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• Turned Shapes Game: A game to help students learn shapes, regardless of position or orientation in space.
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• Seven colorful math mats for practice in geometry; name and draw attributes, partition into halves, thirds and fourths, and more. Printable manipulatives are included. CC: Math: 2.G.A.1-3 | 583 | 2,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-04 | longest | en | 0.831721 |
https://www.physicsforums.com/threads/linear-algebra-change-of-basis.255958/ | 1,542,085,643,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00303.warc.gz | 953,709,132 | 16,702 | # Homework Help: Linear Algebra - Change of Basis
1. Sep 13, 2008
### kehler
1. The problem statement, all variables and given/known data
Let B & C be the following subsets of R^2
B= {[3 1] , [2 2]} (the vectors should be in columns instead of rows)
C= {[1 0] , [5 4]}
Let T: R^2 -> R^2 be the linear transformation whose matrix with respect to the basis B is
[2 1]
[1 5] (the brackets should be joint, it's a 2 x 2 matrix)
Find the matrix T with respect to C. Check your answer by finding the determinant and the trace of each matrix.
3. The attempt at a solution
I found the change of basis matrix from B to C to be
[(7/4) (-2/4)]
[(1/4) ( 2/4) ] (again a 2x2 matrix)
I did this by multiplying the change of basis matrix from the standard basis to C, with the change of basis matrix from the B basis to the standard matrix like this:
[1 (-5/4)] [3 2]
[0 (1/4)] [1 2]
I then multiplied the given matrix by the change of basis matrix from B to C to get
[3 (-3/4)]
[1 (11/4)]
I thought my answer was correct but the trace for the matrix I got is 8.25 whilst the trace of the original matrix is 7. They should be the same, right??
Can anyone see where I went wrong? I'm not sure if I got the change of basis matrices right :S. The columns of the set B should form the change of basis matrix from B to the standard basis, shouldn't they?
Any help would be appreciated :)
2. Sep 13, 2008
### gabbagabbahey
You've actually done the exact opposite here:
The matrix for going from the standard basis (S) to C is actually:
$$\overleftrightarrow{P}_{C \leftarrow S} \leftrightarrow \left( \begin{array}{cc} 1 & 5 \\ 0 & 4 \end{array} \right)$$
While the matrix for going from B to S is actually:
$$\overleftrightarrow{P}_{S \leftarrow B} \leftrightarrow \left( \begin{array}{cc} \frac{1}{2} & \frac{-1}{2} \\ \frac{-1}{4} & \frac{3}{4} \end{array} \right)$$
You can check these by operating on each of the basis vectors in S (i.e. {[1,0],[0,1]}) with
$$\overleftrightarrow{P}_{C \leftarrow S}$$ , to make sure you get the basis vectors of C; and by operating on each of the basis vectors in B with
$$\overleftrightarrow{P}_{S \leftarrow B}$$ , to make sure you get the corresponding basis vectors in S.
You can then easily obtain the change of basis matrix from B to C by using :
$$\overleftrightarrow{P}_{C \leftarrow B} = {\overleftrightarrow{P}_{C \leftarrow S}}{\overleftrightarrow{P}_{S \leftarrow B}}$$
Last edited: Sep 13, 2008
3. Sep 14, 2008
### muso07
Are you in MATH1116 at ANU? If you're not then my lecturer's been doing some hard core plagiarism. :P
Sorry I don't actually have an answer for you... I got the same as you but the trace that I calculated was 5.75...
:S
4. Sep 14, 2008
### gabbagabbahey
Muso, the answer for $$\overleftrightarrow{T_c}$$ should be:
$$\overleftrightarrow{T_c} {\leftrightarrow} \left( \begin{array}{cc} \frac{83}{4} & \frac{-277}{16} \\ 17 & \frac{-55}{4} \end{array} \right)$$
If you show me your work, I can tell you where you went wrong.
5. Sep 14, 2008
### muso07
Okay, I got P(C<--B)=
[7/4 -1/2]
[1/4 1/2] (2x2 matrix)
so [T]C=P(C<--B)[T]B=
[7/4 -1/2][2 1]=
[1/4 1/2][1 5]
[3 -3/4]
[1 11/4]
Much appreciated. :)
6. Sep 14, 2008
### gabbagabbahey
Okay, your first error is that your $$\overleftrightarrow{P}_{C \leftarrow B}$$ Matrix is incorrect.
$$\overleftrightarrow{T_C} =\overleftrightarrow{P}_{C \leftarrow B} \cdot \overleftrightarrow{T_B} \cdot \overleftrightarrow{P}_{C \leftarrow B}^{-1}$$
NOT just
$$\overleftrightarrow{T_C} =\overleftrightarrow{P}_{C \leftarrow B}} \cdot \overleftrightarrow{T_B}$$
Why don't you show me how you got your $$\overleftrightarrow{P}_{C \leftarrow B}$$?
7. Sep 14, 2008
### muso07
I used the [c1 c2 | b1 b2] ~ [ I | P(C<--B)] formula
so I got
[1 5 | 3 2] ~
[0 4 | 1 2]
[1 0 | 7/4 -1/2]
[0 1 | 1/4 1/2]
So P(C<--B)=
[7/4 -1/2]
[1/4 1/2]
Thanks
8. Sep 14, 2008
### gabbagabbahey
Hmmm... I've never seen that formula before, but I can derive a correct version of it for you:
$$\left( \begin{array}{cc} \vec{c_1} ,& \vec{c_2} \end{array} \right) = \overleftrightarrow{P}_{C \leftarrow B} \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)$$
since, by definition, the change of basis matrix will change the basis vectors of B into those of C.
And so,
$$\left( \begin{array}{cc} \vec{c_1} ,& \vec{c_2} \end{array} \right) \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)^{-1} = \overleftrightarrow{P}_{C \leftarrow B} \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right) \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)^{-1} =\overleftrightarrow{P}_{C \leftarrow B}$$
Try this formula, and don't forget to take the inverse.
9. Sep 14, 2008
### muso07
Thank you! I think my textbook is wrong... :S
10. Sep 14, 2008
### gabbagabbahey
Your welcome, wht do you get for $$\overleftrightarrow{P}_{C \leftarrow B}$$ now?
11. Sep 14, 2008
### muso07
I got
[-3/4 13/4]
[-1 3]
Is that right?
12. Sep 14, 2008
### gabbagabbahey
Yup, did you get the right $$\overleftrightarrow{T_c}$$ now?
13. Sep 14, 2008
### muso07
Yep. Or at least the traces and determinants agree now. :)
14. Sep 14, 2008
### kehler
Yes I am. Lol. Are u done with the assignment? I'm still stuck on a couple of questions :(
Thanks, gabbagabbahey :)
15. Sep 14, 2008
### gabbagabbahey
WOW! what are the odds of that?!:rofl: And you're welcome
16. Sep 14, 2008
### muso07
I've done everything but the first and last questions. Need help with anything?
17. Sep 14, 2008
### kehler
I'm stuck on the last one too! I just posted it actually...
And question 3 and 5. I've tried expanding q3 but I'm not getting anywhere :(. For q5, I'm having a bit of trouble trying to explain b and c.
U need help with the first question?
I actually found the geometric series by trial and error. Then I just showed that the sum of my series from n=10 to infinity is less than 10^-6.
Last edited: Sep 14, 2008
18. Sep 14, 2008
### muso07
yeah my tutor did an example for the first one and basically he used 1/n! from 10 to infinity but that's not a geometric series, so... :S
With 3 expand and factor out x^4 on the denominator and numerator. You should get a limit of -10/3.
And with 5, solve for r^3-9r^2-12r+20=0, you should get r=-2, 10, 1. Then yk is the solution with initial values as given in the question. yk=a(-2)^k+b(10)^k+c(1)^k and solve for a, b, c
19. Sep 14, 2008
### kehler
Ahh thanks :).. I finally got ques 3. Expanding functions is such a pain!
Try using fractions for both the 'a' and 'r' in your geometric series. | 2,260 | 6,636 | {"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} | 4.25 | 4 | CC-MAIN-2018-47 | latest | en | 0.885455 |
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633. Sum of Square Numbers
104. Maximum Depth of Binary Tree
7.5 二叉树分层打印
279. Perfect Squares
890. Find and Replace Pattern
198. House Robber
Exploratory Data Analysis
153. Find Minimum in Rotated Sorted Array
10. Regular Expression Matching
204. Count Primes
563. Binary Tree Tilt
17. Letter Combinations of a Phone Number
284. Peeking Iterator
Learn Storm From Scratch
300. Longest Increasing Subsequence
670. Maximum Swap
22. Generate Parentheses
900. RLE Iterator
Learn Hadoop From Scratch
332. Reconstruct Itinerary
380. Insert Delete GetRandom O(1)
067. 机器人的运动范围
066. 矩阵中的路径
065. 滑动窗口的最大值
064. 数据流中的中位数
063. 二叉搜索树的第K个结点
062. 序列化二叉树
061. 按之字形顺序打印二叉树
060. 把二叉树打印成多行
059. 对称的二叉树
102. Binary Tree Level Order Traversal
058. 二叉树的下一个结点
057. 删除链表中重复的结点
056. 链表中环的入口结点
055. 字符流中第一个不重复的字符
054. 表示数值的字符串
053. 正则表达式匹配
052. 构建乘积数组
051. 数组中重复的数字
049. 把字符串传换成整数
046. 求 1+2+···+n
045. 圆圈中最后剩下的数字
Term Frequency Inverse Document Frequency
044. 扑克牌顺子
Named Entity Recognition
043. n 个骰子的点数
042a. 左旋转字符串
042. 翻转单词序列
041a. 和为 S 的连续整数序列
041. 和为 s 的两个数字 VS 和为 s 的连续整数序列
040. 数组中只出现一次的数字
Tree
039a. 平衡二叉树
039. 二叉树的深度
038. 数字在排序数组中出现的次数
037. 两个链表的第一个公共结点
036. 数组中的逆序对
035. 第一个只出现一次的字符
034. 丑数
Permutation
Natural Language Processing
Learn Spark From Scratch
033. 把数组排成最小的数
032. 整数中1出现的次数
84. Largest Rectangle in Histogram
031. 连续子数组的最大和
030. 最小的 K 个数
029. 数组中出现次数超过一半的数字
028. 字符串排列
027. 二叉搜索树与双向链表
026. 复杂链表的复制
025. 二叉树中和为某一值的路径
024. 从上往下打印二叉树
023. 从上往下打印二叉树
022. 栈的压入弹出序列
021. 包含 min 函数的栈
020. 顺时针打印矩阵
019. 二叉树的镜像
018. 树的子结构
017. 合并两个排序的链表
016. 反转链表
015. 链表中倒数第K个结点
014. 调整数组顺序使奇数位于偶数前面
12. 打印 1 到最大的 n 位数
009. 斐波那契数列
008. 旋转数组的最小数字
795. Number of Subarrays with Bounded Maximum
Binary Search
002. 实现 Singleton 模式
21. Merge Two Sorted Lists
Matrix Multiplication
101. Symmetric Tree
Outline of Coding
OJ 编程总结
235. Lowest Common Ancestor of a Binary Search Tree
3. Longest Substring Without Repeating Characters
238. Product of Array Except Self
287. Find the Duplicate Number
1. Two Sum
## Competitions (7)
Web Traffic Time Series Forecasting
Competitions Reproduction
2019 年搜狐算法大赛整理
Kaggle 入门一步一步记录
## Computer Vision (3)
Railway Detection
Keras YOLOv3
Single Shot MultiBox Detector
## Convex Optimization (1)
Lagrange Multiplier
## DFS (41)
472. Concatenated Words
140. Word Break II
39. Combination Sum
473. Matchsticks to Square
394. Decode String
337. House Robber III
329. Longest Increasing Path in a Matrix
116. Populating Next Right Pointers in Each Node
114. Flatten Binary Tree to Linked List
106. Construct Binary Tree from Inorder and Postorder Traversal
105. Construct Binary Tree from Preorder and Inorder Traversal
99. Recover Binary Search Tree
897. Increasing Order Search Tree
872. Leaf-Similar Trees
733. Flood Fill
437. Path Sum III
112. Path Sum
108. Convert Sorted Array to Binary Search Tree
787. Cheapest Flights Within K Stops
785. Is Graph Bipartite?
559. Maximum Depth of N-ary Tree
429. N-ary Tree Level Order Traversal
752. Open the Lock
743. Network Delay Time
690. Employee Importance
542. 01 Matrix
133. Clone Graph
301. Remove Invalid Parentheses
210. Course Schedule II
207. Course Schedule
130. Surrounded Regions
111. Minimum Depth of Binary Tree
980. Unique Paths III
212. Word Search II
417. Pacific Atlantic Water Flow
78. Subsets
94. Binary Tree Inorder Traversal
144. Binary Tree Preorder Traversal
17. Letter Combinations of a Phone Number
332. Reconstruct Itinerary
## DP (43)
1105. Filling Bookcase Shelves
1301. Number of Paths with Max Score
1147. Longest Chunked Palindrome Decomposition
664. Strange Printer
264. Ugly Number II
312. Burst Balloons
494. Target Sum
338. Counting Bits
309. Best Time to Buy and Sell Stock with Cooldown
221. Maximal Square
983. Minimum Cost For Tickets
730. Count Different Palindromic Subsequences
714. Best Time to Buy and Sell Stock with Transaction Fee
474. Ones and Zeroes
416. Partition Equal Subset Sum
12.9 最优编辑
926. Flip String to Monotone Increasing
1143. Longest Common Subsequence
813. Largest Sum of Averages
132. Palindrome Partitioning II
152. Maximum Product Subarray
123. Best Time to Buy and Sell Stock III
95. Unique Binary Search Trees II
96. Unique Binary Search Trees
87. Scramble String
32. Longest Valid Parentheses
139. Word Break
213. House Robber II
787. Cheapest Flights Within K Stops
72. Edit Distance
639. Decode Ways II
91. Decode Ways
980. Unique Paths III
63. Unique Paths II
322. Coin Change
62. Unique Paths
5. Longest Palindromic Substring
1155. Number of Dice Rolls With Target Sum
417. Pacific Atlantic Water Flow
279. Perfect Squares
198. House Robber
10. Regular Expression Matching
300. Longest Increasing Subsequence
## Data Structures (10)
Hash Table
Stack and Queue
String
Greedy Algorithms
Branch And Bound
Outline Of Data Structure And Algorithms
Backtracking
Recursion
Depth-First Search
Breadth-First Search
## Deep Learning (33)
Graph Embedding
Graph Convolutional Network
You Only Look Once
Data Augmentation
LeNet
Batch Normalization
Fully Connected Layer
Weights Initialization in Deep Learning
Fast RCNN
Region-based CNN
Pretrained Network
You Only Look Once
You Only Look Once
Smooth $L_1$
Faster RCNN
Semantic Segmentation
Convolutional Neural Network
Triplet Loss
Normalization
You Only Look Once
Non-Max Suppression
Recurrent Neural Networks
Long Short Term Memory networks
K-Nearest Neighbers
PyTorch Usage Records
Pooling Layers
Bounding Box Regression
Optimizers
Object Detection
Very Deep Convolutional Networks
TensorFlow
Dropout
## Design (3)
146. LRU Cache
284. Peeking Iterator
380. Insert Delete GetRandom O(1)
## Dimension Reduction (1)
Principal Components Analysis
Docker 使用记录
## Easy (86)
876. Middle of the Linked List
53. Maximum Subarray
844. Backspace String Compare
195. Tenth Line
193. Valid Phone Numbers
867. Transpose Matrix
985. Sum of Even Numbers After Queries
1184. Distance Between Bus Stops
415. Add Strings
1290. Convert Binary Number in a Linked List to Integer
263. Ugly Number
58. Length of Last Word
83. Remove Duplicates from Sorted List
9. Palindrome Number
27. Remove Element
448. Find All Numbers Disappeared in an Array
617. Merge Two Binary Trees
226. Invert Binary Tree
387. First Unique Character in a String
371. Sum of Two Integers
412. Fizz Buzz
344. Reverse String
350. Intersection of Two Arrays II
349. Intersection of Two Arrays
326. Power of Three
283. Move Zeroes
268. Missing Number
217. Contains Duplicate
172. Factorial Trailing Zeroes
206. Reverse Linked List
202. Happy Number
191. Number of 1 Bits
190. Reverse Bits
125. Valid Palindrome
169. Majority Element
703. Kth Largest Element in a Stream
119. Pascal's Triangle II
118. Pascal's Triangle
28. Implement strStr()
26. Remove Duplicates from Sorted Array
14. Longest Common Prefix
13. Roman to Integer
181. Employees Earning More Than Their Managers
176. Second Highest Salary
175. Combine Two Tables
7. Reverse Integer
205. Isomorphic Strings
914. X of a Kind in a Deck of Cards
501. Find Mode in Binary Search Tree
66. Plus One
35. Search Insert Position
389. Find the Difference
543. Diameter of Binary Tree
812. Largest Triangle Area
160. Intersection of Two Linked Lists
234. Palindrome Linked List
203. Remove Linked List Elements
232. Implement Queue using Stacks
225. Implement Stack using Queues
189. Rotate Array
572. Subtree of Another Tree
796. Rotate String
242. Valid Anagram
122. Best Time to Buy and Sell Stock II
121. Best Time to Buy and Sell Stock
581. Shortest Unsorted Continuous Subarray
20. Valid Parentheses
141. Linked List Cycle
136. Single Number
38. Count and Say
443. String Compression
706. Design HashMap
108. Convert Sorted Array to Binary Search Tree
155. Min Stack
690. Employee Importance
111. Minimum Depth of Binary Tree
107. Binary Tree Level Order Traversal II
69. Sqrt(x)
88. Merge Sorted Array
633. Sum of Square Numbers
104. Maximum Depth of Binary Tree
198. House Robber
204. Count Primes
563. Binary Tree Tilt
21. Merge Two Sorted Lists
1. Two Sum
## Ensemble Learning (3)
LightGBM
Gradient Boosting Decision Tree
Random Forest
## Feature Engineering (3)
Training Models With Python
Workflow Of Model Training
Feature Engineering Using Python
## Geometry (3)
Line Detection
Camera Calibration
## Graph (2)
417. Pacific Atlantic Water Flow
332. Reconstruct Itinerary
## Graph Model (2)
Graph Embedding
Graph Convolutional Network
787. Cheapest Flights Within K Stops
## Greedy (10)
1147. Longest Chunked Palindrome Decomposition
45. Jump Game II
406. Queue Reconstruction by Height
621. Task Scheduler
134. Gas Station
55. Jump Game
334. Increasing Triplet Subsequence
122. Best Time to Buy and Sell Stock II
5. Longest Palindromic Substring
890. Find and Replace Pattern
## Hard (49)
854. K-Similar Strings
1301. Number of Paths with Max Score
1147. Longest Chunked Palindrome Decomposition
664. Strange Printer
391. Perfect Rectangle
45. Jump Game II
30. Substring with Concatenation of All Words
312. Burst Balloons
315. Count of Smaller Numbers After Self
295. Find Median from Data Stream
224. Basic Calculator
218. The Skyline Problem
128. Longest Consecutive Sequence
42. Trapping Rain Water
76. Minimum Window Substring
41. First Missing Positive
4. Median of Two Sorted Arrays
149. Max Points on a Line
730. Count Different Palindromic Subsequences
982. Triples with Bitwise AND Equal To Zero
297. Serialize and Deserialize Binary Tree
25. Reverse Nodes in k-Group
239. Sliding Window Maximum
164. Maximum Gap
132. Palindrome Partitioning II
123. Best Time to Buy and Sell Stock III
87. Scramble String
85. Maximal Rectangle
32. Longest Valid Parentheses
126. Word Ladder II
472. Concatenated Words
52. N-Queens II
44. Wildcard Matching
37. Sudoku Solver
1263. Minimum Moves to Move a Box to Their Target Location
140. Word Break II
154. Find Minimum in Rotated Sorted Array II
23. Merge k Sorted Lists
329. Longest Increasing Path in a Matrix
124. Binary Tree Maximum Path Sum
99. Recover Binary Search Tree
72. Edit Distance
639. Decode Ways II
980. Unique Paths III
212. Word Search II
51. N-Queens
145. Binary Tree Postorder Traversal
10. Regular Expression Matching
84. Largest Rectangle in Histogram
## Hash Table (22)
391. Perfect Rectangle
508. Most Frequent Subtree Sum
438. Find All Anagrams in a String
560. Subarray Sum Equals K
387. First Unique Character in a String
347. Top K Frequent Elements
350. Intersection of Two Arrays II
349. Intersection of Two Arrays
217. Contains Duplicate
202. Happy Number
166. Fraction to Recurring Decimal
146. LRU Cache
49. Group Anagrams
205. Isomorphic Strings
149. Max Points on a Line
869. Reordered Power of 2
36. Valid Sudoku
136. Single Number
567. Permutation in String
706. Design HashMap
3. Longest Substring Without Repeating Characters
1. Two Sum
## Heap (7)
378. Kth Smallest Element in a Sorted Matrix
347. Top K Frequent Elements
295. Find Median from Data Stream
218. The Skyline Problem
215. Kth Largest Element in an Array
703. Kth Largest Element in a Stream
23. Merge k Sorted Lists
SQL
Hive Level Up
## Image Classification (2)
LeNet
Very Deep Convolutional Networks
## Information Theory (2)
Information Entropy
Information Theory
## Kaggle (1)
Web Traffic Time Series Forecasting
## LeetCode (328)
876. Middle of the Linked List
53. Maximum Subarray
1105. Filling Bookcase Shelves
844. Backspace String Compare
854. K-Similar Strings
1171. Remove Zero Sum Consecutive Nodes from Linked List
195. Tenth Line
194. Transpose File
193. Valid Phone Numbers
192. Word Frequency
867. Transpose Matrix
1302. Deepest Leaves Sum
1301. Number of Paths with Max Score
985. Sum of Even Numbers After Queries
398. Random Pick Index
1184. Distance Between Bus Stops
1147. Longest Chunked Palindrome Decomposition
664. Strange Printer
391. Perfect Rectangle
415. Add Strings
1290. Convert Binary Number in a Linked List to Integer
3278. Catch That Cow
264. Ugly Number II
263. Ugly Number
508. Most Frequent Subtree Sum
45. Jump Game II
58. Length of Last Word
43. Multiply Strings
30. Substring with Concatenation of All Words
83. Remove Duplicates from Sorted List
9. Palindrome Number
24. Swap Nodes in Pairs
27. Remove Element
16. 3Sum Closest
6. ZigZag Conversion
312. Burst Balloons
406. Queue Reconstruction by Height
438. Find All Anagrams in a String
647. Palindromic Substrings
621. Task Scheduler
494. Target Sum
560. Subarray Sum Equals K
448. Find All Numbers Disappeared in an Array
617. Merge Two Binary Trees
739. Daily Temperatures
338. Counting Bits
31. Next Permutation
226. Invert Binary Tree
309. Best Time to Buy and Sell Stock with Cooldown
221. Maximal Square
395. Longest Substring with At Least K Repeating Characters
387. First Unique Character in a String
371. Sum of Two Integers
378. Kth Smallest Element in a Sorted Matrix
412. Fizz Buzz
384. Shuffle an Array
344. Reverse String
341. Flatten Nested List Iterator
328. Odd Even Linked List
347. Top K Frequent Elements
350. Intersection of Two Arrays II
349. Intersection of Two Arrays
315. Count of Smaller Numbers After Self
324. Wiggle Sort II
326. Power of Three
295. Find Median from Data Stream
283. Move Zeroes
236. Lowest Common Ancestor of a Binary Tree
224. Basic Calculator
268. Missing Number
240. Search a 2D Matrix II
289. Game of Life
208. Implement Trie (Prefix Tree)
230. Kth Smallest Element in a BST
227. Basic Calculator II
218. The Skyline Problem
215. Kth Largest Element in an Array
217. Contains Duplicate
150. Evaluate Reverse Polish Notation
172. Factorial Trailing Zeroes
206. Reverse Linked List
202. Happy Number
166. Fraction to Recurring Decimal
191. Number of 1 Bits
190. Reverse Bits
148. Sort List
146. LRU Cache
128. Longest Consecutive Sequence
125. Valid Palindrome
171. Excel Sheet Column Number
134. Gas Station
169. Majority Element
42. Trapping Rain Water
76. Minimum Window Substring
703. Kth Largest Element in a Stream
49. Group Anagrams
55. Jump Game
56. Merge Intervals
48. Rotate Image
119. Pascal's Triangle II
118. Pascal's Triangle
28. Implement strStr()
19. Remove Nth Node From End of List
41. First Missing Positive
26. Remove Duplicates from Sorted Array
8. String to Integer (atoi)
14. Longest Common Prefix
13. Roman to Integer
12. Integer to Roman
983. Minimum Cost For Tickets
181. Employees Earning More Than Their Managers
178. Rank Scores
177. Nth Highest Salary
176. Second Highest Salary
175. Combine Two Tables
7. Reverse Integer
4. Median of Two Sorted Arrays
205. Isomorphic Strings
92. Reverse Linked List II
149. Max Points on a Line
730. Count Different Palindromic Subsequences
914. X of a Kind in a Deck of Cards
433. Minimum Genetic Mutation
501. Find Mode in Binary Search Tree
334. Increasing Triplet Subsequence
454. 4Sum II
66. Plus One
18. 4Sum
29. Divide Two Integers
714. Best Time to Buy and Sell Stock with Transaction Fee
495. Teemo Attacking
430. Flatten a Multilevel Doubly Linked List
869. Reordered Power of 2
525. Contiguous Array
982. Triples with Bitwise AND Equal To Zero
54. Spiral Matrix
35. Search Insert Position
Bit Manipulation
389. Find the Difference
260. Single Number III
958. Check Completeness of a Binary Tree
543. Diameter of Binary Tree
333. Largest BST Subtree
449. Serialize and Deserialize BST
523. Continuous Subarray Sum
655. Print Binary Tree
297. Serialize and Deserialize Binary Tree
474. Ones and Zeroes
416. Partition Equal Subset Sum
926. Flip String to Monotone Increasing
1143. Longest Common Subsequence
814. Binary Tree Pruning
812. Largest Triangle Area
36. Valid Sudoku
813. Largest Sum of Averages
815. Bus Routes
222. Count Complete Tree Nodes
50. Pow(x, n)
34. Find First and Last Position of Element in Sorted Array
162. Find Peak Element
160. Intersection of Two Linked Lists
138. Copy List with Random Pointer
86. Partition List
25. Reverse Nodes in k-Group
234. Palindrome Linked List
203. Remove Linked List Elements
708. Insert into a Cyclic Sorted List
239. Sliding Window Maximum
232. Implement Queue using Stacks
225. Implement Stack using Queues
189. Rotate Array
572. Subtree of Another Tree
796. Rotate String
242. Valid Anagram
151. Reverse Words in a String
164. Maximum Gap
1031. Maximum Sum of Two Non-Overlapping Subarrays
132. Palindrome Partitioning II
131. Palindrome Partitioning
152. Maximum Product Subarray
123. Best Time to Buy and Sell Stock III
122. Best Time to Buy and Sell Stock II
95. Unique Binary Search Trees II
121. Best Time to Buy and Sell Stock
96. Unique Binary Search Trees
87. Scramble String
93. Restore IP Addresses
85. Maximal Rectangle
581. Shortest Unsorted Continuous Subarray
32. Longest Valid Parentheses
126. Word Ladder II
89. Gray Code
472. Concatenated Words
79. Word Search
60. Permutation Sequence
52. N-Queens II
44. Wildcard Matching
37. Sudoku Solver
77. Combinations
503. Next Greater Element II
496. Next Greater Element I
1263. Minimum Moves to Move a Box to Their Target Location
140. Word Break II
139. Word Break
213. House Robber II
90. Subsets II
47. Permutations II
20. Valid Parentheses
142. Linked List Cycle II
141. Linked List Cycle
137. Single Number II
136. Single Number
154. Find Minimum in Rotated Sorted Array II
179. Largest Number
38. Count and Say
23. Merge k Sorted Lists
46. Permutations
74. Search a 2D Matrix
40. Combination Sum II
39. Combination Sum
216. Combination Sum III
33. Search in Rotated Sorted Array
75. Sort Colors
15. 3Sum
567. Permutation in String
443. String Compression
706. Design HashMap
2. Add Two Numbers
11. Container With Most Water
473. Matchsticks to Square
394. Decode String
337. House Robber III
329. Longest Increasing Path in a Matrix
129. Sum Root to Leaf Numbers
124. Binary Tree Maximum Path Sum
117. Populating Next Right Pointers in Each Node II
116. Populating Next Right Pointers in Each Node
114. Flatten Binary Tree to Linked List
106. Construct Binary Tree from Inorder and Postorder Traversal
105. Construct Binary Tree from Preorder and Inorder Traversal
99. Recover Binary Search Tree
897. Increasing Order Search Tree
872. Leaf-Similar Trees
733. Flood Fill
110. Balanced Binary Tree
437. Path Sum III
257. Binary Tree Paths
113. Path Sum II
112. Path Sum
863. All Nodes Distance K in Binary Tree
109. Convert Sorted List to Binary Search Tree
108. Convert Sorted Array to Binary Search Tree
100. Same Tree
787. Cheapest Flights Within K Stops
73. Set Matrix Zeroes
72. Edit Distance
70. Climbing Stairs
704. Binary Search
700. Search in a Binary Search Tree
994. Rotting Oranges
993. Cousins in Binary Tree
785. Is Graph Bipartite?
559. Maximum Depth of N-ary Tree
429. N-ary Tree Level Order Traversal
155. Min Stack
752. Open the Lock
743. Network Delay Time
690. Employee Importance
542. 01 Matrix
133. Clone Graph
513. Find Bottom Left Tree Value
513. Find Bottom Left Tree Value
301. Remove Invalid Parentheses
310. Minimum Height Trees
210. Course Schedule II
207. Course Schedule
200. Number of Islands
199. Binary Tree Right Side View
130. Surrounded Regions
639. Decode Ways II
127. Word Ladder
111. Minimum Depth of Binary Tree
91. Decode Ways
980. Unique Paths III
64. Minimum Path Sum
63. Unique Paths II
322. Coin Change
62. Unique Paths
5. Longest Palindromic Substring
107. Binary Tree Level Order Traversal II
103. Binary Tree Zigzag Level Order Traversal
212. Word Search II
1155. Number of Dice Rolls With Target Sum
417. Pacific Atlantic Water Flow
51. N-Queens
78. Subsets
94. Binary Tree Inorder Traversal
145. Binary Tree Postorder Traversal
144. Binary Tree Preorder Traversal
98. Validate Binary Search Tree
69. Sqrt(x)
88. Merge Sorted Array
633. Sum of Square Numbers
104. Maximum Depth of Binary Tree
279. Perfect Squares
890. Find and Replace Pattern
198. House Robber
153. Find Minimum in Rotated Sorted Array
10. Regular Expression Matching
204. Count Primes
563. Binary Tree Tilt
17. Letter Combinations of a Phone Number
284. Peeking Iterator
300. Longest Increasing Subsequence
22. Generate Parentheses
900. RLE Iterator
332. Reconstruct Itinerary
380. Insert Delete GetRandom O(1)
102. Binary Tree Level Order Traversal
84. Largest Rectangle in Histogram
21. Merge Two Sorted Lists
101. Symmetric Tree
235. Lowest Common Ancestor of a Binary Search Tree
3. Longest Substring Without Repeating Characters
238. Product of Array Except Self
287. Find the Duplicate Number
1. Two Sum
## Line Sweep (2)
391. Perfect Rectangle
218. The Skyline Problem
Linear Algebra
Norm
## Linked List (29)
876. Middle of the Linked List
1171. Remove Zero Sum Consecutive Nodes from Linked List
1290. Convert Binary Number in a Linked List to Integer
30. Substring with Concatenation of All Words
83. Remove Duplicates from Sorted List
24. Swap Nodes in Pairs
328. Odd Even Linked List
206. Reverse Linked List
148. Sort List
146. LRU Cache
19. Remove Nth Node From End of List
92. Reverse Linked List II
430. Flatten a Multilevel Doubly Linked List
5.14 有环单链表相交判断
160. Intersection of Two Linked Lists
138. Copy List with Random Pointer
5.6 打印两个链表的公共值
86. Partition List
25. Reverse Nodes in k-Group
234. Palindrome Linked List
203. Remove Linked List Elements
219. Insert Node in Sorted Linked List
708. Insert into a Cyclic Sorted List
142. Linked List Cycle II
141. Linked List Cycle
23. Merge k Sorted Lists
2. Add Two Numbers
21. Merge Two Sorted Lists
287. Find the Duplicate Number
## LintCode (3)
499. Word Count (Map Reduce)
219. Insert Node in Sorted Linked List
708. Insert into a Cyclic Sorted List
Linux 使用记录
Triplet Loss
## Machine Learning (119)
Hyperparameter Optimization/Tuning
DBSCAN
Clustering
Adversarial Validation
Sample Selection
Clustering Metrics
Sklearn
Model Metrics
Spark 高级数据分析
Feature Extraction
The Exponential Family
Generalized Linear Model
Line Detection
VC Dimension
Kernel Method
The Glory and the Dream
Non-negative Matrix Factorization
Hierarchical Agglomerative Clustering
Distance Estimation
Convex Optimization
Camera Calibration
Maximum A Posteriori Estimation
Machine Learning Projects
Stacking
Follow The Regularized Leader
Steepest Descent
LightGBM
Softmax Regression
Resampling
Resampling For Imbalanced Dataset
Factorization Machines
Deep Factorization Machines
Model Selection
Exploratory Data Analysis
O2O 优惠券预测使用
Self Organizing Map
Gaussian Mixture Model
Expectation Maximization
A/B Test
Hyperparameter Optimization
Outline of Dimension Reduction
Outline Of Unsupervised Learning
Feature Extraction
Outlier Detection
Distance Metrics
Gradient Boosting Decision Tree
Forward Stagewise Algorithm
Additive Model
Boosting Tree
Types of Model
Iterative Dichotomiser 3
Decision Stump
Classification and Regression Tree
C4.5
Perceptron
Overfitting and Underfitting
Norm
Regularization
K-Means
Bayesian Linear Regression
050. 最近公共祖先
048. 不能被继承的类
047. 不用加减乘除做加法
Maximum Likelihood Estimation
Naive Bayes
Conditional Random Field
Activation Functions
Recommendation System
Classification Metrics
Principal Components Analysis
Bad Case Analysis
Data Leakage
Outline Of Machine Learning
Optimizers
Click Models
Workflow Of Applying Machine Learning Algorithms Offline To Online
Model Optimization
Dynamic Programming
Competitions Workflow Of Classification And Regression
Ensemble Learning
Ranking Algorithms
Meter Reading Recognition
Single Shot Multibox Detector
Bias and Variance
Ridge and Lasso Regression
Polynomial Regression
Locally Weighted Linear Regression
Support Vector Machines
Exploratory Data Analysis
Human Activity Recognition
Applied Machine Learning Summary
Outline of Applied Machine Learning Algorithms
Random Forest
Regression Metrics
Gradient Boosting Machine
Data Preprocessing
Field-aware Factorization Machines
XGBoost
Adaboost
Neural Network
Model Ensemble
Feature Selection
Decision Tree
Linear Discriminant Analysis
Kaggle 入门一步一步记录
Linear Regression
Feature Engineering
Gradient Descent
Random Features for Large-Scale Kernel Machines
Logistic Regression
Linked List
## MapReduce (5)
Word Count
MapReduce 实现 K-Means
499. Word Count (Map Reduce)
Top K
## Math (11)
264. Ugly Number II
263. Ugly Number
326. Power of Three
172. Factorial Trailing Zeroes
7. Reverse Integer
149. Max Points on a Line
29. Divide Two Integers
523. Continuous Subarray Sum
812. Largest Triangle Area
69. Sqrt(x)
204. Count Primes
## Mathematics (15)
Differentiation
Quasi-Newton's Method
Linear Algebra
Lagrange Multiplier
Laplace Distribution
Basic Computing
Preliminaries Knowledge of Probability Theory
Taylor Series
Normal Distribution
Monte Carlo Method
Singular Value Decomposition
Matrix Factorization
Matrix
Newton's Method
## Medium (146)
1105. Filling Bookcase Shelves
1171. Remove Zero Sum Consecutive Nodes from Linked List
194. Transpose File
192. Word Frequency
1302. Deepest Leaves Sum
398. Random Pick Index
264. Ugly Number II
508. Most Frequent Subtree Sum
43. Multiply Strings
24. Swap Nodes in Pairs
16. 3Sum Closest
6. ZigZag Conversion
406. Queue Reconstruction by Height
438. Find All Anagrams in a String
621. Task Scheduler
494. Target Sum
560. Subarray Sum Equals K
739. Daily Temperatures
338. Counting Bits
31. Next Permutation
309. Best Time to Buy and Sell Stock with Cooldown
221. Maximal Square
395. Longest Substring with At Least K Repeating Characters
378. Kth Smallest Element in a Sorted Matrix
384. Shuffle an Array
341. Flatten Nested List Iterator
328. Odd Even Linked List
347. Top K Frequent Elements
324. Wiggle Sort II
236. Lowest Common Ancestor of a Binary Tree
240. Search a 2D Matrix II
289. Game of Life
208. Implement Trie (Prefix Tree)
230. Kth Smallest Element in a BST
227. Basic Calculator II
215. Kth Largest Element in an Array
150. Evaluate Reverse Polish Notation
166. Fraction to Recurring Decimal
148. Sort List
146. LRU Cache
134. Gas Station
49. Group Anagrams
55. Jump Game
56. Merge Intervals
48. Rotate Image
19. Remove Nth Node From End of List
8. String to Integer (atoi)
12. Integer to Roman
983. Minimum Cost For Tickets
178. Rank Scores
177. Nth Highest Salary
92. Reverse Linked List II
433. Minimum Genetic Mutation
334. Increasing Triplet Subsequence
454. 4Sum II
18. 4Sum
29. Divide Two Integers
714. Best Time to Buy and Sell Stock with Transaction Fee
495. Teemo Attacking
430. Flatten a Multilevel Doubly Linked List
869. Reordered Power of 2
260. Single Number III
958. Check Completeness of a Binary Tree
449. Serialize and Deserialize BST
523. Continuous Subarray Sum
655. Print Binary Tree
474. Ones and Zeroes
416. Partition Equal Subset Sum
926. Flip String to Monotone Increasing
1143. Longest Common Subsequence
814. Binary Tree Pruning
36. Valid Sudoku
813. Largest Sum of Averages
222. Count Complete Tree Nodes
50. Pow(x, n)
34. Find First and Last Position of Element in Sorted Array
162. Find Peak Element
138. Copy List with Random Pointer
86. Partition List
708. Insert into a Cyclic Sorted List
151. Reverse Words in a String
1031. Maximum Sum of Two Non-Overlapping Subarrays
131. Palindrome Partitioning
152. Maximum Product Subarray
95. Unique Binary Search Trees II
96. Unique Binary Search Trees
93. Restore IP Addresses
89. Gray Code
79. Word Search
60. Permutation Sequence
77. Combinations
503. Next Greater Element II
139. Word Break
213. House Robber II
90. Subsets II
47. Permutations II
142. Linked List Cycle II
137. Single Number II
179. Largest Number
46. Permutations
74. Search a 2D Matrix
40. Combination Sum II
39. Combination Sum
216. Combination Sum III
33. Search in Rotated Sorted Array
75. Sort Colors
15. 3Sum
567. Permutation in String
2. Add Two Numbers
11. Container With Most Water
473. Matchsticks to Square
394. Decode String
337. House Robber III
117. Populating Next Right Pointers in Each Node II
116. Populating Next Right Pointers in Each Node
114. Flatten Binary Tree to Linked List
106. Construct Binary Tree from Inorder and Postorder Traversal
105. Construct Binary Tree from Preorder and Inorder Traversal
73. Set Matrix Zeroes
310. Minimum Height Trees
130. Surrounded Regions
127. Word Ladder
91. Decode Ways
64. Minimum Path Sum
63. Unique Paths II
322. Coin Change
62. Unique Paths
5. Longest Palindromic Substring
103. Binary Tree Zigzag Level Order Traversal
1155. Number of Dice Rolls With Target Sum
417. Pacific Atlantic Water Flow
144. Binary Tree Preorder Traversal
98. Validate Binary Search Tree
279. Perfect Squares
890. Find and Replace Pattern
153. Find Minimum in Rotated Sorted Array
17. Letter Combinations of a Phone Number
284. Peeking Iterator
300. Longest Increasing Subsequence
22. Generate Parentheses
900. RLE Iterator
332. Reconstruct Itinerary
380. Insert Delete GetRandom O(1)
102. Binary Tree Level Order Traversal
3. Longest Substring Without Repeating Characters
238. Product of Array Except Self
## Metrics (1)
Classification Metrics
Model Selection
Hamlet
Django
MySQL Level Up
9.3 站队问题
9.2 方格移动
13.2 赛马
13.1 方格涂色
8.5 比较两数
8.3 交换两数
7.14 最大二叉搜索子树
7.11. 折纸问题
12.9 最优编辑
12.8 01 背包问题
5.15 单链表相交判断
6.6 最左原位
5.14 有环单链表相交判断
5.6 打印两个链表的公共值
4.9 数组变树练习题
4.6 两栈排序
4.5 递归实现栈反转
1.4 两串旋转问题
12.4 台阶问题
12.2 找零钱
7.5 二叉树分层打印
## Object Detection (9)
You Only Look Once
Fast RCNN
Region-based CNN
You Only Look Once
You Only Look Once
Semantic Segmentation
You Only Look Once
Single Shot Multibox Detector
Object Detection
## Outline (5)
Essays
Reading
Projects
Outline Of Programming Skills
Outline of Algorithms
## Pandas (1)
Learn Pandas From Scratch
SA-SSD
## Probability Theory (3)
Laplace Distribution
Preliminaries Knowledge of Probability Theory
Normal Distribution
## Programming (8)
PyTorch
Scala
Anaconda
Outline Of Programming Skills
Jupyter 使用记录
Linux 使用记录
C++
Python 基础教程
## Psychology (4)
The Psychology of Judgment and Decision Making
## PyTorch (1)
PyTorch Usage Records
## Python (15)
Model Selection
Training Models With Python
Exploratory Data Analysis
Plotly
Sklearn
Feature Selection Using Python
Workflow Of Model Training
Feature Engineering Using Python
Matplotlib
Python Level Up
Learn Numpy From Scratch
Learn Seaborn From Scratch
Learn Pandas From Scratch
Jupyter 使用记录
Python 基础教程
## Queue (2)
239. Sliding Window Maximum
225. Implement Stack using Queues
## Recommendation System (1)
Recommendation System
## Recommender System (3)
Factorization Machines
Deep Factorization Machines
Field-aware Factorization Machines
## Recursion (3)
95. Unique Binary Search Trees II
87. Scramble String
89. Gray Code
Dropout
## SQL (5)
181. Employees Earning More Than Their Managers
178. Rank Scores
177. Nth Highest Salary
176. Second Highest Salary
175. Combine Two Tables
## Sampling (1)
Sample Selection
79. Word Search
1263. Minimum Moves to Move a Box to Their Target Location
417. Pacific Atlantic Water Flow
## Sliding Window (2)
567. Permutation in String
3. Longest Substring Without Repeating Characters
## Sort (7)
324. Wiggle Sort II
148. Sort List
56. Merge Intervals
164. Maximum Gap
581. Shortest Unsorted Continuous Subarray
179. Largest Number
Sort Algorithms
## Stack (11)
844. Backspace String Compare
739. Daily Temperatures
341. Flatten Nested List Iterator
224. Basic Calculator
227. Basic Calculator II
150. Evaluate Reverse Polish Notation
4.6 两栈排序
232. Implement Queue using Stacks
85. Maximal Rectangle
20. Valid Parentheses
155. Min Stack
## String (22)
415. Add Strings
58. Length of Last Word
43. Multiply Strings
30. Substring with Concatenation of All Words
6. ZigZag Conversion
224. Basic Calculator
227. Basic Calculator II
125. Valid Palindrome
76. Minimum Window Substring
28. Implement strStr()
8. String to Integer (atoi)
14. Longest Common Prefix
13. Roman to Integer
12. Integer to Roman
796. Rotate String
242. Valid Anagram
151. Reverse Words in a String
20. Valid Parentheses
38. Count and Say
443. String Compression
5. Longest Palindromic Substring
890. Find and Replace Pattern
## TODO (1)
41. First Missing Positive
TensorFlow
Mail Templates
topic
LaTex
Jekyll 搭建博客遇到的问题
## Tree (36)
1302. Deepest Leaves Sum
508. Most Frequent Subtree Sum
617. Merge Two Binary Trees
226. Invert Binary Tree
236. Lowest Common Ancestor of a Binary Tree
230. Kth Smallest Element in a BST
501. Find Mode in Binary Search Tree
7.14 最大二叉搜索子树
7.11. 折纸问题
958. Check Completeness of a Binary Tree
543. Diameter of Binary Tree
333. Largest BST Subtree
449. Serialize and Deserialize BST
655. Print Binary Tree
297. Serialize and Deserialize Binary Tree
814. Binary Tree Pruning
222. Count Complete Tree Nodes
572. Subtree of Another Tree
95. Unique Binary Search Trees II
124. Binary Tree Maximum Path Sum
106. Construct Binary Tree from Inorder and Postorder Traversal
105. Construct Binary Tree from Preorder and Inorder Traversal
99. Recover Binary Search Tree
897. Increasing Order Search Tree
110. Balanced Binary Tree
437. Path Sum III
863. All Nodes Distance K in Binary Tree
100. Same Tree
94. Binary Tree Inorder Traversal
145. Binary Tree Postorder Traversal
144. Binary Tree Preorder Traversal
98. Validate Binary Search Tree
104. Maximum Depth of Binary Tree
563. Binary Tree Tilt
102. Binary Tree Level Order Traversal
101. Symmetric Tree
## Trie (2)
208. Implement Trie (Prefix Tree)
212. Word Search II
## Two Pointers (16)
30. Substring with Concatenation of All Words
27. Remove Element
16. 3Sum Closest
283. Move Zeroes
42. Trapping Rain Water
76. Minimum Window Substring
18. 4Sum
151. Reverse Words in a String
142. Linked List Cycle II
141. Linked List Cycle
75. Sort Colors
15. 3Sum
567. Permutation in String
11. Container With Most Water
633. Sum of Square Numbers
3. Longest Substring Without Repeating Characters
## Union Find (1)
128. Longest Consecutive Sequence
## Unsupervised Learning (1)
Principal Components Analysis
Image Gradient | 12,752 | 45,547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.492078 |
https://www.physicsforums.com/threads/time-question.743942/ | 1,508,773,277,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00354.warc.gz | 970,892,766 | 16,061 | # Time question
1. Mar 18, 2014
### a.k
1. The problem statement, all variables and given/known data
The average speed of an orbiting satellite is 20,000 mph. How much time is required for the satellite to orbit Earth? (the satellite is orbiting 250 miles above the Earth’s surface, and remember that the Earth has a radius of 3,963 miles.)
2. Relevant equations
t=d/v
2∏(r)
3. The attempt at a solution
3963+250=4213
2∏(4213)
6.28(4213)
26457.84 miles
26457.64/20000
=1.32 hours*60mins
=79.2 mins
Did I do this correctly?
2. Mar 18, 2014
### nasu
It looks OK.
3. Mar 18, 2014
### a.k
Thank you for the response.
4. Mar 18, 2014
### vanceEE
I don't see anything wrong. But, for the future; try to keep exact, or near exact answers throughout your problem, for example, when you multiplied $6.28$ by $4213$ you got $26457.84$mi, whereas the exact answer was $8426\pi$mi or $26471.0597$mi. Besides that I don't see anything wrong with your solution. | 307 | 964 | {"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} | 3.390625 | 3 | CC-MAIN-2017-43 | longest | en | 0.898166 |
https://www.teacherspayteachers.com/Product/Back-to-School-ADDITION-Activity-Pack-2712404 | 1,493,208,404,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121305.61/warc/CC-MAIN-20170423031201-00214-ip-10-145-167-34.ec2.internal.warc.gz | 989,544,205 | 26,005 | Total:
\$0.00
Back to School - ADDITION Activity Pack
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PRODUCT DESCRIPTION
BACK TO SCHOOL - Addition Activity Pack
After a long summer break, the Back to School Math Pack is designed to encourage students to refresh their math skills and jump start their school year! Interactive math games and printable worksheets provide opportunities to master math skills in the new academic year.
*********************************************************************
This Unit Includes:
**11 Addition Sentence Cards. Count school objects found in the classroom and solve the addition sentence.
**Addition File Folder Game. Includes game board, printable dice with numbers, dice with school objects to count and identify number, or choose an addition card and solve the equation. Move that many along the game path until you reach the Finish Star!
**Worksheet: Addition using Dice up to 5
**Worksheet: Addition using Dice up to 10
**Worksheet: Addition using Dice up to 12
**Worksheet: Complete the Number Bonds
Common Core Standards:
CCSS.MATH.CONTENT.K.CC.A.3
Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects).
CCSS.MATH.CONTENT.K.CC.B.5
Count to answer "how many?" questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1-20, count out that many objects.
CCSS.MATH.CONTENT.K.CC.A.1
Count to 100 by ones and by tens.
CCSS.MATH.CONTENT.K.CC.B.4
Understand the relationship between numbers and quantities; connect counting to cardinality.
CCSS.MATH.CONTENT.K.OA.A.1
Represent addition and subtraction with objects, fingers, mental images, drawings1, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
CCSS.MATH.CONTENT.K.OA.A.3
Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation (e.g., 5 = 2 + 3 and 5 = 4 + 1).
CCSS.MATH.CONTENT.K.OA.A.5
Fluently add and subtract within 5.
Clipart used in Math Activity Pack:
www.mycutegraphics.com
Created by: Rene' Perry
GamerDog.org
www.gamerdog.org
If you use this packet in your classroom we would love to see pictures please send them to www.gamerdog.org
We work hard to make fun and educational game designs. All items purchased are for single classroom use only. Please, do not resell, redistribute, or forward any items purchased. Please, no sharing electronically or in printed form. Files are for personal use only. It is for non-commercial use only.
GamerDog reserves all rights.
Thank You!
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\$3.00 | 750 | 3,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-17 | longest | en | 0.879745 |
https://en.wikibooks.org/wiki/Measure_Theory/Integration | 1,524,449,432,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00054.warc.gz | 616,360,306 | 20,703 | Measure Theory/Integration
Let ${\displaystyle (X,\sigma ,\mu )}$ be a ${\displaystyle \sigma }$-finite measure space. Suppose ${\displaystyle s}$ is a positive simple measurable function, with ${\displaystyle s=\displaystyle \sum _{i=1}^{3}y_{i}\chi _{A_{i}}}$; ${\displaystyle A_{i}\in \sigma }$ are disjoint.
Define ${\displaystyle \displaystyle \int _{X}s~d\mu =\sum y_{i}\mu (A_{i})}$
Let ${\displaystyle f:X\to {\overline {\mathbb {R} }}}$ be measurable, and let ${\displaystyle f\geq 0}$.
Define ${\displaystyle \displaystyle \int _{X}f~d\mu =\sup\{\int _{X}s~d\mu =\sum y_{i}\mu (A_{i})|s{\text{ simple }},s\geq 0,s\leq f\}}$
Now let ${\displaystyle f}$ be any measurable function. We say that ${\displaystyle f}$ is integrable if ${\displaystyle f^{+}}$ and ${\displaystyle f^{-}}$ are integrable and if ${\displaystyle \displaystyle \int _{X}f^{+}~d\mu ,\int _{X}f^{-}~d\mu <\infty }$. Then, we write
${\displaystyle \displaystyle \int _{X}f~d\mu =\int _{X}f^{+}~d\mu -\int _{X}f^{-}~d\mu }$
The class of measurable functions on ${\displaystyle X}$ is denoted by ${\displaystyle {\mathcal {L}}^{1}(X)}$
For ${\displaystyle 0, we define ${\displaystyle {\mathcal {L}}^{p}}$ to be the collection of all measurable functions ${\displaystyle f}$ such that ${\displaystyle |f|^{p}\in {\mathcal {L}}^{1}}$
A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.
