpy as np
>>> import fastmat as fm
>>> import fastmat.algorithms as fma
>>> # define the dimensions
>>> # and the sparsity
>>> n, k = 512, 3
>>> # define the sampling positions
>>> t = np.linspace(0, 20 * np.pi, n)
>>> # construct the convolution matrix
>>> c = np.cos(2 * t)
>>> C = fm.Circulant(c)
>>> # create the ground truth
>>> x = np.zeros(n)
>>> x[np.random.choice(range(n), k, replace=0)] = 1
>>> b = C * x
>>> # reconstruct it
>>> omp = fma.OMP(C, numMaxSteps=100)
>>> y = omp.process(b)
>>> # test if they are close in the
>>> # domain of C
>>> print(npl.norm(C * y - b))
We describe a sparse deconvolution problem, where the signal is in $$\mathbb{R}^{512}$$ and consists of $$3$$ windowed cosine pulses of the form $$c$$ with circulant displacement. Then we take the convolution and try to recover the location of the pulses using the OMP algorithm.
Note
The algorithm exploits two mathematical shortcuts. First it obviously uses the fast transform of the involved system matrix during the correlation step and second it uses a method to calculate the pseudo inverse after a rank-$$1$$ update of the matrix.
Todo
• optimize einsum-stuff
[1] S. G. Mallat, Z. Zhang, “Matching pursuits with time-frequency dictionaries”, IEEE Transactions on Signal Processing, vol. 41, no. 12, pp. 3397-3415, Dec 1993
Parameters: fmatA : fm.Matrix the system matrix arrB : np.ndarray the measurement vector numMaxSteps : int the desired sparsity order np.ndarray solution array
__init__(*args, **kwargs)
Initialize self. See help(type(self)) for accurate signature.
arrA
arrB
arrC
arrResidual
arrSupport
arrX
arrXtmp
fmatA
fmatC
matPinv
newCols
newIndex
numL
numM
numMaxSteps
numN
numStep
v2
v2n
v2y | 702 | 2,438 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-49 | latest | en | 0.763799 |
http://centerofmathematics.blogspot.co.uk/2017/01/using-math-to-create-something-beautiful.html | 1,527,427,971,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868316.29/warc/CC-MAIN-20180527131037-20180527151037-00007.warc.gz | 50,640,444 | 20,957 | # The Center of Math Blog
DO the math, DON'T overpay. We make high quality, low-cost math resources a reality.
## Friday, January 27, 2017
### Using Math to Create Something Beautiful
Think back to when you were first introduced to functions, thin lines depicting a single value output for each input in a domain.
A Function
For many people, the idea of what a function ‘looks’ like does not change much from this bland depiction of data. However, data can be crafted into something that carries much more information than just inputs and outputs, and in the right hands an enormous and messy set of data can be presented in a powerful way. Indeed, data representation is an important part of any scientific field.
Compacting more data into inputs and outputs provides not only more information, but also a more stunning visualization of data. In a vector field, a single point can contain information about location, strength and direction of a force. The more information a function tracks, the more stunning the display becomes, with 4 or even 5 dimensions represented on a graph of three-dimensional space and color.
Vectors depicting the strength and direction of a magnetic field at discrete points.
A 3D graph, with a 4th color dimension
Vector fields can even represent information that cannot be easily compiled into a simple function, which allows for out-of-the-ordinary occurrences in nature to be studied more carefully.
With developments in technologies that offer efficient data manipulation, the possibilities of what we can do with functions and data are more far-reaching than ever. Anne M. Burns of Long Island University Uses computers to create beautiful representations of functions.
Burns plots complex valued functions as a vector field, seen here.
The advantages of this technique transcend aesthetic purpose, and can be used to find roots of functions at a glance.
Attributing more dimensions to an occurrence is useful and can be beautiful, but what if the object or function in question is impossible to make sense of as it is? It is often handy to project or unfold an N-dimensional surface onto an (N-1)-dimensional surface. Most of the time, in calculus, a three-dimensional surface will be looked at as a two-dimensional projection on the xy, yz, or xz plane in order to set up an integral to find the volume of the object. In theoretical physics, this technique of reducing the dimension of mysterious happenings is used to speculate the nature of the universe. A common example, and perhaps the most accessible way to think of this process is the unfolding of a four-dimensional cube, the tesseract.
The nets of a 3D cube and a 4D hypercube above.
Dali's Corpus Hypercubus (1954)
This way of thinking about higher dimensions caught more than just the eyes of mathematicians and scientists. Salvador Dali, the great surrealist painter, was fascinated by the advances in science during the twentieth-century. In the 1950’s Dali was fascinated by nuclear physics and quantum mechanics, and found inspiration for many paintings in mathematics.
-->
At its core, mathematics does not only seek knowledge, but also pursues beauty in the natural world.
Works Cited
• The Function graphic was found on the page of Maret School's BC calculus page, and is spliced with Charlie Brown of Peanuts, created by Charles M. Schulz.
• The Magnetic field graphic was found on Vassar College's Wordpress blog, under a lecture by Prof. Magnes.
• 4D graph curtesy of user Blue7 on math.stackexchange.
• Find all of Anne M. Burn's Work here.
• The Cube net image was found here
Any unwanted images in this article will be removed at the request of the owner. | 768 | 3,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-22 | latest | en | 0.917933 |
https://community.oracle.com/thread/2102593?tstart=103365 | 1,438,110,311,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042982013.25/warc/CC-MAIN-20150728002302-00169-ip-10-236-191-2.ec2.internal.warc.gz | 866,298,375 | 22,232 | 7 Replies Latest reply: Mar 15, 2007 7:24 PM by 807606
# returning boolean in a recursive method - binary search tree
Ok, I know that this won't work as it is because java can't reason about the program properly and doesn't think that this will always return a boolean: it will though! This method is recursing through a binary search tree and it will always either find the "word" it is looking for, else cur will equal null. This means that the method will terminate and return a boolean. However java can't spot this and I wouldn't expect it to tbh. If anyone can think of a way to make this work, then please tell me :D
By the way this is my first post here, and I hope I can start to help others soon!
``````
/** Recursively searches the tree to see if the word is present */
private boolean presentRec(String word, HashTableNode cur) {
countNodes++; // increase the number of times a node has been compared
if (cur == null) {
// if we have reached the bottom of the tree and the word is not
// found then return false
return false;
} else if (cur.getWord().equals(word)) {
// if we have found the word, then it is in the dictionary so return
// true
return true;
} else if (cur.getWord().compareTo(word) < 0) {
// if the current node is less than the word, then go to the right
presentRec(word, cur.right);
} else {
// otherwise the current node must be greater than the word so we
// need to go to the left
presentRec(word, cur.left);
}
}``````
Message was edited by:
oliverj4455
Message was edited by:
oliverj4455
• ###### 1. Re: returning boolean in a recursive method - binary search tree
Ok, I know that this won't work as it is because java
can't reason about the program properly and doesn't
think that this will always return a boolean: it will
though!
No, it won't.
Your final else doesn't return anything.
Edit: Nor does the else if right before it.
Message was edited by:
jverd
• ###### 2. Re: returning boolean in a recursive method - binary search tree
Ok I know that, but my final else has a recursive call back to the same method, therefore it will actually always return something in the end, always! I just wanted to know if there was any way of telling java this :D or if there is any way of keeping track of the boolean somewhere then returning it right at the end or something. Btw. don't really want to use a global variable for this, but I suppose I could.
• ###### 3. Re: returning boolean in a recursive method - binary search tree
Ok I know that, but my final else has a recursive
call back to the same method, therefore it will
actually always return something in the end, always!
No, it won't.
Your else calls a method. That method returns something. Then you're back to right after invoking that method.
``````boolean foo(int x) {
if (x = 0) {
return 0;
}
else {
foo (x - 1);
// After foo(x - 1), we are here. We don't automagically return
// what the previous method call returned. The fact
// that that method was also foo is irrelevant.
}
} ``````
The fact that that method happens to be the same method you're in now is irrelevant. The multiple recursive invocations are independent. The fact that a deeper call returns something doesn't automagically make all the prior invocations return that thing.
• ###### 4. Re: returning boolean in a recursive method - binary search tree
Ok I feel totally stupid, thanks for pointing that out for me :D. I just forgot that the return didn't just exit all the methods. For anyone who was interested here is my working code:
`````` /** Recursively searches the tree to see if the word is present */
private boolean presentRec(String word, HashTableNode cur) {
countNodes++; // increase the number of times a node has been compared
if (cur == null) {
// if we have reached the bottom of the tree and the word is not
// found then return false
return false;
} else if (cur.getWord().equals(word)) {
// if we have found the word, then it is in the dictionary so return
// true
return true;
} else if (cur.getWord().compareTo(word) < 0) {
// if the current node is less than the word, then go to the right
return presentRec(word, cur.right);
} else {
// otherwise the current node must be greater than the word so we
// need to go to the left
return presentRec(word, cur.left);
}
}``````
• ###### 5. Re: returning boolean in a recursive method - binary search tree
btw i like the word automagically :D
• ###### 6. Re: returning boolean in a recursive method - binary search tree
Good work!
On the lightbulb coming on, that is. I haven't actually read your code, so no comments there. :-)
• ###### 7. Re: returning boolean in a recursive method - binary search tree
I am writing a spell checking program, this is part of a hash table that contains the dictionary. Thought I would use a binary search tree because it is faster than an ordinary linked list for the bucket, well i have finished now, and it is quite efficient. | 1,198 | 4,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2015-32 | latest | en | 0.926779 |
https://testbook.com/question-answer/a-boy-goes-to-his-school-from-his-house-at-a-speed--5f1812f15d78ac0d109534f6 | 1,631,907,136,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00327.warc.gz | 617,949,304 | 31,630 | # A boy goes to his school from his house at a speed of 5 km/hr and returns at a speed of 3 km/hr. If he takes 6 hours in going and coming, the distance between his house and school is:
This question was previously asked in
DMRC CRA Paper I Previous Paper 7 (Held On: 22 Feb 2017 Shift 1)
View all DMRC CRA Papers >
1. 11 km
2. 11.5 km
3. 11.75 km
4. 11.25 km
Option 4 : 11.25 km
Free
Cell
268228
10 Questions 10 Marks 7 Mins
## Detailed Solution
Given,
A boy goes to his school from his house with the speed of = 5 km/hr
The boy return from his school to his house with the speed of = 3 km/hr
Formula:
Speed = distance/time
Calculation:
Let the distance between school and house be x km, then
According to the question
x/5 + x/3 = 6
⇒ (5x + 3x)/15 = 6
⇒ 8x/15 = 6
⇒ x = 6 × (15/8)
⇒ x = 11.25 km
∴ The distance is 11.25 km. | 287 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-39 | latest | en | 0.965933 |
https://nl.mathworks.com/matlabcentral/cody/problems/933-ordinal-numbers/solutions/1804770 | 1,579,789,333,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610919.33/warc/CC-MAIN-20200123131001-20200123160001-00117.warc.gz | 577,419,504 | 15,750 | Cody
# Problem 933. Ordinal numbers
Solution 1804770
Submitted on 3 May 2019 by Joe Blomer
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
assert(isequal(ord(1),'1st'))
2 Pass
assert(isequal(ord(2),'2nd'))
3 Pass
assert(isequal(ord(3),'3rd'))
4 Pass
assert(isequal(ord(4),'4th'))
5 Pass
assert(isequal(ord(5),'5th'))
6 Pass
assert(isequal(ord(10),'10th'))
7 Pass
assert(isequal(ord(11),'11th'))
8 Pass
assert(isequal(ord(12),'12th'))
9 Pass
assert(isequal(ord(13),'13th'))
10 Pass
assert(isequal(ord(14),'14th'))
11 Pass
assert(isequal(ord(15),'15th'))
12 Pass
assert(isequal(ord(82),'82nd'))
13 Pass
assert(isequal(ord(126),'126th'))
14 Pass
assert(isequal(ord(911),'911th'))
15 Pass
assert(isequal(ord(2012),'2012th'))
16 Pass
assert(isequal(ord(4077),'4077th'))
17 Pass
assert(isequal(ord(0),'0th'))
18 Pass
assert(isequal(ord(-101),'-101st')) | 341 | 1,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-05 | latest | en | 0.419436 |
https://brilliantessays.org/manufacturing/ | 1,653,361,520,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00557.warc.gz | 190,691,664 | 14,825 | Select Page
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## Manufacturing
### Project Description:
You have recently become the CFO for Beta Manufacturing, a small cap company that produces auto parts. As you step into your new position, you have decided to compile a report that details all aspects of the business, including: employee tax withholding, facility management, sales data, and product inventory. To complete the task, you will duplicate existing formatting, utilize various conditional logic functions, complete an amortization table with financial functions, visualize data with PivotTables, and lastly import data from another source.
### Steps to Perform:
Step
Instructions
Points Possible
1
Start Excel. Open eApp_Cap2_Manufacturing.xlsx and save the workbook as ACC2343_Final_Manufacturing_LastFirst.
2
Group all the worksheets in the workbook and fill the range A1:F1 from the Insurance worksheet across all worksheets maintaining the formatting. Ungroup the worksheets after the fill is complete and ensure the Insurance worksheet is active.
Hint: -All worksheet grouped
-Value from range A1:F1 filled across all worksheets
-Same format maintained across all worksheets
-Worksheets ungrouped
3
Click cell I5 and enter a function that determines the number of full-time employees, (FT).
Hint: -Function in cell I5 uses the right cells
-Function used the right conditional Count function
4
Enter a function in cell I6 that determines the average salary of all full-time employees. Format the results in Accounting Number Format.
Hint:
Use DAVERAGE(…….).
-Function entered in I6 uses correct cells
-Function calculates average salary of all full-time employees
-Accounting number format applied to result
5
Enter a lookup function in cell E5 that returns the tax deduction amount for the number of dependents listed in the cell C5. Use the table in range H13:I17 to complete the function. The maximum deduction is \$500.00; therefore, employees with more than four dependents will
Hint: -Used suggested type of Lookup function (VLOOKUP) to return correct value.
-Absolute cell reference, and relative cell reference must be used in their places to get the complete marks here.
6
Use Auto Fill to copy the function down, completing column E. Be sure to use the appropriate cell referencing. Format the data in column E with the Accounting Number Format.
Hint: -Auto fill used to copy function down column
-Appropriate cell referencing (absolute cell reference, and relative cell reference must be used in their places to get the complete marks here)
-Data in column E formatted with Accounting Number format
7
Enter a logical function in cell F5 that calculates employee FICA withholding. If the employee is full-time and has at least one dependent, then he or she pays 7% of the annual salary minus any deductions. All other employees pay 5% of the annual salary minus any deductions. Copy the function down through column F. Format the data in column F with Accounting Number Format.
Hint:
= IF (AND (logical 1, logical 2
.))
-Logical function in cell F5 calculate FICA withholding-Full-time (FT) employee with at less 1 dependent pays 7% of the annual salary minus any deductions.
-Other employees pay 5% of the annual salary minus any deductions
-Cell reference I20 and I21 correctly
-Function copied down column F
-Data in column F formatted with Accounting Number format
8
Apply conditional formatting to the range C5:C34 that highlights any dependents that are greater than 3 with Light Red Fill and Dark Red Text.
Hint:
In the Styles group, click Conditional Formatting.
-Conditional formatting applied to range C5:C34
-Dependents greater than 3 highlighted
-Light Red Fill
-Dark Red Text
9
Click cell H10, and enter an AVERAGEIFS function to determine the average salary of full-time employees with at least one dependent. Format the results in Accounting Number Format.
Hint:
AVERAGEIFS (average-range, criteria-range1, criteria
.)
-AVERAGEIFS function used correctly
-Returns average salary of full-time employees with at least one dependent
-Result formatted with Accounting number format
10
Use Advanced Filtering to restrict the data to only display full-time employees with at least one dependent. Place the results in cell A37. Use the criteria in the range H24:M25 to complete the function.
Hint:
On the Data tab, in the Sort & Filter group, click Advanced.
-Criteria in range H24:M25 used to complete function
-Results placed in cell A37
11
Ensure that the Facilities worksheet is active. Use Goal Seek to reduce the monthly payment in cell B6 to the optimal value of \$6000. Complete this task by changing the Loan amount in cell E6.
Hint:
On the Data tab, in the Forecast group, click What-If Analysis.
-Goal seek used properly
-Monthly payment is reduced to S6000.
-Loan amount in cell E6 changed correctly
12
Create the following three scenarios using Scenario Manager. The scenarios should change the cells B7, B8, and E6.
Good
B7 = .0325
B8 = 5
E6 = 275000
Most Likely
B7 = .057
B8 = 5
E6 = 312227.32
B7 = .0700
B8 = 3
E6 = 350000
Create a Scenario Summary Report based on the value in cell B6. Format the new report appropriately.
Hint: On the Data tab, in the Forecast group, click What-If Analysis, Scenario Manager
–Three scenarios created with
-Correct name
-Correct values
-Summary Report is based on the value in cell B6
-New report formatted appropriately
13
Ensure that the Facilities worksheet is active. Enter a reference to the beginning loan balance in cell B12 and enter a reference to the payment amount in cell C12.
Hint: -Facilities worksheet is active
-Reference to beginning loan balance in cell B12
-Reference to payment amount in Cell C12
14
Enter a function in cell D12, based on the payment and loan details, that calculates the amount of interest paid on the first payment. Be sure to use the appropriate absolute, relative, or mixed cell references.
Hint:
The function is =IPMT(rate, per, nper, – pv, [fv], [type]).
-Correct cell referenced in the IPMT function to calculate the amount of interest paid on first payment
-Appropriate cell referencing (absolute cell reference, and relative cell reference must be used in their places to get the complete marks here)
-Function in D12 return correct value
15
Enter a function in cell E12, based on the payment and loan details, that calculates the amount of principal paid on the first payment. Be sure to use the appropriate absolute, relative, or mixed cell references.
Hint:
The function is =PPMT(rate, per, nper, – pv, [fv], [type]).
-Correct cell referenced in the PPMT function to calculate the amount of principal paid on first payment
-Appropriate cell referencing (absolute cell reference, and relative cell reference must be used in their places to get the complete marks here)
-Function in E12 return correct value
16
Enter a formula in cell F12 to calculate the remaining balance after the current payment. The remaining balance is calculated by subtracting the principal payment from the balance in column B.
Hint: -Formula in cell F12 used correct cells
-Displayed correct result
17
Enter a function in cell G12, based on the payment and loan details, that calculates the amount of cumulative interest paid on the first payment. Be sure to use the appropriate absolute, relative, or mixed cell references.
Hint:
The function is =-CUMIPMT(rate, nper, pv, start_period, end_period, type).
-Correct cell referenced in the CUMIPMT function to calculate the amount of cumulative interest paid on the first payment
-Appropriate cell referencing (absolute cell reference, and relative cell reference must be used in their places to get the complete marks here)
-Function in G12 return correct value
18
Enter a function in cell H12, based on the payment and loan details, that calculates the amount of cumulative principal paid on the first payment. Be sure to use the appropriate absolute, relative, or mixed cell references.
Hint:
The function is =-CUMPRINC(rate, nper, pv, start_period, end_period, type).
-Correct cell referenced in the CUMPRINC function to calculate the amount of cumulative principal paid on the first payment
-Appropriate cell referencing (absolute cell reference, and relative cell reference must be used in their places to get the complete marks here)
-Function in H12 return correct value
19
Enter a reference to the remaining balance of payment 1 in cell B13. Use the fill handle to copy the functions created in the prior steps down to complete the amortization table.
Hint: -Value in B13 is a cell reference to Remaining Balance of Payment 1 (That is B13 cell reference F12
-Fill handle used to copy the above functions (functions created in step# 13 to step# 19) down to complete amortization table
20
Ensure the Sales worksheet is active. Enter a function in cell B8 to create a custom transaction number. The transaction number should be comprised of the item number listed in cell C8 combined with the quantity in cell D8 and the first initial of the payment type in cell E8. Use Auto Fill to copy the function down, completing the data in column B.
Hint:
The function is =C8&D8&LEFT(E8,1).
-Sales Worksheet active
-Function created in B8
-Function returns
-item number listed in cell C8
-combined with the quantity in cell D8 and
-the first initial of the payment type in cell E8.
-Auto Fill used to copy the above function down to complete in column B
21
Enter a nested function in cell G8 that displays the word Flag if the Payment Type is Credit and the Amount is greater than or equal to \$4000. Otherwise, the function will display a blank cell. Use Auto Fill to copy the function down, completing the data in column G.
Hint:
The function is =IF (AND (logical 1, logical 2
.), )
For example: IF (AND (E8=”Credit”, logic 2),”Flag”,””)
-Nested function in cell G8 has correct criteria
-Nested function in cell G8 has correct cell references
-Nested function in cell G8, displays word Flag if
-Payment Type is Credit and
– Amount is greater than or equal to \$4000
-Nested function in cell G8, displays a blank cell for other conditions
-Auto Fill used to copy the above function down to complete in column G
22
Create a data validation list in cell D5 that displays Quantity, Payment Type, and Amount.
Hint:
On the Data tab, in the Data Tools group, click Data Validation, Allow, List
-Data list created in cell D5
-List has Quantity, Payment Type, and Amount
23
Type the Trans# 30038C in cell B5, and select Quantity from the validation list in cell D5.
Hint: -Trans# 30038C in cell B5
-Validation List Quantity displayed in cell D5
24
Enter a nested lookup function in cell F5 that evaluates the Trans # in cell B5 as well as the Category in cell D5, and returns the results based on the data in the range A8:F32.
Hint:
Use function INDEX() and MATCH().
-Nested lookup function in cell F5
-Evaluates Trans # in cell B5, Category in cell D5
-Returns results based on date in range A8:F32
25
Create a PivotTable based on the range A7:G32. Place the PivotTable in cell I17 on the current worksheet. Place Payment Type in the Rows box and Amount in the Values box. Format the Amount with Accounting Number Format.
Hint:
On the Insert tab, in the Tables group, select PivotTable.
-Range A7:G32 used to create PivotTable
-PivotTable placed in cell I17 in current worksheet
-Payment Type in the Rows box area
-Amount in the Values box area
-Amount formatted with Accounting Format
26
Insert a PivotChart using the Pie chart type based on the data. Place the upper-left corner of the chart inside cell I22. Format the Legend of the chart to appear at the bottom of the chart area. Format the Data Labels to appear on the Outside end of the chart.
Note, Mac users, select the range I18:J20, on the Insert tab, click Recommended Charts, and then click Pie. Format the legend, and apply the data labels as specified.
Hint:
On the Analyze tab, in the Tools group, click PivotChart.
-Create PivotChart Pie Chart based on above data
-Upper-left corner of the PivotChart Pie Chart placed in cell I22
-Legend of PivotChart formatted to appear at the bottom of Chart area
-Data Labels to appear on the Outside end of the chart
27
Insert a Slicer based on Date. Place the upper-left corner of the Slicer inside cell L8.
Hint:
On the Insert tab, in the Filters group, and click Slicer.
-Slicer based on Date from PivotTable
-Upper-left corner of the Slicer in cell L8
-Slicer has right values
28
Ensure the Inventory worksheet is active. Import the Access database eApp_Cap2_Inventory.accdb into the worksheet starting in cell A3.
Note, Mac users, download and import the delimited Inventory.txt file into the worksheet starting in cell A3.
Hint:
Use the tools on the Data tab to import the file as specified.
On the Data tab, in the Get & Transform Data group, click Get Data, From Database, From Microsoft Access database, Load
-Database file imported to Inventory worksheet
-Table starts from cell A3
-Table has right headings and data
29
Create a footer with your name on the left, the sheet code in the center, and the file name on the right for each worksheet.
Hint: -For each worksheet:
-Name on left side of footer
-Sheet code in the Center
-File name on the right side
30
Save and close the file. Submit
ACC2343_Final_Manufacturing_LastFirst.xlsx.
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6. 24/7 Customer Support: At Eminence Papers, we have put in place a team of experts who answer all customer inquiries promptly. The best part is the ever-availability of the team. Customers can make inquiries anytime. | 3,679 | 15,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-21 | latest | en | 0.788134 |
https://www.thelittleaussiebakery.com/what-is-the-density-of-steel-per-cubic-foot/ | 1,723,539,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00237.warc.gz | 781,451,669 | 11,870 | # What is the density of steel per cubic foot?
## What is the density of steel per cubic foot?
about 490 pounds per cubic foot
Plain steel’s density is about 490 pounds per cubic foot, which can also be expressed as 7.85 g/cm3.
What does 1 cubic foot of steel weigh?
Fact: Most steel weighs about 489 pounds per cubic foot.
What is the density of steel per cubic inch?
0.284 pounds per cubic inches
Calculating the Weight of Carbon Steel Plate To calculate the weight of carbon steel plate, you’ll need to know the thickness, width, length, and quantity. The typical density for carbon steel plate is 0.284 pounds per cubic inches (490 pounds/cubic feet) (imperial) or 7.85 tons/cubic meters (metric).
### What is density of ss304?
Stainless steel 304 and stainless steel 304L are also known as 1.4301 and 1.4307 respectively. Type 304 is the most versatile and widely used stainless steel….Physical Properties.
Property Value
Density 8,000 Kg/m3
Melting Point 1450 °C
Thermal Expansion 17.2 x 10-6 /K
Modulus of Elasticity 193 GPa
Why is steel so dense?
Weight. Because it’s stronger and more durable than aluminum, steel also weighs more than its counterpart. Steel is essentially 250% times denser than aluminum, making it obviously heavier. And due to its high density/weight, it’s less likely to bend under force or heat.
How many m3 are in a tonne of steel?
It’s 0.12738 m^3.
#### How much does steel weigh per cubic cm?
Steel or reinforcing bar embedded in concrete fibre to increase the tensile strength and to provide structural support, in this regard, “How much does Steel weigh per cubic foot”, in general, typically, steel can weighs around 493 pounds per cubic foot, if steel weight measured in kilogram per cubic metre, then it …
What is the density of 316 stainless steel?
7.99 g/cm3
Physical Properties (Room Temperature)
Specific Heat (0-100°C) 500 J.kg-1.°K-1
Thermal Expansion 15.9 μm/μm/°C
Modulus Elasticity 193 GPa
Electrical Resistivity 7.4 μohm/cm
Density 7.99 g/cm3
How do you calculate steel tonnage?
Multiply the total length of bar with the following unit weight. For, 10mm ø bar = 0.617 kg/m. 12mm ø bar = 0.888 kg/m….
1. 10mm ø bar = 0.188 kg/ft.
2. 12mm ø bar = 0.270 kg/ft.
3. 16mm ø bar = 0.480 kg/ft.
4. 20mm ø bar = 0.750 kg/ft.
5. 25mm ø bar = 1.172 kg/ft.
## How do you calculate the density of steel?
– Place the object on an accurate scale and record the mass in your notebook. – Alternatively, you can measure mass by using a balance. Place your object on one side and place weights of known mass on the other side until both sides are balanced. – Make sure object is dry so that absorbed water does not affect the accuracy of the weighing.
What is the average density of steel?
Though the densities vary, the density of steel is about 7700 kg/m3….Density of Steel. Is tool steel dense? The density of steel is in the range of 7.75 and 8.05 g/cm3 (7750 and 8050 kg/m3 or 0.280 and 0.291 lb/in3). Density of carbon steels, alloy steels, tool steels and stainless steels are shown below in g/cm3, kg/m3 and lb/in3.
What is the density of galvanized steel?
The most common is 50 grams of galvanizing and 100 grams on both sides. This will let you know a simple relationship. The density of the galvanized sheet is 7.85 tons/m3, that is to say, the thickness of 1.0 mm is 1.85 Kg, and the thickness of 0.75 mm is 5.4875 Kg.
### What is the density of alloy steel?
The density of steel is in the range of 7.75 and 8.05 g/cm 3 (7750 and 8050 kg/m 3 or 0.280 and 0.291 lb/in 3 ). The theoretical density of mild steel (low-carbon steel) is about 7.87 g/cm 3 (0.284 lb/in 3 ). Density of carbon steels, alloy steels, tool steels and stainless steels are shown below in g/cm 3, kg/m 3 and lb/in 3. | 1,053 | 3,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.863749 |
http://credit-help.biz/bank/47749 | 1,487,931,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171463.89/warc/CC-MAIN-20170219104611-00392-ip-10-171-10-108.ec2.internal.warc.gz | 58,471,855 | 4,057 | # Calculating the Earnings Per Share (EPS) Ratio
Publicly owned businesses, according to generally accepted accounting principles (GAAP), must report earnings per share (EPS) below the net income line in their income statements — giving EPS a certain distinction among financial ratios.
Why is EPS considered so important? Because it gives investors a means of determining the amount the business earned on their stock share investments: EPS tells you how much net income the business earned for each stock share you own.
## The basic EPS ratio
For the example shown in the following figures, the company’s \$32.47 million net income is divided by the 8.5 million shares of stock the business has issued to compute its \$3.82 EPS.
A balance sheet example for a business.
EPS is extraordinarily important to the stockholders of businesses whose stock shares are publicly traded. These stockholders pay close attention to market price per share. They want the net income of the business to be communicated to them on a per share basis so that they can easily compare it with the market price of their stock shares.
The stock shares of privately owned corporations are not actively traded, so there is no readily available market value for the stock shares. Private businesses do not have to report EPS according to GAAP. The thinking behind this exemption is that their stockholders do not focus on per share values and are more interested in the business’s total net income.
## The diluted EPS ratio
The business in the example could be listed on the New York Stock Exchange (NYSE). Assume that its capital stock is being traded at \$70 per share. The Big Board (as it is called) requires that the market cap (total value of the shares issued and outstanding) be at least \$100 million and that it have at least 1.1 million shares available for trading.
With 8.5 million shares trading at \$70 per share, the company’s market cap is \$595 million, well above the NYSE’s minimum. At the end of the year, this corporation has 8.5 million stock shares outstanding. which refers to the number of shares that have been issued and are owned by its stockholders. Thus, its EPS is \$3.82, as just computed.
But here’s a complication: The business is committed to issuing additional capital stock shares in the future for stock options that the company
has granted to its executives, and it has borrowed money on the basis of debt instruments that give the lenders the right to convert the debt into its capital stock.
Under terms of its management stock options and its convertible debt, the business may have to issue 500,000 additional capital stock shares in the future. Dividing net income by the number of shares outstanding plus the number of shares that could be issued in the future gives the following computation of EPS:
\$32,470,000 net income ÷ 9,000,000 capital stock shares issued and potentially issuable = \$3.61 EPS
This second computation, based on the higher number of stock shares, is called the diluted earnings per share. (Diluted means thinned out or spread over a larger number of shares.) The first computation, based on the number of stock shares actually issued and outstanding, is called basic earnings per share. Both are reported at the bottom of the income statement.
So, publicly owned businesses report two EPS figures — unless they have a simple capital structure that does not require the business to issue additional stock shares in the future. Generally, publicly owned corporations have complex capital structures and have to report two EPS figures, as you see in the first figure above.
Sometimes it’s not clear which of the two EPS figures is being used in press releases and in articles in the financial press. You have to be careful to determine which EPS ratio is being used — and which is being used in the calculation of the price/earnings (P/E) ratio. | 791 | 3,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-09 | latest | en | 0.952561 |
http://www.fixya.com/support/t23860984-calculate_present_future_value_casio_fx | 1,532,155,424,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00167.warc.gz | 451,871,589 | 35,564 | Question about Casio FX-115MS Scientific Calculator
# How do I calculate Present and future value with Casio fx-115ms?
Posted by on
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• Casio Master
The calculator does not have a TVM application, so you cannot perform these calculations.
Posted on Mar 19, 2014
Hi there,
Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two.
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Posted on Jan 02, 2017
Hi,
Sorry to disappoint you: This calculator is not programmable.
Good luck.
Posted on Nov 30, 2009
Testimonial: "Very much appreciated, quick response etc. "
change the mode to stat and then start operating it in statistical mode
Posted on Feb 26, 2008
In any scientific calculator log2(n) can be calculated with either ln or log function as follows
Log2(n)= ln(n) / ln(2)
Or
Log2(n)=log(n) / log(2)
both will give nearly the same answers
Posted on Apr 24, 2009
SOURCE: absolutes on FX - 115ES
The absolute value button is "SHIFT", "Abs" (located on the "hyp" key)
Posted on May 03, 2009
This will show you how to do statistics (calculate average, standard deviation, regressions etc.) as to the mode and median refer to the definitions.
TO COMPUTE STANDARD DEVIATION AND 2-VAR STATISTICS.
I assume you know the theory. I will show you the key strokes
For 1-Var statistics
Press [MODE][3:STAT] [1:1-VAR]. You are ready to enter values in the X column.
Enter a number and press [=]. Cursor jumps to second number to enter.
Keep entering numbers and pressing [=] till all numbers are in. Press the [=] key after the last one.
Press [AC] key to clear the screen.
Press [SHIFT] [STAT] (above digit 1.) then [5:Var]. Screen displays the statistical variables 1:n ;2: x bar; 3: x sigma n; 4:x sigma n-1.
Press the number corresponding to the statistical value you want, ex 1:n . The variable appears on screen. Press [=] and it will be displayed.
To display another variable press [SHIFT][STAT][5:Var][ 1,2, 3, or 4] .
To access the sum of squares sigma x^2 and the sum of data sigma x press[SHIFT][STAT][4:SUM] then [1: for sigma x^2] or [2: for sigma x]. Press [SHIFT][STAT][6:MinMax] to access minX and maxX.
For 2-var statistics
To perform 2 variable statistics you press [MODE][3:STAT] and any of the other regression options (except 1:1-Var). A two column template opens where you enter the X and Y values. When finished entering data, press [SHIFT][STAT][5:Var]. to access the different statistics. As I assumed above, you should be able to recognize what each variable means.
Posted on Apr 21, 2011
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## Related Questions:
### Future value
The calculator does not have a TVM function/feature.
Aug 19, 2014 | Casio FX-115ES Scientific Calculator
### What are step by step instructions for finding present value of 20%?
The first step would be to determine the rest of the problem. What's the future value? How long? Are there any payments? If so, how much and how often?
Asking for the present value of 20% is like asking "how do I get to the store?" Which store? From where? Are you driving, walking, riding a bike, or something else?
Sep 15, 2013 | Casio FX991MS Scientific Calculator
### Can this calculator do present and future value?
No, it does not. You will need a graphing calculator such as the Casio FX-9750G Plus or the newer version FX-9750 GII.
Sep 19, 2010 | Casio FX-300MS Calculator
### How to calculate 17^23mod55 in casio scientific calculator fx-991MS?
Using the Windows calculator in scientific view, we get:
17^23 = 19,967,568,900,859,523,802,559,065,713
which is 29 digits long, ie much longer than the capacity of most calculators.
Because both 17 and 23 are prime numbers, it isn't possible to simplify this calculation, so you will need to use find a calculator that has sufficient capacity to be able to handle numbers of this length.
The xnumbers add-in for Microsoft Excel can also be useful for this sort of calculation (do an internet search for it).
Using the Windows calculator, MOD55 of the above number is 18.
Jun 30, 2010 | Casio FX-115MS Plus Calculator
### Trying to solve a e^x problem?
Hello
20.3 [x][SHIFT][e^]0.025 [)] [x] 20 Do not forget closing parenthesis )
key [e^] is the one above [ln]
The answer is 416.277 and not 33.5
Verification [e^]0.025 [)] =1.02531 and 20x20.3=406
Hope it helps
Apr 16, 2009 | Casio FX-115ES Scientific Calculator
### NEED INSTRUCTIONS
Since you're looking for so much, it sounds like you do not have the users manual for the calculator. You can download it here:
http://safemanuals.com/user-guide-instructions-owner-manual/CASIO/FX-270W%20PLUS-_E
John
Jan 19, 2009 | Office Equipment & Supplies
### User manual
This page contains 2 manuals for FX115 calculators, and while the standard seems to be MS, each manual includes several models. This should address what you need.
http://www.retrevo.com/s/Casio-fx-115MS-Calculators-review-manual/id/467ci265/t/1-2/
If you need more help, please post back, otherwise thank you for using and rating FixYa!
Jun 04, 2008 | Casio Office Equipment & Supplies
## Open Questions:
#### Related Topics:
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https://ngsolve.org/docu/latest/i-tutorials/unit-1.1-poisson/poisson.html | 1,610,952,225,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00001.warc.gz | 488,577,715 | 21,068 | This page was generated from unit-1.1-poisson/poisson.ipynb.