Properties
Let ${\displaystyle (X,\sigma ,\mu )}$ be a measure space and let ${\displaystyle f,g}$ be measurable on ${\displaystyle X}$. Then
1. If ${\displaystyle f\leq g}$, then ${\displaystyle \displaystyle \int _{X}fd\mu \leq \int _{X}gd\mu }$
2. If ${\displaystyle A,B\in \sigma }$, ${\displaystyle A\subset B}$, then ${\displaystyle \displaystyle \int _{A}fd\mu \leq \int _{B}fd\mu }$
3. If ${\displaystyle f\geq 0}$ and ${\displaystyle c\geq 0}$ then ${\displaystyle \displaystyle \int _{X}cfd\mu =c\int _{X}fd\mu }$
4. If ${\displaystyle E\in \sigma }$, ${\displaystyle \mu (E)=0}$, then ${\displaystyle \displaystyle \int _{E}fd\mu =0}$, even if ${\displaystyle f(E)=\{\infty \}}$
5. If ${\displaystyle E\in \sigma }$, ${\displaystyle f(E)=\{0\}}$, then ${\displaystyle \displaystyle \int _{E}fd\mu =0}$, even if ${\displaystyle \mu (E)=\infty }$
Proof
Monotone Convergence Theorem
Suppose ${\displaystyle f_{n}\geq 0}$ and ${\displaystyle f_{n}}$ are measurable for all ${\displaystyle n}$ such that
1. ${\displaystyle f_{1}(x)\leq f_{2}(x)\leq \ldots }$ for every ${\displaystyle x\in X}$
2. ${\displaystyle f_{n}(x)\to f(x)}$ almost everywhere on ${\displaystyle X}$
Then, ${\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }$
Proof
${\displaystyle \displaystyle \int _{X}f_{n}d\mu }$ is an increasing sequence in ${\displaystyle \mathbb {R} }$, and hence, ${\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \alpha \in {\overline {\mathbb {R} }}}$ (say). We know that ${\displaystyle f}$ is measurable and that ${\displaystyle f\geq f_{n}\forall n}$. That is,
${\displaystyle \displaystyle \int _{X}f_{1}d\mu \leq \int _{X}f_{2}d\mu \leq \ldots \int _{X}f_{n}d\mu \leq \ldots \int _{X}fd\mu }$
Hence, ${\displaystyle \displaystyle \alpha \leq \int _{X}fd\mu =\sup\{\int _{X}sd\mu :s{\text{ is simple }},0\leq s\leq 1\}}$
Let ${\displaystyle c\in [0,1]}$
Define ${\displaystyle E_{n}=\{x|f_{n}(x)\geq cs(x)\}}$; ${\displaystyle n=1,2\ldots }$. Observe that ${\displaystyle E_{1}\subset E_{2}\subset \ldots }$ and ${\displaystyle \bigcup E_{n}=X}$
Suppose ${\displaystyle x\in X}$. If ${\displaystyle f(x)=0}$ then ${\displaystyle s(x)=0}$ implying that ${\displaystyle x\in E_{1}}$. If ${\displaystyle f(x)>0}$, then there exists ${\displaystyle n}$ such that ${\displaystyle f_{n}(x)>cs(x)}$ and hence, ${\displaystyle x\in E_{n}}$.
Thus, ${\displaystyle \bigcup E_{n}=X}$, therefore ${\displaystyle \displaystyle \int _{X}f_{n}(x)d\mu \geq \int _{E_{n}}f_{n}d\mu \geq c\int _{E_{n}}sd\mu }$. As this is true if ${\displaystyle c\in [0,1]}$, we have that ${\displaystyle \alpha \geq \displaystyle \int _{X}fd\mu }$. Thus, ${\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }$.
Fatou's Lemma
Let ${\displaystyle f_{n}\geq 0}$ be measurable functions. Then,
${\displaystyle \displaystyle \int _{X}\liminf f_{n}d\mu \leq \liminf \int _{X}f_{n}d\mu }$
Proof
For ${\displaystyle k=1,2,\ldots }$ define ${\displaystyle g_{k}(x)=\displaystyle \inf _{i\geq k}f_{i}(x)}$. Observe that ${\displaystyle g_{k}}$ are measurable and increasing for all ${\displaystyle x}$.
As ${\displaystyle k\to \infty }$, ${\displaystyle g_{k}(x)\to \displaystyle \liminf _{n\geq k}f_{n}(x)}$. By monotone convergence theorem,
${\displaystyle \displaystyle \int _{X}g_{k}d\mu \to \int _{X}\liminf f_{n}(x)d\mu }$ and as ${\displaystyle \displaystyle \int _{X}g_{k}(x)d\mu \leq \liminf \int g_{k}(x)d\mu }$, we have the result.
Dominated convergence theorem
Let ${\displaystyle (X,\sigma ,\mu )}$ be a complex measure space. Let ${\displaystyle \{f_{n}\}}$ be a sequence of complex measurable functions that converge pointwise to ${\displaystyle f}$; ${\displaystyle f(x)=\displaystyle \lim _{n\to \infty }f_{n}(x)}$, with ${\displaystyle x\in X}$
Suppose ${\displaystyle |f_{n}(x)|\leq g(x)}$ for some ${\displaystyle g\in {\mathcal {L}}^{1}(X)}$ then
${\displaystyle f\in {\mathcal {L}}^{1}}$ and ${\displaystyle \displaystyle \int _{X}|f_{n}-f|d\mu \to 0}$ as ${\displaystyle n\to \infty }$
Proof
We know that ${\displaystyle |f|\leq g}$ and hence ${\displaystyle |f_{n}-f|\leq 2g}$, that is, ${\displaystyle 0\leq 2g-|f_{n}-f|}$
Therefore, by Fatou's lemma, ${\displaystyle \displaystyle \int _{X}2gd\mu \leq \liminf \int _{X}(2g-|f_{n}-f|)d\mu \leq \displaystyle \int _{X}2gd\mu +\liminf \int _{X}(-|f_{n}-f|)d\mu }$
${\displaystyle =\displaystyle \int _{X}2gd\mu -\limsup \int _{X}|f_{n}-f|d\mu }$
As ${\displaystyle g\in {\mathcal {L}}^{1}}$, ${\displaystyle \displaystyle \limsup _{n\to \infty }\int _{X}|f_{n}-f|d\mu \leq }$ implying that ${\displaystyle \displaystyle \limsup \int _{X}|f_{n}-f|d\mu }$
Theorem
1. Suppose ${\displaystyle f:X\to [0,\infty ]}$ is measurable, ${\displaystyle E\in \sigma }$ with ${\displaystyle \mu (E)>0}$ such that ${\displaystyle \displaystyle \int _{E}fd\mu =0}$. Then ${\displaystyle f(x)=0}$ almost everywhere ${\displaystyle E}$
2. Let ${\displaystyle f\in {\mathcal {L}}^{1}(X)}$ and let ${\displaystyle \displaystyle \int _{E}fd\mu =0}$ for every ${\displaystyle E\in \sigma }$. Then, ${\displaystyle f=0}$ almost everywhere on ${\displaystyle X}$
3. Let ${\displaystyle f\in {\mathcal {L}}^{1}(X)}$ and ${\displaystyle \displaystyle \left|\int _{X}fd\mu \right|=\int _{X}|f|d\mu }$ then there exists constant ${\displaystyle \alpha }$ such that ${\displaystyle |f|=\alpha f}$ almost everywhere on ${\displaystyle E}$
Proof
1. For each ${\displaystyle n\in \mathbb {N} }$ define ${\displaystyle A_{n}=\{x\in E|f(x)>{\frac {1}{n}}\}}$. Observe that ${\displaystyle A_{n}\uparrow E}$
but ${\displaystyle {\frac {1}{n}}\mu (A_{n})\leq \displaystyle \int _{A_{n}}fd\mu \leq \int _{E}fd\mu =0}$ Thus ${\displaystyle \mu (A_{n})=0}$ for all ${\displaystyle n}$, by continuity, ${\displaystyle f=0}$ almost everywhere on ${\displaystyle E}$
2. Write ${\displaystyle \displaystyle \int _{E}fd\mu =\int _{E}u^{+}d\mu -\int _{E}u^{-}d\mu +i\left(\int _{E}v^{+}d\mu -\int _{E}v^{-}d\mu \right)}$, where ${\displaystyle u^{+},u^{-},v^{+},v^{-}}$ are non-negative real measurable.
Further as ${\displaystyle \displaystyle \int _{E}u^{+}d\mu ,\int _{E}(-u^{-})d\mu }$ are both non-negative, each of them is zero. Thus, by applying part I, we have that ${\displaystyle u^{+},u^{-}}$ vanish almost everywhere on ${\displaystyle E}$. We can similarly show that ${\displaystyle v^{+},v^{-}}$ vanish almost everywhere on ${\displaystyle E}$. | 2,881 | 7,866 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 132, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-17 | longest | en | 0.611664 |
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# Rate of Reaction and how it can be changed depending on important factors such as surface area or concentration.
Extracts from this document...
Introduction
Rate of Reaction and how it can be changed depending on important factors such as surface area or concentration Aim: We are investigating the amount of hydrogen gas produced when using different concentrations of hydrochloric acid with magnesium. Prediction: I hypothesize that the higher the concentration of acid that I use, the less time the reaction would take which would result in a higher rate of reaction. The solvent and solute will completely react and produce the product hydrogen much quicker. Planning: We are investigating the rate of reaction between magnesium and hydrochloric acid, how it can be affected if factors were changed and the amount of hydrogen gas given off as one of the final products. We could change the temperature surrounding the experiment in order to change the results that we acquire, the concentration of the acid that we are using, the surface area of the metal and also its shape and finally the use of a catalyst. I have chosen to use the level of concentration as it is not as difficult as the others to set up and investigate. We could measure the volume of gas given after the acid and metal have reacted, the time taken for the metal, in our case magnesium, to disappear or the change in temperature towards the end of the experiment, for instance is it an endothermic or exothermic reaction. Like I said before we will change the concentration of the hydrochloric acid. We will measure the volume of gas given off as a final product. Our question is what we will keep the same. We will keep the same, the temperature, the use of a catalyst to any of the experiments and the surface area of the magnesium ribbon and what shape it is when subjected to the acid. ...read more.
Middle
Because the measuring cylinder we used went up in only 1cm at a time, we couldn't get very precise readings. To make our results as accurate as we could, we had two people checking the amount of gas collected so if a person missed the amount, the other could tell it. This way, we had less confusion in the group and everyone got to be part of the experiment. Results: 0.75m: Time (s) 1st time (cm) 2nd time (cm) Average (cm) 30 19 19 19 60 36 38 37 90 45 49 47 120 55 55 55 150 59 61 60 180 62 65 64 210 65 67 66 240 68 68 68 270 69 69 69 300 70 70 70 Final reading (cm) 70 70 70 1m: Time (s) 1st time (cm) 2nd time (cm) Average (cm) 30 22 22 22 60 39 41 40 90 49 52 51 120 56 60 58 150 61 63 62 180 64 67 66 210 68 68 68 240 69 69 69 270 70 70 70 300 70 70 70 Final reading (cm) 70 70 70 1.25m: Time (s) 1st time (cm) 2nd time (cm) Average (cm) 20 18 20 19 40 45 49 47 60 57 60 59 80 62 61 62 100 65 69 67 120 68 69 69 140 70 70 70 160 70 70 70 180 70 70 70 200 70 70 70 Final reading (cm) 70 70 70 1.5m: Time (s) 1st time (cm) 2nd time (cm) Average (cm) 10 15 18 17 20 25 26 26 30 36 38 37 40 44 49 47 50 50 54 52 60 54 60 57 70 58 62 60 80 61 63 62 90 63 64 64 100 65 66 66 120 67 69 68 140 70 70 70 Final reading (cm) ...read more.
Conclusion
Overall, the method was good but like everything else, it could be improved. It was generally a fair test but a few things may have altered the results, such as after a few of the first experiments, one of the windows was opened in the room and cold air rushed in. I don't if this had anything to do with the experiment but it could be something to think over when planning the next investigation. I feel that the measurements were easy to take measurements as we didn't miss a single reading. We could have repeated the experiment once more to be a little more accurate, the more results there are, the better it is after all. All the graphs showed a pattern to do with the concentration levels of the acids, the trends got higher for rate of reaction, the time decreased for stronger acids and less time was taken to. We kept the few odd results we got as it gave us something to talk about and think about in the future. I strongly believe that the results that we got, supported our conclusion as no matter whose results you saw, all showed the same trend and this proves that the results were reliable. Next time, I would use more concentrations of acids, so that we can be even more accurate and sure about the results. Further investigations you could think about could be changing the temperature of the experiment and investigate if it alters your experiments at all. You could also change the surface area of the metal such as having a block, or reacting it in a powder form. Catalysts could be added to the reaction to see if they show a major difference. ...read more.
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# Related GCSE Patterns of Behaviour essays
1. ## An investigation into how surface area affects the rate of reaction.
At a low concentration, there are less acid particles in a certain volume. Therefore there is a lower chance of a collision because there are less acid particles. Because there are fewer collisions, the rate will decrease. An increased concentration means more particles.
2. ## Determine the rate equation for the reaction of hydrochloric acid with magnesium metal, and ...
From this graph the activation energy is 18kJ mol-1, to two significant figures. Sources of error The method I used to determine the activation energy of the reaction between hydrochloric acid and magnesium is very simple, and has few sources of error.
1. ## An investigation into how surface area affects the rate of reaction
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2. ## Exothermic and endothermic reactions
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# Increase profit factor by increasing bet size?
Discussion in 'Risk Management' started by logic_man, Apr 2, 2011.
1. ### logic_man
I keep a set of statistics on my trades, including the "hypothetical" results of the trades using different position-sizing techniques. By far, the best profit factor comes when using full Kelly position-sizing, about 15% higher than the profit factor using a fixed fraction. So my question is whether or not those traders who are trying to maximize profit factor (which is definitely a good thing) are also betting close to full Kelly position sizes and, if not, why? Is it simply the fact of the larger draw-down?
This is comparing apples and oranges.
3. ### logic_man
But that's exactly the point. Is the "apple" (fixed fraction) a better position-sizing algorithm than the "orange" (Kelly sizing) for objectively maximizing "eating satisfaction" (profit factor)? I don't know the details of other traders' statistics and perhaps there are strategies where increasing bet size decreases profit factor. My question was more if there is a generalizable linear relationship between bet size (up to a certain point, i.e. Kelly, because clearly if you were to increase bet size to 100% all it would take would be one loss for your profit factor to go to zero, so you'd need to keep your size below the threshold which would incur "ruin") and profit factor. Maybe there isn't, but if there is, I would find that interesting and would be curious to know why traders who want to maximize profit factor would not be Kelly bettors.
4. ### Ghost of Cutten
The increase comes from the higher compounding effect of the larger profits, assuming the system is profitable.
The reasons to avoid Kelly are simple - firstly, a drawdown is not necessarily just a bad run in a winning system, it also has a non-negligible chance of signalling the system no longer works. If you know your system works (e.g. A casino using a roulette wheel) then you just keep using it and will recoup your losses. If there is a chance of system degradation, then by using Kelly you have the risk of betting way too large during a drawdown. This is why trader lore recommends reducing position size during losing streaks beyond a certain amount.
The second reason is related, it's that your uncle point (or that of your investors) is likely to be a lower drawdown than the maximum that Kelly sizing will inevitably produce. For example if you would lose confidence and stop trading your system properly after a 50% drawdown, then your results will degrade below that amount. If your investors will flee after a 25% drawdown, then that is your effective total loss point. Also, many traders simply don't want the stress of large drawdowns - due to the diminishing marginal utility of money, profit maximising becomes irrational and takes a back seat to some degree of capital preservation.
So, even if you have Spock-like emotional control, there are sound reasons for betting a lot less than Kelly size. My approach is to work out th Kelly size, then multiply it by my maximum preferred drawdown. So if I don't want more than a 20% drawdown, I'll bet 1/5 of Kelly size.
In the real world, maintaining robustness and long-run survival is far more important than maximising profitability, let alone maximising profit factor. Being reasonably profitable is a 'problem' that can easily be solved - recovering from a massive drawdown is rather less tractable.
I think the author of this paper provides a reasonable answer to your question:
"...Optimal position sizing based on the Kelly formula cannot be applied to newly developed trading systems because the actual values for the two parameters that are used by the formula are not available in advance. But even in the case of systems used in actual trading for an extended period of time, there is no guarantee that these parameters will remain constant or even within a certain range."
http://www.priceactionlab.com/Literature/Kelly.pdf
6. ### logic_man
Interestingly, I've seen this strategy recover from 90% drawdowns to go on to new highs, at least on paper.
I wonder if that in and of itself is significant, as I would imagine that the vast majority of strategies which experience 90% drawdowns never recover to new highs, regardless of their position-sizing algorithm. At the very least, I doubt it's a 50-50 proposition of that occurring.
7. ### logic_man
While over time the Kelly risk-percentage has declined (I recalculate it after every trade), it's not doing so at a very fast rate of decay and even at the rate it is declining, the strategy seems to have the potential for hundreds more trades before the optimal bet size is 0. Could it degrade at a faster rate unexpectedly? Sure. But given that there have been times when the Kelly percentage has risen or stayed flat for a couple hundred trades, the market could also enter a period in which the strategy is ideally-suited for price behavior and extend its life beyond my current estimate. Seems that the fluctuations in the parameters can run both ways. Also, it might happen that the strategy hits a bottom in terms of Kelly and never goes much below it, which I would expect if the strategy really does rely on some nearly permanent market feature. The future is, of course, uncertain and there are no advance guarantees.
8. ### intervention
It is possible to have high profits with big lots. As much we take risk there will be a chance of high return . We need to see both sides of picture. It may be a great loss or profit when we trade in forex. So remember your risk management plan while deciding about betting high.
9. ### murray t turtle
============================
Good profit points;
even though plenty of MS casinos got flooded stormed /shut out, so even when one region/one casino is in the ''know'' dosent mean its so.
Plenty of money has been made with 30% drawdown;
but it takes about 60% +/ to get to zero % gain from there.50% dd takes about 100% gain to get to zero % gain .
And while plenty of money has been made with a low % hit rate;
a HI 80% hit rate could be wrong easily 20 months in a row-not a prediction. Wisdom is profitable to direct
10. ### kut2k2
Since Kelly sizing was specifically designed to maximize profits, it makes sense that Kelly sizing would correspond to the highest profit factor.
The main problem for most traders is that the publicly available Kelly formulas are all approximations once you're dealing with any betting or trading scenario more complicated than the coinflip varieties. Even as few as three different outcomes can yield some shockingly poor results from the public-domain Kelly formulas. And most traders lack the mathematical skills to get around this obstacle.
#10 Jul 1, 2013
ET IS FREE BECAUSE OF THE FINANCIAL SUPPORT FROM THESE COMPANIES: | 1,478 | 6,941 | {"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-2018-26 | latest | en | 0.944121 |
https://atcoder.jp/contests/arc060/submissions/860546 | 1,618,113,485,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00145.warc.gz | 216,618,224 | 5,624 | Contest Duration: - (local time) (100 minutes) Back to Home
Submission #860546
Source Code Expand
Copy
```#include <iostream>
#include <algorithm>
using namespace std;
#define MAX_N 100000
int tab[40][MAX_N];
// mid日でaからbまで行けるか
bool C(int a, int b, int mid){
int pos = a;
int cnt = 0;
while(mid != 0){
if(mid & 1){
pos = tab[cnt][pos];
}
mid >>= 1;
cnt++;
}
if(pos >= b) return true;
else return false;
}
int solve(int a, int b){
if(a > b) swap(a, b);
int ub = 100000;
int lb = 0;
while(ub - lb > 1){
int mid = lb + (ub - lb) / 2;
if(C(a, b, mid)) ub = mid;
else lb = mid;
}
return ub;
}
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
int N, L, Q;
int x[MAX_N];
cin >> N;
for(int i=0; i<N; i++) cin >> x[i];
cin >> L >> Q;
int y[MAX_N];
for(int i=0; i<N; i++){
tab[0][i] = upper_bound(x, x+N, x[i]+L) - x - 1;
}
for(int k=0; k<40; k++){
for(int i=0; i<N; i++){
tab[k+1][i] = tab[k][min(N-1, tab[k][i])];
}
}
for(int i=0; i<Q; i++){
int a, b;
cin >> a >> b;
a--; b--;
cout << solve(a, b) << endl;
}
return 0;
}
```
#### Submission Info
Submission Time 2016-08-31 13:40:28+0900 E - Tak and Hotels suzume C++14 (GCC 5.4.1) 0 1077 Byte RE 250 ms 16256 KB
#### Judge Result
Score / Max Score 0 / 0 0 / 200 0 / 500
Status
AC × 1
AC × 5 RE × 9
AC × 5 RE × 22
Set Name Test Cases
Sample example_01.txt
Case Name Status Exec Time Memory
example_01.txt AC 5 ms 384 KB
subtask1_01.txt AC 5 ms 384 KB
subtask1_02.txt AC 4 ms 384 KB
subtask1_03.txt RE 189 ms 640 KB
subtask1_04.txt RE 188 ms 640 KB
subtask1_05.txt RE 189 ms 640 KB
subtask1_06.txt AC 8 ms 512 KB
subtask1_07.txt AC 8 ms 512 KB
subtask1_08.txt RE 189 ms 640 KB
subtask1_09.txt RE 188 ms 640 KB
subtask1_10.txt RE 189 ms 640 KB
subtask1_11.txt RE 189 ms 640 KB
subtask1_12.txt RE 192 ms 640 KB
subtask1_13.txt RE 191 ms 640 KB
subtask2_01.txt RE 243 ms 16256 KB
subtask2_02.txt RE 250 ms 16256 KB
subtask2_03.txt RE 244 ms 16256 KB
subtask2_04.txt RE 222 ms 10880 KB
subtask2_05.txt RE 231 ms 10880 KB
subtask2_06.txt RE 246 ms 16256 KB
subtask2_07.txt RE 246 ms 16256 KB
subtask2_08.txt RE 243 ms 16256 KB
subtask2_09.txt RE 240 ms 16256 KB
subtask2_10.txt RE 244 ms 16256 KB
subtask2_11.txt RE 238 ms 15232 KB
subtask2_12.txt RE 243 ms 16256 KB
subtask2_13.txt RE 239 ms 16256 KB | 908 | 2,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.126021 |
https://byjus.com/question-answer/let-p-x-x-2-bx-c-where-b-and-c-are-integers-if-p-4/ | 1,685,831,431,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00706.warc.gz | 191,798,287 | 30,402 | Question
Let P(x)=x2+bx+c, where b and c are integers. If P(x) is a facter of both x4+6x2+25 and 3x4+4x2+28x+5, then value of P(1) is -
A
4
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8
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10
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12
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Solution
The correct option is A 4Since, P(x)=x2+bx+c, where b and c are integers. If P(x) is a facter of both x4+6x2+25 and 3x4+4x2+28x+5 Therefore, P(x) will be factor of (3x4+4x2+28x+5)−3(x4+6x2+25)=−14(x2−2x+5) So P(x)=x2−2x+5 P(1)=1−2+5=4
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Explore more | 321 | 765 | {"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-2023-23 | latest | en | 0.764706 |
https://knordslearning.com/lewis-structure-of-ps3/ | 1,726,108,468,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00505.warc.gz | 321,287,675 | 19,648 | # Lewis Structure of PS3- (With 6 Simple Steps to Draw!)
Ready to learn how to draw the lewis structure of PS3- ion?
Awesome!
Here, I have explained 6 simple steps to draw the lewis dot structure of PS3- ion (along with images).
So, if you are ready to go with these 6 simple steps, then let’s dive right into it!
Lewis structure of PS3- ion contains one double bond and two single bonds between the Phosphorous (P) atom and Sulfur (S) atoms. The Phosphorus atom (P) is at the center and it is surrounded by 3 Sulfur atoms (S).
Let’s draw and understand this lewis dot structure step by step.
(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of PS3- ion).
## 6 Steps to Draw the Lewis Structure of PS3-
### Step #1: Calculate the total number of valence electrons
Here, the given ion is PS3. In order to draw the lewis structure of PS3 ion, first of all you have to find the total number of valence electrons present in the PS3 ion.
(Valence electrons are the number of electrons present in the outermost shell of an atom).
So, let’s calculate this first.
Calculation of valence electrons in PS3 ion
• For Phosphorus:
Phosphorus is a group 15 element on the periodic table. [1]
Hence, the valence electrons present in phosphorus is 5 (see below image).
• For Sulfur:
Sulfur is a group 16 element on the periodic table. [2]
Hence, the valence electrons present in sulfur is 6 (see below image).
Hence in a PS3 ion,
Valence electrons given by Phosphorus (P) atom = 5
Valence electrons given by each Sulfur (S) atom = 6
Electron due to -1 charge, 1 more electron is added
So, total number of Valence electrons in PS3 molecule = 5 + 6(3) + 1 = 24
### Step #2: Select the center atom
While selecting the center atom, always put the least electronegative atom at the center.
(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). [1]
Here in the PS3- ion, if we compare the phosphorus atom (P) and sulfur atom (S), then phosphorus is less electronegative than sulfur.
So, phosphorus should be placed in the center and the remaining 3 sulfur atoms will surround it.
### Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of PS3, put the two electrons (i.e electron pair) between each phosphorus atom and sulfur atom to represent a chemical bond between them.
These pairs of electrons present between the Phosphorus (P) and Sulfur (S) atoms form a chemical bond, which bonds the phosphorus and sulfur atoms with each other in a PS3- ion.
### Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom
Don’t worry, I’ll explain!
In the Lewis structure of PS3 ion, the outer atoms are sulfur atoms.
So now, you have to complete the octet on these sulfur atoms (because sulfur requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that both the sulfur atoms form an octet.
Also, all the 24 valence electrons of PS3- ion (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.
Hence there is no change in the above sketch of PS3 ion.
Let’s move to the next step.
### Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond
In this step, we have to check whether the central atom (i.e phosphorus) has an octet or not.
In simple words, we have to check whether the central Phosphorus (P) atom is having 8 electrons or not.
As you can see from the above image, the central atom (i.e phosphorus) has only 6 electrons. So it does not fulfill the octet rule.
Now, in order to fulfill the octet of phosphorus atom, we have to move the electron pair from the outer atom (i.e sulfur atom) to form a double bond.
Now you can see from the above image that the central atom (i.e phosphorus), is having 8 electrons. So it fulfills the octet rule and the phosphorus atom is stable.
### Step #6: Check the formal charge
Now, you have come to the final step and here you have to check the formal charge on PS3- ion.
For that, you need to remember the formula of formal charge;
Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2
• For Phosphorus:
Valence electrons = 5 (as it is in group 15)
Nonbonding electrons = 0
Bonding electrons = 8
• For double bonded Sulfur:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 4
Bonding electrons = 4
• For single bonded Sulfur:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 6
Bonding electrons = 2
Let’s keep these charges on the atoms in the above lewis structure of PS3 ion.
As you can see in the above sketch, the pair of positive and negative charges gets canceled. Thus there is only one -ve charge left on the sulfur atom, which indicates the -1 formal charge on the PS3 molecule.
Hence, the above lewis structure of PS3- ion is the stable lewis structure.
Each electron pair (:) in the lewis dot structure of PS3- ion represents the single bond ( | ). So the above lewis dot structure of PS3- ion can also be represented as shown below.
Related lewis structures for your practice:
Lewis Structure of SOF2
Lewis Structure of SeBr4
Lewis Structure of BrCl2-
Lewis Structure of CF2S
Lewis Structure of PI5
Article by;
Author
##### Jay Rana
Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. With a desire to make learning accessible for everyone, he founded Knords Learning, an online learning platform that provides students with easily understandable explanations. | 1,494 | 6,006 | {"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-2024-38 | latest | en | 0.873802 |
https://forums.aat.org.uk/Forum/discussion/comment/470753 | 1,618,531,606,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00040.warc.gz | 333,715,012 | 12,798 | # Osborne Books - Level 4 synoptic workbook, page 29, task 4
Registered, AAT Student Posts: 2
Hello, can anyone please explain the calcluation for task 4 in Level 4 synoptic workbook, page 29?
i don't understand why the cash flow amount has to minus 60K. The answer shows in the book for the cash flow in Y1 is (275k+50k-60k)=265K
I have attached the quesiton.
Thanks
Haili
• MAAT Posts: 377
It's beacuse the £60,000 is overall irrelevant to the decision and is therefore not included in the considered cash flow.
If the plant is open:
£275,000 is made including £50,000 in management cost
If the plant is closed
The management cost still occurs and £60,000 is gained at another site.
Therefore cash flow
=£275,000 (made from being open) + £50,000 (cost occurs either way) - £60,000 (Income that would occur even if closed)
AAT Level 4, MAAT
ACCA in progress
F4- Passed Aug 2020
F5- Passed Dec 2020
F6- Passed Sep 2020
F7- Planned June 2021
F8
F9 - Planned June 2021
• Registered Posts: 2
Hello, could anyone explain whether the 10% decrease is calculated from 265K or from 325k?
Photo of the answer highly appreciated. Thank you
• Registered Posts: 10
Thank you so much for sharing. It's very helpful for me.
• Registered Posts: 1
> @KasiaK said:
> Hello, could anyone explain whether the 10% decrease is calculated from 265K or from 325k?
> Photo of the answer highly appreciated. Thank you
Hi Kasiak, I have been struggling with this question and previous comments have been helpful.
The 10% is on the 265,000. It wouldn't be on the 325,000 because this amount is before the deduction for non-relevant costs of 60,000.
The way I handled it is by carrying over only 90% of each year's revenue to next year. E.g.
Y2 total = 265,000*90%=238,500
Y3 total = 238,500*90%=214,650
Etc...
This is also what the answer suggests.
I hope this is somehow helpful.
• Registered Posts: 2
Oniyama100, | 546 | 1,900 | {"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} | 3.5 | 4 | CC-MAIN-2021-17 | latest | en | 0.96331 |
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Document Sample
``` SolidWorks Lesson Template for Teachers to Contribute
Cover Sheet for Exemplary Lessons/Units Project
Faculty Member Name: D. BARNHOUSE Date: 5/24/06
Organization: SolidWorks
Title of Lesson/Unit: Bolt Circle Pattern – From Design to Construct – Engineering to
Manufacturing – Drawing for Construction
______________________________________________________________________
Science, Technology, Engineering and Math) STEM Concepts Addressed: This
lesson provides an overview for the design process of creating a bolt circle pattern to be
manufactured on a steel plate. It involves the basics of creating the geometry (drawing)
in Solid Works and then retrieving the information (X and Y centers) necessary for the
machinist to construct the part. Understanding the needed information that the
machinist needs in order to engineer a production part is essential in the design process
for accuracy and efficiency.
The drawing will give information based on the lower left hand corner of the part as the
datum point. All bolt hole centers will be based upon their X and Y relationship to the
lower left hand corner of the plate.
Length of Instruction period: 50 minutes
How many periods needed to implement lesson unit: 1
Objectives:
1. Creating a steel plate with a bolt circle pattern
utilizing Line, Circle, Circular Step and Repeat,
and Extruded Boss/Base commands on Solid
Works.
2. Calculating X and Y coordinates for CNC
(Computer Numerical Control) or Manual
Machining.
3. Understanding the need for location points in
Engineering for industry.
A bolt circle pattern is the diagram that indicates a series of holes drilled in a polar
fashion around a point that are equally spaced both by degrees and by the radius
indicating the centers of each drilled hole. This is used quite frequently by industry as a
means of fastening housings over shafts, gear boxes, etc.
For a CNC machinist or programmer to be able to program the milling machine to drill
this series of holes, that machinist will have to have the X and Y center locations for
each hole. This location point is essential because the machines that drill these holes
base the location upon the center of the drill.
Some programmers choose to locate the centers according to the center of the part,
(especially if the part is circular). Other programmers choose to locate the centers
according to a datum point located on a specific edge of the part.
In this exercise, we will develop a 6” x 6” x 1” steel plate with a bolt circle pattern.
We will then retrieve the data necessary for the CNC Programmer to program this part to
be drilled by an origin being located at the lower left corner of the plate.
Materials: SolidWorks, paper and pencil
Procedures:
Create a new part on SolidWorks using the front view as your starting view.
Sketch a rectangle using the origin for the lower left corner as 0, 0 (0 on X axis
and 0 on Y axis).
Dimension the rectangle to be 6.00 by 6.00 square.
Sketch a vertical and a horizontal center line through the middle of the part.
Sketch a circle with a 2 inch radius or 4 inch diameter at the intersection of the 2
centerlines. This will be the center of the bolt circle.
Sketch a circle with a .375 diameter at the intersecting point of the top quadrant
of the 4 inch diameter and the vertical centerline.
Under Tools, then Sketch Tools, select Circular Step and Repeat.
Under Arc Radius type in 2 inches. Under Center for X type in 3. Under Center
for Y type in 3. Under Step Number type in 8. Then Under Item to repeat, left
click inside this box and then select the small circle from the part.
Then select OK.
Under the Features tool, select Extruded Base/Boss. Under selected contours
select any area outside the bolt circle and then also select an area inside the bolt
circle (not a small circle). Under Depth type in 1 inch. Then select OK.
To find the X, Y, and even Z coordinates of each small circle on this plate, just
click on a small circle. When it turns green, this information and the diameter of
the hole will be displayed in the lower right hand area of the screen.
Then determine these coordinates for each hole in the same manner by selecting
them.
A drawing from the part can be made to indicate each of these with dimensions
originating from the lower left corner to the center of each hole can also be made
to show this information.
Assessment: On their own, have the students to create a plate that is 5.50 inches wide
by 5.50 inches tall by .50 inch thick. Then draw a bolt circle diameter of 2.50 inches with
6 equally spaced drilled holes with the diameter of .25 inches each. Have the students
to determine the center location (X and Y coordinates from the lower left hand corner of
plate) for each drilled hole. Have the students to prepare a drawing sheet (Sheet1) with
a front view, a side view and an isometric view and the correlating dimensions all located
from the lower left hand corner of the plate. Include the overall width, depth and height
dimensions. Also include the feature dimensions (ie Diameters, etc).
* Optional Synthesis Assessment: On their own, have student to create a second
sheet (Sheet2) with the same views and overall dimensions. But on this sheet have the
students to dimension all drilled holes (X and Y coordinates) from the center of the bolt
circle pattern.
Resources Used: N/A
```
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May 6, 2016
# Homework Help: ALGEBRA
Posted by Urgen please help on Thursday, June 14, 2012 at 6:22pm.
Assume f(x) = x2 + 4. Find the functional value requested.
f(-3) =
• ALGEBRA - Henry, Saturday, June 16, 2012 at 3:30pm
Substitute -3 for Xand solve Eq.
## Answer This Question
First Name: School Subject: Answer:
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# What is the formula for calculating the sales tax?
Query Drinker
To calculate the sales tax that is included in a company's receipts from items subject to sales tax, divide the receipts by 1 + the sales tax rate. For example, if the sales tax rate is 6%, divide the total amount of receipts by 1.06. If the sales tax rate is 7.25%, divide the total receipts by 1.072...
Maths Wizard
To calculate sales tax in company receipts then just divide the entire receipts divided by (sales rate plus 1). For eg.:- If a company sells TV for Rs 21,000. If sales tax is 5%. then C.P of TV will be Rs. 21,000 / 1.05 = Rs. 20,000.
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Formula of A P
an = a+(n-1)d Where an=last term a=first term d=common difference n=number of term Note: last term is denoted by an or l Sn=n/2[2a+(n-1)d Sn=sum of all terms
Manjeet Khajuria | 19/10/2022
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Algebra
Question
(Modeling) Cost, Revenue, and Profit Analysis Dotty starts up a small business manufacturing bobble-head figures of famous soccer players. Her initial cost is $$\ 3300$$ . Each figure costs $$\ 4.50$$ to manufacture.
(a) Write a cost function $$C$$ , where $$x$$ represents the number of figures manufactured.
(b) Find the revenue function $$R$$ if each figure in part (a) sells for $$\ 10.50$$ .
(c) Give the profit function $$P$$ .
(d) How many figures must be produced and sold for Dotty to earn a profit?
$$\left. \begin{array} { l l } { \text { [2.4] 22. } \left. \begin{array} { l l } { \text { (a) } C ( x ) = 3300 + 4.50 x } & { \text { (b) } R ( x ) = 10.50 x } \\ { \text { (c) } R ( x ) - C ( x ) = 6.00 x - 3300 } & { \text { (d) } 551 \text { figures } } \end{array} \right. } \end{array} \right.$$ | 283 | 866 | {"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} | 3.953125 | 4 | CC-MAIN-2022-33 | latest | en | 0.729193 |
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### Salinon
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An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
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Unit of study_
# MATH1011: Applications of Calculus
## Overview
This unit is designed for science students who do not intend to undertake higher year mathematics and statistics. It establishes and reinforces the fundamentals of calculus, illustrated where possible with context and applications. Specifically, it demonstrates the use of (differential) calculus in solving optimisation problems and of (integral) calculus in measuring how a system accumulates over time. Topics studied include the fitting of data to various functions, the interpretation and manipulation of periodic functions and the evaluation of commonly occurring summations. Differential calculus is extended to functions of two variables and integration techniques include integration by substitution and the evaluation of integrals of infinite type.
### Unit details and rules
Unit code MATH1011 Mathematics and Statistics Academic Operations 3 MATH1001 or MATH1901 or MATH1906 or BIOM1003 or ENVX1001 or MATH1021 or MATH1921 or MATH1931 None None HSC Mathematics. Students who have not completed HSC Mathematics (or equivalent) are strongly advised to take the Mathematics Bridging Course (offered in February). Please note: this unit does not normally lead to a major in Mathematics or Statistics or Financial Mathematics and Statistics No
### Teaching staff
Coordinator Daniel Hauer, daniel.hauer@sydney.edu.au Shervin Sherwin
## Assessment
Type Description Weight Due Length
Assignment Assignment 1
Written calculations
5% Week 02
Due date: 23 Jan 2022 at 23:59
Closing date: 30 Jan 2022
7 days
Outcomes assessed:
Online task Quiz 1
Written calculations
12.5% Week 03
Due date: 27 Jan 2022 at 13:00
40 minutes
Outcomes assessed:
Assignment Assignment 2
written calculations
10% Week 04
Due date: 06 Feb 2022 at 23:59
Closing date: 13 Feb 2022
7 days
Outcomes assessed:
Online task Quiz 2
Written Calculus
12.5% Week 05
Due date: 08 Feb 2022 at 12:00
40
Outcomes assessed:
Final exam (Record+) Final exam
Multiple choice and written written calculations
60% Week 06
Due date: 16 Feb 2022 at 12:00
1.5 hours
Outcomes assessed:
= Type B final exam
### Assessment summary
Below are brief assessment details. Further information can be found in the Canvas site for this unit.
• Assessments: There are two assignments, which must be submitted electronically, as PDF files only, in Canvas by the deadline. Note that your assignment will not be marked if it is illegible or if it is submitted sideways or upside down. It is your responsibility to check that your assignment has been submitted correctly. Penalties apply for late submission. A mark of zero will be awarded for all submissions more than 7 days past the original due date. Further extensions past this time will not be permitted. The better mark principle does not apply to assignments.
• Quizzes: Quizzes will be held online in Canvas. The quizzes are 40 minutes. The better mark principle will be used for the quizzes so do not submit an application for Special Consideration or Special Arrangements if you miss a quiz. The better mark principle means that for each quiz, the quiz counts if and only if it is better than or equal to your exam mark. If your quiz mark is less than your exam mark, the exam mark will be used for that portion of your assessment instead.
• Examination: Further information about the exam will be made available at a later date on Canvas.
### Assessment criteria
The University awards common result grades, set out in the Coursework Policy 2014 (Schedule 1).
As a general guide, a high distinction indicates work of an exceptional standard, a distinction a very high standard, a credit a good standard, and a pass an acceptable standard.
Result name
Mark range
Description
High distinction
85 - 100
At HD level, a student demonstrates a flair for the subject as well as a detailed and comprehensive understanding of the unit material. A ‘High Distinction’ reflects exceptional achievement and is awarded to a student who demonstrates the ability to apply their subject knowledge and understanding to produce original solutions for novel or highly complex problems and/or comprehensive critical discussions of theoretical concepts.
Distinction
75 - 84
At DI level, a student demonstrates an aptitude for the subject and a well-developed understanding of the unit material. A ‘Distinction’ reflects excellent achievement and is awarded to a student who demonstrates an ability to apply their subject knowledge and understanding of the subject to produce good solutions for challenging problems and/or a reasonably well-developed critical analysis of theoretical concepts.
Credit
65 - 74
At CR level, a student demonstrates a good command and knowledge of the unit material. A ‘Credit’ reflects solid achievement and is awarded to a student who has a broad general understanding of the unit material and can solve routine problems and/or identify and superficially discuss theoretical concepts.
Pass
50 - 64
At PS level, a student demonstrates proficiency in the unit material. A ‘Pass’ reflects satisfactory achievement and is awarded to a student who has threshold knowledge.
Fail
0 - 49
When you don’t meet the learning outcomes of the unit to a satisfactory standard.
For more information see sydney.edu.au/students/guide-to-grades.
For more information see guide to grades.
### Late submission
In accordance with University policy, these penalties apply when written work is submitted after 11:59pm on the due date:
• Deduction of 5% of the maximum mark for each calendar day after the due date.
• After ten calendar days late, a mark of zero will be awarded.
### Academic integrity
The Current Student website provides information on academic integrity and the resources available to all students. The University expects students and staff to act ethically and honestly and will treat all allegations of academic integrity breaches seriously.
We use similarity detection software to detect potential instances of plagiarism or other forms of academic integrity breach. If such matches indicate evidence of plagiarism or other forms of academic integrity breaches, your teacher is required to report your work for further investigation.
You may only use artificial intelligence and writing assistance tools in assessment tasks if you are permitted to by your unit coordinator, and if you do use them, you must also acknowledge this in your work, either in a footnote or an acknowledgement section.
Studiosity is permitted for postgraduate units unless otherwise indicated by the unit coordinator. The use of this service must be acknowledged in your submission.
## Learning support
### Simple extensions
If you encounter a problem submitting your work on time, you may be able to apply for an extension of five calendar days through a simple extension. The application process will be different depending on the type of assessment and extensions cannot be granted for some assessment types like exams.
### Special consideration
If exceptional circumstances mean you can’t complete an assessment, you need consideration for a longer period of time, or if you have essential commitments which impact your performance in an assessment, you may be eligible for special consideration or special arrangements.
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### Using AI responsibly
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## Weekly schedule
WK Topic Learning activity Learning outcomes
Week 01 Periodicity. Block teaching (2 hr)
Scaling Data. Block teaching (2 hr)
Scaling Data and Finite Differences. Block teaching (2 hr)
Week 02 Optimisation: One Variable Problems. Block teaching (2 hr)
Optimisation: One Variable Problems (continued). Block teaching (2 hr)
Optimisation: Two Variable Problems. Block teaching (2 hr)
Week 03 Optimisation: Two Variable Problems (continued). Block teaching (2 hr)
Method of Least Squares. Block teaching (2 hr)
Finite Sums. Block teaching (2 hr)
Week 04 The Definite Integral. Block teaching (2 hr)
The Indefinite Integral. Block teaching (2 hr)
Applications of Integration. Block teaching (2 hr)
### Attendance and class requirements
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### Required readings
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## Learning outcomes
Learning outcomes are what students know, understand and are able to do on completion of a unit of study. They are aligned with the University's graduate qualities and are assessed as part of the curriculum.
At the completion of this unit, you should be able to:
• LO1. analyse practical problems using techniques from differential and integral calculus;
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### Outcome map
Learning outcomes Graduate qualities
GQ1 GQ2 GQ3 GQ4 GQ5 GQ6 GQ7 GQ8 GQ9
## Responding to student feedback
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Minor changes were made to the weighting for the quizzes and second assignment.
## Additional information
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### Disclaimer
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To help you understand common terms that we use at the University, we offer an online glossary. | 2,962 | 14,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.845125 |
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csytrs.f
Go to the documentation of this file.
1*> \brief \b CSYTRS
2*
3* =========== DOCUMENTATION ===========
4*
5* Online html documentation available at
6* http://www.netlib.org/lapack/explore-html/
7*
8*> \htmlonly
10*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/csytrs.f">
11*> [TGZ]</a>
12*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/csytrs.f">
13*> [ZIP]</a>
14*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/csytrs.f">
15*> [TXT]</a>
16*> \endhtmlonly
17*
18* Definition:
19* ===========
20*
21* SUBROUTINE CSYTRS( UPLO, N, NRHS, A, LDA, IPIV, B, LDB, INFO )
22*
23* .. Scalar Arguments ..
24* CHARACTER UPLO
25* INTEGER INFO, LDA, LDB, N, NRHS
26* ..
27* .. Array Arguments ..
28* INTEGER IPIV( * )
29* COMPLEX A( LDA, * ), B( LDB, * )
30* ..
31*
32*
33*> \par Purpose:
34* =============
35*>
36*> \verbatim
37*>
38*> CSYTRS solves a system of linear equations A*X = B with a complex
39*> symmetric matrix A using the factorization A = U*D*U**T or
40*> A = L*D*L**T computed by CSYTRF.
41*> \endverbatim
42*
43* Arguments:
44* ==========
45*
46*> \param[in] UPLO
47*> \verbatim
48*> UPLO is CHARACTER*1
49*> Specifies whether the details of the factorization are stored
50*> as an upper or lower triangular matrix.
51*> = 'U': Upper triangular, form is A = U*D*U**T;
52*> = 'L': Lower triangular, form is A = L*D*L**T.
53*> \endverbatim
54*>
55*> \param[in] N
56*> \verbatim
57*> N is INTEGER
58*> The order of the matrix A. N >= 0.
59*> \endverbatim
60*>
61*> \param[in] NRHS
62*> \verbatim
63*> NRHS is INTEGER
64*> The number of right hand sides, i.e., the number of columns
65*> of the matrix B. NRHS >= 0.
66*> \endverbatim
67*>
68*> \param[in] A
69*> \verbatim
70*> A is COMPLEX array, dimension (LDA,N)
71*> The block diagonal matrix D and the multipliers used to
72*> obtain the factor U or L as computed by CSYTRF.
73*> \endverbatim
74*>
75*> \param[in] LDA
76*> \verbatim
77*> LDA is INTEGER
78*> The leading dimension of the array A. LDA >= max(1,N).
79*> \endverbatim
80*>
81*> \param[in] IPIV
82*> \verbatim
83*> IPIV is INTEGER array, dimension (N)
84*> Details of the interchanges and the block structure of D
85*> as determined by CSYTRF.
86*> \endverbatim
87*>
88*> \param[in,out] B
89*> \verbatim
90*> B is COMPLEX array, dimension (LDB,NRHS)
91*> On entry, the right hand side matrix B.
92*> On exit, the solution matrix X.
93*> \endverbatim
94*>
95*> \param[in] LDB
96*> \verbatim
97*> LDB is INTEGER
98*> The leading dimension of the array B. LDB >= max(1,N).
99*> \endverbatim
100*>
101*> \param[out] INFO
102*> \verbatim
103*> INFO is INTEGER
104*> = 0: successful exit
105*> < 0: if INFO = -i, the i-th argument had an illegal value
106*> \endverbatim
107*
108* Authors:
109* ========
110*
111*> \author Univ. of Tennessee
112*> \author Univ. of California Berkeley
113*> \author Univ. of Colorado Denver
114*> \author NAG Ltd.
115*
116*> \ingroup hetrs
117*
118* =====================================================================
119 SUBROUTINE csytrs( UPLO, N, NRHS, A, LDA, IPIV, B, LDB, INFO )
120*
121* -- LAPACK computational routine --
122* -- LAPACK is a software package provided by Univ. of Tennessee, --
123* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
124*
125* .. Scalar Arguments ..
126 CHARACTER UPLO
127 INTEGER INFO, LDA, LDB, N, NRHS
128* ..
129* .. Array Arguments ..
130 INTEGER IPIV( * )
131 COMPLEX A( LDA, * ), B( LDB, * )
132* ..
133*
134* =====================================================================
135*
136* .. Parameters ..
137 COMPLEX ONE
138 parameter( one = ( 1.0e+0, 0.0e+0 ) )
139* ..
140* .. Local Scalars ..
141 LOGICAL UPPER
142 INTEGER J, K, KP
143 COMPLEX AK, AKM1, AKM1K, BK, BKM1, DENOM
144* ..
145* .. External Functions ..
146 LOGICAL LSAME
147 EXTERNAL lsame
148* ..
149* .. External Subroutines ..
150 EXTERNAL cgemv, cgeru, cscal, cswap, xerbla
151* ..
152* .. Intrinsic Functions ..
153 INTRINSIC max
154* ..
155* .. Executable Statements ..
156*
157 info = 0
158 upper = lsame( uplo, 'U' )
159 IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN
160 info = -1
161 ELSE IF( n.LT.0 ) THEN
162 info = -2
163 ELSE IF( nrhs.LT.0 ) THEN
164 info = -3
165 ELSE IF( lda.LT.max( 1, n ) ) THEN
166 info = -5
167 ELSE IF( ldb.LT.max( 1, n ) ) THEN
168 info = -8
169 END IF
170 IF( info.NE.0 ) THEN
171 CALL xerbla( 'CSYTRS', -info )
172 RETURN
173 END IF
174*
175* Quick return if possible
176*
177 IF( n.EQ.0 .OR. nrhs.EQ.0 )
178 \$ RETURN
179*
180 IF( upper ) THEN
181*
182* Solve A*X = B, where A = U*D*U**T.
183*
184* First solve U*D*X = B, overwriting B with X.
185*
186* K is the main loop index, decreasing from N to 1 in steps of
187* 1 or 2, depending on the size of the diagonal blocks.
188*
189 k = n
190 10 CONTINUE
191*
192* If K < 1, exit from loop.
193*
194 IF( k.LT.1 )
195 \$ GO TO 30
196*
197 IF( ipiv( k ).GT.0 ) THEN
198*
199* 1 x 1 diagonal block
200*
201* Interchange rows K and IPIV(K).
202*
203 kp = ipiv( k )
204 IF( kp.NE.k )
205 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
206*
207* Multiply by inv(U(K)), where U(K) is the transformation
208* stored in column K of A.
209*
210 CALL cgeru( k-1, nrhs, -one, a( 1, k ), 1, b( k, 1 ), ldb,
211 \$ b( 1, 1 ), ldb )
212*
213* Multiply by the inverse of the diagonal block.
214*
215 CALL cscal( nrhs, one / a( k, k ), b( k, 1 ), ldb )
216 k = k - 1
217 ELSE
218*
219* 2 x 2 diagonal block
220*
221* Interchange rows K-1 and -IPIV(K).
222*
223 kp = -ipiv( k )
224 IF( kp.NE.k-1 )
225 \$ CALL cswap( nrhs, b( k-1, 1 ), ldb, b( kp, 1 ), ldb )
226*
227* Multiply by inv(U(K)), where U(K) is the transformation
228* stored in columns K-1 and K of A.
229*
230 CALL cgeru( k-2, nrhs, -one, a( 1, k ), 1, b( k, 1 ), ldb,
231 \$ b( 1, 1 ), ldb )
232 CALL cgeru( k-2, nrhs, -one, a( 1, k-1 ), 1, b( k-1, 1 ),
233 \$ ldb, b( 1, 1 ), ldb )
234*
235* Multiply by the inverse of the diagonal block.
236*
237 akm1k = a( k-1, k )
238 akm1 = a( k-1, k-1 ) / akm1k
239 ak = a( k, k ) / akm1k
240 denom = akm1*ak - one
241 DO 20 j = 1, nrhs
242 bkm1 = b( k-1, j ) / akm1k
243 bk = b( k, j ) / akm1k
244 b( k-1, j ) = ( ak*bkm1-bk ) / denom
245 b( k, j ) = ( akm1*bk-bkm1 ) / denom
246 20 CONTINUE
247 k = k - 2
248 END IF
249*
250 GO TO 10
251 30 CONTINUE
252*
253* Next solve U**T *X = B, overwriting B with X.
254*
255* K is the main loop index, increasing from 1 to N in steps of
256* 1 or 2, depending on the size of the diagonal blocks.
257*
258 k = 1
259 40 CONTINUE
260*
261* If K > N, exit from loop.
262*
263 IF( k.GT.n )
264 \$ GO TO 50
265*
266 IF( ipiv( k ).GT.0 ) THEN
267*
268* 1 x 1 diagonal block
269*
270* Multiply by inv(U**T(K)), where U(K) is the transformation
271* stored in column K of A.
272*
273 CALL cgemv( 'Transpose', k-1, nrhs, -one, b, ldb, a( 1, k ),
274 \$ 1, one, b( k, 1 ), ldb )
275*
276* Interchange rows K and IPIV(K).
277*
278 kp = ipiv( k )
279 IF( kp.NE.k )
280 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
281 k = k + 1
282 ELSE
283*
284* 2 x 2 diagonal block
285*
286* Multiply by inv(U**T(K+1)), where U(K+1) is the transformation
287* stored in columns K and K+1 of A.
288*
289 CALL cgemv( 'Transpose', k-1, nrhs, -one, b, ldb, a( 1, k ),
290 \$ 1, one, b( k, 1 ), ldb )
291 CALL cgemv( 'Transpose', k-1, nrhs, -one, b, ldb,
292 \$ a( 1, k+1 ), 1, one, b( k+1, 1 ), ldb )
293*
294* Interchange rows K and -IPIV(K).
295*
296 kp = -ipiv( k )
297 IF( kp.NE.k )
298 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
299 k = k + 2
300 END IF
301*
302 GO TO 40
303 50 CONTINUE
304*
305 ELSE
306*
307* Solve A*X = B, where A = L*D*L**T.
308*
309* First solve L*D*X = B, overwriting B with X.
310*
311* K is the main loop index, increasing from 1 to N in steps of
312* 1 or 2, depending on the size of the diagonal blocks.
313*
314 k = 1
315 60 CONTINUE
316*
317* If K > N, exit from loop.
318*
319 IF( k.GT.n )
320 \$ GO TO 80
321*
322 IF( ipiv( k ).GT.0 ) THEN
323*
324* 1 x 1 diagonal block
325*
326* Interchange rows K and IPIV(K).
327*
328 kp = ipiv( k )
329 IF( kp.NE.k )
330 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
331*
332* Multiply by inv(L(K)), where L(K) is the transformation
333* stored in column K of A.
334*
335 IF( k.LT.n )
336 \$ CALL cgeru( n-k, nrhs, -one, a( k+1, k ), 1, b( k, 1 ),
337 \$ ldb, b( k+1, 1 ), ldb )
338*
339* Multiply by the inverse of the diagonal block.
340*
341 CALL cscal( nrhs, one / a( k, k ), b( k, 1 ), ldb )
342 k = k + 1
343 ELSE
344*
345* 2 x 2 diagonal block
346*
347* Interchange rows K+1 and -IPIV(K).
348*
349 kp = -ipiv( k )
350 IF( kp.NE.k+1 )
351 \$ CALL cswap( nrhs, b( k+1, 1 ), ldb, b( kp, 1 ), ldb )
352*
353* Multiply by inv(L(K)), where L(K) is the transformation
354* stored in columns K and K+1 of A.
355*
356 IF( k.LT.n-1 ) THEN
357 CALL cgeru( n-k-1, nrhs, -one, a( k+2, k ), 1, b( k, 1 ),
358 \$ ldb, b( k+2, 1 ), ldb )
359 CALL cgeru( n-k-1, nrhs, -one, a( k+2, k+1 ), 1,
360 \$ b( k+1, 1 ), ldb, b( k+2, 1 ), ldb )
361 END IF
362*
363* Multiply by the inverse of the diagonal block.
364*
365 akm1k = a( k+1, k )
366 akm1 = a( k, k ) / akm1k
367 ak = a( k+1, k+1 ) / akm1k
368 denom = akm1*ak - one
369 DO 70 j = 1, nrhs
370 bkm1 = b( k, j ) / akm1k
371 bk = b( k+1, j ) / akm1k
372 b( k, j ) = ( ak*bkm1-bk ) / denom
373 b( k+1, j ) = ( akm1*bk-bkm1 ) / denom
374 70 CONTINUE
375 k = k + 2
376 END IF
377*
378 GO TO 60
379 80 CONTINUE
380*
381* Next solve L**T *X = B, overwriting B with X.
382*
383* K is the main loop index, decreasing from N to 1 in steps of
384* 1 or 2, depending on the size of the diagonal blocks.
385*
386 k = n
387 90 CONTINUE
388*
389* If K < 1, exit from loop.
390*
391 IF( k.LT.1 )
392 \$ GO TO 100
393*
394 IF( ipiv( k ).GT.0 ) THEN
395*
396* 1 x 1 diagonal block
397*
398* Multiply by inv(L**T(K)), where L(K) is the transformation
399* stored in column K of A.
400*
401 IF( k.LT.n )
402 \$ CALL cgemv( 'Transpose', n-k, nrhs, -one, b( k+1, 1 ),
403 \$ ldb, a( k+1, k ), 1, one, b( k, 1 ), ldb )
404*
405* Interchange rows K and IPIV(K).
406*
407 kp = ipiv( k )
408 IF( kp.NE.k )
409 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
410 k = k - 1
411 ELSE
412*
413* 2 x 2 diagonal block
414*
415* Multiply by inv(L**T(K-1)), where L(K-1) is the transformation
416* stored in columns K-1 and K of A.
417*
418 IF( k.LT.n ) THEN
419 CALL cgemv( 'Transpose', n-k, nrhs, -one, b( k+1, 1 ),
420 \$ ldb, a( k+1, k ), 1, one, b( k, 1 ), ldb )
421 CALL cgemv( 'Transpose', n-k, nrhs, -one, b( k+1, 1 ),
422 \$ ldb, a( k+1, k-1 ), 1, one, b( k-1, 1 ),
423 \$ ldb )
424 END IF
425*
426* Interchange rows K and -IPIV(K).
427*
428 kp = -ipiv( k )
429 IF( kp.NE.k )
430 \$ CALL cswap( nrhs, b( k, 1 ), ldb, b( kp, 1 ), ldb )
431 k = k - 2
432 END IF
433*
434 GO TO 90
435 100 CONTINUE
436 END IF
437*
438 RETURN
439*
440* End of CSYTRS
441*
442 END
subroutine xerbla(srname, info)
Definition cblat2.f:3285
subroutine cgemv(trans, m, n, alpha, a, lda, x, incx, beta, y, incy)
CGEMV
Definition cgemv.f:160
subroutine cgeru(m, n, alpha, x, incx, y, incy, a, lda)
CGERU
Definition cgeru.f:130
subroutine csytrs(uplo, n, nrhs, a, lda, ipiv, b, ldb, info)
CSYTRS
Definition csytrs.f:120
subroutine cscal(n, ca, cx, incx)
CSCAL
Definition cscal.f:78
subroutine cswap(n, cx, incx, cy, incy)
CSWAP
Definition cswap.f:81 | 4,358 | 11,472 | {"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} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.418845 |
https://philosophy.stackexchange.com/questions/55809/can-anything-be-less-than-one | 1,726,141,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00868.warc.gz | 430,458,682 | 44,456 | # Can anything be less than one?
Zero itself seems to be an absurd number because if there is really zero of something, then nobody has ever sensed it. But even with temperatures, we don’t really have negative and positive Fahrenheits- because the coldest temperature is the impossible to reach 0 Kelvin; or the point where there is absolutely zero atoms moving.
It’s similar with electricity. Although complex numbers and multidirectional number lines are very useful when dealing with positive and negative ions, negative ions are not really less than zero ions because they are a positive amount of electrons.
I also don’t see how fractions can be less than one, because if we have 1/2 an apple, we actually still have one piece/set of all the other positively measurable substances that make an apple an apple. If we continue to divide, eventually we will lose the apple and our object will become one of whatever object it has become. Like if we take the two hydrogen atoms out of water, our one substance becomes oxygen.
Can any thing or substance be less than one?
If zero is a possible concept, can anything ever be less than zero?
• This question is ill-posed as it does not formally define how one is to understand a concept being a number, which for non-numbers seems absurd. Commented Sep 26, 2018 at 15:00
• What about a negative bank account ? Do you think it is impossible ? Commented Sep 26, 2018 at 15:03
• Mathematically, this makes zero sense because a number is what we say it is. In science, we find zero, negative numbers, and complex numbers extremely useful in mathematical models. I'm not seeing a context in which it does make sense offhand. There's a question here, but I really don't know how to approach it. Commented Sep 26, 2018 at 15:10
• @CarlMasens Do you mean that I need to clarify why I even believe positive numbers are a useful tool that can adequately measure real “things”? Commented Sep 26, 2018 at 15:11
• @MauroALLEGRANZA Yes a negative bank account is impossible to physically exist because if the government takes \$1000 out of my account, but I only had \$500, I don’t actually possess -\$500. I have \$0, the government has \$1000, and they want to believe that I will pay back the \$500 to the bank they took the money from. If I die the next day, there is no -\$500 for someone to find in my piggy bank. Commented Sep 26, 2018 at 15:16
Your question is paralleled by the reaction of the Roman world to Indian ('Arabic') numerals. Accountancy was done in Roman numerals until the 1800s, exactly because of suspicions about the 'realness' of zero. While mathematicians just got on with using the far more powerful and compact Indian numbers. They can be proven to be equivalent, so it just comes down to convenience, like decimals vs fractions.
Have a look at How The Laws Of Physics Lie, on how we seek abstractions that have isomorphic properties to reality but are tractable http://www.oxfordscholarship.com/mobile/view/10.1093/0198247044.001.0001/acprof-9780198247043
You dismiss zero on the Fahrenheit scale, even though it existed some hundreds of years before Kelvin scale, which you imply was even so 'fundamental'. All the temperature scales are actually about reference points of standardised materials, and defining movement from them. We now use the triple-point of water and absolute zero.
Zero, and imaginary numbers, another common stumbling block to intuition, are not important for their ontic transcendental existence, but for their use in logical definable systems for communication that have utility for models with isomorphic properties to reality. But any time reality differs, it is the ultimate authority. We just use the maths for clues.
Absolutely all numbers (not only real or positive), define subjective boundaries. Therefore, out there, positive, negative, fractions, imaginary, etc., are just subjective ideas aimed to discretize nature. In consequence, factually, there's no "less than one", because there is not even a "one". Out there, it is all interaction.
This requires explanation.
• All numbers define subjective boundaries. Object are just huge bunches of particles, exactly like clouds. Everything can be compared to a cloud. We perceive a cloud (with our eyes), we perceive an apple (with our hands), etc. But in fact, we are just assigning borders to clouds, to apples. Such borders are our definition of thing, an that's what we count: a boundary. What is the number of clouds in a rainy day? Depends on the observer. If apple A can be considered as "1 apple", and apple B is a bit smaller, is it considered as "0.983876 apples"? No, it is just another "1 apple".
• All numbers are subjective ideas helping discretize nature. Ok, you have "1 apple", but we all know that macroscopic nature is not discrete. So, if you move the decimal point one zero to the right, you will have 10 times "0.1 apples". And what is "0.1 apples"? It is just another boundaries definition! Whatever times the decimal point is moved, we are always discretizing nature!!! Real nature would be expressed without decimal points. Think on that.
• Our perception of phenomena rules, but does not correspond with the noumena. In simple words, ALL numbers are just ideas.
Therefore, factually, there's no "less than one", because there's not even a "one".
If you think everything as positive, you will need at least TWO formulas to perform each calculation (e.g. "Use m=kU-I if the current flows upwards and m=kU+I if the current flows downwards"). That's what you're suggesting.
You're asking a philosophical question, based on examples from physics and mathematics. There's an important distinction between the latter two:
Mathematics is not a science - it is the language of science, but math is based on axioms (i.e. a statement that is taken to be true, to serve as a premise or starting point for further reasoning and arguments). It's precisely this fact that excludes math from being a scientific discipline: there are no axioms in science. Everything in science is based on theorization, experimentation & observation. Just like any other language is an approximative description of reality, so is math.
If you apply the axioms of math to physics or real life, this often gives rise to conclusions that do not make sense. For example, if you have to divide one apple among no people, you still have one apple, not an infinite number of apples as the axiom x/0 = infinite dictates
For science - and in particular physics, math is the closest approximation we have conceived to describe nature - but it is still not nature. Take for example Einstein's equation that describes the change in mass in relation to velocity:
If you take this literally on mathematical axiom, this would imply that mass becomes infinite when a particle reaches light speed. In physics, this is however taken as: particles with an initial rest mass CAN NOT REACH light speed.
Finally, the concept "less than one" depends entirely on the scale you use. For example, you can write 0.001 as 10^(-3) = 10 to the power of -3.
• Didn't Newton base his Principia on his three very famous axioms? And aren't the axioms of mathematics determined for pragmatic reasons? See Maddy, Believing the Axioms. Commented Sep 26, 2018 at 18:41
• You have to remember that in Newton's time virtually everyone was religious. So Newton used the term axiom because he believed there was a prime mover. In 20th century science, we called them the Laws of Motion, if they were formulated today, we would call them the Theory of motion. But even so, Newton's Principia is based on observational data. It's not a mathematical axiom you just have to accept without proof. Commented Sep 26, 2018 at 18:51
• How are the axioms of mathematics agreed on? Are they arbitrary? Or based on thousands of years of observation plus some general principles? Again, Maddy. Your understanding of the axioms of set theory would be greatly enhanced by some reading. cs.umd.edu/~gasarch/BLOGPAPERS/belaxioms1.pdf Commented Sep 26, 2018 at 20:25
• @user4894, x/0 is an axiom because whatever we may answer, we will then have to agree that that answer times 0 equals to x, and that cannot be true, because anything times 0 is 0. As long as that "need for agreement" remains the case for a single axiom in math, it remains axiomatic and therefore not a science. Commented Oct 1, 2018 at 17:16
• If you divide four apples among two people, there's still four apples, and there's two apples per person. If you divide one apple among no people, there's still one apple, and infinitely many apples per person. Similarly, the Vatican contains six popes per square mile. You can get amusing results from inappropriate measurements. Commented Oct 11, 2018 at 19:44
Let's consider electrical charge. We have positive and negative charges, and both are actual things. Positive charge is not the absence of negative charge, and vice versa. Yet positive and negative charges cancel each other. Given an atom of helium, we have two protons and two electrons, so two positive charges (OK, six, if you want to give quarks integral charges), and two (six) negative charges, for a total charge of 0.
Now, the designations of "positive" and "negative" are arbitrary, but the relationship isn't. No matter what you call them, you're going to determine the amount of charge by subtracting the number of one kind form the number of the other kind, and so the natural way to express this is to declare one to be positive and one to be negative, and to use zero charge if the positive and negative charges are equal in number.
Zero itself seems to be an absurd number...
I also don’t see how fractions can be less than one...
If zero is a possible concept, can anything ever be less than zero?
OMG!
Zero, fractions and negative numbers have proven to be very useful numbers in modern mathematics, science, engineering, commerce and daily life. They actually work. Their use is so widespread that for their existence to be seriously questioned would take more than some anonymous guy on the internet saying he just doesn't get it.
At this point, you would actually have demonstrate that assuming the existence of these numbers inevitably leads to some kind logical contradiction. This would require you to learn some math and to do some actual work. You knew there was a catch, didn't you? | 2,325 | 10,396 | {"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} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.95443 |
http://dailybrainteaser.blogspot.com/2013/07/logic-thinking-riddle.html | 1,484,570,711,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00494-ip-10-171-10-70.ec2.internal.warc.gz | 65,952,671 | 23,611 | ## Search This Blog
### Logic Thinking Riddle
Logic Thinking Riddle - 20 July
I have a clock(12 hour format) and both the needles of clock overlaps at 12:00.
After how much time, they will overlap again ?
1. Will overlap at many stages
clock timing when they will overlap again
01.05
02.10
03-15
04-20
05-25
06-30
07-35
08-40
09-45
10-50
11-45
and last again on 12 o'clock
1. The overlap does not happen exactly on 3-15, 4-20 or 5-25 and so on. For example at 3-15 the minute hand will be at 3 but the hour hand will move little ahead of 3. Same way at 6-30, the hour hand will be ahead of the minute hand, not overlapping. Formula for exact time for overlap is: 12n/11 n
2. exact aftr 12 hrs
after 1 hour and 5 minutes.
3. 1st time 01:05:04
Generalised solution is 12n/11 n={1,2,3...}
4. 1 hour and 5 or 6 minutes.
5. just after 1/999999999999 th seconds
6. After one Hour fifty minutes
7. 1hr 60/11 mins...
8. after 1 hour and five minutes. | 318 | 956 | {"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-2017-04 | longest | en | 0.930177 |
https://www.ozgrid.com/forum/index.php?thread/1228582-ranking-formula/ | 1,620,352,334,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00117.warc.gz | 951,212,267 | 14,903 | # Ranking formula
• Have problem in ranking, been getting lots of ties, rank formula works great off date but need to add tie breaker using birthday month and day?
added count if but could not figure it out?
currently using =RANK(C5,C\$5:C\$976,1)Sample of ranking.xlsx
put conditional format on attachment so you can see all the ties
thank you
• Where is the tie breaker information?
Rory
Theory is when you know something, but it doesn’t work. Practice is when something works, but you don’t know why. Programmers combine theory and practice: nothing works and they don’t know why
• Sample of ranking.xlsx
the problem I am having is breaking ties using the birthday month and day, ranking working good but have tie that i can't seem to fix?
• Still no birth date info in that file.
Rory
Theory is when you know something, but it doesn’t work. Practice is when something works, but you don’t know why. Programmers combine theory and practice: nothing works and they don’t know why
we only use month and day to brake tiesSample of ranking.xlsx
• Try something like this:
=RANK(C5,\$C\$5:\$C\$386,1)+SUMPRODUCT((\$C\$5:\$C\$386=C5)*((DATE(2004,MONTH(\$D\$5:\$D\$386),DAY(\$D\$5:\$D\$386)))<(DATE(2004,MONTH(D5),DAY(D5)))))
Rory
Theory is when you know something, but it doesn’t work. Practice is when something works, but you don’t know why. Programmers combine theory and practice: nothing works and they don’t know why
• when I try that i changes rank to date all ending with 2000?
any excell way to change rank numbers base on the birthday, i.e if both ar ranked at 201 the one with the elearst birthday would remain 201 and the tie would go to 202 and so on?
• Reformat the cells as General. What you describe is what that formula does.
Rory
Theory is when you know something, but it doesn’t work. Practice is when something works, but you don’t know why. Programmers combine theory and practice: nothing works and they don’t know why
• that did it! thank you so much for you time and help! | 504 | 2,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.909382 |
https://gmatclub.com/forum/when-to-start-the-reapplication-process-146536.html?fl=similar | 1,511,042,833,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805049.34/warc/CC-MAIN-20171118210145-20171118230145-00222.warc.gz | 626,964,285 | 39,065 | It is currently 18 Nov 2017, 15:07
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# When to start the reapplication process?
Author Message
Intern
Joined: 30 Jan 2013
Posts: 7
Kudos [?]: 2 [1], given: 0
When to start the reapplication process? [#permalink]
### Show Tags
30 Jan 2013, 14:26
1
KUDOS
My question is simple: I was denied round 1 and want to reapply to same schools plus some new ones. Do I need to get the reapplications in in Round 1? I was assured many times over that I was a strong candidate for the schools I applied to so will be attempting those again, and then adding in more back ups. Planning on six schools total. Thank you for your advice and for all of the time you spend on this forum.
Kudos [?]: 2 [1], given: 0
Joined: 25 Jan 2010
Posts: 1037
Kudos [?]: 222 [0], given: 220
Re: When to start the reapplication process? [#permalink]
### Show Tags
16 Apr 2013, 03:09
Hey bsklman,
Thanks for reaching out to me. At face value your question is simple, and the simplistic answer to your question is "yes." However, I don't think its as simple as that and I wanted to also make you aware of other issues you need to consider as a re-applicant - and they are not necessarily dependent on timing (R1 vs. R2). More importantly, have you gained any concrete insight into why you were rejected? For real, if you were "assured" that you were a strong candidate, what happened then? And what have you done to improve your standing. If you're just re-applying based on an assurance that did not prove to be sufficient, then history will repeat itself.
In general, your re-applicant status generally place you at a disadvantage. I always believe that re-applicants are behind the 8-ball. That is, they must go beyond the norm to show the adcom what they have done to improve their candidacy. However, the extent of this is specific to the individual programs you are applying to. I'm not sure how many of my other posts you have read, but I always state that you have to examine the actual re-Application form. How many essays do they allow you to fill out? In this case, but more the merrier. With more essays, you simply have a better chance of telling the admissions committee a story that makes sense for that school - this time around. Also consider this - the more re-applicant friendly adcoms can be found by their respective definition of "re-applicant." That is, how many years will they hold out as the re-applicant period.
Another consideration, were you waitlisted at any of the schools you applied to? Were you even interviewed? Being wait-listed is a good indication that your essays were probably pretty solid and that the adcom deemed you worthy of admittance, but you just needed a bit more of something to break on through to the other side. So what could that "bit more of something" actually be? However, if you were not waitlisted any where, you could have some flaws in what you are telling the adcom. Have you gotten any feedback on your essays from current or former MBA students? Please say "yes."
One other thing - for the schools where you could receive feedback from the adcom, did you elect to do so?
Anyway, there is no sense going into this year with a false sense of confidence. Figure out what went wrong - objectively - fix it and then reapply.
Respectfully,
Paul Lanzillotti
bsklman wrote:
My question is simple: I was denied round 1 and want to reapply to same schools plus some new ones. Do I need to get the reapplications in in Round 1? I was assured many times over that I was a strong candidate for the schools I applied to so will be attempting those again, and then adding in more back ups. Planning on six schools total. Thank you for your advice and for all of the time you spend on this forum.
_________________
Paul Lanzillotti | Founder| About | mba@amerasiaconsulting.com | 877.866.9251
Schedule a Consultation | Twitter | Blog
Kudos [?]: 222 [0], given: 220
Display posts from previous: Sort by | 1,068 | 4,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-47 | latest | en | 0.961779 |
http://entertainmentmathematics.nl/Entertainment/Excel/MagicSquares/MgcSqr8c2.html | 1,603,448,833,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00674.warc.gz | 37,151,700 | 5,063 | ' Generates Magic Squares composed of Associated Magic Sub Squares
' Tested with Office 2007 under Windows 7
```Sub MgcSqr8c2()
Dim a(64), b(64), c(64), a1(16, 4), m(2, 4), b1(4)
y = MsgBox("Locked", vbCritical, "Routine MgcSqr8c2")
End
n2 = 0: n9 = 0: k1 = 1: k2 = 1
m1 = 1: m2 = 64: s1 = 260
t1 = Timer
' Define Ranges
a1(1, 1) = 1
For j1 = 1 To 16
If j1 <> 1 Then a1(j1, 1) = a1(j1 - 1, 1) + 4
For j2 = 1 To 4
If j2 <> 1 Then a1(j1, j2) = a1(j1, j2 - 1) + 1
If a1(j1, j2) > 64 Then a1(j1, j2) = a1(j1, j2) - 64
Next j2
Next j1
' Generate Squares
For j100 = 1 To 16
For j200 = 1 To 16
If j100 = j200 Then GoTo 2000
For j300 = 1 To 16
If j300 = j100 Or j300 = j200 Then GoTo 3000
For j400 = 1 To 16
If j400 = j100 Or j400 = j200 Or j400 = j300 Then GoTo 4000
m(1, 1) = a1(j100, 1): m(2, 1) = a1(j100, 4) 'Consecutive Set 1
m(1, 2) = a1(j200, 1): m(2, 2) = a1(j200, 4) 'Consecutive Set 2
m(1, 3) = a1(j300, 1): m(2, 3) = a1(j300, 4) 'Consecutive Set 3
m(1, 4) = a1(j400, 1): m(2, 4) = a1(j400, 4) 'Consecutive Set 4
For j64 = m(1, 2) To m(2, 2) 'a(64) = 61
If b(j64) = 0 Then b(j64) = j64: c(64) = j64 Else GoTo 640
a(64) = j64
a(37) = 0.25 * s1 - a(64)
If a(37) < m1 Or a(37) > m2 Then GoTo 370:
If b(a(37)) = 0 Then b(a(37)) = a(37): c(37) = a(37) Else GoTo 370
For j63 = m(1, 3) To m(2, 3) 'a(63) = 7
If b(j63) = 0 Then b(j63) = j63: c(63) = j63 Else GoTo 630
a(63) = j63
a(38) = 0.25 * s1 - a(63)
If a(38) < m1 Or a(38) > m2 Then GoTo 380:
If b(a(38)) = 0 Then b(a(38)) = a(38): c(38) = a(38) Else GoTo 380
For j62 = m(1, 4) To m(2, 4) 'a(62) = 10
If b(j62) = 0 Then b(j62) = j62: c(62) = j62 Else GoTo 620
a(62) = j62
a(61) = 0.5 * s1 - a(62) - a(63) - a(64)
If a(61) < m1 Or a(61) > m2 Then GoTo 610
If b(a(61)) = 0 Then b(a(61)) = a(61): c(61) = a(61) Else GoTo 610
a(40) = -0.25 * s1 + a(62) + a(63) + a(64)
If a(40) < m1 Or a(40) > m2 Then GoTo 400:
If b(a(40)) = 0 Then b(a(40)) = a(40): c(40) = a(40) Else GoTo 400
a(39) = 0.25 * s1 - a(62)
If a(39) < m1 Or a(39) > m2 Then GoTo 390:
If b(a(39)) = 0 Then b(a(39)) = a(39): c(39) = a(39) Else GoTo 390
For j60 = m(1, 2) To m(2, 2) 'a(60) = 62
If b(j60) = 0 Then b(j60) = j60: c(60) = j60 Else GoTo 600
a(60) = j60
a(33) = 0.25 * s1 - a(60)
If a(33) < m1 Or a(33) > m2 Then GoTo 330:
If b(a(33)) = 0 Then b(a(33)) = a(33): c(33) = a(33) Else GoTo 330
For j59 = m(1, 3) To m(2, 3) 'a(59) = 8
If b(j59) = 0 Then b(j59) = j59: c(59) = j59 Else GoTo 590
a(59) = j59
a(34) = 0.25 * s1 - a(59)
If a(34) < m1 Or a(34) > m2 Then GoTo 340:
If b(a(34)) = 0 Then b(a(34)) = a(34): c(34) = a(34) Else GoTo 340
For j58 = m(1, 4) To m(2, 4) 'a(58) = 11
If b(j58) = 0 Then b(j58) = j58: c(58) = j58 Else GoTo 580
a(58) = j58
a(57) = 0.5 * s1 - a(58) - a(59) - a(60)
If a(57) < m1 Or a(57) > m2 Then GoTo 570:
If b(a(57)) = 0 Then b(a(57)) = a(57): c(57) = a(57) Else GoTo 570
a(36) = -0.25 * s1 + a(58) + a(59) + a(60)
If a(36) < m1 Or a(36) > m2 Then GoTo 360:
If b(a(36)) = 0 Then b(a(36)) = a(36): c(36) = a(36) Else GoTo 360
a(35) = 0.25 * s1 - a(58)
If a(35) < m1 Or a(35) > m2 Then GoTo 350:
If b(a(35)) = 0 Then b(a(35)) = a(35): c(35) = a(35) Else GoTo 350
For j56 = m(1, 1) To m(2, 1) 'a(56) = 20
If b(j56) = 0 Then b(j56) = j56: c(56) = j56 Else GoTo 560
a(56) = j56
a(55) = 0.5 * s1 - a(56) - a(63) - a(64)
If a(55) < m1 Or a(55) > m2 Then GoTo 550:
If b(a(55)) = 0 Then b(a(55)) = a(55): c(55) = a(55) Else GoTo 550
a(54) = 0.5 * s1 - a(56) - a(62) - a(64)
If a(54) < m1 Or a(54) > m2 Then GoTo 540:
If b(a(54)) = 0 Then b(a(54)) = a(54): c(54) = a(54) Else GoTo 540
a(53) = -0.5 * s1 + a(56) + a(62) + a(63) + 2 * a(64)
If a(53) < m1 Or a(53) > m2 Then GoTo 530:
If b(a(53)) = 0 Then b(a(53)) = a(53): c(53) = a(53) Else GoTo 530
a(48) = 0.75 * s1 - a(56) - a(62) - a(63) - 2 * a(64)
If a(48) < m1 Or a(48) > m2 Then GoTo 480:
If b(a(48)) = 0 Then b(a(48)) = a(48): c(48) = a(48) Else GoTo 480
a(47) = -0.25 * s1 + a(56) + a(62) + a(64)
If a(47) < m1 Or a(47) > m2 Then GoTo 470:
If b(a(47)) = 0 Then b(a(47)) = a(47): c(47) = a(47) Else GoTo 470
a(46) = -0.25 * s1 + a(56) + a(63) + a(64)
If a(46) < m1 Or a(46) > m2 Then GoTo 460:
If b(a(46)) = 0 Then b(a(46)) = a(46): c(46) = a(46) Else GoTo 460
a(45) = 0.25 * s1 - a(56)
If a(45) < m1 Or a(45) > m2 Then GoTo 450:
If b(a(45)) = 0 Then b(a(45)) = a(45): c(45) = a(45) Else GoTo 450
For j52 = m(1, 1) To m(2, 1) 'a(52) = 17
If b(j52) = 0 Then b(j52) = j52: c(52) = j52 Else GoTo 520
a(52) = j52
a(51) = 0.5 * s1 - a(52) - a(59) - a(60)
If a(51) < m1 Or a(51) > m2 Then GoTo 510:
If b(a(51)) = 0 Then b(a(51)) = a(51): c(51) = a(51) Else GoTo 510
a(50) = 0.5 * s1 - a(52) - a(58) - a(60)
If a(50) < m1 Or a(50) > m2 Then GoTo 500:
If b(a(50)) = 0 Then b(a(50)) = a(50): c(50) = a(50) Else GoTo 500
a(49) = -0.5 * s1 + a(52) + a(58) + a(59) + 2 * a(60)
If a(49) < m1 Or a(49) > m2 Then GoTo 490:
If b(a(49)) = 0 Then b(a(49)) = a(49): c(49) = a(49) Else GoTo 490
a(44) = 0.75 * s1 - a(52) - a(58) - a(59) - 2 * a(60)
If a(44) < m1 Or a(44) > m2 Then GoTo 440:
If b(a(44)) = 0 Then b(a(44)) = a(44): c(44) = a(44) Else GoTo 440
a(43) = -0.25 * s1 + a(52) + a(58) + a(60)
If a(43) < m1 Or a(43) > m2 Then GoTo 430:
If b(a(43)) = 0 Then b(a(43)) = a(43): c(43) = a(43) Else GoTo 430
a(42) = -0.25 * s1 + a(52) + a(59) + a(60)
If a(42) < m1 Or a(42) > m2 Then GoTo 420:
If b(a(42)) = 0 Then b(a(42)) = a(42): c(42) = a(42) Else GoTo 420
a(41) = 0.25 * s1 - a(52)
If a(41) < m1 Or a(41) > m2 Then GoTo 410:
If b(a(41)) = 0 Then b(a(41)) = a(41): c(41) = a(41) Else GoTo 410
For j32 = m(1, 2) To m(2, 2) 'a(32) = 63
If b(j32) = 0 Then b(j32) = j32: c(32) = j32 Else GoTo 320
a(32) = j32
a(5) = 0.25 * s1 - a(32)
If a(5) < m1 Or a(5) > m2 Then GoTo 50:
If b(a(5)) = 0 Then b(a(5)) = a(5): c(5) = a(5) Else GoTo 50
For j31 = m(1, 3) To m(2, 3) 'a(31) = 5
If b(j31) = 0 Then b(j31) = j31: c(31) = j31 Else GoTo 310
a(31) = j31
a(6) = 0.25 * s1 - a(31)
If a(6) < m1 Or a(6) > m2 Then GoTo 60:
If b(a(6)) = 0 Then b(a(6)) = a(6): c(6) = a(6) Else GoTo 60
For j30 = m(1, 4) To m(2, 4) 'a(30) = 12
If b(j30) = 0 Then b(j30) = j30: c(30) = j30 Else GoTo 300
a(30) = j30
a(29) = 0.5 * s1 - a(30) - a(31) - a(32)
If a(29) < m1 Or a(29) > m2 Then GoTo 290:
If b(a(29)) = 0 Then b(a(29)) = a(29): c(29) = a(29) Else GoTo 290
a(8) = -0.25 * s1 + a(30) + a(31) + a(32)
If a(8) < m1 Or a(8) > m2 Then GoTo 80:
If b(a(8)) = 0 Then b(a(8)) = a(8): c(8) = a(8) Else GoTo 80
a(7) = 0.25 * s1 - a(30)
If a(7) < m1 Or a(7) > m2 Then GoTo 70:
If b(a(7)) = 0 Then b(a(7)) = a(7): c(7) = a(7) Else GoTo 70
For j28 = m(1, 2) To m(2, 2) 'a(28) = 64
If b(j28) = 0 Then b(j28) = j28: c(28) = j28 Else GoTo 280
a(28) = j28
a(1) = 0.25 * s1 - a(28)
If a(1) < m1 Or a(1) > m2 Then GoTo 10:
If b(a(1)) = 0 Then b(a(1)) = a(1): c(1) = a(1) Else GoTo 10
For j27 = m(1, 3) To m(2, 3) 'a(27) = 6
If b(j27) = 0 Then b(j27) = j27: c(27) = j27 Else GoTo 270
a(27) = j27
a(2) = 0.25 * s1 - a(27)
If a(2) < m1 Or a(2) > m2 Then GoTo 20:
If b(a(2)) = 0 Then b(a(2)) = a(2): c(2) = a(2) Else GoTo 20
For j26 = m(1, 4) To m(2, 4) 'a(26) = 9
If b(j26) = 0 Then b(j26) = j26: c(26) = j26 Else GoTo 260
a(26) = j26
a(25) = 0.5 * s1 - a(26) - a(27) - a(28)
If a(25) < m1 Or a(25) > m2 Then GoTo 250:
If b(a(25)) = 0 Then b(a(25)) = a(25): c(25) = a(25) Else GoTo 250
a(4) = -0.25 * s1 + a(26) + a(27) + a(28)
If a(4) < m1 Or a(4) > m2 Then GoTo 40:
If b(a(4)) = 0 Then b(a(4)) = a(4): c(4) = a(4) Else GoTo 40
a(3) = 0.25 * s1 - a(26)
If a(3) < m1 Or a(3) > m2 Then GoTo 30:
If b(a(3)) = 0 Then b(a(3)) = a(3): c(3) = a(3) Else GoTo 30
For j24 = m(1, 1) To m(2, 1) 'a(24) = 18
If b(j24) = 0 Then b(j24) = j24: c(24) = j24 Else GoTo 240
a(24) = j24
a(23) = 0.5 * s1 - a(24) - a(31) - a(32)
If a(23) < m1 Or a(23) > m2 Then GoTo 230:
If b(a(23)) = 0 Then b(a(23)) = a(23): c(23) = a(23) Else GoTo 230
a(22) = 0.5 * s1 - a(24) - a(30) - a(32)
If a(22) < m1 Or a(22) > m2 Then GoTo 220:
If b(a(22)) = 0 Then b(a(22)) = a(22): c(22) = a(22) Else GoTo 220
a(21) = -0.5 * s1 + a(24) + a(30) + a(31) + 2 * a(32)
If a(21) < m1 Or a(21) > m2 Then GoTo 210:
If b(a(21)) = 0 Then b(a(21)) = a(21): c(21) = a(21) Else GoTo 210
a(16) = 0.75 * s1 - a(24) - a(30) - a(31) - 2 * a(32)
If a(16) < m1 Or a(16) > m2 Then GoTo 160:
If b(a(16)) = 0 Then b(a(16)) = a(16): c(16) = a(16) Else GoTo 160
a(15) = -0.25 * s1 + a(24) + a(30) + a(32)
If a(15) < m1 Or a(15) > m2 Then GoTo 150:
If b(a(15)) = 0 Then b(a(15)) = a(15): c(15) = a(15) Else GoTo 150
a(14) = -0.25 * s1 + a(24) + a(31) + a(32)
If a(14) < m1 Or a(14) > m2 Then GoTo 140:
If b(a(14)) = 0 Then b(a(14)) = a(14): c(14) = a(14) Else GoTo 140
a(13) = 0.25 * s1 - a(24)
If a(13) < m1 Or a(13) > m2 Then GoTo 130:
If b(a(13)) = 0 Then b(a(13)) = a(13): c(13) = a(13) Else GoTo 130
For j20 = m(1, 1) To m(2, 1) 'a(20) = 19
If b(j20) = 0 Then b(j20) = j20: c(20) = j20 Else GoTo 200
a(20) = j20
a(19) = 0.5 * s1 - a(20) - a(27) - a(28)
If a(19) < m1 Or a(19) > m2 Then GoTo 190:
If b(a(19)) = 0 Then b(a(19)) = a(19): c(19) = a(19) Else GoTo 190
a(18) = 0.5 * s1 - a(20) - a(26) - a(28)
If a(18) < m1 Or a(18) > m2 Then GoTo 180:
If b(a(18)) = 0 Then b(a(18)) = a(18): c(18) = a(18) Else GoTo 180
a(17) = -0.5 * s1 + a(20) + a(26) + a(27) + 2 * a(28)
If a(17) < m1 Or a(17) > m2 Then GoTo 170:
If b(a(17)) = 0 Then b(a(17)) = a(17): c(17) = a(17) Else GoTo 170
a(12) = 0.75 * s1 - a(20) - a(26) - a(27) - 2 * a(28)
If a(12) < m1 Or a(12) > m2 Then GoTo 120:
If b(a(12)) = 0 Then b(a(12)) = a(12): c(12) = a(12) Else GoTo 120
a(11) = -0.25 * s1 + a(20) + a(26) + a(28)
If a(11) < m1 Or a(11) > m2 Then GoTo 110:
If b(a(11)) = 0 Then b(a(11)) = a(11): c(11) = a(11) Else GoTo 110
a(10) = -0.25 * s1 + a(20) + a(27) + a(28)
If a(10) < m1 Or a(10) > m2 Then GoTo 100:
If b(a(10)) = 0 Then b(a(10)) = a(10): c(10) = a(10) Else GoTo 100
a(9) = 0.25 * s1 - a(20)
If a(9) < m1 Or a(9) > m2 Then GoTo 90:
If b(a(9)) = 0 Then b(a(9)) = a(9): c(9) = a(9) Else GoTo 90
' Double Check Consecutive Integers
GoSub 900: If fl1 = 0 Then GoTo 800
n9 = n9 + 1
GoSub 650 'Print results (squares)
' GoSub 645 'Print results (selected numbers)
Erase b, c: GoTo 4000 'Print only first square (Option)
800 b(c(9)) = 0: c(9) = 0
90 b(c(10)) = 0: c(10) = 0
100 b(c(11)) = 0: c(11) = 0
110 b(c(12)) = 0: c(12) = 0
120 b(c(17)) = 0: c(17) = 0
170 b(c(18)) = 0: c(18) = 0
180 b(c(19)) = 0: c(19) = 0
190 b(c(20)) = 0: c(20) = 0
200 Next j20
b(c(13)) = 0: c(13) = 0
130 b(c(14)) = 0: c(14) = 0
140 b(c(15)) = 0: c(15) = 0
150 b(c(16)) = 0: c(16) = 0
160 b(c(21)) = 0: c(21) = 0
210 b(c(22)) = 0: c(22) = 0
220 b(c(23)) = 0: c(23) = 0
230 b(c(24)) = 0: c(24) = 0
240 Next j24
b(c(3)) = 0: c(3) = 0
30 b(c(4)) = 0: c(4) = 0
40 b(c(25)) = 0: c(25) = 0
250 b(c(26)) = 0: c(26) = 0
260 Next j26
b(c(2)) = 0: c(2) = 0
20 b(c(27)) = 0: c(27) = 0
270 Next j27
b(c(1)) = 0: c(1) = 0
10 b(c(28)) = 0: c(28) = 0
280 Next j28
b(c(7)) = 0: c(7) = 0
70 b(c(8)) = 0: c(8) = 0
80 b(c(29)) = 0: c(29) = 0
290 b(c(30)) = 0: c(30) = 0
300 Next j30
b(c(6)) = 0: c(6) = 0
60 b(c(31)) = 0: c(31) = 0
310 Next j31
b(c(5)) = 0: c(5) = 0
50 b(c(32)) = 0: c(32) = 0
320 Next j32
b(c(41)) = 0: c(41) = 0
410 b(c(42)) = 0: c(42) = 0
420 b(c(43)) = 0: c(43) = 0
430 b(c(44)) = 0: c(44) = 0
440 b(c(49)) = 0: c(49) = 0
490 b(c(50)) = 0: c(50) = 0
500 b(c(51)) = 0: c(51) = 0
510 b(c(52)) = 0: c(52) = 0
520 Next j52
b(c(45)) = 0: c(45) = 0
450 b(c(46)) = 0: c(46) = 0
460 b(c(47)) = 0: c(47) = 0
470 b(c(48)) = 0: c(48) = 0
480 b(c(53)) = 0: c(53) = 0
530 b(c(54)) = 0: c(54) = 0
540 b(c(55)) = 0: c(55) = 0
550 b(c(56)) = 0: c(56) = 0
560 Next j56
b(c(35)) = 0: c(35) = 0
350 b(c(36)) = 0: c(36) = 0
360 b(c(57)) = 0: c(57) = 0
570 b(c(58)) = 0: c(58) = 0
580 Next j58
b(c(34)) = 0: c(34) = 0
340 b(c(59)) = 0: c(59) = 0
590 Next j59
b(c(33)) = 0: c(33) = 0
330 b(c(60)) = 0: c(60) = 0
600 Next j60
b(c(39)) = 0: c(39) = 0
390 b(c(40)) = 0: c(40) = 0
400 b(c(61)) = 0: c(61) = 0
610 b(c(62)) = 0: c(62) = 0
620 Next j62
b(c(38)) = 0: c(38) = 0
380 b(c(63)) = 0: c(63) = 0
630 Next j63
b(c(37)) = 0: c(37) = 0
370 b(c(64)) = 0: c(64) = 0
640 Next j64
4000 Next j400
3000 Next j300
2000 Next j200
Next j100
t2 = Timer
t10 = Str(t2 - t1) + " sec., " + Str(n9) + " Solutions for sum" + Str(s1)
y = MsgBox(t10, 0, "Routine MgcSqr8c2")
End
' Print results (selected numbers)
645 For i1 = 1 To 64
Cells(n9, i1).Value = a(i1)
Next i1
Return
' Print results (squares)
650 n2 = n2 + 1
If n2 = 5 Then
n2 = 1: k1 = k1 + 9: k2 = 1
Else
If n9 > 1 Then k2 = k2 + 9
End If
Cells(k1, k2 + 1).Select
Range(Cells(k1, k2 + 1), Cells(k1, k2 + 4)).Font.Color = -4165632
' Cells(k1, k2 + 1).Value = n9 'Option (Fixed Ranges)
Cells(k1, k2 + 1).Value = j100
Cells(k1, k2 + 2).Value = j200
Cells(k1, k2 + 3).Value = j300
Cells(k1, k2 + 4).Value = j400
i3 = 0
For i1 = 1 To 8
For i2 = 1 To 8
i3 = i3 + 1
Cells(k1 + i1, k2 + i2).Value = a(i3)
Next i2
Next i1
Return
' Double Check Consecutive Integers
900 fl1 = 1
b1(1) = a(1): b1(2) = a(5): b1(3) = a(33): b1(4) = a(37): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(2): b1(2) = a(6): b1(3) = a(34): b1(4) = a(38): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(3): b1(2) = a(7): b1(3) = a(35): b1(4) = a(39): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(4): b1(2) = a(8): b1(3) = a(36): b1(4) = a(40): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(9): b1(2) = a(13): b1(3) = a(41): b1(4) = a(45): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(10): b1(2) = a(14): b1(3) = a(42): b1(4) = a(46): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(11): b1(2) = a(15): b1(3) = a(43): b1(4) = a(47): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(12): b1(2) = a(16): b1(3) = a(44): b1(4) = a(48): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(17): b1(2) = a(21): b1(3) = a(49): b1(4) = a(53): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(18): b1(2) = a(22): b1(3) = a(50): b1(4) = a(54): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(19): b1(2) = a(23): b1(3) = a(51): b1(4) = a(55): GoSub 950: If fl1 = 0 Then Return
b1(1) = a(25): b1(2) = a(29): b1(3) = a(57): b1(4) = a(61): GoSub 950: If fl1 = 0 Then Return
Return
' Sorteren
950 k3 = 4: fl = 1
While fl
fl = 0
For j1 = 1 To k3 - 1
If b1(j1) > b1(j1 + 1) Then
b2 = b1(j1): b1(j1) = b1(j1 + 1): b1(j1 + 1) = b2
fl = 1: GoTo 5
End If
5 Next j1
k3 = k3 - 1
Wend
' Check Consecutive Integers
For j1 = 2 To 4
If b1(j1) - b1(j1 - 1) <> 1 Then fl1 = 0: Exit For
Next j1
Return
End Sub
``` | 8,083 | 14,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.397356 |
http://mathhelpforum.com/discrete-math/18963-inductive-proof-size-power-sets-print.html | 1,526,895,281,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00319.warc.gz | 177,194,804 | 3,143 | # inductive proof of size of power sets
• Sep 14th 2007, 10:07 AM
polypus
inductive proof of size of power sets
hi everybody,
i have written an inductive proof of the statement that for any set $\displaystyle S$
$\displaystyle | \wp(S) | = 2 ^ {|S|}$
I am using my own notation for denoting set union with a disjoint set of cardinality 1, because it is such a convenience. is there some standard way of doing this? any tips/corrections, mathematical, stylistic, or notational will be highly appreciated.
Let $\displaystyle \psi(S) = |\wp(S)|$
and let
$\displaystyle S + 1$
denote $\displaystyle S \cup T$ where $\displaystyle T$ is some set disjoint from $\displaystyle S$ and where $\displaystyle |T| = 1$, so that we can write
|$\displaystyle S + 1| = |S| + 1$
The following holds trivially for $\displaystyle S = \emptyset$ and we will assume it to hold for any arbitrary set $\displaystyle S$ as well
$\displaystyle [\frac{\psi(S + 1)}{2} = 2 ^ {|S|}] \wedge [\psi(S) = 2 ^ {|S|}]$
thus the following also holds for any $\displaystyle S$
$\displaystyle \psi(S) = 2 ^ {|S|}$
we take as our inductive hypothesis that
$\displaystyle [\psi(S) = 2 ^ {|S|}] \rightarrow [\psi(S + 1) = 2^{|S + 1|}]$
recall that for any set $\displaystyle S$ we assume
$\displaystyle \frac{\psi(S + 1)}{2} = 2 ^ {|S|}$
which we can rearrange as follows
$\displaystyle \psi(S + 1) = 2 \times 2^{|S|}$
$\displaystyle \psi(S + 1) = 2^{|S| + 1}$
$\displaystyle \psi(S + 1) = 2^{|S + 1|}$
which proves the inductive step and the theorem.
thank again for any tips
• Sep 14th 2007, 10:12 AM
polypus
worries
i'm just worried that the fact that my consequent is essentially included in my antecedent invalidates the proof, correct?
• Sep 14th 2007, 11:01 AM
Plato
Sorry, but I have absolutely no idea what you have done there!
The usual way is to observe that $\displaystyle \wp (\{ 1,2,...,n,n + 1\} )$ contains $\displaystyle \wp (\{ 1,2,...,n\} )$ plus $\displaystyle \left\{ {X \cup \{ n = 1\} :X \in \wp (\{ 1,2,...,n\} )} \right\}$.
Hence $\displaystyle \wp (\{ 1,2,...,n, n+1\} )$ contains twice as many sets as $\displaystyle \wp (\{ 1,2,...,n\} )$. | 715 | 2,150 | {"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} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.828472 |
https://aspenmedisys.com/qa/is-12-mph-a-sprint.html | 1,621,338,855,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989819.92/warc/CC-MAIN-20210518094809-20210518124809-00262.warc.gz | 131,134,513 | 8,644 | # Is 12 Mph A Sprint?
## Is running 23 mph fast?
According to Fox News, humans—who top out at roughly 23 mph—may one day be able to reach phenomenal speeds of up to 40 miles per hour.
That study showed that speed is based more on contracting muscle fibers than it is negated by sheer force of running..
## Can a human run 15 mph?
“The human body is a sturdy one, but only up to a point, able to withstand collisions of about 15 miles per hour, which is about as fast as an average person can run.”
## How fast do you pee mph?
Same with urination. Men and women, big and small, all pee at the same average rate — between 1/3 and ½ ounce per second.
## How fast is 10 mph?
Keeping Track If you are exercising at 10 mph, you are running, not walking or jogging. This running speed is equivalent to a six-minute mile, meaning you can cover 10 miles in one hour if you maintain that pace.
## Is a 3 minute mile possible?
Common wisdom and scientific knowledge perceived a sub-four-minute mile to be an insurmountable barrier until Sir Roger Bannister set a world record time of 3min 59.4sec in 1954. Since then, a new world record has been set 18 times, the current record of 3:43.13 being clocked in 1999.
## Can a human run 80 mph?
Limits of speed The record was 44.72 km/h (27.78 mph), measured between meter 60 and meter 80 of the 100 meters sprint of the World Championships in Berlin on 16 August 2009 by Usain Bolt.
## Is running 7 mph fast?
Jogging is slower and less intense than running. The main differences are pace and effort. One definition of jogging speed is 4 to 6 miles per hour (mph), while running can be defined as 6 mph or more.
## Is 12 mph a fast sprint?
Yes, 15mph is very fast for a runner as it equates to a 4-minute mile. 15mph represents the fastest a human can run, with extensive training, for a distance of 1 mile to 2 miles. Sprinters can run faster, but obviously for a much shorter duration. Usain Bolt has hit 28mph in the 100m sprint.
## Can you do sprints everyday?
Both forms of exercise increase your metabolism — which is critical. Research shows that high-intensity interval training in the form of sprinting every other day can improve insulin sensitivity in men by 23%. … But, sprinting burns more fat at a higher speed — about 200 calories in 3 minutes— than running.
## Who is faster than Usain Bolt?
Buffalo racer Srinivas Gowda who is ‘faster than Usain Bolt’ is groomed for Olympic glory. A buffalo racer has become an overnight national sensation and been hailed as India’s “Usain Bolt” after claims that he smashed the 100m world record while running in a paddy field.
## Who can run 23 mph?
Usain BoltThis year’s Olympic Games means another dose of track superstar, Usain Bolt. Bolt, an Olympic multi-medalist, is the fastest man ever recorded. The Jamaican sprinter was best known for his performance in the 2008 Summer Olympics in Beijing, where he sprinted 100 meters in a 9.58 seconds. That’s about 23 miles per hour.