1.1 First NGSolve example¶
Let us solve the Poisson problem of finding $$u$$ satisfying
\begin{split}\begin{aligned} -\Delta u & = f && \text { in the unit square}, \\ u & = 0 && \text{ on the bottom and right parts of the boundary}, \\ \frac{\partial u }{\partial n } & = 0 && \text{ on the remaining boundary parts}. \end{aligned}\end{split}
Quick steps to solution:¶
1. Import NGSolve and Netgen Python modules:¶
[1]:
import netgen.gui
from ngsolve import *
from netgen.geom2d import unit_square
2. Generate an unstructured mesh¶
[2]:
mesh = Mesh(unit_square.GenerateMesh(maxh=0.2))
mesh.nv, mesh.ne # number of vertices & elements
[2]:
(39, 56)
• Here we prescribed a maximal mesh-size of 0.2 using the maxh flag.
• The mesh can be viewed by switching to the Mesh tab in the Netgen GUI.
3. Declare a finite element space:¶
[3]:
fes = H1(mesh, order=2, dirichlet="bottom|right")
fes.ndof # number of unknowns in this space
[3]:
133
Python’s help system displays further documentation.
[4]:
help(fes)
Help on H1 in module ngsolve.comp object:
class H1(FESpace)
| An H1-conforming finite element space.
|
| The H1 finite element space consists of continuous and
| element-wise polynomial functions. It uses a hierarchical (=modal)
| basis built from integrated Legendre polynomials on tensor-product elements,
| and Jaboci polynomials on simplicial elements.
|
| Boundary values are well defined. The function can be used directly on the
| boundary, using the trace operator is optional.
|
| The H1 space supports variable order, which can be set individually for edges,
| faces and cells.
|
| Internal degrees of freedom are declared as local dofs and are eliminated
| if static condensation is on.
|
| The wirebasket consists of all vertex dofs. Optionally, one can include the
| first (the quadratic bubble) edge basis function, or all edge basis functions
| into the wirebasket.
|
| Keyword arguments can be:
| order: int = 1
| order of finite element space
| complex: bool = False
| Set if FESpace should be complex
| dirichlet: regexpr
| Regular expression string defining the dirichlet boundary.
| More than one boundary can be combined by the | operator,
| i.e.: dirichlet = 'top|right'
| dirichlet_bbnd: regexpr
| Regular expression string defining the dirichlet bboundary,
| i.e. points in 2D and edges in 3D.
| More than one boundary can be combined by the | operator,
| i.e.: dirichlet_bbnd = 'top|right'
| dirichlet_bbbnd: regexpr
| Regular expression string defining the dirichlet bbboundary,
| i.e. points in 3D.
| More than one boundary can be combined by the | operator,
| i.e.: dirichlet_bbbnd = 'top|right'
| definedon: Region or regexpr
| FESpace is only defined on specific Region, created with mesh.Materials('regexpr')
| or mesh.Boundaries('regexpr'). If given a regexpr, the region is assumed to be
| mesh.Materials('regexpr').
| dim: int = 1
| Create multi dimensional FESpace (i.e. [H1]^3)
| dgjumps: bool = False
| Enable discontinuous space for DG methods, this flag is needed for DG methods,
| since the dofs have a different coupling then and this changes the sparsity
| pattern of matrices.
| low_order_space: bool = True
| Generate a lowest order space together with the high-order space,
| needed for some preconditioners.
| order_policy: ORDER_POLICY = ORDER_POLICY.OLDSTYLE
| CONSTANT .. use the same fixed order for all elements,
| NODAL ..... use the same order for nodes of same shape,
| VARIBLE ... use an individual order for each edge, face and cell,
| OLDSTYLE .. as it used to be for the last decade
| wb_withedges: bool = true(3D) / false(2D)
| use lowest-order edge dofs for BDDC wirebasket
| wb_fulledges: bool = false
| use all edge dofs for BDDC wirebasket
|
| Method resolution order:
| H1
| FESpace
| NGS_Object
| pybind11_builtins.pybind11_object
| builtins.object
|
| Methods defined here:
|
| __getstate__(...)
| __getstate__(self: ngsolve.comp.FESpace) -> tuple
|
| __init__(...)
| __init__(self: ngsolve.comp.H1, mesh: ngsolve.comp.Mesh, autoupdate: bool = False, **kwargs) -> None
|
| __setstate__(...)
| __setstate__(self: ngsolve.comp.H1, arg0: tuple) -> None
|
| ----------------------------------------------------------------------
| Static methods defined here:
|
| __flags_doc__(...) from builtins.PyCapsule
| __flags_doc__() -> dict
|
| ----------------------------------------------------------------------
| Data descriptors defined here:
|
| __dict__
|
| ----------------------------------------------------------------------
| Methods inherited from FESpace:
|
| ApplyM(...)
| ApplyM(self: ngsolve.comp.FESpace, vec: ngsolve.la.BaseVector, rho: ngsolve.fem.CoefficientFunction = None, definedon: ngsolve.comp.Region = None) -> None
|
| Apply mass-matrix. Available only for L2-like spaces
|
| ConvertL2Operator(...)
| ConvertL2Operator(self: ngsolve.comp.FESpace, l2space: ngsolve.comp.FESpace) -> BaseMatrix
|
| CouplingType(...)
| CouplingType(self: ngsolve.comp.FESpace, dofnr: int) -> ngsolve.comp.COUPLING_TYPE
|
|
| Get coupling type of a degree of freedom.
|
| Parameters:
|
| dofnr : int
| input dof number
|
| Elements(...)
| Elements(self: ngsolve.comp.FESpace, VOL_or_BND: ngsolve.comp.VorB = <VorB.VOL: 0>) -> ngsolve.comp.FESpaceElementRange
|
|
| Returns an iterable range of elements.
|
| Parameters:
|
| VOL_or_BND : ngsolve.comp.VorB
| input VOL, BND, BBND,...
|
| FinalizeUpdate(...)
| FinalizeUpdate(self: ngsolve.comp.FESpace) -> None
|
| finalize update
|
| FreeDofs(...)
| FreeDofs(self: ngsolve.comp.FESpace, coupling: bool = False) -> pyngcore.BitArray
|
|
|
| Return BitArray of free (non-Dirichlet) dofs\n
| coupling=False ... all free dofs including local dofs\n
| coupling=True .... only element-boundary free dofs
|
| Parameters:
|
| coupling : bool
| input coupling
|
| GetDofNrs(...)
| GetDofNrs(*args, **kwargs)
| Overloaded function.
|
| 1. GetDofNrs(self: ngsolve.comp.FESpace, ei: ngsolve.comp.ElementId) -> tuple
|
|
|
| Parameters:
|
| ei : ngsolve.comp.ElementId
| input element id
|
|
|
| 2. GetDofNrs(self: ngsolve.comp.FESpace, ni: ngsolve.comp.NodeId) -> tuple
|
|
|
| Parameters:
|
| ni : ngsolve.comp.NodeId
| input node id
|
| GetDofs(...)
| GetDofs(self: ngsolve.comp.FESpace, region: ngsolve.comp.Region) -> pyngcore.BitArray
|
|
| Returns all degrees of freedom in given region.
|
| Parameters:
|
| region : ngsolve.comp.Region
| input region
|
| GetFE(...)
| GetFE(self: ngsolve.comp.FESpace, ei: ngsolve.comp.ElementId) -> object
|
|
| Get the finite element to corresponding element id.
|
| Parameters:
|
| ei : ngsolve.comp.ElementId
| input element id
|
| GetOrder(...)
| GetOrder(self: ngsolve.comp.FESpace, nodeid: ngsolve.comp.NodeId) -> int
|
| return order of node.
| by now, only isotropic order is supported here
|
| GetTrace(...)
| GetTrace(self: ngsolve.comp.FESpace, arg0: ngsolve.comp.FESpace, arg1: ngsolve.la.BaseVector, arg2: ngsolve.la.BaseVector, arg3: bool) -> None
|
| GetTraceTrans(...)
| GetTraceTrans(self: ngsolve.comp.FESpace, arg0: ngsolve.comp.FESpace, arg1: ngsolve.la.BaseVector, arg2: ngsolve.la.BaseVector, arg3: bool) -> None
|
| HideAllDofs(...)
| HideAllDofs(self: ngsolve.comp.FESpace, component: object = <ngsolve.ngstd.DummyArgument>) -> None
|
| set all visible coupling types to HIDDEN_DOFs (will be overwritten by any Update())
|
| InvM(...)
| InvM(self: ngsolve.comp.FESpace, rho: ngsolve.fem.CoefficientFunction = None) -> BaseMatrix
|
| Mass(...)
| Mass(self: ngsolve.comp.FESpace, rho: ngsolve.fem.CoefficientFunction = None, definedon: Optional[ngsolve.comp.Region] = None) -> BaseMatrix
|
| ParallelDofs(...)
| ParallelDofs(self: ngsolve.comp.FESpace) -> ngsolve.la.ParallelDofs
|
| Return dof-identification for MPI-distributed meshes
|
| Prolongation(...)
| Prolongation(self: ngsolve.comp.FESpace) -> ngmg::Prolongation
|
| Return prolongation operator for use in multi-grid
|
| Range(...)
| Range(self: ngsolve.comp.FESpace, arg0: int) -> ngsolve.la.DofRange
|
| deprecated, will be only available for ProductSpace
|
| SetCouplingType(...)
| SetCouplingType(*args, **kwargs)
| Overloaded function.
|
| 1. SetCouplingType(self: ngsolve.comp.FESpace, dofnr: int, coupling_type: ngsolve.comp.COUPLING_TYPE) -> None
|
|
| Set coupling type of a degree of freedom.
|
| Parameters:
|
| dofnr : int
| input dof number
|
| coupling_type : ngsolve.comp.COUPLING_TYPE
| input coupling type
|
|
|
| 2. SetCouplingType(self: ngsolve.comp.FESpace, dofnrs: ngsolve.ngstd.IntRange, coupling_type: ngsolve.comp.COUPLING_TYPE) -> None
|
|
| Set coupling type for interval of dofs.
|
| Parameters:
|
| dofnrs : Range
| range of dofs
|
| coupling_type : ngsolve.comp.COUPLING_TYPE
| input coupling type
|
| SetDefinedOn(...)
| SetDefinedOn(self: ngsolve.comp.FESpace, region: ngsolve.comp.Region) -> None
|
|
| Set the regions on which the FESpace is defined.
|
| Parameters:
|
| region : ngsolve.comp.Region
| input region
|
| SetOrder(...)
| SetOrder(*args, **kwargs)
| Overloaded function.
|
| 1. SetOrder(self: ngsolve.comp.FESpace, element_type: ngsolve.fem.ET, order: int) -> None
|
|
|
| Parameters:
|
| element_type : ngsolve.fem.ET
| input element type
|
| order : object
| input polynomial order
|
|
| 2. SetOrder(self: ngsolve.comp.FESpace, nodeid: ngsolve.comp.NodeId, order: int) -> None
|
|
|
| Parameters:
|
| nodeid : ngsolve.comp.NodeId
| input node id
|
| order : int
| input polynomial order
|
| SolveM(...)
| SolveM(self: ngsolve.comp.FESpace, vec: ngsolve.la.BaseVector, rho: ngsolve.fem.CoefficientFunction = None, definedon: ngsolve.comp.Region = None) -> None
|
|
| Solve with the mass-matrix. Available only for L2-like spaces.
|
| Parameters:
|
| vec : ngsolve.la.BaseVector
| input right hand side vector
|
| rho : ngsolve.fem.CoefficientFunction
| input CF
|
| TestFunction(...)
| TestFunction(self: ngsolve.comp.FESpace) -> object
|
| Return a proxy to be used as a testfunction for :any:Symbolic Integrators<symbolic-integrators>
|
| TnT(...)
| TnT(self: ngsolve.comp.FESpace) -> Tuple[object, object]
|
| Return a tuple of trial and testfunction
|
| TraceOperator(...)
| TraceOperator(self: ngsolve.comp.FESpace, tracespace: ngsolve.comp.FESpace, average: bool) -> BaseMatrix
|
| TrialFunction(...)
| TrialFunction(self: ngsolve.comp.FESpace) -> object
|
| Return a proxy to be used as a trialfunction in :any:Symbolic Integrators<symbolic-integrators>
|
| Update(...)
| Update(self: ngsolve.comp.FESpace) -> None
|
| update space after mesh-refinement
|
| UpdateDofTables(...)
| UpdateDofTables(self: ngsolve.comp.FESpace) -> None
|
| update dof-tables after changing polynomial order distribution
|
| __eq__(...)
| __eq__(self: ngsolve.comp.FESpace, space: ngsolve.comp.FESpace) -> bool
|
| __mul__(...)
| __mul__(self: ngsolve.comp.FESpace, arg0: ngsolve.comp.FESpace) -> ngcomp::CompoundFESpace
|
| __pow__(...)
| __pow__(self: ngsolve.comp.FESpace, arg0: int) -> ngcomp::CompoundFESpaceAllSame
|
| __str__(...)
| __str__(self: ngsolve.comp.FESpace) -> str
|
| __timing__(...)
| __timing__(self: ngsolve.comp.FESpace) -> object
|
| ----------------------------------------------------------------------
| Static methods inherited from FESpace:
|
| __special_treated_flags__(...) from builtins.PyCapsule
| __special_treated_flags__() -> dict
|
| ----------------------------------------------------------------------
| Readonly properties inherited from FESpace:
|
| components
| deprecated, will be only available for ProductSpace
|
| couplingtype
|
| dim
| multi-dim of FESpace
|
| globalorder
| query global order of space
|
| is_complex
|
| loembedding
|
| lospace
|
| mesh
| mesh on which the FESpace is created
|
| ndof
| number of degrees of freedom
|
| ndofglobal
| global number of dofs on MPI-distributed mesh
|
| type
| type of finite element space
|
| ----------------------------------------------------------------------
| Data and other attributes inherited from FESpace:
|
| __hash__ = None
|
| ----------------------------------------------------------------------
| Readonly properties inherited from NGS_Object:
|
| __memory__
|
| ----------------------------------------------------------------------
| Data descriptors inherited from NGS_Object:
|
| name
|
| ----------------------------------------------------------------------
| Static methods inherited from pybind11_builtins.pybind11_object:
|
| __new__(*args, **kwargs) from pybind11_builtins.pybind11_type
| Create and return a new object. See help(type) for accurate signature.
4. Declare test function, trial function, and grid function¶
• Test and trial function are symbolic objects - called ProxyFunctions - that help you construct bilinear forms (and have no space to hold solutions).
• GridFunctions, on the other hand, represent functions in the finite element space and contains memory to hold coefficient vectors.
[5]:
u = fes.TrialFunction() # symbolic object
v = fes.TestFunction() # symbolic object
gfu = GridFunction(fes) # solution
Alternately, you can get both the trial and test variables at once:
[6]:
u, v = fes.TnT()
5. Define and assemble linear and bilinear forms:¶
[7]:
a = BilinearForm(fes, symmetric=True)
a += grad(u)*grad(v)*dx
a.Assemble()
f = LinearForm(fes)
f += x*v*dx
f.Assemble()
[7]:
<ngsolve.comp.LinearForm at 0x7f2133b63d30>
You can examine the linear system in more detail:
[8]:
print(f.vec)
0.000333333
0.00888574
0.00633333
0.00074628
0.00399953
0.00769352
0.0104008
0.0115294
0.0150217
0.0167909
0.015258
0.0182768
0.0211091
0.0106045
0.00711336
0.00320035
0.000735854
0.000876249
0.000529403
0.00282218
0.0124495
0.0182096
0.0224435
0.0204234
0.0317338
0.0270543
0.0291603
0.0187131
0.0217393
0.00854715
0.00505989
0.0038006
0.00883394
0.0205505
0.0163797
0.0213793
0.0150785
0.0170279
0.0191555
-6.66667e-05
-3.33333e-05
-0.000526369
-0.000575033
-0.00109143
-0.0008
-0.000766667
-8.08556e-05
-2.24397e-05
-0.000120589
-0.000249038
-0.000218716
-0.000440093
-0.000380259
-0.000572665
-0.000711098
-0.000487162
-0.00080946
-0.000916707
-0.000914728
-0.000944161
-0.000759221
-0.00110619
-0.00127157
-0.000646306
-0.0014197
-0.00131979
-0.000648788
-0.00123697
-0.0012392
-0.0015957
-0.00146355
-0.000469258
-0.0014082
-0.00106589
-0.000437144
-0.000880349
-0.000856527
-0.000190932
-0.000699086
-0.000461922
-0.000315546
-0.000220954
-1.64401e-05
-9.82846e-05
-8.35917e-05
-2.64702e-05
-9.73025e-05
-0.000122662
-2.64702e-05
-0.000105881
-0.000353264
-0.00021487
-0.00066398
-0.000468351
-0.00059592
-0.000843865
-0.000796408
-0.000722285
-0.000922806
-0.00105484
-0.000955488
-0.00104202
-0.00117171
-0.00100951
-0.000963748
-0.00114996
-0.000965717
-0.00092025
-0.00104152
-0.00098591
-0.000825307
-0.000865806
-0.000574259
-0.000880372
-0.000710639
-0.000885225
-0.000308496
-0.000477574
-0.000191583
-0.00032424
-0.000250767
-0.000328592
-0.000559968
-0.00047995
-0.000773918
-0.00082872
-0.000774775
-0.000675938
-0.000761632
-0.000801479
-0.000841885
-0.000775647
-0.000743459
[9]:
print(a.mat)
Row 0: 0: 1 4: -0.5 19: -0.5 39: -0.0833333 40: -0.0833333 50: 0.166667
Row 1: 1: 0.828441 7: -0.208968 8: -0.209694 23: -0.40978 41: -0.0340869 42: -0.0342097 43: -0.069777 59: 0.0689149 61: 0.0691587
Row 2: 2: 1 11: -0.5 12: -0.5 44: -0.0833333 45: -0.0833333 69: 0.166667
Row 3: 3: 0.870927 15: -0.344377 16: -0.336878 30: -0.189673 46: -0.0173948 47: -0.0142173 48: -0.113542 81: 0.074791 83: 0.0703636
Row 4: 0: -0.5 4: 1.88347 5: -0.397227 19: -0.177276 20: -0.808965 39: 0 40: 0.0833333 49: -0.0535736 50: -0.164587 51: -0.0957505 53: 0.119778 90: 0.1108
Row 5: 4: -0.397227 5: 1.77564 6: -0.326749 20: -0.345183 21: -0.706484 49: -0.028366 51: 0.0945705 52: -0.0526905 53: -0.131261 54: -0.0836226 56: 0.107149 92: 0.0942211
Row 6: 5: -0.326749 6: 1.7517 7: -0.273913 21: -0.432506 22: -0.718531 52: -0.0380314 54: 0.0924896 55: -0.0573097 56: -0.116904 57: -0.0797051 58: 0.102962 95: 0.0964985
Row 7: 1: -0.208968 6: -0.273913 7: 1.82559 22: -0.374026 23: -0.968685 41: -0.0835417 43: 0.11837 55: -0.0420363 57: 0.0876885 58: -0.123558 59: -0.0551294 98: 0.0982072
Row 8: 1: -0.209694 8: 1.8809 9: -0.350939 23: -1.05698 24: -0.263291 42: -0.0831255 43: 0.118074 60: -0.0322109 61: -0.0466199 62: -0.151527 64: 0.0907008 101: 0.104708
Row 9: 8: -0.350939 9: 1.74429 10: -0.262838 24: -0.64648 25: -0.484029 60: -0.0557926 62: 0.114282 63: -0.0342301 64: -0.104931 65: -0.0957605 67: 0.0780365 102: 0.0983955
Row 10: 9: -0.262838 10: 1.77715 11: -0.267845 25: -0.84644 26: -0.400025 63: -0.0697744 65: 0.113581 66: -0.0327463 67: -0.0777309 68: -0.11594 70: 0.0773871 105: 0.105223
Row 11: 2: -0.5 10: -0.267845 11: 1.9829 12: -0.115438 26: -1.09961 44: 0 45: 0.0833333 66: -0.0708467 68: 0.115488 69: -0.195755 70: -0.0638806 72: 0.131662
Row 12: 2: -0.5 11: -0.115438 12: 1.81266 13: -0.253469 26: -0.239874 27: -0.703881 44: 0.0833333 45: 0 69: -0.11965 70: 0.0555561 71: -0.0432149 72: -0.0933383 73: -0.0459076 75: 0.0854598 108: 0.0777613
Row 13: 12: -0.253469 13: 1.79007 14: -0.404897 27: -0.775929 28: -0.355771 71: -0.0598976 73: 0.102142 74: -0.0406718 75: -0.0608683 76: -0.136907 78: 0.108155 110: 0.0880473
Row 14: 13: -0.404897 14: 1.83152 15: -0.218833 28: -0.309609 29: -0.898185 74: -0.0388314 76: 0.106314 77: -0.0683139 78: -0.148866 79: -0.0492423 80: 0.104786 112: 0.0941536
Row 15: 3: -0.344377 14: -0.218833 15: 1.78023 29: -0.432178 30: -0.784846 46: -0.0795022 48: 0.136898 77: -0.0424951 79: 0.0789672 80: -0.0877775 81: -0.0869308 116: 0.0808401
Row 16: 3: -0.336878 16: 1.81347 17: -0.20561 30: -0.908731 31: -0.36225 47: -0.0873491 48: 0.143495 82: -0.033994 83: -0.0825273 84: -0.0983745 86: 0.0682624 118: 0.0904872
Row 17: 16: -0.20561 17: 1.94288 18: -0.368349 31: -1.16439 32: -0.204537 82: -0.0846663 84: 0.118935 85: -0.0154075 86: -0.0529503 87: -0.17079 89: 0.076799 120: 0.12808
Row 18: 17: -0.368349 18: 2.04376 19: -0.553239 32: -1.12218 85: -0.0781072 87: 0.139499 88: -0.108922 89: -0.153598 91: 0.201129
Row 19: 0: -0.5 4: -0.177276 18: -0.553239 19: 1.87497 20: -0.545937 32: -0.098519 39: 0.0833333 40: 0 50: -0.124342 51: 0.0705543 88: 0.0154075 89: 0.076799 90: -0.0613734 91: -0.142188 93: 0.0818086
Row 20: 4: -0.808965 5: -0.345183 19: -0.545937 20: 3.54246 21: -0.647227 32: -0.584699 34: -0.610448 49: 0.0819397 50: 0.122262 51: -0.0693744 53: -0.10714 54: 0.082731 90: -0.146695 91: 0.115423 92: -0.077413 93: -0.103474 94: -0.0863131 97: 0.102553 122: 0.0855011
Row 21: 5: -0.706484 6: -0.432506 20: -0.647227 21: 3.50609 22: -0.571505 33: -0.714154 34: -0.43421 52: 0.0907219 53: 0.118623 54: -0.091598 56: -0.102619 57: 0.0839811 92: -0.107223 94: 0.0964712 95: -0.0880292 96: -0.0755245 97: -0.119354 100: 0.0992989 124: 0.0952513
Row 22: 6: -0.718531 7: -0.374026 21: -0.571505 22: 3.51428 23: -0.708085 24: -0.536887 33: -0.60525 55: 0.099346 56: 0.112374 57: -0.0919644 58: -0.111917 59: 0.0749089 95: -0.107744 96: 0.0906212 98: -0.0555055 99: -0.118983 100: -0.0995997 101: 0.0986109 103: 0.109853
Row 23: 1: -0.40978 7: -0.968685 8: -1.05698 22: -0.708085 23: 3.76596 24: -0.62243 41: 0.117629 42: 0.117335 43: -0.166667 58: 0.132513 59: -0.0886944 61: -0.0901615 62: 0.148989 98: -0.125692 99: 0.111193 101: -0.156444
Row 24: 8: -0.263291 9: -0.64648 22: -0.536887 23: -0.62243 24: 3.433 25: -0.459506 33: -0.354764 37: -0.549644 60: 0.0880035 61: 0.0676228 62: -0.111744 64: -0.0782661 65: 0.0980093 98: 0.0829906 99: -0.0835425 100: 0.0900331 101: -0.0468751 102: -0.0942456 103: -0.103593 104: -0.0539003 107: 0.0728206 125: 0.072687
Row 25: 9: -0.484029 10: -0.84644 24: -0.459506 25: 3.57375 26: -0.517151 35: -0.379385 37: -0.887236 63: 0.104004 64: 0.0924966 65: -0.11583 67: -0.0772401 68: 0.114309 102: -0.124075 104: 0.108163 105: -0.111385 106: -0.113421 107: -0.0536744 109: 0.0832674 129: 0.0933844
Row 26: 10: -0.400025 11: -1.09961 12: -0.239874 25: -0.517151 26: 3.71385 27: -0.788273 35: -0.66891 66: 0.103593 67: 0.0769345 68: -0.113857 69: 0.148738 70: -0.0690626 72: -0.198236 73: 0.0894771 105: -0.0965358 106: 0.105793 108: -0.0525365 109: -0.0887462 111: 0.0944382
Row 27: 12: -0.703881 13: -0.775929 26: -0.788273 27: 3.68527 28: -0.733256 35: -0.683932 71: 0.103112 72: 0.159913 73: -0.145712 75: -0.107402 76: 0.133611 108: -0.124052 109: 0.0955182 110: -0.133459 111: -0.103586 113: 0.122057
Row 28: 13: -0.355771 14: -0.309609 27: -0.733256 28: 3.48151 29: -0.790531 35: -0.449095 36: -0.434461 38: -0.408789 74: 0.0795032 75: 0.0828109 76: -0.103019 78: -0.10339 79: 0.0754885 110: -0.0450785 111: 0.0844769 112: -0.0400872 113: -0.0863546 114: -0.108835 115: -0.0934871 117: 0.0963537 130: 0.0767268 131: 0.0848918
Row 29: 14: -0.898185 15: -0.432178 28: -0.790531 29: 3.74275 30: -0.98463 36: -0.637229 77: 0.110809 78: 0.144102 79: -0.105213 80: -0.135473 81: 0.0966939 112: -0.133883 114: 0.121537 116: -0.0832396 117: -0.165982 119: 0.150651
Row 30: 3: -0.189673 15: -0.784846 16: -0.908731 29: -0.98463 30: 3.81689 31: -0.851669 36: -0.0973365 46: 0.096897 47: 0.101566 48: -0.166851 80: 0.118465 81: -0.0845541 83: -0.0722726 84: 0.122161 116: -0.0628419 117: 0.108482 118: -0.0687924 119: -0.180835 121: 0.0885758
Row 31: 16: -0.36225 17: -1.16439 30: -0.851669 31: 3.85451 32: -0.517495 36: -0.958711 82: 0.11866 83: 0.0844363 84: -0.142722 86: -0.0722565 87: 0.147661 118: -0.100344 119: 0.157852 120: -0.195221 121: -0.131876 123: 0.133809
Row 32: 17: -0.204537 18: -1.12218 19: -0.098519 20: -0.584699 31: -0.517495 32: 3.69368 34: -0.800343 36: -0.365907 85: 0.0935147 86: 0.0569444 87: -0.11637 88: 0.0935147 89: 0 90: 0.0972688 91: -0.174364 93: -0.0930223 94: 0.0932033 120: -0.0398024 121: 0.0691071 122: -0.0718724 123: -0.120182 127: 0.11206
Row 33: 21: -0.714154 22: -0.60525 24: -0.354764 33: 3.57925 34: -0.624553 37: -0.928035 38: -0.352491 95: 0.0992748 96: -0.0975796 97: 0.11733 99: 0.0913324 100: -0.0897323 103: -0.135259 104: 0.103054 124: -0.105243 125: -0.0419266 126: -0.126801 128: 0.0920044 132: 0.0935455
Row 34: 20: -0.610448 21: -0.43421 32: -0.800343 33: -0.624553 34: 3.5401 36: -0.341534 38: -0.729009 92: 0.0904152 93: 0.114688 94: -0.103361 96: 0.0824828 97: -0.10053 122: -0.0897809 123: 0.108484 124: -0.083047 126: 0.104656 127: -0.140852 128: -0.0724453 131: 0.0892903
Row 35: 25: -0.379385 26: -0.66891 27: -0.683932 28: -0.449095 35: 3.53739 37: -0.674877 38: -0.68119 105: 0.102697 106: -0.119567 107: 0.0801006 108: 0.0988275 109: -0.0900393 110: 0.09049 111: -0.0753288 113: -0.136673 115: 0.121032 129: -0.0640386 130: -0.103918 132: 0.0964175
Row 36: 28: -0.434461 29: -0.637229 30: -0.0973365 31: -0.958711 32: -0.365907 34: -0.341534 36: 3.57061 38: -0.735427 112: 0.079817 114: -0.113163 115: 0.105756 116: 0.0652415 117: -0.0388535 118: 0.078649 119: -0.127668 120: 0.106943 121: -0.0258068 122: 0.0761522 123: -0.122111 127: -0.101772 128: 0.0825421 131: -0.0657272
Row 37: 24: -0.549644 25: -0.887236 33: -0.928035 35: -0.674877 37: 3.68681 38: -0.64702 102: 0.119925 103: 0.128998 104: -0.157316 106: 0.127194 107: -0.0992469 125: -0.108677 126: 0.134351 129: -0.118715 130: 0.104 132: -0.130514
Row 38: 28: -0.408789 33: -0.352491 34: -0.729009 35: -0.68119 36: -0.735427 37: -0.64702 38: 3.55392 113: 0.100971 114: 0.100462 115: -0.133301 124: 0.0930384 125: 0.0779166 126: -0.112206 127: 0.130564 128: -0.102101 129: 0.0893687 130: -0.0768082 131: -0.108455 132: -0.0594486
Row 39: 0: -0.0833333 4: 0 19: 0.0833333 39: 0.0416667 40: 0 50: -0.0208333
Row 40: 0: -0.0833333 4: 0.0833333 19: 0 39: 0 40: 0.0416667 50: -0.0208333
Row 41: 1: -0.0340869 7: -0.0835417 23: 0.117629 41: 0.0381141 43: -0.0208854 59: -0.00852172
Row 42: 1: -0.0342097 8: -0.0831255 23: 0.117335 42: 0.038071 43: -0.0207814 61: -0.00855243
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6. Solve the system:¶
[10]:
gfu.vec.data = \
a.mat.Inverse(freedofs=fes.FreeDofs()) * f.vec
Draw(gfu)
The Dirichlet boundary condition constrains some degrees of freedom. The argument fes.FreeDofs() indicates that only the remaining “free” degrees of freedom should participate in the linear solve.
You can examine the coefficient vector of solution if needed:
[11]:
print(gfu.vec)
0
0
0
0.0923044
0
0
0
0
0
0
0
0
0.0578972
0.0863393
0.0954128
0.0944891
0.0888235
0.0780439
0.059628
0.0330628
0.0374521
0.0370446
0.0317995
0.0185123
0.0383632
0.0427969
0.047351
0.0760518
0.0895602
0.0939297
0.0914143
0.0833171
0.0654761
0.0574569
0.0647326
0.0723792
0.0852186
0.0626708
0.0776805
0
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0
0
0.0208108
0
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0
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0
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0
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0
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Ways to interact with NGSolve¶
• A jupyter notebook (like this one) gives you one way to interact with NGSolve. When you have a complex sequence of tasks to perform, the notebook may not be adequate.
• You can write an entire python module in a text editor and call python on the command line. (A script of the above is provided in poisson.py.) python3 poisson.py
• If you want the Netgen GUI, then use netgen on the command line: netgen poisson.py You can then ask for a python shell from the GUI’s menu options (Solve -> Python shell). | 19,644 | 41,481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-04 | latest | en | 0.738443 |
https://questioncove.com/updates/53950627e4b01c619df16d47 | 1,503,557,919,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133042.90/warc/CC-MAIN-20170824062820-20170824082820-00550.warc.gz | 815,891,639 | 2,701 | OpenStudy (anonymous):
Dean bought a nightstand that was marked down 20% from an original price of $25. If he paid 15% sales tax, what was the total cost of the nightstand? OpenStudy (xguardians): .8 (80% of original price) *$25 (Original Price) = 20$(Step One) 20$ (Sale Price) * .15 (Tax Price) = $3 (Tax Amount) 20$ (Sale Price) + $3 (Tax Amount) =$23 (Final Sale) This should be the answer, if the tax comes after the discount. Have a nice day!
OpenStudy (anonymous):
wowww you r really good, thank u sooo much
OpenStudy (xguardians):
No Problem, Goodluck!
OpenStudy (anonymous):
A store in Oak Grove bought a computer for \$100 and marked it up 125% from the original cost. Later on, Jack purchased the computer and paid Oak Grove sales tax of 8%. How much, including tax, did he pay for the computer? | 224 | 813 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-34 | latest | en | 0.953635 |
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Topic: Incorporating ABS Activitie
Replies: 0
Jeff Witmer Posts: 2 Registered: 12/6/04
Incorporating ABS Activitie
Posted: Jan 6, 1997 11:57 PM
Subject: Time:12:59 PM
OFFICE MEMO Incorporating ABS Activities into a... Date:1/7/97
Recently Chuck Biehl wrote to the list asking about ABS activities. The
following outline shows one way to fit Activity Based Statistics activities
into an introductory statistics course for which the primary text is Moore and
McCabe. This is based on my teaching of such a course last year.
I used "Getting to Know the Class" on the second day of the course. When we
were in Chapter 1 of M/McC (describing distributions), I used "Matching Graphs
to Variables" as an activity for reviewing how to read a histogram and think
about how the shape of a histogram is related to features in the data. I
could also have used "Matching Statistics to Graphs" here. After discussing
boxplots I used the "Living Boxplot" activity.
When we were in Chapter 2 (relationships, correlation, regression) I used a
versions of "Getting Rid of the Jitters," "Matching Descriptions to
Scatterplots," and "The Regression Effect."
The "How to Ask Questions" activity fits with Chapter 3 (producing data). I
used "Capture/Recapture" when discussing sampling and I used "Random
Rectangles" to aid our discussion of bias.
Chapter 4 in M/McC deals with probability. Here I used "Dueling Dice." Once
the class knew something about probability, I used "Randomized Response
Sampling." I used "Spinning Pennies" to set up the idea of a sampling
distribution and "Cents and the Central Limit Theorem" when we got to Chapter
5 (on the distribution of counts and of means from samples). I also used "The
Central Limit Theorem and the Law of Large Numbers" here.
"What is a Confidence Interval Anyway" fits with an introduction to confidence
intervals, which are covered in Chapter 6 in M/McC. I used "Introduction to
Hypothesis Testing" when we started Chapter 7 (hypothesis testing).
Toward the end of the semester I used "Gummy Bears in Space" when we were
discussing ANOVA.
Of course, there are many other choices of activities that one could make. I
would have used more activities, but I only had 13 weeks for the course. I
Jeff Witmer
Oberlin College
Claimer: I am one of the authors of ABS, so my comments should not be taken as
those of an impartial observer. | 636 | 2,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-17 | longest | en | 0.933679 |
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Ajuda do Scilab >> Dicas de Conversão de Matlab para Scilab > Matlab-Scilab equivalents > I > isa (Matlab function)
# isa (Matlab function)
Detect an object of a given type
### Matlab/Scilab equivalent
There is no equivalent function for Matlab isa function in Scilab but it can be replaced by equivalent syntaxes as shown is examples.
Equivalence table Matlab Scilab a = isa(x,"logical") a = type(x)==4; b = isa(x,"char") b = type(x)==10; c = isa(x,"numeric") c = or(type(x)==[1,5,8]); d = isa(x,"int8") d = typeof(x)=='int8'; e = isa(x,"uint8") e = typeof(x)=='uint8'; f = isa(x,"int16") f = typeof(x)=="int16"; g = isa(x,"uint16") g = typeof(x)=="uint16"; h = isa(x,"int32") h = typeof(x)=="int32"; k = isa(x,"uint32") k = typeof(x)=="uint32"; l = isa(x,"single") l = type(x)==1; m = isa(x,"double") m = type(x)==1; n = isa(x,"cell") n = typeof(x)=="ce"; o = isa(x,"struct") o = typeof(x)=="st"; p = isa(x,"function_handle") p = type(x)==13; q = isa(x,"sparse") q = type(x)==5; r = isa(x,"lti") r = typeof(x)=="state-space"; | 384 | 1,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-16 | longest | en | 0.583759 |
https://www.zanestan.es/37101.html!/ | 1,600,457,679,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00095.warc.gz | 1,060,998,579 | 5,835 | # conveyor belt length calculator
## Conveyor Belt Calculations - Bright Hub Engineering
Determine the gauge, or thickness, of your conveyor belt. Belt gauges generally run from between 0.1 inches to 1.3 inches, depending of the application. You can either measure directly or ask your belt manufacturer for the belt specifications. For example, you might have a belt that is 0.5 inches thick.