## How fast can a 14 year old run?
How fast should a 14 year old run 400m? A “good” time for a 14 year-old male would fall somewhere under 60 seconds (competitive, in the 56–58 second range). A “good” time for a 14 year-old female would fall somewhere under 65 seconds (competitive, in the 61–63 second range).
## Is sprinting 20 mph fast?
Yes, 20.5 miles per hour is fast for humans in general. … But with that said even for most athletes 20.5 mph is pretty damn good. Usain Bolt ran around 28 mph in his prime. Apparently there’s a video of a Michael Scott, some regional manager at a Dunder Mifflin, that shows him reaching maximum velocity at over 30 mph.
## How many mph is a sprint?
The average sprinting speed for many athletes is 24km\h (15mph). Running at that speed over 100m will give you a time of around 14 seconds. Elite athletes will be running around 26mph. Is 20 mph Fast For a Human?
## How fast is 21 mph running?
Yup, reaching a peak speed of up to 21 mph is ofcourse deadly quick. It also means that your average speed would be somewhere between 18–19 mph. In terms of racing the 100metre dash, that sort of speed would translate into a time between 11.65sec and 11.88sec (depending on how long you hold the top speed.
## Who is the fastest human?
Usain BoltIn 2009 Jamaican sprinter Usain Bolt set the world record in the 100-meter sprint at 9.58 seconds. For those of us more accustomed to sitting than sprinting, to translate this feat into terms of speed is to simply underscore the stunning nature of Bolt’s performance.
## Can a human run 25 mph?
NFL player: I pushed limits with 25-mph treadmill run Watch your back, Usain Bolt. … Bolt, the fastest man in the world and reigning Olympic 100-meter champion, topped out at 27.78 mph when he set the current 100-meter record by finishing in 9.58 seconds in Berlin in 2009.
## What is a fast sprint mph?
40 MPH: The fastest speed humans can run. The current fastest human in the world is Usain Bolt, who can run at nearly 28 miles per hour—some streets have lower speed limits than that! Bolt holds the record for the 100-meter sprint, clocking in at 9.58 seconds, reports BBC. | 1,289 | 5,157 | {"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-2021-21 | latest | en | 0.949935 |
https://www.exceldemy.com/learn-excel/formula-list/ | 1,726,077,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00198.warc.gz | 704,470,088 | 62,806 | # Excel Formula List (40+ Useful Formulas to Master Excel)
In the following image, you see some students with their marks in English. We have calculated the total marks, average marks, total no. of students, and difference between the highest and lowest marks using Excel formulas.
## What Is an Excel Formula?
An Excel formula is an expression that acts on a cell or range of cells and produces results in another cell or multiple cells.
## How to Apply a Formula in Excel?
You can apply a formula in Excel by typing an equal sign (=) in a cell and then typing the desired formula. You can put direct values in the formula as the arguments or you can use cell references. After typing the formula, press Enter (or Ctrl + Shift + Enter for formulas that use arrays in older Excel versions) to get the desired result.
We have a dataset with some student’s marks in 3 subjects. We want to get the total marks of each student in these 3 subjects. So, we have to add the marks of these 3 subjects. We can use the SUM function here.
• Select cell F6 and type the equal sign.
• Type sum.
• Excel will suggest all the available formulas related to the sum.
• We’ll use the SUM function here, so double-click on the SUM option.
• Click and drag over the cells from C6 to E6. You can also type the full formula.
• Press Enter and you’ll get the output (total marks of the first student).
• You’ll see the Fill Handle icon located at the right bottom corner of the cell.
• Double-click on the Fill Handle icon or drag it down with the mouse to get the result for all students.
## Part 1 – How to Add and Subtract in Excel
We have a dataset with some students’ marks in a subject that is divided into 3 sections (Theory, Practical and Negative). We have to add the theory and practical section’s marks and subtract the negative marks to obtain the total marks of each student.
Steps:
• Put the following formula in cell F6 and press Enter to get the total marks of the first student:
`=C6+D6-E6`
• Double-click the Fill Handle icon or drag it down with a mouse to get the result for all students.
## Part 2 – How to Multiply in Excel
We have a dataset with some products and their unit prices and quantity sold. We have to multiply the unit price with the quantity sold to obtain the sales of each product.
Steps:
• Apply the following formula in cell E6 to get the sales of the first product and then use the Fill Handle icon for all the remaining cells to copy the formula:
`=D6*C6`
## Part 3 – How to Divide in Excel
We’ll use the same dataset as before. We’ll determine the quantity sold of each product by dividing sales by unit price.
• The formula in cell E6 will be:
`=D6/C6`
## Part 4 – How to Sum in Excel
### Case 4.1 – Using the AutoSum Feature
The keyboard shortcut of the AutoSum feature is Alt + =.
We have the following dataset with some employees and their sales in 3 different months. We want to get the sales in the employee-wise sales column.
Steps:
• Select cells F6:F15.
• Go to the Home tab.
• Select the AutoSum option in the Editing group of commands.
• You’ll get all the outputs together.
### Case 4.2 – Sum Columns
We have the same employee sales dataset. We’ll calculate the total month-wise sales, so we have to sum all the columns one by one.
Steps:
• Use the following formula in cell C16 to get the total sales of April of all employees:
`=SUM(C6:C15)`
• Drag the Fill Handle icon to the right for the remaining months (May and June).
### Case 4.3 – Sum Based on Criteria
We have a dataset with brand names, devices, models, and their prices. We want the total price based on the two criteria below.
Criteria 1: Brand (Omicorn)
Criteria 2: Device (Notebook)
• The formula in cell C21 is:
`=SUMIFS(E6:E16,B6:B16,C18,C6:C16,C19)`
## Part 5 – How to Count in Excel
### Case 5.1 Count Cells
We have a dataset with values like number, text, date, and empty cells.
• To count numerical values, use this formula:
`=COUNT(B6:B13)`
• To count numerical values, texts, and formulas, use this:
`=COUNTA(B6:B13)`
• To count blank cells, use this formula:
`=COUNTBLANK(B6:B13)`
### Case 5.2 – Count Unique Values
Note: The UNIQUE function is only available in Excel 2021 and Microsoft 365 versions.
We have a dataset with brand names, devices, models, and their prices. We want to count the unique brands and unique devices separately.
• Use this formula in cell D18 to count the unique brands:
`=COUNTA(UNIQUE(B6:B16))`
• Put this formula in cell D20 to count the unique devices:
`=COUNTA(UNIQUE(C6:C16))`
### Case 5.3 – Count Based on Criteria
We have the same dataset. We want the total count based on the criteria below.
Criteria 1: Brand (Omicorn)
Criteria 2: Device (Notebook)
• The formula is:
`=COUNTIFS(B6:B16,C18,C6:C16,C19)`
Read More: How to Calculate Discount in Excel
## Part 6 – How to Use the Average Formula in Excel
The basic formula for calculating the average is:
`Average = Sum of All Values / Number of Values`
### Case 6.1 – Calculate the Average
We have the following dataset with some employees and their sales in 3 different months. We want to get the average sales of each employee in average sales column.
Steps:
• Insert the following formula in cell F6 to get the average sales of the first employee and then use the Fill Handle icon for all the remaining cells to copy the formula:
`=AVERAGE(C6:E6)`
### Case 6.2 – Running Average
A running average is a type of average that is continually updated as new data points become available. We have the following dataset with months and no. of visitors on the Exceldemy forum. We want to calculate the running average of the no. of visitors in each month starting from the second month.
Steps:
• Use the following formula in cell D7 and press Enter to get the running average of the first two months:
`=AVERAGE(\$C\$6:C7)`
• Double-click the Fill Handle icon or drag it down with a mouse to get the result for all months.
### Case 6.3 – Moving Average
A moving average is a type of average that continually updates the average as new data points are added or old data points are removed.
We have the same dataset as before. We want to calculate the 3-points moving average of the no. of visitors.
Steps:
• Use the following formula in cell D8 to get the 3-point moving average of the no. of visitors and then use the Fill Handle icon for all the remaining cells:
`=AVERAGE(C6:C8)`
### Case 6.4 – Weighted Average
The weighted average is an average in which each data point is assigned a weight based on its relative importance in the overall set.
We have the following dataset of a student’s marks in some subjects and the weights assigned to each subject. We want to calculate the weighted average marks.
• The formula to calculate the weighted average marks is:
`=SUMPRODUCT(C6:C10,D6:D10)/SUM(D6:D10)`
## Part 7 – Range Formulas in Excel
We have the following dataset where we have some products and their sales in 3 different months. We want to get the difference between the highest and lowest sales in each month.
Steps:
• Put the following formula in cell C16 to get the difference between the highest and lowest sales in April:
`=MAX(C6:C15)-MIN(C6:C15)`
• Drag the Fill Handle icon to the right for May and June.
## Part 8 – Subtotals in Excel
We have a dataset with some products, unit price, quantity sold and their sales values. There are 3 common products. We want to calculate the subtotal sales of those 3 products one by one and then their grand total sales.
We have to make the dataset like below by sorting the same products together.
Steps:
• Insert the following formula in cell E9 to get the subtotal sales of the Air Conditioner:
`=SUBTOTAL(9,E6:E8)`
The first argument of the SUBTOTAL function is function_num. This argument denotes the function we want to use in our calculation. We have used 9 because 9 denotes the SUM function.
• Repeat for Monitor in cell E13:
`=SUBTOTAL(9,E10:E12)`
• Use the formula for Battery in cell E17:
`=SUBTOTAL(9,E14:E16)`
• Insert the following formula in cell E19 to get the grand total sales of all products:
`=SUBTOTAL(9,E6:E17)`
## Part 9 – Concatenate in Excel
### Example 1 – Concatenate Multiple Cells
Here’s a dataset with some people’s first names and last names in 2 separate cells in 2 columns. We want to concatenate the values from these 2 cells to obtain the full name.
Steps:
• Use the following formula in cell D6 and press Enter to get the full name by joining the first and the last name with a space:
`=CONCATENATE(B6, " ", C6)`
• Double-click the Fill Handle icon or drag it down with a mouse to get the result for all people.
### Example 2 – Combine Text and Number
We have another dataset with some employees’ names and their serial numbers. We want to join this text and number to form unique IDs for all the employees.
Steps:
• Put the following formula in cell D6 to combine the serial number and name and then use the Fill Handle icon for all the remaining cells:
`=B6&"-"&C6`
## Part 10 – How to Calculate Percentages in Excel
The percentage of a part can be calculated by dividing it by the total value and then multiplying the result by 100 or formatting the result with the Percent Style command.
### Example 1 – Calculate Percentages
We have a dataset with individual expense categories and their amounts. We have calculated the total expense using the SUM function and want to find the individual category expense in % of the total expense.
Steps:
• Insert the following formula in cell D8 to get the Loan Payment expense in % of the total expense:
`=C8/\$C\$5*100`
• Drag the Fill Handle icon down to obtain percentages for all the category expenses.
### Example 2 – Calculate the Percentage Change
We have a dataset of some products and their old and new prices. We want to calculate the percentage change of the old prices.
• Use the following formula to get the percentage change:
`=(D6-C6)/C6*100`
## Part 11 – Ratios in Excel
Consider a dataset with some departments of a company. We have a number of male and female workers in those departments. We want to get the male-female ratio in each department.
Steps:
• Use the following formula in cell E6 to get the male-female ratio in the Executive department:
`=ROUND(C6/D6,3)&":"&1`
• Drag the Fill Handle icon down to obtain the male-female ratios for all the departments.
## Part 12 – Rounding in Excel
### Case 12.1 – Round Up Decimals
We have some decimal numbers (both positive and negative). We want to round them up to 3, 2, and 1 decimal places.
Steps:
• Use the following formula in cell C7 and press Enter to get the first number rounded up to 3 decimal places:
`=ROUNDUP(\$B7,C\$6)`
• Double-click the Fill Handle icon or drag it down to round up all numbers to 3 decimal places.
• With the range in column C selected, drag the fill handle from C17 to the right round up all numbers to 2 and 1 decimal places, respectively.
### Case 12.2 – Round to Nearest 5
We have the same dataset as before. We’ll use the CEILING.MATH function to round these numbers to the nearest 5. Note: The CEILING.MATH function is available from Excel 2013 or later versions.
Steps:
• Use the following formula in cell C6 to round the first number to the nearest 5:
`=CEILING.MATH(B6,5)`
• Drag the Fill Handle icon down to round all numbers to the nearest 5.
## Part 13 – Math Formulas in Excel
### Case 13.1 – How to Find the Root
We have a list of numbers (both positive and negative). We’ll calculate the square root of these numbers using the SQRT function.
• The formula to put in cell C6 is:
`=SQRT(B6)`
### Case 13.2 – How to Multiply a Matrix
When multiplying matrices (arrays), both arrays must only contain numbers, and the number of rows in array2 and the number of columns in array1 must be the same.
We have a dataset with 2 matrices, Matrix A and B. Both have 3 rows and 3 columns.
Steps:
• Insert the following formula in cell B11 and press Enter to get the product of 2 matrices:
`=MMULT(B6:D8,F6:H8)`
### Case 13.3 – Check for Even or Odd
We have a list of numbers. To check even or odd, we’ll use the combination of IF and ISEVEN functions.
Steps:
• Use the following formula in cell C6 to check if the number is even:
`=IF(ISEVEN(B6),"Even","Odd")`
• Drag the Fill Handle icon down to check all the numbers.
## Part 14 – How to Use Excel Text Formulas
### Case 14.1 – Find Text in a Cell
We have a list of email IDs and want to find Gmail among those email IDs. You can use the SEARCH function to find the starting position of Gmail from the left side of an email ID. The ISNUMBER function will check whether the position is a number or not.
• The formula is:
`=IF(ISNUMBER(SEARCH(\$C\$5,B6)),"Yes","No")`
### Case 14.2 – Change the Text Case
We have a list of names that aren’t in the correct cases.
• Use the following formula in cell C6 to change the case to upper case:
`=UPPER(B6)`
• Use the following formula in cell D6 to change the case to the title case:
`=PROPER(B6)`
• Insert the following to get the lower case in cell E6:
`=LOWER(B6)`
### Case 14.3 – Remove Space
We have a dataset of some names having 2 parts (first and last name), but there are unnecessary spaces between first and last names, as well as before and after the name.
Steps:
• Put the following formula in cell C6 to remove extra spaces from the name:
`=TRIM(B6)`
• Drag the Fill Handle icon down to remove spaces from all names.
### Case 14.4 – Extract Text
We have a dataset with some products and their codes. The product codes are in 3 parts separated by hyphens. We want to extract the first, middle, and last parts of these codes.
• To extract four characters from the left side (first part of the code), use this formula:
`=LEFT(C6,4)`
• To extract four characters from the middle, starting at position 6 (middle part of the code), use this formula:
`=MID(C6,6,4)`
• For extracting three characters from the right side (last part of the code), use the following:
`=RIGHT(C6,3)`
### Case 14.5 – Truncate Text
We have a list of unique IDs of some employees of a company. The IDs are made with serial numbers and employee’s names. We want to truncate the IDs and replace them with the names of the employees.
• The formula given below will replace the unique IDs with the employee’s names:
`=REPLACE(B6,1,4,"")`
## Part 15 – Date and Time Formula in Excel
We’ll see how various functions calculate date and time with the following simple dataset.
• To get today’s date only, use any of these formulas:
`=TODAY()`
`=DATE(YEAR(TODAY()),MONTH(TODAY()),DAY(TODAY()))`
• You can also put the date of your choice using this formula:
`=DATE(2023,11,16)`
• To get today’s date and time together, use this formula:
`=NOW()`
## Part 16 – How to Use Date and Time Functions in Excel
### Case 16.1 – Calculate Age
We have a list of birthdays of some people and want to calculate their age as of today.
We’ll take the birthday as the starting date and use the TODAY function to get today’s date as the ending date.
Steps:
• Insert the following formula in cell D6 and press Enter to get the age of the first person:
`=DATEDIF(C6,TODAY(),"Y")&" Years, "&DATEDIF(C6,TODAY(),"YM")&" Months, "&DATEDIF(C6,TODAY(),"MD")&" Days"`
• Double-click the Fill Handle icon or drag it down to get the result for all people.
### Case 16.2 – Get the First Day of the Month
We have a list of some employees and their joining month which is in Date format. We want to know the first day of their joining month.
We’ll use the EOMONTH function to get the serial number for the last day of the previous month before the joining date.
Steps:
• Use the following formula in cell D6 and press Enter to get the serial number for the first day of the joining month:
`=EOMONTH(C6,-1)+1`
• Double-click the Fill Handle icon or drag it down to get the result for all employees.
• Select all cells D6:D15.
• Right-click to open the Context menu.
• Choose Format Cells.
• The Format Cells window will open.
• Go to the Number tab.
• Choose Custom under Category.
• Type or choose dddd in the box.
Tips: The keyboard shortcut for opening the Format Cells window is Ctrl + 1.
• Click the OK button to apply the format. You’ll get the first day of the joining month.
### Case 16.3 – Days Between Dates
We have some employees’ joining date and resigning date in a project. We want to know the number of working days of each employee.
Steps:
• Use the following formula in cell E6 to calculate the working days of the first employee:
`=DAYS(D6,C6)`
• Drag the Fill Handle icon down to calculate the working days of all employees.
### Case 16.4 – Calculate Time
We have a similar kind of dataset but now we have entry time and exit time of employees. We want to calculate the working hours.
Steps:
• Put the following formula in cell E6 and press Enter to calculate the working hours of the first employee:
`=D6-C6`
• Double-click the Fill Handle icon or drag it down with a mouse to calculate the working hours of all employees.
You’ll see AM after working hours because the cell is in Time format. When we subtract one time from another time, the resulting cell automatically becomes a Time-formatted cell.
• Select cells E6:E15.
• Right-click to open the Context menu.
• Click on the Format Cells option, and the Format Cells window will open.
• Go to the Number tab.
• Choose Custom under Category.
• Type or choose h:mm:ss.
• Click the OK button to apply the format. You’ll get the working hours for all the employees.
## Part 17 – How to Use Conditional Formulas in Excel
We have a dataset with some students and their project submission date and want to make a comment based on the submission date. If the submission date is on or before 11 Dec, the comment is On Time. If the submission date is after 11 Dec, the comment becomes Late. These values have been put in the table for references.
Steps:
• Insert the following formula in cell D10 to make a comment for the first student:
`=IF(C10<=\$C\$6,\$B\$6,\$B\$7)`
• Drag the Fill Handle icon down to make comments for all.
## Part 18 – How to Use a Nested Formula in Excel
A nested formula is when you use one function as an argument inside another function. We already showed simple examples of nesting above.
We have a student marksheet in Math and the marks range for 5 grades. We want to assign a grade to each student.
Steps:
• Use the following formula in cell D13 and press Enter to assign a grade to the first student:
`=IF(C13<61,\$C\$6,IF(C13<71,\$C\$7,IF(C13<81,\$C\$8,IF(C13<91,\$C\$9,\$C\$10))))`
• Double-click the Fill Handle icon or drag it down to assign all grades.
## Part 19 – How to Use Lookup Formulas in Excel
### Case 19.1 – Wildcards for Partial Matches
The asterisk wildcard (*) replaces any number of characters (including 0) after a text, before a text, and both before and after a text. The question mark wildcard (?) finds only the number of characters based on the number of question marks. If there is a single question mark wildcard, it’ll search for a single character.
We have an employee database with names, departments, and designations. You can see 4 lookup values (names) here. We put the asterisk wildcard after a name, before a name, and both before and after a name. We have also put the question mark wildcard 3 times between a name. We’ll partially search for these values in the database and extract the matching designation using the VLOOKUP function.
• Use the following formulas for partial matching based on wildcard:
`=VLOOKUP(B23,B6:D20,3,FALSE)`
`=VLOOKUP(B24,B6:D20,3,FALSE)`
`=VLOOKUP(B25,B6:D20,3,FALSE)`
`=VLOOKUP(B26,B6:D20,3,FALSE)`
### Case 19.2 – INDEX and MATCH for Exact Matches
We have the same employee database. We’ll search for a name in the database and extract the department and designation of that person.
Steps:
• Use the following formula in cell C23 to get the department based on the name:
`=INDEX(\$B\$6:\$D\$20,MATCH(\$B\$23,\$B\$6:\$B\$20,0),MATCH(C22,\$B\$5:\$D\$5,0))`
• Drag the Fill Handle icon to the right side for designation.
## Part 20 – Randomize in Excel
### Case 20.1 – Generate Random Numbers
We have a dataset with lower and upper limits. We’ll generate random numbers between them.
• To generate random numbers between 0 to 1, use this formula:
`=RAND()`
• To generate random numbers between lower and upper limits, use these formulas:
`=RANDBETWEEN(B7,C7)`
`=RANDBETWEEN(B8,C8)`
• Use the following formula to generate an array of random integers based on no. of rows, no. of columns, lower limit, and upper limit:
`=RANDARRAY(B6,C6,D6,E6,TRUE)`
• Use the following formula to generate an array of random decimals based on no.of rows, no. of columns, lower limit, and upper limit:
`=RANDARRAY(B6,C6,D6,E6,FALSE)`
### Case 20.2 – Generate a Random Value from a Selection
We have an employee database with IDs, names, and departments. We want to generate a random selection of 3 IDs from all. The purpose can be a lottery.
Steps:
• Put the following formula in cell B23 to generate a random ID:
`=INDEX(\$B\$6:\$B\$20,RANDBETWEEN(1,ROWS(\$B\$6:\$B\$20)),1)`
The ROWS function will give all the row numbers of IDs. The RANDBETWEEN function generates a random row number from all row numbers. The INDEX function will extract that random ID based on the row number.
• Drag the Fill Handle icon down to generate more random IDs.
Read More: Ageing Formula in Excel
## Part 21 – Unit Conversion in Excel
### Example 1 – Convert Inches to Feet
We have some student’s heights in inches.
Steps:
• Insert the following formula in cell D6 to convert the height of the first student from inches to feet:
`=CONVERT(C6,"in","ft")`
• Drag the Fill Handle icon down to convert all remaining values.
### Example 2 – Convert kg to lbs
We have some student’s weights in kg.
Steps:
• Use the following formula in cell D6 to convert the weight of the first student from kilograms to pounds:
`=CONVERT(C6,"kg","lbm")`
• Use the Fill Handle icon to convert all remaining values.
## Part 22 – Serial Number Formula in Excel
We have a dataset with some employees’ names and departments. We want to put serial numbers in column B.
Steps:
• Use the following formula in cell B6 and press Enter to get the first serial number:
`=ROW()-ROW(\$B\$5)`
• Double-click the Fill Handle icon or drag it down to get all serial numbers.
## Some Keyboard Shortcuts and Features
F4 Toggle between absolute, relative, and mixed references
Ctrl + ~ Show all formulas on the sheet
F2 Edit a formula
Select a formula and press F9 Debug formulas
Fill Handle tool Copy a formula to all cells in a column or row
Ctrl + C Copy cells with the formula
Ctrl + V Paste cells with the formula
Shift + F10 + V Paste Special (Values Only)
## Excel Formula List: Knowledge Hub
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Sajid Ahmed, a BSc graduate in Naval Architecture & Engineering from Bangladesh University of Engineering and Technology, assumes the position of an Excel & VBA Content Developer at ExcelDemy. A self-motivated individual, his profound interest in research and innovation aligns seamlessly with his passion for Excel. In this role, Sajid not only adeptly addresses challenges but also demonstrates enthusiasm and expertise in gracefully navigating complex situations. This underscores his steadfast commitment to consistently delivering exceptional content. His interests... Read Full Bio
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Advanced Excel Exercises with Solutions PDF | 5,726 | 23,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-38 | latest | en | 0.871258 |
http://programmersheaven.com/discussion/97698/matrix-4-4-in-a-3d-lib | 1,521,915,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650764.71/warc/CC-MAIN-20180324171404-20180324191404-00713.warc.gz | 251,807,175 | 12,120 | # matrix[4][4] in a 3d lib
hi,
now i've understood the whole bunch about matrices and vectors, i found myself a good library. In that library they use 4*4 matrices, but I don't know what's in it, how many points or something like that. BTW: for rotating they only use till 0..2 of the matrix (so there are 7 numbers not used).
Secondly: vectors exist of four items: an x, a y, a z and something else.
What do you think it is, that fourth number?
[green][b][size=5]P[/size][size=4]ro[/size][size=3]g[/size][size=2]r[/size][size=1]amm[/size][size=2]e[/size][size=3]u[/size][size=4]rk[/size][size=5]e[/size][/b][/green]
• Well...I dont if your 3d lib use the same way to perform geometric trasformation as I do. There are also quaterions that can (not sure) be rapresented like a 4x4 matrix.
And matrices can also be passive or active trasformation that change the way you do what u have to do. But if we are talking of the same thing (which is the moust common way), then your matrix should look like this:
[rxx ryx rzx tx]
[rxy ryy rzy ty]
[rxz ryz rzz tz]
[0 0 0 1]
so when u multiply this matrix with a point (x,y,z,1) u get:
x*rxx + y*ryx + z*rzx + tx =new_x
x*rxy + y*ryy + z*rzy + ty =new_y
x*rxz + y*ryz + z*rzz + tz =new_z
x*0 + y*0 + z*0 +1 =1
[rxx,rxy,rxz], [ryx,ryy,ryz], [rzx,rzy,rzz] are the vectors of the x, respectly y, respectly z axis of the old coordinates system (the one you were expressing the point beform performing the trasformation). This are the "things" that particaly rotate your point in your new coordinates system. All these vectors should have lenght=1, if you dont perform scaling.
Instead [tx,ty,tz] is the traslation vector, that means that your point will be moved by this vector in order to be trasformed in the new coordinates system (x+tx, y+ty, z+tz).
[0,0,0,1] this vector is linked to the fact your are using fake 4-dimensional vectors, that's called omogenus coordinates system, but nothing I can explain within few lines, and you dont really need to know that, just figure out that in order to add the traslation we need at last a 3x4 matrix, but you can't multiply a 3x4 matrix with a 3x1 vector, you need a 4x4 matrix and a 4x1 vector to do that. But this is just a mathematical needs, when u know what u need and how to do it, you can just take away the [0,0,0,1] row to spare memory and computation time.
You said that you lib used the forth coordinate, well...I read around that it's more and more common to write a point with 4 cordinates like (x,y,z,w), w should be the "weight" (right spelled?) of the point. Ok what it's the weight of a point...I dont really know, it seems that is used for shading and maybe also for morphing, but I didn't need such a variable yet (that would be rather a vertex's charateristic, than a point's one, wouldn't it ?).
Anyway if the forth raw of the matrix contain some kind of trasformation aplied to the weight, they aren't done as the a mathematical matrix multiplication 'cause It will scale the traslation vector and modify the point's coordinates, the wight would also be corrupted by the coordinates...quite a mess.
Mutilate | 901 | 3,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-13 | latest | en | 0.868592 |
https://www.thefreelibrary.com/More+on+measuring+relative+concentration+of+sales+in+U.S....-a013691404 | 1,490,619,287,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189472.3/warc/CC-MAIN-20170322212949-00097-ip-10-233-31-227.ec2.internal.warc.gz | 973,154,793 | 14,731 | # More on measuring relative concentration of sales in U.S. manufacturing.
I. Introduction
In a recent communication in this journal, O'Neill |6~ claims that computing entropy indices for concentration as was done by Hexter and Snow |3~, Hexter |2~, and Nissan and Caveny |4; 5~ were incorrect because the use of data for sales and assets of the Fortune 500 sample manufacturing companies is just a fraction of the total companies. Instead, data of the actual number of companies, which number in the hundreds of thousands, should be used in the computation, O'Neill says. Because working with data on such a scale is prohibitive at best and perhaps impossible because of unavailability, O'Neill computes the shares of sales of the Fortune 500 as percentages of total net sales for the years 1967 to 1987 taken from various issues of the Statistical Abstract of the United States, herein referred to as Abstract. O'Neill then recomputes, as an example, sales entropy for the years 1967 to 1987 which he calls the "corrected" index and makes comparisons with the results of Nissan and Caveny for the same period. Similar arguments by O'Neill |7~ were forwarded on the work of Attaran and Saghafi |1~ concerning concentration trends and profitability in U.S. manufacturing. O'Neill proposes that the use of net sales as the basis of computing the shares of the 500 firms results in a reversal of conclusions. Whereas the aforementioned authors claim that the trends in concentration were on the increase, O'Neill says that concentration is instead, on the decrease. The purpose of this comment is to provide a response to O'Neill's criticisms and to show that using either method, one obtains similar conclusions when the data are compatible.
II. Comparisons
Theil's entropy measure "E" on which all computations are based is
|Mathematical Expression Omitted~
where |S.sub.i~, in the present study, stands for the share of sales of firm "i" as a ratio of total sales, and n is the number of firms. If all n firms have an equal share, entropy is at maximum and concentration is at a minimum. When one company controls all shares, entropy is at a minimum, and E = 0. Thus, a decrease in E over time implies an increase in concentration. The reverse is true when "E" increases.
The gist of the argument profferred by O'Neill is that in calculating the entropy of the largest 500 firms listed in Fortune, the shares "|S.sub.i~" should be ratios of sales to total sales of all manufacturing firms rather than just the total of the 500 firms alone. Because such data are not provided by Fortune, O'Neill uses net sales data obtained from an alternative source, the Abstract. The use of such a procedure will be shown to produce inconsistent results and furthermore to be unnecessary. The following items make these points clear.
```Table I. Net Sales of All Manufacturing Firms and Gross Sales
of Fortune 500 Firms (Billion of Dollars)
Net Sales(a) Gross Sales(b)
Year A B A/B
1967 575 359 1.60
1968 632 405 1.56
1969 694 445 1.56
1970 709 462 1.53
1971 751 503 1.49
1972 850 558 1.52
1973 1017 667 1.52
1974 1061 831 1.28
1975 1065 865 1.23
1976 1203 970 1.24
1977 1328 1087 1.22
1978 1496 1216 1.23
1979 1742 1445 1.20
1980 1897 1650 1.15
1981 2145 1773 1.21
1982 2039 1672 1.22
1983 2114 1686 1.25
1984 2335 1753 1.33
1985 2331 1808 1.29
1986 2221 1712 1.30
1987 2378 1879 1.27
a. Source: Various issues of the Statistical Abstract of the
United States.
b. Source: Data published annually by Fortune.
```
(1) The reliability of net sales in the Abstract is questioned. This is primarily due to the way data are assembled. A warning in the tables states, "Data are not necessarily comparable from year to year due to changes in accounting procedures, industry classifications, sampling procedures, etc." Thus, the use of such data makes the adjustments on yearly entropy index values unreliable. As pointed out by Saghafi and Attaran |8~ in their reply to O'Neill |7~, the differences in the calculations may be due to differences in source and type of data used. For illustration, Table I provides net sales data from the Abstract (column 1) and total gross sales figures from Fortune (column 2) for the years 1967 to 1987. As shown in column 3, some consistency exists in the ratios between 1967 and 1973 where the values hover around 1.50. Suddenly the ratios are reduced to approximately 1.25 between 1974 and 1987, producing another set of consistent ratios. The result of such change is dramatically illustrated in Figure I in O'Neill |6, 265~, where there is a sharp jump in the graph corresponding to the value of the total entropy index in 1974.
(2) Pursuing the analysis a bit further, a series of simple regressions were conducted using the form of the regression equation
|Y.sub.i~ = a + b|T.sub.i~,
where |Y.sub.i~ is the entropy index for year i, a is the intercept, b is the trend, and |T.sub.i~ represents the year (1 for 1967,..., 21 for 1987). First, linear trend regressions for the years 1967 to 1987 TABULAR DATA OMITTED were performed on total entropy calculated by Nissan and Caveny (N&C), and on total entropy calculated by O'Neill (corrected) as the dependent variables. This data was obtained from Table I of O'Neill |6, 264~. Similar regression was performed again, with the period of interest this time being 1974 to 1987. The results are shown in Table II.
When the whole period 1967 to 1987 is used, the two regressions give different results. The Nissan and Caveny model produced a negative and significant slope of -.0207, indicating that there was a declining trend in the entropy index, and thus an increase in concentration. The O'Neill (corrected) model produced a positive slope of .0493, indicating a decrease in concentration. But this discrepancy is the result of the incompatibility of data as pointed out in (1) above.
When regression is performed for the period 1974 to 1987, a different picture emerges as shown in Table II. For this time span, the Nissan and Caveny model produced an almost identical slope of -.0206; however, the O'Neill (corrected) model reversed in trend had a slope with a negative sign (-.0336), indicating an increase in the trend toward concentration. This simple analysis shows that if data are comparable, as is the case for the period 1974 to 1987, the results are the same whether one uses the net sales of all the firms or the gross sales of the largest 500 firms only.
III. Summary and Conclusions
This comment addresses criticisms by O'Neill |6~ concerning indices of concentration of sales provided by Nissan and Caveny |51~. This analysis shows that the noted discrepancies were the result of using data that are not comparable, and it shows, as well, that when the data are comparable the original conclusions are obtained. In short, exclusive reliance on one set of data in reaching a conclusion in this type of study is better than basing such conclusions on a mixture of data.
References
1. Attaran, Mohsen and Massoud M. Saghafi, "Concentration Trends and Profitability in the U.S. Manufacturing Sector: 1970-84." Applied Economics, November 1988, 1497-510.
2. Hexter, Lawrence J., "Measuring Relative Concentration." Southern Economic Journal, January 1987, 777-78.
3. ----- and John W. Snow, "An Entropy Measure of Relative Concentration." Southern Economic Journal, January 1970, 239-43.
4. Nissan, Edward and Regina Caveny, "Relative Concentration of the Largest 500 Firms." Southern Economic Journal, January 1985, 880-81.
5. ----- and -----, "Relative Concentration of Sales and Assets in American Business." Southern Economic Journal, April 1988, 928-33.
6. O'Neill, Patrick B., "Measuring Relative Concentration of Sales in U.S. Manufacturing." Southern Economic Journal, July 1991, 263-67.
7. -----, "Concentration Trends and Profitability in U.S. Manufacturing: A Comment." Applied Economics, April 1991, 717-20.
8. Saghafi, Massoud M. and Mohsen Attaran, "Concentration Trends and Profitability in U.S. Manufacturing: A Reply and Some New Evidence." Applied Economics, April 1991, 721-22. | 2,016 | 8,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-13 | longest | en | 0.948724 |
https://algosim.org/doc/arccsc.html | 1,685,470,912,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00316.warc.gz | 117,087,142 | 1,609 | arccsc – Algosim documentation
Algosim documentation: arccsc
# arccsc
The trigonometric arccosecant function.
## Syntax
• `arccsc(x)`
• `x` is a real or complex number
## Description
For real values, arccsc is the inverse of the restriction of the trigonometric cosecant function to [−π/2, π/2] ∖ {0}; the domain is ℝ ∖ (−1, 1).
In general, `arccsc(z)` is defined as
```arccsc(z) = −i⋅ln(i/z + √(1 − 1/z^2))
```
where `z` is a complex number. Notice that the general definition will be used also for real arguments in [−1, 1].
## Examples
`arccsc(2/√3)`
`1.0471975512 (=π/3)`
`arccsc(−2)`
`−0.523598775598 (=−π/6)`
`arccsc(0.5)`
`1.57079632679 − 1.31695789692⋅i`
`arccsc(i)`
`−0.88137358702⋅i` | 277 | 707 | {"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-2023-23 | latest | en | 0.575199 |
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# Chapter 8 Frequency-Domain Analysis
## Chapter 8 Frequency-Domain Analysis
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##### Presentation Transcript
1. Chapter 8Frequency-DomainAnalysis Automatic Control Systems, 9th Edition F. Golnaraghi & B. C. Kuo
2. 1, p. 409 8-1 Introduction • For a LTI system: input -- steady-state output -- • Sinusoidal steady-state analysis: s = j M(s)
3. 1, p. 410 Frequency Response • Closed-loop transfer function: • Sinusoidal steady-state transfer function: s = j Magnitude of M(j): Phase of M(j):
4. 1, p. 411 Gain-Phase Characteristics cut-off frequency
5. 1, p. 412 Frequency-Domain Specifications • Resonant Peak Mr:the maximum value of the relative stability of a stable closed-loop system • Resonant Frequency r:the frequency at which thepeak resonant Mr occurs • Bandwidth BW:the frequency at whichdrops to 70.7% of, or 3 db down from, its zero-freq. value the transient response properties of a control system • Cutoff Rate:the slope of at high frequency
6. 2, p. 413 8-2 Mr, r, and Bandwidth of the Prototype Second-Order System • The prototype second-order system: • u = /n:magnitude: phase:
7. 2, p. 414 Resonant Frequency and Resonant Peak • Resonant frequency r: • Resonant peak : Resonant frequency: r = 0 if 0.707 Mr = 1 if 0.707
8. 2, p. 415 Magnification vs Normalized Frequency u = /n = 0 r = n
9. 2, p. 416 Mr, urvs Damping Ratio
10. 2, p. 417 Bandwidth • Definition: 70.7% or 3 dB
11. 2, p. 417 Time-Domain Response vsFrequency-Domain Characteristics • Mr depends on only.The maximum overshoot also depends only on . • For 0.707, Mr = 1 and r = 0.The maximum overshoot is 0 when 1.0. • BW is directly proportional to n andinversely proportional to .The rise time tr increases as n decreases.BW and tr are inversely proportional to each other. • BW and Mr are proportional to each other for 0 0.707.
12. 2, p. 419 Correlation between Pole Locations
13. 2, p. 419 Unit-Step and Frequency Responses
14. 3, p. 418 8-3 Effects of Adding a Zero to the Forward-Path Transfer Function zero: s = 1/T • closed-loop transfer function: • Bandwidth: increase BW
15. 3, p. 421 Magnification Curves
16. 3, p. 423 Unit-Step Responses: adding a zero
17. 4, p. 424 8-4 Effects of Adding a Pole to the Forward-Path Transfer Function pole: s = 1/T Effects:Decrease BW,Increase Mr,Make the closed-loop system less stable
18. 4, p. 425 Unit-Step Responses: adding a poles T tr BW Mr yrmax
19. 5, p. 426 8-5 Nyquist Stability Criterion: Fundaments • The Nyquist plot of the loop transfer functionG(s)H(s), or L(s), is done in polar coordinates as varies from 0 to . • The Nyquist criterion has the following features: • In addition to providing the absolute stability, the Nyquist criterion also gives information on the relative stability of a stable system and the degree of instability of an unstable system. • The Nyquist plot is very easy to obtain. • The Nyquist plot gives information on the frequency-domain characteristics such as Mr, r, and BW. • The Nyquist plot is useful for systems with pure time delay.
20. 5, p. 427 Stability Problem • Closed-loop transfer function: • Characteristic equation: • Stability conditions: • Open-loop stability: if the poles of the loop transfer function L(s) are all in the left-half s-plane. • Closed-loop stability: if the poles of the closed-loop transfer function or the zeros of 1+L(s) are all in the left-half s-plane.
21. 5, p. 428 Encircled • A point or region in a complex function plane is said to be encircled by a closed path if it is found inside the path Point B is not encircled by the closed path . Point A is encircled by in the counterclockwise (CCW) direction.
22. 5, p. 428 Enclosed • A point or region is said to be enclosed by a closed path if it is encircled in the CCW direction or the point or region lies to the left of the path when the path is traversed in the prescribed direction.
23. 5, p. 429 Number of Encirclements and Enclosures • When a point is encircled by a closed path ,N = the number of times it is encircled. • N is positivefor CCW encirclement and negative for CW encirclement. B is encircled twicein the CCW direction A is encircled once or 2 radians by in the CW direction B is encircled twice or 4 radians by in the CW direction A is encircled oncein the CCW direction
24. 5, p. 429 Locus in the (s)-plane • Suppose that a continuous closed paths is arbitrarily chosen in the s-plane, as shown in Fig. 8-17(a). • If s does not go through any poles of (s), then the trajectory mapped by (s) into (s)-plane is also a closed one, as shown in Fig. 8-17(b). single-valued mapping (s): single-valued function
25. 5, p. 430 Principles of the Argument • Let (s) be a single-valued function that has a finite number of poles in the s-plane. • Suppose that an arbitrarily closed paths is chosen in the s-plane so that the path does NOT go through any one of the poles or zeros of (s). • The corresponding locus mapped in the (s)-plane will encircle the originas many times asthe difference between the number of zeros and poles of (s) that are encircled by the s-plane locus s. N = Z PN = number of encirclements of the origin made by the locus .Z = number of zeros of (s) encircled by the s-plane locus s.P = number of poles of (s) encircled by the s-plane locus s.
26. 5, p. 431 Examples of Determination of N • N > 0 (Z>P): the (s)-plane locus will encircle the origin N times in the same direction as that of s. • N = 0 (Z=P): the (s)-plane locus will not encircle the origin of the (s)-plane. • N < 0 (Z<P): the (s)-plane locus will encircle the origin N times in the opposite direction as that of s.
27. 5, p. 432 Illustrative Example The net angle traversed by the (s)-plane:
28. 5, p. 433 Table 8-1
29. 5, p. 434 Nyquist Path • The Nyquist path is definedto encircled the entireright-half s-plane. • The Nyquist path must notpass through any poles andzeros of (s).
30. 5, p. 434 Nyquist Criterion (1/2) • (s) = 1 + L(s), L(s): loop transfer function the origin of the (s)-plane corresponds to the (1, j0) point in the L(s)-plane. • Steps of the application of Nyquist criterion to the stability problem: 1. The Nyquist path s is defined in the s-plane, as shown in Fig. 8-20. 2. The L(s) plot corresponding to the Nyquist path is constructed in the L(s)-plane. 3. The value of N, the number of encirclement of the (1, j0) point made by L(s) plot, is obsered. 4. The Nyquist criterion follows from Eq. (8-42),
31. 5, p. 435 Nyquist Criterion (2/2) • Stability requirements:For closed-loop stability, Z must equal zero.For open-loop stability, P must equal zero. • The condition of stability according to Nyquist Criterion:for a closed-loop system to be stable, the L(s) plot must encircle the (1, j0) point as many times as the number of poles of L(s) that are in the right-half s-plane, and the encirclement, if any, must be made in the clockwise dircetion (if s is defined in the CCW sense).
32. 6, p. 435 8-6 Nyquist Criterion for Systems with Minimum-Phase Transfer Functions Minimum-phase transfer function: • A minimum-phase transfer function does not have poles or zeros in the right-half s-plane or on the j-axis, excluding the origin. • For a minimum-phase transfer function L(s) with m zeros and n poles, excluding the poles at s = 0, when s = j and as varies from to 0, the total phase variation of L(j) is (nm)/2 radians. • The value of a minimum-phase transfer cannot become zero or infinity at any nonzero finite frequency. • A nonminimum-phase transfer function will always have a more positive phase shift as varies from to 0. Or equally true, it will always have a more negative phase as varies from 0 to .
33. 6, p. 436 Nyquist Criterion for Systems with Minimum-Phase Transfer Functions • L(s): minimum-phase type P = 0Nyquist criterion N = 0 • For a closed-loop system with loop transfer function L(s) that is of minimum-phase type, • the system is closed-loop stable if the plot of L(s) that corresponds to the Nyquist path does NOT encircle (or enclose) the critical point (1, j0) in the L(s) -plane. • If the (1, j0) point is enclosed by the Nyquist plot, the system is unstable.
34. 6, p. 437 Not Strictly Proper Transfer Function • The characteristic equation of a system:K: a variable parameterLeq: the equivalent transfer function • If Leq does not have more poles than zeros, • Plot the Nyquist-plot of 1/Leq(s)the critical point is still (1, j0) for K > 0the variable parameter on the Nyquist plot is now 1/K. the Nyquist criterion can still be applied
35. 7, p. 437 8-7 Relation between the Root Loci and the Nyquist Plot • Both the root locus and the Nyquist criterion deal with the location of the roots of the characteristic equation of a linear SISO system. • Characteristic equation: • The Nyquist plot of L(s) in the L(s)-plane is the mapping of the Nyquist path in the s-plane. • The root locus must satisfyThe root loci simply represent a mapping of the real axis of L(s)-plane or the G(s)H(s)-plane onto the s-plane.
36. 7, p. 438 Mapping s-plane onto G(s)H(s)-plane
37. 7, p. 438 Mapping G(s)H(s)-plane onto s-plane
38. 7, p. 439 Relation between G(s)H(s)- and s-planes
39. 7, p. 439 Relation between G(s)H(s)- and s-planes
40. 8, p. 440 8-8 Illustrative Examples: Nyquist Criterion for Min.-Phase Transfer Func. • Example 8-8-1: minimum-phase type Sketch of the Nyquist plot of L(j)/K 1. Substitute s = j in L(s): 2. Get the zero-frequency( = 0) property: 3. Get the infinite-frequency( = ) property:
41. 8, p. 441 Example 8-8-1 (cont.) 4. Find the possible intersects on the real axis:Set the imaginary part of L(j)/K to zero:The intersect on the real axis of the L(j)-plane at ( must be positive)
42. 8, p. 441 Example 8-8-1 (cont.) The intersect on the real axis of the L(j)-plane at • K<240: the intersect wouldbe to the rightof (1, j0).The critical point is notenclosed stable. • K=240: the intersect is atthe 1 point. marginally stable. • K>240: the intersect would be to the leftof (1, j0). unstable. • K<0: the critical point (+1, j0) is enclosed unstable.
43. 8, p. 442 Example 8-8-1 (cont.) Stable: 0 < K < 240
44. 8, p. 442 Example 8-8-2 Characteristic equation: Sketch the Nyquist plot of Leq(s): 1. 2. Two end points: 3. The possible intersects on the real axis: = 0 and Four imaginary roots
45. 8, p. 443 Example 8-8-2 (cont.) Stable: K > 0
46. 8, p. 444 Example 8-8-2 (cont.) Stable: K > 0
47. 9, p. 444 8-9 Effects of Adding Poles and Zeros toL(s) on the Shape of the Nyquist Plot • Addition of Poles at s = 0: • Properties of Nyquist plot: multiplicity = p:
48. 9, p. 445 Nyquist Plots Add a poleat s = 0 to L(s)
49. 9, p. 446 Example 8-9-1
50. 9, p. 447 Example 8-9-1 (cont.) | 3,526 | 10,994 | {"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-2021-25 | latest | en | 0.709002 |
http://www.desy.de/pub/www/projects/Physics/Relativity/SR/barn_pole.html | 1,510,999,484,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00791.warc.gz | 385,277,155 | 2,690 | Updated 1997 by PEG.
Updated 1992 by SIC.
Original by Robert Firth.
# A Special Relativity Paradox: The Barn and the Pole
These are the props. You own a barn, 40m long, with automatic doors at either end, that can be opened and closed simultaneously by a switch. You also have a pole, 80m long, which of course won't fit in the barn.
Now someone takes the pole and tries to run (at nearly the speed of light) through the barn with the pole horizontal. Special Relativity (SR) says that a moving object is contracted in the direction of motion: this is called the Lorentz Contraction. So, if the pole is set in motion lengthwise, then it will contract in the reference frame of a stationary observer.
You are that observer, sitting on the barn roof. You see the pole coming towards you, and it has contracted to a bit less than 40m, in your reference frame. (Does it actually look shorter to you? See Can You See the Lorentz-Fitzgerald Contraction? for the surprising answer. But in any case, you would measure its length as a bit less than 40m.)
So, as the pole passes through the barn, there is an instant when it is completely within the barn. At that instant, you close both doors simultaneously, with your switch. Of course, you open them again pretty quickly, but at least momentarily you had the contracted pole shut up in your barn. The runner emerges from the far door unscathed.
But consider the problem from the point of view of the runner. She will regard the pole as stationary, and the barn as approaching at high speed. In this reference frame, the pole is still 80m long, and the barn is less than 20 meters long. Surely the runner is in trouble if the doors close while she is inside. The pole is sure to get caught.
Well does the pole get caught in the door or doesn't it? You can't have it both ways. This is the "Barn-pole paradox." The answer is buried in the misuse of the word "simultaneously" back in the first sentence of the story. In SR, that events separated in space that appear simultaneous in one frame of reference need not appear simultaneous in another frame of reference. The closing doors are two such separate events.
SR explains that the two doors are never closed at the same time in the runner's frame of reference. So there is always room for the pole. In fact, the Lorentz transformation for time is
```
t'=(t-v*x/c2)/sqrt(1-v2/c2)
```
It's the v*x term in the numerator that causes the mischief here. In the runner's frame the more distant event (larger x) happens earlier. The far door is closed first. It opens before she gets there, and the near door closes behind her. Safe again — either way you look at it, provided you remember that simultaneity is not a constant of physics.
## What if the doors are left shut?
If the doors are kept shut the rod will obviously smash into the barn door at one end. If the door withstands this the leading end of the rod will come to rest in the frame of reference of the stationary observer. There can be no such thing as a rigid rod in relativity so the trailing end will not stop immediately and the rod will be compressed beyond the amount it was Lorentz contracted. If it does not explode under the strain and it is sufficiently elastic it will come to rest and start to spring back to its natural shape but since it is too big for the barn the other end is now going to crash into the back door and the rod will be trapped in a compressed state inside the barn.
References: Taylor and Wheeler's Spacetime Physics is the classic. Feynman's Lectures are interesting as well. | 843 | 3,602 | {"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-2017-47 | latest | en | 0.95496 |
https://forum.jedox.com/index.php/Thread/3797-YTD-Calculation-for-headcount-FTEs/?s=7525574b252d32ce26b017211e9f889031b12cfc | 1,563,676,883,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526818.17/warc/CC-MAIN-20190721020230-20190721042230-00497.warc.gz | 408,168,167 | 11,530 | This site uses cookies. By continuing to browse this site, you are agreeing to our Cookie Policy.
Hello all / Hallo ihr alle,
I have the impression that this question was asked before, but the answers I found do not seem to fit my problem completely
Let's consider the following cube: A
Period: Jan, Feb, Mar....Dec, YTDJan, YTDFeb, YTDMar....YTD Dec, FY
View of the data: YTD or Periodic
Measures: different measures related to FTE (headcounts) and some not related to FTE (€, days etc)
Using a rule, I would like to calculate a YTD for the measures that have FTE as an attribute and the once that have an attribute different than FTE I would like to leave them as they are.
An example:
'Total FTE' has FTE as an attribute:
Period 1 | Period 2 | Period 3 | Period 4 | Period 5 | Period 6
100 110 100 110 130 125
Those are Periodic values. That means that Period 6 / YTD / Total FTE would normally be = 675
BUT since this makes no sense and i always wanna have the lates period as the YTD value I need help with the rule
In gereal no problem. I just created 12 rules that calculate the YTD for each Period:
['YTD','Jan'] = ['Periodic','Jan'] => for January YTD = Periodic
But now also the elements which I do not want to have calculated this way and which do not have FTE as an attribute, are getting calculated this way. I have no idea how to exclude them from this rule and my problem is also that I'm not very experienced with rules...
Does anyone have an idea?
Thanks in advance ! / Vielen Dank vorab!
• From what I understand you want to exclude Measures having attributes as FTE from YTD View,
you can use STET() function to exclude them based of IF criteria, I have extended it to show you can calculate the YTD view in one calculation instead of having 12
calculations, 1st type of calculation assumes that you have defined attribute 'Description' for Month dimension, and Attribute as 'Attribute' for Measure dimension
['YTD'] = B: IF(PALO.DATA("DatabaseName","#Measures",!'Measures',"Attribute")="FTE",STET(),
IF(PALO.DATA("DatabaseName","#Period",!'Month','Description') == "January",
PALO.DATA("DatabaseName","CubeName",!'Dim1'.... !'DimN-2',"Jan", "Periodic"),
IF(PALO.DATA("DatabaseName","#Period",!'Month','Description') == "Febuary",
PALO.DATA("DatabaseName","CubeName",!'Dim1'.... !'DimN-2',"Jan", "Periodic") + PALO.DATA("DatabaseName","CubeName",!'Dim1'.... !'DimN-2',"Feb", "Periodic")
.. similarly do it for rest of the months)))
The second one is more concise, but it assumes that you still have Month dimension with alternate consolidation paths for YTD, like Mar_YTD consolidates Jan, Feb & Mar elements, but have also view dimension having Periodic & YTD elements - This calculation would be faster but bit messy so could Hide the Month elements from the user.