## Conveyor Horsepower Calculator - Superior Industries
Conveyor Horsepower Calculator Sara Hoidahl 2017-08-28T21:17:11+00:00 Conveyor Length (center-to-center) 25 feet 50 feet 75 feet 100 feet 150 feet 200 feet 250 feet 300 feet 350 feet 400 feet 450 feet 500 feet 550 feet 600 feet 650 feet 700 feet
## Belt Length and Center Distance Calculator
d = Pitch Diameter Small Pulley. C = Center Distance. L = Belt Pitch Length
## Flat Belt Length Distance Calculator | Engineers Edge ...
Flat Belt Length Distance Calculator V and Flat Belt Design Guide Menu. This engineering calculator will determine the distance between two known diameters pulleys centers. Flat Belt Length and Pulley Center Distance Calculation and Equations
## Conveyor Calculators - Superior Industries
Conveyor Lift - Stockpile Volume - Conveyor Horsepower - Maximum Belt Capacity - Idler Selector Find conveyor equipment calculators to help figure specs. +1 (320) 589-2406 | (800) 321-1558 | [email protected]
## Belt Conveyors for Bulk Materials Practical Calculations
3. Belt Length Correction, L c: Short belt conveyors require relatively more power to overcome the resistance to friction than long ones and therefore an adjustment is made to calculate the effective tension. L C = L + 70 m (metric) L C = L + 230 ft (imperial) Therefore, the belt length …
## How to Calculate Conveyor Belt Speed | Sciencing
This calculation provides the linear distance traversed by a point on the conveyor belt in one minute. Calculate the distance traveled for one hour. For example, a roller with a diameter of 2 inches, has a circumference of 2 x 3.14159 or 6.28 inches.
## Calculator - Batco Manufacturing
what belt speed is ideal for your conveyor; how much ground length is necessary to ensure your conveyor will reach the desired bin height; The Batco Calculator is easy to use and is designed to be practical. Use the Batco Calculator as a tool to help you with your next conveyor purchase! Click "Conveyor Length" or "Ground Length" on the ...
## Pulley Belt Calculations - The Engineering Mindset
May 26, 2019· These calculations are mainly for existing installations but you could also put it towards your design if you want to calculate what the length needs to be of the belt, as long as you already know the distance between the two pulleys and the size of the pulleys.
## How to Measure a Conveyor Belt: Center-to-Center Distance ...
If you need help calculating the length of your replacement conveyor belt, Accurate Industrial's experts can walk you through the process. Fill out our request form to receive a belt quote. If you have additional questions, call us at call us at (800) 684-2358
## Belt Length Calculator
Belt length = ((D L + D S) * π / 2) + (2 * L) + (D L - D S) 2 / (4 * L) where: D L is the diameter of the large pulley. D S is the diameter of the smaller pulley. L is s the distance between the pulley axles. With our belt length calculator, you can also calculate the distance between the centers of two pulleys for a given belt length.
## How Do I Calculate Conveyor Belt Tension? (with pictures)
The first factor to calculate when figuring out conveyor belt tension is the tension needed to move the empty belt, or TC. You'll need to know the weight of the conveyor belt components, or CW, the belt length, or L, and the friction factor for normal empty belt operation, or F1.
## Conveyor Belt Length Calculation | How To Calculate Length ...
Jun 06, 2020· Conveyor belt roll length Calculation||How to calculate length of belt roll||Rolled conveyor length - Duration: 10:24. SUPER FAST STUDY & Experiment 9,439 views 10:24
## Timing Belt Calculator - Belt Length Calculator | B&B ...
Belt Length Calculator. Please Select the Following Reset. Desired Center Distance (in) Set. Attribute: Value: Speed Ratio : Pitch Diameter - Large Pulley (A) (in) Pitch Diameter - Small Pulley (B) (in) Desired Belt Pitch Length (in) Belt Pitch Length (in) Number of Teeth - Belt ...
## How to Calculate Belt Length - Bolton Engineering
How to Calculate Belt Length How to Determine which Fan Belt you require. If the centre distance (C) is known, belt length or pitch length can be determined as follows: So if you know the distance between the center points of the pulleys, and their diameters, you can calculate the length of belt …
## How To Calculate Conveyor Belt Length On A Roll - Industry ...
Calculation method for the length of the roll conveyor belt 73. Therefore, the length of the roll conveyor belt can be calculated according to the empirical formula. 3. Factors influencing the deviation. It can be seen from the above examples that the calculation is based on the empirical formula.
## Formula for rolled up conveyor belting - Mining ...
Aug 15, 2008· I was wondering if anyone has a formula for figuring out the length of a conveyor belt rolled up on a spool. RE: Formula for rolled up conveyor belting minerk (Mechanical) 11 Aug 08 08:23.
## Belt Calculator | Conveyor Belt Maintenance | Shipp Belting
Belt Speed Calculator These calculation tools are to provide product selection ONLY and final application suitability is the sole responsibility of the user. Belt Speed Calculator Slave Belt Speed Calculator Dura-Drive Belt Speed Calculator Belt Speed
## CONVEYOR HANDBOOK - hcmuaf.edu.vn
protect the carcass and give the conveyor belt an economical life span. The general properties and the application usage of the more economical available reinforcement fabrics and rubber compounds are discussed in this section. REINFORCEMENTS Fabrics Fabrics that are commonly used as reinforcement in conveyor belts are shown in Table 1 of this
## Belt Length Calculator | Distance Between Pulleys Calculator
Belt Length Calculator | Distance Between Pulleys Calculator <<< SEE ALL ONLINE CALCULATORS . Practical Application: Broken Belt, Unkown Length. Have you ever been out in the field or to a customers site only to discover a broken belt? You take a closer look only to realize that the belt is old and worn and you can't make out a single word on it.
## Roll Length Calculator - FLEXcon
Roll Length Calculator. Enter any three of the four values below and click "Calculate". The remaining value will calculate for you.
## How to Calculate Belt Length - Bolton Engineering
Jan 06, 2015· How to Calculate Belt Length How to Determine which Fan Belt you require. If the centre distance (C) is known, belt length or pitch length can be determined as follows: So if you know the distance between the center points of the pulleys, and their diameters, you can calculate the length of belt you need using this formula.
## How To Calculate Conveyor Belt Length - Industry knowledge ...
Calculate the length of each belt and then add up. Is the length of the conveyor belt, but because the conveyor has a certain thickness. Degree, when calculating the length of the belt conveyor, we take the thickness of the belt. The length of the circumference of the heart line is taken as the length of the belt conveyor, ie:
## Calculation methods – conveyor belts
Conveyor length l T β Belt speed v m/s Belt sag y B mm Drum deflection y Tr mm Arc of contact at drive drum and idler β ° Opening angle at drive drum γ ° Incline (+) or decline (–) angle of conveyor α, δ ° Elongation at fitting ε % Drive efficiency η – Density of material conveyed ρ s kg/m3 Designation Symbol Unit Terminology 2
## V-Belt Length Calculator | John Henry Foster
V-Belt Length Calculator. Our free V-Belt Length Calculator can be used to select the correct belt for your system or product. V-Belt Length Calculator, John Henry Foster.
## Understanding Conveyor Belt Calculations | Sparks Belting
Common Calculations for Proper Design Belt Length . When the head and tail pulley are the same size: L=(D+d)/2 x 3.1416+2C When one pulley is larger than the other pulley: L=(D+d)/2 x 3.1416+2C+(D-d) 2 /4c Belt …
## Conveyor Belt Equations
The required take-up length is calculated as follows, where . S Sp = take-up length (m) L = conveyor length (m) ε = belt elongation, elastic and permanent (%) As a rough guideline, use 1,5 % elongation for textile belts. and 0,2 % for steel cord belts. Note: For long-distance conveyors, dynamic start-up calculations
## Conveyor belt roll length Calculation||How to calculate ...
Sep 16, 2019· Hello friend In this video we see how to calculate length of rolled industrial conveyor belt By simple formula in Hindi If this video is helpful to you please subscribe my channel. | 2,007 | 8,928 | {"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} | 3.203125 | 3 | CC-MAIN-2020-40 | latest | en | 0.801937 |
https://www.thestatesman.com/world/what-is-pi-day-1502601197.html | 1,679,422,933,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00667.warc.gz | 1,177,298,663 | 31,318 | What is Pi Day? Who started it, and when?
Google Doodle celebrates Pi Day with a pie recipe by celebrated pastry chef Dominique Ansel; NASA has shared its annual Pi in the Sky challenge
SNS | New Delhi | Updated :
Observed on March 14 (3/14) every year, Pi Day is an annual celebration of the mathematical constant π (pi), which calculates to over 1 trillion digits beyond its decimal point. While its digits go on infinitely, Pi is most popularly known by the first three: 3.14, which is why March 14 has come to be known as the Pi Day.
Said to have been first recognised in 1988 by physicist Larry Shaw, Pi Day is now celebrated across the word in different ways.
While many celebrate the day with a slice of their favourite pie, thanks to the number’s similar sounding name, there are others who dedicate the day to the genius of mathematics.
Wednesday’s Google Doodle has a beautifully baked pie cooked and built by award-winning pastry chef and creator of the Cronut® Dominique Ansel. It pays homage to the mathematical constant by representing the pi formula (circumference divided by diameter) using a pie.
Google has also shared Chef Dominique’s unique with the viewers.
Pi helps determine the circumference or the surface area of a round celestial body. It represents the ratio between a circle’s circumference to its diameter, and is a fundamental element of many mathematical fields.
NASA celebrates Pi Day by sharing a series of cosmic calculations to be solved by the public. Its ‘Pi in the Sky’ challenge, created by Jet Propulsion Laboratory’s Education Office, features math problems that illustrate how pi can be used to learn about all kinds of curious features of the universe, including earthquakes on Mars, helium rain on Jupiter, and planets orbiting other stars.
In its fifth year, the annual “Pi in the Sky” illustrated math challenge features pi-related space problems that people can do at home.
“All of the problems in the ‘Pi in the Sky’ challenge are real problems that JPL scientists and engineers solve using pi,” the NASA website quotes Ota Lutz, a senior education specialist at JPL who helped create the Pi Day Challenge.
This year’s NASA Pi Day challenge features the agency’s InSight Mars lander, Kepler space telescope and Juno spacecraft at Jupiter — plus a visit from an interstellar object. The solutions to the problems posted online on March 9 will be published on March 15.
In previous years, Pi in the Sky featured problems about how NASA space probes gather information about objects in our solar system, how scientists search for planets around other stars, and how astronomers predict the occurrence of solar eclipses.
“The Pi in the Sky problems give people a little glimpse into what goes on at JPL,” Lutz said. “And that’s empowering, because it shows people that they can understand some of the magic that goes into space exploration.” | 601 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-14 | longest | en | 0.952869 |
https://www.coursehero.com/file/6622158/ClassNotesComplete/ | 1,529,495,643,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00159.warc.gz | 789,193,379 | 128,642 | ClassNotesComplete
# ClassNotesComplete - STA 3032(7661 Engineering Statistics...
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STA 3032 (7661) Engineering Statistics Rob Gordon University of Florida Fall 2011 Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 1 / 251
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Introduction: Sampling & Descriptive Statistics Definition A population is the entire collection of objects or outcomes about which information is sought. Examples: Entire United States Entire State of Florida All UF students Definition A sample is a subset of the population, containing the objects or outcomes that are actually observed. Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 2 / 251
Introduction: Sampling & Descriptive Statistics A common question might be: “ How do I know if a sample is truly representative of its population? Ideally, the best way to accomplish this goal is to select the members of the sample in the most unbiased possible way. Throughout the rest of this course we will assume our samples will follow the definition of a simple random sample: Definition A simple random sample of size n is a sample chosen by a method in which each collection of N population items is equally likely to comprise a sample (as in a lottery). Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 3 / 251
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Introduction: Sampling & Descriptive Statistics “Summary Statistics” help the important features of a sample stand out. Definition Let x 1 , x 2 , . . . , x n denote numbers in a sample. The value of the sample mean is ¯ x = 1 n n X i =1 x i . Definition The value of the sample variance s 2 = 1 n - 1 n X i =1 ( x i - ¯ x ) 2 = 1 n - 1 n X i =1 x 2 i - n ¯ x 2 ! Definition The value of the sample standard deviation is s = s 2 . Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 4 / 251
Introduction: Sampling & Descriptive Statistics Definition Outliers are points in the sample that are much smaller/larger than the rest. Outliers often result from data entry errors (e.g. incorrect decimal place) and can present many problems for statisticians (more on this later). Caution: Only delete an outlier if it exists due to error! Definition The sample median is numerical value separating the higher half of a sample from the lower half. ˜ x = ( x n +1 2 , if n is odd 1 2 ( x n / 2 + x n / 2+1 ) , if n is even. Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 5 / 251
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Introduction: Sampling & Descriptive Statistics The median divides the sample in halves, while quartiles divide the data into quarters. Let Q 1 = 1st quartile = # greater than 25% of all data points. Q 2 = 2nd quartile = # greater than 50% of all data points. Q 3 = 3rd quartile = # greater than 75% of all data points. Note: Sometimes quartiles are not numbers in the sample. Definition Q 1 = x 0 . 25( n +1) if 0 . 25( n + 1) is an integer avg of values above and below otherwise. Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 6 / 251
Introduction: Sampling & Descriptive Statistics Example 1 Sample = { 1, 2, 3, 4, 5, 6, 7 } Q 1 = x 0 . 25(7+1) = x 2 = 2 Example 2 Sample = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } Q 1 = x 0 . 25(10) = x 2 . 5 = 1 2 ( x 2 + x 3 ) = 1 2 (2 + 3) = 2 . 5 Rob Gordon (University of Florida) STA 3032 (7661) Fall 2011 7 / 251
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Introduction: Sampling & Descriptive Statistics Similarly, Definition Q 2 = median Q 3 = x 0 . 75( n +1) if 0 . 75( n + 1) is an integer avg of values above and below otherwise.
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http://mathhelpforum.com/discrete-math/111192-cartesian-product-proofs.html | 1,496,151,424,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615105.83/warc/CC-MAIN-20170530124450-20170530144450-00123.warc.gz | 289,910,845 | 11,668 | 1. ## cartesian product proofs
A X (B U C)= (A X B) U (A X C)
proof:
Assume (x,y) is an element of A X (B U C). This means x is an element of A and y is an element of B or y is an element of C. Since (x,y) can be x as an element of A and y as an element of B, (x,y) is an element of A X B. Since (x,y) can also have x as an element of A and y as an element of C, (x,y) is an element of A X C.
A X (B-C)= (A X B) - (A X C)
not sure
A X (B intersect C)= (A X B) intersect (A X C)
A X (B-C)= (A X B) - (A X C)
not sure
Here is one. They all work about the same way.
$\begin{gathered}
(x,y) \in A \times (B\backslash C) \hfill \\
x \in A \wedge y \in B \wedge y \notin C \hfill \\
\left( {x \in A \wedge y \in B} \right) \wedge \left( {x \in A \wedge y \notin C} \right) \hfill \\
(x,y) \in A \times B \wedge (x,y) \notin A \times C \hfill \\
(x,y) \in \left( {A \times B} \right)\backslash \left( {A \times C} \right) \hfill \\
\end{gathered}$
3. is a slash equal to the difference
also is the first 1right
4. i dont understand
i dont understand
Maybe you should try to get some one-on-one help from your instructor.
6. A X (B U C)= (A X B) U (A X C)
Proof:
(x,y) E A X (B U C).
x E A and y E (B U C)
X E A and (y E B U y E C)
(x E A and y E B) or (x E A and y E C)
(x,y) E A X B or (x,y) E A X C
(x,y) E (A X B) or (A X C)
and the argument can be reversed
the only thing I cant justify is how I go from step 3 to step 4 and how I go from step 4 to step 3.
E= element of
7. ## cartesiian proof
Proof:
(x,y) E A X (B intersect C)
x E A and y E (B intersect C)
x E A and y E B and y E C
(x E A and y E B) and (x E A and y E C)
(x,y) E A X B and (x,y) E A X C
(x,y) E (A X B) intersect (A X C)
again i am only not sure how to justify going from step 3 to step 4 and how to go from step 4 to step 3
the argument reverses to prove they are equal
E= element of
,
,
,
,
,
,
,
,
,
,
,
,
,
,
# cartessian product of intersection
Click on a term to search for related topics. | 729 | 1,995 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-22 | longest | en | 0.820339 |
https://intecexpo.com/faqs/what-is-the-conversion-factor-from-gigaliters-to-liters/ | 1,674,949,811,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00559.warc.gz | 339,946,283 | 8,130 | # What Is the Conversion Factor From Gigaliters to Liters?
November 10, 2022
The SI prefix “giga” stands for a factor of 109 or in exponential notation 1E9. So 1 gigaliter = 109 liters.
## Can u convert G to L?
How to convert grams to liters. To convert a gram measure to a liter measure, divide the weight by 1,000 times the density of the ingredient or material. Thus, the weight in liters is equal to grams divided by 1,000 times the density of the ingredient or material.
## How many liters are there in one Gigaliter in scientific notation?
How many liters are in a gigaliter? The answer is: One gigaliter is 1000000000 liters.
## How many liters is a Teraliter?
How many liters are in a teraliter? The answer is: One teraliter equals 1000000000000 liters.
## How do you convert volume to litres?
Cubic meters and liters are two common metric units of volume. 1 cubic meter is 1000 liters. The easiest way to convert cubic meters to liters is to move the decimal point three places to the right. In other words, multiply a value in cubic meters by 1000 to get the result in liters.
## How do you convert grams to liters per percent?
To convert mg/l to percent, divide mg/l by 1000 and multiply by 100.
## Are g and L equal?
Grams and liters are common units. A gram is a unit of mass, roughly equivalent to a paper clip, while a liter is a unit of volume and is a common allotment of liquids such as beverages or gasoline.
## What is volume measured by?
Volume is the measure of the three-dimensional space occupied by matter or enclosed by a surface, measured in cubic units. The SI unit of volume is the cubic meter (m3), which is a derived unit. Liter (L) is a special term for cubic decimeter (dm3).
## Is 250ml half a Litre?
So 250ml corresponds to a quarter liter. And 500 ml equals half a liter. To convert liters to milliliters, multiply the liter value by 1000.
## How do you convert from liters to milliliters?
You can just move the decimal point. Move it right three places to convert liters to milliliters (e.g. 5.442 L = 5443 ml) or left three places to convert milliliters to liters (e.g. 45 ml = 0.045 L).
## Is a 1000ml 1 liter?
Is 1 liter the same as 1000 ml? Yes, 1L = 1000ml. Although both liters (L) and milliliters (mL) represent the same amount, their values are different.
## How do you calculate liters?
The first thing you need to do is multiply the length by the width by the height. This gives the number of cubic millimeters. To calculate the number of liters, then divide that number by a million.
### References:
1. https://mste.illinois.edu/dildine/tcd_files/metric/volume.htm
2. https://www.inchcalculator.com/convert/gram-to-liter/
3. https://hextobinary.com/unit/volume/from/gigaliter/to/litre
4. https://hextobinary.com/unit/volume/from/teraliter/to/litre
6. https://www.thoughtco.com/cubic-meters-to-liters-example-problem-609385
7. https://www.quora.com/How-can-I-convert-mg-L-to-percentage
8. https://www.easyunitconverter.com/grams-to-liters
9. https://sciencing.com/convert-gram-liters-4869894.html
10. https://support.edx.org/hc/en-us/articles/360000038208-Scientific-notation-and-metric-affixes | 844 | 3,169 | {"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} | 3.828125 | 4 | CC-MAIN-2023-06 | latest | en | 0.891534 |
https://www.examrace.com/NIFT/NIFT-Model-Questions/Physics-Questions/Physics-MCQs-Practice-Test-30.html | 1,603,923,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00629.warc.gz | 722,869,466 | 7,383 | NIFT: Physics MCQs (Practice_Test 30 of 35)
Get unlimited access to the best preparation resource for IAS : fully solved questions with step-by-step explanation- practice your way to success.
1. The hypercharge Y of an elementary particle is equal to
1. Q, the charge on the particle
2. B, the Baryon number
3. S, the strangeness quantum number
4. B + S
2. The particle W-was predicted in 1963 as member of the Baryon decoupled and discovered in 1964. Its spin J, isospin I and strangeness quantum number S are J I S
1. ½ 0 − 1
2. 3/2 1 − 2
3. 3/2 0 − 3
4. 3/2 0 + 3
3. Given that Eb1 = 250 V at Ec1 = -50V, Eb2 = 180 V at Ec2 = -30 V and plate current being kept the same, the of a triode is
1. 6
2. 5
3. 3.5
4. 5.5
4. Pure silicon is simultaneously doped with boron to a concentration of 1020 atoms per cubic centimeter and with phosphorus to a concentration of 7 × 1019 atoms per cubic centimeter. The number of holes per unit volume of silicon will be approximately
1. 1020
2. 17 × 1019
3. 7 × 1019
4. 3 × 1019
5. The output characteristic of a bipolar junction transistor is shown in the following diagram, in the diagram, the Active region A, cut-off region B, Saturation region C and Burnout region D are respectively the regions labeled as
1. n/a
2. N/A
3. N/A
4. N/A
6. Which one of the following diagram correctly indicates the widths of the space charge regions of the given pn junction?
1. n/a
2. N/A
3. N/A
4. N/A
7. An electrical property (X) of a solid varies with temperature (T, in Kelvin) as shown schematically in the given diagram. This is an illustration of the variation with temperature of the
1. resistivity of pure silicon
2. resistivity of pure annealed cope
3. conductivity of pure silicon
4. conductivity of p-type silicon
8. The current (J) voltage (V) characteristic of a p-n diode is as shown in the given figure. The knee voltage is at the point labeled
1. C
2. A
3. O
4. B
9. Match List-I with List-II and select the correct answer.
List-I List-II Emitter-base junction currentBase currentEmitter-Collector currentBase-Collector Driff currentDiffusion currentRecombinationjunction current current
• A
• B
• C
• D
• 1
• 2
• 3
• 4
• 1
• 3
• 2
• 4
• 2
• 3
• 1
• 4
• 3
• 2
• 4
• 1
10. The number 6.25 in the decimal notation, when converted to the binary notation will read as
1. 110.100
2. 101.010
3. 110.010
4. 101.001
Developed by: | 757 | 2,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-45 | latest | en | 0.833765 |
https://forum.ansys.com/forums/topic/the-reaction-rate-for-a-wall-surface-reaction/ | 1,695,346,504,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00604.warc.gz | 289,551,031 | 32,264 | ## Fluids
Topics relate to Fluent, CFX, Turbogrid and more
#### The reaction rate for a wall surface reaction
• WEI LI
Subscriber
Hello Ansys experts,
I am modeling a wall surface reaction. Consider the chemical reaction ongoing on a surface is A (species in fluid) + B (site species) --> C (solid species);
If I understand correctly, Ansys Fluent computes the reaction rate by R=k*c(A)*c(B), where k is the reaction constant, c(A) is the molar concentration in fluid in the unit [kmol/m^3] and c(B) represents the site density in the unit [kmol/m^2]. Neither the backward reaction nor coverage dependence was considered in this context.
k and c(B) are both just inputs from me, and they are not changing throughout the simulation period if I understand correctly. Notice that the temperature exponent and activation energy for reactions are both zero in the current simulation.
My question is
1. how Ansys Fluent selects c(A). Is it the molar concentration at the cell adjacent to the reaction surface?
2. If my understanding above is correct?
Thank you and regards
• Swathi V. V.
Ansys Employee
Hi ,
Please refer to the following user manaul page to know how rate of reaction is computed for surface reaction. Please refer to the equation 7-54.in 7.1.2. Wall Surface Reactions and Chemical Vapor Deposition (ansys.com)
C(A) is the Molar concentration of surface absorbed gaseous species on the wall
• WEI LI
Subscriber
Hi Swathi,
Thank you for your reply. However, I am still not quite sure about the difference between the molar concentration of surface-absorbed gaseous species on the wall and that at the cell adjacent to the wall.
The below image shows the molar concentration of A near the target wall. Can I say c(A)=5.7e-16 kmol/m^3?
Thank you and regards
• WEI LI
Subscriber
Also, I think it should be [Gi] instead of [Ci] in the equation 7-54.in 7.1.2. Wall Surface Reactions and Chemical Vapor Deposition (ansys.com), right? | 473 | 1,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-40 | latest | en | 0.899282 |
http://www.greatschools.org/students/academic-skills/536-preparing-for-fifth-grade.gs | 1,429,418,597,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246637445.19/warc/CC-MAIN-20150417045717-00140-ip-10-235-10-82.ec2.internal.warc.gz | 554,136,936 | 24,880 | # What to expect in fifth grade
By Miriam Myers
In fifth grade, your child will build on the skills she developed in fourth grade and study many of the same subjects at a higher level.
She will read many types of literature and informational material, think critically about what she reads, and discuss it with the teacher and other students. She will write for many different purposes and learn new writing techniques, including making effective transitions, using dialogue to advance the plot and creating a mood, such as suspense. In math she will work with fractions, decimals, negative numbers and very large numbers (in the billions). She will also focus on multiplying and dividing multi-digit numbers.
Tonya Breland, our teacher consultant, explains: "While preparing your child for fifth grade, keep in mind her growing understanding of more complex issues in society. This is a great time to get involved in community service activities to help foster character development, a sense of giving and responsibility. "
### Everyday math
To prepare your child for fifth-grade math, involve her in solving math problems you encounter in real life. When you go to a store together let her pay and have her check that she received the correct change. She can estimate what the grocery store bill or restaurant bill will be. She can measure the distance you'll travel on a family trip or help measure a room when you are hanging a picture or rearranging furniture.
Children pass through a range of social, academic and developmental stages, each at their own pace. Below are rough guidelines of what to look forward to in the year ahead.
### Physical and social skills you can expect of your fifth grader:
• Develop increasing independence
• Improve problem-solving skills
• Acquire more advanced listening and responding skills
• Need for more sleep
• Enjoy organizing and classifying objects and ideas
• Be able to read and concentrate for long periods of time
• Read complex text fluently and with good comprehension
• Research a topic using a variety of sources and use the features of books (e.g., index, glossary, appendix) to find information
• Identify the conflict, climax and resolution in a story
• Write an organized, multi-paragraph composition in sequential order with a central idea
• Use problem-solving strategies to solve real-world math problems
• Add and subtract fractions and decimals
• Identify and describe three-dimensional shapes, and find their volumes and surface areas
• Use long division to divide large numbers by multi-digit divisors (86,554 / 392)
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"this is a good article but i was expecting a list of school supplies for 5th grade "
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06/16/2009:
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06/10/2009: | 867 | 3,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-18 | latest | en | 0.956998 |
https://en.wikipedia.org/wiki/Lenz_law | 1,493,279,961,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121893.62/warc/CC-MAIN-20170423031201-00214-ip-10-145-167-34.ec2.internal.warc.gz | 783,031,387 | 20,856 | # Lenz's law
(Redirected from Lenz law)
Lenz's law (pronounced /ˈlɛnts/), named after the physicist Emil Lenz who formulated it in 1834,[1] says:
The direction of current induced in a conductor by a changing magnetic field due to Faraday's law of induction will be such that it will create a field that opposes the change that produced it.
${\displaystyle {\mathcal {E}}=-{\frac {\partial \Phi }{\partial t}},}$
which indicates that the induced voltage (${\displaystyle {\mathcal {E}}}$) and the change in magnetic flux (${\displaystyle \partial \Phi }$) have opposite signs.[2] It is a qualitative law that specifies the direction of induced current but says nothing about its magnitude. Lenz's Law explains the direction of many effects in electromagnetism, such as the direction of voltage induced in an inductor or wire loop by a changing current, or why eddy currents exert a drag force on moving objects in a magnetic field.
Lenz's law can be seen as analogous to Newton's third law in classic mechanics.[3]
For a rigorous mathematical treatment, see electromagnetic induction and Maxwell's equations.
## Opposing currents
If a change in the magnetic field of current i1 induces another electric current, i2, the direction of i2 is opposite that of the change in i1. If these currents are in two coaxial circular conductors 1 and 2 respectively, and both are initially 0, then the currents i1 and i2 must counter-rotate. The opposing currents will repel each other as a result.
Lenz's law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux and to exert a mechanical force opposing the motion.
### Example
Currents bound inside the atoms of strong magnets can create counter-rotating currents in a copper or aluminum pipe. This is shown by dropping the magnet through the pipe. The descent of the magnet inside the pipe is observably slower than when dropped outside the pipe.
When a voltage is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced voltage is such that it produces a current whose magnetic field opposes the change which produces it. The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant. In the examples below, if the flux is increasing, the induced field acts in opposition to it. If it is decreasing, the induced field acts in the direction of the applied field to oppose the change.
## Detailed interaction of charges in these currents
Aluminium ring moved by electromagnetic induction, thus demonstrating Lenz's law.
Experiment showing Lenz's law with two aluminium rings on a scales-like device set up on a pivot so as to freely move in the horizontal plane. One ring is fully enclosed, while the other has an opening, not forming a complete circle. When we place a bar magnet near the fully enclosed ring, the ring is repulsed by it. However, when the system comes to a rest, and we remove the bar magnet, then the ring is attracted by it. In the first case, the induced current created in the ring resists the increase of magnetic flux caused by the proximity of the magnet, while in the latter, taking the magnet out of the ring decreases the magnetic flux, inducing such current whose magnetic field resists the decrease of flux. This phenomenon is absent when we repeat the experiment with the ring that isn't enclosed by inserting and removing the magnet bar. The induced currents in this ring can't enclose themselves in the ring, and have a very weak field that cannot resist the change of the magnetic flux.
In electromagnetism, when charges move along electric field lines work is done on them, whether it involves storing potential energy (negative work) or increasing kinetic energy (positive work).
When net positive work is applied to a charge q1, it gains speed and momentum. The net work on q1 thereby generates a magnetic field whose strength (in units of magnetic flux density (1 tesla = 1 volt-second per square meter)) is proportional to the speed increase of q1. This magnetic field can interact with a neighboring charge q2, passing on this momentum to it, and in return, q1 loses momentum.
The charge q2 can also act on q1 in a similar manner, by which it returns some of the momentum that it received from q1. This back-and-forth component of momentum contributes to magnetic inductance. The closer that q1 and q2 are, the greater the effect. When q2 is inside a conductive medium such as a thick slab made of copper or aluminum, it more readily responds to the force applied to it by q1. The energy of q1 is not instantly consumed as heat generated by the current of q2 but is also stored in two opposing magnetic fields. The energy density of magnetic fields tends to vary with the square of the magnetic field's intensity; however, in the case of magnetically non-linear materials such as ferromagnets and superconductors, this relationship breaks down.
## Field energy
The electric field stores energy. The energy density of the electric field is given by:
${\displaystyle u={\frac {1}{2}}\varepsilon |\mathbf {E} |^{2}\,,}$
In general the incremental amount of work per unit volume ${\displaystyle \delta W}$ needed to cause a small change of magnetic flux density ${\displaystyle \delta }$B is:
${\displaystyle \delta W=\mathbf {H} \cdot \delta \mathbf {B} .}$
## Conservation of momentum
Momentum must be conserved in the process, so if q1 is pushed in one direction, then q2 ought to be pushed in the other direction by the same force at the same time. However, the situation becomes more complicated when the finite speed of electromagnetic wave propagation is introduced (see retarded potential). This means that for a brief period the total momentum of the two charges is not conserved, implying that the difference should be accounted for by momentum in the fields, as asserted by Richard P. Feynman.[4] Famous 19th century electrodynamicist James Clerk Maxwell called this the "electromagnetic momentum".[5] Yet, such a treatment of fields may be necessary when Lenz's law is applied to opposite charges. It is normally assumed that the charges in question have the same sign. If they do not, such as a proton and an electron, the interaction is different. An electron generating a magnetic field would generate an EMF that causes a proton to accelerate in the same direction as the electron. At first, this might seem to violate the law of conservation of momentum, but such an interaction is seen to conserve momentum if the momentum of electromagnetic fields is taken into account.
## References
1. ^ Lenz, E. (1834), "Ueber die Bestimmung der Richtung der durch elektodynamische Vertheilung erregten galvanischen Ströme", Annalen der Physik und Chemie, 107 (31), pp. 483–494. A partial translation of the paper is available in Magie, W. M. (1963), A Source Book in Physics, Harvard: Cambridge MA, pp. 511–513.
2. ^ Giancoli, Douglas C. (1998). Physics: principles with applications (5th ed.). p. 624.
3. ^ Schmitt, Ron. Electromagnetics explained. 2002. Retrieved 16 July 2010.
4. ^ The Feynman Lectures on Physics: Volume I, Chapter 10, page 9.
5. ^ Maxwell, James C. A treatise on electricity and magnetism, Volume 2. Retrieved 16 July 2010. | 1,646 | 7,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-17 | latest | en | 0.931314 |
https://onlinetyari.com/blog/rrb-group-d-exam-analysis-2018-all-shift-3rd-october/ | 1,670,276,361,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00384.warc.gz | 462,121,804 | 25,808 | # RRB Group D Exam Analysis 2018 (All Shift): 3rd October
0
1565
RRB Group D Exam Analysis 2018 (All Shift): The Railway Recruitment Board has accelerated to Day 13 spot to complete the journey of the group D recruitment process. The day trigged off in multiple shifts and here is the complete analysis with the questions asked of all shifts
## RRB Group D Exam Analysis 2018 (All Shift): 3rd October
As we all know, the exam was composed of 4 sections i.e. Mathematics, General Intelligence & Reasoning, General Science, and General Awareness. THE OVERALL DIFFICULTY LEVEL OF RRB Group D CBT Exam was Moderate. Let us dig a bit deeper and understand the construct of the paper section-wise
### Mathematics
• S.I, CI: 3 Questions
• Pipe & Cistern: 1 Questions
• Profit/ Loss: 2 Questions
• Time, Speed and Distance: 2 Questions
• Average: 1 Question
• Trigonometry: 3 Questions
• Percentage: 2 Questions
• DI (Pie Chart): 2 Questions
• Ages: 1 Question
• Time and Work: 2 Questions
• Geometry (circle): 1 Question
• Simplification : 3 Questions
### Reasoning
• Syllogism – 4 Questions
• Analogy – 3 Questions
• Series – 2 Questions
• Odd one Out – 2 Question
• Venn Diagram – 2 Questions
• Mirror Image – 2 Questions
• Direction – 1 Question
• Statement & Conclusion – 4 Questions
• Coding – Decoding – 3 Questions
• Calendar – 2 Question
• Blood Relation – 2 Question
• Counting Figure – 2 Question
• Miscellaneous
### General Science
• Physics: 7-8 Questions
• Chemistry: 6-7 Questions
• Biology: 10-11 Questions
### Current Affairs Questions (All Shifts)
1. Who is the Brand Ambassador of India in Peru?
Sandeep Chakravorty
2. Who is the author of “Why I am Hindu”?
Shashi Tharoor
3. What is the new capital of Andhra Pradesh?
Amravati
4. Who is the Chief Minister of Punjab?
Amarinder Singh
5. Who was defeated in the First Battle of Panipat?
Lodi
6. Who has won the 32nd Santosh Trophy?
West Bengal
7. Left Hand Day is celebrated on ?
13th August
8. Who is the new chairman of Nasscom?
Devjaani Ghosh
9. Sachin Tendulkar has launched which new app ?
100 Mb Cricket app
10. Who is hosting the 39th National Games in India?
Meghalaya
11. Neeraj Chopra belongs to which sports?
Athletics
12. Which Movie got the Best Movie title in IIFA 2017?
Neerja
13. Who is the Governor of Orissa?
Professor Ganeshi Lal
14. What is the name of the lake between Hyderabad and Secunderabad?
Hussain Sagar lake
15. Who is the Chief Secretary of Delhi Government?
Anshu Prakash
16. What is the original name of Savita Ambedkar?
Maisaheb
17. In which section of the constitution the scheduled caste and scheduled tribe is mentioned?
Article 341 and Article 342
18. When was United Nations established?
24 October 1945
19. Who wrote Satyarth Prakash?
Swami Dayanand Saraswati
20. Which is the third Highest National Award in India? | 792 | 2,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.788133 |
https://metanumbers.com/461049 | 1,631,827,943,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00589.warc.gz | 455,370,958 | 7,430 | # 461049 (number)
461,049 (four hundred sixty-one thousand forty-nine) is an odd six-digits composite number following 461048 and preceding 461050. In scientific notation, it is written as 4.61049 × 105. The sum of its digits is 24. It has a total of 3 prime factors and 8 positive divisors. There are 305,760 positive integers (up to 461049) that are relatively prime to 461049.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 24
• Digital Root 6
## Name
Short name 461 thousand 49 four hundred sixty-one thousand forty-nine
## Notation
Scientific notation 4.61049 × 105 461.049 × 103
## Prime Factorization of 461049
Prime Factorization 3 × 313 × 491
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 461049 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 461,049 is 3 × 313 × 491. Since it has a total of 3 prime factors, 461,049 is a composite number.
## Divisors of 461049
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 617952 Sum of all the positive divisors of n s(n) 156903 Sum of the proper positive divisors of n A(n) 77244 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 679.006 Returns the nth root of the product of n divisors H(n) 5.96874 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 461,049 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 461,049) is 617,952, the average is 77,244.
## Other Arithmetic Functions (n = 461049)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 305760 Total number of positive integers not greater than n that are coprime to n λ(n) 76440 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 38459 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 305,760 positive integers (less than 461,049) that are coprime with 461,049. And there are approximately 38,459 prime numbers less than or equal to 461,049.
## Divisibility of 461049
m n mod m 2 3 4 5 6 7 8 9 1 0 1 4 3 1 1 6
The number 461,049 is divisible by 3.
## Classification of 461049
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (461049)
Base System Value
2 Binary 1110000100011111001
3 Ternary 212102102220
4 Quaternary 1300203321
5 Quinary 104223144
6 Senary 13514253
8 Octal 1604371
10 Decimal 461049
12 Duodecimal 1a2989
20 Vigesimal 2hcc9
36 Base36 9vqx
## Basic calculations (n = 461049)
### Multiplication
n×y
n×2 922098 1383147 1844196 2305245
### Division
n÷y
n÷2 230524 153683 115262 92209.8
### Exponentiation
ny
n2 212566180401 98003424907700649 45184381050270476520801 20832213698846152929438780249
### Nth Root
y√n
2√n 679.006 77.2531 26.0577 13.5753
## 461049 as geometric shapes
### Circle
Diameter 922098 2.89686e+06 6.67796e+11
### Sphere
Volume 4.10516e+17 2.67119e+12 2.89686e+06
### Square
Length = n
Perimeter 1.8442e+06 2.12566e+11 652022
### Cube
Length = n
Surface area 1.2754e+12 9.80034e+16 798560
### Equilateral Triangle
Length = n
Perimeter 1.38315e+06 9.20439e+10 399280
### Triangular Pyramid
Length = n
Surface area 3.68175e+11 1.15498e+16 376445
## Cryptographic Hash Functions
md5 830c41786711cc5804850d594af3b413 09fca9e74866dc449d92a55d58501b4056a63ba2 f726e56c5e7fbb0ba2f04c1d791b98eb3639e4b6e1a9fe1762d7d9bd4ab94239 f7b1f15fc39e1e5dd1967f516faa22650e1adf0fe5a7280e560906f83667021498415840f8e6b74fdd837853242e2b43ccede60d064d7c035d65f5ceb72737da 45f994a3f100245b46dea32e8f8b2a46517074ee | 1,469 | 4,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-39 | latest | en | 0.80283 |
https://www.physicsforums.com/threads/error-analysis-calculating-error-from-radiation-counts.385759/ | 1,713,656,922,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00794.warc.gz | 872,900,347 | 15,444 | # Error Analysis: Calculating Error from Radiation Counts
• hartraft
In summary, the conversation discusses calculating errors from a set of radiation counts. The error in a single count is the square root of the number of counts, but when taking multiple counts to find an average, the error is estimated using the square root of the average count. This can be compared to the theoretical error measure to determine the likelihood of the experiment's results.
hartraft
Not sure if this is the best place for this post, if it isn't any recommendations would be appreciated.
My qustion concerns calculating the error from a set of radiation counts. I understand that the error in a single radiation count is $$\sqrt{N}$$ where N is the number of counts. My question is what happens if i have taken multiple counts with the intention of using an average.
For example if i have three counts of 225, 197 and 211 with errors of 15, 14, and 14 respectively, the average count is 211, what is the error? Do i use the standard method of combining errors even this these are all measurements of the same variable?
Any help would be appreciated
Average of counts is the unbiased estimator for N in this case. So the estimator of square root of N is the square root of average counts. In other words, if you want to estimate the error you use square root of average counts, as population of the counts getting large we are able to reach the theoretical error measure. By comparing the theoretical error with measures of error got from experienced you can know how likely the experiment result to happen.
## 1. What is error analysis in the context of calculating error from radiation counts?
Error analysis in this context refers to the process of evaluating the accuracy and precision of measurements of radiation counts. It involves identifying and quantifying sources of error and determining the overall uncertainty in the calculated result.
## 2. How do you calculate error from radiation counts?
The error from radiation counts can be calculated by taking the square root of the sum of the squared uncertainties in the individual measurements. This is known as the root mean square (RMS) error. Alternatively, the percent error can be calculated by dividing the absolute value of the difference between the measured value and the accepted value by the accepted value and multiplying by 100.
## 3. What are the sources of error in calculating radiation counts?
There are several potential sources of error in calculating radiation counts, including instrumental errors, environmental factors such as background radiation, and human error in taking and recording measurements. Other sources of error may also include statistical fluctuations in the data and systematic errors in the experimental design.
## 4. How can error analysis improve the accuracy of radiation count measurements?
Error analysis allows for the identification and quantification of sources of error in radiation count measurements. By understanding and accounting for these sources of error, the accuracy of the measurements can be improved. Additionally, error analysis can help to determine the overall uncertainty in the calculated result, providing a more comprehensive understanding of the measured value.
## 5. How does the type of radiation affect the error in radiation count measurements?
The type of radiation can affect the error in radiation count measurements due to differences in the properties of the radiation, such as energy, penetration depth, and detection efficiency. For example, gamma radiation may have a higher detection efficiency compared to alpha radiation, resulting in potentially lower error in counts for gamma radiation measurements.
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2K | 875 | 4,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.926706 |
https://docest.com/doc/320279/exam-tips-from-graham | 1,721,009,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00851.warc.gz | 187,164,292 | 9,724 | Maths EmporiumGuides to Life:
GCSE Mathematics examinations
Before the examination:
Equipment. The front of the examination paper says you should have a ruler graduated in centimetres and millimetres, a protractor, a pair of compasses, a pen which works, a sharp HB pencil, an eraser and a calculator. Chances are that you will need to use all of these in the examination, so make sure you have them with you in a suitably modest pencil case. Gather them together in plenty of time – don’t be trying to find all these items the night before, or on the morning of, the examination.
Ruler: best to have one that’s transparent with cm and mm graduations. Leave your gran’s wooden rulergraduated in fractions of inches in the drawer where you found it. Flexible curve rulers(as pictured right) are not permitted and there’s no reason you’d need one; the same goes for French curves.
Protractor: A 180 version (left) will be sufficient, but 360 protractors (right) are permitted. Those that convert degrees in a circle to percentages (for example90=25%) are not permitted in the exam.
Pair of compasses: Some people are very insistent that the correct term “pair of compasses is used”. So use
This pair: Not this pair:
You’re likely to need a pair of compasses for any questions to do with constructions, loci orbearings. Make sure your pencil isn’t too thickto be used with the pair of compasses you have. Get some practice drawing circles and arcs before the examination.
Pen: The front of the exam paper says to use black ink or ball-point pen. I expect few students use fountain pensin the exam these days but everyone can, and should, have a black ballpoint pen. Examination scripts are all scanned to be marked on computer screens and black ink shows up best; don’t risk anything being missed by the scanners using blue ink. As for green or pink (not unknown), remember you’re in an exam – take things seriously. And red? – unless you’re marking the exam (which you’re not), avoid red at all costs.
Take more than one black pen with you, just in case something goes wrong with the first one. And another, just in case something goes wrong with the second one.
Pencil: Use an HB pencil – not an H, which is too hard and might be too faint for scanners to pick up and not a B, which is too soft and may not be precise enough for graphs and diagrams. HB is the Goldilocks of pencils – just right.Use pencils for sketches, graphs and diagrams. Use black ink for everything else.
Erasers: Used for rubbing out pencil marks, which is why we used to call them rubbers. You will need one in case your sketches, graphs or diagrams don’t come out perfectly first time.
Calculators: Most commercially available calculators can be used in GCSE examinations, including graphical calculators, but only a basic scientific calculator is required. The Casio fx991-ex Classwiz may be used.
If you have any doubts, findout in plenty of time from your teacher whether your calculator can be used in the examination. Don’t try to find out the night before, or on the morning of, the calculator exam – people at the exam board who might know the answers can be hard to reach on exam days.
Make sure your calculator is charged up and that you have some spare batteries if needed, or even a spare calculator. Make sure you know how it works and that it is set for calculations in degrees (not radians or gradians – you can learn about those functions once you have passed your GCSE).
Tracing paper: The front of the examination paper says that tracing paper may be used. Some schools will provide this, though it doesn’t mean you will need to use it. Tracing paper should not be appended to scripts at the end, so don’t use it as paper for working out.
Highlighter pens: You may take these into the exam if they are useful for highlighting key words in questions. However, you shouldn’t use them for any part of your answers; they might not scanvery well or might obscure the answers you are intending to highlight.
Anything else:If it’s not on the list, it can’t be taken into the exam hall. So no mirrors (or bits of card that act like mirrors), no multiplication squares, no slide rules or log tables (for the old folk), no isometric paper, no stencils.
If it’s not on a list of things that aren’t allowed, don’t assume itis allowed. There’s nothing in the regulations that says you can’t take an elephant into the exam hall, but most people know not to try. There are exceptions for students with special requirements to take in special equipment, but they need to be agreed well in advance by your school or college.
Food and drink: Some schools and colleges will allow you to take this in, some won’t. It’s only two hours, you won’t actually expire without such sustenance – but if it’s important to you, find out from your institution their policy well in advance.
Mobile phones: Does this need to be mentioned? Leave them outside the exam hall. The Joint Council for Qualifications (JCQ) produces a poster which you might see outside the exam hall – no i-pods, i-pads, iphones or iwatches. Maybe best not to take in anything beginning with i (tough on anyone who uses an inhaler, perhaps, but you can’t be too careful).
No android versions of those either,and certainly no actual androids (though we probably would be secretly impressed if you pulled that off).
During the examination:
Examiners:Always remember that examiners are human beings, not mind-reading robots. They are looking to give you marks where they can, so make that as easy as possible for them to do. Write your calculations, working and answers clearly and logically so examiners can follow your arguments. The examination is not the time to showcase a new kind of notation or fancy handwriting styles.
Working: Your teachers have no doubt told you this, but make sure you write down all your working out; you’d be surprised how many marks you can lose by not doing so. There is no need to be embarrassed by anything you might write; all papers are marked anonymously and all papers are burned after six months. Remember, though, that your school or college might opt to ask for your exam script as part of any review process, so don’t go mad.
If you need to cross anything out, just put a single line through it – don’t use all the ink in your pen (and a lot of time) utterly obliterating what you have written.
Make sure answers are realistic – for example, gas bills don’t tend to run into thousands of pounds, no-one walks 400 miles in a day and people are not 10 metres tall.
The Exam Paper:The paper has been designed so that you should have more than enough space to answer each question. Don’t write below the line at the end of the question – that part will not be scanned and examiners will not see anything you write there.
You should only ask for extra paper if it is absolutely necessary, for example if you have to completely re-write a question. Make sure any extra pages are securely attached to your exam paper. Do not use extra pages for working out; all working should be written on the examination paper.
Make sure you check the last page or back page – it often has a question on it and this shouldn’t be missed. If it saysBLANK PAGE, that’s been put there so you know that we know it’s blank and there hasn’t been a printing error.
Formula sheets:There won’t be a formula sheet with the examination paper. You may have been used to seeing one when trying out past papers, but now you will need to know them. Make sure in advance that you know what formulas you are expected to remember and be able to use. You will be given the formulas for the volume and/or surface area of a cone and/or a sphere if it is required.
Everything in the examination paper will be correct – it will have been revised and proof-read many times before you see it. Don’t waste time asking a teacher or an invigilator to ring the exam board if you don’t understand the question – instead, spend the time trying to figure it out. Apart from anything, by the time an invigilator has got through to the exam board on exam day and then got back to you, the exam will probably be over.
If you can’t proceed, whether you think the question is wrong or that you just don’t know what to do, leave it and come back to it at the end if you have time.
Time: Depending on which paper you are taking, you’ll have between one and two hours to complete all the questions on the examination paper; this will seem like no time at all to most students, although to a few it will seem like an eternity.
Whatever it feels like, use all the time you have – you have nothing else you can do in the exam hall except to try to answer the questions on the examination paper. Every minute counts and every mark you achieve might be the one which takes you from a grade 3 to a grade 4 or a grade8 to a grade9.
Any time you have left at the end should be used for going through your paper looking for daft errors. For example, check you have copied any final answer of your working into the answer line correctly; make sure instances of simple addition and multiplication are correct; make sure you have given a reason for your answer if you have been asked for one. Make sure that if you have been asked the best buy on a packet of soap powder, for example, you have actually said which one it is.
If you have time, have another look at questions you weren’t able to answer – perhaps part(a) isn’t that difficult after all. And if it is, perhaps you can answer part (b) anyway.
Every minute counts – so don’t use them up producing elaborate (or indeed simple) doodles, or writinglong notes to examiners.Crucially, don’t spend any time thinking about how funny you will be on Twitter after the exam. Just get on with it.
All of the above has been based on what past students have done or haven’t done – if you can learn from their mistakes, it might make the difference between the grade you want and the one you don’t. Good luck!
Soon enough it will be all over and you will be allowed out…
1 | 2,208 | 10,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.92933 |
https://www.lidolearning.com/questions/m-bb-rsaggarwal7-ch10-ex10a-q1-sum6/415/ | 1,679,428,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00050.warc.gz | 1,004,049,979 | 12,406 | # RS Aggarwal Solutions Class 7 Mathematics Solutions for Percentage Exercise 10A in Chapter 10 - Percentage
Question 4 Percentage Exercise 10A
(4/15)
Solution:-
In order to convert a fraction into a percentage multiply the fraction by 100 and put the percent sign
%.
= (4/15) × 100
= (4/3) × 20
= 26(2/3) %
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A326362 Number of maximal intersecting antichains of nonempty, non-singleton subsets of {1..n}. 12
%I
%S 1,1,1,2,16,163,11742
%N Number of maximal intersecting antichains of nonempty, non-singleton subsets of {1..n}.
%C A set system (set of sets) is an antichain if no part is a subset of any other, and is intersecting if no two parts are disjoint.
%F For n > 1, a(n) = A326363(n) - n - 1 = A007363(n + 1) - n.
%e The a(4) = 16 maximal intersecting antichains:
%e {{1,2,3,4}}
%e {{1,2},{1,3},{2,3}}
%e {{1,2},{1,4},{2,4}}
%e {{1,3},{1,4},{3,4}}
%e {{2,3},{2,4},{3,4}}
%e {{1,2},{1,3,4},{2,3,4}}
%e {{1,3},{1,2,4},{2,3,4}}
%e {{1,4},{1,2,3},{2,3,4}}
%e {{2,3},{1,2,4},{1,3,4}}
%e {{2,4},{1,2,3},{1,3,4}}
%e {{3,4},{1,2,3},{1,2,4}}
%e {{1,2},{1,3},{1,4},{2,3,4}}
%e {{1,2},{2,3},{2,4},{1,3,4}}
%e {{1,3},{2,3},{3,4},{1,2,4}}
%e {{1,4},{2,4},{3,4},{1,2,3}}
%e {{1,2,3},{1,2,4},{1,3,4},{2,3,4}}
%t stableSets[u_,Q_]:=If[Length[u]==0,{{}},With[{w=First[u]},Join[stableSets[DeleteCases[u,w],Q],Prepend[#,w]&/@stableSets[DeleteCases[u,r_/;r==w||Q[r,w]||Q[w,r]],Q]]]];
%t fasmax[y_]:=Complement[y,Union@@(Most[Subsets[#]]&/@y)];
%t Table[Length[fasmax[stableSets[Subsets[Range[n],{2,n}],Or[Intersection[#1,#2]=={},SubsetQ[#1,#2]]&]]],{n,0,5}]
%Y Antichains of nonempty, non-singleton sets are A307249.
%Y Minimal covering antichains are A046165.
%Y Maximal intersecting antichains are A007363.
%Y Maximal antichains of nonempty sets are A326359.
%Y Cf. A000372, A003182, A006126, A006602, A014466, A051185, A058891, A261005, A305000, A305844, A326358, A326360, A326361, A326363.
%K nonn,more
%O 0,4
%A _Gus Wiseman_, Jul 01 2019
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Last modified April 5 23:20 EDT 2020. Contains 333260 sequences. (Running on oeis4.) | 873 | 2,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-16 | latest | en | 0.524424 |
https://ch.mathworks.com/matlabcentral/cody/problems/44092-find-the-minimal-value-in-n-n-matrix/solutions/1609341 | 1,600,518,073,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191780.21/warc/CC-MAIN-20200919110805-20200919140805-00576.warc.gz | 331,146,011 | 16,425 | Cody
# Problem 44092. Find the minimal value in N*N Matrix
Solution 1609341
Submitted on 14 Aug 2018
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
A =[100 200 2 5 0.1]; B_correct = 0.1; assert(isequal(Min_Val(A),B_correct))
B = 0.1000
2 Fail
A =[10 3 2; 5 0 1;7 8 2]; B_correct = 0; assert(isequal(Min_Val(A),B_correct))
B = 5 0 1
Assertion failed.
3 Pass
A =1; B_correct =A; assert(isequal(Min_Val(A),B_correct))
B = 1 | 198 | 554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.604206 |
http://www.haskell.org/haskellwiki/index.php?title=Recursive_function_theory&oldid=3900 | 1,406,108,083,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997877670.44/warc/CC-MAIN-20140722025757-00060-ip-10-33-131-23.ec2.internal.warc.gz | 430,314,255 | 9,111 | # Recursive function theory
## 2 Plans towards a programming language
Well-known concepts are taken from [Mon:MatLog], but several new notations (only notations, not concepts) are introduced to reflect all concepts described in [Mon:MatLog], and some simplification are made (by allowing zero-arity generalizations). These are plans to achive formalizations that can allow us in the future to incarnate the main concepts of recursive function theory in a programming language.
## 3 Primitive recursive functions
### 3.1 Type system
$\left\lfloor0\right\rfloor = \mathbb N$
$\begin{matrix}\left\lfloor n + 1\right\rfloor = \underbrace{\mathbb N\times\dots\times\mathbb N}\to\mathbb N\\\;\;\;\;\;\;\;\;n+1\end{matrix}$
### 3.2 Base functions
#### 3.2.1 Constant
$\mathbf 0 : \left\lfloor0\right\rfloor$
$\mathbf 0 = 0$
Question: is the well-known $\mathbf{zero}\;x = 0$ approach superfluous? Can we avoid it and use the more simple and indirect $\mathbf{0} = 0$ approach?
Are these approaches equivalent? Is the latter (more simple) one as powerful as the former one? Could we define a $\mathbf{zero}\;x = 0$ using the $\mathbf{0} = 0$ approach? Let us try:
$\mathbf{zero} = \underline\mathbf K^0_1 \mathbf0$
This looks like working, but raises new questions: what about generalizing operations (here composition) to deal with zero-arity cases in an appropriate way? E.g.
$\underline\mathbf\dot K^0_n c\left\langle\right\rangle$
$\underline\mathbf K^0_n c$
where $c : \left\lfloor0\right\rfloor, n \in \mathbb N$ can be regarded as n-ary functions throwing all their n arguments away and returning c.
Does it take a generalization to allow such cases, or can it be inferred?
#### 3.2.2 Successor function
$\mathbf s : \left\lfloor1\right\rfloor$
$\mathbf s = \lambda x . x + 1$
#### 3.2.3 Projection functions
For all $0\leq i:
$\mathbf U^m_i : \left\lfloor m\right\rfloor$
$\mathbf U^m_i x_0\dots x_i \dots x_{m-1} = x_i$
### 3.3 Operations
#### 3.3.1 Composition
$\underline\mathbf\dot K^m_n : \left\lfloor m\right\rfloor \times \left\lfloor n\right\rfloor^m \to \left\lfloor n\right\rfloor$
$\underline\mathbf\dot K^m_n f \left\langle g_0,\dots, g_{m-1}\right\rangle x_0 \dots x_{n-1} = f \left(g_0 x_0 \dots x_{n-1}\right) \dots \left(g_{m-1} x_0 \dots x_{n-1}\right)$
This resembles to the $\mathbf\Phi^n_m$ combinator of Combinatory logic (as described in [HasFeyCr:CombLog1, 171]). If we prefer avoiding the notion of the nested tuple, and use a more homogenous style (somewhat resembling to currying):
$\begin{matrix}\underline\mathbf K^m_n : \left\lfloor m\right\rfloor \times \underbrace{\left\lfloor n\right\rfloor\times\dots\times\left\lfloor n\right\rfloor} \to \left\lfloor n\right\rfloor\\\;\;\;\;\;\;\;\;\;\;m\end{matrix}$
Let underbrace not mislead us -- it does not mean any bracing.
$\underline\mathbf K^m_n f g_0\dots g_{m-1} x_0 \dots x_{n-1} = f \left(g_0 x_0 \dots x_{n-1}\right) \dots \left(g_{m-1} x_0 \dots x_{n-1}\right)$
remembering us to
$\underline\mathbf K^m_n f g_0\dots g_{m-1} x_0 \dots x_{n-1} = \mathbf \Phi^n_m f g_0 \dots g_{m-1} x_0 \dots x_{n-1}$
#### 3.3.2 Primitive recursion
$\underline\mathbf R^m : \left\lfloor m\right\rfloor \times \left\lfloor m+2\right\rfloor \to \left\lfloor m+1\right\rfloor$
$\underline\mathbf R^m f h = g\;\mathrm{where}$
$g x_0 \dots x_{m-1} 0 = f x_0 \dots x_{m-1}$
$g x_0 \dots x_{m-1} \left(\mathbf s y\right) = h x_0 \dots x_{m-1} y \left(g x_0\dots x_{m-1} y\right)$
The last equation resembles to the $\mathbf S_n$ combinator of Combinatory logic (as described in [HasFeyCr:CombLog1, 169]):
$g x_0 \dots x_{m-1} \left(\mathbf s y\right) = \mathbf S_{m+1} h g x_0 \dots x_{m-1} y$
## 4 General recursive functions
Everything seen above, and the new concepts:
### 4.1 Type system
$\widehat{\,m\,} = \left\{ f : \left\lfloor m+1\right\rfloor\;\vert\;f \mathrm{\ is\ special}\right\}$
See the definition of being special [Mon:MathLog, 45]. This property ensures, that minimalization does not lead us out of the world of total functions. Its definition is the rather straightforward formalization of this expectation.
$\mathbf{special}^m f \equiv \forall x_0, \dots, x_{m-1} \in \mathbb N\;\;\exists y \in \mathbb N\;\;f x_0 \dots x_{m-1} y = 0$
It resembles to the concept of inverse -- more exactly, to the existence part.
### 4.2 Operations
#### 4.2.1 Minimalization
$\underline\mu^m : \widehat m \to \left\lfloor m\right\rfloor$
$\underline\mu^m f x_0 \dots x_{m-1} = \min \left\{y\in\mathbb N\;\vert\;f x_0 \dots x_{m-1} y = 0\right\}$
Minimalization does not lead us out of the word of total functions, if we use it only for special functions -- the property of being special is defined exactly for this purpose [Mon:MatLog, 45]. As we can see, minimalization is a concept resembling somehow to the concept of inverse.
Existence of the required minimum value of the set -- a sufficient and satisfactory condition for this is that the set is never empty. And this is equivalent to the statement
$\forall x_0, \dots, x_{m-1} \in \mathbb N\;\;\exists y \in \mathbb N\;\;f x_0 \dots x_{m-1} y = 0$
## 5 Partial recursive functions
Everything seen above, but new constructs are provided, too.
### 5.1 Type system
$\begin{matrix}\left\lceil n + 1\right\rceil = \underbrace{\mathbb N\times\dots\times\mathbb N}\supset\!\to\mathbb N\\\;\;\;\;\;\;n+1\end{matrix}$
Question: is there any sense to define $\left\lceil0\right\rceil$ in another way than simply $\left\lceil0\right\rceil = \left\lfloor0\right\rfloor = \mathbb N$? Partial constant? Is
$\left\lceil0\right\rceil = \mathbb N$
or
$\left\lceil0\right\rceil = \mathbb N\;\cup\;\left\{\bot\right\}$?
### 5.2 Operations
$\overline\mathbf\dot K^m_n : \left\lceil m\right\rceil \times \left\lceil n\right\rceil^m \to \left\lceil n\right\rceil$
$\overline\mathbf R^m : \left\lceil m\right\rceil \times \left\lceil m+2\right\rceil \to \left\lceil m+1\right\rceil$
$\overline\mu^m : \left\lceil m+1\right\rceil \to \left\lceil m\right\rceil$
Their definitions are straightforward extension of the corresponding total function based definitions.
Remark: these operations take partial functions as arguments, but they are total operation themselves in the sense that they always yield a result -- at worst an empty function.
## 6 Bibliography
[HasFeyCr:CombLog1]
Curry, Haskell B; Feys, Robert; Craig, William: Combinatory Logic. Volume I. North-Holland Publishing Company, Amsterdam, 1958.
[Mon:MathLog]
Monk, J. Donald: Mathematical Logic. Springer-Verlag, New York * Heidelberg * Berlin, 1976. | 2,209 | 6,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 42, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-23 | longest | en | 0.731986 |
https://www.physicsforums.com/threads/cauchy-bounded.173335/ | 1,477,194,179,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00315-ip-10-171-6-4.ec2.internal.warc.gz | 982,739,077 | 16,248 | # Cauchy => bounded?
1. Jun 9, 2007
### pivoxa15
1. The problem statement, all variables and given/known data
Thm: If a sequence is Cauchy than that sequence is bounded.
However Take the partial sums of the series (sigma,n->infinity)(1/n). The partial sums form a series which is Cauchy. But the series diverges so the sequence of partial sums is unbounded.
Sequence of partial sums is Cauchy b/c d(1/x,1/(x+1))=1/(k(k+1)) -> 0 as k->infinity
Have I done something wrong?
Last edited: Jun 9, 2007
2. Jun 9, 2007
### matt grime
Why do you think the sequences of partial sums are cauchy? They aren't.
I cannot begin to decipher what b/c d(..... means.
If S_n means the sum to n terms, what makes you think that given e>0, there is an n(e) such that for all n,m>n(e) we have |S_n - S_m| < e?
It is easy to show this is impossible, using the same idea as to show that the series itself diverges. Remember that
1+1/2+1/3+1/4+1/5+... > 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8+1/8+1/8+...) > 1+1+1+1+...
Last edited: Jun 9, 2007
3. Jun 9, 2007
### siddharth
Is b/c d(.., "because" , the "distance" ...?
As matt suggested, I think you should be less cryptic in your notation.
4. Jun 9, 2007
### AKG
Yes, that's not the definition of Cauchy. A sequence a(n) is Cauchy if for all E > 0, there exist N natural such that for all m and n > N, |a(n) - a(m)| < E. All you've shown is that for all E > 0, there exists some natural N such that |a(N) - a(N+1)| < E, or something like that.
5. Jun 9, 2007
### pivoxa15
oh... I thought it was obvious. I actually used b/c in my exam recently to indicate 'because'. Although I used it in the middle of a sentence so hopefully might be less confusing than here.
6. Jun 10, 2007
### matt grime
If you'd've put a 'The' at the start of the sentence to actually make it into a sentence that might have helped. Anyway, have you understood the explanation of your mistake?
7. Jun 10, 2007
### pivoxa15
I think I have. Each term in the sequence is a series summed over a finite number of terms. However the larger the number of terms, the larger the sequence gets. Very much unlike Cauchy seqences where a recquirement is that each term in the seqence decreases in magnitude the further out the sequence gets.
8. Jun 10, 2007
### matt grime
it is perfectly possible to be strictly increasing and still bounded/cauchy
nope, there is nothing that states the terms have to decrease in magnitude. It is the difference between terms in the sequence that is important. | 758 | 2,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-44 | longest | en | 0.938948 |
https://www.jamesrmeyer.com/ffgit/godel_sb | 1,725,727,573,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00227.warc.gz | 796,770,736 | 24,210 | Logic and
Language
Copyright James R Meyer 2012 - 2024 https://www.jamesrmeyer.com
# Gödel’s Substitution Function
Page last updated 16 Feb 2024
A formal system is a language system consisting of an alphabet, a set of axioms that is a set of formal strings composed from that alphabet, and a set of rules of inference that determine what formal strings are valid strings, and a set of rules of inference that determine what strings can be constructed from the axiom strings.
In the incompleteness proof by Kurt Gödel, a proponent of intelligent design (see Statements by Kurt Gödel), one of the principal ideas is his Gödel numbering system. This consists of a function in a meta-language (a language that talks about a sub-language such as the formal system) that defines a correspondence between symbol strings of a formal system and natural numbers.
## Infinitely many possible Gödel Numbering Systems
Now, while Gödel chooses one particular Gödel encoding system, there are, in fact infinitely many possible different Gödel numbering encoding systems. Gödel’s numbering system works by matching every symbol of the formal system to some number - and there are infinitely many ways of doing this matching. For example, Gödel’s system matches up the symbols as:
0 ⇔ 1, f ⇔ 3, ~ ⇔ 5, ∨ ⇔ 7, ∀ ⇔ 9, ( ⇔ 11, ) ⇔ 13
and variables as 17, 19, 23 etc. (Footnote: In Gödel’s own numbering system, variables that consist of more than one symbol (e.g. xss0) are treated as though they are one symbol. In Gödel’s formal system, there are higher type variables, where type n variables match to 17n, 19n, 23n, … . This is ignored in the above for the sake of simplicity; it makes no difference to the logical argument.)
One could use another matching, such as:
0 ⇔ 3, f ⇔ 11, ~ ⇔ 17, ∨ ⇔ 23, ∀ ⇔ 41, ( ⇔ 53, ) ⇔ 71
and variables as 101, 107, 113 etc (every second prime number over 100)
As well as that matching of symbols to numbers, Gödel’s numbering system also matches up each such matching number to consecutively larger prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, …
for example, the string:
~ (x1fff 0)
matches in Gödel’s system to the series of numbers:
5, 11, 17, 7, 3, 3, 3, 3, 1, 13
and this, in Gödel’s system, gives rise to the number:
25 · 311 · 517 · 77 · 113 · 133 · 173 · 193 · 231 · 2913 (note: the dot · represents multiplication)
but it does not have to be done in that way.
For example, the one could have a system that leaves out every second prime number, giving the series:
2, 5, 11, 17, 23, …
so for the above example we would have:
25 · 511 · 1117 · 177 · 233 · 313 · 413 · 473 · 591 · 6713
which is a completely different number.
## Correspondences defined by a Gödel Numbering System
So what are the correspondences defined by a Gödel numbering system?
A Gödel numbering system defines a correspondence from
symbol strings of the formal system to natural numbers.
This correspondence is given by a function called a Gödel numbering function, and the natural numbers defined by this correspondence are called Gödel numbers. However, to be precise, the numbers that Gödel uses for the correspondence are actually of two types:
1. Gödel numbers are those given by the Gödel numbering function on symbol strings of the formal system.
2. For variables of the formal system, Gödel uses the exponent of Gödel numbers to correspond to the variables of the formal system - for example, while the Gödel number of a variable might be 217, Gödel uses the number 17 to correspond to the variable of the formal system, rather than 217. For convenience we will call such numbers Gödel variable numbers. (Footnote: It might be noted that while Gödel uses ‘Gödel variable numbers’ for numbers corresponding to variables, he could have used the actual Gödel numbers for variables instead. Presumably the reason he did not was that it would have entailed further complexity in his definitions of various relations. )
In Gödel’s proof, various relations of natural numbers are defined, and Gödel asserts that these number-theoretic relations correspond to relationships between symbol strings of the formal system - but that applies only when the natural numbers are defined to be Gödel numbers (or Gödel variable numbers). This is fundamental to the correspondence defined in Gödel’s proof, so we will repeat it in order to emphasize it:
Given a relation between natural numbers, there may be a corresponding relationship between symbol strings of the formal system - but when those natural numbers are not defined to be Gödel numbers of symbol strings of the formal system, nor Gödel variable numbers of variables of the formal system, there cannot be such a corresponding relationship.
### Caveats of Gödel numbering
However, there is a drawback to this Gödel numbering: number-theoretic expressions can be generated where there can be numbers which have the characteristic of a Gödel number (in that its prime factors all have exponents that belong to a specific predefined set), but where that generation has absolutely nothing to do with an association to an expression of the formal system. And that is why examining of a number-theoretic expression without taking account of how such numbers were generated may lead to an erroneous conclusion that a number in an expression has an inherent association to an expression of the formal system.
But this difficulty does not apply to a purely number-theoretic expression that contains a purely numerical function where its free variable have been substituted by specific values, and whose output has the characteristic of a Gödel number - in contrast to the same expression but which only contains the numerical value of the function’s output, rather than the function itself. This is because it can immediately be seen that this characteristic of a Gödel number has originated not externally to the system by Gödel numbering, but within the system itself, and hence has no inherent association to some expression of the formal system - in other words, that function is a function of the formal system, not a function of the meta-language, hence it cannot be a Gödel numbering function, nor can it be equivalent to a Gödel numbering function, even though it generates numbers that have the characteristic of a Gödel number in that its prime factors all have exponents that belong to a specific predefined set.
Although this is a fundamental point, Gödel’s proof manages to ignore it, which results in an illogical confusion as will be detailed below.
## The ‘Substitution’ Function
Gödel’s ‘substitution’ function (Gödel’s relation 31) which is Sb(x, v, y), is a function of natural numbers that, for numerical values for xv and y, returns some numerical value. (Footnote: See Functions 27-31: Defining numbers that correspond to substitution in the formal system and Relation 31 in Gödel’s paper for the formal presentation of this function in Gödel’s paper.) In the following, bold capital letters indicate symbol strings of the formal system, whereas bold lower letters indicate numbers of the meta-language (not numbers of the formal system). Gödel asserts that Sb(x, v, y) is a function that corresponds, by Gödel numbering to the expression that results from the substitution of a free variable V in a formal system formula X by some symbol string Y of the formal system, where x and y are Gödel numbers (where GN(u) is the Gödel numbering function x = GN(X) and y = GN(Y) and v is a Gödel variable number corresponding to a formal system variable V). The following explains the essence of the operation of this function.
If x takes the value:
2a · 3b · 5c · … · p17 · qw · ry · sz · …
where pqr and s are consecutive prime numbers,
and y takes the value:
29 · 37 · 59,
and v takes the value 17, then the function will return the value:
2a · 3b · 5c · … · p9 · q7 · r9 · sw · tv · …
Here the exponent 17 in p17 has been replaced by 9, and the exponent of the next two prime numbers above p have been replaced by 7 and 9 respectively, and the exponents of the remaining prime numbers have been ‘shifted’ up by three ‘places’.
The idea is that, if
• the number x in the function is a number that is a Gödel number that corresponds to a formal system formula with a free variable, and
• the number 17 corresponds to that free variable by the Gödel numbering system (here, for simplicity, we assume that the free variable symbol v only occurs once in the formula), and
• if y is a number that is a Gödel number that corresponds to a formal system symbol string, then the function is said to correspond to the substitution of the free variable of a formula by that symbol string of the formal system. In the above example, this is by the symbol string ‘∀∨∀’, since 29 · 37 · 59 is the Gödel number (by the Gödel number encoding in Gödel’s paper) of that symbol string.
Note that if the variable is not free in the formula, there will be no substitution, and the function then simply has the value x; and if there is more than one occurrence of the free variable in the formula, there will be more than one occurrence of the substitution. If y is not a valid Gödel number (i.e: not a Gödel number of any formal system symbol string), then the numerical value that the function outputs is not a Gödel number since it does not have any inherent correspondence to the notion of the substitution of the variable of the formula by some numerical value. This follows, since the notion of the correspondence of Sb(x, v, y) by Gödel numbering is such a one-to-one correspondence only when the domain of x and y is Gödel numbers, and the domain of v is Gödel variable numbers.