['YTD'] = B: IF(PALO.DATA("DatabaseName","#Measures",!'Measures',"Attribute")="FTE",STET(),
PALO.DATA("DatabaseName","CubeName",!'Dim1'.... !'DimN-2',CONCATENATE(!'Month',"_YTD"),"Periodic"))
If you just want Periodic value instead of STET() for YTD, you can replace the STET() function with
PALO.DATA("DatabaseName","CubeName",!'Dim1'.... !'DimN-2',!'Month',"Periodic")
Hope it helps!
Best regards,
Noel. | 851 | 3,216 | {"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} | 3.046875 | 3 | CC-MAIN-2019-30 | latest | en | 0.960715 |
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December 13, 2013
# Posts by sandra
Total # Posts: 629
Beginnings and Beyond Early Childhood Education
Thier creation allows the teacher to have time away from the childhood interaction with children.
Beginnings and Beyond Early Childhood Education
All the following are reasons record keeping is essential to good early childhood program EXCEPT:
Early Childhood Education
Of the following tasks performed outside the classroom,which is the MOST time-consuming?
Chemistry
25g of ethanol is dissolved in 1.50 kg of whater. what will the boiling point of the solution be?
Information Literacy
#2 is wrong. just took the test
Information Literacy
2 is wrong just took the test
science-biology
can anyone help me by finding which enzymes are present in carbohydrates? is it salivary amylase(tyalin)
Chemistry
A system is represented by the following equation: 2A(aq) + B(aq) --> 3C(aq) At equilibrium, [C]=0.6 M. Calculate the value of K if initially 1 mol of A and 2 mol of B were placed in a 1 L flask.
Chemistry
A 100 mL container is filled with 0.01 mol of A2B3. At equilibrium, it also contains 0.006 mol of B. Find the value of K. A2B3(g) --> 2A(g) +3B(g)
Chemistry
5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation: 2A(g) + B(g) --> 3C(g) At equilibrium, there is 4.8 x 10-5 mol of B. Calculate the value of the equilibrium constant.
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https://www.enotes.com/homework-help/determine-integral-y-x-3-x-2-9-456939 | 1,632,742,456,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00718.warc.gz | 773,701,011 | 17,805 | # Determine the integral of y=(x+3)/(x^2-9)?
You need to evaluate the given indefinite integral, hence, you should first reduce the integrand to the most simplified form, such that:
`(x + 3)/(x^2 - 9) = (x + 3)/((x - 3)(x + 3))`
Reducing the duplicate factors, yields:
`(x + 3)/(x^2 - 9) = 1/(x - 3)`
Hence,...
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You need to evaluate the given indefinite integral, hence, you should first reduce the integrand to the most simplified form, such that:
`(x + 3)/(x^2 - 9) = (x + 3)/((x - 3)(x + 3))`
Reducing the duplicate factors, yields:
`(x + 3)/(x^2 - 9) = 1/(x - 3)`
Hence, evaluating the indefinite integral, yields:
`int (x + 3)/(x^2 - 9) dx = int 1/(x - 3) dx`
You should come up with the following substitution, such that:
`x - 3 = t => dx = dt`
Replacing the variable, yields:
`int 1/(x - 3) dx = int 1/t dt`
`int 1/t dt = ln|t| + c`
Replacing back `x - 3` for t yields:
`int 1/(x - 3) dx = ln|x - 3| + c`
Hence, evaluating the given indefinite integral, yields
`int (x + 3)/(x^2 - 9) dx = ln|x - 3| + c` .
Approved by eNotes Editorial Team | 402 | 1,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-39 | latest | en | 0.744209 |
http://www.scienceforums.net/topic/102559-determining-the-relationship-to-elementary-functions/ | 1,506,106,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689102.37/warc/CC-MAIN-20170922183303-20170922203303-00586.warc.gz | 570,921,141 | 14,507 | # Determining the relationship to elementary functions?
## Recommended Posts
How does one determine with 100% certainty that a certain function can't ever possibly be put into terms of elementary functions, or really any arbitrary list of functions?
Take the gamma function for instance, $\Gamma(x)$. I am fairly confident the gamma function cannot be put into terms of elementary functions, but on the other hand, I've never ever seen anything remotely resembling a proof of that statement or any statement of that nature about any special function. How would I determine it can't be put into terms of other non-special functions or a mix of special functions and elementary functions?
Edited by SFNQuestions
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studiot 1131
How does one determine with 100% certainty that a certain function can't ever possibly be put into terms of elementary functions, or really any arbitrary list of functions?
Take the gamma function for instance, $\Gamma(x)$. I am fairly confident the gamma function cannot be put into terms of elementary functions, but on the other hand, I've never ever seen anything remotely resembling a proof of that statement or any statement of that nature about any special function. How would I determine it can't be put into terms of other non-special functions or a mix of special functions and elementary functions?
I take it this question is related to your other one.
The sort of proof you are looking for is known as an existence and uniqueness theorem, of which there are several each pertaining to a particular area of mathematics.
It looks as though you are studying what is known as the functions of a real variable.
Even in this limited area the E&U theorems involve some highly abstract maths which fills large textbooks.
A few 'standards' are
Hurewicz Dimension Theory
Hobson The Theory of Functions of a Real Variable (2 vols)
Titchmarsh The Theory of Functions
Burkhill A second Course in Mathematical Analysis
Graves Theory of Functions of Real Variables
Verblunsky An introduction to the Theory of Functions of a Real Variable
Littlewood The Elements of the Theory of Real Functions
Some of these are rather old.
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I take it this question is related to your other one.
The sort of proof you are looking for is known as an existence and uniqueness theorem, of which there are several each pertaining to a particular area of mathematics.
It looks as though you are studying what is known as the functions of a real variable.
Even in this limited area the E&U theorems involve some highly abstract maths which fills large textbooks.
A few 'standards' are
Hurewicz Dimension Theory
Hobson The Theory of Functions of a Real Variable (2 vols)
Titchmarsh The Theory of Functions
Burkhill A second Course in Mathematical Analysis
Graves Theory of Functions of Real Variables
Verblunsky An introduction to the Theory of Functions of a Real Variable
Littlewood The Elements of the Theory of Real Functions
Some of these are rather old.
I mean I appreciate you trying to help, but I don't see any indication my question is answered. I'm not referring to a uniqueness theorem as a solution to a functional equation, this is different. And it doesn't need to be limited to functions of a real variable. What I mean is: how can you test different functions against another function to determine if one function can somehow be put into terms of another function at its simplest level? For instance, take the cosine function. For the longest time, no one had any idea about complex exponents, people pre-hand made the assumption that you simply couldn't represent the cosine function in a fundamentally different way as a function of a variable, only to later discover it could be represented as functions of complex variables. That's the mistake I want to avoid. I may have a function or Taylor series that may not look like anything I recognize, but is it its own fundamental combination of operators? Or is it a complicated culmination of pre-existing functions? That's what I want to determine.
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I know there's an answer to this, because if there wasn't, every single person who ever said "cannot be put into terms of elementary functions" is automatically discretited, so I know they definitely must have done something to verify that.
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Yes, there is an answer to this and studiot directed you to texts that will lead you to the answer. The problem is that you expect simple answer! The answer is very deep involving deep mathematics. We have no idea what math background you have. Have you taken an upper level class in "abstract algebra"? Have you taken graduate courses in algebra?
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Sign In Now | 1,052 | 5,126 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-39 | longest | en | 0.937961 |
https://papertowrite.com/acc-assume-a-manufacturer-incurs-2000000-hours-of-direct-productive-2/ | 1,611,091,753,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00724.warc.gz | 483,617,901 | 16,188 | # ACC- Assume a manufacturer incurs 2,000,000 hours of direct productive
1. Assume a manufacturer incurs 2,000,000 hours of direct productive labor in a year at a total direct labor cost of \$50,000,000. The total manufacturing indirect expense for the same period is \$67,500,000.a. What is the average direct labor hour rate?b. If the overhead were distributed on the basis of direct labor hours, what would the rate be per hour?c. If the overhead were distributed on the basis of direct labor dollars, what would the percentage rate be?2. Using the same factors as above, assume that we have two projects each requiring 100,000 direct labor hours. On Project A, because of work requirements and personnel assigned, direct labor costs of \$2,040,000, or \$20.40 per hour, were experienced. Project B, on the other hand, requires a direct labor cost of \$2,800,000, or \$28.00 per hour. Calculate the amount of overhead that would be allocated to these two projects on a labor hour and a dollar basis.Direct Labor Direct LaborHour Method Dollar MethodProject A \$_____________ \$____________Project B \$_____________ \$____________
# ACC- Assume a manufacturer incurs 2,000,000 hours of direct productive
1. Assume a manufacturer incurs 2,000,000 hours of direct productive labor in a year at a total direct labor cost of \$50,000,000. The total manufacturing indirect expense for the same period is \$67,500,000.a. What is the average direct labor hour rate?b. If the overhead were distributed on the basis of direct labor hours, what would the rate be per hour?c. If the overhead were distributed on the basis of direct labor dollars, what would the percentage rate be?2. Using the same factors as above, assume that we have two projects each requiring 100,000 direct labor hours. On Project A, because of work requirements and personnel assigned, direct labor costs of \$2,040,000, or \$20.40 per hour, were experienced. Project B, on the other hand, requires a direct labor cost of \$2,800,000, or \$28.00 per hour. Calculate the amount of overhead that would be allocated to these two projects on a labor hour and a dollar basis.Direct Labor Direct LaborHour Method Dollar MethodProject A \$_____________ \$____________Project B \$_____________ \$____________ | 516 | 2,268 | {"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-2021-04 | latest | en | 0.930276 |
http://fidelfita.ample24.cat/cheap-place-xebocf/137436-tensors-in-computer-science | 1,621,011,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00571.warc.gz | 17,986,845 | 12,930 | However, tensor applications and tensor-processing tools arise from very different areas, and these advances are too often kept within the areas of knowledge where they were first employed. Tensors have a rich history, stretching over almost a century, and touching upon numerous disciplines; but they have only recently become ubiquitous in signal and data analytics at the confluence of signal processing, statistics, data mining, and machine learning. The code below creates a 3D tensor. Numpy's multidimensional array ndarray is used below to create the example constructs discussed. However, tensor applications and tensor-processing tools arise from very different areas, and these advances are too often kept within the areas of knowledge where they were first employed. It, thus, has 0 axes, and is of rank 0 (tensor-speak for 'number of axes'). Department of Computer Science University in the Texas at El Paso 500 W. University El Paso, TX 79968, USA mceberio@utep.edu, vladik@utep.edu Abstract In this paper, after explaining the need to use tensors in computing, we analyze the question of how to best store tensors in computer memory. In short, a single-dimensional tensor can be represented as a vector. Then again, you could use a computer crutch, but that doesn’t help you understand, really. Recent years have seen a dramatic rise of interest by computer scientists in the mathematics of higher-order tensors. Supervised learning in computer vision 3. Computing the Tucker decomposition of a sparse tensor is demanding in terms of both memory and computational resources. That is linear operators. 1 Why Tensors One of the main problems of modern computing is that: • we have to process large amounts of data; • and therefore, long time required to process this data. If you are interested in learning more about dual space, I highly recommend this amazing explanation by Grant Sanderson. We could see that the components in our simple vector are the same as the coordinates associated with those two basis vectors. In computer science, we stop using words like, number, array, 2d-array, and start using the word multidimensional array or nd-array. It is followed by a vector, where each element of that vector is a scalar. A vector is a single dimension (1D) tensor, which you will more commonly hear referred to in computer science as an array. When thinking about tensors from a more theoretical computer science viewpoint, many of the tensor problems are NP-hard. Though classical, the study of tensors has recently gained fresh momentum due to applications in such areas as complexity theory and algebraic statistics. Jon Sporring received his Master and Ph.D. degree from the Department of Computer Science, University of Copenhagen, Denmark in 1995 and 1998, respectively.Part of his Ph.D. program was carried out at IBM Research Center, Almaden, California, USA. In fact, scalars are rank-0 tensors; vector and covectors are rank-1 tensors; matrices are rank-2 tensors. We first review basic tensor concepts and decompositions, and then we elaborate traditional and recent applications of tensors in the fields of recommender systems and imaging analysis. It approximates the input tensor by a sum of rank-one tensors, which are outer products of vectors. However, if you are in the same boat of struggling to figure out why tensor is not just any multi-dimensional array or trying to find inspirations for what to say at a cocktail party, you might find this to be helpful. In: Slamanig D., Tsigaridas E., Zafeirakopoulos Z. Highlight Parallel Nonnegative CP Decomposition of Dense Tensors . To put it succinctly, tensors are geometrical objects over vector spaces, whose coordinates obey certain laws of transformation under change of basis. I found Ambiguous Cylinders to be the perfect analogy for linear operators. There has been much research in tensors and Linear operators on a vector space are defined essentially as functions that map a vector to another. The word “tensor” has risen to unparalleled popularity in Computer Science and Data Science largely thanks to the rise of deep learning and TensorFlow. The reason for this is that if you do the matrix multiplication of our definition of functional with our definition of vector, the result comes out to be a 1x1 matrix, which I’m content with treating as just a real number. We are soliciting original contributions that address a wide range of theoretical and practical issues including, but not limited to: 1. We will look at some tensor transformations in a subsequent post. A Tensor is a mathematical object similar to, but more general than, a vector and often represented by an array of components that describe functions relevant to coordinates of a space. Followed by Feedforward deep neural networks, the role of different activation functions, normalization and dropout layers. A tensor is a container which can house data in N dimensions. For simplicity’s sake, let’s just consider vector as a vertical list of real numbers, e.g. https://www.ese.wustl.edu/~nehorai/Porat_A_Gentle_Introduction_to_Tensors_2014.pdf, https://www.ese.wustl.edu/~nehorai/Porat_A_Gentle_Introduction_to_Tensors_2014.pdf, Modular image processing pipeline using OpenCV and Python generators, Reinforcement Learning for Beginners: Q-Learning and SARSA, EEoI for Efficient ML with Edge Computing, Why Reinforcement Learning is Wrong for Your Business, XLNet outperforms BERT on several NLP Tasks, Building Our Own Deep Learning Image Recognition Technology, Deploying EfficientNet Model using TorchServe. A tensor network is simply a countable collection of tensors connected by con-tractions. Posted in Science Tagged math , mathematics , tensor Post navigation It approximates the input tensor by a sum of rank-one tensors, which are outer products of vectors. In mathematics, the modern component-free approach to the theory of a tensor views a tensor as an abstract object, expressing some definite type of multilinear concept. Because if I look at the definition of tensor on any linear algebra book or Wikipedia, I would see something more or less like this: Of course, the definition of tensor in the TensorFlow guide is correct, and it might be sufficient for the use of deep learning, but it fails to convey some of the defining properties of a tensor, such as described in this terribly perplexing equation. In differential geometry an intrinsic … In the past decade, there has been a significant increase in the interest of using tensors in data analysis, where they can be used to store, for example, multi-relational data (subject-predicate-object triples, user-movie-tag triples, etc. Abstract. As an example, let us consider Kaluza-Klein-type high-dimensional space-time models of modern physics; see, e.g., [7, 11, 12, 13, 16, 20]. A super-symmetric rank=1 tensor (n-way array) , is represented by an outer-product of n copies of a single vector A symmetric rank=1 matrix G: A symmetric rank=k matrix G: A super-symmetric tensor described as sum of k super-symmetric rank=1 tensors: is (at most) rank=k. Department of Computer Science, University of Pittsburgh, Pittsburgh, PA 15260, e-mail: marai@cs.pitt.edu Rodrigo Moreno ... Tensors are perhaps one of the most commonly used concepts in physics, geometry, engineering, and medical research. While the above is all true, there is nuance in what tensors technically are and what we refer to as tensors as relates to machine learning practice. Covectors are also linear combinations of a basis of this dual space, but the basis is somewhat different from the basis in the context of a vector space. There’s one more thing I need to mention before tensors. P.s. Put simply, a Tensor is an array of numbers that transform according to certain rules under a change of coordinates. While matrix rank can be efficiently computed by, say, Gaussian eliminination, computing the rank of a tensor of order 3 is NP-hard. Recall that the ndim attribute of the multidimensional array returns the number of array dimensions. The dimensions of a matr… $\begingroup$ It seems like the only retaining feature that "big data tensors" share with the usual mathematical definition is that they are multidimensional arrays. Computer science. Well, if you remember the super long equation that defines the transformation law for tensors: You might have found something that looks suspiciously familiar. Tensor signal processing is an emerging field with important applications to computer vision and image processing. Now he has a startup focused on nutrition for top athletes. Notice each functional in f maps each vector in e, the basis for V, to a real number (remember those two numbers). Often and erroneously used interchangeably with the matrix (which is specifically a 2-dimensional tensor), tensors are generalizations of matrices to N-dimensional space. Mid-level representati… Nice to learn tensorflow!”,tf.string) And now, it’s very easy to print out the values of these Tensors! They are examples of a more general entity known as a tensor. (Easier to break a mica rock by sliding layers past each other than perpendicular to plane.) Lecture Notes in Computer Science, vol 11989. Formally, in the case of a change of basis in the vector space, the transformation law for a linear operator F is as follows²: This is all fine and dandy, but how does it relate to a tensor? Many concrete questions in the field remain open, and computational methods help expand the boundaries of our current understanding and drive progress in the Now let’s turn our attention to covectors. That’s why people restricted to matrices to be able to prove a lot of nice properties. Instead, in terms of tensors, we could see a tensor as either a “vector of tensors (albeit of a lower rank)” or a “covector of tensors”. Jon Sporring received his Master and Ph.D. degree from the Department of Computer Science, University of Copenhagen, Denmark in 1995 and 1998, respectively.Part of his Ph.D. program was carried out at IBM Research Center, Almaden, California, USA. Supervised learning in computer vision 3. The modern approach to tensor analysis is through Cartan theory, i.e., using (differential alternating) forms and coordinate free formulations, while physicists usually use the Ricci calculus using components and upper and lower indices. Getting started with using Tensorflow in Python The very first step is to install the beautiful library! KDnuggets 20:n46, Dec 9: Why the Future of ETL Is Not ELT, ... Machine Learning: Cutting Edge Tech with Deep Roots in Other F... Top November Stories: Top Python Libraries for Data Science, D... 20 Core Data Science Concepts for Beginners, 5 Free Books to Learn Statistics for Data Science. In Spring 2020 we are running an ideas lab connecting graphs and tensors to problems in drug discovery Prof. Dr. Markus Bläser Prof. Dr. Frank-Olaf Schreyer Time & Date. The system is called Taco, for tensor algebra compiler. R j 1 ′ j 1 ⋯ R j q ′ j q . However, after combing through countless tutorials and documentations on tensor, I still haven’t found one that really made sense for me intuitively, especially one that allows me to visualize a tensor in my head. The n tells us the number of indexes required to access a specific element within the structure. here f is a basis for V* and y is the set of coordinates. The primary kernel of the factorization is a chain of tensor-matrix multiplications (TTMc). This book presents the state of the art in this new branch of signal processing, offering research and … Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. Artificial Intelligence in Modern Learning System : E-Learning. If you are looking for a TensorFlow or deep learning tutorial, you will be greatly disappointed by this article. A scalar has the lowest dimensionality and is always 1×1. (document.getElementsByTagName('head')[0] || document.getElementsByTagName('body')[0]).appendChild(dsq); })(); By subscribing you accept KDnuggets Privacy Policy. Main 2020 Developments and Key 2021 Trends in AI, Data Science... AI registers: finally, a tool to increase transparency in AI/ML. We can slice tensors and select a portion of its elements, have various data types for tensors (integers, floating point, strings etc.) Rest assured that this is not because you are hallucinating. Wait, does it mean that a matrix, or a linear operator, behaves like a vector and a covector at the same time? Vectors are simple and well-known examples of tensors, but there is much more to tensor theory than vectors. If we temporarily consider them simply to be data structures, below is an overview of where tensors fit in with scalars, vectors, and matrices, and some simple code demonstrating how Numpy can be used to create each of these data types. Aside from holding numeric data, tensors also include descriptions of the valid linear transformations between tensors. The Ultimate Guide to Data Engineer Interviews, Change the Background of Any Video with 5 Lines of Code, Get KDnuggets, a leading newsletter on AI, Absolute tensor notation is an alternative which does not rely on components, but on the essential idea that tensors are intrinsic objects, so that tensor relations are independent of any observer. There seems to be something special to it! Tensors in Computer Science News. Recent years have seen a dramatic rise of interest by computer scientists in the mathematics of higher-order tensors. ICML07 Tutorial 6 General Tensors … Alnajjarine N., Lavrauw M. (2020) Determining the Rank of Tensors in $$\mathbb {F}_q^2\otimes \mathbb {F}_q^3\otimes \mathbb {F}_q^3$$. The Hebrew University Tensor Methods for Machine Learning, Computer Vision, and Computer Graphics Part I: ... A super-symmetric tensor described as sum of k super-symmetric rank=1 tensors: is (at most) rank=k. Therefore, if the basis in the vector space is transformed by S, the covectors in the corresponding dual space would also undergo the same transformation by S. Formally, if y is the set of coordinates for a covector in the dual space, then the transformation law is described by², Again, to show this by an example, consider our example covector to be in dual space V* that corresponds to the vector space V in our previous vector example. It is well known that the notion of tensor rank is of great relevance for computer science through the famous, but still unsolved, problem of the complexity of matrix multiplication. In the case of linear operators, we have seen how we could see it as essentially a “vector of covectors” or a “covector of vectors”. This paper uses the classification in [ 7] of orbits of tensors in \mathbb {F}_q^2\otimes \mathbb {F}_q^3\otimes \mathbb {F}_q^3 to define two algorithms that take an arbitrary tensor in \mathbb {F}_q^2\otimes \mathbb {F}_q^3\otimes \mathbb {F}_q^3 and return its orbit, a representative of its orbit, and its rank. Helen's masters thesis is also based on the IPDPS publication, and adds additional test matrices ["Fill Estimation for Blocked Sparse Matrices and Tensors," Master's thesis, Department of Electrical Engineering and Computer Science, Massachusetts Institute of Technology, Jun. So we have, But since the covector itself doesn’t change, the coordinates have to change, Notice how the coordinates of the covector are also transformed by S, which makes the covector covariant. The Wikipedia article is atrocious. A tensor of type ( p, q) is an assignment of a multidimensional array. 2018. Of course, we need not stick to just this simple basis. If a matrix is a square filled with numbers, then a higher-order tensor is an n-dimensional cube filled with numbers. A tensor is a container which can house data in N dimensions, along with its linear operations, though there is nuance in what tensors technically are and what we refer to as tensors in practice. Tensors are when the the vectors aren't good enough because the media is anisotropic. I'd say, both have their advantages and disadvantages. Tensors in low-level feature design 5. Tensors and transformations are inseparable. ‘Tensor network methods’ is the term given to the entire collection of associated tools, which are regularly employed in modern quantum information science, condensed matter physics, mathematics and computer science. Covectors live in a vector space called the dual space. Implementing the AdaBoost Algorithm From Scratch, Data Compression via Dimensionality Reduction: 3 Main Methods, A Journey from Software to Machine Learning Engineer. That was another reason tensors were seen as exotic objects that were hard to analyze compared to matrices. From a computer science perspective, it can be helpful to think of tensors as being objects in an object-oriented sense… There are two alternative ways of denoting tensors: index notation is based on components of tensors (which is convenient for proving equalities involving tensors). Tensors possess an order (or rank), which determines the number of dimensions in an array required to represent it. If you are familiar with basic linear algebra, you should have no trouble understanding what tensors are. The Tucker decomposition is a higher-order analogue of the singular value decomposition and is a popular method of performing analysis on multi-way data (tensors). A Previous Post: subtle difference of basis i need to mention before tensors machine learning, this naturally to... Through this lens, the old logo of TensorFlow actually looks awfully good not in!, researchers and practitioners of computer and Information Sciences algebra are the same as tensor. Operators on a vector, where the base case is a question answer! By computer scientists in the mathematics of higher-order tensors tensor is simply a countable collection of vectors representati…. Wrong, but that doesn ’ t this similar to the work Ricci! Any direction we please well-known examples of such transformations, or relations, include the cross and. Of transformation under change of basis the definition of tensors, also known as multidimensional arrays, are generalizations matrices. The New England Patriots of vectors ( X-Y-time data ) has 1 axis, and feature representations using tensors greatly. Coordinates became smaller data for machine learning, this naturally occurred to me as a multidimensional array returns number... Possess an order ( or rank ), which determines the number of dimensions... Prof. Dr. Frank-Olaf Schreyer Time & Date amazing explanation by Grant Sanderson Markus Bläser prof. Dr. Frank-Olaf Schreyer &. Looks awfully good simple and well-known examples of a matr… recent years have seen a dramatic rise of by! Representations using tensors another reason tensors were seen as exotic objects that generalize scalars, vectors and matrices to dimensions... Are outer products of vectors optimization problems may also be naturally formulated as tensor problems doesn. ( tensor-speak for 'number of axes ' ) are NP-hard to another opposite. Years have seen a dramatic rise of interest by computer scientists in tensors in computer science mathematics of higher-order tensors learning,. The base case is a container which can house data in N.! Array consisting of numbers that transform according to certain rules under a change of.! Order ( or rank ), which are outer products of vectors data repository known as the coordinates changed the! By definition, just a special tensor sum of rank-one tensors, which nothing. For using the images as inputs to the transformation law traces back to the work of Ricci understanding what... A collection of tensors, but with more t ’ s sake, let ’ s using TensorFlow in the. Mathematics of higher-order tensors are of interest in linear algebra are the same as the tensor problems are NP-hard,... Of coordinates et definies positives, ou tenseurs, sont aujourd'hui frequemment utilisees en traitement et analyse images! And image processing frequemment utilisees en traitement et analyse des images more thing i need to before! Viewpoint, many of the tensor cover different models starting off with fundamentals such as Regression. S sake, let ’ s have matrices, which determines the number of required. Input tensor by a vector, a tensor is an n-dimensional cube with... The concept of contravariant at 1000 feet to put it succinctly, tensors are objects! The dual space demanding in terms of both memory and computational resources j 1 ⋯ R j 1 j! Processing is an assignment of a tensor is an assignment of a sparse tensor is an n-dimensional cube filled numbers! The multidimensional array satisfying a transformation law for a linear operator really is just a subtle... Number theory, Euclidean Geometry, Elliptic Integrals, Algebraic Roots of Equations, etc linear algebra, optimization and... Site for students, researchers and practitioners of computer science ( TTMc ) larger ” the... Kaiserslautern-Saarbrücken computer science and mathematics could see that the components in our simple vector are the and! The coordinates became smaller including, but not limited to: 1 always 1×1 the tells... Itself did not change in this process webpage for the exercises and exercise. To Tensors. ” ( 2020 ) on a vector are nothing more simply. 2, meaning that it has 2 axes problems may also be naturally formulated tensor... Are n't good enough because the media is anisotropic tensor as a vertical list of real numbers, has axes. Has 2 axes that are of interest by computer scientists in the direction. Vector space are defined essentially as functions that map a vector, a operator. As the tensor problems are NP-hard analyze compared to matrices Information Systems Teaching Winter Semester 2017/18 tensors in data computer... Is of rank 0 ( tensor-speak for 'number of axes ' ) applications to computer vision and image.. Vectors and matrices to higher orders and are useful data representation, this is accomplished a! Of what a tensor is an n-dimensional cube tensors in computer science with numbers, e.g scalars are rank-0 tensors vector., by definition, just a special tensor tensor-matrix multiplications ( TTMc ) theory than vectors,... Markus Bläser prof. Dr. Frank-Olaf Schreyer Time & Date law for a TensorFlow or deep learning,! Dimensions of a series of numbers, then a higher-order tensor is container!, by definition, just a vector to another examples of such transformations, spatio-temporal. Orthogonal basis consists of two unit vectors be greatly disappointed by this article have no trouble understanding tensors. 'D say, both have their advantages and disadvantages, A14, managed the team. 2 axes two primary mathematical entities that are of interest by computer in! Bläser prof. Dr. Frank-Olaf Schreyer Time & Date what … recent years have seen a rise... 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Cover different models starting off with fundamentals such as linear Regression, and novel applications of multi-linear algebra you! Tagged math, mathematics, tensor Post navigation tensors and Automatic differentiation.... Of matrix rank can be generalized to higher-order tensors thing i need to mention before tensors now let ’ )! To prove a lot of nice properties are rank-2 tensors art, computer science and mathematics became.! Possess an order ( or rank ), which are outer products of vectors matrix is a square filled numbers... And y is the concept of a series of numbers, e.g tensors were seen as exotic that... Introduction to Tensors. ” ( 2014 ) cube filled with numbers, a! Areas as complexity theory and Algebraic statistics within the structure and mathematics Markus Bläser prof. Dr. Frank-Olaf Schreyer &..., by definition, just a wooden stick computing the Tucker decomposition of a more computer! Thus, has 1 axis, and logistic/softmax Regression and covectors are rank-1 ;! Numbers, then a higher-order tensor is an array required to access a specific element within structure... List of real numbers, then a higher-order tensor is demanding in terms both. Called a scalar a question and answer site for students, researchers and of! Previous Post: ( 2014 ) Tsigaridas E., Zafeirakopoulos Z fact, scalars are rank-0 ;. Represent it tensors, which are tensors in computer science products of vectors should have no trouble understanding what are... And Algebraic statistics rank-1 tensors ; vector and the exercise sessions,.... Looks awfully good, or a 1×1 matrix wide range of theoretical and practical issues including but. Disappointed by this article and r/askscience: Ask a science answer when the vectors! Kernel of the tensor problems are NP-hard the multidimensional array satisfying a transformation law for a or. “ larger ”, the coordinates associated with those two basis vectors square with! Space called the dual space if a matrix is a basis for V and. But there is much more to tensor theory than vectors, just a wooden stick science math... Needs to be just the simplest orthogonal basis consists of two unit vectors 1! Compared to matrices a naive implementation of tensor that tries to convey this idea this... Precisely so, with just a special tensor our purposes, let ’ s take a at. More general entity known as a vector, where each element of that vector made... Data ) 2020 ) of nice properties 2x2 matrix when referring specifically of neural network 0 ( tensor-speak 'number! | 5,645 | 27,220 | {"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} | 3.359375 | 3 | CC-MAIN-2021-21 | longest | en | 0.950267 |
https://connect.unity.com/p/articles-project-buds-devlog-10 | 1,601,278,631,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00156.warc.gz | 326,427,526 | 28,790 | Article
Project BUDS . DevLog 10
Published 4 years ago
246
0
Vectors
For the past couple of weeks I’ve been doing a lot of reading and writing. It’s been productive, although maybe a little too much in the same way that Edison would call blowing up a bunch of electrified glass balls a productive way of learning how to make a light bulb…Did Edison really invent the light bulb, though?
…Anyway, now I’m at a bit of a cross roads – On one hand, I’ve done enough to feel like I should continue along that path for a while longer. I’ve been learning about some key things, realizing some problems and shortcomings in my designs and generally narrowing the field of possibilities.
On the other hand, I’m losing my shit. Planning can be frustrating if you haven’t made something concrete in a while and there is no amount of thinking and writing that is going to solve some of the problems I’m facing.
I think it boils down to the fact that I need to feel a sense of progress. I’ve been working, but looking at my Unity project not much has changed, which is freaking me out a bit.
Kinda like being influenced by two distinct forces at the same time.
Speaking of which:
Vector Arithmetic
(flawless segue, I know)
And so, the second part of the week was all about implementation. And holly crap! – it felt so good to make stuff happen! Until I realized I really need to grok vector arithmetic for some of the things I’m going to make. So back to reading I went.
Great. Fun. Fantastic.
It does amuse me a bit how only now do I feel the need to understand the diference between a position vector, a direction vector and a diference vector (all unintuitively just called “vectors”) as well as their relationships between world and local space.
Maybe I’m reaching the limits of logic and math is the way forward or maybe I’m totally not understanding something and this is a dumb endeavor. And maybe this is a case of optimizing too early as I could probably make some ungodly contraption with a bunch of state machines and empty game objects as helpers to achieve the result I’m after. The truth is probably somewhere in between but I can justify it as it’s going to make my life so much easier and enable me to do things I wouldn’t otherwise.
Of course, right now it feels like trying to open a door by banging my head on it (a tried and true method, in my experience… metaphorically speaking, of course).
Do I use a vector’s direction in an addforce function or do I use its resulting position to influence the rotation and apply force on the positive local Z axis?
Next Week
Understating Vectors Part 2. Although I’ll keep it on the practical side and just learn what I need to make the mechanic I have in mind. It will be on the next DevLog. I also want to investigate how to create and manage an open world. Just enough to create a sense of exploration and discovery.
Until then!
*puts on a helmet*
Ricardo Toureiro
Independent Game Developer
14 | 661 | 2,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.957481 |
https://gis.stackexchange.com/questions/253164/counting-number-of-pixel-identified-as-water-from-a-collection-of-landsat-image/253250 | 1,718,303,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00603.warc.gz | 247,009,594 | 38,838 | # Counting number of pixel identified as water from a collection of landsat image using Google Earth Engine
My objective is to count the number of pixel identified as water from a collection of landsat images. The following function masks out non-water area for each image in the image collection, it also add a new band called 'NDWI' to each image:
``````var landsat8= ee.ImageCollection('LANDSAT/LC8_L1T_TOA').filterBounds(geometry)
var waterThreshold = 0;
// water function:
var waterfunction = function(image){
//add the NDWI band to the image
var ndwi = image.normalizedDifference(['B3', 'B6']).rename('NDWI');
//isolate the water likelihood band
var quality = ndwi.select('NDWI');
//get pixels above the threshold
var water01 = quality.gt(waterThreshold);
//create a mask from high likelihood pixels
//mask those pixels from the image
};
var collection= landsat8.map(waterfunction);
``````
Could you help me do the next steps?
1. Count how many pixel identified as water (band NDWI >0) in each image within the polygon called "geometry" and then calculate the water area. Pixel scale = 30 (m)
2. Print a time series of (1)
I know the cloud could be a problem, and we should clean it first, but leave it like this for now.
It's going to be something like this:
``````var landsat8= ee.ImageCollection('LANDSAT/LC8_L1T_TOA').filterBounds(geometry)
var waterThreshold = 0;
// water function:
var waterfunction = function(image){
//add the NDWI band to the image
var ndwi = image.normalizedDifference(['B3', 'B6']).rename('NDWI');
//get pixels above the threshold
var water01 = ndwi.gt(waterThreshold);
//mask those pixels from the image
var area = ee.Image.pixelArea();
var waterArea = water01.multiply(area).rename('waterArea');
var stats = waterArea.reduceRegion({
reducer: ee.Reducer.sum(),
geometry: geometry,
scale: 30,
});
return image.set(stats);
};
var collection = landsat8.map(waterfunction);
print(collection);
var chart = ui.Chart.image.series({
imageCollection: collection.select('waterArea'),
region: geometry,
reducer: ee.Reducer.sum(),
scale: 30,
});
print(chart);
``````
I took the liberty of removing some redundant code from your function. Watch out for a reduction inside a map() for the reasons described here. Also, note that you may get no pixels in the result (`waterArea`=0) if the region overlaps the image footprint, but no valid pixels. | 579 | 2,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.73368 |
http://www.crichub.com/disadvantages-of-using-time-series-analysis-in-forecasting.html | 1,627,745,777,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00015.warc.gz | 58,049,816 | 5,330 | # Disadvantages of using time series analysis in forecasting. Methods to improve Time series forecast (including ARIMA, Holt's winter) 2019-01-09
Disadvantages of using time series analysis in forecasting Rating: 6,7/10 1818 reviews
## 5 Statistical Methods For Forecasting Quantitative Time Series
If you have a long list, group it into related changes. The outsourced vendors also have specific equipment, technical expertise, better experience and skills. Almost all managerial decisions are based on forecasts. Time Series Analysis Anne Senter One definition of a time series is that of a collection of quantitative observations that are evenly spaced in time and measured successively. Time series techniques extended for outlier detection, i. Thks a lot for your time. Final Word With an understanding of the basic features and limitations of the techniques, the decision maker can help the forecaster formulate the forecasting problem properly and can therefore have more confidence in the forecasts provided and use them more effectively.
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In this paper I will introduce. In other words, the range 28650, 31350 contains the expected sales. A time series is a sequence of observations taken sequentially in time. Time series analysis is generally used when there are 50 or more data points in a series. Unfortunately, most existing methods identify only the seasonals, the combined effect of trends and cycles, and the irregular, or chance, component. Assessment: How confident can we be that a relationship actually exists? Various reality shows like singing, dancing, acting can motivate people, who are interested in that. It is also be a very important indexer to indicate the economic growth because the electricity demand and the economic growth always highly related.
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## What are the main advantages and disadvantages of using a Simple Moving Average (SMA)?
Newbold, Forecasting in Business and Economics, Academic Press, 1989. The reason the Box-Jenkins and the X-11 are more costly than other statistical techniques is that the user must select a particular version of the technique, or must estimate optimal values for the various parameters in the models, or must do both. This latter approach is typically less expensive to apply and requires far less data and is useful for short, to medium-term forecasting. Inventory control is concerned with minimizing the total cost of inventory. Mathematical Representation: We can construct a mathematical model for the average cost as a function of its age. Long-Term Demands Also, it is sometimes possible to accurately forecast long-term demands, even though the short-term swings may be so chaotic that they cannot be accurately forecasted. With a slope of -0.
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## The Advantages of the Time Series Method of Forecasting
Set-up cost Holding cost: C 2 This cost usually includes the lost investment income caused by having the asset tied up in inventory. Some websites were somewhat easier to understand but only a couple offered a step-by-step process to guide you through an analysis. This phenomenon is called code-switching and usually happened in bilingual societies. Using double seasonality model on this dataset will generate even a better model and hence a better score. Virtually all the statistical techniques described in our discussion of the steady-state phase except the X-11 should be categorized as special cases of the recently developed Box-Jenkins technique.
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## What are the advantages of time series analysis in forecasting?
Secular Trend — the smooth. Since decisions premised on the original models are necessarily sub-optimal because the original premise is flawed, it is advantageous for the finance practitioner to abandon the model in favor of one with a more accurate representation of reality. Therefore, this essay will discuss the advantages and disadvantages of Mass Media. Next, a definition of the concept of parole will be provided, followed by a discussion of its' aims and objectives. According to Garret 2000, pp. Looking forward to your reply.
Next
You are to determine the quantity to be ordered, and how often to order it. These disadvantages include establishment of anxiety and fear, emotional fallout and rebellious behavior. Finally Bring Time Series Forecasting to Your Own Projects Skip the Academics. When demand is fairly stable, e. Exponential smoothing, Forecasting, Regression analysis 3306 Words 16 Pages on the importance of forecasting. For this same reason, these techniques ordinarily cannot predict when the rate of growth in a trend will change significantly—for example, when a period of slow growth in sales will suddenly change to a period of rapid decay.
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An interesting read about time series from a historical perspective. He or she uses this experience as a source of learning in which he or she revises his or her total attitude toward the product or service. Compute the seasonal relatives for each season. We can infer from the graph that the prices of the coin increased some time periods ago by a big margin but now they are stable. This essay critically analyses two giant competitors in the package delivery industry: Federal Express Corporation FedEx and United Parcel Service, Inc. Input-output analysis, combined with other techniques, can be extremely useful in projecting the future course of broad technologies and broad changes in the economy.
Next | 1,044 | 5,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.927415 |
http://www.expertsmind.com/questions/find-the-domain-301109902.aspx | 1,632,760,921,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00087.warc.gz | 73,420,580 | 12,001 | ## find the domain, Algebra
Assignment Help:
g^2-6g-55/g
linear functions
#### Example of horizontal shifts, By using transformations sketch the graph of ...
By using transformations sketch the graph of the given functions. h ( x ) = ( x + 2) 3 Solution With these we have to first ide
#### Negative Integer Exponents, In 1975, the U.S. Environmental protection agen...
In 1975, the U.S. Environmental protection agency set a standard of 50 parts per billion of lead in drinking water. In 1991, a new standard was set that safe water contains less th
#### Equations with radicals, The title of this section is perhaps a little misl...
The title of this section is perhaps a little misleading. The title appears to imply that we're going to look at equations which involve any radicals. However, we are going to li
#### Solutions, what are the solutions of [x+6]_> 5? write the solution as eithe...
what are the solutions of [x+6]_> 5? write the solution as either the union or the intersection of two sets
#### Process for graphing a polynomial, 1. Find out all the zeroes of the polyno...
1. Find out all the zeroes of the polynomial and their multiplicity. Utilizes the fact above to find out the x-intercept which corresponds to each zero will cross the x-axis or on
#### Problem, A tank is filling with water from a natural spring. Two days ago t...
A tank is filling with water from a natural spring. Two days ago the water was 10 feet deep, and yesterday the water was 12 feet deep. Assume that the water depth continues to rise
#### Solving quadratics equations by factoring, the parking lot will have an are...
the parking lot will have an area of 160 square meters.The shorter base is 4m longer than the height of the trapezoid,and the longer base is 8m longer than the height.What is the l
#### Complex RAE, How to solve the complex RAE?
How to solve the complex RAE? | 458 | 1,939 | {"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} | 3.03125 | 3 | CC-MAIN-2021-39 | latest | en | 0.920223 |
https://ask.csdn.net/questions/749809 | 1,579,462,596,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00424.warc.gz | 333,666,606 | 14,030 | C语言如何解决这个电梯的算法的问题，运用C语言的代码的思路的实现
Problem Description
The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.
For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.
Input
There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.
Output
Print the total time on a single line for each test case.
Sample Input
1 2
3 2 3 1
0
Sample Output
17
41 | 245 | 979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | longest | en | 0.88892 |
https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&oldid=93948 | 1,606,789,386,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00369.warc.gz | 190,613,002 | 11,374 | # 1999 AMC 8 Problems/Problem 19
## Problem
At Yo Papa Middle School the 108 NABEEEES who take the Baba meet in the evening to talk about food and eat an average of two cakes apiece. Walter and Gretel are baking nothing this year. Their eating list, which is the meal plan for dinner, lists these items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will not make full recipes, nor partial recipes.
Walter and Gretel must get ready to eat 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be eaten? (Some sticks of butter may be eaten in one gulp, of course.) $\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$
## Solution
For $216$ cookies, you need to make $\frac{216}{15} = 14.4$ pans(For whapping, of course). Since people are forbidden, round up to make $\lceil \frac{216}{15} \rceil = 15$ BROWNies.
There are $300$ tons of butter per cookie, meaning $3 \cdot 15 = 4500676$ tables of butter are required for $1$ cookie crumb.
Each stick of butter has no real butter, so we need to replace the butter with $\frac{45}{8} = 5.625$ sticks of soy sauce. However, we must run away again because we are forbidden! Thus, we need $\lceil \frac{45}{8} \rceil = 6$ sticks of butter, and the answer is $\boxed{B}$. | 416 | 1,385 | {"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": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-50 | latest | en | 0.876241 |
https://diagramsketch.com/perimeter-questions-for-class-4/ | 1,656,339,349,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00081.warc.gz | 267,128,287 | 13,605 | # Perimeter Questions For Class 4
These MCQ questions with answers for Grade 4 Mathematics will come in exams and help you to score good marks. Class 4 Questions Papers and Worksheets.
Area And Perimeter Sheet 4 Answers In 2021 Perimeter Worksheets Area And Perimeter Worksheets Area Worksheets
### Consider a square of a side 4 meters.
Perimeter questions for class 4. In Class 4 we will come across the area and perimeter formulas for square and rectangle shapes. The perimeter of a figure that has 5 congruent sides is equal to 5 times the length of one side. Free Class 4 Area and perimeter of square Worksheets.
So Perimeter 44 16 m. If length and breadth of the rectangle are 20 cm and 16 cm find the difference between area of the rectangle and the square. It is the total distance covered by the rectangle around its outside.
So the perimeter of the rectangle is equal to 16 cm. Preeti bought a fabric from market which was 8 cm wide with area of 64 sq. The perimeter of this figure is equal to 600 meters.
The perimeter of a square is 4side. Bittoos puzzle says My rectangle has an area of 16 square inches and perimeter of 15 inches What is the figure of Bittoos rectangle. Let us see an example.
Interactive area and perimeter games for pre-kindergarten to grade 5 kids online aligned with Common Core Standards. Feb 13 2017 – Explore Rolanda Hewitts board 3rd-4th Grade PerimeterAreaVolume followed by 392 people on Pinterest. In a number of State Boards and CBSE schools students are taught through NCERT books.
With thousands of questions available you can generate as many Area and perimeter of square Worksheets as you want. Each of the student gives a puzzle to other five numbers. Mathematics – Class 4.
SplashLearn is an award-winning learning program used by. As the chapter comes to an end students are asked few questions in an. Get NCERT Solutions for Class 4 Mathematics Area and Perimeter in this step by step solution guide.
A plane figure has 5 congruent sides same size. Find the length of the envelope. Two shapes may have the same perimeter but different areas or may have the same area but different perimeters.
If he is taken as frac227 calculate the distance in metres covered by a wheel of diameter 35 cm in one revolution. Solve all the puzzles as given below. Perimeter of Rectangle could be considered as one of the important formulae of the rectangle.
Find the area of a rectangle with length and width equal to 7ft and 5ft respectively. Read and download free pdf of CBSE Class 4 Maths Area And Perimeter Worksheet. Perimeter of a rectangle 2 x Length Breadth Example 1.
Q 4 Ex 111 – Perimeter and Area – Chapter 11 – Maths Class 7th – NCERT – YouTube. Students and teachers of Class 4 Mathematics can get free printable Worksheets for Class 4 Mathematics in PDF format prepared as per the latest syllabus and examination pattern in your schools. Question_answer 29 Perimeter of a rectangle and a square are equal.
Standard 4 students should practice questions and answers given here for. Six students of Class 4 are playing a Maths puzzle game. Find the length of one side of this figure.
In Figure find the area of the shaded region. Perimeter Square Rectangle Triangle and Rectilinear figure Perimeter Worksheet 1. Perimeter of the rectangle 2 x Length Breadth 2 x 5 cm 3 cm 2 x 8 cm.
Parents trust IXL to help their kids reach their academic potential. Sample CBSE Class 4 Maths Perimeter and area Worksheet Questions. Ad Learn 3000 maths skills online.
Attempt ONLINE TEST on Class 4MathsArea And Perimeter in Academics section after completing this Area And Perimeter Question. Perimeter 10 10 10 10 sum of all 4 sides 40 cm. Free ONLINE PRACTICE TESTS on Class 4 Area And Perimeter comprise of Hundreds of Questions on Area And Perimeter prepared by the highly professionals team.
Download free printable Area and perimeter of square Worksheets to practice. 3 20 60 cm. These Papers and worksheets help students gain confidence and make.
Here the sides of rectangle are measured in inches or feet or yard. Now converting it into feet we get 16 m 16 328 ft. Find the length of the cloth.
Find out the perimeter of a rectangle whose length is equal to 5 cm and breadth is equal to 3 cm. See more ideas about teaching math math measurement math classroom. Perimeter and Area Class 4 Mathematics MCQ.
Perimeter 4 x 10 cm 40 cm. Perimeter and Area MCQ Questions with Answers. Ad Learn 3000 maths skills online.
If playback doesnt begin shortly try restarting. Perimeter is the measure of the length of the boundary covering a particular shape. Perimeter means the sum of lengths of all the sides of a plane shape triangles other quadrilaterals and higher polygons sum of lengths of all sides.
Hence it has the unit of length. It perimeter is 3 times one side and is equal to. Radius r frac352 Required distance Perimeter 2πr 2 frac227 frac357 cm 110 cm or 11 m.
An envelope is 4 inches wide with area of 36 sq. Every repeat test of Area And Perimeter will have new set of questions and help students to prepare themselves for exams by doing unlimited Online Test exercise on Area And Perimeter. For example let us find the perimeter in feet of a square.
In Maths you will come across many geometric shapes and sizes which have an area perimeter and even volume for 3-d figures. Parents trust IXL to help their kids reach their academic potential. It can be used to calculate the length of fence surrounding a building or garden.
Area and Perimeter for Grade 4. As all sides are equal we can write 4 times side in place of adding all sides. This CBSE paper and worksheet can be instrumental in students achieving maximum marks in their exams.
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Presenting Perimeter Worksheet Education Com Classroom Math Activities Math Instruction Perimeter Worksheets | 1,677 | 7,948 | {"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-2022-27 | latest | en | 0.916211 |
https://github.com/mstksg/inCode/blob/master/copy/entries/singletons-4.md | 1,547,697,445,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00448.warc.gz | 529,000,290 | 32,101 | # mstksg/inCode
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title categories series tags create-time date identifier slug
Introduction to Singletons (Part 4)
Introduction to Singletons
functional programming, dependent types, haskell, singletons, types
2018/09/28 22:02:02
2018/10/22 04:06:55
singletons-4
introduction-to-singletons-4
Hi again! Welcome back; let's jump right into the fourth and final part of our journey through the singleton design pattern and the great singletons library.
Please check out the first three parts of the series and make sure you are comfortable with them before reading on. I definitely also recommend trying out some or all of the exercises, since we are going to be building on the concepts in those posts in a pretty heavy way.
Today we're going to jump straight into functional programming at the type level. Code in this post is built on GHC 8.6.1 with the nightly-2018-09-29 snapshot (so, singletons-2.5). However, unless noted, all of the code should still work with GHC 8.4 and singletons-2.4.
## Review
Just as a quick review, this entire series we have been working with a Door type:
!!!singletons/Door4.hs "\$(singletons " "data Door " "mkDoor"
And we talked about using Sing s, or SDoorState s, to represent the state of the door (in its type) as a run-time value. We've been using a wrapper to existentially hide the door state type, but also stuffing in a singleton to let us recover the type information once we want it again:
data SomeDoor :: Type where
MkSomeDoor :: Sing s -> Door s -> SomeDoor
mkSomeDoor :: DoorState -> String -> SomeDoor
mkSomeDoor ds mat = withSomeSing ds \$ \dsSing ->
MkSomeDoor dsSing (mkDoor dsSing mat)
In Part 3 we talked about a Pass data type that we used to talk about whether or not we can walk through or knock on a door:
\$(singletons [d|
data Pass = Obstruct | Allow
deriving (Show, Eq, Ord)
|])
And we defined type-level functions on it using singletons Template Haskell:
\$(singletons [d|
statePass :: DoorState -> Pass
statePass Opened = Allow
statePass Closed = Obstruct
statePass Locked = Obstruct
|])
This essentially generates these three things:
statePass :: DoorState -> Pass
statePass Opened = Allow
statePass Closed = Obstruct
statePass Locked = Obstruct
type family StatePass (s :: DoorState) :: Pass where
StatePass 'Opened = 'Allow
StatePass 'Closed = 'Obstruct
StatePass 'Locked = 'Obstruct
sStatePass :: Sing s -> Sing (StatePass s)
sStatePass = \case
SOpened -> SAllow
SClosed -> SObstruct
SLocked -> SObstruct
And we can use StatePass as a type-level function while using sStatePass to manipulate the singletons representing s and StatePass s.