## The Use of the ‘Substitution’ Function in Gödel’s Proof
In Gödel’s proof, after having defined the function Sb, it turns out that his actual use of the substitution function in his proof is as a combination of the function Sb and another function, the Z function (Gödel’s relation 17), that is, in the form Sb(x, v, Z(w)).
The function Z(w) is a function of natural numbers that, given the number w, returns the value:
23 · 33 · 53 · … · p3 · q3 · r1
where there are exactly w consecutive prime numbers with the exponent 3. (In Gödel’s encoding, 3 corresponds to the successor symbol s, and 1 to the zero symbol 0).
Consider now the Gödel numbering function, which we will call GN(u). This is a meta-language function that given the symbol string in the form sss…0, where there are u s’s returns the value
23 · 33 · 53 · … · p3 · q3 · r1
where there are exactly u consecutive prime numbers with the exponent 3. (In Gödel’s encoding, 3 corresponds to the successor symbol s, and 1 to the zero symbol 0).
Certainly there is a similarity between the function Z(w) and the function GN(u) - but being similar should never be confused with being identical. And since there are infinitely many different Gödel numbering functions, there are infinitely many functions like Z(w) that are similar to some Gödel numbering function.
When Gödel uses the composite function Sb(x, v, Z(w)) in Proposition VI of his proof, he requires that the first variable x and the third variable w must have the same value - as in Sb(x, v, Z( x)).
But, crucially, for the function Sb(x, v, x) to be a number-theoretic function that corresponds by Gödel numbering to a function in the meta-language whose input is a formal system expression, the value given for third variable x of the function must be declared, outside of the formal system, to be a Gödel number, that is, it must be declared in the meta-language that x = GN(X), where X is some formal expression. This means that Gödel has to assume that Sb(x, v, Z( x)) is precisely equivalent to the entirety of the requirements of a correspondence by Gödel numbering, which is:
Sb(x, v, y), where y = GN(Y) and x = GN(X) and x = Y
since the free variables of Sb must be Gödel numbers.
But the free variable x of Z( x) is a variable of the formal system with the domain of natural numbers, whereas Y must be an expression of the formal system. They belonging to different language systems, and the notion that they can be equivalent as Gödel claims is nonsensical. The claim of equivalence is an absurd conflation of two different levels of language, a cheap trick to achieve a superficial appearance of self-reference within the formal system.
Really, nothing more needs to be said - Gödel’s illogical assumptions render his proof completely invalid. But we can achieve the same result in different ways, as shown below, just in case the reader might suspect that we have somehow misled them.
### The Gödel number of a Gödel number
With the function GN(X), i.e, “the Gödel number of the expression X” there is a perfectly logical and valid specific number for a specific formal system expression X, so that there must be some number n that corresponds to X by Gödel numbering. And while the formal system cannot express the Gödel numbering function, we can say that there is a valid formal system expression that has a given numerical value for the first term x in Sb(x, v, Z( x)).
But the same does not apply for the third term Z( x)). Gödel wants us to assume that this generates the notion of GN(GN(X)), i.e, “the Gödel number of the Gödel number of X”, and Gödel also wants us to assume that this is a logically valid concept that preserves the correspondences given by Gödel numbering.
But the inner GN must give an output that is a number that is not an expression of the formal system, whereas the input of the outer GN must be an input that is an expression of the formal system. And this means that there is no number that is a logically valid output for the notion GN(GN(X)).
We can also see this by ensuring that the symbols of the meta-language for numbers are not the same as for the formal system. For example, if the numbers of the meta-language are always in the format 0, 1, 2, 3, …, awhile the numbers in the formal system are in the format 0, f 0, ff 0, fff 0, …, then it is obvious that the output of the GN function cannot also be the input of the GN function And this means that there is no number that is a logically valid output for the notion GN(GN(X)).
Now, while one could force the formats for numbers to be the same in both systems, that would not somehow magically make the conflation of meta-language and sub-language logically valid. in short, that would be a fudge that should fool no-one. The entire notion of Gödel numbering is that, for every relationship between expressions A and B of the formal system there is a corresponding relation between numbers a and b, where a = GN(A) and b = GN(B) (and similarly for more than two expressions). But Gödel’s construction is an illogical distortion of the correspondence between formal system expressions and Gödel numbers.
In summary, Gödel makes fundamental errors and manages to confuse different systems by his incorrect and illogical assumptions regarding his Sb(x, v, Z(w)) function; this confusion renders his proof invalid. See also Gödel’s Proposition V and the paper The Fundamental Flaw in Gödel’s Proof of his Incompleteness Theorem where it is shown in detail that Gödel incorrectly assumes that there is an equivalence between the function Z and the Gödel numbering function, and why this renders his result logically impossible. There is now a section to the paper that gives a brief summary of the underlying illogical assumption that the proof relies on, so that the reader can see in a few pages that the proof is flawed. See also the webpage Gödel’s Proposition V: The Z function which describes the contradictions in Gödel’s use of the Z function.
#### Correspondences by Gödel numbering
We can also show that there isn’t even a correspondence through Gödel numbering between the Z function and the GN function. We suppose that there might be a function in the meta-language that corresponds to Gödel’s Z function. If we have two expressions C and D of the formal system, and if c and d are defined outside of the formal system to be Gödel numbers (i.e, c = GN[C] and d = GN[D]) then, according to the conventional assertions of Gödel numbering, given the formula d = Z(c), this should mean that the Z function is a number-theoretic function that corresponds to some function in the meta-language that outputs the formal expression D when given an input that is the expression C.
Now, considering the relation d = Z(c), if a correspondence is asserted through Gödel numbering, then according to conventional assertions regarding Gödel numbering, there should be some corresponding function F of the meta-language such that D = F[C]. But if this F could be the GN function, then we would have:
D = GN[C]
but it is also the case that:
c = GN[C]
which would give us that:
c = D
but c is an expression of the meta-language, whereas D is an expression of the formal system, so that this is an illogical conflation of levels of language.
Therefore the function F cannot be the Godel numbering function, and hence the output of Z(c) cannot be a Gödel number that corresponds to an expression of the formal language. In other words, besides the absurdity of Gödel’s claim of equivalence of the Z function and the GN function, there isn’t even a correspondence through Gödel numbering between those two functions.
The reason for this is that, on the one hand, there is an implicit claim that the Gödel numbering system is a function that is in a meta-language to two separate language systems, one being number-theoretic relations and the other the formal system. But the definition of the Gödel numbering function itself necessarily uses number-theoretic relations, and in fact part of the definition of the Z function forms part of the definition of the Gödel numbering function.
##### Example for specific expressions C and D
For example, if C is the formal expression f 0, then:
c = GN(C) = 23 * 31 = 24
and
d = Z(c) = 23 · 33 · 53 · … · p3 · q1, where p is the 24th largest prime and q is the 25th largest prime, (Footnote: This number is too large to be written here in standard notation.)
and the Z function here is a number-theoretic function that, according to the conventional assertions of Gödel numbering, should correspond to a function in the meta-language that outputs the formal expression D from the input as the formal expression C, where D is ffffffffffffffffffffffff 0, and d = GN[D].
As above, regarding the relation d = Z(c), if such a correspondence is asserted through Gödel numbering, that indicates that there is some corresponding function F of the meta-language such that D = F[C]. And if this F could be the GN function, then we would have:
D = GN[C], i.e, that ffffffffffffffffffffffff 0 = GN[ f 0],
but:
c = GN[C]
indicating that
c = D
i.e,
24 = ffffffffffffffffffffffff 0
but “24” is an expression of the meta-language (since GN[ f 0] = 23 * 31 = 24), whereas “ffffffffffffffffffffffff 0” is an expression of the formal system.
The crucial point here is that an expression such as “24 = ffffffffffffffffffffffff 0” has no inherent meaning of itself, a meaning is only established by asserting which language system it belongs to - and in this case, it does not belong to a logically valid language system, since the left side belongs to the meta-language, while the right side belongs to the formal system.
### The Use of the ‘Substitution’ Function by Nagel-Newman and Hofstadter
As a point of interest, Nagel-Newman’s book, Gödel’s Proof, (Footnote: PDF E Nagel and J Newman: Gödel’s Proof New York University Press, revised edition, 2001. ISBN: 0814758169.) also refers to this substitution function. Nagel-Newman also makes erroneous assumptions in the treatment of that function, but the error in the use of the function is subtly different. See Nagel-Newman for an overview of Nagel-Newman’s book.
Similarly, Douglas Hofstadter also refers to a substitution function in his book, Gödel, Escher, Bach, (Footnote: Douglas Hofstadter. ‘Gödel, Escher, Bach’. Basic Books, 1999, ISBN‑13: 978‑0465026562 Gödel, Escher, Bach - Hofstadter: details.) and he also makes erroneous assumptions regarding that function. See Gödel, Escher, Bach for an overview of Hofstadter’s book.
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https://www.jamesrmeyer.com | 5,038 | 21,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.883474 |
https://www.electionguard.vote/spec/0.95.0/9_Verifier_construction/ | 1,623,850,226,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00086.warc.gz | 685,809,761 | 11,705 | # Verifier Construction
While it is desirable for anyone who may construct an ElectionGuard verifier to have as complete understanding as possible of the ElectionGuard design, this section isolates the items which must be verified and maps the variables used in the specification equations herein to the labels provided in the artifacts produced in an election. 1
## Implementation details
There are four operations which must be performed – all on very large integer values: modular addition, modular multiplication, modular exponentiation, and SHA-256 hash computations. These operations can be performed using a programming language that provides native support, by importing tools to perform these large integer operations, or by implementing these operations from scratch.
To compute
$(a+b) \bmod n$
one can compute
$((a \bmod n)+(b \bmod n)) \bmod n$
However, this is rarely beneficial. If it is known that $$a,b \in Z_n$$, then one can choose to avoid the division normally inherent in the modular reduction and just use
$(a+b) \bmod n=a+b$
(if $$a+b \lt n$$) or
$a+b-n$
(if $$a+b \ge n$$).
### Modular Multiplication
To compute
$(a \times b) \bmod n$
one can compute
$((a \bmod n) \times(b \bmod n)) \bmod n$
Unless it is already known that $$a,b \in Z_n$$, it is usually beneficial to perform modular reduction on these intermediate values before computing the product. However, it is still necessary to perform modular reduction on the result of the multiplication.
### Modular Exponentiation
To compute
$a^b \bmod n$
one can compute
$(a \bmod n)^b \bmod n$
but one should not perform a modular reduction on the exponent.2 One should, however, never simply attempt to compute the exponentiation $$a^b$$ before performing a modular reduction as the number $$a^b$$ would likely contain more digits then there are particles in the universe. Instead, one should use a special-purpose modular exponentiation method such as repeated squaring which prevents intermediate values from growing excessively large. Some languages include a native modular exponentiation primitive, but when this is not available a specialized modular exponentiation tool can be imported or written from scratch.
### Hash Computation
Hashes are computed using the SHA-256 hash function in NIST (2015) Secure Hash Standard (SHS) which is published in FIPS 180-4 and can be found in https://csrc.nist.gov/publications/detail/fips/180/4/final.
For the purposes of SHA-256 hash computations, all inputs – whether textural or numeric – are represented as utf-8 encoded strings. Numbers are represented as strings in base ten. The hash function expects a single-dimensional array of input elements that are hashed iteratively, rather than concatenated together. Each element in the hash is separated by the pipe character (“|”). When dealing with multi-dimensional arrays, the elements are hashed recursively in the order in which they are input into the hash function. For instance, calling
$H(1,2,[3,4,5],6)$
is a function call with 4 parameters, where the 3rd parameter is itself an array. The hash function will process arguments 1 and 2, when it gets to argument 3 it will traverse the array (and hash the values 3, 4 ,5) before hashing the final fourth argument (whose value is 6). When hashing an array element that is empty, the array is instead replaced with the word “null” as a placeholder.
## Parameter Validation
Important
An ElectionGuard version 1 election verifier may assume that the baseline parameters match the parameters provided above. However, it is recommended that the below parameters be checked against the parameters of each election to accommodate the possibility of different parameters in future versions of ElectionGuard.3
Variable Meaning Folder File Level
$$p$$ 4096-bit modulus root constants.json large_prime
$$q$$ 256-bit prime order of subgroup $$Z_p^*$$ of encryptions root constants.json small_prime
$$r$$ co-factor of $$q$$ root constants.json cofactor
$$g$$ generator of order $$q$$ multiplicative subgroup of $$Z_p^*$$ root contstants.json generator
$$n$$ number of guardians root context.json number_of_guardians
$$k$$ minimum number of guardians required to decrypt tallies and produce verification data root context.json quorum
$$Q$$ base hash value formed by $$p$$, $$q$$, $$g$$, $$n$$, $$k$$, date, and jurisdictional information root context.json crypto_base_hash
## Guardian Public Key Validation
Important
An election verifier must confirm the following for each guardian $$T_i$$ and for each $$j \in Z_k$$:
(A) The challenge $$c_{i,j}$$ is correctly computed as
$c_{i,j}=H(Q,K_{i,j},h_{i,j} ) \bmod q$
(B) The equation
$g^{u_{i,j} } \bmod p=h_{i,j} K_{i,j}^{c_{i,j} } \bmod p$
is satisfied.
Variable Meaning Folder File Level
$$K_{i,j}$$ public form of each random coefficient $$a_{i,j}$$ coefficients every file in this folder large_prime
$$h_(i,j)$$ coefficient commitments coefficients every file in this folder coefficient_proofs $$\rightarrow$$ [Item] $$\rightarrow$$ commitment
$$c_i$$ challenge value coefficients every file in this folder coefficient_proofs $$\rightarrow$$ [Item] $$\rightarrow$$ challenge
$$u_{i,j}$$ response coefficients every file in this folder coefficient_proofs $$\rightarrow$$ [Item] $$\rightarrow$$ response
## Election Public Key Validation
Important
An election verifier must verify the correct computation of the joint election public key and extended base hash.
(A) Joint election public key
$K=\prod_{i=1}^n K_i \bmod p$
(B) Extended base hash
$\bar{Q} = H(Q,K_1,0,K_1,1,K_1,2,\ldots,K_{1,k-1},K_2,0,K_2,1,K_2,2,\ldots,K_{2,k-1},\ldots,K_{n,0},K_{n,1},K_{n,2},\ldots,K_{n,k-1})$
Variable Meaning Folder File Level
$$\bar{Q}$$ extended base hash Root context.json crypto_extended_base_hash
$$K$$ joint election public key Root context.json elgamal_public_key
## Correctness of Selection Encryptions
Important
An election verifier must confirm the following for each possible selection on a ballot:
The given values $$\alpha$$, $$\beta$$, $$a_0$$, $$b_0$$, $$a_1$$, and $$b_1$$ are all in the set $$Z_p^r$$. (A value $$x$$ is in $$Z_p^r$$ if and only if $$x$$ is an integer such that $$0 \le x \lt p$$ and $$x^q \bmod p=1$$ is satisfied.)
(J) The challenge $$c$$ is correctly computed as
$c=H(\bar{Q},(\alpha,\beta),(a_0,b_0 ),(a_1,b_1 ))$
(K) The given values $$c_0$$, $$c_1$$, $$v_0$$, and $$v_1$$ are each in the set $$Z_q$$. (A value $$x$$ is in $$Z_q$$ if and only if $$x$$ is an integer such that 0 \le x \lt q.)
(L) The equation
$c=(c_0+c_1 ) \bmod q$
is satisfied.
(M) The equation
$g^{v_0} \bmod p= a_0 \alpha^{c_0} \bmod p$
is satisfied.
(N) The equation
$g^{v_1} \bmod p=a_1 \alpha^{c_1} \bmod p$
is satisfied.
(O) The equation
$K^{v_0} \bmod p=b_0 \beta^{c_0} \bmod p$
is satisfied.
(P) The equation
$g^{c_1} K^{v_1} \bmod p=b_1 \beta^{c_1} \bmod p$
is satisfied.
Variable Meaning Folder File Level
$$(\alpha,\beta)$$ encryption of vote encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ ciphertext $$\rightarrow$$ pad, data
(a_0,b_0) commitment to vote being an encryption of zero encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_zero_pad, proof_zero_data
$$(a_1,b_1)$$ commitment to vote being an encryption of one encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_one_pad, proof_one_data
$$c$$ challenge encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_challenge
$$c_0$$ derived challenge to encryption of zero encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_zero_challenge
$$c_1$$ derived challenge to encryption of one encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_one_challenge
$$v_0$$ response to zero challenge encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_zero_response
$$v_1$$ response to one challenge encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ proof_one_response
## Adherence to Vote Limits
Important
An election verifier must confirm the following for each contest on the ballot:
(H) The number of placeholder positions matches the contest’s selection limit $$L$$.
(I) The contest total $$(A,B)$$ satisfies
$A=\prod_i \alpha_i \bmod p$
and
$B=\prod_i \beta_i \bmod p$
where the $$(\alpha_i,\beta_i )$$ represent all possible selections (including placeholder selections) for the contest.
(J) The given value $$V$$ is in $$Z_q$$.
(K) The given values a and b are each in $$Z_p^r$$.
(L) The challenge value $$C$$ is correctly computed as
$C=H(\bar{Q},(A,B),(a,b))$
(M) The equation
$g^V \bmod p=(aA^C ) \bmod p$
is satisfied.
(N) The equation
$((g^{LC} K)^V ) \bmod p=(bB^C ) \bmod p$
is satisfied.
Variable Meaning Folder File Level
$$(\alpha_i,\beta_i)$$ encryption of $$i^th$$ vote in contest encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ ciphertext $$\rightarrow$$ pad, data
$$(A,B)$$ encryption of total votes in contest encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ total $$\rightarrow$$ pad, data
$$(a,b)$$ commitment to vote being an encryption of one encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ total $$\rightarrow$$ proof $$\rightarrow$$ pad, data
$$C$$ selection limit challenge encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ challenge
$$L$$ contest selection limit root description.json contests $$\rightarrow$$ [Item] $$\rightarrow$$ votes_allowed
$$V$$ response to selection limit challenge encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ response
## Validation of Ballot Chaining
Important
An election verifier must confirm that each of the values in the running hash is correctly computed. Specifically, an election verifier must confirm each of the following.
(D) The equation $$H_0=H(\bar{Q})$$ is satisfied.
(E) For each ballot
$B_i, H_i=H(H_{i-1},D,T,B_i)$
is satisfied.
(F) The closing hash
$\bar{H} =H(H_l,\text{“CLOSE"})$
is correctly computed from the final tracking code $$H_l$$.
Variable Meaning Folder File Level
$$H_i$$ running hash of ballots produced encrypted_ballots every file in this folder tracking_hash (also previous_tracking_hash)
## Correctness of Ballot Aggregation
Important
An election verifier must confirm for each (non-placeholder) option in each contest in the ballot coding file that the aggregate encryption $$(A,B)$$ satisfies
$A=\prod_j\alpha_j$
and
$\beta=\prod_j \beta_j$
where the $$(\alpha_j,\beta_j )$$ are the corresponding encryptions on all cast ballots in the election record.
Variable Meaning Folder File Level
$$(\alpha_j,\beta_j)$$ encryption of vote encrypted_ballots every file in this folder contests $$\rightarrow$$ [Item] $$\rightarrow$$ ballot_selections $$\rightarrow$$ [Item] $$\rightarrow$$ ciphertext $$\rightarrow$$ pad, data
$$(A,B)$$ encrypted aggregate total of votes in contest root tally.json [prefix] $$\rightarrow$$ message $$\rightarrow$$ pad, data
## Correctness of Partial Decryptions
Important
An election verifier must then confirm for each (non-placeholder) option in each contest in the ballot coding file the following for each decrypting trustee $$T_i$$.
The given value $$v_i$$ is in the set $$Z_q$$.
The given values $$a_i$$ and $$b_i$$ are both in the set $$Z_q^r$$.
The challenge value $$c_i$$ satisfies
$c_i=H(\bar{Q},(A,B),(a_i,b_i ),M_i )$
The equation
$g^{v_i} \bmod p=(a_i K_i^{c_i} ) \bmod p$
is satisfied.
The equation $$A^{v_i} \bmod p=(b_i M_i^{c_i} ) \bmod p$$ is satisfied.
## Correctness of Partial Decryptions
Important
An election verifier must then confirm for each (non-placeholder) option in each contest in the ballot coding file the following for each decrypting trustee $$T_i$$.
(F) The given value $$v_i$$ is in the set of $$Z_q$$
(G) The given values $$a_i$$ and $$b_i$$ are both in the set $$Z_q^r$$.
(H) The challenge value $$c_i$$ satisfies
$c_i=H(Q ̅,(A,B),(a_i,b_i ),M_i )$
(I) The equation
$g^{v_i} \bmod p=(a_i K_i^{c_i} ) \bmod p$
is satisfied.
(J) The equation
$A^{v_i} \bmod p=(b_i M_i^{c_i} ) \bmod p$
is satisfied.
Variable Meaning Folder File Level
$$(A,B)$$ encrypted aggregate total of votes in contest root tally.json [prefix] $$\rightarrow$$ message $$\rightarrow$$ pad, data
$$M_i$$ partial decryption of $$(A,B)$$ by guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ share
$$(a_i,b_i)$$ commitment by guardian $$T_i$$ to partial decryption of $$(A,B)$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ pad, data
$$c_i$$ challenge to partial decryption of guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ challenge
$$v_i$$ response to challenge of guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ proof $$\rightarrow$$ response
## Correctness of Substitute Data for Missing Guardians
Note
This is only necessary if some guardians are missing during tallying
Important
An election verifier must confirm for each (non-placeholder) option in each contest in the ballot coding file the following for each missing trustee $$T_i$$ and for each surrogate trustee $$T_\ell$$.
(A) The given value $$v_{i,l}$$ is in the set $$Z_q$$.
(B) The given values $$a_{i,l}$$ and $$b_{i,l}$$ are both in the set $$Z_q^r$$.
(C) The challenge value $$c_{i,l}$$ satisfies
$c_(i,l)=H(Q ̅,(A,B),(a_(i,l),b_(i,l) ),M_(i,l) )$
(D) The equation
$g^{v_{i,l}} \bmod p=\left(a_{i,l}\cdot \left(\prod_{j=0}^{k-1} K_{i,j}^{l^j} \right)^{c_{i,l} } \right) \bmod p$
is satisfied.
(E) The equation
$A^{v_{i,\ell}} \bmod p= \left( b_{i,\ell} M_{i,l}^{c_{i,l}} \right) \bmod p$
is satisfied.
Variable Meaning Folder File Level
$$(A,B)$$ encrypted aggregate total of votes in contest root tally.json [prefix] $$\rightarrow$$ message $$\rightarrow$$ pad, data
$$M_(i,\ell)$$ share of guardian $$T_\ell$$ of missing partial decryption of $$(A,B)$$ by guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ substitute_proof $$\rightarrow$$ [item] $$\rightarrow$$ share
$$(a_{i,l},b_{i,l})$$ commitment by guardian $$T_\ell$$ to share of partial decryption for missing guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ substitute_proof $$\rightarrow$$ [item] $$\rightarrow$$ pad, data
$$c_{i,\ell}$$ challenge to guardian $$T_\ell$$ share of missing partial decryption of guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ substitute_proof $$\rightarrow$$ [item] $$\rightarrow$$ challenge
$$v_{i,l}$$ response to challenge of guardian $$T_\ell$$ share of partial decryption of guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ substitute_proof $$\rightarrow$$ [item] $$\rightarrow$$ response
## Correctness of Construction of Replacement Partial Decryptions
Note
This is only necessary if some guardians are missing during tallying
Important
An election verifier should confirm that for each trustee $$T_\ell$$ serving to help compute a missing share of a tally, that its Lagrange coefficient $$w_\ell$$ is correctly computed by confirming the equation
$\left(\prod_{j\in(U-{\ell})}j\right) \bmod q=\left(w_l⋅\left(\prod_{j\in (U-{l})} (j-\ell) \right)\right) \bmod q$
An election verifier should then confirm the correct missing tally share for each (non-placeholder) option in each contest in the ballot coding file for each missing trustee $$T_i$$ as
$M_i=\prod_{\ell \in U}\left(M_{i,\ell}\right)^{w_\ell} \bmod p$
Variable Meaning Folder File Level
$$w_\ell$$ coefficient for use with shares of guardian $$T_\ell$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [item] $$\rightarrow$$ coefficient
$$M_{i,\ell}$$ share of guardian $$T_\ell$$ of missing partial decryption of $$(A,B)$$ by guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ substitute_proof $$\rightarrow$$ [item] $$\rightarrow$$ share
$$M_i$$ partial decryption of $$(A,B)$$ by guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ share
## Validation of Correct Decryption of Tallies
Important
An election verifier should confirm the following equations for each (non-placeholder) option in each contest in the ballot coding file.
(C)
$B=\left(M \cdot \left(\prod_{i=1}^n M_i \right)\right) \bmod p$
(D)
$M=g^t \bmod p$
An election verifier should also confirm that the text labels listed in the election record match the corresponding text labels in the ballot coding file.
Variable Meaning Folder File Level
$$M_i$$ partial decryption of $$(A,B)$$ by guardian $$T_i$$ root tally.json [prefix] $$\rightarrow$$ shares $$\rightarrow$$ [Item] $$\rightarrow$$ share
(M) full decryption of $$(A,B)$$ root tally.json [prefix] $$\rightarrow$$ value
$$t$$ tally value root tally.json [prefix] $$\rightarrow$$ tally
# Validation of Correct Decryption of Spoiled Ballots
Important
An election verifier should confirm the correct decryption of each spoiled ballot using the same process that was used to confirm the election tallies.
An election verifier should also confirm that for each decrypted spoiled ballot, the selections listed in text match the corresponding text in the ballot coding file.
Validation of Correct Decryption of Spoiled Ballots is a repeat of verification steps 8 through 11 for each spoiled ballot instead of for the aggregate ballot that contains encrypted tallies
1. Special thanks to Rainbow Huang (@rainbowhuanguw) for her help in producing this mapping.
2. In general, if $$n$$ is prime, one can compute $$a^b \bmod n$$ as $$(a \bmod n)^{(b \bmod (n-1)} ) \bmod n$$. But within this application, the efficiency benefits of performing a modular reduction on an exponent are limited, and the risk of confusion or error from doing so likely exceeds the benefit. In the particular instance of this specification, if $$a \in Z_p^r$$, then one can compute $$a^b \bmod p$$ as $$a^{(b mod q)} \bmod p$$. This has greater efficiency benefits, but the risk of confusion or error still likely exceed the efficiency benefit.
3. If alternative parameters are allowed, election verifiers must confirm that $$p$$, $$q$$, $$r$$, $$g$$, and \$\bar{g} are such that both $$p$$ and $$q$$ are prime (this may be done probabilistically using the Miller-Rabin algorithm), that $$p-1=qr$$ is satisfied, that $$q$$ is not a divisor of $$r$$, and $$1 \lt g \lt p$$, that $$g^q \bmod p=1$$, that $$g \bar{g} \bmod p=1$$, and that generation of the parameters is consistent with the cited standard. | 5,227 | 19,922 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | latest | en | 0.875974 |
https://www.mrexcel.com/board/threads/multiply-2-vectors-with-booleans-and-numbers-and-take-the-max-as-result.1190053/ | 1,695,501,663,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00590.warc.gz | 1,022,756,376 | 17,429 | # Multiply 2 Vectors With Booleans and Numbers and Take The MAX as Result
#### rkaczano
##### Board Regular
The formula below is attempting to multiply Booleans in \$D\$30:\$D\$33 with numbers in AD30:AG30. And I want to take the MAX of that result. However I am getting a funny answer and the MAX seems to be taking the highest value in AD30:AG30 irrespective of the Booleans.
The problem is that there is nothing implicitly multiplying the two arrays together. And a SUMPRODUCT does not work as I would need to calculate the MAX prior to the summing. I also tried an MMULT but could not get this to work either.
Any idea how to refine this.
### Excel Facts
Format cells as time
Select range and press Ctrl+Shift+2 to format cells as time. (Shift 2 is the @ sign).
Try:
Excel Formula:
``=MAX(FILTER(AD30:AG33,D30:D33=TRUE))``
Great thanks!
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00:00 - 00:59 the question is to solve the following equation 15 + x is equal to 5 x + 3 sorry equation is 15 + x is equal to X + 32 we will take all the variables on one side and the constants on if I start my 3 on both sides with printed 15 + X - 3 is equal to 5 x + 3 - 3 the soul the come 12 + x is equal to the office abstract on both the sides India 12 + X - x is equal to 5 x minus x and thus will become 12 explain cancel this will be equal to 4 x 9 with divider equation before we will get 12 upon 4 is equal to a
01:00 - 01:59 describe the search the value of excess 12843 so this is required answer
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निम्नलिखित समीकरणों को हल करें: <br> `5x=15` Solve the following equations: <br> `5x+15=20` Solve the following equations : <br> x - 3 = 15 निम्न समीकरणों को हल कीजिए - <br> `10x-5-7x=5x+15-8` Solve the following simutaneous equation:<br> `5x+2y = 13`, `4x-3y =15`. নীচের সমীকরণটি সমাধান করো অর্থাৎ সমীকরণের বীজ নির্ণয় করো : `x+3=15` নীচের সমীকরণটি সমাধান করো অর্থাৎ সমীকরণের বীজ নির্ণয় করো : `x+3=15` Solve the following simultaneous equations.`5x+2y=-3,x+5y=4`. खालील एकसामयिक समीकरणे सोडवा. <br> `5x+2y=-3, x+5y= 4` Solve the following simultaneous equations. <br> 5x + 2y = -3, x + 5y = 4 निम्न समीकरणों को हल कीजिए : <br> 5x - 3 = 7 Solve the following equations :<br>`5x - 3 = 7` निम्नलिखित समीकरण को हल कीजिए : `5x-3=7` निम्नलिखित समीकरणों को हल कीजिए :<br>`(7x+3)/(5x)=2` Solve the following equation : `(5x)/2 + 3 = 13`
Very Important Questions
FAQs on Linear Equations In One Variable
Introduction
Equation: a statement of equality which contains one or more unknown quality or variable ( literals) is called an equation
Linear Equation: an equation involving only linear polynomials is called a linear equation
Solution: a value of the variable which when substituted for the variable in on equation makes
L.H.S. =R.H.S.
is said to satisfy the equation and is called a solution or a root of the equation .
Solving an equation: solving an equation means determining its roots
Solving equations having variable terms on one side and number(s) on the other side
Transposition method
Solve :
(3t-2)/4 - (2t+3)/3 = 2/3 - t
Solve :
2/5x -5/3x =1/15
Cross-multiplication method for solving equations of the form
(ax+b)/(cx+d) = m/n
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Physics 6C Ch28worksheet Solutions
# Physics 6C Ch28worksheet Solutions - PHYSICS 6C Ch.28...
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PHYSICS 6C Ch.28 Worksheet Solutions clas.ucsb.edu/staff/vince/ 1) A flashlight emits 1.5W of power. Assuming a frequency of 5.2x10 14 Hz for the light, determine the number of photons given off per second. The energy of each photon is given by E=hf, or E=(6.63x10 -34 J-s)(5.2x10 14 Hz)=3.45x10 -19 J Dividing 1.5W by this energy gives 4.4x10 18 photons per second. 2) The work function of gold is 4.58 eV. What frequency of light must be used to eject electrons from a gold surface with a maximum kinetic energy of 6.48x10 -19 J? Is this light visible to the human eye? Converting to Joules, (4.58eV)(1.6x10 -19 J/eV)=7.33x10 -19 J This is the energy necessary to just barely eject the electrons from gold. We need to add the extra kinetic energy to get the energy of the photon. E photon =1.38x10 -18 J. The frequency is given by E=hf. f photon =2.1x10 15 Hz This light is in the ultraviolet range of the spectrum and cannot be seen by humans. 3) Find the wavelength of the photon required to excite a hydrogen atom from the n=2 state to the n=5 state.
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in expansion 1 + x raise to power and the coefficient of p and q are the term are equal prove p + q is equal to 1 + 2
Asked by urvithakur013 | 07 Feb, 2024, 08:56: PM
In the expansion of (1 + x)^n, the coefficients of pth and qth terms are equal.
i.e. p = q or p + q = n + 2
Hence, p + q = n + 2 since p is not equal to q.
Answered by Renu Varma | 08 Feb, 2024, 10:33: AM
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32,861 results
1. ## Mathematics
4 times x is equal to x increased by 6. x A.1 B.2 C.3 D.4
2. ## Math
A flower garden has three times as many red roses as pink roses. Twice the number of red roses is equal to four times the number of pink roses increased by ten. How many red roses are there? How many pink roses are there?
3. ## Math
If each edge of a cube is increased by 2 inches, the A. volume is increased by 8 cubic inches B. area of each face is increased by 4 square inches C. diagonal of each face is increased by 2 inches D. sum of the edges is increased by 24 inches is the answer
4. ## algebra
If four times a certain number increased by 6 is equal to 94, what is the number? If x represents the number, then which of the following equations could be used to solve the problem? 4x + 6 = 94 or 4(x + 6) = 94
5. ## Math
Hi everyone, my name is Emerson. I was wondering if anybody could help me on a test. I'm stuck on some, but I think I know some others. Thank you! 1. Translate the phrase "nine more than two times a number" in to an algebraic expression. A. 9x + 2 B. 2x +
6. ## Math
What is the relationship between 1.0 and 0.1? A) 0.1 is 10 times as much as 1.0 B) 1.0 is 1/10 of 0.1 C) 0.1 is 1/10 of 1.0 D) 1.0 is equal to 0.1 My first elimination are c and d and I know that 0.1 is 10 times as much as 1.0. Now choice b is also true.
7. ## Algebra
A number w increased by 6 is less than or equal to 17
8. ## Geometry
Regular pentagons A and B are similar. The apothem of Pentagon A equals the radius of Pentagon B. Compare the areas. The area of Pentagon A is equal to 1.49 times the area of Pentagon B. The area of Pentagon B is equal to 1.49 times the area of Pentagon A.
9. ## mathematics
6. In a GP the 3rd term is equal to four times the first term and the sixth term is equal to three times the fourth term plus 32. Determine the sequence.
10. ## history
What effect did Samuel Slater's actions have on the U.S. economy? Tariff revenues on imported clothing increased. The efficiency of American textile production increased. The price of clothing produced in the United States increased. Wages for workers in
11. ## math
If each edge of a cube is increased by 2 inches, the A. volume is increased by 8 cubic inches B. area of each face is increased by 4 square inches D. sum of the edges is increased by 24 inches
12. ## chemistry
what is the initial temperature (in degrees C) of a system that has the pressure decreased by 10 times while the volume increased by 5 times with a final temperature of 150 K? a)27 b)75 c)300 d)-198 e) none of the above Please explain--I need to know how
13. ## American History: Iranian Revolution
What was one Impact of the Iranian Revolution on the United States? A) Higher oil prices B) Increased terrorism C) Increased taxes D) Increased military spending I think that it's D. Can anyone verify?
14. ## ALGEBRA 1
1. The sum of a number and 2 is 6 less than twice that number. 2. A rectangular garden has a width that is 8 feet less than twice the length. Find the dimensions if the perimeter is 20 feet. 3. Complementary angles sum up to equal 90 degrees. Find the
15. ## Math
Three times a number increased by seven is less than eleven
16. ## Algebra
Find a positive real number such that its square is equal to 4 times the number, increased by 60.
17. ## Algebriac Expressions
2 less than 7 times a number = 2) Sarah's age decreased by 2 = 3) One third as many books = 4) Earns \$13 per hour = 5) 4 years younger than Marcus 6) 2 times a number, decreased by 4 7) 9 less than the product of 15 and a number 8) The quotient of a number
18. ## Algebra
Seven times the smaller of 2 numbers plus 9 times the larger is 178. When 10 times the larger is increased by 11 times the smaller the result is 230. Find the numbers.
19. ## Physics
What happens to the force between the charges if the distance between the charges is increased to two times its original value? Value of the force is increased by two. Value of the force is decreased by two. Value of the force is decreased by four. Value
1. What was one impact of the Iranian Revolution on the United States? a. Higher oil prices*** b. Increased terrorisim c. Increased taxes d. Increased military spending
21. ## Physics
A singer moves a note up one octave. How has the sound changed? The amplitude has decreased The wavelength has increased The speed has increased The frequency has increased
22. ## algebra one
four times the sum of a number and two increased by three is at least twenty seven
23. ## Math
1. The sum of a number and 2 is 6 less than twice that number. 2. A rectangular garden has a width that is 8 feet less than twice the length. Find the dimensions if the perimeter is 20 feet. 3. Complementary angles sum up to equal 90 degrees. Find the
24. ## physics
7)In calculating the force of gravity between two objects, if the mass of one object increased by 4 and the other by 2, how many times would the force of gravity increase? I don't really understand but I got 6 times 8)In calculating the force of gravity
25. ## Calculus
Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3. 1. For 0 is
26. ## Calculus
Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3. 1. For 0 is
27. ## Math problem
The area of a triangle is equal to 1/2 the hight times the base. A triangle has a height of 10 and a base of 20. If the height of the triangle is increased by 10% and the base is decreased by 10%, then the area of the new triangle formed___________. ?
28. ## Social Studies
2. What is a positive affect of competition? A. Increased value for consumers B. Increased wages for workers C. Increased profits for producers D. Increased incentives for investors Is it A or C
29. ## math
7. The tennis team is selling key chains as a fundraiser. If its goal is to raise is at least 180, how many key chains must it sell at \$2.25 each to meet that goal? Write and solve an inequality. a. 2.25k (greater than or equal to) 180; k (greater than or
30. ## mathematics
evng .my teacher gave me this as a homework but i dnt undertand. 1.In a GP the 3rd term is equal to four times the first term and the sixth term is equal to three times the fourth term plus 32. Determine the sequence.
31. ## Math
Tanker boats cannot be in the bay when the depth of water is less or equal to 25 feet. Set up an inequality and solve it graphically to determine all points in time, t, on the interval 0 less than or equal to t less than or equal to 24 when tankers cannot
32. ## algebra 2
1. 5x+2>3x+10 =x>4 2. 8+2x< or equal too 6x-20 x> or equal too 7 3. 4(1-3n)-14>4(2n+3)-9n =? 4-12n-14>8n+12-9n 5. x/4-9 -p>-63 p or equal too 6 -t> or equal to 72 t< or equal too 72 9. -y36 10. define a variable, write an inequality, and solve each
33. ## maths
A flower garden has three times as many red roses as pink roses. Twice the number of red roses is equal to four times the number of pink roses increased by ten. How many red roses are there? How many pink roses are there?
34. ## Sequence
In a GP the 3rd term is equal to four times the first term and the sixth term is equal to three times the fourth term plus 32. determine the sequence.
35. ## physics
TWO EQUAL FORCES HAVE A RESULTANT EQUAL TO 1 1/2 TIMES THE EITHER FORCE . AT WHAT ANGLE THEY ARE INCLINED TO EACH OTHER
36. ## math
if you threw a 6-sided die 48 times, about how many times would you expect it to land on a number greater than or equal to 4.
37. ## math
if you threw a 6-sided die 48 times, about how many times would you expect to land on a number greater than or equal to 4? how do i find the answer?explain.
38. ## precalculus
A study investigated whether injecting mice with a certain hormone would affect the time it takes them to learn a task. Learning times of 12 mice with the hormone injected were compared with those of 12 mice without the hormone injected. The mean of the
39. ## Shiv jyoti public school..jalandhar..punjab
If wire has resiatence 0.45 ohm. What will be the length of another wire if length is increased by4 times and area decreased by 3 times?
40. ## math
A. One board is one-third the length of another. Six times the sum of the length of the short board and -10 is equal to the length of the longer board decreased by 11 inches. Find the length of the longer board. B. The length of a rectangle is 4 feet more
41. ## Math
a standard number cube is rolled 180 times. Predict how many times a 3 or 5 will be the result. 31 times 60 times 57 times 30 times I chose 31 times Is that correct i would appreciate your help
42. ## algebra
how do i translate this to a inequalities 2 times a number increased by 28 is less then or equal to 6 times that number 2x + 28
43. ## word expressions
2 times a number, increased by 28 is less than or equal to 6 times that number is this how i should write this 2*x+28 28 x > 7
44. ## math,algebra
can someone correct this for me: Translate the following statements into inequalities. Let x represent the number in each case. 2 times a number, increased by 28, is less than or equal to 6 times that number. My answer: 2x+28
45. ## math
show work 2 times a number increased by 28 is less than or equal to 6 times a number?
46. ## math
how to find two consecutive positive even integers such that the square of first increased by 2 is equal to 3 times the second.
47. ## math correction
Translate the following statements into inequalities. Let x represent the number in each case. 2 times a number, increased by 28, is less than or equal to 6 times that number.
48. ## math
five times the number increased by thirteen is less than or equal to twice the nummmber decreased by twenty find the solution
49. ## Algebra
Which situation can be represented by 3x + 2 = 8x β 13? Thirteen less than three times a number is the same as eight times the number, x, increased by 2. What is the value of x? Three times a number decreased by 2 is the same as 8 times the number
50. ## Matrices Word Problem
The stopping distance of a car traveling 25mph is 61.7 feet, and for a car traveling 35 mph it is 106 feet. The stopping distance in feet can be described by the equation y = axΒ² +bx, where x is the speed in mph. (a) Find the values of a and b. (b)Use
51. ## Algebra 2
Fifty four decreased by three times a number is equal to the number increased by 10
52. ## Electricity
2. According to Coulombβs Law, what is the change of force between two electrically charged bodies when the distance between them is increased by 3 times? a. Decrease 9 times b. Increase 3 times c. Decrease 3 times d. Increase 9 times
53. ## Math
x=2 the product of 4 and 3 times x 12x x increased by 2 x+2 the quotient of x and 2 x/2 7 increased by x x+7 the quotient of 6 and x x/6 5 times x divided by 2 5x/2
54. ## Algebra
Please help. I have tried many times to come up with something after reading the material but I am completely stumped. A cubic container, with sides of length, x inches, has a volume equal to x^3 cubic inches. The height of the container was decreased and
55. ## math
write the verbal sentence as an equation.then solve the equation Three less than 11 times a number is equal to 9 plus 5 times the number. twelve less than -9 times a number is equal to 8 minus 4 times the number
56. ## Physics
In calculating the force of gravity between two objects, if the mass of one objects increased by 4 and the other by 2, how many times would the force of gravity increase? Ok well Im horrible at physics but the way I understood it was that you would take 4
57. ## algebra
three times a number decreased by 7 is equal to the number increased by 39. what is the number?
58. ## algebra
If the dimensions of the rectangular prism are increased 4 times, the surface area will increase 8 times.
59. ## physics
we can say that the force of gravity is equal to mass times acceleration were the acceleration is equal to gravity sense gravity is an acceleration because of newtons second law force = mass times acceleration hence Fg = mass times acceleration Fg = mass
60. ## math
can anyone help me with these 3 problems I am trying to check my daughters hoime work it has bee na few years. 1. The sum of n and twenty -two, multiplied by three, is 78 2. the product of z and 17 is less than or equal to the difference of twenty -one
61. ## algebra
The sides of an rectangle was 5cm by 9cm. When both dimension were increased by an equal amounts the area of the rectangle increased by 120 cm^2 find the dimensions of the new rectangle.
62. ## statistics
I am trying to determine if the dice that I have is fair, so I will toss the dice 300 times and measure how many times each side appears. Here are my results: Side : Frequency 1 : 43 2 : 49 3 : 56 4 : 45 5 : 66 6 : 41 Looking at the data, is the dice
63. ## algebra 1
1)write an inequality a. the sum of three times a number and 2 lies between 7 and 14 b.seven less than 4 times a number is at most 44 and at least -24. 2)which is not a solution of -4 is less than or equal to 2-6x less than or equal to 8. 3) 2x-1
64. ## Mathbio in 12th science
2)a stone of mass 1kg tied to one end of a string of length 1m whirld in a horizontal circle with a costant speed 2 radian/second tension of the string is. A)0.5n b)1n c)2n d)4n 3)the orbital velocity of a satelight close to the earth is v then the orbital
65. ## social studies
How has democracy impacted Central America? A. increased dependence on Spain B. increased economic prosperity and stability C. increased control by the United States D. increased restrictions on trade
66. ## math question
-The product of all the positive factors of 20 ( including 20) can be written2^x.5^y. Find x+y. -The number 18 is equal to 2 times the sum of its digits: 1+8=9. 2x9=18. There are 4 two digit numbers that are equal to 4 times the sum of their digitss . Find
67. ## Math
How would you solve for these word problems in one variable inequality form? 1. A number, n, decreased by 6 is less than or equal to 4. 2.One half of a number, increased by 9 is less than 33. 3.Six less than or equal to the sum of 4 and -2x.
68. ## Math
The dimensions are 20 ft by 50 ft the garden is to be increased by equal amounts. The new area will increase by 1800 Sq ft. SO the new dimensions would be 40 x70 this would equal 2800 Sq ft???????
69. ## Algebra
Write each statement as an inequality. Then give 3 solutions of the inequality. 7 minus 5 is less than x 8 is greater than or equal to u H is greater than 5 times 8 P times 7 is less than or equal to 63
70. ## 9th math algebra
The stars indicate the answers i think are correct. 1. What number is a solution of the inequality? m > 13/3 a.3 b.5 * c.2 d.-9 2. Write the inequality in words. 3n < 52 (1 point) a.fifty-two less than three times n *b.Three times n is less than fifty-two.
71. ## Algebra
One number is 4 more than 3 times another. Four times the larger decreased by 5 times the smaller is equal to the smaller subtracted from 32. Find the numbers.
72. ## Math Check
Solve the inequality. So i have done this problem at least three times and I keep getting the same answer. But in the back of the book the correct answer for this problem is x
73. ## Math
Write an algebraic expression for each of the following verbal phrases. _______________ 6) eleven times the sum of twelve and five times x _______________ 9) the product of eight and x increased by fifteen _______________ 10) four less than the quotient of
74. ## math
Write a variable expression for each word phrase. 1. two times the number of windows 2. 81 increased by n 3.eleven more than a number \$15 times the number of hours
75. ## Math
Can someone please help me write this statement out in formula form. There should be two formulas. Question.... Three times the larger of two numbers is equal to four times the smaller. The sum of the numbers is 21. Find the numbers. If we let X = larger
76. ## geometry
find the positive difference between to consecutive even integers such that 14 times the first is equal to 12 times the second
77. ## Science
At 0Β°c and 760mm hg pressure, a gas occupies a volume of 100cm cube. The Kelvin temperature of the gas is increased by one - fifth while the pressure is increased one and a half times. Calculate the final volume of the gas.
78. ## Math
The sum of two numbers is nine. Four times the smaller is equal to two times the larger.
79. ## Alg 2
Expressions, Equations and Inequalities I am struggling can someone please explain how to get the correct answer choice thank you! 1. Which of the following is equal to (3x/5-7)+7 5x/3 3x/5 3x-35/5 (3x-7)+35 5. name the property of real numbers illustrated
80. ## Statistics
Multiple Choice: A die is rolled 600 times. The face with six spots appears 112 times. Is the die biased towards that face, or is this just chance variation? Answer the question in the steps outlined in Problems 1-6. 1) The null hypothesis is: a) The die
81. ## algebra 1 help pplz this is an emergency
fifteen less than six times a number b is equal to seven more than four times b. Find b.
82. ## Math
A study investigated whether injecting mice with a certain hormone would affect the time it takes them to learn a task. Learning times of 12 mice with the hormone injected were compared with those of 12 mice without the hormone injected. The mean of the
83. ## Math
A study investigated whether injecting mice with a certain hormone would affect the time it takes them to learn a task. Learning times of 12 mice with the hormone injected were compared with those of 12 mice without the hormone injected. The mean of the
84. ## math
Translate 17 increased bt 4 times d
85. ## Math
There are 4 boys and 2 girls in the Science Club. The members draw straws to determine which two members will give the demonstration at the science fair. Which simulation could be used to determine the probability that at least one of the demonstrators
86. ## Math
In 1900, half of the babies born in America were born with blue eyes. Which simulation could be used to find the probability that 3 out of 4 babies born had blue eyes? A. Rolling a 6-sided number cube 3 times B. Spinning a spinner with 3 equal sections 4
87. ## math
the product of 16 and c is increased by 6 to equal the qoutient 25 and 9. what is c?
88. ## Chemistry
For the equilibrium system below, which of the following would decrease the concentration of ππΆπ3(π)? ππΆπ3(π) + πΆπ2(π) β ππΆπ5(π) + 45ππ½ a. Increased temperature b. Decreased pressure c. Increased
89. ## physics
centripetal acceleration= friction m*w^2 r = mu*mg changing 50 rpm to rad/sec mu=w^2 r/g=(50*2PI/60)^2 * .12 which is not your answer. Check my thinking. ok I thought centripetal force was equal to m ac were ac is the centripetal acceleration which is
90. ## Science
[I need help finding the missing words] An increase in population could result from increased birth rate, increased _____________, or immigration. Increased life span.
91. ## US HISTORY 7B
WHAT WAS ON IMPACT OF THE IRANIAN REVOLUTION ON THE UNITED STATES? HIGHER OIL PRICES*** INCREASED TERRORISM INCREASED TAXES INCREASED MILITARY SPENDING
92. ## history
What was one impact of the Iranian Revolution on the United States? A. higher oil prices B. increased terrorism C. increased taxes D. increased military spending
93. ## Math
The AC voltage coming out of a standard wall plug in the United States can be expressed as the function V(t) = 120sin(120*pi*t), where t represents time (in seconds) and V is the voltage. We will let t be greater than or equal to 0 for simplicity. A. Make
94. ## more math
sqrt50/sqrt2 (sqrt50)/(sqrt2)*(sqrt2/sqrt2)= (sqrt50*sqrt2)/2 =sqrt100/2=10/2=5 Check my thinking. thanks... what does this equal: 3sqrt2+4sqrt2 3 times the square root of 2 PLUS 4 times the square root of 2. This would equal 7 times the square root of 2,
95. ## english
12. βhas trebledβ (line 15) means: (A) is 3 times less important (B) is one third as important (C) has become 30 times more important (D) has increased threefold i think D
96. ## math
three times a number increased by seven is less than eleven
97. ## Math
in 1900, half of the babies born in America were born with blue eyes. Which simulation could be used to find the probability that 3 out of 4 babies born had blue eyes? A.) Toss a coin two times B.) Spin a spinner with 4 equal sections two times C.) Roll a
98. ## Algebra
The result of multiplying 2 by x plus y times 3 plus 5 times z is equal to 10.
99. ## state & local government
1. Why have the powers of the governor increased in modern times?
100. ## Math
Ten times increased by 12 is 32.find the number | 5,725 | 21,216 | {"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} | 3.8125 | 4 | CC-MAIN-2020-34 | latest | en | 0.956742 |
https://stackoverflow.com/questions/54346263/tensorflow-gradient-getting-all-nan-values | 1,618,285,394,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00370.warc.gz | 623,253,882 | 38,305 | # tensorflow gradient - getting all nan values
I am using python 3 with anaconda, and tensorflow 1.12 with eager eval.
I am using it to create a triplet loss function for a siamese network, and need to calculate distance between different data samples.
I created a function in order to create the distance calculation, but no matter what I do, when I try to calculate it's gradient with respect to the networks output, It keeps giving me all nan gradient.
This is the code:
``````def matrix_row_wise_norm(matrix):
import tensorflow as tf
tensor = tf.expand_dims(matrix, -1)
tensor = tf.transpose(tensor, [0, 2, 1]) - tf.transpose(tensor, [2, 0, 1])
norm = tf.norm(tensor, axis=2)
return norm
``````
In the loss function I am using
``````def loss(y_true, p_pred):
t.watch(y_pred)
distance_matrix = matrix_row_wise_norm(y_pred)
``````
And the grad is all `nan`s. I checked that `y_pred` is made of legit values - and it does. I tried to create a gradient of `y_pred * 2` with respect to itself and got legitimate gradient values.
What am I missing here? Is the indexing in the creation of the distance matrix problematic?
edit:
the dtype of both `y_pred` and `loss` is `tf.float32`
edit: found an open bug report in tf - could this be the issue?
edit:
When I change the norm axis to 0 or 1, I am getting legitimate values and nothing goes to `nan`. The operation I am getting using norm with `axis=2` is the pairwise distance between the pairs of rows in the matrix, I suspected this might have something to do with 0 distance between a row to itself, so I clipped the values with min value of 1e-7 without any luck.
Thanks
• I had same problem, please check `dtype` of `y_pred` and `loss`. – Ankish Bansal Jan 24 '19 at 12:16
• @AnkishBansal - thanks for the reply, both are tf.float32 – thebeancounter Jan 24 '19 at 12:33
• What are each of the axes of your matrix? My only guess is that `norm(tensor, axis=2)` or the transpose and subtract operation above it does not have a gradient. I've run into that issue before with custom loss functions and, I think, reshaping? Non-differentiable operations seem to kill the gradient computation. – Engineero Jan 24 '19 at 14:52
• @Engineero - what i do here, is to take a matrix, each row is a vector, I am trying to create pairwise distance between all the vectors, and getting this by duplicating the vectors, transposing, subtracting and using norm, How could this not have a gradient? – thebeancounter Jan 24 '19 at 14:54
Seems that tf.norm suffers from numeric instability as explained here
They also suggest using l2 norm that is more numeric stable, So I tried that, also getting nan values, thanks to 0 gradients. So I used those together with gradient clipping, so far so good, the loss function is working and manages to converge.
``````def last_attempt(y_true, y_pred):
import tensorflow as tf
import numpy as np
loss = tf.zeros(1)
for i in range(y_pred.shape[0]):
dist = tf.gather(y_pred, [i], axis=0)
y = y_true.numpy().squeeze()
norm = tf.map_fn(tf.nn.l2_loss, dist-y_pred)
d = norm.numpy()
d[np.where(y != y[i])] = 0.0
max_pos = tf.gather(norm, np.argmax(d))
d = norm.numpy()
d[np.where(y == y[i])] = np.inf
min_neg = tf.gather(norm, np.argmin(d))
loss += tf.clip_by_value(max_pos - min_neg + tf.constant(1, dtype=tf.float32),
1e-8, 1e1)
return loss
``````
There is much room for optimizing that function, here is a reference to my other SO question - working on that.
• Can we just add an epsilon inside the norm to make a safe norm like that in this answer? – Lerner Zhang Aug 22 '20 at 15:19 | 944 | 3,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | latest | en | 0.865793 |
enetcl.com | 1,726,076,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00432.warc.gz | 208,881,084 | 32,641 | Search
## Why are light bulbs connected in series with resistors?
To facilitate you to better understand the article, in understanding why the light bulbs series resistance, I would like to give you a little bit of knowledge about resistance.
First of all, what is resistance? Why do lights need resistors? What effect does the resistor have on the LED ball? So the next thing to talk about is what resistance has to do with the light bulb.
Simply put, resistance is the resistance that an electric current encounters in a circuit, or the ability of an object to obstruct that current.
The LED bulb is not necessarily equal to the power supply voltage. The general power supply voltage will be slightly higher than the light bulb voltage. At this point need to add series resistance to reduce voltage and current limit to maintain normal work to prevent LED lights burn out. So What’s the resistance.
1. Low power red and yellow, its resistance = voltage/current = 1.8/0.02 = 90Ω.
2. Low Power Blue Green, its resistance = voltage/current = 3.2/0.02 = 160Ω.
3. High power red and yellow, its resistance = voltage/current = 2.2/0.35 = 6Ω.
4. High Power Blue Green, its resistance = voltage/current = 3.4/0.35 = 10Ω.
Resistance is a physical quantity, in physics, the amount by which a conductor impedes the flow of current. Its English name is resistance, commonly abbreviated as R. It is a basic property of a conductor, and the size of the conductor, materials, and temperature. The greater the resistance of the conductor, the greater the resistance of the conductor to the current.
ENE TECH LED All In One Solar Streetlight
### Different conductors have different resistances.
The basic unit of resistance is the ohm, denoted by the Greek letter “Omega”. Ohm’s law states that the relationship between voltage, current, and resistance is i = u/r, or R = U/i. The resistance value of a resistance element is generally related to temperature, material, length, and cross-sectional area.
The physical quantity to measure the effect of temperature on resistance is the temperature coefficient, which is defined as the percentage change in the resistance value for every 1 °C increase in temperature.
The main physical characteristic of resistance is to convert electrical energy into heat energy, it can also be said that it is an energy-consuming element, the current through it to produce internal energy. The resistance usually acts as a voltage and current divider in a circuit. For signals, both AC and DC signals can pass through resistors.
However, in daily work, led lamp beads are more series resistant, than led lamp beads why do series resistant? There are three main reasons.
### First, to better protect the LED lights.
What do you mean? Because LED has a working voltage limit, exceeds working voltage, positive PN junction current is too large, easy to burn out LED, so need a series resistance to protect LED.
In the early days, in 1962, LEDs emitted only low-light red light, and then other monochromatic lights. Today’s LEDs emit visible, infrared, and ultraviolet light, the luminosity is also raised to high brightness or even super brightness.
And the LED lamp bead uses for the first time as the indicator light, the display board, the general lighting, the medical device, and so on. The LED bulb, on the other hand, is essentially a light-emitting device that emits energy through a combination of electrons and holes to efficiently convert electrical energy into light energy, but because the LED bulb has a working voltage limit that exceeds the working voltage, positive PN junction current is too large, easy to burn LED lights, so you need to series a resistor.
### Second, the LED lights add resistance to reduce the operating current.
LED light beads do not have to add resistance, if the power supply is constant current can not use resistance. But because LED Nonlinear elements, sensitive to power and voltage, can not be controlled in a voltage manner, mainly by controlling the brightness of LED lights in a current manner, adding resistance is to limit the current adding resistance will affect the efficiency of LED light pearlescent conversion, low-power LED lights can ignore the loss of resistance, and high-power LED lights are usually directly used constant current power supply.
### Third, LED lights plus resistance, in the final analysis, is because if you do not add resistance, LED lights because the working current will cause PN junction heating serious, burn out PN junction.
Too high a voltage will lead to too high a current, and the optical decay will increase. Shortens the life of light bulbs.
Speaking of which, do you understand why the LED bulb needs resistance?
Led beads are a low-current, low-voltage electronic component. But our market is usually 220 volts or 110 volts in Hong Kong. And according to the voltage and current characteristics, the number of LED lights in series, voltage added parallel, and the current increases.
Therefore, to better match the driving power supply, general LED lights are used in series. For example, an LED bead 3528 is about 3 volts, and 103528 beads in series are 10 times 3 volts, 30 volts. So, in this case, we can use a series of LED lights, and then add resistance to use.
Why, after all, are LED lights connected in series with resistors? Is to limit the current, is to protect LED lights, is to avoid LED lights because the current is too large to make the PN junction heating serious, burned PN junction.
As a result, the components on which we attach resistors to light bulbs are often called current limiting resistors. Of course, this is just a more popular name, to let us better understand the meaning of the LED lamp bead series resistance.
Finally, does this knowledge help you? If you need help or want to know more about LED lights, please feel free to leave us a message in the discussion area or contact us in the following ways. Best wishes to you
Website: www. enetcl.com
Whatsapp: +86 158 7620 7215
Email: sales1@enetcl.com
## Don’t Be A Stranger…
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Information of news
• Author: mitsumi
• Date: 28-06-2021, 15:16
28-06-2021, 15:16
### 5s training - Statistical View and Tutorials
Category: Tutorials
MP4 | Video: h264, 1280x720 | Audio: AAC, 44100 Hz
Language: English | Size: 351 MB | Duration: 53m
What you'll learn
5s Statistical analysis
How to develop Graphs and templates in Excel
Explanation of what each S means numerically
How to Quantify 5s
Requirements
Quality engineers, manufacturing engineers, process engineers engineer in general.
Basic Excel Understanding
Statistical Background
Description
We can expect to learn the various statistical methods of analyzing 5s.
In this course we will go through a tutorials for each S of the cycle and explain in detail the affects it has on(Sort,Set,Shine,Standardize and Sustain).
This will be analysed in order to make better continuous improvement contributions.
In this course i will be addressing:
- Each step from a blank excel sheet to the fully populated and calculated data will be shown
- Each step describe what is happening as we analyse along
- Fully aware of how to quantify 5s
- Basic lean manufacturing and quality techniques shown and discussed.
- Basic excel technique shown
- Understanding of setting up your own quality control systems
- Foundational mathematics technique and graphs described
Hope you learn a lot more as I describe various scenarios and the analyses of each S at its independent states.
I also describe the profits and loses in real terms and explain how to avoid them or at least be aware of pitfalls.
Its not a strategy, its an analysis and a way of quantify a culture.
I Greatly Appreciate your participation in these lecture and hope you gain a lot of knowledge from them
Remember , Enjoy
Always feel free to ask questions
Who this course is for:
5s
Quality
Manufacturing
Data
Process Engineering
Excel
Microsoft Office
Data Science
Screenshots
Site BBcode/HTML Code:
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2. / Drive Calculation For A Ball Mill Portable
# Drive Calculation For A Ball Mill Portable
### Drive Calculation For A Ball Mill Operation
Drive calculation for a ball mill - phu-universum.pl Variables in Ball Mill Operation. In ball milling of dry solids the main independent variables are mill.
### Drive Calculation For A Ball Mill Jar
Drive calculation for a ball mill jar. 2018 11 17 Please find below two calculators for sizing mills using the Bond and Rowland methods Ball mill sizing Calculator for ball mill s in a single stage circuit Rod ball mill sizing Calculator for rod mill s as first stage of the circuit and ball mill s.
### Drive Calculation For A Ball Mill
Calculation for ball mill driveschoren . calculation for ball mill drive volejbalmelnik.cz drive calculation for a ball mill How To Calculate The Primary Drive For A Ball Mill What To calculate ball mill drive hp ball mill designpower calculation the ball mill motor power requirement calculated above as 1400 hp is the power that must be applied at the mill drive in order to grind the tonnage.
### Ball Mill Gearbox Drive Calculation
An Experimental Investigation into Torsional Vibration in Ball Mills. A small batch ball mill was investigated using a servo motor as both the drive and a torsional exciter. The rig consisted of an electric motor, a 20 1 gearbox, drive shaft and a 0.3 m diameter by . damping was assumed in these calculations. .
### Ball Mill Gearbox Drive Calculation
Ball mill gearbox drive calculation ball mill gearbox drive calculation Popular Searches. Shaft calculation - Villanova University. For horizontal mills - FL. 6 The CPU drive concept Function1 Power input is achieved by the sun pinion of the first stage. In the first and second stage three planet wheels in mesh with the sun.
### Ball Mill Parameter Selection & Calculation
Aug 30, 2019 Rho s — loose density of grinding medium, t m3. Forged steel balls P=s=4.5-4.8t m3 cast steel balls P=4.3-4.6t m3 rolling steel balls P=6.0-6.8t m3 steel segments P=4.3-4.6t m3_-filling ratio of grinding medium, When wet grinding lattice ball mill pi = 40 – 45 overflow ball mill phi = 40 rod mill.
### Ball Mill Design Power Calculation
Jun 19, 2015 The ball mill motor power requirement calculated above as 1400 HP is the power that must be applied at the mill drive in order to grind the tonnage of feed from one size distribution. The following shows how the size or select the matching mill required to draw this power is calculated from known tables ‘the old fashion way’.
### How To Size A Ball Mill Design Calculator & Formula
May 15, 2015 A) Total Apparent Volumetric Charge Filling – including balls and excess slurry on top of the ball charge, plus the interstitial voids in between the balls – expressed as a percentage of the net internal mill volume (inside liners). B) Overflow Discharge Mills operating at low ball fillings – slurry may accumulate on top of the ball charge causing, the Total Charge Filling Level to be.
### How To Calculate The Primary Drive For A Ball Mill
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### How To Make A Portable Ball Mill
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### Variables In Ball Mill Operation Paul O Abbe
A Slice Mill of 72” diameter by 12” wide would replicate the result of a normal production, mill 72” in diameter as 120” long. A Slice Mill is the same diameter as the production mill but shorter in length. Click to request a ball mill quote online or call 630-350-3012 to speak with an expert at Paul O. Abbe to help you determine.
### Torque Factor Of Ball Mill Grinding Mill China
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Ball Mill Group No.3 Calculation Feed capacity → Mill dimensions Filling ratio (30-45 ), Mill dimensions → Bulk volume of the balls ball size D2 = k Dp (where k.
### Ball Mill Ball Mills Wet & Dry Grinding Dove
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### Grinding Ball Mill Girth Gear Root Clearance
Apr 27, 2020 How do you measure the root clearance between a girth gear and a pinion of a grinding ball mill and what is the recommended root clearance. This ball mill uses helical gears(i.e. 284 teeth girth gear with a 24 teeth pinion).
### Small Ball Mills For Sale 911metallurgist
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### 2000 Ton Capacity Ball Mill Media Charge Calculation
Apr 12, 2013 Calculation of Power at Ball Mill Shaft for hours per metric or short ton) at 35 ball charge 6 Mill drive process calculation for ball mills percentage filling grinding media charge calculation in cement ball mills. ball mill capacity calculations in Panama.
### Ball Mill Reduction Ratio
Calculation Of The Ball Mill Reduction Ratio Henan. For overflow ball mills the charge should not exceed 45 of the mill volume For grate discharge mills the charge should occupy about 50 of the mill volume Bond developed a relationship that can be used to determine the percent charge by volume as a function of the vertical height above the charge He and the radius of the mill R ie.
### How Much Are Popup Control Rooms For Gravel Ball Mills
Crusher plant control room 32211 giorgiozambrino.ch. crusher plant control room 32211 groundnut wet ball mills. Crusher plant control room beckersmuehlede Rock Crusher Control Room Control room of a crushing plant needs with portable crushing plants plate feeder overbelt magnet 150kw EMD all conveyers electrical control room with radial crusher control cone crushers There is an option for.
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### 3 Easy Steps To Calculate Ball Mill Capacity
Mar 09, 2016 Ball mill drum dimensions. Specify the length and capacity of the drum. If not available, specify the length and diameter of the drum. Step 3 (the final). Grinding bodies. Specify the weight of grinding bodies. You can simply specify a fill ratio for the drum if you do not know the exact weight of grinding bodies. Done!.
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540 RPM Drive. Build Quote. STATIONARY ROLLER MILL. From \$19,443. 20 or 30 inch roller mill. Electric or PTO Drive. Build Quote. PORTABLE ROLLER MILL. From \$22,327. 20 or 30 inch roller mill. 540 RPM Drive. Optional hopper assembly. The cookie is used to calculate visitor, session, campaign data and keep track of site usage for the site.
### Patterson Industries Ball & Pebble Mills
PATTERSON Mills are equipped with motor, ring gear and pinion drive, magnetic brake and inching device. Type DJ Jacketed Mills provide controlled temperatures ranging from below 0 to 500 F by the circulation of brine or other refrigerants for low temperatures, or hot water, steam, or thermal fluid (hot oil) for high temperatures.
### Ball Mill Relining Standard Operating Procedure Pdf Free
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### Ball Mill&rod Mill Grinding System Used In Mineral Milling
Prominer provides ball mill rod mill grinding system which is widely used in various types of ores' beneficiation, electricity, cement and chemical industries. It can carry out dry or wet pulverizing and can meet demand for sustainable large-scale production.
### Wet Grinding Ball Mill With Vfd Top Project Machinery
We have wet grinding ball mill with vfd ,Mill Speed is one variable that can often be easily changed with a variable frequency drive (VFD). The starting point for mill speed calculations is the critical speed. Critical speed (CS) is the speed at which the grinding media will centrifuge against the wall of the cylinder. Obviously no milling will occur when the media is pinned against.
### Measurement And Calculation Bible For Cement Plant
All process calculations of pyro section. Engineer Bilal. \$27. Process Measurement Training. Procedure of using portable instruments_ Anemometer, Pitot-Tube, Temperature probes, Radiation Pyrometer and Gas analyzer. Engineer Bilal. \$37. Ball Mill Calculation Training. Process calculations of Ball mill, Separator, Fans etc. | 2,395 | 10,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-05 | longest | en | 0.797585 |
https://www.sscadda.com/reasoning-rrb-ntpc-7th-jan-2020 | 1,620,860,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00398.warc.gz | 1,034,266,933 | 25,565 | # Reasoning Quiz For RRB NTPC : 7th January 2020
Q1. A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series.
WFB, TGD, QHG, ?
(a) NIL
(b) NIK
(c) NLK
(d) NJL
Q2.Identify the diagram that best represents the relationship among classes given below Tennis fans, Cricket players, Students
Q3.Which answer figure will complete the pattern in the question figure?
Q4. Arrange the following words as per order in the dictionary
1. Dyke 2. Dwindle 3. Dwell 4. Dye
(a) 3,2,4,1
(b) 1,3,4,2
(c) 2,1,3,4
(d) 3,4,2,1
Q5. In the following problem, ‘=’ stands for ‘÷’, ‘+’ stands for ‘–’, ‘x’ stands for ‘=’, ‘–’ stands for ‘+’ and ‘÷’ stands for ‘x’. Find the correct equation.
(a) 8 ÷ 4 + 1 5 = 6 x 4
(b) 4 x 6 ÷ 4 + 4 = 7
(c) 5 ÷ 3 – 25 + 20 = 20 x 39
(d) 96 ÷ 2 x 6 ÷ 105 + 1
Q6. How many squares are there in the square figure ABCD?
(a) 16
(b) 17
(c) 26
(d) 30
Q7. A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Q8. X is the husband of Y. W is the daughter of X. Z is the husband of W. N is the daughter of Z. What is the relationship of N to Y?
(a) Cousin
(b) Niece
(c) Daughter
(d) Grand daughter
Q9. Find missing pair of letters from given responses to replace?
(a) HR
(b) HS
(c) HV
(d) HU
Q10. X walked 20 feet from A to B in the East direction. Then X turned to the right and walked 6 feet. Again, X turned to the right and walked 28 feet. How far is X from A?
(a) 28
(b) 9
(c) 10
(d) 27
### Solutions:
S6. Ans.(d)
S7. Ans.(c)
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Are you sure you want to skip this step? | 1,122 | 3,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-21 | latest | en | 0.850897 |
http://www.capitalbudgetingtechniques.com/present-value-of-multiple-cash-flows/ | 1,516,404,708,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888302.37/warc/CC-MAIN-20180119224212-20180120004212-00051.warc.gz | 424,261,570 | 18,254 | # Present Value of Multiple Cash Flows
We come across many cases where we have to determine the present value of series of multiple cash flows. There are two ways we can calculate present value of multiple cash flows. Either we discount back individual cash flow at a time, or we can just calculate the present values individually and add them up.
Example:
Suppose if we want \$10,000 in one year and \$15,000 more in two years. If we can earn 8% on this money, how much we need to invest today to exactly earn this much in the future? In other words, what is the present value of two cash flows at 8%.
Present value of \$15,000 in 2 years at 8 percent is:
\$15000/1.082 =\$12860.082
Present value of \$100 in 1 years at 8% is:
\$10,000/1.08 =\$9259.259
The total present value is :
\$12860.082+\$9259.259=\$22119.341
Present value of multiple cash flowshttps://i1.wp.com/www.capitalbudgetingtechniques.com/wp-content/uploads/2010/07/Present-Value-of-Multiple-Cash-Flows.png?fit=244%2C207https://i1.wp.com/www.capitalbudgetingtechniques.com/wp-content/uploads/2010/07/Present-Value-of-Multiple-Cash-Flows.png?resize=125%2C125NPVCapital budgeting basicsPresent Value of Multiple Cash Flows We come across many cases where we have to determine the present value of series of multiple cash flows. There are two ways we can calculate present value of multiple cash flows. Either we discount back individual cash flow at a time, or we... | 376 | 1,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-05 | latest | en | 0.896413 |
https://www.physicsforums.com/threads/numerical-calculation-about-curve-length.182636/ | 1,519,397,524,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814787.54/warc/CC-MAIN-20180223134825-20180223154825-00224.warc.gz | 924,691,276 | 15,627 | # Numerical calculation about curve length
1. Sep 2, 2007
### xyz3003
I think I have returned all my math back to teachers without any refund.
y=f(x);
h=xb-xa, which is very small.
My Q is to calculate curve length rather than area numerically.
But let me use area as example to show you what i want.
to calculate area between xa to xb, we have 2 ways:
1) area=(f(xa)+f(xb))*h/2; (trapezoid?)