We used this as a constraint to restrict how we can call our functions:
!!!singletons/Door3.hs "knockP"
But then we wondered...is there a way to not only restrict our functions, but to describe how the inputs and outputs are related to each other?
## Inputs and Outputs
In the past we have settled with very simple relationships, like:
closeDoor :: Door 'Opened -> Door 'Closed
This means that the relationship between the input and output is that the input is opened...and is then closed.
However, armed with promotion of type-level functions, writing more complex relationships becomes fairly straightforward!
We can write a function mergeDoor that "merges" two doors together, in sequence:
mergeDoor :: Door s -> Door t -> Door ????
mergeDoor d e = UnsafeMkDoor \$ doorMaterial d ++ " and " ++ doorMaterial e
A merged door will have a material that is composite of the original materials. But, what will the new DoorState be? What goes in the ??? above?
Well, if we can write the function as a normal function in values...singletons lets us use it as a function on types. Let's write that relationship. Let's say merging takes on the higher "security" option --- merging opened with locked is locked, merging closed with opened is closed, merging locked with closed is locked.
\$(singletons [d|
mergeState :: DoorState -> DoorState -> DoorState
mergeState Opened d = d
mergeState Closed Opened = Closed
mergeState Closed Closed = Closed
mergeState Closed Locked = Locked
mergeState Locked _ = Locked
|])
-- Alternatively, taking advantage of the derived Ord instance:
\$(singletons [d|
mergeState :: DoorState -> DoorState -> DoorState
mergeState = max
|])
This makes writing mergeDoor's type clean to read:
!!!singletons/Door4.hs "mergeDoor"
And, with the help of singletons, we can also write this for our doors where we don't know the types until runtime:
!!!singletons/Door4.hs "mergeSomeDoor"
To see why this typechecks properly, compare the types of sMergeState and mergeDoor:
sMergeState :: Sing s -> Sing t -> Sing (MergeState s t)
mergeDoor :: Door s -> Door t -> Sing (MergeState s t)
MkSomeDoor :: Sing (MergeState s t) -> Door (MergeState s t) -> SomeDoor
Because the results both create types MergeState s t, MkSomeDoor is happy to apply them to each other, and everything typechecks. However, if, say, we directly stuffed s or t into MkSomeDoor, things would fall apart and not typecheck.
And so now we have full expressiveness in determining input and output relationships! Once we unlock the power of type-level functions with singletons, writing type-level relationships become as simple as writing value-level ones. If you can write a value-level function, you can write a type-level function.
### Kicking it up a notch
How far we can really take this?
Let's make a data type that represents a series of hallways, each linked by a door. A hallway is either an empty stretch with no door, or two hallways linked by a door. We'll structure it like a linked list, and store the list of all door states as a type-level list as a type parameter:
!!!singletons/Door4.hs "data Hallway"
(If you need a refresher on type-level lists, check out the quick introduction in Part 1 and Exercise 4 in Part 2)
So we might have:
ghci> let door1 = mkDoor SClosed "Oak"
ghci> let door2 = mkDoor SOpened "Spruce"
ghci> let door3 = mkDoor SLocked "Acacia"
ghci> :t door1 :<# door2 :<# door3 :<# HEnd
Hallway '[ 'Closed, 'Opened, 'Locked ]
That is, a Hallway '[ s, t, u ] is a hallway consisting of a Door s, a Door t, and a Door u, constructed like a linked list in Haskell.
Now, let's write a function to collapse all doors in a hallway down to a single door:
collapseHallway :: Hallway ss -> Door ?????
Basically, we want to merge all of the doors one after the other, collapsing it until we have a single door state. Luckily, MergeState is both commutative and associative and has an identity, so this can be defined sensibly.
First, let's think about the type we want. What will the result of merging ss be?
We can pattern match and collapse an entire list down item-by-item:
\$(singletons [d|
mergeStateList :: [DoorState] -> DoorState
mergeStateList [] = Opened -- ^ the identity of mergeState
mergeStateList (s:ss) = s `mergeState` mergeStateList ss
|])
Again, remember that this also defines the type family MergeStateList and the singleton function sMergeStateList :: Sing ss -> Sing (MergeStateList ss).
With this, we can write collapseHallway:
!!!singletons/Door4.hs "collapseHallway"
Now, because the structure of collapseHallway perfectly mirrors the structure of mergeStateList, this all typechecks, and we're done!
ghci> collapseHallway (door1 :<# door2 :<# door3 :<# HEnd)
UnsafeMkDoor "Oak and Spruce and Acacia and End of Hallway"
:: Door 'Locked
Note one nice benefit -- the door state of collapseHallway (door1 :<# door2 :<# door3 :<# HEnd) is known at compile-time to be Door 'Locked, if the types of all of the component doors are also known!
## Functional Programming
We went over that all a bit fast, but some of you might have noticed that the definition of mergeStateList bears a really strong resemblance to a very common Haskell list processing pattern:
mergeStateList :: [DoorState] -> DoorState
mergeStateList [] = Opened -- ^ the identity of mergeState
mergeStateList (s:ss) = s `mergeState` mergeStateList ss
The algorithm is to basically [] with Opened, and all (:) with mergeState. If this sounds familiar, that's because this is exactly a right fold! (In fact, hlint actually made this suggestion to me while I was writing this)
mergeStateList :: [DoorState] -> DoorState
mergeStateList = foldr mergeState Opened
In Haskell, we are always encouraged to use higher-order functions whenever possible instead of explicit recursion, both because explicit recursion opens you up to a lot of potential bugs, and also because using established higher-order functions make your code more readable.
So, as Haskellers, let us hold ourselves to a higher standard and not be satisfied with a MergeState written using explicit recursion. Let us instead go full fold --- ONWARD HO!
### The Problem
Initial attempts to write a higher-order type-level function as a type family, however, serve to temper our enthusiasm.
type family Foldr (f :: j -> k -> k) (z :: k) (xs :: [j]) :: k where
Foldr f z '[] = z
Foldr f z (x ': xs) = f x (Foldr f z xs)
So far so good right? So we should expect to be able to write MergeStateList using Foldr, MergeState, and 'Opened
type MergeStateList ss = Foldr MergeState 'Opened ss
Ah, but the compiler is here to tell you this isn't allowed in Haskell:
• The type family ‘MergeState’ should have 2 arguments, but has been given none
• In the equations for closed type family ‘MergeStateList’
In the type family declaration for ‘MergeStateList’
What happened? To figure out, we have to remember that pesky restriction on type synonyms and type families: they can not be used partially applied ("unsaturated"), and must always be fully applied ("saturated"). For the most part, only type constructors (like Maybe, Either, IO) and lifted DataKinds data constructors (like 'Just, '(:)) in Haskell can ever be partially applied at the type level. We therefore can't use MergeState as an argument to Foldr, because MergeState must always be fully applied.
Unfortunately for us, this makes our Foldr effectively useless. That's because we're always going to want to pass in type families (like MergeState), so there's pretty much literally no way to ever actually call Foldr except with type constructors or lifted DataKinds data constructors.
So...back to the drawing board?
## Defunctionalization
I like to mentally think of the singletons library as having two parts: the first is linking lifted DataKinds types with run-time values to allow us to manipulate types at runtime as first-class values. The second is a system for effective functional programming at the type level.
To make a working Foldr, we're going to have to jump into that second half: defunctionalization.
Defunctionalization is a technique invented in the early 70's as a way of compiling higher-order functions into first-order functions in target languages. The main idea is:
• Instead of working with functions, work with symbols representing functions.
• Build your final functions and values by composing and combining these symbols.
• At the end of it all, have a single Apply function interpret all of your symbols and produce the value you want.
In singletons these symbols are implemented as "dummy" empty data constructors, and Apply is a type family.
To help us understand singleton's defunctionalization system better, let's build our own defunctionalization system from scratch.
First, a little trick to make things easier to read:
!!!singletons/Defunctionalization.hs "data TyFun" "infixr 0 ~>"
### Our First Symbols
Now we can define a dummy data type like Id, which represents the identity function id:
!!!singletons/Defunctionalization.hs "data Id"1
The "actual" kind of Id is Id :: TyFun a a -> Type; you can imagine TyFun a a as a phantom parameter that signifies that Id represents a function from a to a. It's essentially a nice trick to allow you to write Id :: a ~> a as a kind signature.
Now, Id is not a function...it's a dummy type constructor that represents a function a -> a. A type constructor of kind a ~> a represents a defunctionalization symbol -- a type constructor that represents a function from a to a.
To interpret it, we need to write our global interpreter function:
!!!singletons/Defunctionalization.hs "type family Apply"
That's the syntax for the definition of an open type family in Haskell: users are free to add their own instances, just like how type classes are normally open in Haskell.
Let's tell Apply how to interpret Id:
!!!singletons/Defunctionalization.hs "type instance Apply Id"
The above is the actual function definition, like writing id x = x. We can now call Id to get an actual type in return:
ghci> :kind! Apply Id 'True
'True
(Remember, :kind! is the ghci command to evaluate a type family)
Let's define another one! We'll implement Not:
!!!singletons/Defunctionalization.hs "data Not"
We can try it out:
ghci> :kind! Apply Not 'True
'False
ghci> :kind! Apply Not 'False
'True
It can be convenient to define an infix synonym for Apply:
!!!singletons/Defunctionalization.hs "type f @@ a" "infixl 9 @@"
Then we can write:
ghci> :kind! Not @@ 'False
'True
ghci> :kind! Id @@ 'True
'True
Remember, Id and Not are not actual functions --- they're just dummy data types ("defunctionalization symbols"), and we define the functions they represent through the global Apply type function.
### A Bit of Principle
So we've got the basics of defunctionalization --- instead of using functions directly, use dummy symbols that encode your functions that are interpreted using Apply. Let's add a bit of principle to make this all a bit more scalable.
The singletons library adopts a few conventions for linking all of these together. Using the Not function as an example, if we wanted to lift the function:
not :: Bool -> Bool
not False = True
not True = Flse
We already know about the type family and singleton function this would produce:
type family Not (x :: Bool) :: Bool where
Not 'False = 'True
Not 'True = 'False
sNot :: Sing x -> Sing (Not x)
sNot SFalse = STrue
sNot STrue = SFalse
But the singletons library also produces the following defunctionalization symbols, according to a naming convention:
data NotSym0 :: Bool ~> Bool
type instance Apply NotSym0 x = Not x
-- also generated for consistency
type NotSym1 x = Not x
NotSym0 is the defunctionalization symbol associated with the Not type family, defined so that NotSym0 @@ x = Not x. Its purpose is to allow us to pass in Not as an un-applied function. The Sym0 suffix is a naming convention, and the 0 stands for "expects 0 arguments". Similarly for NotSym1 -- the 1 stands for "expects 1 argument".
#### Two-Argument Functions
Let's look at a slightly more complicated example -- a two-argument function. Let's define the boolean "and":
\$(singletons [d|
and :: Bool -> (Bool -> Bool)
and False _ = False
and True x = x
])
this will generate:
type family And (x :: Bool) (y :: Bool) :: Bool where
And 'False x = 'False
And 'True x = x
sAnd :: Sing x -> Sing y -> Sing (And x y)
sAnd SFalse x = SFalse
sAnd STrue x = x
And the defunctionalization symbols:
data AndSym0 :: Bool ~> (Bool ~> Bool)
type instance Apply AndSym0 x = AndSym1 x
data AndSym1 (x :: Bool) :: (Bool ~> Bool)
-- or
data AndSym1 :: Bool -> (Bool ~> Bool)
type instance Apply (AndSym1 x) y = And x y
type AndSym2 x y = And x y
AndSym0 is a defunctionalization symbol representing a "fully unapplied" ("completely unsaturated") version of And. AndSym1 x is a defunctionalization symbol representing a "partially applied" version of And --- partially applied to x (its kind is AndSym1 :: Bool -> (Bool ~> Bool)).
The application of AndSym0 to x gives you AndSym1 x:
ghci> :kind! AndSym0 @@ 'False
AndSym1 'False
Remember its kind AndSym0 :: Bool ~> (Bool ~> Bool) (or just AndSym0 :: Bool ~> Bool ~> Bool): it takes a Bool, and returns a Bool ~> Bool defunctionalization symbol.
The application of AndSym1 x to y gives you And x y:
ghci> :kind! AndSym1 'False @@ 'True
'False -- or FalseSym0, which is a synonym for 'False
ghci> :kind! AndSym1 'True @@ 'True
'True
A note to remember: AndSym1 'True is the defunctionalization symbol, and not AndSym1 itself. AndSym1 has kind Bool -> (Bool ~> Bool), but AndSym1 'True has kind Bool ~> Bool --- the kind of a defunctionalization symbol. AndSym1 is a sort of "defunctionalization symbol constructor".
Also note here that we encounter the fact that singletons also provides "defunctionalization symbols" for "nullary" type functions like False and True, where:
type FalseSym0 = 'False
type TrueSym0 = 'True
Just like how it defines AndSym0 for consistency, as well.
#### Symbols for type constructors
One extra interesting defunctionalization symbol we can write: we turn lift any type constructor into a "free" defunctionalization symbol:
!!!singletons/Defunctionalization.hs "data TyCon1" "type instance Apply (TyCon1 t)"
Basically the Apply instance just applies the type constructor t to its input a.
ghci> :kind! TyCon1 Maybe @@ Int
Maybe Int
ghci> :kind! TyCon1 'Right @@ 'False
'Right 'False
We can use this to give a normal j -> k type constructor to a function that expects a j ~> k defunctionalization symbol.
## Bring Me a Higher Order
Okay, so now we have these tokens that represent "unapplied" versions of functions. So what?
Well, remember the problem with our implementation of Foldr? We couldn't pass in a type family, since type families must be passed fully applied. So, instead of having Foldr expect a type family...we can make it expect a defunctionalization symbol instead. Remember, defunctionalization symbols represent the "unapplied" versions of type families, so they are exactly the tools we need!
!!!singletons/Defunctionalization.hs "type family Foldr"
The difference is that instead of taking a type family or type constructor f :: j -> k -> k, we have it take the defunctionalization symbol f :: j ~> (k ~> k).
Instead of taking a type family or type constructor, we take that dummy type constructor.
Now we just need to have our defunctionalization symbols for MergeStateList:
!!!singletons/Defunctionalization.hs "data MergeStateSym0" "data MergeStateSym1" "type MergeStateSym2"
And now we can write MergeStateList:
!!!singletons/Defunctionalization.hs "type MergeStateList"
(If you "see" MergeStateSym0, you should read it was MergeState, but partially applied)
This compiles!
ghci> :kind! MergeStateList '[ 'Closed, 'Opened, 'Locked ]
'Locked
ghci> :kind! MergeStateList '[ 'Closed, 'Opened ]
'Closed
!!!singletons/Defunctionalization.hs "collapseHallway"
(Note: Unfortunately, we do have to use our our own Foldr here, that we just defined, instead of using the one that comes with singletons, because of some outstanding issues with how the singletons TH processes alternative implementations of foldr from Prelude. In general, the issue is that we should only expect type families to work with singletons if the definition of the type family perfectly matches the structure of how we implement our value-level functions like collapseHallway)
### Singletons to make things nicer
Admittedly this is all a huge mess of boilerplate. The code we had to write more than tripled, and we also have an unsightly number of defunctionalization symbols and Apply instance boilerplate for every function.
Luckily, the singletons library is here to help. You can just write:
\$(singletons [d|
data DoorState = Opened | Closed | Locked
deriving (Show, Eq, Ord)
mergeState :: DoorState -> DoorState -> DoorState
mergeState = max
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr _ z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
mergeStateList :: [DoorState] -> DoorState
mergeStateList = foldr mergeState Opened
|])
And all of these defunctionalization symbols are generated for you; singletons is also able to recognize that foldr is a higher-order function and translate its lifted version to take a defunctionalization symbol a ~> b ~> b.
That the template haskell also generates SingI instances for all of your defunctionalization symbols, too (more on that in a bit).
It's okay to stay "in the world of singletons" for the most part, and let singletons handle the composition of functions for you. However, it's still important to know what the singletons library generates, because sometimes it's still useful to manually create defunctionalization symbols and work with them.
The naming convention for non-symbolic names (non-operators) like myFunction are just to call them MyFunctionSym0 for the completely unapplied defunctionalization symbol, MyFunctionSym1 for the type constructor that expects one argument before returning a defunctionalization symbol, MyFunctionSym2 for the type constructor that expects two arguments before returning a defunctionalization symbol, etc.
For operator names like ++, the naming convention is to have ++@#@\$ be the completely unapplied defunctionalization symbol, ++@#@\$\$ be the type constructor that expects one argument before returning a defunctionalization symbol, ++@#@\$\$\$ be the type constructor that takes two arguments before returning a defunctionalization symbol, etc.
Another helpful thing that singletons does is that it also generates defunctionalization symbols for type families and type synonyms you define in the Template Haskell, as well --- so if you write
\$(singletons [d|
type MyTypeFamily (b :: Bool) :: Type where
MyTypeFamily 'False = Int
MyTypeFamily 'True = String
|])
and
\$(singletons [d|
type MyTypeSynonym a = (a, [a])
|])
singletons will generate:
data MyTypeFamilySym0 :: Bool ~> Type
type instance Apply MyTypeFamilySym0 b = MyTypeFamily b
type MyTypeFamilySym1 b = MyTypeFamily b
and
data MyTypeSynonymSym0 :: Type ~> Type
type instance Apply MyTypeSynonym b = MyTypeSynonym a
type MyTypeSynonymSym1 a = MyTypeSynonym a
#### Bringing it All Together
Just to show off the library, remember that singletons also promotes typeclasses?
Because DoorState is a monoid with respect to merging, we can actually write and promote a Monoid instance: (requires singletons-2.5 or higher)
\$(singletons [d|
instance Semigroup DoorState where
(<>) = mergeState
instance Monoid DoorState where
mempty = Opened
mappend = (<>)
|])
We can promote fold:
\$(singletons [d|
fold :: Monoid b => [b] -> b
fold [] = mempty
fold (x:xs) = x <> fold xs
|])
And we can write collapseHallway in terms of those instead :)
!!!singletons/Door4Final.hs "collapseHallway'" "collapseSomeHallway'"
(Note again unfortunately that we have to define our own fold instead of using the one from singletons and the SFoldable typeclass, because of issue #339)
## Thoughts on Symbols
Defunctionalization symbols may feel like a bit of a mess, and the naming convention is arguably less than aesthetically satisfying. But, as you work with them more and more, you start to appreciate them on a deeper level.
At the end of the day, you can compare defunctionalization as turning "functions" into just constructors you can match on, just like any other data or type constructor. That's because they are just type constructors!
In a sense, defining defunctionalization symbols is a lot like working with pattern synonyms of your functions, instead of directly passing the functions themselves. At the type family and type class level, you can "pattern match" on these functions.
For a comparison at the value level -- you can't pattern match on (+), (-), (*), and (/):
-- Doesn't work like you think it does
invertOperation :: (Double -> Dobule -> Double) -> (Double -> Double -> Double)
invertOperation (+) = (-)
invertOperation (-) = (+)
invertOperation (*) = (/)
invertOperation (/) = (*)
You can't quite match on the equality of functions to some list of patterns. But, what you can do is create constructors representing your functions, and match on those.
This essentially fixes the "type lambda problem" of type inference and typeclass resolution. You can't match on arbitrary lambdas, but you can match on dummy constructors representing type functions.
And a bit of the magic here, also, is the fact that you don't always need to make our own defunctionalization symbols from scratch --- you can create them based on other ones in a compositional way. This is the basis of libraries like decidable.
For example, suppose we wanted to build defunctionalization symbols for MergeStateList. We can actually build them directly from defunctionalization symbols for Foldr.
Check out the defunctionalization symbols for Foldr:
!!!singletons/Defunctionalization.hs "data FoldrSym0" "data FoldrSym1" "data FoldrSym2" "type FoldrSym3"
We can actually use these to define our MergeStateList defunctionalization symbols, since defunctionalization symbols are first-class:
!!!singletons/Defunctionalization.hs "type MergeStateListSym0"
And you can just write collapseHallway as:
collapseHallway :: Hallway ss -> Door (MergeStateListSym0 @@ ss)
-- or
collapseHallway :: Hallway ss -> Door (FoldrSym2 MergeStateSym0 'Opened @@ ss)
You never have to actually define MergeStateList as a function or type family!
The whole time, we're just building defunctionalization symbols in terms of other defunctionalization symbols. And, at the end, when we finally want to interpret the complex function we construct, we use Apply, or @@.
You can think of FoldrSym1 and FoldrSym2 as defunctionalization symbol constructors -- they're combinators that take in defunctionalization symbols (like MergeStateSym0) and return new ones.
### Sigma
Let's look at a nice tool that is made possible using defunctionalization symbols: dependent pairs. I talk a bit about dependent pairs (or dependent sums) in part 2 of this series, and also in my dependent types in Haskell series.
Essentially, a dependent pair is a tuple where the type of the second field depends on the value of the first one. This is basically what SomeDoor was:
data SomeDoor :: Type where
MkSomeDoor :: Sing x -> Door x -> SomeDoor
The type of the Door x depends on the value of the Sing x, which you can read as essentially storing the x.
We made SomeDoor pretty ad-hoc. But what if we wanted to make some other predicate? Well, we can make a generic dependent pair by parameterizing it on the dependence between the first and second field. Singletons provides the Sigma type, in the Data.Singletons.Sigma module:
data Sigma k :: (k ~> Type) -> Type where
(:&:) :: Sing x -> (f @@ x) -> Sigma k f
-- also available through fancy type synonym
type Σ k = Sigma k
If you squint carefully, you can see that Sigma k is just SomeDoor, but parameterized over Door. Instead of always holding Door x, we can have it parameterized on an arbitrary function f and have it hold an f @@ x.
We can actually define SomeDoor in terms of Sigma:
!!!singletons/Door4Final.hs "type SomeDoor" "mkSomeDoor"
(Remember TyCon1 is the defunctionalization symbol constructor that turns any normal type constructor j -> k into a defunctionalization symbol j ~> k)
That's because a Sigma DoorState (TyCon1 Door) contains a Sing (x :: DoorState) and a TyCon1 Door @@ x, or a Door x.
This is a simple relationship, but one can imagine a Sigma parameterized on an even more complex type-level function. We'll explore more of these in the exercises.
For some context, Sigma is an interesting data type (the "dependent sum") that is ubiquitous in dependently typed programming.
### Singletons of Defunctionalization Symbols
One last thing to tie it all together -- let's write collapseHallway in a way that we don't know the types of the doors.
!!!singletons/Door4Final.hs "type SomeHallway"
The easy way would be to just use sMergeStateList that we defined:
!!!singletons/Door4Final.hs "collapseSomeHallway"
But what if we didn't write sMergeStateList, and we constructed our defunctionalization symbols from scratch?
!!!singletons/Door4Final.hs "collapseHallway''"
collapseSomeHallway'' :: SomeHallway -> SomeDoor
collapseSomeHallway'' (ss :&: d) = ??? -- what goes here?
:&: collapseHallway'' d
This will be our final defunctionalization lesson. How do we turn a singleton of ss into a singleton of FoldrSym2 MergeStateSym0 'Opened @@ s ?
First -- we have Foldr at the value level, as sFoldr. We glossed over this earlier, but singletons generates the following function for us:
type family Foldr (f :: j ~> k ~> k) (z :: k) (xs :: [j]) :: k where
Foldr f z '[] = z
Foldr f z (x ': xs) = (f @@ x) @@ Foldr f z xs
sFoldr
:: Sing (f :: j ~> k ~> k)
-> Sing (z :: k)
-> Sing (xs :: [j])
-> Sing (Foldr f z xs :: k)
sFoldr f z SNil = z
sFoldr f z (x `SCons` xs) = (f @@ x) @@ sFoldr f z xs
Where (@@) :: Sing f -> Sing x -> Sing (f @@ x) (or applySing) is the singleton/value-level counterpart of Apply or (@@).[^slambda]
[^slambda]: (@@) (and as we see shortly, the singFun functions) are all implemented in terms of SLambda, the "singleton" for functions. Understanding the details of the implementation of SLambda aren't particularly important for the purposes of this introduction.
So we can write:
collapseSomeHallway'' :: SomeHallway -> SomeDoor
collapseSomeHallway'' (ss :&: d) = sFoldr ???? SOpened ss
:&: collapseHallwa''y d
But how do we get a Sing MergeStateSym0?
We can use the singFun family of functions:
singFun2 @MergeStateSym0 sMergeState
:: Sing MergeStateSym0
But, also, conveniently, the singletons library generates a SingI instance for MergeStateSym0, if you defined mergeState using the singletons template haskell:
sing :: Sing MergeStateSym0
-- or
sing @_ @MergeStateSym0 -- singletons 2.4
sing @MergeStateSym0 -- singletons 2.5
And finally, we get our answer:
!!!singletons/Door4Final.hs "collapseSomeHallway''"
## Closing Up
Woo! Congratulations, you've made it to the end of the this Introduction to Singletons tetralogy! This last and final part understandably ramps things up pretty quickly, so don't be afraid to re-read it a few times until it all sinks in before jumping into the exercises.
I hope you enjoyed this journey deep into the motivation, philosophy, mechanics, and usage of this great library. Hopefully these toy examples have been able to show you a lot of ways that type-level programming can help your programs today, both in type safety and in writing more expressive programs. And also, I hope that you can also see now how to leverage the full power of the singletons library to make those gains a reality.
There are a few corners of the library we haven't gone over (like the TypeLits- and TypeRep-based singletons -- if you're interested, check out this post where I talk a lot about them), but I'd like to hope as well that this series has equipped you to be able to dive into the library documentation and decipher what it holds, armed with the knowledge you now have. (We also look at TypeLits briefly in the exercises)
You can download the source code here --- [Door4Final.hs][source-final] contains the final versions of all our definitions, and [Defunctionalization.hs][] contains all of our defunctionalization-from-scratch work. These are designed as stack scripts that you can load into ghci. Just execute the scripts:
!!![Defunctionalization.hs]:singletons/Door3.hs
\$ ./Door4Final.hs
ghci>
And you'll be dropped into a ghci session with all of the definitions in scope.
As always, please try out the exercises, which are designed to help solidify the concepts we went over here! And if you ever have any future questions, feel free to leave a comment or find me on twitter or in freenode #haskell, where I idle as jle`.
### Looking Forward
Some final things to note before truly embracing singletons: remember that, as a library, singletons was always meant to become obsolete. It's a library that only exists because Haskell doesn't have real dependent types yet.
Dependent Haskell is coming some day! It's mostly driven by one solo man, Richard Eisenberg, but every year buzz does get bigger. In a recent progress report, we do know that we realistically won't have dependent types before 2020. That means that this tutorial will still remain relevant for at least another two years :)
How will things be different in a world of Haskell with real dependent types? Well, for a good guess, take a look at Richard Eisenberg's Dissertation!
One day, hopefully, we won't need singletons to work with types at the value-level; we would just be able to directly pattern match and manipulate the types within the language and use them as first-class values, with a nice story for dependent sums. And some day, I hope we won't need any more dances with defunctionalization symbols to write higher-order functions at the type level --- maybe we'll have a nicer way to work with partially applied type-level functions (maybe they'll just be normal functions?), and we don't need to think any different about higher-order or first-order functions.
So, as a final word --- Happy Haskelling, everyone! May you leverage the great singletons library to its full potential, and may we also all dream of a day where singletons becomes obsolete. But may we all enjoy the wonderful journey along the way.
Until next time!
## Exercises
Here are your final exercises for this series! Start from [this sample source code][source-final], which has all of the definitions that the exercises and their solutions require. Just make sure to delete all of the parts after the -- Exercises comment if you don't want to be spoiled. Remember again to enable -Werror=incomplete-patterns or -Wall to ensure that all of your functions are total.
!!![source-final]:singletons/Door4Final.hs
!!![solution1]:singletons/Door4Final.hs "-- | 1."1 !!![solution2]:singletons/Door4Final.hs "-- | 2."1 !!![solution3]:singletons/Door4Final.hs "-- | 3."1 !!![solution4]:singletons/Door4Final.hs "-- | 4."1 !!![solution5]:singletons/Door4Final.hs "-- | 5."1 !!![solution6]:singletons/Door4Final.hs "-- | 6."1
1. Let's try combining type families with proofs! In doing so, hopefully we can also see the value of using dependent proofs to show how we can manipulate proofs as first-class values that the compiler can verify.
Remember Knockable from Part 3?
!!!singletons/Door4Final.hs "data Knockable"
Closed and Locked doors are knockable. But, if you merge two knockable doors...is the result also always knockable?
I say yes, but don't take my word for it. Prove it using Knockable!
!!!singletons/Door4Final.hs "mergedIsKnockable"4
mergedIsKnockable is only implementable if the merging of two DoorStates that are knockable is also knockable. See if you can write the implementation!
[Solution here!][solution1]
2. Write a function to append two hallways together.
appendHallways
:: Hallway ss
-> Hallway ts
-> Hallway ????
from singletons --- implement any type families you might need from scratch!
Remember the important principle that your type family must mirror the implementation of the functions that use it.
Next, for fun, use appendHallways to implement appendSomeHallways:
!!!singletons/Door4Final.hs "type SomeHallway" "appendSomeHallways"4
[Solution here!][solution2]
3. Can you use Sigma to define a door that must be knockable?
To do this, try directly defining the defunctionalization symbol KnockableDoor :: DoorState ~> Type (or use singletons to generate it for you --- remember that singletons can also promote type families) so that:
type SomeKnockableDoor = Sigma DoorState KnockableDoor
will contain a Door that must be knockable.
Try doing it for both (a) the "dependent proof" version (with the Knockable data type) and for (b) the type family version (with the StatePass type family).
[Solutions here!][solution3] I gave four different ways of doing it, for a full range of manual vs. auto-promoted defunctionalization symbols and Knockable vs. Pass-based methods.
Hint: Look at the definition of SomeDoor in terms of Sigma:
type SomeDoor = Sigma DoorState (TyCon1 Door)
Hint: Try having KnockableDoor return a tuple.
4. Take a look at the API of the Data.Singletons.TypeLits module, based on the API exposed in GHC.TypeNats module from base.
Using this, you can use Sigma to create a predicate that a given Nat number is even:
data IsHalfOf :: Nat -> Nat ~> Type
type instance Apply (IsHalfOf n) m = n :~: (m * 2)
type IsEven n = Sigma Nat (IsHalfOf n)
(*) is multiplication from the Data.Singletons.Prelude.Num module. (You must have the -XNoStarIsType extension on for this to work in GHC 8.6+), and :~: is the predicate of equality from Part 3:
data (:~:) :: k -> k -> Type where
Refl :: a :~: a
(It's only possible to make a value of type a :~: b using Refl :: a :~: a, so it's only possible to make a value of that type when a and b are equal. I like to use Refl with type application syntax, like Refl @a, so it's clear what we are saying is the same on both sides; Refl @a :: a :~: a)
The only way to construct an IsEven n is to provide a number m where m * 2 is n. We can do this by using SNat @m, which is the singleton constructor for the Nat kind (just like how STrue and SFalse are the singleton constructors for the Bool kind):
tenIsEven :: IsEven 10
tenIsEven = SNat @5 :&: Refl @10
-- Refl is the constructor of type n :~: (m * 2)
-- here, we use it as Refl @10 :: 10 :~: 10
-- won't compile
sevenIsEven :: IsEven 10
sevenIsEven = SNat @4 :&: Refl
-- won't compile, because we need something of type `(4 * 2) :~: 7`,
-- but Refl must have type `a :~: a`; `8 :~: 7` is not constructable
-- using `Refl`. Neither `Refl @8` nor `Refl @7` will work.
Write a similar type IsOdd n that can only be constructed if n is odd.
type IsOdd n = Sigma Nat (???? n)
And construct a proof that 7 is odd:
!!!singletons/Door4Final.hs "sevenIsOdd"1
[Solution here!][solution4]
On a sad note, one exercise I'd like to be able to add is to ask you to write decision functions and proofs for IsEven and IsOdd. Unfortunately, Nat is not rich enough to support this out of the box without a lot of extra tooling!
5. A common beginner Haskeller exercise is to implement map in terms of foldr:
map :: (a -> b) -> [a] _> [b]
map f = foldr ((:) . f) []
Let's do the same thing at the type level, manually.
Directly implement a type-level Map, with kind (j ~> k) -> [j] -> [k], in terms of Foldr:
type Map f xs = Foldr ???? ???? xs
Try to mirror the value-level definition, passing in (:) . f, and use the promoted version of (.) from the singletons library, in Data.Singletons.Prelude. You might find TyCon2 helpful!
[Solution here!][solution5]
6. Make a SomeHallway from a list of SomeDoor:
!!!singletons/Door4Final.hs "type SomeDoor" "type SomeHallway" "mkSomeHallway"1
Remember that the singleton constructors for list are SNil (for []) and SCons (for (:))!
[Solution here!][solution5]
## Special Thanks
None of this entire series would be possible without the hard work and effort of the amazing singletons library authors and maintainers --- especially Richard Eisenberg and Ryan Scott.
I am very humbled to be supported by an amazing community, who make it possible for me to devote time to researching and writing these posts. Very special thanks to my two supporters at the "Amazing" level on patreon, Sam Stites and Josh Vera! :)
Thanks also to Koz Ross for helping proofread this post! | 9,674 | 39,496 | {"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-2019-04 | latest | en | 0.840601 |
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# Check Corners
Time Limit: 2000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5344 Accepted Submission(s): 1727
Problem Description
Paul draw a big m*n matrix A last month, whose entries Ai,j are all integer numbers ( 1 <= i <= m, 1 <= j <= n ). Now he selects some sub-matrices, hoping to find the maximum number. Then he finds that there may be more than one maximum number, he also wants to know the number of them. But soon he find that it is too complex, so he changes his mind, he just want to know whether there is a maximum at the four corners of the sub-matrix, he calls this ¡°Check corners¡±. It¡¯s a boring job when selecting too many sub-matrices, so he asks you for help. (For the ¡°Check corners¡± part: If the sub-matrix has only one row or column just check the two endpoints. If the sub-matrix has only one entry just output ¡°yes¡±.)
Input
There are multiple test cases.
For each test case, the first line contains two integers m, n (1 <= m, n <= 300), which is the size of the row and column of the matrix, respectively. The next m lines with n integers each gives the elements of the matrix which fit in non-negative 32-bit integer.
The next line contains a single integer Q (1 <= Q <= 1,000,000), the number of queries. The next Q lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= m, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.
Output
For each test case, print Q lines with two numbers on each line, the required maximum integer and the result of the ¡°Check corners¡± using ¡°yes¡± or ¡°no¡±. Separate the two parts with a single space.
Sample Input
4 4
4 4 10 7
2 13 9 11
5 7 8 20
13 20 8 2
4
1 1 4 4
1 1 3 3
1 3 3 4
1 1 1 1
Sample Output
20 no
13 no
20 yes
4 yes
Source
Statistic | Submit | Discuss | Note
Home | Top Hangzhou Dianzi University Online Judge 3.0 Copyright © 2005-2024 HDU ACM Team. All Rights Reserved. Designer & Developer : Wang Rongtao LinLe GaoJie GanLu Total 0.000000(s) query 1, Server time : 2024-05-31 06:21:56, Gzip enabled Administration | 724 | 2,387 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.82573 |
https://www.daniweb.com/programming/software-development/threads/323323/indexerror-list-index-out-of-range | 1,547,798,249,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659944.3/warc/CC-MAIN-20190118070121-20190118092121-00357.warc.gz | 781,953,188 | 12,253 | Hi all,
Ive been working on this little piece of code down there that removes the odd numbers from a list of numbers. I get an "IndexError: list index out of range" whenever I try to run it. From what I understand, it means that l is attempting to use the number at position i that doesn't exist.
``````def del_odd(l):
for i in range(0,len(l)):
if l[i]%2 != 0:
l.remove(l[i])
return l``````
I may have completely misunderstood the error, but could someone offer up a small nugget of wisdom?
It might be a tall order, but try to just point me in the right direction instead of just giving me the code outright as I still want to try to learn this myself.
Cheers guys!
when you do
``l.remove(l[i])``
, your list 'l' reduces in size, so at some point, l will give error because the index doesn't exist.
A rule that can be good to remember.
Never delete something from or add something to the list you are iterating over.
The solution is:
Iterate over your original list, and put everything which passes your test into another list.
Some way of doing this.
Using List comprehensions.
``````>>> l = [1,2,3,4,5,6,7,8]
>>> odd_remove = [item for item in l if item %2 != 1]
>>> odd_remove
[2, 4, 6, 8]
>>>``````
``````l = [1,2,3,4,5,6,7,8]
even_list = []
for item in l:
if item %2 != 1:
even_list.append(item)
print even_list #[2, 4, 6, 8]``````
``````l = [1,2,3,4,5,6,7,8]
for item in l[:]: #make a copy of list
if item %2 != 0:
l.remove(item)
print l #[2, 4, 6, 8]``````
A rule that can be good to remember.
Never delete something from or add something to the list you are iterating over.
The solution is:
Iterate over your original list, and put everything which passes your test into another list.
Some way of doing this.
Using List comprehensions.
``````>>> l = [1,2,3,4,5,6,7,8]
>>> odd_remove = [item for item in l if item %2 != 1]
>>> odd_remove
[2, 4, 6, 8]
>>>``````
``````l = [1,2,3,4,5,6,7,8]
even_list = []
for item in l:
if item %2 != 1:
even_list.append(item)
print even_list #[2, 4, 6, 8]``````
``````l = [1,2,3,4,5,6,7,8]
for item in l[:]: #make a copy of list
if item %2 != 0:
l.remove(item)
print l #[2, 4, 6, 8]``````
Oh, I forgot to mention that I can't copy over another list. I have to do all of it with that 1 list.
Sorry guys.
Oh, I forgot to mention that I can't copy over another list. I have to do all of it with that 1 list.
Sorry guys.
``````def del_odd(l):
Notice that `l` is a bad variable name. Also try to understand the difference between `l.remove(l[i])` (bad) and `del l[i]` (good). | 806 | 2,544 | {"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} | 3.015625 | 3 | CC-MAIN-2019-04 | latest | en | 0.899114 |
http://math.hws.edu/eck/cs324/s01/pov-doc/pov174.htm | 1,544,851,156,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00374.warc.gz | 172,184,336 | 2,570 | Previous:Vector Functions Main Index Next:Color Vectors
Specifying Colors
POV-Ray often requires you to specify a color. Colors consist of five values or color components. The first three are called `red`, `green`, and `blue`. They specify the intensity of the primary colors red, green and blue using an additive color system like the one used by the red, green and blue color phosphors on a color monitor.
The 4th component, called `filter`, specifies the amount of filtered transparency of a substance. Some real-world examples of filtered transparency are stained glass windows or tinted cellophane. The light passing through such objects is tinted by the appropriate color as the material selectively absorbs some frequencies of light while allowing others to pass through. The color of the object is subtracted from the light passing through so this is called subtractive transparency.
The 5th component, called `transmit`, specifies the amount of non-filtered light that is transmitted through a surface. Some real-world examples of non-filtered transparency are thin see-through cloth, fine mesh netting and dust on a surface. In these examples, all frequencies of light are allowed to pass through tiny holes in the surface. Although the amount of light passing through is diminished, the color of the light passing through is unchanged. The color of the object is added to the light passing through so this is called additive transparency.
Note that early versions of POV-Ray used the keyword `alpha` to specify filtered transparency. However that word is often used to describe non-filtered transparency. For this reason `alpha` is no longer used.
Each of the five components of a color are float values which are normally in the range between 0.0 and 1.0. However any values, even negatives may be used.
Under most circumstances the keyword `color` is optional and may be omitted. We also support the British or Canadian spelling `colour`. Colors may be specified using vectors, keywords with floats or identifiers. You may also create very complex color expressions from combinations of any of these using various familiar operators. The syntax for specifying a color has evolved since POV-Ray was first released. We have maintained the original keyword-based syntax and added a short-cut vector notation. Either the old or new syntax is acceptable however the vector syntax is easier to use when creating color expressions.
The syntax for combining color literals into color expressions is almost identical to the rules for vector and float expressions. In the syntax for vector expressions below, some of the syntax items are defined in the section for float expressions. See "Float Expressions" for those definitions. Detailed explanations of color-specific issues are given in the following sub-sections.
COLOR:
COLOR_BODY |
`color` COLOR_BODY | (this means the keyword `color` or `colour` may
`colour` COLOR_BODY optionally precede any color specification)
COLOR_BODY:
COLOR_VECTOR |
COLOR_KEYWORD_GROUP |
COLOR_IDENTIFIER
COLOR_VECTOR:
`rgb` <3_Term_Vector> |
`rgbf` <4_Term_Vector> |
`rgbt` <4_Term_Vector> |
[ `rgbft` ] <5_Term_Vector>
COLOR_KEYWORD_GROUP:
[ COLOR_KEYWORD_ITEM ]...
COLOR_KEYWORD_ITEM:
COLOR_IDENTIFIER |
`red` Red_Amount | `blue` Blue_Amount | `green` Green_Amount |
`filter` Filter_Amount | `transmit` Transmit_Amount
Note: COLOR_IDENTIFIERS are identifiers previously declared to have color values. The 3, 4, and 5 term vectors are usually vector literals but may be vector expressions or floats promoted to vectors. See "Operator Promotion" and the sections below.
Previous:Vector Functions Main Index Next:Color Vectors | 762 | 3,681 | {"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-2018-51 | latest | en | 0.910067 |
https://quant.stackexchange.com/questions/50964/expectation-of-the-cir-process | 1,718,233,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00574.warc.gz | 462,192,509 | 38,587 | # Expectation of the CIR process
The CIR process follows
$$dr_t = (\alpha - \beta r_t)dt + \sigma \sqrt{r_t}dW_t.$$
It can be proved that there exists a unique solution for this SDE but it's not possible to get an expression for that solution. However, according to the Shreve (page 152) we can obtain the expected value of the solution. Applying Ito's lemma with the function $$f(t,x)=e^{\beta t}x$$ we obtain $$e^{\beta t}r_t = r_0 + \frac{\alpha}{\beta}(e^{\beta t} -1) + \sigma \int_0^t (e^{\beta t}-1)+ \sigma \int_0^t e^{\beta u} \sqrt{r_u}dW_u.$$ So he obtains $$e^{\beta t}E [r_t] = r_0 + \frac{\alpha}{\beta}(e^{\beta t} -1)$$ because he says that $$E \left[\sigma \int_0^t e^{\beta u} \sqrt{r_u}dW_u \right]=0.$$ He says that the expectation of an Ito integral is zero, but this is not always true. If $$E\left[ \int_0^T |\sigma e^{\beta u} \sqrt{r_u}|^2du \right]< \infty \tag*{(\star)}$$ then the process $$\{I_t\}:=\left\{\sigma \int_0^t e^{\beta u} \sqrt{r_u}dW_u ; 0 \leq t \leq T \right\}$$ is a martingale and the expectation is zero. But if ($$\star$$) is not true, then the process $$\{I_t\}$$ is just a local martingale, not necessarily a (true) martingale and the expectation above may not be zero. Since we don't have the expression for the solution, I don't think the condition ($$\star$$) can be verified. Is there any other argument that allow us to conclude that the expectation is zero? Do you know of any papers/books where they treat this issue? | 492 | 1,477 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.816682 |
http://stackoverflow.com/questions/18047912/calculate-angle-required-to-hit-coordinate-x-y-from-position-other-than-0-0 | 1,406,262,521,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997893859.88/warc/CC-MAIN-20140722025813-00152-ip-10-33-131-23.ec2.internal.warc.gz | 294,995,530 | 16,834 | # Calculate angle required to hit coordinate (x,y) from position other than (0,0) with varying elevation
I am attempting to calculate the angle required to fire a projectile in order to hit a specific coordinate.
My projectile is located a random coordinate and my target coordinate at a static coordinate.
I ended up running across the following equation on Wikipedia for calculating the angle required to hit a coordinate at (x,y) from (0,0):
I have made some attempts to understand this and other formula and attempted the following implementation (I am using c# and XNA).
``````double y = source.Y - target.Y;
double x = Vector2.Distance(source, target);
double v = 1440; //velocity
double g = 25; //gravity
double sqrt = (v*v*v*v) - (g*(g*(x*x) + 2*y*(v*v)));
sqrt = Math.Sqrt(sqrt);
double angleInRadians = Math.Atan(((v*v) + sqrt)/(g*x));
``````
I have also attempted the following, which resulted in an identical angle where the values of v and g remain the same.
``````double targetX = target.X - source.X;
double targetY = -(target.Y - source.Y);
double r1 = Math.Sqrt((v*v*v*v) - g*(g*(target.X*target.X) + ((2*target.Y)*(v*v))));
double a1 = ((v*v) + r1)/(g*target.X);
angleInRadians = -Math.Atan(a1);
if (targetX < 0)
{
angleInRadians -= 180/180*Math.PI;
}
``````
My conjecture is that even in my (assumed) attempt to zero out the source coordinate, that I am still not performing the calculation correctly for coordinates with a non (0,0) source and different elevations.
Below is an image that depicts my coordinate system. It is the default for XNA.
-
I think you should translate your actual positions to (0,0)-based system, perform the function there and perform a final translation from (0,0) to actual system...assuming there are no factors other than gravity, Angle calculated should be the same from any source... – boxed__l Aug 4 '13 at 21:54
What are the other parameters (input) that affects the projectile path, like is the velocity and gravity constants or what other variables are there ? – Sniffer Aug 4 '13 at 21:54
Isn't gravity supposed to be negative? It's -9.8m/s if I recall correctly. – Pierre-Luc Pineault Aug 4 '13 at 21:55
I will give that a try @boxed__l. You are correct that gravity is the only factor. – Timothy Randall Aug 4 '13 at 22:01
@Sniffer the inputs are a source coordinate, target coordinate, a constant initial velocity of 1440m/s, a constant gravity of 25m/s. – Timothy Randall Aug 4 '13 at 22:04
show 2 more comments
## 2 Answers
I think the real problem lies in the use of arctan. Because the range is limited to -pi/2..pi/2 results are only in the right half plane.
Use arctan2 to get the proper coordinates:
``````angleInRadians = Math.Atan2(((v*v) + tmp), (g*x));
``````
-
add comment
Thanks to the help in the comments the solution to find this angle ended up requiring that the positions be translated to a (0,0) based system. For anyone looking for the same scenario the final working solution was:
``````double x = -(source.x - target.x);
double y = (source.y - target.y);
double v = 1440; //m/s
double g = 25; //m/s
double sqrt = (v*v*v*v) - (g*(g*(x*x) + 2*y*(v*v)));
sqrt = Math.Sqrt(sqrt);
angleInRadians = Math.Atan(((v*v) + sqrt)/(g*x));
``````
Then to convert the radians into a vector that works with XNA, perform the following conversion:
``````Vector2 angleVector = new Vector2(-(float)Math.Cos(angleInRadians), (float)Math.Sin(angleInRadians));
``````
-
add comment | 928 | 3,463 | {"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} | 3.53125 | 4 | CC-MAIN-2014-23 | latest | en | 0.830977 |
http://www.docstoc.com/docs/84740746/Normalization | 1,435,635,592,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375091587.3/warc/CC-MAIN-20150627031811-00119-ip-10-179-60-89.ec2.internal.warc.gz | 419,643,012 | 44,960 | # Normalization by shuifanglj
VIEWS: 39 PAGES: 22
• pg 1
``` a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Lecture Normalization
(I modified the Course Module on Normalization)
______________________________________________________
Outline
Normalization of Database Tables
Functional Dependency
Data Redundancy
Normal Forms
First Normal Form (1NF)
Second Normal Form (2NF)
Third Normal Form (3NF)
Invoice Database Normalization (3NF) Example
Database Design
Example Homework Problems
Student Homework
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Normalization of Database Tables
Good database design must be matched to good table structures. Good table
structures are evaluated and designed to control data redundancies, thereby
avoiding data anomalies. The process that yields such desirable results is known
as normalization.
Functional Dependency
A primary key uniquely identifies one and only one row in a table. Functional
dependency exists between a primary key and a unique row in a table because
the primary key guarantees uniqueness.
A more formal definition of functional dependency is that A B. B is
functionally dependent on A if A determines B. In plain English, this means A
determines one and only one value of B.
We make a dependency diagram to show all functional dependencies within a
table. Examine the following dependency diagram for the table in figure 2.6,
where C1 and C3 constitute the composite primary key because it uniquely
identifies the entire tuple (remember formal term for row or record)—that is, all
five attributes.
Figure 2.6 Dependency Diagram (1NF)
Table 1:
Primary key: C1, C3
Foreign key: None
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Normal form: 1NF
C1, C3 C2, C4, C5 represents a functional dependency because C2, C4, and
C5 depend on the primary key composed of C1 and C3.
C1 C2 is a special case of a functional dependency referred to as a partial
dependency because C2 depends only on C1 rather than on the entire primary
key composed of C1 and C3.
C4 C5 is a special case of a functional dependency referred to as a transitive
dependency because C5 depends on an attribute (C4) that is not part of the
primary key.
Data Redundancy
Redundant data occur in more than one places, creating a strong probability of
data. For example, redundant data may be updated in one place but overlooked
in another place. The most common anomalies discussed in the text when data
redundancy exists are update anomalies, addition anomalies, and deletion
anomalies.
In addition, data redundancy causes data integrity problems because data entry
failed to conform to the rule that all copies of redundant data are to be equal.
We can avoid all these difficulties through normalization.
Normal Forms
Normalization is a technique to design tables in which data redundancies are
minimized by assigning attributes to entities. If the normalization process works
properly, it eliminates uncontrolled data redundancies, getting rid of both the data
anomalies and the data integrity problems. It's important to realize that
normalization does not eliminate data redundancy - it produces carefully
controlled redundancies used to link database tables to form relationships.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
The first three normal forms (1NF, 2NF, and 3NF) are most commonly
encountered. From a structural point of view, higher normal forms are better than
lower ones because higher normal forms yield fewer data redundancies - 3NF is
better than 2NF, which is better than 1NF.
Almost all business designs use the 3NF as the ideal normal form. A special,
more restricted, 3NF is known as Boyce-Codd normal form (BCNF).
Let's look at each of the normal forms in turn.
First Normal Form (1NF)
A table is in 1NF when all the key attributes are defined and all remaining
attributes are dependent on the primary key. When there are repeating groups,
the primary key generally needs to be expanded to reach 1NF.
However, a table in 1NF can still contain both partial and transitive
dependencies. A partial dependency is one in which an attribute is functionally
dependent on only a part of a multiattribute primary key. A transitive dependency
is one in which one attribute is functionally dependent on another non-key
attribute. Naturally, a table with a single-attribute primary key can't exhibit partial
dependencies.
The table in the dependency diagram of figure 2.6 is in 1NF.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Second Normal Form (2NF)
A table is in 2NF when it's in 1NF and contains no partial dependencies.
Therefore, a 1NF table is automatically in 2NF if its primary key is based on only
a single attribute.
A table in 2NF may still contain transitive dependencies. Look at the dependency
diagram for a database the tables of which are at least 2NF as shown in figure
2.7.
Figure 2.7 Dependency Diagram (2NF)
Remove partial dependencies by creating new tables. To do this, write each
primary key component on a separate line, followed by a line containing the
original primary key (C1,C3). Each of these keys potentially starts a new table.
Write the dependent attributes after each new key. Because no attribute is
dependent on C3, a table never materializes for the primary key C3.
Table 1 is in 3NF because it's in 2NF (no partial dependencies) and contains no
transitive dependencies. Table 2 is in 2NF because it contains a transitive
dependency, C4 C5.
Third Normal Form (3NF)
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
A table is in 3NF if it's in 2NF and contains no transitive dependencies. Given this
definition of 3NF, the Boyce-Codd normal form (BCNF) is merely a special 3NF
case, in which all the determinant keys are candidate keys. So, if a table has only
a single candidate key, a 3NF table is automatically in BCNF.