2) area=(f(xa)+4*f(xm)+f(xb))*h/6; here xm=(xa+xb)/2; (parabola?)
As my test, second one is much better than first.
for curve length:
1) len=square root( (f(xb)-f(xa))*(f(xb)-f(xa)) + h*h);
actually, it is distance from (xa, f(xa)) to (xb, f(xb)).
do you know second way to calculate curve length as in area sample above, simple, easy-to-use and better?
any links or explainations are highly appreciated.
thanks.
.
.
2. Sep 2, 2007
### arildno
Well, in regions where the curvature is slowly varying, you might interpolate with circular arc segments instead.
Just a suggestion..
3. Sep 3, 2007
### AlephZero
Start from an integral representing the length of the curve
$$\int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$$
or $$\int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)}\,dt$$ for a curve defined by parametric equations.
Evaluate the integrals with your favorite numerical formula.
4. Sep 5, 2007
### xyz3003
I just want to know the "favorite numerical formula", if which is better than summing line distances from one point to another.
5. Sep 6, 2007
### HallsofIvy
As I said before, in most cases, Simpson's rule is most efficient.
6. Sep 6, 2007
### JonF
I like the trapezoidal rule because it does a good enough job for most things and is much less tedious than simpons. | 536 | 1,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-09 | latest | en | 0.868414 |
https://howmany.cc/convert/convert-185-42-cm-to-inches/ | 1,686,180,640,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00302.warc.gz | 339,037,616 | 15,892 | # Convert 185.42 cm to inches
## How many inches is in a centimeter?
When you are looking to convert 185.42 cm to the equivalent of inches, first, you should be aware of how many inches 1 cm is equal to.
This is how I will be specific: one centimeter is equivalent to 0.3937 inches.
## Facts about centimeter
A centimeter is a common unit of length in the metric system.
It equals to 10 millimeters.
This unit is used in CGS system, maps, home repaire and all areas in our life.
A single centimeter is roughly equivalent to 39.37 inches.
## Inch Definition
The inch is a unit of length in the UK and the US customary systems of measurement. An inch is equal to 1/12 of a foot or 1/36 yard.
## How do you convert 1 cm into inches?
To convert inches from 1cm, multiply 1cm times 0.3937.
This makes it easier for you to calculate 185.42 cm to inches.
Therefore, 1 cm to inches = 1 x 0.3937 = 0.3937 inches.
Based on this, you can answer the following question very lightly and simply.
• What is 1 cm to inches?
• What is cm to inches conversion formula?
• How many inches equals 1 cm?
• What is 1 cm in inches equal?
### What is 185.42 cm converted to inches?
You have fully understood cm to inches by the above.
Here is the exact algorithm:
Value in inches = value in cm × 0.3937
So, 185.42 cm to inches = 185.42 cm × 0.3937 = 72.999854 inches
This formula can be used to answer the related questions:
• What is 185.42 cm in inches?
• How to convert cm to inches?
• How can I change cm into inches?
• How to calculate cm to inches?
• How many inches is 185.42 cm equal to?
cm inches 184.62 cm 72.684894 inches 184.72 cm 72.724264 inches 184.82 cm 72.763634 inches 184.92 cm 72.803004 inches 185.02 cm 72.842374 inches 185.12 cm 72.881744 inches 185.22 cm 72.921114 inches 185.32 cm 72.960484 inches 185.42 cm 72.999854 inches 185.52 cm 73.039224 inches 185.62 cm 73.078594 inches 185.72 cm 73.117964 inches 185.82 cm 73.157334 inches 185.92 cm 73.196704 inches 186.02 cm 73.236074 inches 186.12 cm 73.275444 inches 186.22 cm 73.314814 inches | 623 | 2,061 | {"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} | 3.59375 | 4 | CC-MAIN-2023-23 | latest | en | 0.847007 |
http://bootmath.com/countable-subset-and-monotonic-function.html | 1,534,310,990,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209884.38/warc/CC-MAIN-20180815043905-20180815063905-00546.warc.gz | 61,207,405 | 6,834 | # Countable subset and monotonic function
let E be subset of R which has no isloated points(or C does not have any isolated point of E) and C be countable subset of R does there exist a monotonic function on E which is continuous only at points in E-C?
The problem is from Royden 4th edition page 109.
I know the proof in case E is an open bounded interval only.
#### Solutions Collecting From Web of "Countable subset and monotonic function"
Let
$$L=\left\{x\in C:\exists y_x\in\Bbb R\big(y_x<x\text{ and }(y_x,x)\cap E=\varnothing\big)\right\}\;;$$
this is the set of points in $C$ that are not limits from the left of points in $E$. Another way to say it is that $x\in L$ if and only if $x\in C$ and $x>\sup_{\Bbb R}\{y\in E:y<x\}$. Let $Y=\{y_x:x\in L\}$; $C\cup Y$ is countable, so let
$$C\cup Y=\{x_n:n\in\Bbb N\}\;.$$
(I’m assuming that $C$ is countably infinite; if $C$ is finite, the problem is fairly trivial.) Let
$$f:E\cup Y\to\Bbb R:x\mapsto\sum_{x_n\le x}\frac1{2^n}\;.$$
Finally, let
$$g:E\to\Bbb R:x\mapsto\begin{cases} f(y_{x_n}),&\text{if }x=x_n\in Y\\ f(x),&\text{otherwise}\;. \end{cases}$$
The definition of $f$ ensures that $g$ is discontinuous from the left at every point of $C\setminus L$ and continuous everywhere else, and the modification to get $g$ ensures that $g$ is discontinuous from the right at every point of $L$ without affecting continuity at any other point of $E$. Thus, $g$ is discontinuous precisely at the points of $C$.
Yes, list $E\cap C = \{x_n:n=1,2..\}$. For each $x$ let $N_x=\{n:x_n<x\}$. Define $f(x)=\sum_{n\in N_x}2^{-n}$.
I do not quite see what is the role of $E$ in this question, we could define $f$ as above, initially disregarding $E$ (and using $C = \{x_n:n=1,2..\}$) and then later restricting this function to $E$.
Edit. This may not be “discontinuous enough” at points of $C$, as discussed in comments below. So define also $M_x=\{n:x_n\le x\}$, define $g(x)=\sum_{n\in M_x}2^{-n}$, and define $h(x)=f(x)+g(x)$. I hope $h$ works, if not then I do not understand the question and may need to read all over again.
Edit. $h$ is continuous at points of $E\setminus C$ (verify:). But $h$ (and any function) would be continuous at those points of $C$ that are isolated in $E$, that is at any $c\in C$ which has a neighborhood which misses all other points in $E$. It is ok if $c$ is isolated in $C$,but not in $E$, in that case $h$ would be discontinuous at $c$. Indeed $c=x_m$ for some $m$ (according to the above definition of $f$). So, $2^{-m}$ is not one of the members of the sum that defines $f(c)=f(x_m)=\sum_{n\in N_{x_m}}2^{-n}$ (this is since $m\not\in N_{x_m}$, since in the definition of $N_x$ we have strict inequality, but it is not true that $x_m<x_m$). On the other hand, if $x>x_m$, then $2^{-m}$ is one of the members of the sum that defines $f(x)=\sum_{n\in N_{x}}2^{-n}$, this is because $m\in N_x$, since $x_m<x$. It follows that if $x>x_m$ then $f(x)\ge f(x_m)+2^{-m}$, so $\lim_{x\to x_m^+} f(x)\ge f(x_m)+2^{-m}$, here we are assuming the there are points in $E$ to the right of $x_m$, arbitrarily close to $x_m$ (that is, we are assuming that $x_m$ is not isolated from the right in $E$). So $f$ has a jump at $x_m$ (to the right of $x_m$) so $f$ is discontinuous (from the right) at $x_m$. Similarly, if $x_m$ is not isolated from the left in $E$ then one may show that $\lim_{x\to x_m^-}f(x)\le f(x_m)-2^{-m}$, so $g$ has a jump at $x_m$ (at the left of $x_m$). So, if $x_m$ is not isolated in $E$ then it is either not isolated from the left, or not isolated from the right, or both, so either $g$ of $f$ or both are discontinuous at $x_m$, so $h$ is discontinuous at $x_m$. You should be able to verify all these details yourself, once you know what you are trying to prove. | 1,254 | 3,777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-34 | latest | en | 0.816647 |
https://www.mathinee.com/content/problem/174787 | 1,620,773,525,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00214.warc.gz | 907,638,508 | 1,749 | The user-friendly version of this content is available here.
The following content is copyright (c) 2009-2013 by Goods of the Mind, LLC.
This problem trains for: AMC-10, AMC-12, AIME.
How many non-degenerate triangles have integer length sides, an area of 108 square units and a circle of radius 12 can be inscribed in them?
The area of a triangle is related to the radius of the inscribed circle and to the perimeter of the triangle:
The conditions set by the problem are equivalent to finding all triangles with integer length sides and perimeter equal to:
Therefore, if the lengths of the sides are denoted by a, b and c, then:
Since the minimum length of a side must be 1 (the triangle is non-degenerate) we make the change of variable:
in order to have all integer values start at 0.
Now, the Diophantine equation looks like:
and, by the "stars and bars" method, its number of distinct solutions is: | 212 | 914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-21 | latest | en | 0.925664 |
https://www.codecogs.com/library/maths/approximation/regression/stiefel.php | 1,695,319,284,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00844.warc.gz | 788,893,932 | 9,642 | I have forgotten
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# Stiefel
Approximates a given function using the Stiefel-Remes method.
Controller: CodeCogs
C++
Excel
## Class Stiefel
This algorithm gives the best approximation of a continuous function with real argument in the sense of maximum norm. It uses the Stiefel-Remes method.
An important detail when using this class is that the abscissas array given as argument to the constructor needs to be sorted in ascending order and its elements need to be equally spaced, meaning that:
$X[i]&space;-&space;X[i&space;-&space;1]&space;=&space;\alpha&space;\in&space;\mathcal{R}&space;\mbox{&space;a&space;constant&space;}&space;\qquad&space;1&space;\leq&space;i&space;\leq&space;N&space;-&space;1$
where N is the size of the array. In other words, the associated function needs to be discrete.
Below you will find the regression graphs for a set of points obtained by evaluating the function $\inline&space;&space;f(x)&space;=&space;\sin(2x)&space;/&space;x$, displayed in light blue, at particular abscissas. The Stiefel-Remes polynomial, displayed in red, has been calculated using this class. In the first graph there had been chosen a number of 12 points and degree of polynomial 9, while in the second 36 points were considered with the polynomial degree 16. You may notice the root mean squared error in each of the cases.
### Warning
Choosing too high a degree, numerical rounding errors lead to high-frequency contamination of the results, which become increasingly evident with large values of x. This is visible with the following graph.
## References:
• "Procedures Algol en Analyse Numerique", Editions du CNRS, 1967
### Example 1
#include <codecogs/maths/regression/stiefel.h>
#include <cmath>
#include <iostream>
#include <iomanip>
using namespace std;
#define PI 3.1415
#define N 12
int main()
{
// Declare and initialize two arrays to hold the coordinates of the initial data points
double x[N], y[N];
// Generate the points
double xx = PI, step = 4 * PI / (N - 1);
for (int i = 0; i < N; ++i, xx += step) {
x[i] = xx;
y[i] = sin(2 * xx) / xx;
}
// Initialize the Stiefel-Remes approximation routine with known data points
Maths::Regression::Stiefel A(N, x, y, 7);
// Interrogate Stiefel polynomial to find approximated values
int N_out = 20;
xx = PI, step = (3 * PI) / (N_out - 1);
for (int i = 0; i < N_out; ++i, xx += step)
cout << "x = " << setw(7) << xx << " y = " << setw(13) << A.getValue(xx) << endl;
return 0;
}
Output:
x = 3.1415 y = 0.11495
x = 3.63753 y = 0.154609
x = 4.13355 y = 0.0875854
x = 4.62958 y = 0.00476885
x = 5.12561 y = -0.0516197
x = 5.62163 y = -0.0705339
x = 6.11766 y = -0.058554
x = 6.61368 y = -0.0298593
x = 7.10971 y = 0.000714345
x = 7.60574 y = 0.0219604
x = 8.10176 y = 0.0283645
x = 8.59779 y = 0.0203744
x = 9.09382 y = 0.00330991
x = 9.58984 y = -0.0146569
x = 10.0859 y = -0.0250193
x = 10.5819 y = -0.0215192
x = 11.0779 y = -0.00236588
x = 11.5739 y = 0.028343
x = 12.07 y = 0.0604278
x = 12.566 y = 0.079094
Vince Cole (August 2005)
### Authors
Lucian Bentea (August 2005)
##### Source Code
Source code is available when you buy a Commercial licence.
Not a member, then Register with CodeCogs. Already a Member, then Login.
## Members of Stiefel
#### Stiefel
Stiefel( int n double* x double* y int degree )[constructor]
Initializes the necessary data for following evaluations of the polynomial.
n The number of points x The x-coordinates of the points y The y-coordinates of the points degree The order of the regression
#### GetValue
doublegetValue( double x )
Returns the approximated ordinate at the given abscissa.
x The x-coordinate of the position of interest
## Stiefel Once
doubleStiefel_once( int n double* x double* y int degree double a )
This function implements the Forsythe class for one off calculations, thereby avoid the need to instantiate the Forsythe class yourself.
### Example 2
The following graph performs a regression on the following values, using a 3rd order polynomial:
x = 1 y = 0.22
x = 2 y = 0.04
x = 3 y = -0.13
x = 4 y = -0.17
x = 5 y = -0.04
x = 6 y = 0.09
x = 7 y = 0.11
There is an error with your graph parameters for Stiefel_once with options n=7 x="1 2 3 4 5 6 7" y="0.22 0.04 -0.13 -0.17 -0.04 0.09 0.11" degree=3 a=1:7 .input
Error Message:Function Stiefel_once failed. Ensure that: Invalid C++
### Parameters
n The number of initial points in the arrays x and y x The x-coordinates for the initial points y The y-coordinates for the initial points degree The number of polynomials to use in the approximation a The x-coordinate for the output point
### Returns
the interpolated y-coordinate that corresponds to a.
##### Source Code
Source code is available when you buy a Commercial licence.
Not a member, then Register with CodeCogs. Already a Member, then Login. | 1,561 | 4,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 2, "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} | 3.03125 | 3 | CC-MAIN-2023-40 | latest | en | 0.758448 |
https://apps.apple.com/us/app/gravitational-force-calculator/id1396933116 | 1,723,267,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00553.warc.gz | 73,649,856 | 178,097 | Gravitational Force Calculator 4+
• \$2.99
Description
The Gravitational Force Calculator is a physics/math calculator that calculates the gravitational force between two objects, considering their mass and the distance between them.
Features:
- Instant calculation
- Results are copyable to other apps
- Formulas are included as references
- Supports up to 16 decimal places
- Supports various units for each input
Newton’s Law of Gravity states that every point mass attracts every single other point mass by a force pointing along the line intersecting both points. This Newton’s Law of Gravity Calculator is used to calculate and find the gravitational force between two objects, considering their mass and the distance between them.
Formulas:
1. Gravitational Force:
F = Gm₁m₂/ r²
2. Mass of Object 1:
m₁ = Fr² /Gm₂
3. Mass of Object 2:
m₂ = Fr²/ Gm₁
4. Distance between the Objects:
r = √(Gm₁ m₂/F)
Where,
G = Universal Gravitational Constant = 6.6726 x 10⁻¹¹N-m²/kg²
m₁ = Mass of Object 1
m₂ = Mass of Object 2
r = Distance Between the Objects.
Gravity, or gravitation, is a natural phenomenon by which all things with mass are brought toward (or gravitate toward) one another, including objects ranging from electrons and atoms to planets, stars, and galaxies. Since energy and mass are equivalent, all forms of energy (including photons and light) cause gravitation and are under its influence. On Earth, gravity gives weight to physical objects, and the Moon's gravity causes ocean tides. The gravitational attraction of the original gaseous matter present in the Universe caused it to begin coalescing, forming stars – and for the stars to group together into galaxies – so gravity is responsible for many of the large-scale structures in the Universe. Gravity has an infinite range, although its effects become increasingly weaker on farther objects.
Gravity is most accurately described by the general theory of relativity (proposed by Albert Einstein in 1915), which describes gravity not as a force but as a consequence of the curvature of spacetime caused by the uneven distribution of mass. The most extreme example of this curvature of spacetime is a black hole, from which nothing—not even light—can escape once past the black hole's event horizon. However, for most applications, gravity is well approximated by Newton's law of universal gravitation, which describes gravity as a force that causes any two bodies to be attracted to each other, with the force proportional to the product of their masses and inversely proportional to the square of the distance between them.
Thanks for your support, and do visit nitrio.com for more apps for your iOS devices.
What’s New
Version 1.2
- Updated for the newest devices.
- Minor UI update.
- Minor bugs fixed.
App Privacy
The developer, Nitrio, indicated that the app’s privacy practices may include handling of data as described below. For more information, see the developer’s privacy policy.
Data Not Collected
The developer does not collect any data from this app.
Privacy practices may vary, for example, based on the features you use or your age. Learn More | 687 | 3,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.936604 |
https://stats.stackexchange.com/questions/409392/unusual-markov-inequality-for-normal-distribution | 1,585,653,458,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00175.warc.gz | 722,344,028 | 31,427 | # Unusual Markov inequality for normal distribution
I'm trying to answer the following question from Larry Wassermans book on statistical inference.
My question is how did they arrive at the Markov bound, it does not seem like the normal form of the Markov inequality i.e. $$P(Z > t) \leq \frac{\mathbb{E}(Z)}{t}$$. Also how could we analytically calculate $$\mathbb{E}(|Z|^k)$$.
• Apply the "normal form" of the equality to the random variable $|Z|^k.$ – whuber May 21 '19 at 12:04
First note that the Markov inequality applies to non-negative random variables; you can't apply it directly to $$Z$$.
Let $$Y=|Z|^k$$ for $$k>0$$, which is non-negative. Now apply the Markov inequality to $$Y$$, then relate that back to a probability statement about $$|Z|$$. (Edit: as whuber already noted in comments; not sure how I missed that)
As for evaluating $$E(|Z|^k)$$, there's several ways to approach it. Let's take a fairly straightforward one -- just write the integral, use the symmetry of $$|Z|$$ to write it in terms of an integral on the positive half line, and then make an obvious substitution to get a gamma function. | 295 | 1,125 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-16 | latest | en | 0.912073 |
https://www.coursehero.com/file/135867/p02-046/ | 1,516,470,364,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889677.76/warc/CC-MAIN-20180120162254-20180120182254-00366.warc.gz | 881,325,990 | 22,498 | p02_046
# Fundamentals of Physics,Vol 1 (Chapters 1 - 20)
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Unformatted text preview: 46. We neglect air resistance, which justifies setting a = -g = -9.8 m/s2 (taking down as the -y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to y = 0. (a) With y0 = h and v0 replaced with -v0 , Eq. 2-16 leads to v= (-v0 ) - 2g (y - y0 ) = 2 2 v0 + 2gh . The positive root is taken because the problem asks for the speed (the magnitude of the velocity). (b) We use the quadratic formula to solve Eq. 2-15 for t, with v0 replaced with -v0 , -v0 + 1 y = -v0 t - gt2 = t = 2 (-v0 ) - 2gy g 2 where the positive root is chosen to yield t > 0. With y = 0 and y0 = h, this becomes t= 2 v0 + 2gh - v0 . g (c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a). An important perspective related to this is treated later in the book (in the context of energy conservation) . (d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +v0 in the equation where we had put in -v0 in part (b). The details follow: v0 + 1 y = v0 t - gt2 = t = 2 2 v0 - 2gy g with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain t= 2 v0 + 2gh + v0 . g ...
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http://www.careerride.com/mcq/surds-and-indices-quantitative-aptitude-mcq-questions-18.aspx | 1,493,138,423,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120694.49/warc/CC-MAIN-20170423031200-00472-ip-10-145-167-34.ec2.internal.warc.gz | 480,373,781 | 10,682 | # Surds and Indices - Quantitative Aptitude (MCQ) questions
1) What is value of M in (p/q)2M+2 = (q/p)9-M
- Published on 13 Apr 17
a. 6
b. 5
c. -7/2
d. -11
Answer Explanation ANSWER: -11 Explanation:
2) What is value of 22(-2)
- Published on 12 Apr 17
a. -16
b. -42
c. 42
d. 1/16
Answer Explanation ANSWER: 42 Explanation:
3) What is value of 2(-2)2
- Published on 12 Apr 17
a. 1/16
b. 1/8
c. 32
d. 16
Answer Explanation ANSWER: 16 Explanation:
4) There are four expressions given below separated by ‘=’ sign. Which one of the 4 expressions is not the same as the other three?
62 x 6 = (61)3 = 23+33 = 83 - 142 - 102
- Published on 11 Apr 17
a. (61)3
b. 83 - 142 - 102
c. 23+33
d. All are same
Answer Explanation ANSWER: 23+33 Explanation:
5) Find the value of X.
(15)3.5 * (15)x = 158
- Published on 05 May 16
a. 2.29
b. 2.75
c. 4.25
d. 4.5
Answer Explanation ANSWER: 4.5 Explanation: Let (15)3.5 * (15)x = 158Then, (15)3.5 + x = (15)83.5 + x = 8x = (8 – 3.5)x = 4.5
6) If 5a = 625, then the value of 5(a – 2) is:
- Published on 07 Apr 16
a. 5
b. 25
c. 125
d. 625
Answer Explanation ANSWER: 25 Explanation: 5a = 625 5a = 54 a = 4Therefore,5(a – 2) = 5(4 – 2) = 52 = 25
7) (1000)6/1015 = ?
- Published on 22 Mar 16
a. 10
b. 100
c. 1000
d. 10000
Answer Explanation ANSWER: 1000 Explanation: (1000)6/1015 = (1000)6 / 1015 = (103)6/1015= 10(3*6)/1015 = 1018/1015 = (10)(18 – 15) = 103 = 1000
8) 40 feet rope is cut into 2. One piece is 18 feet longer than the other. What is the length of the shorter piece?
- Published on 12 Jun 15
a. 11
b. 12
c. 18
d. 22
Answer Explanation ANSWER: 11 Explanation: No explanation is available for this question!
9) Akash and Akshay are business partners. Akash invests Rs 35,000 for a period of 8 months and Akshay invests Rs 42,000 for a period of 10 months. Out of a total profit of Rs 31,570 what is Akash’s share?
- Published on 12 Jun 15
a. Rs 12,420
b. Rs 18,040
c. Rs 18,942
d. Rs 12,628
Answer Explanation ANSWER: Rs 12,628 Explanation: For a given business profit is directly dependent upon the capital invested and the time of investment.=> Ratio of shares of Akash and Akshay becomes: (35,000*8)/(42,000*10) = 2/3=>% of profit belonging to Akash: 2/(3+2)*(31,570)=>Rs 12,768
1 | 927 | 2,276 | {"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} | 3.9375 | 4 | CC-MAIN-2017-17 | longest | en | 0.645023 |
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Whodunit Part 2: Places, Places (Posted on 2004-08-16)
The day following the robbery, Bill Bonche was out searching for the five suspects. He decided to go to the city's database on the computer and looked on the city radar. He located the robbed house and saw the five suspects and the streets they returned to (one lives on Dale Avenue). However, it was raining out, and parts of the notes he took were ruined by the water.
With the remaining information, figure out what street the suspects live on, their house number, their phone number, and what color their house is.
1. The suspect who lives on Morgan Street has a 1 in their phone number. The red house has two of the same numbers in its house number.
2. If you add up all of the digits in Chip Circle's house number, the total is three less than the total of the digits in the gray house's number, which is four greater than that of the house whose phone number is 693-9737.
3. If you multiply the last two digits of the yellow house's phone number, the total is equal to the sum of the last two digits in the house with the house number 1036's phone number.
4. The blue house's house number isn't the highest number, but is larger than both the house with the phone number 697-2215, which is the white house, and the house on Monarch Rd.
5. The house on Sandy Lane doesn't have a repeated digit in its house number, has a phone number beginning in 693, and is not the blue house.
6. One paper was completely wiped out except for the lists of house numbers (1036, 2137, 3321, 3778, 7978) and phone numbers (693-1314, 693-9737, 696-7123, 697-2215, 697-4479).
No Solution Yet Submitted by matt runchey Rating: 3.5455 (11 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Whodunit Part 2: Places, Places | Comment 11 of 20 |
<table border="black">
<tr>
<td>Monarch Rd.</td>
<td>3321</td>
<td>693-9737</td>
<td>Red</td>
</tr>
<tr>
<td>Chip Circle</td>
<td>1036</td>
<td>697-2215</td>
<td>White</td>
</tr>
<tr>
<td>Sandy Lane</td>
<td>2137</td>
<td>693-1314</td>
<td>Grey</td>
</tr>
<tr>
<td>Morgan St.</td>
<td>7978</td>
<td>696-7123</td>
<td>Yellow</td>
</tr>
<tr>
<td>Dale Avenue</td>
<td>3778</td>
<td>697-4479</td>
<td>Blue</td>
</tr>
</table>
Posted by Jammy Sehgal on 2004-08-19 16:06:17
Search: Search body:
Forums (0) | 697 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-09 | latest | en | 0.945254 |
https://www.jiskha.com/questions/102266/please-check-my-answer-According-to-JCAHO-standards-the-delinquency-rate-for-incomplete | 1,566,406,298,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316075.15/warc/CC-MAIN-20190821152344-20190821174344-00218.warc.gz | 865,509,905 | 4,772 | # JCAHO
According to JCAHO standards, the delinquency rate for incomplete medical records should be no greater than 20%
I said True
1. 👍 0
2. 👎 0
3. 👁 115
1. What does your text say? I should think it should be MUCH lower than 20%.
1. 👍 0
2. 👎 0
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# CEO, Global Sugars; Setting up a Beet sugar plant in civil
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Joined: 13 May 2010
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CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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04 Jun 2010, 03:10
1
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CEO, Global Sugars; Setting up a Beet sugar plant in civil war-torn Kanjing province is not imprudent, despite the opposition from critics of the strategy. We opened a Beet sugar plant in Yangon presidency, amidst massive left sponsored revolution, ten years ago and that plant has been our trump card generating successful profits ever since.
Which of the following is the author of the argument above most reasonably intending the reader to conclude?
A. Wars are profitable for the author's particular business.
B. Kanjing province is a more stable nation than Yangon presidency.
C. Critics of the proposed plant in Yangon presidency are likely to be biased.
D. The proposed plant in Kanjing province will generate profits, despite war.
E. The proposed plant in Kanjing province will be more successful than in Yangon presidency.
I have marked
[Reveal] Spoiler:
d
If you have any questions
New!
Manager
Joined: 14 Apr 2010
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05 Jun 2010, 19:53
D for me
SVP
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### Show Tags
08 Jun 2010, 13:23
1 more for D.
VP
Joined: 15 Jul 2004
Posts: 1452
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
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10 Jul 2010, 23:17
Whats the OA?
Senior Manager
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10 Jul 2010, 23:31
D
Manager
Joined: 09 Jul 2010
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11 Jul 2010, 05:06
D
_________________
consider cudos if you like my post
Intern
Joined: 14 Jun 2010
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### Show Tags
11 Jul 2010, 05:45
D.
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11 Jul 2010, 09:15
D
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11 Jul 2010, 12:55
D for me as well.
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13 Jul 2010, 05:32
D for me.
i remember a question of similar type from OG.
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02 Aug 2010, 14:23
D .......but OA ?
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02 Aug 2010, 14:23
I have seen this question somewhere remember OA to be D .........
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03 Aug 2010, 06:35
I go for D as well. Hope the OA is D too
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03 Aug 2010, 11:27
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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31 Mar 2012, 18:55
What is the OA
IMO D is d correct answer by Fact test.
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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05 Jun 2012, 03:35
dont think it's a 700 level
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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06 Jun 2012, 21:07
It is definitely not a 700 level question.
Despite facing revolutions in the yongan region, the plant went on to generate profits for the company. Likewise the author is confident that despite war in Kanjing province, the proposed plant would bring in profits.
Clear D
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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24 Jun 2012, 03:48
Straight D. For the author, the plan is not so imprudent so long as it generates profits. It's all about the bottom-line when it comes to business.
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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24 Jun 2012, 21:38
Clear D, as profits is mentioned in the argument and AC also talks about profits..
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink]
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08 Jul 2012, 04:39
CEO, Global Sugars; Setting up a Beet sugar plant in civil war-torn Kanjing province is not imprudent, despite the opposition from critics of the strategy. We opened a Beet sugar plant in Yangon presidency, amidst massive left sponsored revolution, ten years ago and that plant has been our trump card generating successful profits ever since.
Which of the following is the author of the argument above most reasonably intending the reader to conclude?
This is a "Main Point" question. The correct answer choice should encapsulate the author's main point.
"Main Point" questions are a subset of "Must be True" questions
In Main point questions, even if an answer choice is "Must be True" according to the stimulus - if it fails to capture the main point, it cannot be correct.
A. Wars are profitable for the author's particular business.
The main point here is about the plant in Kanjing province.
B. Kanjing province is a more stable nation than Yangon presidency.
This is no where stated in the stimulus
C. Critics of the proposed plant in Yangon presidency are likely to be biased.
This is no where stated in the stimulus. In the stimulus, the critics have just opposed the strategy. We ave no information whether they are biasedor not
D. The proposed plant in Kanjing province will generate profits, despite war.
This is what the CEO of Global Sugars finally wants to convey , giving an analogy with the plant in Yangon presidency.
E. The proposed plant in Kanjing province will be more successful than in Yangon presidency.
This is no where stated in the stimulus
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Re: CEO, Global Sugars; Setting up a Beet sugar plant in civil [#permalink] 08 Jul 2012, 04:39
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Display posts from previous: Sort by | 2,627 | 8,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-22 | latest | en | 0.907546 |
https://geniebook.com/us/tuition/secondary-3/physics/kinematics-distance-displacement | 1,716,196,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00690.warc.gz | 231,951,007 | 67,169 | # Kinematics - Distance & Displacement
• Distance & Displacement
• Speed & Velocity
• Acceleration
## Scalars and Vectors
Scalars and vectors are two types of physical quantities.
Scalar Quantities Vector Quantities
Have only magnitude Have both magnitude and direction
## Distance & Displacement
Distance is the total length covered by a moving object along the path taken. The direction of motion does not matter.
Displacement is the straight-line distance covered by a moving object regardless of the path taken. It takes into account only the starting and the ending points.
The direction of the motion, relative to a reference point, needs to be specified. It mainly needs the starting point to be specified.
Let us consider two points A and B. The car travels the distance as shown in the figure below.
The distance travelled by the car is 40 km.
The displacement of the car is 19 km to the right of A.
Let us consider another example.
We have a car that moves from A to B and then from B to A.
The distance travelled is 38 km.
The displacement of the car is 0 km.
The displacement is a straight-line distance between the starting and ending point. The starting point and the ending point are the same, so the displacement of the car is zero. Since the displacement is zero, the direction is not required.
Let us consider another scenario.
There are three points A, B and C. First, the car moves from point A to B and then it returns from B to C.
The distance travelled is 31 km.
The displacement of the car is 5 km to the left of A.
### Positive & Negative Displacement
For an object moving in a straight line, we can assign one direction from the reference point as the positive direction.
The opposite direction will be the negative direction.
There are no fixed rules for assigning directions. Positive and negative directions are assigned purely for convenience.
Let us consider the following example to understand this better.
We have three points A, B and C. The car starts from Point A and reaches Point B. Then it changes direction and moves from Point B to Point C.
Taking right as the positive direction,
Displacement of the car is – 5 km.
The displacement is negative because the car is moving in the left direction. If the car were moving in the right direction, then it would have been positive as we have considered right as the positive direction.
Let us consider the following:
Taking left as the positive direction,
Then the displacement of the car will be 5 km.
## Practice Questions
Question 1:
A boy walks 35 m to the east and then 13 m to the west. What is the distance travelled by him and the magnitude of his displacement?
Solution:
Distance Travelled
= 35 m + 13 m
= 48 m
Displacement
= 35 m − 13 m
= 22 m east
Since the question asked only for the magnitude of the displacement, it is 22 m. We need not mention the direction of the displacement.
Question 2:
A girl runs 60 m to the west and then 80 m to the north. What is the distance travelled by her and the magnitude of her displacement?
Solution:
Distance Travelled
= 60 m + 80 m
= 140 m
Displacement
$= \sqrt {60² + 80²}$
$= \sqrt {3600 + 6400}$
$= \sqrt {10000}$
= 100 m NW direction
Since the question asked only for the magnitude of the displacement, we write 100 m.
## Speed
Speed is the distance moved per unit of time.
$\small \displaystyle{\textsf{Speed} = \frac {\textsf{Distance Travelled} }{\textsf{Time Taken}}}$
It is a scalar quantity.
SI unit of speed is metre per second (m s⁻¹)
Another commonly used unit of speed is kilometre per hour (km h⁻¹) but it is not an SI unit.
Question 3:
A woman must reach the airport which is 91 km away from her house in two hours. She has a tea break for half an hour and spends one-fifth of an hour stationary in a traffic jam. At what average speed must she travel during the rest of the time if she wants to complete the journey in two hours?
Solution:
Time left to travel
= 2 h - 0.5 h - 0.2 h
= 1.3 h
Average speed
=
$\displaystyle{\frac {91 km}{1.3 h}}$
=
70 km h⁻¹
## Velocity
Velocity is the rate of change of displacement.
$\displaystyle{\textsf{Velocity} = \frac{\textsf{Displacement}}{\textsf{Time Taken}}}$
It is a vector quantity.
SI unit of velocity is metre per second (m s⁻¹)
Velocity has both magnitude and direction.
If the direction of travel has changed, velocity will change even if the speed remains the same.
We can see from the above figure, even though the speed of the car has not changed, the direction of the car has changed. Thus, velocity has changed.
## Speed & Velocity
If the whole journey took 2 hours, then the average speed is:
Average speed = $\displaystyle{\frac {40}{2}}$
= 20 km h⁻¹
If the journey took 2 hours, and taking right as positive, what is the velocity of the car?
Average velocity = $\frac {19}{2}$
= 9.5 km h⁻¹
(Since it is already mentioned that right is positive, we are not mentioning the direction of the velocity.)
Let us see another example.
If the journey took 2 hours, what is the average speed and average velocity?
Average Speed = $\frac {19\;+\;19}{2}$
= 19 km h⁻¹
Average Velocity = $\frac {19\;-\;19}{2}$
= 0 km h⁻¹
( It is zero because displacement is zero. The starting and the ending point are the same.)
Question 4:
Sally walks 10 m to the south in 4 s followed by another 12 m to the east in 6 s. What is her average speed and the magnitude of her velocity?
Solution:
Option 1 is the correct answer
Average Speed = $\frac {10\;+\;12}{4\;+\;6}$
= 2.2 m s⁻¹
Taking East and South as the positive direction,
Displacement = $\sqrt {{10^2} + {12^2}}$
Displacement = 15.620 m
Velocity = $\frac {15.620}{4\;+\;6}$
= 1.56 m s⁻¹
Question 5:
John runs a circular path of radius 60 m. He completes one round in 2 minutes. Find the distance travelled by him and his average velocity.
Solution:
Option 2 is the correct answer
Distance ran = $2\pi r$
= $2 \times π \times 60$
= 377 m
Displacement = 0 m since he returned to his original position.
Average Velocity = 0 m s⁻¹
## Acceleration
Acceleration is the rate of change of velocity. In other words, it means how fast the velocity is changing.
$Acceleration = \frac {Change \;In \;Velocity} {Time \;Taken}$
It is a vector quantity.
SI unit of acceleration is metre per second per second (m s⁻²)
A body accelerates when its velocity changes.
Since the magnitude of the velocity has increased, the car has undergone acceleration.
Let us see another example.
We see in the above figure that even though the magnitude of velocity has not changed, the direction of the velocity has changed. Hence, the velocity has changed.
Since the car has undergone velocity change, the car is said to have accelerated.
This acceleration is dealing with the change in direction and not the magnitude. This is called directional acceleration.
Let us see another scenario.
The car in the above figure is said to be accelerated since both the speed and the direction have changed.
Thus, we see that acceleration can come in two forms -
• Change in speed
• Change in direction
### Acceleration
Acceleration, when it is uniform, can be calculated using the formula below.
$a = \frac {v\;-\;u}{∆t}$
Where,
$a$ = acceleration;
$v$ = final velocity;
$u$ = initial velocity;
$∆t$ = time interval.
For the O-level syllabus, the calculation of acceleration only deals with the change in the magnitude of the velocity and not the change in the direction of the velocity.
Question 6:
A train decelerates from 50 m s⁻¹ to 25 m s⁻¹ in 5 s. What is its acceleration?
Solution:
$a = \frac {v\;-\;u}{∆t}$
Here,
$v$ = 25 m s⁻¹
$u$ = 50 m s⁻¹
$∆t$ = 5 s
So,
$a$ = $\frac {25 \;-\;50}{5}$
= − 5 m s⁻²
What is the deceleration?
Deceleration = 5 m s⁻²
## Conclusion
In this article, we have covered the following topics as per the Secondary 3 Physics syllabus.
• Distance & Displacement
• Speed & Velocity
• Acceleration
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Let’s refresh the page! | 2,237 | 8,960 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.912549 |
https://artofproblemsolving.com/wiki/index.php/Arrangement_Restriction_Theorem | 1,669,838,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00199.warc.gz | 139,608,243 | 9,861 | # Arrangement Restriction Theorem
The Arrangement Restriction Theorem is discovered by aops-g5-gethsemanea2 and is not an alternative to the Georgeooga-Harryooga Theorem because in this theorem the only situation that is not allowed is that all $k$ objects are together.
## Definition
If there are $n$ objects to be arranged and $k$ of them should not be beside each other altogether, then the number of ways to arrange them is $n! - (n - k + 1)!k!$.
## Proof/Derivation
If there are no restrictions, then we have $n!$. But, if we put $k$ objects beside each other, we have $(n-k+1)!k!$ because we can count the $k$ objects as one object and just rearrange them.
So, by complementary counting, we get $n! - (n - k + 1)!k!$.
## Problem
Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl cannot be altogether in the line.
With these conditions, how many different ways can you arrange these kids in a line?
Problem by Math4Life2020, edited by aops-g5-gethsemanea2
### Solution
By the Arrangement Restriction Theorem, we get $\frac{8! - (8 - 3 + 1)!3!}{2} = \boxed{18000}$ because Fred and George are indistinguishable.
Solution by aops-g5-gethsemanea2
## Testimonials
I like this theorem, but not as much as the Georgeooga-Harryooga Theorem or the Wooga Looga Theorem ~ ilp
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# What is the radius of 4x^2 + 8y^2 + 14x + 16y + 25 = 0
Student
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What is the radius of 4x^2 + 8y^2 + 14x + 16y + 25 = 0
Posted by greyapple on September 1, 2012 at 6:05 PM via web and tagged with circle, math, radius
College Teacher
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The equation 4x^2 + 8y^2 + 14x + 16y + 25 = 0 is not that of a circle. In the equation of a circle, the coefficient of x^2 and the coefficient of y^2 have to be equal. 4x^2 + 8y^2 + 14x + 16y + 25 = 0 is the equation of an ellipse and it is not possible to determine the radius.
As 4x^2 + 8y^2 + 14x + 16y + 25 = 0 does not represent a circle its radius cannot be determined.
Posted by justaguide on September 1, 2012 at 6:08 PM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students. | 319 | 905 | {"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} | 3.671875 | 4 | CC-MAIN-2013-48 | latest | en | 0.906058 |
https://www.unitconverters.net/heat-flux-density/horsepower-metric-square-foot-to-btu-it-second-square-foot.htm | 1,660,230,787,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00469.warc.gz | 915,985,006 | 3,145 | Home / Heat Flux Density Conversion / Convert Horsepower (metric)/square Foot to Btu (IT)/second/square Foot
# Convert Horsepower (metric)/square Foot to Btu (IT)/second/square Foot
Please provide values below to convert horsepower (metric)/square foot to Btu (IT)/second/square foot, or vice versa.
From: horsepower (metric)/square foot To: Btu (IT)/second/square foot
### Horsepower (metric)/square Foot to Btu (IT)/second/square Foot Conversion Table
Horsepower (metric)/square FootBtu (IT)/second/square Foot
0.01 horsepower (metric)/square foot0.0069711831 Btu (IT)/second/square foot
0.1 horsepower (metric)/square foot0.0697118307 Btu (IT)/second/square foot
1 horsepower (metric)/square foot0.6971183072 Btu (IT)/second/square foot
2 horsepower (metric)/square foot1.3942366144 Btu (IT)/second/square foot
3 horsepower (metric)/square foot2.0913549216 Btu (IT)/second/square foot
5 horsepower (metric)/square foot3.485591536 Btu (IT)/second/square foot
10 horsepower (metric)/square foot6.971183072 Btu (IT)/second/square foot
20 horsepower (metric)/square foot13.9423661441 Btu (IT)/second/square foot
50 horsepower (metric)/square foot34.8559153602 Btu (IT)/second/square foot
100 horsepower (metric)/square foot69.7118307204 Btu (IT)/second/square foot
1000 horsepower (metric)/square foot697.1183072042 Btu (IT)/second/square foot
### How to Convert Horsepower (metric)/square Foot to Btu (IT)/second/square Foot
1 horsepower (metric)/square foot = 0.6971183072 Btu (IT)/second/square foot
1 Btu (IT)/second/square foot = 1.4344767447 horsepower (metric)/square foot
Example: convert 15 horsepower (metric)/square foot to Btu (IT)/second/square foot:
15 horsepower (metric)/square foot = 15 × 0.6971183072 Btu (IT)/second/square foot = 10.4567746081 Btu (IT)/second/square foot | 527 | 1,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-33 | latest | en | 0.587465 |
http://www.nixtu.info/2011/09/infinite-lists-in-python.html | 1,722,728,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00524.warc.gz | 48,942,524 | 11,011 | I wrote books about Webpack and React. Check them out!
## Saturday, September 24, 2011
### Infinite Lists in Python
In my previous post I discussed infinite lists in Haskell. As it happens the concept may be useful beyond it. As we saw before, using them may yield nice and concise definitions for series. In this post I'm going to show you one way to port the concept to Python.
Initial Port
In Haskell we define naive infinite lists using [1,2..] kind of syntax. I decided to use a class based syntax to mimic this. This particular snippet would be written as InfList(1, 2) in this scheme. The same works for negative numbers as well. You could write InfList(-1, -2) and expect that to work as well.
So what does "InfList(1, 2)" actually do? Consider the following example showing some initial tests of mine:
As you can see once you instantiate the object, you'll have access to its content just like in the case of a regular list. It calculates the values needed on demand. In other words it's lazy.
The example above also showcases the use of lambdas (fibo) and Python's slice syntax. The implementation infers the rule used to generate the series by default. If you want, you can provide a custom lambda that can be used instead. The slice syntax allows you to mimic Haskell's "take".
Besides the aforementioned functionality I ended up implementing __unicode__ just to make it easier to see the first items of the list.
I've listed the actual implementation below. Overall it ended up being surprisingly simple.
Extra Functions
Just having a way to access list items isn't enough. In my previous post I used "zip" and "concat" to combine multiple infinite lists. It's also handy to be able to map and filter the lists. In order to do this I ended up implementing little wrappers for these operations. I've listed the implementation and some tests below.
As you can see there's some overhead. I do think these kind of functions yield some extra power to the concept overall. There are many ways to combine these.
Conclusion
I hope this post gave you some idea on how to implement infinite lists in your favorite programming language. I've no doubt it would be pretty easy to port the concept to say JavaScript, PHP or Ruby. I think the main advantage of the concept is the concise syntax it allows. Pythonistas might want to check out the itertools module for more powerful tools such as this. | 520 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.935345 |
http://www.mathcasts.org/mtwiki/GlossaryT/Glossary | 1,582,956,201,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148671.99/warc/CC-MAIN-20200229053151-20200229083151-00538.warc.gz | 210,221,004 | 8,406 | # Glossary in Brief
absolute value. A number's distance from zero on the number line. The absolute value of -4 is 4; the absolute value of 4 is 4.
algorithm. An organized procedure for performing a given type of calculation or solving a given type of problem. An example is long division.
arithmetic sequence. A sequence of elements, a_1,\,a_2,\,a_3,\,... , such that the difference of successive terms is a constant a_{i+1} - a_{i} = k ; for example, the sequence {2, 5, 8, 11, 14, . . .} where the common difference is 3.
asymptotes. Straight lines that have the property of becoming and staying arbitrarily close to the curve as the distance from the origin increases to infinity. For example, the x- axis is the only asymptote to the graph of \frac{sin(x)}{x} .
axiom. A basic assumption about a mathematical system from which theorems can be deduced. For example, the system could be the points and lines in the plane. Then an axiom would be that given any two distinct points in the plane, there is a unique line through them.
binomial. In algebra, an expression consisting of the sum or difference of two monomials (see the definition of monomial), such as 4a-8b.
binomial distribution. In probability, a binomial distribution gives the probabilities of k outcomes A (or n-k outcomes B) in n independent trials for a two-outcome experiment in which the possible outcomes are denoted A and B.
binomial theorem. In mathematics, a theorem that specifies the complete expansion of a binomial raised to any positive integer power.
box-and-whisker plot. A graphical method for showing the median, quartiles, and extremes of data. A box plot shows where the data are spread out and where they are concentrated.
complex numbers. Numbers that have the form a+bi where a and b are real numbers and i satisfies the equation i^2=-1 . Multiplication is denoted by ( a + bi )( c + di ) = ( ac - bd ) + ( ad + bc ) i , and addition is denoted by ( a + bi ) + ( c + di ) = ( a + c ) + ( b + d ) i .
congruent. Two shapes in the plane or in space are congruent if there is a rigid motion that identifies one with the other (see the definition of rigid motion).
conjecture. An educated guess.
coordinate system. A rule of correspondence by which two or more quantities locate points unambiguously and which satisfies the further property that points unambiguously determine the quantities; for example, the usual Cartesian coordinates x, y in the plane.
cosine. cos(\theta) is the x- coordinate of the point on the unit circle so that the ray connecting the point with the origin makes an angle of θ with the positive x- axis. When θ is an angle of a right triangle, then cos(θ) is the ratio of the adjacent side with the hypotenuse.
dilation. In geometry, a transformation D of the plane or space is a dilation at a point P if it takes P to itself, preserves angles, multiplies distances from P by a positive real number r, and takes every ray through P onto itself. In case P is the origin for a Cartesian coordinate system in the plane, then the dilation D maps the point ( x, y ) to the point ( rx, ry ).
dimensional analysis. A method of manipulating unit measures algebraically to determine the proper units for a quantity computed algebraically. For example, velocity has units of the form length over time (e.g., meters per second [m/sec] ), and acceleration has units of velocity over time; so it follows that acceleration has units ( [(m/sec)/sec] = [m/sec^2] ).
expanded form. The expanded form of an algebraic expression is the equivalent expression without parentheses. For example, the expanded form of ( a + b )2 is (a+b)^2=a^2+2ab+b^2 .
exponent. The power to which a number or variable is raised (the exponent may be any real number).
exponential function. A function commonly used to study growth and decay. It has the form y = ax with a positive.
factors. Any of two or more quantities that are multiplied together. In the expression 3.712 x 11.315, the factors are 3.712 and 11.315.
function. A correspondence in which values of one variable determine the values of another.
geometric sequence. A sequence in which there is a common ratio between successive terms. Each successive term of a geometric sequence is found by multiplying the preceding term by the common ratio. For example, in the sequence {1, 3, 9, 27, 81, . . .} the common ratio is 3.
histogram. A vertical block graph with no spaces between the blocks. It is used to represent frequency data in statistics.
inequality. A relationship between two quantities indicating that one is strictly less than or less than or equal to the other.
integers. The set consisting of the positive and negative whole numbers and zero; for example, {. . . -2, -1, 0, 1, 2 . . .}.
irrational number. A number that cannot be represented as an exact ratio of two integers. For example, the square root of 2 or p .
linear expression. An expression of the form ax+b where x is variable and a and b are constants; or in more variables, an expression of the form ax + by + c, ax + by + cz + d, etc.
linear equation. An equation containing linear expressions.
logarithm. The inverse of exponentiation; for example, a^{log_ax}=x .
mean. In statistics, the average obtained by dividing the sum of two or more quantities by the number of these quantities.
median. In statistics, the quantity designating the middle value in a set of numbers.
mode. In statistics, the value that occurs most frequently in a given series of numbers.
monomial. In the variables x, y, z, a monomial is an expression of the form axm yn zk , in which m, n, and k are nonnegative integers and a is a constant (e.g., 5x2 , 3x2 y or 7x3yz2 ).
nonstandard unit. Unit of measurement expressed in terms of objects (such as paper clips, sticks of gum, shoes, etc.).
parallel. Given distinct lines in the plane that are infinite in both directions, the lines are parallel if they never meet. Two distinct lines in the coordinate plane are parallel if and only if they have the same slope.
permutation. A permutation of the set of numbers {1, 2, . . . , n } is a reordering of these numbers.
polar coordinates. The coordinate system for the plane based on r , θ , the distance from the origin and θ , and the angle between the positive x- axis and the ray from the origin to the point.
polar equation. Any relation between the polar coordinates (r, θ ) of a set of points (e.g., r = 2cosθ is the polar equation of a circle).
polynomial. In algebra, a sum of monomials; for example, x^2 + 2xy + y^2 .
prime. A natural number p greater than 1 is prime if and only if the only positive integer factors of p are 1 and p. The first seven primes are 2, 3, 5, 7, 11, 13, 17.
quadratic function. A function given by a polynomial of degree 2.
random variable. A function on a probability space.
range. In statistics, the difference between the greatest and smallest values in a data set. In mathematics, the image of a function.
ratio. A comparison expressed as a fraction. For example, there is a ratio of three boys to two girls in a class (3/2, 3:2).
rational numbers. Numbers that can be expressed as the quotient of two integers; for example, 7/3, 5/11, -5/13, 7 = 7/1.
real numbers. All rational and irrational numbers.
reflection. The reflection through a line in the plane or a plane in space is the transformation that takes each point in the plane to its mirror image with respect to the line or its mirror image with respect to the plane in space. It produces a mirror image of a geometric figure.
rigid motion. A transformation of the plane or space, which preserves distance and angles.
root extraction. Finding a number that can be used as a factor a given number of times to produce the original number; for example, the fifth root of 32 = 2 because 2 x 2 x 2 x 2 x 2 = 32).
rotation. A rotation in the plane through an angle θ and about a point P is a rigid motion T fixing P so that if Q is distinct from P, then the angle between the lines PQ and PT(Q) is always θ . A rotation through an angle θ in space is a rigid motion T fixing the points of a line l so that it is a rotation through θ in the plane perpendicular to l through some point on l.
scalar matrix. A matrix whose diagonal elements are all equal while the non diagonal elements are all 0. The identity matrix is an example.
scatterplot. A graph of the points representing a collection of data.
scientific notation. A shorthand way of writing very large or very small numbers. A number expressed in scientific notation is expressed as a decimal number between 1 and 10 multiplied by a power of 10 (e.g., 7000 = 7 x 103 or 0.0000019 = 1.9 x 10-6 ).
similarity. In geometry, two shapes R and S are similar if there is a dilation D (see the definition of dilation) that takes S to a shape congruent to R. It follows that R and S are similar if they are congruent after one of them is expanded or shrunk.
sine. sin(\theta) is the y- coordinate of the point on the unit circle so that the ray connecting the point with the origin makes an angle of θ with the positive x- axis. When θ is an angle of a right triangle, then sin(θ) is the ratio of the opposite side with the hypotenuse.
square root. The square roots of n are all the numbers m so that m2 = n. The square roots of 16 are 4 and -4. The square roots of -16 are 4 i and -4 i .
standard deviation. A statistic that measures the dispersion of a sample.
symmetry. A symmetry of a shape S in the plane or space is a rigid motion T that takes S onto itself (T(S) = S). For example, reflection through a diagonal and a rotation through a right angle about the center are both symmetries of the square.
system of linear equations. Set of equations of the first degree (e.g., x + y = 7 and x - y = 1 ). A solution of a set of linear equations is a set of numbers a, b, c, . . . so that when the variables are replaced by the numbers all the equations are satisfied. For example, in the equations above, x = 4 and y = 3 is a solution.
translation. A rigid motion of the plane or space of the form X goes to X + V for a fixed vector V.
transversal. In geometry, given two or more lines in the plane a transversal is a line distinct from the original lines and intersects each of the given lines in a single point.
unit fraction. A fraction whose numerator is 1 (e.g., 1/ p , 1/3, 1/x). Every nonzero number may be written as a unit fraction since, for n not equal to 0, n = 1/(1/ n ).
variable. A placeholder in algebraic expressions; for example, in 3x + y = 23, x and y are variables.
vector. Quantity that has magnitude (length) and direction. It may be represented as a directed line segment.
zeros of a function. The points at which the value of a function is zero. | 2,618 | 10,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-10 | latest | en | 0.909766 |
http://mathhelpforum.com/differential-equations/129901-separation-variables-question.html | 1,524,773,023,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948464.28/warc/CC-MAIN-20180426183626-20180426203626-00572.warc.gz | 196,825,394 | 9,394 | Thread: Separation of Variables question!
1. Separation of Variables question!
v (dv/dx) = - (1/250) vē - g
solve the equation for v as a function of x
this is for upward motion using quadratic model of air resistance
thanks
2. Originally Posted by acu04385
v (dv/dx) = - (1/250) vē - g
solve the equation for v as a function of x
this is for upward motion using quadratic model of air resistance
thanks
Is $\displaystyle g$ a constant?
If so...
$\displaystyle v\,\frac{dv}{dx} = -\frac{1}{250}v^2 - g$
$\displaystyle \frac{dv}{dx} = -\frac{1}{250}v - gv^{-1}$
$\displaystyle \frac{dv}{dx} = -\frac{v}{250} - \frac{g}{v}$
$\displaystyle \frac{dv}{dx} = \frac{-v^2 - 250g}{250v}$
$\displaystyle \frac{dv}{dx} = -\frac{v^2 + 250g}{250v}$
$\displaystyle \frac{dx}{dv} = -\frac{250v}{v^2 + 250g}$
$\displaystyle \frac{dx}{dv} = -250v(v^2 + 250g)^{-1}$.
You can now solve this with a $\displaystyle u$ substitution. | 326 | 927 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-17 | latest | en | 0.646953 |
https://jp.mathworks.com/matlabcentral/fileexchange/34639-decimatepoly?s_tid=prof_contriblnk | 1,601,165,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00662.warc.gz | 462,882,515 | 20,830 | File Exchange
## DecimatePoly
version 1.2.1.0 (30.2 KB) by
Simplify a 2D closed contour/polygon to within user-defined boundary offset tolerance
Updated 04 Sep 2020
Editor's Note: This file was selected as MATLAB Central Pick of the Week
Have you ever worked with 2D boundary shape representations and noticed that the same shape can be represented with reasonable accuracy by a much smaller number of points? Did you ever need to reduce the complexity of a 2D shape while retaining as many salient features as possible? If so then DecimatePoly is the function you have been looking for.
In the attached .ZIP folder you will find the primary, self-contained function titled DecimatePoly. Additionally, in the Demos folder you will find three files demonstrating the capability and utility of this function.
DecimatePoly_demo1: Simplify a 2D contour extracted from a binary image. Three binary images are included as examples. This demo requires Image Processing Toolbox to run.
DecimatePoly_demo2: Simplifying complex shapes while retaining geometrically and visually salient features.
DecimatePoly_demo3: Use `DecimatePoly.m` to improve the run-time of in-polygon tests at the cost of minor misclassification errors.
### Cite As
Anton Semechko (2020). DecimatePoly (https://github.com/AntonSemechko/DecimatePoly), GitHub. Retrieved .
Erik Husby
A Python implementation of this function may be found here:
https://github.com/husby036/DecimatePoly
Besides the small issue I've mentioned in my previous comments (which has been corrected in the translation), this is a direct translation from MATLAB to Python.
Erik Husby
EDIT to my previous comment:
"Here, `t` is an N-by-1 array. `t<0` is only true when all elements of `t` are less than *zero*, so negative `t` values may pass. Similarly, `t` values greater than one may pass."
Erik Husby
On lines 79 and 80 of DecimatePoly.m, I believe the following statements don't work as intended:
if t<0, t(t<0)=0; end %#ok<*BDSCI>
if t>1, t(t>1)=1; end
Here, `t` is an N-by-1 array. `t<0` is only true when all elements of `t` are less than one, so negative `t` values may pass. Similarly, `t` values greater than one may pass.
From the context of the two statements and the fact that we're calculating distances, I believe the author means to "clip" the values of `t` to the range (0, 1). If this is true, the conditionals should be taken out. Those two lines would be replaced as:
t(t<0)=0;
t(t>1)=1;
If I'm correct that this is an error, lines 175 and 176 display the same flawed logic. However, `t` is a 1-by-2 array within the `RecomputeErrors` function, so the statements are not erroneous but only highly redundant.
Very useful function. I would enjoy an option on deciding if you could actually save output statistics on variables instead of directly printing them (useful when I use it inside other routines).
Mattias Karlsson
Great function. It's nice that it is self contained too.
Catarina
Hi Anton,
Your code was all I needed to solve problems in my codes due to too many useless points. Thanks a lot!
Andrii
Anton Semechko
http://www.insight-journal.org/browse/publication/829
Andrii
It is for building extraction from aerial images. After segmentation process boundaries are zigzag-shaped with many intrusions. I would like to simplify polygons but to maintain their essential shape. It means buildings should have orthogonal angles and small extrusions should be cut off. So I am looking for some implementation algorithm in MatLab. Like this: http://img812.imageshack.us/img812/3284/exampleus.png
Anton Semechko
@ Andrii, what is it you want to achieve by orthogonalizing a set of input points? Could you clarify what this operation means or perhaps provide a reference?
Andrii
Hi Anton. I have tried your function.
It works very well.
which can simplify and after orthogonalize input set of points?
6 May 2019 1.2.1.0 - migrated to GitHub 11 Feb 2018 1.2.0.0 Modified code according to suggestions made by Georgios Gkantzounis and Erik Husby 23 Jan 2012 1.1.0.0 updated description
##### MATLAB Release Compatibility
Created with R2015a
Compatible with any release
##### Platform Compatibility
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var a, b, c, d, e, f, g, h, i, j : real;x : real;Sgm, StandarDeviasi : Real;Kc : Real;kb : Real;begina := StrToFloat(Edit1.Text);b := StrToFloat(Edit2.Text);c := StrToFloat(Edit3.Text);d := StrToFloat(Edit4.Text);e := StrToFloat(Edit5.Text);f := StrToFloat(Edit6.Text);g := StrToFloat(Edit7.Text);h := StrToFloat(Edit8.Text);i := StrToFloat(Edit9.Text); j := StrToFloat(Edit10.Text);//menghitung nilai rata-ratax := (a+b+c+d+e+f+g+h+i+j)/10;Label11.Caption := 'Nilai Rata-Ratanya Adalah';Edit11.Text := FloatToStr(x);//menghitung standar deviasiSgm := 0;Sgm := sgm+sqr(a-x)+sqr(b-x)+sqr(c-x)+sqr(d-x)+sqr(e-x)+sqr(f-x)+sqr(g-x)+sqr(h-x)+sqr(i-x)+sqr(j-x);StandarDeviasi := sqrt(sgm/9);Label12.Caption := 'StandarDeviasinya Adalah';Edit12.Text :=FloatToStr(StandarDeviasi);//menghitung nilai terkecilkc := a;if b < kc then kc := belse kc := kc;if c < kc then kc := celse kc := kc;if d < kc then kc := delse kc := kc;if e < kc then kc := eelse kc := kc;if f < kc then kc := f else kc := kc;if g < kc then kc := gelse kc := kc;if h < kc then kc := helse kc := kc;if i < kc then kc := ielse kc := kc;if j < kc then kc := jelsekc := kc;Label13.Caption := 'Nilai TerkecilAdalah';
Edit13.Text := FloatToStr(Kc);//menghitung nilai terbesarkb := a;if b > kb then kb := belse kb := kb;if c > kb then kb := celse kb := kb;if d > kb then kb := delse kb := kb;if e > kb then kb := eelse kb := kb ;if f > kb then kb := f else kb := kb;if g >kb then kb := gelse kb := kb;if h > kb then kb := helse kb := kb;if i > kb then kb := ielse kb := kb;if j > kb then kb := jelse kb := kb;Label14.Caption := 'Nilai Terbesarnya Adalah';Edit14.Text := FloatToStr(kb);form1.color:=cllime;end;procedure TForm1.Button2Click(Sender: TObject);beginEdit1.Clear;Edit2.Clear; SAMAPAI BERAPA YGDIBUTUHKANLabel11.Caption := '';Label12.Caption := '';Label13.Caption := '';Label14.Caption := '';form1.color:=clyellow;
LINGKARAN
var r : real;Luas, Keliling, Volume : real;beginr := StrToFloat(Edit1.Text);Keliling := 2*pi*r;// Luas := pi*r*rLuas := pi* sqr(r);// Volume := (4 * pi * r * r * r)/3;Volume := (4 * pi * sqr(r) * r)/3;Edit2.Text := FloatToStr(Keliling);Edit3.Text := FloatToStr(Luas);Edit4.Text := FloatToStr(Volume);// Label2.Font := "Palatino Linotype";// Label2.Font.Style := 'Bold';
Label2.Font.Size := 12;Label2.Font.Color := clRed;// Edit2.Font.Size := 12;
DISKON BELANJA
var Belanja, Diskon : Real;NH : Char;beginBelanja := StrToFloat(Edit1.Text);if Belanja < 100000 thenBeginDiskon := 0.1*Belanja;Belanja := Belanja - Diskon;endelseBeginDiskon := 0.2*Belanja;Belanja := Belanja - Diskon;end;Label2.Caption := 'Anda cukupmembayar ';Edit2.Text := FloatToStr(Belanja);Edit1.SetFocus;
HARI DALAM BULAN
var Kode_Bulan : integer;beginKode_Bulan := StrToInt(Edit1.Text);case Kode_Bulan of 1,3,5,7,8,10,12 : Label2.Caption :='Jumlah hari adalah 31 hari';4,6,9,11 : Label2.Caption :='Jumlah hari adalah 30 hari';2 : Label2.Caption :='Jumlah hari adalah 28 atau 29 hari';elseLabel2.Caption := 'Andamemasukkan kode bulan yangsalah...!';end;// Edit1.SetFocus;
TAMBAH BAGI KALI
var //x, y, a : byte;x : real;y : real;a : real;beginx := StrToFloat(Edit1.Text);y := StrToFloat(Edit2.Text);a := x * y;Edit3.Text := FloatToStr(a);end;procedure TForm1.Button2Click(Sender: TObject);var //x, y, a : byte;x : byte;
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# Against Ranked Teams (Football) - Week 9
#### SWC75
##### Bored Historian
This year I will be again charting how teams have done “against ranked teams”, which is something that always comes up at the end of the year when the arguments over who should be in the 4 team national championship playoff takes place. As usual, this is not meant to end those arguments, just to give those making them something to look at in forming their opinions.
I’m going to do it using the procedure I started a couple years ago. The original procedure was that I would first grant teams playing ranked teams a certain number of points based on who they are playing. If you played the #1 team, (I use the AP- writer’s poll for continuity), you get 25 points. If you are playing the #2 team, you get 24 points, etc. down to getting 1 point for playing the #25 team. I call these “schedule points”. Here’s a chart to make it simple:
1-25, 2-24, 3-23, 4-22, 5-21, 6-20, 7-19, 8-18, 9-17, 10-16, 11-15, 12-14, 13-13, 14-12, 15-11, 16-10, 17-9, 18-8, 19-7, 20-6, 21-5, 22-4, 23-3, 24-2, 25-1.
(Of course you can always just subtract the ranking from 26. That will work every time.)
Then I look at the point differential in the actual game. If you defeated the ranked team, you add the number of points you won by to the schedule points. If you lost to that team, you subtract the margin you lost by from the schedule points. (I call these ‘game’ points.) The difference between this procedure and the original one if that I didn’t count any deficit: if you got to zero, you got zero. This year I’m going to try to record negative points, too, so a team can get full ‘credit’ for a really bad performance.
WEEK 9
Michigan beat #8 Notre Dame 45-14 = +49 points
Ohio State beat #13 Wisconsin 38-7 = +44 points
Kansas State beat #5 Oklahoma 48-41 = +28 points
Auburn lost to #2 Louisiana State 20-23 = +21 points
Texas Christian beat #15 Texas 37-27 =- +21 points
Louisiana State beat #9 Auburn 23-20 = +20 points
Washington State lost to #11 Oregon 35-37 = +13 points
UCLA beat #24 Arizona State 42-32 = +12 points
Oklahoma State beat #23 Iowa State 34-27 = +10 points
Houston lost to #16 Southern Methodist 31-34 = +7 points
Michigan State lost to #6 Penn State 7-28 = -1 points
Wisconsin lost to #3 Ohio State 7-38 = -8 points
Northwestern lost to #20 Iowa 0-20 = -14 points
Arkansas lost to #1 Alabama 7-48 = -16 points
California lost to #12 Utah 0-35 = -21 points
South Alabama lost to #21 Appalachian State 3-30 = -22 points
Notre Dame lost to #19 Michigan 14-45 = -24 points
Boston College lost to #4 Clemson 7-59 = -30 points
Maryland lost to #17 Minnesota 10-52 = -33 points
Comments: The games are getting bigger and more of the better teams are rising to the top of the standings. Poor Clemson may only have one more game against a ranked team, (Wake Forest, if they keep winning). But they had a good day when K-State took out Oklahoma. South Alabama could have used Ryan Alexander.
TOP 25
Louisiana State +77 points
Michigan +72 points
Ohio State +69 points
Southern California +65 points
Auburn +64 points
Florida +37 points
South Carolina +32 points
Washington +29 points
Clemson +28 points
Kansas State +28 points
U of Miami +28 points
Wisconsin +28 points
Utah +27 points
Texas +26 points
Arizona State +24 points
North Carolina +24 points
Oklahoma State +23 points
Oklahoma +22 points
Alabama +21 points
Texas A&M +21 points
Texas Christian +21 points
Pittsburgh +18 points
Penn State +17 points
Army +16 points
Notre Dame +16 points
The worst team so far is Rutgers with -99 points.
Syracuse has -10 points
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http://forum.allaboutcircuits.com/threads/transposition-of-formula.4830/ | 1,474,836,546,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660350.13/warc/CC-MAIN-20160924173740-00171-ip-10-143-35-109.ec2.internal.warc.gz | 98,322,177 | 13,331 | # Transposition of Formula
Discussion in 'Math' started by Steve1992, Feb 17, 2007.
1. ### Steve1992 Thread Starter Senior Member
Apr 7, 2006
100
0
In context to an non-inverting op-amp (from text book)
Vin = Vo x Ri/Ri + Rf
Rearranging:
Vo = (1 + Rf/Ri) x Vin
I believe Ri is the common factor from the first equation, but I dont know how to get to the second equation.
If any one can give me the sequence of steps?
2. ### hgmjr Moderator
Jan 28, 2005
9,030
214
First, the equation you have shown should have parenthesis around the two terms in the denominator like below.
Vin = Vo x Ri/(Ri + Rf)
Start by multiplying both sides of the equal sign by (Ri + Rf).
Next, multiply both sides of the equal sign by Ri.
Then, you should have the expression
Vo = Vin*(Ri+Rf)/Ri
Next, expand the fraction to Vin*(Ri/Ri + RF/Ri)
Since Ri/Ri = 1 then you can substitute to get:
Vo = Vin*(1+(Rf/Ri)) | 274 | 903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-40 | longest | en | 0.892833 |
http://de.metamath.org/mpeuni/syl2an2.html | 1,656,505,317,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00177.warc.gz | 16,953,275 | 3,251 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > syl2an2 Structured version Visualization version GIF version
Theorem syl2an2 871
Description: syl2an 493 with antecedents in standard conjunction form. (Contributed by Alan Sare, 27-Aug-2016.)
Hypotheses
Ref Expression
syl2an2.1 (𝜑𝜓)
syl2an2.2 ((𝜒𝜑) → 𝜃)
syl2an2.3 ((𝜓𝜃) → 𝜏)
Assertion
Ref Expression
syl2an2 ((𝜒𝜑) → 𝜏)
Proof of Theorem syl2an2
StepHypRef Expression
1 syl2an2.1 . . 3 (𝜑𝜓)
2 syl2an2.2 . . 3 ((𝜒𝜑) → 𝜃)
3 syl2an2.3 . . 3 ((𝜓𝜃) → 𝜏)
41, 2, 3syl2an 493 . 2 ((𝜑 ∧ (𝜒𝜑)) → 𝜏)
54anabss7 858 1 ((𝜒𝜑) → 𝜏)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 196 df-an 385 This theorem is referenced by: elrab3t 3330 reusv2lem3 4797 fvmpt2d 6202 fmptco 6303 fseqdom 8732 hashimarn 13085 divalgmod 14967 lcmfunsnlem2 15191 lcmflefac 15199 cncongr2 15220 esum2dlem 29481 bj-restsnss 32217 bj-restsnss2 32218 k0004lem3 37467 usgr1v 40482 cplgr2vpr 40655 vtxdg0e 40689 wlknewwlksn 41084 wwlksnextwrd 41103 wwlksnwwlksnon 41121 clwlkclwwlklem2a4 41206
Copyright terms: Public domain W3C validator | 644 | 1,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-27 | longest | en | 0.211934 |
https://www.proprofs.com/quiz-school/story.php?title=ota4mzm4ddw3 | 1,675,204,332,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00328.warc.gz | 975,907,217 | 67,518 | # Avid 2014 Fall ACP
31 Questions | Attempts: 60
Settings
• 1.
1.Calculate the GPA of the following grades:1st Period- 92%2nd Period- 74%3rd Period- 81%4th Period- 90%5th Period- 68%6th Period- 84%7th Period- 100%8th Period- 89%
• 2.
What does GPA stand for?
• 3.
True or False: A long term goal is a goal that takes over a year to accomplish.
• A.
True
• B.
False
• 4.
In Costas Levels of Inquiry, a Level 1 question
• A.
• B.
• C.
• 5.
In Costas Levels of Inquiry, a Level 2 question
• A.
• B.
• C.
• 6.
In Costas Levels of Inquiry, a Level 3 question
• A.
• B.
• C.
• 7.
Which of the following is a key word of the Level Two Inquiry Process?
• A.
Compare
• B.
Contrast
• C.
Analyze
• D.
All of the above
• 8.
Which of the following is a key word of the Level One Inquiry Process?
• A.
Contrast
• B.
Analyze
• C.
Define
• D.
Evaluate
• 9.
Which of the following is a key word of the Level Three Inquiry Process?
• A.
Analyze
• B.
Hypothesize
• C.
Observe
• D.
None of the above
• 10.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:Predict what will happen to_______ as _______ is changed.
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 11.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:What is…?
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 12.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:Pretend you are a character in the story. Rewrite the episode from your point of view.
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 13.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:Give me an example of…
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 14.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:Describe in your own words...
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 15.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:List the…
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 16.
MATCH THE QUESTION OR STATEMENT WITH THE CORRECT LEVEL OF INQUIRY:Compare and Contrast…
• A.
Level 1
• B.
Level 2
• C.
Level 3
• 17.
True or False: Your grades in middle school play an important role in getting accepted to college.
• A.
True
• B.
False
• 18.
True or False: When calculating your GPA, you always divide your total number of points by 8, no matter how many classes you are enrolled in.
• A.
True
• B.
False
• 19.
The purpose of teambuilding activities is:
• A.
To have fun
• B.
To have an easy day in class
• C.
To get energy out
• D.
To collaborate, problem solve and appreciate our team members
• 20.
POC stands for:
• A.
Point of Confidence
• B.
Point of Confusion
• C.
Point of Collaboration
• D.
Point of College
• 21.
During tutorials, we take 3 column notes. What is the second column labeled?
• A.
Steps
• B.
• C.
Notes
• D.
POC
• 22.
During tutorials, we give our classmates the answers to their POC.
• A.
True
• B.
False
• 23.
The purpose of tutorials is
• A.
To create a deeper understanding in materials covered in your core classes
• B.
To develop skills to become self-directed learners
• C.
There is no purpose
• D.
A and B
• 24.
True or False: A short term goal is a goal that takes less than a year to complete, such as a day, a month or a year.
• A.
True
• B.
False
• 25.
The definition of accountability is
• A.
The ability to count
• B.
Counting money
• C. | 1,063 | 3,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.646908 |
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