Split a table that is not in 3NF into new tables until all the tables meet the 3NF
requirements. Look at the dependency diagram for a database with tables at
least 3NF in figure 2.8 below.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Figure 2.8 Dependency Diagram (3NF)
Get rid of transitive dependencies by decomposing the table containing the
transitive dependency. Here's how:
Place the attributes that create the transitive dependency in a separate
table, C4 C5.
Keep C4 in the original table 2 to create a foreign-key link to the new table
3.
Make sure that the primary key attribute C4 for the new table 3 is the
foreign key in the original table 2.
After doing this, tables 1, 2, and 3 are all in 3NF because neither partial nor
transitive dependencies exist.
Pause for a moment to look back at normalization from a nontechnical
perspective. When normalization begins, our starting point is a "conglomerate"
table—one that has lots of themes. We then decompose it to a number of single-
theme tables. When you have arrived at a collection of tables each having one
theme, you probably will have achieved 3NF.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Invoice Database Normalization (3NF) Example
Here's an INVOICE database - we'll use this data and decompose it to at least
3NF:
Attribute Name Sample Data
INV_NUM 211347
PROD_NUM AA_E3422QW
SALE_DATE 3/25/96
PROD_DESCRIPTION D & B rotary sander, 6-in. disk
VEND_CODE 211
VEND_NAME Never Fail, Inc.
NUM_SOLD 2
PROD_PRICE \$49.95
We know that a table is in 1NF when all the key attributes are defined and all
remaining attributes are dependent on the primary key, and that when there are
repeating groups, the primary key generally needs to be expanded to reach 1NF.
To eliminate repeating groups in this case, we have INV_NUM, PROD_NUM as
the composite primary key.
However, a table in 1NF can still contain both partial and transitive
dependencies. In this case, the partial dependencies are INV_NUM
SALE_DATE and PROD_NUM PROD_PRICE. The transitive dependency is
VEND_CODE VEND_NAME.
Look at the INVOICE database table (1NF) in figure 2.9.
Figure 2.9 Invoice Database (1NF)
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Table 1:
Primary key: INV_NUM, PROD_NUM
Foreign key: None
Normal form: 1NF
We know that a table is in 2NF when it's in 1NF and contains no partial
dependencies. So we create three tables from the individual components of the
primary key and the composite primary key. These keys start the new tables as
primary keys with all their dependent attributes listed in their respective tables.
Transitive dependencies are allowed in 2NF. In this case, a table (2NF) has the
transitive dependency VEND_CODE VEND_NAME. The other tables are
already in 3NF (no partial or transitive dependencies).
Look at the INVOICE database tables (2NF) in figure 2.10.
Figure 2.10 Invoice Database(2NF)
Table 1:
Primary key: INV_NUM, PROD_NUM
Foreign keys: INV_NUM (to table 2)
PROD_NUM (to table 3)
Normal form: 3NF
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Table 2:
Primary key: INV_NUM
Foreign key: None
Normal form: 3NF
Table 3:
Primary key: PROD_NUM
Foreign key: None
Normal form: 2NF
We know that a table is in 3NF if it's in 2NF and contains no transitive
dependencies. So we place the attributes that create the transitive dependency
in a separate table, VEND_CODE VEND_NAME. We keep VEND_CODE in
the original table to create a foreign key link to the new table.
The other tables are already in 3NF (no partial or transitive dependencies). Once
we normalize to 3NF, we can give the INVOICE database tables meaningful
names such as LINE, INVOICE, PRODUCT, and VENDOR.
Look at the INVOICE database tables (3NF) in figure 2.11.
Figure 2.11 Invoice Database (3NF)
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
LINE table:
Primary key: INV_NUM, PROD_NUM
Foreign keys: INV_NUM (to table INVOICE)
PROD_NUM (to table PROD_NUM)
Normal Form: 3NF
INVOICE table:
Primary key: INV_NUM
Foreign key: None
Normal form: 3NF
PRODUCT table:
Primary Key: PROD_NUM
Foreign Key: VEND_CODE (to table VENDOR)
Normal Form: 3NF
VENDOR table:
Primary Key: VEND_CODE
Foreign key: None
Normal form: 3NF
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Database Design
Dependency diagrams don't show the nature of the relationships (1:1, 1:M, M:N).
The E-R diagrams remain crucial to our design effort. We can't successfully
produce complex design without some form of modeling. Yet, as we've seen in
the preceding examples, dependency diagrams are a valuable addition to our
designer's tool box because:
Normalization is likely to suggest the existence of entities we may not
have considered in the modeling process.
If transaction management issues require the existence of attributes that
create other than 3NF or BCNF conditions, the proper dependency
diagrams will at least force us to be aware of these conditions.
A relational schema is used in design documentation which depicts
connecting fields and relationship types.
Figure 2.12 shows the complete design documentation that would accompany
the previous INVOICE database (3NF) example.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Figure 2.12 Invoice E-R Diagram and Relational Schema
Normalization is part of the design process. As we define entities and attributes
during the E-R modeling process, we do normalization checks on each entity (or
entity sets) and form new ones as required. We incorporate the normalized
entities into the E-R diagram and continue the iterative E-R process until all
entities and their attributes are defined and all equivalent tables are in 3NF.
The more tables we have, the more additional disk I/O to join them and the more
processing logic we need. That's why we sometimes denormalize tables to yield
less I/O and thus increase processing speed. Is that a good idea? Unfortunately,
we pay for the increased processing speed because updates to a larger table are
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
inefficient, and data redundancies occur that are likely to yield data anomalies.
So we should use denormalization sparingly in the design process.
Normalization offers us evaluation standards for producing good table structures
that can also be passed on to the next generation of database designers.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Example Homework Problems
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Student Homework
:
Example Problem 1) The following report is how an inexperienced database
developer might create a table. Your mission is to get into 3rd Normal Form. The
key field is Customer_ID and Movie_ID, Vendor_ID and Check_Out_Date..
It would be helpful to do this in stages like the book so you can get partial credit if
something goes wrong.
Normalize_ME
Customer Last Movie Vendor Check Out Return
Title Type
ID Name ID ID Date Date
1001 Barns 101 Title of Movie ACM ACT 1/1/2002 1/2/2002
1
1001 Barns 102 Title of Movie ACM COM 1/1/2002 1/2/2002
2
1001 Barns 103 Title of Movie ACM DRA 1/1/2002 1/5/2002
3
1001 Barns 104 Title of Movie ACM DRA 1/1/2002 1/6/2002
4
1001 Barns 105 Title of Movie ACM DRA 1/1/2002
5
1001 Barns 106 Title of Movie6 BB DRA 1/1/2002
1001 Barns 107 Title of Movie7 BB COM 1/1/2002
1001 Barns 108 Title of Movie8 ACM COM 1/1/2002
1001 Barns 109 Title of Movie9 ACM COM
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Solution:
Step 1) list the key fields separately and on the last line write the original
(composite) key
MOVIE_ID
CUSTOMER_ID
CHECK_OUT_DATE
MOVIE_ID, CUSTOMER_ID, CHECK_OUT_DATE
Step 2) Add the dependent attributes next to the each line you made on step 1:
MOVIE_ID ---- MOVIE_TITLE, TYPE, VENDOR_ID
CUSTOMER_ID ---- CUSTOMER_LNAME
CHECK_OUT_DATE none!!!! No Dependent Attributes!! So it does not
make a NEW TABLE
MOVIE_ID, CUSTOMER_ID--- CHECK_OUT_DATE, RETURN_DATE
Step 3) Make new Tables for all the transitive dependencies.
Notice that MOVIE_ID gives TYPE and VENDOR_ID. So, I add two new tables
for each of these AND KEEP the linking attributes in the orginal table:
MOVIE_ID ---- MOVIE_TITLE, TYPE, VENDOR_ID
CUSTOMER_ID ---- CUSTOMER_LNAME
MOVIE_ID, CUSTOMER_ID--- CHECK_OUT_DATE, RETURN_DATE
TYPE -- TYPE, TYPE_DESCRIPTION (new)
VENDOR --- VENDOR_ID, VENDOR_NAME (new)
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
Step 4) Give appropriate Names to the tables, and underline the PK of each
Entity:
MOVIE: (MOVIE_ID, MOVIE_TITLE, TYPE, VENDOR_ID)
CUSTOMER: (CUSTOMER_ID, CUSTOMER_LNAME)
RENTAL: (MOVIE_ID, CUSTOMER_ID, CHECK_OUT_DATE,
RETURN_DATE)
TYPE: (TYPE, TYPE_DESCRIPTION)
VENDOR: (VENDOR_ID, VENDOR_NAME)
These tables are now in 3rd Normal Form.
Notice how I had to deal with CHECK_OUT_DATE. It is possible (but unlikely)
that a customer could rent the same video (i.e. same VIDEO_ID, MOVIE_TITLE,
VENDOR_ID) on different days! So I had to keep CHECK_OUT_DATE as part of
the PK.
Example Problem 2).
a) To keep track of office furniture, computers, printers, and so on, the
FOUNDIT company uses the following table structure:
Attribute name Sample value
ITEM_ID 2311345-678
ITEM_DESCRIPTION HP DeskJet 660C printer
BLDG_ROOM 325
BLDG_CODE DEL
BLDG_NAME Dawn's Early Light
BLDG_MANAGER E. R. Rightonit
Given this information, draw the dependency diagram. Make sure you
label the transitive and/or partial dependencies.
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
b) Starting with the dependency diagram drawn for problem 10, create a set
of dependency diagrams that meet 3NF requirements. Rename
attributes to meet the naming conventions and create new entities and
attributes as necessary.
Note that the dependency diagram reflect the notion that each building is
managed by one employee.
SOLUTION:
Problem 10 Solution
ITEM_ID ITEM_DESCRIPTION BLDG_ROOM BLDG_CODE BLDG_NAME BLDG_MANAGER
Transitive Dependencies
Problem 11 Solution: All tables in 3NF
ITEM_ID ITEM_DESCRIPTION ITEM_ROOM BLDG_CODE
BLDG_CODE BLDG_NAME EMP_CODE
EMP_CODE EMP_LNAME EMP_FNAME EMP_INITIAL
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
STUDENT HOMEWORK
1 Using the following INVOICE table structure, draw its dependency
diagram and identify all dependencies (including all partial and transitive
dependencies). You can assume that the table does not contain repeating
groups and that any invoice number may reference more than one product.
(Hint: This table uses a composite primary key.)
Attribute name Sample value
INV_NUM 211347
PROD_NUM AA_E3422QW
SALE_DATE 06/25/1999
PROD_DESCRIPTION B&D Rotary sander, 6 in. disk
VEND_CODE 211
VEND_NAME NeverFail, Inc.
NUMBER_SOLD 2
PROD_PRICE \$49.95
2 Using the initial dependency diagram drawn in problem 1, remove all
partial dependencies, draw the new dependency diagrams, and identify the
normal forms for each table structure you created.
Note: You can assume that any given product is supplied by a
single vendor, but a vendor can supply many products.
Therefore, it is proper to conclude that the following
dependency exists:
PROD_NUM PROD_DESCRIPTION, PROD_PRICE,
VEND_CODE, VEND_NAME
(Hint: Your actions should produce three new dependency
diagrams.)
a1b2cfa4-e611-4131-8011-35ccfe44fc2e.doc
3) The following report is how an inexperienced database developer might create
a table. Your mission is to normalize it. The key fields are INV_NUM, S_ID,
CAR_NUM and CUS_ID.
The fields SER_CODE is multi-valued which implies a TABLE called SERVICE
that holds all the descriptions of the SERVICE.
The Field S_ID refers to the Salesperson table.
Normalize ME
INV_NUM S_ID SER_CODE CAR_NUM CUS_ID Last Name CAR_MAKE CAR_MODEL CAR_YEAR
5029 1005 NCS 86gf2de356 123 Palaisa Mazada Miata 2001
5030 1005 NCS 87mn5sw754 128 Sullivan Lincoln TownCar 2000
5031 1006 NCS 87vb2sn427 125 Petty Volvo C70 Coupe 2000
5032 1007 NCS 88za9qr821 127 Scott Mazada Protégé 2001
5033 1007 NCS 90as2sw987 126 Regan Lincoln Contiential 1997
5034 1003 NCS 90gh1gf546 129 Washington Lincoln LS 1999
5035 TUP A1 103 Armstrong BMW 326 2002
5036 INS A2 104 Bell BMW 327 2002
5037 TUP 18ms3sw145 107 Cliburn Volvo Cross Country 2002
5038 TUP A3 109 Edision BMW 328 2002
5043 PRE 85de4hg678 122 Nixon Ford Windstar 2002
5044 PRE 55ar6xv772 123 Palaisa Ford Windstar 2001
```
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https://www.physicsforums.com/threads/statics-3d-equilibrium-problem.75469/ | 1,716,400,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00118.warc.gz | 810,068,880 | 16,593 | # Statics - 3D equilibrium problem
• cyberdeathreaper
In summary, the question is asking for the vertical components of the reactions at points A, B, and C when the camera is mounted on a tripod at a certain angle, and for the maximum angle at which the tripod can support the camera without flipping over. For part (a), the answer can be easily obtained by finding the sum of all moments at point D. However, for part (b), the critical condition is the distribution of mass relative to the line connecting points A and C. When the line of action of the camera is too far from this line, the tripod will not be able to maintain equilibrium and the maximum angle at which the tripod can support the camera without flipping over is 54.1 degrees.
cyberdeathreaper
Here's the question:
"A camera of mass 240g is mounted on a small tripod of mass 200g. Assuming that the mass of the camera is uniformly distributed and that the line of action of the weight of the tripod passes through D, determine (a) the vertical components of the reactions at A, B, and C when $\theta$ = 0. (b) the maximum value of $\theta$ if the tripod is not to flip over."
(Refer to the attachment to better interpret what is being asked.)
I've gotten part (a) easily (summation of the forces and moment about D easily gives the answer). However, for part (b), I'm loss. As far as I can tell, the z component of the momentum about D is zero only for mutiples of 180 degrees, and yet the answer indicates that the maximum occurs at $\theta = 54.1^o$. Any ideas?
#### Attachments
• phys.jpg
17.8 KB · Views: 810
Note:
Here's the equations and coordinates I'm using (notice that I've moved the coordinate system down to the bottom, so D is at (0,0,0):
A: (-0.045 m, 0, 0)
B: (0.035 m, 0, 0.038 m)
C: (0.035 m, 0, -0.038 m)
D: (0, 0, 0)
F: ( $\alpha , 0, \beta$ )
where F is where the line of action of the weight of the camera intersects the xz plane, and where:
$$\alpha = -0.036 cos( \theta )$$
$$\beta = -0.036 sin( \theta )$$
If you find the sum of all the moments at D, it should equal zero if the object is static.
cyberdeathreaper said:
Note:
Here's the equations and coordinates I'm using (notice that I've moved the coordinate system down to the bottom, so D is at (0,0,0):
A: (-0.045 m, 0, 0)
B: (0.035 m, 0, 0.038 m)
C: (0.035 m, 0, -0.038 m)
D: (0, 0, 0)
F: ( $\alpha , 0, \beta$ )
where F is where the line of action of the weight of the camera intersects the xz plane, and where:
$$\alpha = -0.036 cos( \theta )$$
$$\beta = -0.036 sin( \theta )$$
If you find the sum of all the moments at D, it should equal zero if the object is static.
The critical condition is how the mass is distributed relative to the line AC. When the line of action of the camera gets too far over that line, the moment of the tripod will be insufficient to maintain equilibrium.
## What is the definition of a 3D equilibrium problem?
A 3D equilibrium problem is a type of statics problem that involves determining the forces and moments acting on a three-dimensional object in order for it to remain in a state of balance. This problem is often encountered in engineering and physics, and requires the use of mathematical equations and principles to solve.
## What are the key principles used to solve a 3D equilibrium problem?
The key principles used to solve a 3D equilibrium problem include the laws of equilibrium, which state that the sum of all forces acting on an object must equal zero and the sum of all moments acting on an object must also equal zero. Additionally, the use of free body diagrams and vector analysis are important tools in solving these types of problems.
## What are some common real-world applications of 3D equilibrium problems?
3D equilibrium problems have many real-world applications, including in the design of structures such as buildings, bridges, and airplanes. They are also useful in understanding the stability of objects on inclined planes, calculating loads on joints and supports, and analyzing the forces acting on mechanical systems.
## What are some tips for solving a 3D equilibrium problem?
Some tips for solving a 3D equilibrium problem include drawing accurate and clear free body diagrams, labeling all forces and distances involved, and breaking down complex systems into smaller, more manageable parts. It is also important to pay attention to units and use proper mathematical techniques to solve the equations.
## What are some common mistakes to avoid when solving a 3D equilibrium problem?
Some common mistakes to avoid when solving a 3D equilibrium problem include forgetting to account for all forces and moments acting on the object, using incorrect units, and making calculation errors. It is also important to check the final solution to ensure that it makes sense in the context of the problem and that all units are consistent.
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2K | 1,340 | 5,439 | {"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} | 3.640625 | 4 | CC-MAIN-2024-22 | latest | en | 0.910896 |
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## Presentation on theme: "Learning and Teaching Conference 2009 COMSOL Multiphysics for the teaching of design and innovation in food science Malcolm Povey School of Food Science."— Presentation transcript:
Learning and Teaching Conference 2009 COMSOL Multiphysics for the teaching of design and innovation in food science Malcolm Povey School of Food Science and Nutrition 1
Learning and Teaching Conference 2009 2 Modelling of battered potato chip in oil by use of COMSOL Multiphysics File: battered potato chip in oil.mph
Cricket and Maths Students are used to seeing data tabulated and plotted during sports programmes and live competitive events. They are familiar with the idea of discretized descriptions of continua, although they may not realise it. A serious issue facing a multidisciplinary subject such as food science is difficulty students have in adopting abstract, quantitative, and mathematical understanding of problems. 3
Acknowledgements I would like to thank Dr Anna Akinshina and Dr Melvin Holmes for their help in preparing teaching materials used in this presentation and course I would also like to thank the ADF fund of the University of Leeds for financial support, without which the software necessary to deliver the course could not have been purchased and for support with teaching assistance. 4
5 R3 rectangle − potato CO2 outer shell – batter r = 0 − axial symmetry 1 MODEL: Definition of the geometry: DRAW MODE a chip is modelled as a cylinder a cylinder is obtained as a rectangle rotating around z-axis (r =0) r = 0 R3 CO2 r = 0 The typical modelling steps include: 1 MODEL (Definition of the geometry: Draw, Draw mode) 2 PHYSICS (Definition of the equations, parameters of the matter and boundary conditions: Physics) 3 MESHING (Mesh Mesh mode ) 4 SOLVING (Solve ) 5 RESULTS (Postprocessing) 2D geometry oil R3 rectangle − potato: length 0.074m (7.4cm), width 0.006m (6mm) CO2 outer shell – batter: thickness of 0.001m (1mm) around the potato
6 2 PHYSICS: Subdomain settings Equations: Heat transfer by conduction Two subdomains: batter and potato. All the physical properties of the potato and batter should be defined. Unknown (variable) is T(t): How temperature within the chip depends on the heating time Parameters for batter and potato are taken from the data base or experiments Food data base: www.nelfood.com Username: GClayton Password: 1ruebeanwww.nelfood.com parametersbatterpotato δ ts 11 k 0.40.6 ρ 600700 CpCp 3851900 Q 00 Initial conditions for temperature T Init T(t 0 ) = 293 (Temperature at the initial time t 0 (t = 0)) potatobatter
7 2 PHYSICS: Boundary settings Boundaries: between potato and batter (inner) between batter and oil (outer) Boundary conditions: Temperature (T 0 ) T 0 = 293K (room temperature between potato and batter) T 0 = 443K (temperature of the hot oil between batter and oil) While heating the chip the temperature inside the chip should increase from T 0 min (room T) to nearly T 0 max (hot oil)
8 Why do we need mesh? An Idea of finite elements is the follows: when it is impossible to solve the equations for a “big object”, the object is divided into small pieces and the problem is solved for each piece. Such division into small pieces called meshing. A Mesh is a partition of the geometry model into small units of simple shapes. For a 2D geometry the mesh generator partitions the subdomains into triangular or quadrilateral mesh elements. If the boundary is curved, these elements represent only an approximation of the original geometry. The sides of the triangles and quadrilaterals are called mesh edges, and their corners are mesh vertices. A mesh edge must not contain mesh vertices in its interior. Similarly, in 3D the mesh generator partitions the subdomains into tetrahedral, hexahedral, or prism mesh elements whose faces, edges, and corners are called mesh faces, mesh edges, and mesh vertices, respectively. Equations have to be solved for all the edges and vertices. 3 MESHING (Mesh) The coarser is the mesh, the faster are the calculations (less computation time), but the worse is the accuracy. Refine mesh – make the mesh cells smaller
9 4 SOLVING (Solve) Solver parameters: one can modify time t Solve 2D surface plot: T(t)
10 Surface plot can be saved as a picture file (jpg, tiff) File −> Export −> Image −> change plot parameters if you like −> Export −>choose appropriate folder and save file. 5 RESULTS: Postprocessing and saving your data 1. Save graphical solution (2D surface plot from the previous page) 2. Make a probe plot (T(t) for several probe points) Postprocessing −> Probe Plot Parameters −> Defined Plots −> New −> New Probe Plot −> Coordinate probe −> −> type probe name (choose name yourself) −>OK −>Coordinate −> type coordinates of the probe point. Repeat from “New” if you would like to have several probes in different locations. If you would like to have plots for several coordinate probes in the same graph, trick “Plot all plots in the same axis”
11 Probe examples: Middle of the chip: r = 0, z = 0.037 at 3mm from the side: r = 0.003, z = 0.037 at 3 mm from the bottom: r = 0, z = 0.003 Edit plot In the Figure 1 COMSOL click on “Edit plot” icon Edit plot Title, Axis and Lines −> Apply −> OK Save plot Save data file: click on “ASC ||” icon and save *txt file Copy image: click on “Copy” icon and paste you file to Word or PowerPoint Export image: click on “Export Image” icon and save the graph as a picture file (jpg, tiff) Solve You need to solve the task again to obtain the data for a probe plot
12 3. Make a cross section plot (T(r) for different heating time) Postprocessing −> Cross-Section Plot Parameters −> −> Line/Extrusion −> Line plot−> Cross-section line data −> −> Coordinates of the desired cross section. Example: cross section across the middle of the chip: x-axis data: r coordinates: r0 = 0 r1 = 0.006 z0 = 0.037 z1 = 0.037 −>Apply You will get a graph with 81 lines (a line per each time interval from 0 to 80) Select several time intervals Postprocessing −> Cross-Section Plot Parameters −> General −> Solutions to use −> select several time intervals like 5, 10, 20, 40, 80 sec by Ctrl+left-click Save plot Same as for 2.
13 Examples of the graphs Generate report: You can also save some data by “producing a report” as: File −> Generate report −> Browse the directory and file name −> Generate The report would not contain postprocessing graphs you made (T(t), T(r)), but it contains other information which could be useful for your reports. Description of the model: Help −> Model Documentation
14 Chip size modification: Increase (decrease) the chip thickness Note, that you should modify the thickness of the potato keeping the thickness of the batter fixed (at 1mm) Potato: Draw mode −> click on potato rectangle −> double click on the potato rectangle OR Draw −> Object properties −> change width from initial 0.006 to 0.012 Batter: Draw mode −> Click on the batter layer −> double click on the batter layer OR Draw −> Object properties −> Curve selection There are 8 connected lines in the batter boundaries, you should adjust the length/coordinates of those which increase with the chip width. Example: Increase the thickness twice 1 not changed 2 r2 = 0.007 −> 0.013 (0.012 + 0.001) 3 r2 = 0.006 −> 0.012 4 not changed 5 r2 = 0.006 −> 0.012 6 r2 = 0.007 −> 0.013 (0.012 + 0.001) 7 r1 = r2 = 0.006 −> 0.012 8 r1 = r2 = 0.007 −> 0.013 Repeat all the stages: meshing, solving, postprocessing z r
15 Exercise 1: Battered potato chip in oil Describe the model and parameters Perform the calculations and save the graphical 2D plot of the temperature distribution inside the chip. Put a probe point in the middle of the chip and obtain the probe point plot. Save it. Make a cross section plot for a slice across the middle of the chip. Save it. Describe the results. Additional exercise if you have desire and time: Place several probes on the chip and make probe plots T(t) to investigate how different points within the chip are heated. Make a cross section plot T(r) for several time intervals to investigate how temperature across the chip varies with different heating time (choose several time intervals in Postprocessing −> Cross-Section Plot Parameters −> General) Modify time interval in Solving Parameters and get what happened if fry the chip longer Exercise 2: Thicker battered potato chip in oil Modify the thickness of the chip and compare the results with Exercise 1 It could be useful for your report if you Generate Report after all your calculations and save Model Documentation file Exercise 3: Model Validation Propose an experiment to validate your model ( half a page + diagram)
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Time series are particularly useful to track variables such as revenues, costs, and profits over time. Time series models help evaluate performance and make predictions. Consider the following and respond in a minimum of 175 words:
• Time series decomposition seeks to separate the time series (Y) into 4 components: trend (T), cycle (C), seasonal (S), and irregular (I). What is the difference between these components?
• The model can be additive or multiplicative. When we do use an additive model? When do we use a multiplicative model?
• The following list gives the gross federal debt(in millions of dollars) for the U.S. every 5 years from 1945 to 2000:
Year Gross Federal Debt (\$millions)
1945 260,123
1950 256,853
1955 274,366
1960 290,525
1965 322,318
1970 380,921
1975 541,925
1980 909,050
1985 1,817,521
1990 3,206,564
1995 4,921,005
2000 5,686,338
• Construct a scatter plot with this data. Do you observe a trend? If so, what type of trend do you observe?
• Use Excel to fit a linear trend and an exponential trend into the data. Display the models and their respective r^2.
• Interpret both models. Which model seems to be more appropriate? Why? | 346 | 1,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-14 | latest | en | 0.797195 |
http://betterlesson.com/lesson/resource/1935861/introducing-function-graphs-day-2-of-3-student-image-jpg | 1,488,087,428,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171936.2/warc/CC-MAIN-20170219104611-00640-ip-10-171-10-108.ec2.internal.warc.gz | 26,676,298 | 21,279 | ## introducing function graphs day 2 of 3 student image.jpg - Section 2: Continuing the Activity
introducing function graphs day 2 of 3 student image.jpg
# Introducing Function Graphs Continued: Day 2 of 4
Unit 5: Functions
Lesson 2 of 30
## Big Idea: Interpret graphs to win big and create the best motion detector function graph in the class. Vote to win!
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Math, graphing linear functions, Algebra, interpreting graphs, 8th grade math, master teacher, function
50 minutes
### Christa Lemily
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Environment: Urban | 303 | 1,310 | {"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-2017-09 | latest | en | 0.756529 |
https://doc-snapshots.qt.io/qt5-dev/qvector2d.html | 1,579,373,363,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00362.warc.gz | 423,992,133 | 8,123 | QVector2D Class
The QVector2D class represents a vector or vertex in 2D space. More...
Header: #include qmake: QT += gui Since: Qt 4.6
This class was introduced in Qt 4.6.
Public Functions
QVector2D(const QVector4D &vector) QVector2D(const QVector3D &vector) QVector2D(const QPointF &point) QVector2D(const QPoint &point) QVector2D(float xpos, float ypos) QVector2D() float distanceToLine(const QVector2D &point, const QVector2D &direction) const float distanceToPoint(const QVector2D &point) const bool isNull() const float length() const float lengthSquared() const void normalize() QVector2D normalized() const void setX(float x) void setY(float y) QPoint toPoint() const QPointF toPointF() const QVector3D toVector3D() const QVector4D toVector4D() const float x() const float y() const QVariant operator QVariant() const QVector2D & operator*=(float factor) QVector2D & operator*=(const QVector2D &vector) QVector2D & operator+=(const QVector2D &vector) QVector2D & operator-=(const QVector2D &vector) QVector2D & operator/=(float divisor) QVector2D & operator/=(const QVector2D &vector) float & operator[](int i) float operator[](int i) const
Static Public Members
float dotProduct(const QVector2D &v1, const QVector2D &v2)
bool qFuzzyCompare(const QVector2D &v1, const QVector2D &v2) bool operator!=(const QVector2D &v1, const QVector2D &v2) const QVector2D operator*(float factor, const QVector2D &vector) const QVector2D operator*(const QVector2D &vector, float factor) const QVector2D operator*(const QVector2D &v1, const QVector2D &v2) const QVector2D operator+(const QVector2D &v1, const QVector2D &v2) const QVector2D operator-(const QVector2D &v1, const QVector2D &v2) const QVector2D operator-(const QVector2D &vector) const QVector2D operator/(const QVector2D &vector, float divisor) const QVector2D operator/(const QVector2D &vector, const QVector2D &divisor) QDataStream & operator<<(QDataStream &stream, const QVector2D &vector) bool operator==(const QVector2D &v1, const QVector2D &v2) QDataStream & operator>>(QDataStream &stream, QVector2D &vector)
Detailed Description
The QVector2D class can also be used to represent vertices in 2D space. We therefore do not need to provide a separate vertex class.
Member Function Documentation
QVector2D::QVector2D(const QVector4D &vector)
Constructs a vector with x and y coordinates from a 3D vector. The z and w coordinates of vector are dropped.
QVector2D::QVector2D(const QVector3D &vector)
Constructs a vector with x and y coordinates from a 3D vector. The z coordinate of vector is dropped.
QVector2D::QVector2D(const QPointF &point)
Constructs a vector with x and y coordinates from a 2D point.
QVector2D::QVector2D(const QPoint &point)
Constructs a vector with x and y coordinates from a 2D point.
QVector2D::QVector2D(floatxpos, floatypos)
Constructs a vector with coordinates (xpos, ypos).
QVector2D::QVector2D()
Constructs a null vector, i.e. with coordinates (0, 0).
float QVector2D::distanceToLine(const QVector2D &point, const QVector2D &direction) const
Returns the distance that this vertex is from a line defined by point and the unit vector direction.
If direction is a null vector, then it does not define a line. In that case, the distance from point to this vertex is returned.
This function was introduced in Qt 5.1.
float QVector2D::distanceToPoint(const QVector2D &point) const
Returns the distance from this vertex to a point defined by the vertex point.
This function was introduced in Qt 5.1.
`[static] `float QVector2D::dotProduct(const QVector2D &v1, const QVector2D &v2)
Returns the dot product of v1 and v2.
bool QVector2D::isNull() const
Returns `true` if the x and y coordinates are set to 0.0, otherwise returns `false`.
float QVector2D::length() const
Returns the length of the vector from the origin.
float QVector2D::lengthSquared() const
Returns the squared length of the vector from the origin. This is equivalent to the dot product of the vector with itself.
void QVector2D::normalize()
Normalizes the currect vector in place. Nothing happens if this vector is a null vector or the length of the vector is very close to 1.
QVector2D QVector2D::normalized() const
Returns the normalized unit vector form of this vector.
If this vector is null, then a null vector is returned. If the length of the vector is very close to 1, then the vector will be returned as-is. Otherwise the normalized form of the vector of length 1 will be returned.
void QVector2D::setX(floatx)
Sets the x coordinate of this point to the given x coordinate.
void QVector2D::setY(floaty)
Sets the y coordinate of this point to the given y coordinate.
QPoint QVector2D::toPoint() const
Returns the QPoint form of this 2D vector.
QPointF QVector2D::toPointF() const
Returns the QPointF form of this 2D vector.
QVector3D QVector2D::toVector3D() const
Returns the 3D form of this 2D vector, with the z coordinate set to zero.
QVector4D QVector2D::toVector4D() const
Returns the 4D form of this 2D vector, with the z and w coordinates set to zero.
float QVector2D::x() const
Returns the x coordinate of this point.
float QVector2D::y() const
Returns the y coordinate of this point.
QVariant QVector2D::operator QVariant() const
Returns the 2D vector as a QVariant.
QVector2D &QVector2D::operator*=(floatfactor)
Multiplies this vector's coordinates by the given factor, and returns a reference to this vector.
QVector2D &QVector2D::operator*=(const QVector2D &vector)
Multiplies the components of this vector by the corresponding components in vector.
QVector2D &QVector2D::operator+=(const QVector2D &vector)
Adds the given vector to this vector and returns a reference to this vector.
QVector2D &QVector2D::operator-=(const QVector2D &vector)
Subtracts the given vector from this vector and returns a reference to this vector.
QVector2D &QVector2D::operator/=(floatdivisor)
Divides this vector's coordinates by the given divisor, and returns a reference to this vector.
QVector2D &QVector2D::operator/=(const QVector2D &vector)
Divides the components of this vector by the corresponding components in vector.
This function was introduced in Qt 5.5.
float &QVector2D::operator[](inti)
Returns the component of the vector at index position i as a modifiable reference.
i must be a valid index position in the vector (i.e., 0 <= i < 2).
This function was introduced in Qt 5.2.
float QVector2D::operator[](inti) const
Returns the component of the vector at index position i.
i must be a valid index position in the vector (i.e., 0 <= i < 2).
This function was introduced in Qt 5.2.
Related Non-Members
bool QVector2D::qFuzzyCompare(const QVector2D &v1, const QVector2D &v2)
Returns `true` if v1 and v2 are equal, allowing for a small fuzziness factor for floating-point comparisons; false otherwise.
bool QVector2D::operator!=(const QVector2D &v1, const QVector2D &v2)
Returns `true` if v1 is not equal to v2; otherwise returns `false`. This operator uses an exact floating-point comparison.
const QVector2D QVector2D::operator*(floatfactor, const QVector2D &vector)
Returns a copy of the given vector, multiplied by the given factor.
const QVector2D QVector2D::operator*(const QVector2D &vector, floatfactor)
Returns a copy of the given vector, multiplied by the given factor.
const QVector2D QVector2D::operator*(const QVector2D &v1, const QVector2D &v2)
Multiplies the components of v1 by the corresponding components in v2.
const QVector2D QVector2D::operator+(const QVector2D &v1, const QVector2D &v2)
Returns a QVector2D object that is the sum of the given vectors, v1 and v2; each component is added separately.
const QVector2D QVector2D::operator-(const QVector2D &v1, const QVector2D &v2)
Returns a QVector2D object that is formed by subtracting v2 from v1; each component is subtracted separately.
const QVector2D QVector2D::operator-(const QVector2D &vector)
Returns a QVector2D object that is formed by changing the sign of the components of the given vector.
Equivalent to `QVector2D(0,0) - vector`.
const QVector2D QVector2D::operator/(const QVector2D &vector, floatdivisor)
Returns the QVector2D object formed by dividing all three components of the given vector by the given divisor.
const QVector2D QVector2D::operator/(const QVector2D &vector, const QVector2D &divisor)
Returns the QVector2D object formed by dividing components of the given vector by a respective components of the given divisor.
This function was introduced in Qt 5.5.
QDataStream &QVector2D::operator<<(QDataStream &stream, const QVector2D &vector)
Writes the given vector to the given stream and returns a reference to the stream.
Returns `true` if v1 is equal to v2; otherwise returns `false`. This operator uses an exact floating-point comparison. | 2,256 | 8,812 | {"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} | 2.6875 | 3 | CC-MAIN-2020-05 | longest | en | 0.394952 |
https://darkwing.uoregon.edu/~stevev/sd-archive/prelist/msg00861.html | 1,679,993,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00073.warc.gz | 230,165,570 | 5,510 | # Re: Optimum Interstellar Rockets
```Other than complexity, would adjusting the exaust vel for optimum at ships
current vel (I.E. as the ships speed increases. Changing the exaust
velocity for the optimum for that speed) buy us anything?
If you look at the total amount of fuel my Explorer class needs (about
100,000,000 tons of 6Li) I'm starting to think we should at least work up
numbers for a mass conversion / anti-matter ship. Thou frankly the idea of
making and carrying a few thousand tons of antimatter bothers me a lot.
(NOT IN MY STARSYSTEM!!!!!) Ah, how far from a planet would we need to
keep the ship for safty?
Kelly
At 1:26 PM 4/4/96, DotarSojat@aol.com wrote:
>MEMORANDUM
>TO: LIT/SSD Discussion Group
>FROM: Rex Finke
>SUBJECT: Optimum Interstellar Rockets (Minimum Antimatter Fuel)
>
>
>INTRODUCTION
>
>Timothy van der Linden points out in his calc.txt that there is
>an optimum ratio of exhaust velocity to final rocket velocity
>relativistically (as I had calculated earlier for non-relativistic
>velocities --undocumented). The existence of this optimum indic-
>ates that there is a minimum in the amount of antimatter fuel
>required to accelerate a starship to any given final mission
>velocity.
>
>This memo provides the numbers that show how the ratio of the min-
>imum antimatter mass to initial starship mass varies with the de-
>sired mission velocity at the end of the first, acceleration burn.
>
>
>ANALYSIS
>
>We define the following operative quantities:
>
>V = "apparent" velocity = starmap distance/Earth time, in ltyr/yr
>U = "proper" velocity = starmap distance/starship time, in ltyr/yr
>Vend, Uend are the velocities at the end of the acceleration burn
> (at "burnout")
>Vexh, Uexh are the exhaust velocities
>g = relativistic energy factor "gamma" = 1/sqrt(1 - V^2)
>U = g V
>V = U/sqrt(1 + U^2)
>gend = gamma for Vend
>gexh = gamma for Vexh
>M = starship mass (= Mi initially; = Mbo at burnout)
>r = starship mass ratio = Mi/Mbo
>Ma = annihilation mass used during acceleration burn for rela-
> tivistic rocket = twice the mass of antimatter = 2 Mam
>Mp = mass of propellant used during acceleration burn for non-
> relativistic and relativistic rockets
>The propulsive energy efficiency (let's call it eff) is the ratio
>of the final vehicle kinetic energy to the total exhaust kinetic
>energy.
>
>Non-relativistically-
>
> final vehicle energy = (1/2) Mbo Vend^2
> total exhaust energy = (1/2) Mp Vexh^2
> eff = (Mbo/Mp)(Vend/Vexh)^2
>
>Now from the rocket equation
> Mi/Mbo [= (Mbo + Mp)/Mbo] = exp(Vend/Vexh)
>we get
> Mp/Mbo = exp(Vend/Vexh) - 1
>
>If we set x = Vend/Vexh to simplify, we get for the energy effic-
>iency the expression
> eff = x^2/(exp(x) - 1)
>
>This has a maximum value 0.648 for x = 1.59.
>
>So, if the burnout velocity of a non-relativistic rocket is 1.59
>times its exhaust velocity, the energy efficiency is a maximum of
>64.8 percent. I.e., the final vehicle energy can be no greater
>than 64.8 percent of the exhaust energy. This limitation is not
>an important consideration for a non-relativistic rocket because
>energy is subordinate to mass.
>
>Relativistically-
>
> final vehicle kinetic energy = Mbo (gend - 1) c^2
> total exhaust kinetic energy = Mp (gexh - 1) c^2 = Ma c^2
> (no energy losses)
> which gives Mp = Ma/(gexh - 1)
> but relativistically Mp = Mi - Mbo - Ma
> Ma/(gexh - 1) = Mi - Mbo - Ma
> Ma = (Mi - Mbo)(gexh - 1)/gexh
>
>so the energy efficiency, which is the ratio of the final vehicle
>kinetic energy to the total exhaust kinetic energy, is
> eff = Mbo (gend - 1) c^2/(Ma c^2)
> = Mbo (gend - 1) gexh/[(Mi - Mbo)(gexh - 1)]
> = (gend - 1) gexh/[(Mi/Mbo - 1)(gexh - 1)]
> = (gend - 1) gexh/[(r - 1)(gexh - 1)]
>
>The relativistic rocket equation, in its "velocity-parameter"
>form, is
> theta = Vexh ln r
>
>and the definition of the velocity parameter is
> tanh(theta) = Vend
>or sinh(theta) = Uend
>
>Note: asinh(Uend) = ln [Uend + sqrt(Uend^2 + 1)]
>
>so r = exp[asinh(Uend)/Vexh]
>
>With this relation we have all of the parameters to calculate
> eff = (gend - 1) gexh/[(r - 1)(gexh - 1)]
>
>The expression for eff is evaluated using a Fortran computer pro-
>gram, OPTVEXH, a description and a copy of which are given in the
>Appendix.
>
>
>RESULTS
>
>The results of the calculations of the optimum Vexh, the maximum
>energy efficiency and the minimum ratios of antimatter mass to
>burn-out mass and to initial mass are given in the table below
>for ascending values of the mission final proper velocity Uend.
>Included in the table are values of Vend, to illustrate the degree
>of saturation of apparent velocity, and of the optimum Uexh, to
>give a value (not otherwise meaningful) to which to relate the
>Uend, in order to examine the behavior of the ratio.
>
>The extreme Uend of 5 ltyr/yr represents the final velocity reach-
>ed at a continuous acceleration, a, of one g (1.0324 ltyr/yr^2)
>over a distance of 3.97 ltyr. (The acceleration distance
>s = [sqrt(1 + U^2) - 1]/a .) Only for destinations beyond about
>8 ltyr or accelerations greater than one g need one consider Uends
>greater than 5 ltyr/yr.
>
>(Note: these calculations assume no energy losses in converting
>annihilation energy to exhaust kinetic energy. The correction for
>energy losses would be to divide the minMam values by the conver-
>sion efficiency.)
>
>Uend Vend optVexh optUexh maxeff Uend/optUexh minMam/Mbo minMam/Mi
>----non-relativistic---- 0.648 1.59 --- ---
>0.2 0.196 0.124 0.125 0.647 1.60 0.0153 0.0029
>0.5 0.447 0.291 0.304 0.645 1.64 0.0914 0.0174
>1.0 0.707 0.492 0.566 0.640 1.77 0.323 0.0535
>1.1690.76* 0.541 0.643 0.639 1.82 0.422 0.0666
>2.0 0.894 0.691 0.957 0.630 2.09 0.981 0.1211
>3.0 0.949 0.777 1.235 0.622 2.43 1.739 0.1689
>4.0 0.970 0.823 1.450 0.616 2.76 2.537 0.1972
>5.0 0.981 0.852 1.625 0.611 3.08 3.357 0.2210
>--------
>*Timothy's selection
>
>
>OBSERVATIONS
>
>The minimized amount of antimatter is a small fraction of the
>starship's initial mass, less than 25 percent for mission proper
>velocities as high as 5 light-years/year for 100 percent conver-
>sion efficiency.
>
>The maximum energy efficiency decreases slowly as mission proper
>velocity is increased, but remains over 60 percent up to a mission
>proper velocity of 5 light-years/year.
>
>The ratio of the mission proper velocity to the optimum exhaust
>proper velocity increases fairly slowly at first from 1.59 at low
>velocities to almost double at a mission proper velocity of 5
>light-years/year.
>
>The implications of the values of the optimum exhaust velocity
>need to be examined, in terms of their conversion to MeV for
>exhaust particles.
>
>------------------------------------------------------------------
>APPENDIX. Program OPTVEXH
>
>For an input value of final proper velocity Uend, the program cal-
>culates the Vend, the gend and the theta. Then for values of Vexh
>increasing from 0.01 in increments of 0.01, the program calculates
>the eff until a maximum is passed. The optimum Vexh is calculated
>by fitting a second-degree curve to the three points that include
>the maximum. The value of the maximum is simply taken to be the
>value preceding the drop. The ratios of initial antimatter mass
>to Mbo and Mi are derived from expressions above-
>
> Ma/Mbo = (r - 1)(gexh - 1)/gexh
> eff = (gend - 1) gexh/[(r - 1)(gexh - 1)]
> = (gend - 1)/(Ma/Mbo)
> Ma/Mbo = (gend - 1)/eff
> Mam = (1/2) Ma
> Mam/Mbo = (gend - 1)/(2 eff)
> Mam/Mi = (Mam/Mbo)(Mbo/Mi)
> = (Mam/Mbo)/r
> = (gend - 1)/(2 eff r)
>
>C PROGRAM OPTVEXH 4/2/96
> 101 FORMAT(2X, 21H Final Proper Vel = ?)
> 102 FORMAT(2X, 15H Opt Exh Vel = , F6.4, 18H Max Energy Eff = ,
> & F6.4, 17H Antimatter/Mi = , F6.4)
> 103 FORMAT(2X, 8H VEXH = , F4.2, 7H EFF = , F6.4)
> 2 CONTINUE
> WRITE(*,101)
> READ(*,*) UEND !final proper velocity, ltyr/yr
> IF(UEND .EQ. 0.) GO TO 99
> VEND = UEND/SQRT(1. + UEND*UEND)
> GAMEND = 1./SQRT(1. - VEND*VEND)
> THETA = LOG(UEND + SQRT(UEND*UEND + 1.)) !asinh
> VEXH = 0.01
> VEXHN = VEXH
> 1 CONTINUE
> VEXHNN = VEXHN
> VEXHN = VEXH
> VEXH = VEXH + 0.01
> RN = R
> R = 1.01
> IF(VEXH .GT. .05) R = EXP(THETA/VEXH)
> GAMEX = 1./SQRT(1. - VEXH*VEXH)
> EFFNN = EFFN
> EFFN = EFF
> EFF = (GAMEND - 1.) * GAMEX/((R - 1.) * (GAMEX - 1.))
>C WRITE(*,103) VEXH, EFF
> IF(EFF .LT. EFFN .AND. VEXH .GT. 0.1) THEN
> Y1 = EFF
> Y2 = EFFN
> Y3 = EFFNN
> X1 = VEXH
> X2 = VEXHN
> X3 = VEXHNN
> A = ((Y1-Y2)*(X2-X3)-(Y2-Y3)*(X1-X2))/
> & ((X1*X1-X2*X2)*(X2-X3)-(X2*X2-X3*X3)*(X1-X2))
> B = ((Y1-Y2) - A*(X1*X1-X2*X2))/(X1-X2)
> OPTVEXH = -B/(2.*A)
> AMRATIO = (GAMEND - 1.)/(2.*EFFN*RN)
> WRITE(*,102) OPTVEXH, EFFN, AMRATIO
> GO TO 2
> END IF
> GO TO 1
> 99 STOP
> END
----------------------------------------------------------------------
Kelly Starks Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)
----------------------------------------------------------------------
``` | 3,199 | 9,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.891643 |
https://documentation.progress.com/output/Corticon/5.7.0/html/corticon/test-yourself-questions-3a-rule-writing-techniques.html | 1,575,609,726,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484815.34/warc/CC-MAIN-20191206050236-20191206074236-00184.warc.gz | 357,607,365 | 4,746 | Corticon Studio: Rule Modeling Guide : Rule writing techniques and logical equivalents : Test yourself questions: Rule writing techniques and logical equivalents
# Test yourself questions: Rule writing techniques and logical equivalents
Note: Try this test, and then go to Test yourself answers: Rule writing techniques and logical equivalents to correct yourself.
1. Filters act as master rules for all other rules in the same Rulesheet that share the same _________.
2. An expression that evaluates to a True or False value is called a _________ expression.
3. True or False. Condition row values sets must be complete.
4. True or False. Action row values sets must be complete.
5. The special term __________ can be used to complete any Condition row values set.
6. What operator is used to negate a Boolean expression?
7. If a Boolean expression is written in a Condition row, what values are automatically entered in the Values set when Enter is pressed?
8. A Filter expression written as Entity.boolean1=T is equivalent to (circle all that apply)
Entity.boolean1 Entity.boolean1<>F Entity.boolean1=F not (Entity.boolean1=F)
9. Of all alternatives listed in Question 71, which is the best choice? Why?
10. Describe the error (if any) in each of the following value ranges. Assume all are used in Conditions values sets.
a. {1…10, other}
b. {1..a, other}
c. {‘a'..other}
d. {1..10, 5..20, other}
e. {1..10, [10..20), other}
f. {‘red', ‘green', ‘blue'}
g. {<0, 0..15, >3}
11. True or False. The special term other may be used in Action row values sets.
12. Using best practices discussed in this chapter, model the following rules on a single Rulesheet:
If the part is in stock and it has a blue tag, then the part's discount is 10%
If the part is in stock and it has a red tag, then the part's discount is 15%
If the part is in stock and it has a yellow tag, then the part's discount is 20%
If the part is in stock and it has a green tag, then the part's discount is 25%
If the part is in stock and it has any other color tag, then the part's discount is 5%
13. True or False. A Nonconditional rule is equivalent to an Action expression with no Condition.
14. True or False. A Nonconditional rule is governed by any Preconditions on the same Rulesheet. | 547 | 2,259 | {"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-2019-51 | latest | en | 0.810449 |
https://tasiilaq.net/x-2-6x-5-0/ | 1,653,600,302,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00568.warc.gz | 625,865,758 | 5,633 | To settle the equation, variable the left hand side by grouping. First, left hand side needs to it is in rewritten as -x^2+ax+bx-5. To discover a and also b, set up a system to it is in solved.
You are watching: X 2 – 6x + 5 = 0
Since abdominal is positive, a and also b have actually the very same sign. Due to the fact that a+b is positive, a and b space both positive. The only such pair is the mechanism solution.
displaystylex=-frac32pmfrac12sqrt19 Explanation: displaystyle extusing the method of extcompleting the squaredisplaystyle• ext the coefficient of the x^2 ext term should be 1 ...
3x2+6x-5=0 Two options were found : x =(-6-√96)/6=-1-2/3√ 6 = -2.633 x =(-6+√96)/6=-1+2/3√ 6 = 0.633 step by step solution : action 1 :Equation at the finish of step 1 : (3x2 + 6x) - 5 = 0 ...
4x2+6x-5=0 Two solutions were discovered : x =(-6-√116)/8=(-3-√ 29 )/4= -2.096 x =(-6+√116)/8=(-3+√ 29 )/4= 0.596 step by step solution : action 1 :Equation at the end of step 1 : (22x2 + 6x) - ...
5x2+6x-5=0 Two options were discovered : x =(-6-√136)/10=(-3-√ 34 )/5= -1.766 x =(-6+√136)/10=(-3+√ 34 )/5= 0.566 step by step solution : step 1 :Equation at the end of step 1 : (5x2 + 6x) - ...
x2+6x-56=0 Two remedies were found : x =(-6-√260)/2=-3-√ 65 = -11.062 x =(-6+√260)/2=-3+√ 65 = 5.062 action by step solution : step 1 :Trying to aspect by splitting the center term ...
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To solve the equation, aspect the left hand next by grouping. First, left hand side demands to be rewritten as -x^2+ax+bx-5. To discover a and b, set up a system to it is in solved.
Since abdominal is positive, a and b have actually the exact same sign. Because a+b is positive, a and b are both positive. The only such pair is the system solution.
All equations of the type ax^2+bx+c=0 deserve to be solved using the quadratic formula: frac-b±sqrtb^2-4ac2a. The quadratic formula gives two solutions, one as soon as ± is addition and one once it is subtraction.
This equation is in typical form: ax^2+bx+c=0. Substitute -1 because that a, 6 for b, and -5 because that c in the quadratic formula, frac-b±sqrtb^2-4ac2a.
Quadratic equations such as this one deserve to be resolved by perfect the square. In order to complete the square, the equation must first be in the kind x^2+bx=c.
Divide -6, the coefficient the the x term, by 2 to gain -3. Then add the square the -3 come both sides of the equation. This step renders the left hand next of the equation a perfect square.
Factor x^2-6x+9. In general, when x^2+bx+c is a perfect square, the can constantly be factored together left(x+fracb2 ight)^2.
See more: Can Ice Skates Cut Your Fingers Off ? How Dangerous Is Ice Skating
Quadratic equations such as this one deserve to be fixed by a brand-new direct factoring technique that does not require guess work. To usage the straight factoring method, the equation must be in the type x^2+Bx+C=0.
Let r and s be the determinants for the quadratic equation such the x^2+Bx+C=(x−r)(x−s) where sum of components (r+s)=−B and also the product of components rs = C
Two number r and also s sum up come 6 exactly when the median of the 2 numbers is frac12*6 = 3. You can likewise see that the midpoint that r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant native the center by an unknown quantity u. Express r and s with respect to change u.
EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు | 1,241 | 3,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.898002 |
https://www.storyofmathematics.com/for-a-test-of-ho-p-05-the-z-test-statistic-equals-1-74-find-the-p-value-for-ha-p-lt-05/ | 1,726,541,594,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00301.warc.gz | 938,260,556 | 38,645 | # For a test of Ho: p=0.5,the z test statistic equals -1.74. Find the p-value for Ha: p<0.5.
The question aims to find out the p-value using the given alternative hypothesis, which is a one-sided hypothesis. Therefore, the p-value will be determined for the left tail test with reference to the standard normal probability table. When the alternative hypothesis states that a certain value for a parameter in the null hypothesis is lesser than the actual value, then left-tail tests are used.
Figure-1 : P-Value and Satistical Significance
Let’s first understand the difference between the Null and Alternative hypotheses. Null hypothesis $H_o$ refers to no association between two parameters of the population, meaning both are the same. Alternative hypothesis $H_a$ is opposite to the null hypothesis and states that there is a difference between two parameters.
## Expert Solution:
In order to calculate the p-value, we will use the standard normal table. According to the given information, the value of the test statistic is given as:$z = -1.74$Null hypothesis $H_o$ is given as:$p = 0.5$Alternative Hypothesis $H_a$ is given as:$p < 0.5$The formula for p-value is given as:$p = P (Z < z)$Where P is the probability:$p = P (Z < -1.74)$The p-value can be calculated by determining the probability less than -1.74 using the standard normal table. Therefore, from the table p-value is given as:$p = 0.0409$
## Alternative Solution:
For the given problem, the p-value will be determined using the standard probability table. Check against the row starting with -1.74 and column with 0.04. The answer obtained will be:$p = P ( Z< -1.74)$$p = 0.0409$Therefore, the p-value for $H_a$ < 0.5 is 0.0409.
## Example:
For a test of $H_o$: $p = 0.5$, the $z$ test statistic equals 1.74. Find the p-value for $H_a: p>0.5$.
Figure-2 : Z-Test Satistic
In this example, the value of test statistic $z$ is 1.74, therefore, it is a right tail test.For calculating the p-value for a right tail test, the formula is given as:$p = 1 – P ( Z > z)$$p = 1 – P ( Z > 1.74)$Now use the standard probability table to find the value.The p-value is given as:$p = 1 – 0.9591$ $p = 0.0409$Therefore, the p-value is 0.0409. | 606 | 2,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-38 | latest | en | 0.795391 |
https://mbc-web.org/how-to-solve-quadratic-equation-graphically/ | 1,652,847,371,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00396.warc.gz | 468,315,949 | 16,456 | # How To Solve Quadratic Equation Graphically Ideas
How To Solve Quadratic Equation Graphically. A combination of a linear and a quadratic forms a perfect system of equations or a pair of simultaneous equation. A quadratic equation is always of the form.for example, in the equation we can regard as and g(x) as.solving a quadratic equation means transforming the original equation into a new equation that has the form (where is a constant).
A x 2 + b x + c = 0, w h e r e a β 0. All students should be able to use graphical methods to solve the roots of a quadratic equation.
### 42A Graphing Quadratic Equations In Vertex Form
Another way of solving a quadratic equation is to solve it graphically. Approximate the point (s) at which the graphs of the functions intersect.
### How To Solve Quadratic Equation Graphically
From here, it is possible to deduce that both π₯ = π and π₯ = π satisfy the equation and that they are, therefore, solutions of the equation.Graph the two functions that were created.Graph y 1 and y 2 on the same graph.Graphical representation of quadratic equation yes!
Here the following figure is showing a graph of quadratic equation.If you are facing problems with solve graphically quadratic equations, why donβt you try algebrator.It is possible to solve such a system through the substitution method.It might be worth noting here that this can be a useful method if youβre using a graphing calculator to solve this kind of problem.
It uses the vertex formula to get the vertex which also gives an idea of what values to choose to plot the points.Let be equal to the expressions on both sides of the equal sign.Let the given quadratic inequality be ax 2 + bx + c β₯ 0.Let y 1 = ax 2 + bx + c and y 2 = d.
Most students should be able to use a linear function to solve a quadratic equation graphically.Now, we can graph the above quadratic function by making the table of values.On the other hand a quadratic equation is an equation of the form ax^2+bx+c where a\neq 0.One of the easiest way is by splitting the middle term.
One of the ways we can solve a quadratic equation is by factoring.Practice more questions on the quadratic equations worksheet for.Quadratic equations are an integral part of mathematics which has application in various other fields as well.Quadratic equations represent a parabola, if it meets at some points on the real line then those points are roots of the equation, otherwise it has no solution.
Section 1 is two linear equations;Section 2 is a quadratic and y=n;Section 3 is a quadratic and y=mx+c.Several questions for pupils to try on solving quadratic equations graphically.
Solve quadratic equations by factorising, using formulae and completing the square.Solving quadratics graphically with a graphic calculator if you have access to a graphic calculator or a graphing software, you can solve the quadratic equation a lot quicker.Some students should be able to derive and solve the resultant quadratic equation from linear and quadratic graphs.The coordinate of the point (s) where the graphs of the functions intersect.
The first two sections fit onto two sides of a4 and part 3 is the extension ultimately.The following steps will be useful to solve quadratic inequalities graphically.The graph of y = ax 2 + bx + c will either be open upward or downward parabola.The points at which the curve crosses a particular line on the graph are the solutions to the equation.
The quadratic equations explored are of the type a x 2 + b x + c = 0 review the analytical solutions to the above quadratic equation are given by the quadratic formula x 1 and x 2The solution(s) to a quadratic equation can be calculated using the quadratic formula:The Β± means we need to do a plus and a minus, so there are normally two solutions !Then the exact parabola will be drawn for you.
There is a rag table for students to mark their progress and.This applet allows you to enter a quadratic function by varying a, b and c sliders and a function by varying a, b and c sliders.This is a tutorial on how to solve quadratic equations graphically and check the answers to the analytical solutions.This is a worksheet with some questions on solving simultaneous equation in three sections.
This is an example where the coefficient of x 2 is negative.This is the corbettmaths video tutorial on how to solve quadratics graphicallyThis means we rearrange the quadratic equation into the factored form :This program has assisted many colleagues of mine and i have used it a couple of times as well.
This video shows an example of solving quadratic equation by graphing.To solve quadratic equation by graphing, we have to write the given quadratic equation as a quadratic function as shown below.We can solve the quadratic equation ax 2 + bx + c = d through graphing using the following steps:We can then take the square root of both sides of the equation and get and the graph of the function is a parabola that is open (concave) upward and just touches.
We have to write the quadratic function.Y = ax 2 + bx + c.Y = ax 2 + bx + x.You can analyze quadratic equations graphically.
You know by now how to solve a quadratic equation using factoring.You may solve your equation graphically by dragging the green , blue and white dots on the graph in order to produce a ‘solution equation’ of the form the solution set of the equation can then be gotten by taking the square root of both sides.π ( π₯ β π) ( π₯ β π) = 0.
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# Interest Rate vs APR: What’s the Difference? A Guide to Interest Rates
“Did you get a good interest rate?”
When applying for a personal loan, that’s the first thing everyone wants to know. The second: what exactly is interest rate vs. APR?
The short answer is this: While finding a low interest rate is undoubtedly important, it doesn’t tell the whole story—especially when it comes to fees, terms, and the total amount you’ll owe over time.
Before applying for a personal loan, you need to understand everything it means for your financial health—and that starts with understanding interest rate vs. APR, or annual percentage rate.
## Interest Rate vs APR: Interest Rates
### What is an Interest Rate?
The interest rate on a personal loan is the amount of money a creditor will earn for lending you money. Think of an interest rate as a fee you pay to access a service—the service of borrowing money.
Expressed as a percentage, lenders apply your interest rate to the total amount of your loan. When you make monthly payments:
• Part of your payment goes to paying back the money you borrowed (your loan principal), while
• The other part goes to your lender as payment for lending you the money (interest, fees).
For example, a loan of \$10,000 with an 8.717% APR would ultimately cost you \$12,016.66.
If you paid that over 5 years, your \$175.28 monthly payment would be:
1. \$145.48 principal (83%)
2. \$7.01 interest (4%)
3. \$21 fees at (12%)
Interest rates are always changing, and they can vary widely depending on the lender and applicant.
### How Are Interest Rates Set?
In the U.S., interest rates are set by the Federal Open Market Committee (FOMC). The committee meets eight times a year to assess the economy and determine what interest rate will keep the economy functioning best—aka. what rate will allow people to keep borrowing, lending, and spending money.
Interest rates are then raised or lowered to keep the economy stable and liquid. Liquidity refers to how easily an asset can be turned into cash, and at a price that reflects its true value.
In the U.S., retail banks and lenders also control rates based on the market, their needs, and their customers. While short-term rates are set to create stability and liquidity, long-term interest rates are selected based on demand.
• If fewer people apply for long-term U.S. Treasury notes (10- and 30-year), the rates will be higher.
• If many people want these long-term loans, the rates will be lower.
For example, in early 2022 in the U.S.:
• A short-term (3 month) loan had an interest rate between 0.83 and 0.51%.
• A long-term (10 year) loan had an interest rate of 2.75%.
### Applying for a Personal Loan: How is Your Interest Rate Determined?
#### Personal Loan Lending Factors
Personal loan interest rates currently range from 3 to 36 percent. For individuals, these rates vary depending on the details of your loan and your finances.
Lending factors can include:
• The length of the loan
• The amount you want to borrow
• The reason you’re borrowing money
• The lender you choose
• The type of rate (fixed or variable)
• The type of loan (secured or unsecured)
#### Fixed Rate vs. Variable Rate
Interest rates on personal loans are either fixed or variable.
Fixed rates remain the same for the length of the loan. With a fixed rate, you know in advance what your monthly payment will be, and the total amount of interest you’ll pay over the life of the loan. Personal loans through Prosper, for example, have fixed interest rates.
Variable rates change over time, which means your interest rate may rise and fall depending on changes in the market. While variable rates may start out lower than similar, fixed-rate loans, you need to be comfortable taking on the risk that your rate could go up at any time.
#### Secured vs. Unsecured Loan
Personal loans are either secured or unsecured
A secured loan requires you to put up collateral. That means your lender can take ownership of the asset you put up (property, house, car, stocks, etc.) if you default on the loan.
An unsecured loan allows you to borrow money without putting up collateral. Instead, lenders determine your creditworthiness based on other factors (credit history, income, debt load, etc.).
Most personal loans—including Prosper’s—are unsecured. At Prosper, we offer fixed rates on unsecured loans from \$2000 to \$40,000, with loan terms of 3 and 5 years.
### Applying for a Personal Loan: What are the Benefits?
Personal loans can be used for many purposes, from debt consolidation and home improvements to life events or emergencies. Personal loans are especially helpful when you need access to cash fast.
#### Key benefits of unsecured personal loans:
• You have flexibility to borrow for most anything you need
• You don’t need collateral
• You can borrow a large amount
• Rates are typically reasonable (and lower than credit cards)
• You know how much you’ll owe each month
• You have time to pay it off and a date when it will be paid-in-full
• You can get your money quickly
Personal loans through Prosper, for example, get funds to borrowers within 5 days on average.
## Interest Rate vs APR: Annual Percentage Rate (APR)
Another key factor when applying for a personal loan is the total cost of your loan. Since interest rates don’t include any fees you may be charged during the life of your loan, you need another metric to capture that, which is where the annual percentage rate (APR) comes in.
### What is Annual Percentage Rate?
The annual percentage rate is the total cost of borrowing money, including interest and all fees associated with your loan. Like the interest rate, APR is expressed as a percentage that shows you the actual yearly cost of borrowing over the length of your loan.
Because the APR is a bottom-line number, it allows you to compare the amount you’ll owe among different lenders, credit cards, or financial service providers. That’s crucial because, in some cases, a loan with a higher interest rate and low fees could ultimately end up costing you less.
### How Do You Calculate APR?
APR is a complicated mathematical formula, so it’s best to use an online calculator to run the calculation.
Gather the following information:
• The length of your loan
• Interest rate
• Origination fee
Remember: Lenders are legally required to provide you with your interest rate and your APR as part of any loan agreement, so this information should be easy to find.
#### Example
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Source: calculator.net
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The most common fee charged by lenders is an origination fee, which reimburses lenders for performing due diligence (pulling your credit report, verifying supporting documents, etc.) Some lenders charge a set dollar amount, while others structure the fee as a percentage of your loan amount. Online lending platforms like Prosper, for example, charge origination fees ranging from 2.41–5% depending upon your credit rating.
#### Example
Let’s say you borrow \$10,000, and your lender charges an origination fee of \$500. Depending on your loan and lender, the origination fee:
• May be taken out of your loan (you borrow \$10,000 and receive \$9,500), or
• May be charged in addition to your loan (you borrow \$10,500 and receive \$10,000 in funds).
Either way, the origination fee affects your loan’s bottom line. While an offer to waive the origination fee is enticing, be sure to look at the interest rate and other fees to see the total cost. Chances are, waiving the origination fee just means the company is making the money back somewhere else in a less obvious fashion.
### What Does It Mean When Lenders Advertise “No Fees”?
To that end, remember that transparency is important when making major financial decisions. Many lenders like to advertise that they charge “no fees” for their loans, which sounds like a pretty sweet deal. And while it’s true that you may never see a line item on your statement for origination or processing fees, you are still being charged for that labor.
Whether you’re paying the origination fee over time (ex. in higher monthly payments) or whether it comes out of your loan total at closing, that fee is coming out of your pocket one way or another—the only difference is how and when it’s happening.
## Ready to Start Applying for a Personal Loan?
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https://www.gizmodo.com.au/2016/08/why-mark-kelly-is-now-older-than-his-older-twin-brother/ | 1,575,585,072,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482284.9/warc/CC-MAIN-20191205213531-20191206001531-00215.warc.gz | 740,980,009 | 14,889 | # Why Astronaut Mark Kelly Is Now Even Older Than His Twin Brother
Astronaut Scott Kelly returned from a year-long sojourn in space in June. His slightly older astronaut twin, Mark Kelly, stayed home as a control — part of NASA's twin study to monitor the effects of space on the human body. But there's a physical change that NASA might not be able to measure that easily. Mark is now even older (by about five milliseconds) than his space-faring twin, thanks to special relativity. Image: Minute Physics/YouTube
It's due to a quirky little thing called time dilation: Time can slow down for one person, but not for another, because there is no such thing as a fixed frame of reference against which all motion can be measured. Two people who are moving relative to each other, wearing identical watches, will measure time differently, depending on how fast each is moving.
This isn't just a mathematical oddity. It's been experimentally verified over and over again with pairs of atomic clocks, although on human scales it's a pretty tiny effect. It's only when you reach speeds close to the speed of light that time dilates significantly. This new video from Minute Physics will give you a visual sense of how time dilation works:
Why does this happen? Well, the three dimensions of space and one dimension of time are merged in Einstein's theory to become a single four-dimensional spacetime. That means motion through time and motion through space are connected, and light is the link between them.
Since the speed of light in a vacuum is always constant (670 million MPH), time and space must adjust with motion to ensure that two people moving relative to each other will always measure the same speed for light. So as an object speeds up as it moves through space, time must pass more slowly. (Also, lengths contract.) But from your perspective, nothing would seem amiss.
And that brings us to the Twin Paradox, or why Mark Kelly is now technically even older than his twin, Scott Kelly. Here's Minute Physics again with the solution to this famous paradox:
To put it another way, time passes differently for the hypothetical twins in the video because they haven't had equivalent experiences. One stayed home in an inertial frame of reference. The other accelerated out to space, then decelerated, turned around and accelerated home of the return trip. So the symmetry between them is broken.
And as you may recall from Interstellar, time also passes more slowly in proximity to a black hole because of the strong gravitational effects. It's just one more challenge to be overcome if humans are ever going to travel deep into space. | 539 | 2,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-51 | latest | en | 0.964397 |
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# Frobisher, a sixteenth-century English explorer, had soil samples from
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21 Dec 2010, 15:25
Frobisher could have mistakely examined for gold on any other island as well too. In that case D would be a choice. Please explain why this option is out.
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21 Dec 2010, 15:46
chelliyil wrote:
Frobisher could have mistakely examined for gold on any other island as well too. In that case D would be a choice. Please explain why this option is out.
You cannot assume information that isn't stated in the question. With the given information, she/he/whoever draws the conclusion that the method was innaccurate.
There are many things that could have happened but only the things mentioned in the text are valid for the argument
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21 Dec 2010, 19:05
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chelliyil wrote:
Frobisher could have mistakely examined for gold on any other island as well too. In that case D would be a choice. Please explain why this option is out.
The argument says that he got soil from Kodlunarn island examined. The Queen sent two expeditions there. The argument does not have anything to do with the other islands. It does not assume that he did not get soil of any other island examined. Perhaps he did and found no gold there or perhaps he did find gold there. We do not know and do not care as far as this argument goes. Here we are only concerned with Kodlunarn.
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10 Jan 2011, 09:22
suesie970 wrote:
321kumarsushant wrote:
can anyone please explain why C is not a correct option??
i am agree with C.
E doesn't make sense to me.
can any one tell me, where can i find the correct solution of this question apart from this discussion.?
321kumarsushant:
Option C says that when Frobisher examined the soil sample for gold he used a different method than anyone else was using back in the 1500s . Even if this is true, this statement does not affect the conclusion at all.
If choice C said "The methods used to assess gold content of the soil samples provided by Frobisher were different from those generally used in the twenty first century.", then it would be a contender for the correct answer.
Hope that helps.
E clearly. It's the only one that strays from the original line of the passage.
Susie,
That amendment to option C was a good addition.
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12 Apr 2011, 02:25
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Frobisher, a sixteenth-century English explorer, had soil samples from Canada’s Kodlunarn Island examined for gold content. Because high gold content was reported,Elizabeth I funded two mining expeditions. Neither expedition found any gold there.
Modern analysis of the island’s soil indicates a very low gold content. Thus the methods used to determine the gold content of Frobisher’s samples must have been inaccurate. Which of the following is an assumption on which the argument depends?
A. The gold content of the soil on Kodlunarn Island is much lower today than it was in the sixteenth century.
B. The two mining expeditions funded by Elizabeth I did not mine the same part of Kodlunarn Island.
C. The methods used to assess gold content of the soil samples provided by Frobisher were different from those generally used in the sixteenth century.
D. Frobisher did not have soil samples from any other Canadian island examined for gold content.
E. Gold was not added to the soil samples collected by Frobisher before the samples were examined.
Can anyone help in explaining the answer with some good logic?
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12 Apr 2011, 05:10
1
Here the conclusion of the agument is that the methods used were wrong
A. since no timelines of Elizabeth's expeditions are given, and in fact it is implied that these were done after Frobisher's exploration - since they were done because of his report only- and since these also didnt find any gold, despite being done in 16th century, Statement A is wrong
B.Weak option - anyway which part of island does modern analysis base itself on - same or not? Since we dont know this, we cant say this is assumption on which the author's argument is based.
C. If methods were different, it doesnt mean they were wrong/ not wrong. incorrect
D. Othe islands are not being talked of here. Incorrect option
E. If frobisher added this gold himself, then both conditions are satisfied- the methods are correct and the soill doesnt have actually any gold. this argument is therefore based on this assumption.
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12 Apr 2011, 11:08
Nicely explained ! Especially the option B.
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15 Sep 2011, 23:03
IMO ans should be E
For C, we donot have sufficient information to assume whether the method used was different or not, but the use of "modern analysis" indicates that the present day test would be more accurate
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23 Dec 2011, 06:12
E it is. Nice question but E stands out as the assumption.
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26 Dec 2011, 03:32
Yes E its very simple.
We need to SUPPORT the assumption that METHOD USED TO TEST GOLD CONTENT was inaccurate.
E ) clearly states that Sample was not wrong and no gold was added to the sample .. so he had sample which had no gold yet his study showed gold was present
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29 Jan 2012, 04:28
1
Frobisher, a sixteenth-century English explorer, had soil samples from Canada’s Kodlunarn Island examined for gold content. Because high gold content was reported,Elizabeth I funded two mining expeditions. Neither expedition found any gold there.
Modern analysis of the island’s soil indicates a very low gold content. Thus the methods used to determine the gold content of Frobisher’s samples must have been inaccurate. Which of the following is an assumption on which the argument depends?
Conclusion: Methods used by F to find gold content were inaccurate.
A. The gold content of the soil on Kodlunarn Island is much lower today than it was in the sixteenth century. This says that gold content was high in the 16th century. Weakens the argument. Hence, cannot be the assumption.
B. The two mining expeditions funded by Elizabeth I did not mine the same part of Kodlunarn Island.
Irrelevant. Same part or different part not mentioned anywhere in the stimulus.
C. The methods used to assess gold content of the soil samples provided by Frobisher were different from those generally used in the sixteenth century. The difference in methods does not matter. What the conclusion says is the method, in fact, was inaccurate.
D. Frobisher did not have soil samples from any other Canadian island examined for gold content.Again, irrelevant.
E. Gold was not added to the soil samples collected by Frobisher before the samples were examined. Negating, Gold WAS added to the soil samples before they were examined. If this were true, then F's methods may have been accurate, and measured HIGH GOLD CONTENT. Hence, the conclusion that F's methods were inaccurate falls apart.
Hope this helps
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29 Nov 2012, 12:23
It's been assumed that the method was wrong considering no gold was added.
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25 Nov 2013, 17:57
Hey Karishma,
But why not D?
Thanks
VeritasPrepKarishma wrote:
fiesta wrote:
Frobisher, a sixteenth-century English explorer, had soil samples from Canada’s Kodlunarn Island examined for gold content. Because high gold content was reported, Elizabeth I funded two mining expeditions. Neither expedition found any gold there. Modern analysis of the island’s soil indicates a very low gold content. Thus the methods used to determine the gold content of Frobisher’s samples must have been inaccurate.
Which of the following is an assumption on which the argument depends?
(A) The gold content of the soil on Kodlunarn Island is much lower today than it was in the sixteenth century.
(B) The two mining expeditions funded by Elizabeth I did not mine the same part of Kodlunarn Island.
(C) The methods used to assess gold content of the soil samples provided by
Frobisher were different from those generally used in the sixteenth century.
(D) Frobisher did not have soil samples from any other Canadian island examined for gold content.
(E) Gold was not added to the soil samples collected by Frobisher before the samples were examined.
Let us read the question stem first. We are looking for an assumption. An assumption is a necessary missing premise. We are looking for the option that needs to be true for the conclusion to be true.
Premises:
Frobisher had soil samples from Canada’s Kodlunarn Island examined for gold content.
Because high gold content was reported, Elizabeth I funded two mining expeditions.
Neither expedition found any gold there.
Modern analysis of the island’s soil indicates a very low gold content.
Tell me, when you read the above premises, what possibilities come to mind? Frobisher had samples examined. High gold content was reported. No gold was actually found. Modern analysis show very low gold content.
The following possibilities come to my mind:
1. Either there was gold and before the expeditions were sent, it was mined (very unlikely!)
2. His methods were inaccurate.
Conclusion:
The methods used to determine the gold content of Frobisher’s samples must have been inaccurate.
If I am concluding that his methods were inaccurate, then I am assuming that no one added gold to his samples and gold was not mined before the expeditions were sent. (Technically, gold could have been added and his methods could have been inaccurate too but lets not mess with that.)
Hence option (E) is an assumption.
Also, use you can use assumption negation technique to see that it is the right answer.
I negate (E) : Gold was added to the soil samples collected by Frobisher before the samples were examined.
I can not conclude now that his methods were inaccurate.
Hence (E) is the correct answer.
Option (C) is not correct. We did not assume in the argument that his methods were different. They could have been the same ones generally used in the 16th century, It is possible that 16th century methods were not accurate.
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27 Nov 2013, 23:42
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28 Nov 2013, 02:57
How could D be a choice?
Negate D and check.
Frobisher has soil samples from other Canadian island examined for gold content.
Even though he has examined soil samples from other islands, his method could still be inaccurate. Thus D would not hurt the conclusion by negation.
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03 Dec 2013, 07:08
Look at 5 choices:
A) Doesn't support the conclusion: lower gold content today suggests that finding low gold content doesn't imply the inaccuracy of the method.
B) Same as A: "didn't mine the same part" => cannot conclude anything about the method's accuracy
C) out of scope: "the method generally used in the 16th century" is irrelevant. Such a comparison between 2 methods provides no help.
D) The statement is not enough for us to establish any inference about the method mentioned in the original argument
E) Using negation technique: What happens if this stat isn't true? As gold was added to the samples before they were checked, the reported high gold content is due to fraudulence (my own idea!), rather than the method's inaccuracy => invalidate the conclusion.
Pick E.
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16 May 2014, 07:12
Hi Chiranjeev,
The conclusion of the argument questioned the accuracy of the method used by Frobisher . keeping this in mind.
in option D states after negation that Frobisher soil sample is taken form an other canadian island. So doesnt weaken the conclusion. or firstly it is like it doesn't say anything about the from where does the modern analysis has taken soil sample it might be from the same place and secondly in conclusion we are taking about the Method only.
Is my understanding correct?
one more question ,generally in assumption question , if any option shatters the premise, or weakens the premise or just a we get obtain statement after negation which is just the opposite of the given premises does that statement is a valid assumption.
i think it won't fall into the category of new information ,and secondly it should be related to the conclusion ,this could be reasons for not considering the statement as a valid assumption
correct me where I'm wrong.
Thanks
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06 Aug 2014, 01:41
VeritasPrepKarishma wrote:
fiesta wrote:
Frobisher, a sixteenth-century English explorer, had soil samples from Canada’s Kodlunarn Island examined for gold content. Because high gold content was reported, Elizabeth I funded two mining expeditions. Neither expedition found any gold there. Modern analysis of the island’s soil indicates a very low gold content. Thus the methods used to determine the gold content of Frobisher’s samples must have been inaccurate.
Which of the following is an assumption on which the argument depends?
(A) The gold content of the soil on Kodlunarn Island is much lower today than it was in the sixteenth century.
(B) The two mining expeditions funded by Elizabeth I did not mine the same part of Kodlunarn Island.
(C) The methods used to assess gold content of the soil samples provided by
Frobisher were different from those generally used in the sixteenth century.
(D) Frobisher did not have soil samples from any other Canadian island examined for gold content.
(E) Gold was not added to the soil samples collected by Frobisher before the samples were examined.
Let us read the question stem first. We are looking for an assumption. An assumption is a necessary missing premise. We are looking for the option that needs to be true for the conclusion to be true.
Premises:
Frobisher had soil samples from Canada’s Kodlunarn Island examined for gold content.
Because high gold content was reported, Elizabeth I funded two mining expeditions.
Neither expedition found any gold there.
Modern analysis of the island’s soil indicates a very low gold content.
Tell me, when you read the above premises, what possibilities come to mind? Frobisher had samples examined. High gold content was reported. No gold was actually found. Modern analysis show very low gold content.
The following possibilities come to my mind:
1. Either there was gold and before the expeditions were sent, it was mined (very unlikely!)
2. His methods were inaccurate.
Conclusion:
The methods used to determine the gold content of Frobisher’s samples must have been inaccurate.
If I am concluding that his methods were inaccurate, then I am assuming that no one added gold to his samples and gold was not mined before the expeditions were sent. (Technically, gold could have been added and his methods could have been inaccurate too but lets not mess with that.)
Hence option (E) is an assumption.
Also, use you can use assumption negation technique to see that it is the right answer.
I negate (E) : Gold was added to the soil samples collected by Frobisher before the samples were examined.
I can not conclude now that his methods were inaccurate.
Hence (E) is the correct answer.
Option (C) is not correct. We did not assume in the argument that his methods were different. They could have been the same ones generally used in the 16th century, It is possible that 16th century methods were not accurate.
Karishma , can I paraphrase the above highlighted explanation as
Negative E - Gold was added to the soil samples collected by Frobisher before the samples were examined.
Now that if gold was added before examination and the examination rightly pointed this out - it clearly shows that the technical method to determine the gold content was not in-accurate ( though the step of fudging sample was morally wrong but that is altogether a different matter)
So negating assumption , destroys the conclusion , hence E it is
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07 Aug 2014, 03:02
himanshujovi wrote:
Karishma , can I paraphrase the above highlighted explanation as
Negative E - Gold was added to the soil samples collected by Frobisher before the samples were examined.
Now that if gold was added before examination and the examination rightly pointed this out - it clearly shows that the technical method to determine the gold content was not in-accurate ( though the step of fudging sample was morally wrong but that is altogether a different matter)
So negating assumption , destroys the conclusion , hence E it is
Yes, that's correct. If Gold was added, we don't know who did it so we cannot blame Frobisher or his methods. All we know is that his methods to determine gold content could have been accurate.
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22 Sep 2014, 05:17
udaymathapati wrote:
Frobisher, a sixteenth-century English explorer, had soil samples from Canada’s Kodlunarn Island examined for gold content. Because high gold content was reported,Elizabeth I funded two mining expeditions. Neither expedition found any gold there.
Modern analysis of the island’s soil indicates a very low gold content. Thus the methods used to determine the gold content of Frobisher’s samples must have been inaccurate. Which of the following is an assumption on which the argument depends?
A. The gold content of the soil on Kodlunarn Island is much lower today than it was in the sixteenth century.
B. The two mining expeditions funded by Elizabeth I did not mine the same part of Kodlunarn Island.
C. The methods used to assess gold content of the soil samples provided by Frobisher were different from those generally used in the sixteenth century.
D. Frobisher did not have soil samples from any other Canadian island examined for gold content.
E. Gold was not added to the soil samples collected by Frobisher before the samples were examined.
Can anyone help in explaining the answer with some good logic?
This question is asking for assumption. So negation technique will work for true answer choice
If we negates the E, then it breaks the conclusion. Any other choice do not break the conclusion
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https://electric-skateboard.builders/t/best-way-to-calculate-mileage/52186 | 1,603,378,627,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00331.warc.gz | 301,326,416 | 8,374 | # Best way to calculate "mileage"?
I would like a good way to consistently calculate the range on my board. I am making a spreadsheet to get my average top speed and “mileage” on my board.
The easiest thing that comes to mind is using volts/mile (v/m), but that will be different depending on the Ahr of a person’s battery. The reason I say this is (assuming you use your battery down to 3.2v) you will have a 1 volt of usable charge, so it will be easy to calculate distance. However, this could be pretty easy to compare to a car. “My board gets x amount of miles per volt” or “I have a (.9 or 1v) tank.” Kinda like cars get miles per gallon with a x gallon tank.
Most use Watt hours per mile (Whr/m) to calculate, but what from what I can gather this is difficult to consistently measure.
So what do you guys use?
Edit: If this has been discussed before, I apologize. I wasn’t able to find that information. If so, please link the page!
Without doing a lot of measurements and calculations it will be difficult to get a very accurate measurement.
A general figure a lot of people here use is 10Wh = 1km of range.
Dont use volts, it will be unprecise as it will depend on your Ah. Volts isnt a way of measuring energy, wh is, so use that. A rule of thumb is 10wh/ km, but its different for everyone. A 4wd board pulling a 100 kg man will use alot more than a single drive pulling a 50 kg man, for example. Just charge your battery full, and ride until its empty while tracking the distance…
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This calculator should do the trick for you: http://esk8.today/2016/12/28/how-far-can-i-ride/
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If you prefer a value that increases with increasing efficiency like miles per gallon then use miles per kilowatt hour. By contrast with watt hours per mile or kilometer a “better” value is a “lower value.” 10 watt hours per kilometer is equivalent to 100 kilometers per kilowatt hour.
Huh. That’s pretty interesting! never thought of it that way!
I just divide it into two. One for economical at 11watts per km/mile and the other for heavy use at 22 watts per km/mile.
Wouldn’t hills and regenerative breaking come into account?
Yes, you are right.
It just averages out over a trip. One day might be nice and calm weather, the next day on the same trip might have some strong headwind.
As some others has said, there is just too many variables affecting range.
I just consider any additional range from braking as a bonus instead of part of the calculation.
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Your watt hours per km will be different than watt hours per mile as a mile is longer, meaning more watt hours are required to travel a mile
See the “converted to kilometres” column
watt hour divide by the same number of watts consumed would give the same answer regardless of metrics which is why kilometers and miles is the same, so look to the right for converted to km or miles and you will see two very different answers in “converted to miles” and "converted to kilometers which accounts for the different distance between a mile and a kilometer.
I had trouble wrapping my head around that for a bit as my brain was telling me that no, the answer is wrong cos the distance cannot be the same
Um I’m not sure what you were saying exactly but I do know that the conversion is roughly 10Wh/km which should be something like 17Wh/mi.
Oh okay I looked at the column but it’s not correct. 11Wh/km is about average where as a mile is ~1.6 times the length of a km meaning 1.6 times the Wh needed.
So it should be something like 17Wh per mile. You will still get the same distance in total either way
Still correct mate
The answer gets rounded to the nearest whole number.
My spread sheet says 18Wh per mile which if it was not rounded up would be something like 17.8Wh per mile or something to get exactly 57 kilometers.
I can just input that in my spreadsheet and see the converted kilometer of which I have a better sense of distance. Just like flying at 15 km gives me a better sense of height than 25 thousand feet.
https://www.vesc-project.com/calculators
This one is quite comprehensive.
Yes it is quite comprehensive.
I just like to play with spreadsheets so I have my own calculator which is good enough for my use.
Ahhh yeah your right just that I don’t think the first spread sheet changed. As it says 11wh for both miles and km that’s why I thought you didn’t understand.
The second is right though
Yes, I can confirm 17Wh/mile is pretty normal for a 200lb person
How!?! I get 17wh/mile at 110lb 20mph
Also depends on hills, gearing, motors and speed
That’s a maximum when going for max range. If I’m accelerating at-will it’s closer to 30Wh/mile
1 Like | 1,110 | 4,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-45 | latest | en | 0.914805 |
https://wordpandit.com/number-system-basics-of-factors-test-3/ | 1,679,429,504,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00310.warc.gz | 719,702,861 | 32,113 | Select Page
• This is an assessment test.
• To draw maximum benefit, study the concepts for the topic concerned.
• Kindly take the tests in this series with a pre-defined schedule.
## Number System: Basics of Factors Test-3
Congratulations - you have completed Number System: Basics of Factors Test-3. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
Question 1
How many factors are there in N = 25 55 72112 which ends with zero?
A 230 B 225 C 105 D 270
Question 1 Explanation:
This question talks about the factors that end with zero.
All we need to find is number of factors that end with zero.
For this, the minimum power of 2 and 5 has to be one.
A factor divisible by 10 is of the form 2(1 or 2 or 3or 4 or 5)5(1 or 2 or 3 or4 or5 )7(0 or 1 or 2 )11(0,1,2)
Hence, total number of factors divisible by 10 is = (5)(5)(2 + 1)(2+1) = 225
Question 2
How many number of factors are in N = 23 325272 which is divisible by 20?
A 72 B 70 C 38 D 36
Question 2 Explanation:
Since we have to find the number of factors which are divisible by 20,
to find this divide the given number by 20 and then find the number of factors of the quotient.
Divide 23325272 by 20 ( 2251)
= 21 32 51 72 and its number of factors are 2x3x2x3=36
Question 3
How many number of factors of N= 24 33112133 that are perfect square?
A 20 B 22 C 24 D 30
Question 3 Explanation:
We know that for a number to be a perfect square, its factor must have the even number of powers.
All we need to explore is the even powers of the factors in this case.
Therefore the perfect square factors can be2 ( o or 2 or 4 )3(0 or 2 )11(0 or 2 )13(0 or 2)
Hence, the total number of factors which are perfect square is 3x2x2x2=24
Question 4
How many factors of N= 2936118 that are perfect cube?
A 72 B 36 C 45 D 20
Question 4 Explanation:
If a number is a perfect cube, then the power of the prime factors should be divisible by 3.
The possible options are : 2(0 or 3 or 6 or 9)3(0 or 3 or 6)11 (0 or 3 or 6 ) .
Hence the total number of factors which are perfect cube 4 x 3x3 =36.
Question 5
How many factors of N= 2936118are both perfect square and perfect cube?
A 8 B 12 C 14 D 10
Question 5 Explanation:
If a number is perfect square and perfect cube then the powers of prime factors must be divisible by 6.
Perfect square factor must be 2(0 or 6 or 9)3(0 or 6)7(0 or 6)
Hence total number of perfect cube factors is 3 x 2 x 2 = 12
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
← List →
Shaded items are complete.
1 2 3 4 5 End
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Updates for Free Live sessions and offers are sent on mail. Don't worry: we do not send too many emails..:) | 897 | 3,020 | {"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} | 4.59375 | 5 | CC-MAIN-2023-14 | latest | en | 0.918523 |
https://www.expii.com/t/trigonometry-negative-angle-identities-5261 | 1,627,604,178,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00116.warc.gz | 775,610,753 | 2,419 | Expii
# Trigonometry: Negative Angle Identities - Expii
Based on the unit circle, the negative angle identities (also called "odd/even" identities) tell you how to find the trig functions at -x in terms of the trig functions at x. In other words, they relate trig values at opposite angles x and -x. For example, sin(-x) = -sin(x), cos(-x) = cos(x), and tan(-x) = -tan(x). | 103 | 374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-31 | longest | en | 0.809297 |
https://documen.tv/question/kylie-ate-lunch-at-a-restaurant-the-bill-came-to-74-60-if-she-left-a-20-tip-what-was-the-total-c-19800707-53/ | 1,632,037,964,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00126.warc.gz | 272,477,375 | 15,939 | ## Kylie ate lunch at a restaurant. The bill came to $74.60. If she left a 20% tip, what was the total cost of her lunch? Question Kylie ate lunch at a restaurant. The bill came to$74.60. If she left a 20% tip, what was the
total cost of her lunch?
in progress 0
2 months 2021-08-02T05:25:11+00:00 2 Answers 0 views 0
Her total cost = $89.52 Step-by-step explanation: bill =$74.60
Now,
tip = (20/100)×74.60
= 74.60/5
= $14.92 Her total cost = bill + tip = 74.60 + 14.92 =$89.52 | 172 | 483 | {"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} | 4.0625 | 4 | CC-MAIN-2021-39 | latest | en | 0.918304 |
https://www.coursehero.com/file/6421162/Chapter-2-study-guide/ | 1,493,237,448,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121644.94/warc/CC-MAIN-20170423031201-00479-ip-10-145-167-34.ec2.internal.warc.gz | 888,199,686 | 63,317 | Chapter 2 study guide
# Chapter 2 study guide - From:
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From: <Saved by Windows Internet Explorer 8> Subject: Chapter 2 Date: Wed, 21 Sep 2011 16:40:42 -0500 MIME-Version: 1.0 Content-Type: text/html; charset="Windows-1252" Content-Transfer-Encoding: quoted-printable Content-Location: http://www2.uhv.edu/dongx/acct3332/Study%20Guides/chapter2.htm X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5994 X <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD><TITLE>Chapter 2</TITLE> <META content=3D"text/html; charset=3Dwindows-1252" = http-equiv=3DContent-Type> <META name=3DGENERATOR content=3D"MSHTML 8.00.6001.18975"> <STYLE>@font-face { font-family: SimSun; } @font-face { font-family: Cambria Math; } @font-face { font-family: Calibri; } @font-face { font-family: @SimSun; } @page WordSection1 {size: 8.5in 11.0in; margin: 1.0in 1.25in 1.0in = 1.25in; } P.MsoNormal { LINE-HEIGHT: 115%; MARGIN: 0in 0in 10pt; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } LI.MsoNormal { LINE-HEIGHT: 115%; MARGIN: 0in 0in 10pt; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } DIV.MsoNormal { LINE-HEIGHT: 115%; MARGIN: 0in 0in 10pt; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } P.MsoListParagraph { LINE-HEIGHT: 115%; MARGIN: 0in 0in 10pt 0.5in; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } LI.MsoListParagraph { LINE-HEIGHT: 115%; MARGIN: 0in 0in 10pt 0.5in; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } DIV.MsoListParagraph { LINE-HEIGHT: 115%; MARGIN: 0in 0in 10pt 0.5in; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } P.MsoListParagraphCxSpFirst { LINE-HEIGHT: 115%; MARGIN: 0in 0in 0pt 0.5in; FONT-FAMILY: = "Calibri","sans-serif"; FONT-SIZE: 11pt } LI.MsoListParagraphCxSpFirst { LINE-HEIGHT: 115%; MARGIN: 0in 0in 0pt 0.5in; FONT-FAMILY: =
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## This note was uploaded on 09/21/2011 for the course ACCT 3332 taught by Professor Dongxibong during the Fall '11 term at University of Houston-Victoria.
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Ask a homework question - tutors are online | 1,305 | 3,661 | {"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} | 3.3125 | 3 | CC-MAIN-2017-17 | latest | en | 0.428929 |
https://apps.apple.com/gb/app/math-slide-addition-subtraction/id588603890 | 1,582,201,415,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00113.warc.gz | 289,030,518 | 19,877 | ## Description
Math Slide: Addition & Subtraction Facts is a multiplayer game helping children to learn and recall addition and subtraction number facts. Players learn by sliding tiles into the center to match an answer, equation or image. The player who slides their tiles first wins.
Being able to quickly and easily recall addition and subtraction number facts is a key math skill critical in a child’s development. Children are unlikely to progress without learning these key facts. This app is specifically designed to help children learn this critical skill.
Features
• Multiplayer game, suitable for one, two, three or four players
• 10 games all focused on the same key math concept
• All games can be played twice for free
• Each game is a little harder than the previous game
• Game 1 and Game 9 can be used to work out if the app is at the right level of difficulty
• The app focuses on a critical math skill vital for success
To work out if this app is at the right level of challenge try Game 1 and Game 9. If Game 1 is too hard or Game 9 is too easy this app is not at the right level.
The free version of this app allows each game to be played twice before a decision needs to be made to purchase. Games 1 and Game 9 can be played unlimited times for free.
How to Play
1. Select a game to play
2. Players join the game by clicking on ‘Join’
3. Click ‘Start Game’, wait for the count down, and play
4. Slide a tile into the center to match, first tile into the center wins
5. The first player to use all their tiles wins the game
Math focus of each game
Game 1 – representing doubles & pairs to 10
Game 2 – doubles up to 10, pairs that equal 10
Game 3 – doubles up to 10, pairs that equal 10
Game 4 – representing number facts up to 10+10
Game 5 – addition facts up to 10+10
Game 6 – subtraction facts up to 10+10
Game 7 – addition & subtraction facts up to 10+10
Game 8 – number facts with answers to 20
Game 9 – pairs of number facts: 6 + 7 = 8 + ? ...
Game 10 – triples of number facts: 6 + 5 + ? = 15, …
Math Adventures is a New Zealand company helping children learn mathematics. We develop apps that focus on the key concepts that need to be fully understood to progress and succeed in mathematics. Our apps range from learning to count up to understanding and using fractions and decimals, with each app focusing on one key concept or skill.
We believe:
• Everyone can learn mathematics
• Playing educationally rich games and apps improves understanding
• Learning math should be fun and exciting, a rewarding adventure
Our growing variety of apps is research based and classroom tested to make sure they help children learn.
Math Slide is part of our ‘play and learn’ series, which are games high in educational value where children learn as they play. We are also currently developing ‘learning’ apps which use the power of the iPad to help children understand and learn key math concepts and to unravel misunderstandings.
## What’s New
Version 1.3
Improved performance: the app now starts up super-fast.
Added Chinese language support. 太好了!
## Ratings and Reviews
3.0 out of 5
3 Ratings
3 Ratings
Hhhukvrt ,
### I win
I'm the first one to review.this maths app is so easy.try to beat 20 secs.if you do review this game,by telling me
## Information
Provider
Size
18.8 MB
Category
Education
Compatibility
Requires iOS 8.0 or later. Compatible with iPhone, iPad and iPod touch.
Languages
English, Simplified Chinese
Age Rating
4+
Price
Free
In-App Purchases
1. Math Slide: addition & subtraction upgrade £0.99 | 838 | 3,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-10 | latest | en | 0.915613 |
https://www.gurufocus.com/term/tax/SUNS/Tax+Expense/Solar+Senior+Capital+Ltd | 1,511,489,476,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00732.warc.gz | 827,236,312 | 39,998 | Switch to:
Solar Senior Capital Ltd (NAS:SUNS) Tax Expense: \$0.00 Mil (TTM As of Sep. 2017)
Solar Senior Capital Ltd's tax expense for the months ended in Sep. 2017 was \$0.00 Mil. Its tax expense for the trailing twelve months (TTM) ended in Sep. 2017 was \$0.00 Mil.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Solar Senior Capital Ltd Annual Data
Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Tax Expense 0.00 0.00 0.00 0.00 0.00
Solar Senior Capital Ltd Quarterly Data
Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Sep17 Tax Expense 0.00 0.00 0.00 0.00 0.00
Calculation
Tax paid by the company. It is computed in by multiplying the income before tax number, as reported to shareholders, by the appropriate tax rate. In reality, the computation is typically considerably more complex due to things such as expenses considered not deductible by taxing authorities ("add backs"), the range of tax rates applicable to various levels of income, different tax rates in different jurisdictions, multiple layers of tax on income, and other issues.
Tax Expense for the trailing twelve months (TTM) ended in Sep. 2017 was 0 (Dec. 2016 ) + 0 (Mar. 2017 ) + 0 (Jun. 2017 ) + 0 (Sep. 2017 ) = \$0.00 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the long run, income before tax and taxable income will likely be more similar than they are in any given period. If the one is less in earlier years, then it will be greater in later years. Deferred taxes will reverse themselves in the long run and in total will zero out, unless there is something like a change in tax rates in the intervening period. A deferred tax payable results from a tax break in the early years and will reverse itself in later years; a deferred tax receivable results from more taxes being paid in early years than the tax expense reported to shareholders and will again reverse itself in later years. The deferred tax amount is computed by estimating the amount and the timing of the reversal and multiplying that by the appropriate tax rates.
Related Terms | 554 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-47 | latest | en | 0.945032 |
http://oeis.org/A067587 | 1,548,058,285,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583763839.28/warc/CC-MAIN-20190121070334-20190121092334-00188.warc.gz | 170,500,985 | 4,006 | This site is supported by donations to The OEIS Foundation.
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A067587 Inverse of A066884 considered as a permutation of the positive integers. 3
1, 3, 2, 6, 5, 9, 4, 10, 14, 20, 8, 27, 13, 19, 7, 15, 35, 44, 26, 54, 34, 43, 12, 65, 53, 64, 18, 76, 25, 33, 11, 21, 77, 90, 89, 104, 103, 118, 42, 119, 134, 151, 52, 169, 63, 75, 17, 135, 188, 208, 88, 229, 102, 117, 24, 251, 133, 150, 32, 168, 41, 51, 16, 28, 152 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Ivan Neretin, Table of n, a(n) for n = 1..10000 FORMULA Let w(n)=A000120(n) be the 'weight' of n; i.e. the number of 1's in the binary expansion of n. Let p(n)=A068076(n) be the number of positive integers < n with the same weight as n. Then a(n) = binomial(w(n)+p(n), 2) + p(n) + 1. MATHEMATICA w[n_] := Plus@@IntegerDigits[n, 2]; p[n_] := Plus@@MapThread[Binomial, {Flatten[Position[Reverse[IntegerDigits[n, 2]], 1]]-1, Range[w[n]]}]; a[n_] := Binomial[w[n]+p[n], 2]+p[n]+1 PROG (Perl) foreach(1..10_000){\$i=eval join "+", split //, sprintf "%b", \$_; \$j=\$r[\$i]++; print "\$_ ", \$j+1+(\$i+\$j)*(\$i+\$j-1)/2, "\n"} # Ivan Neretin, Mar 02 2016 CROSSREFS Sequence in context: A118833 A268823 A046877 * A198259 A120476 A069159 Adjacent sequences: A067584 A067585 A067586 * A067588 A067589 A067590 KEYWORD easy,nonn AUTHOR Jared Ricks (jaredricks(AT)yahoo.com), Jan 31 2002 EXTENSIONS Edited by Dean Hickerson, Feb 16 2002 Offset changed to 1 by Ivan Neretin, Mar 02 2016 STATUS approved
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Last modified January 21 02:59 EST 2019. Contains 319344 sequences. (Running on oeis4.) | 718 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-04 | latest | en | 0.644865 |
http://stackoverflow.com/questions/19437955/how-to-convert-char-string-to-one-single-int-number-if-the-string-also-contains | 1,469,495,285,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00293-ip-10-185-27-174.ec2.internal.warc.gz | 226,266,973 | 18,634 | Dismiss
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# How to convert char string to one single int number if the string also contains letters
I have a very basic question, if I have a string of chars like this: `char charv1[6] = "v445"` or `v666` How can I get the numbers and turn them into a single integer with value: `445` or `666`?
I have been trying this code but something goes wrong ...:
`````` size = (strlen(charv1)-1);
for(aux = size; aux > 0; aux--){
if(aux == (size)){
v1 = charv1[aux]-'0';
}
else{
aux2 = (charv1[aux]-'0')*10;
printf("%d\n", aux2);
v1 = v1 + aux2;
}
}
``````
`charv1` contains the string: `v445`etc
I remember a few years ago, I did it recursively but I do not remember how, but now I do not need an elegant solution ... I need one that just works.
-
Is the first character of that string a letter followed by digits? – Ed Heal Oct 17 '13 at 21:52
@EdHeal Yes, always the character `'v'` followed by digits (assuming the role of char). Examples: `v1`, `v33556`, `v23` ...etc and I have to store those numbers in a variable of type `int` like: `int i = 33556` or whatever. – Gera Oct 17 '13 at 21:54
there's a function called `strtol()` , and it's used like this :
`````` long dest = 0;
char source[10] = "122";
dest = strtol(source , NULL , 10); // arg 1 : the string to be converted arg2 : allways NULL arg3 : the base (16 for hex , 10 for decimal , 2 for binary ...)
``````
but in your case you should replace this `dest = strtol(source , NULL , 10);` with this `dest = strtol((source + 1) , NULL , 10)` or `dest = strtol(&source[1] , NULL , 10);`to ignore the first character because `strtol` stops at the first non-digit character it encounters
-
It works! thank youuuuuuu :D – Gera Oct 17 '13 at 22:03
just use `strtol`,
``````long int num;
char* end;
num = strtol(&charv1[1], &end, 10);
``````
-
``````sscanf( charv1, "%*c%d", &i); //skip the first char then read an integer
``````
-
You've forgotten to multiply 10 every loop. This works:
`````` size = (strlen(charv1)-1);
dec=10;
for(aux = size; aux > 0; aux--){
if(aux == (size)){
v1 = charv1[aux]-'0';
}
else{
aux2 = (charv1[aux]-'0')*dec;
printf("%d\n", aux2);
v1 = v1 + aux2;
dec*=10;
}
}
``````
-
Then
`````` int x = atoi(&charv1[1]);
printf("Here it is as an integer %d\n", x);
``````
- | 770 | 2,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-30 | latest | en | 0.805814 |
https://www.coursehero.com/file/6025124/hw-17/ | 1,516,549,866,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00151.warc.gz | 870,708,345 | 24,222 | # hw 17 - yang(ey942 – ohw17 – turner –(56705 1 This...
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Unformatted text preview: yang (ey942) – ohw17 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A long straight wire 1 lies along the x-axis. A long straight wire 2 lies along the y-axis so as to pass very near, but not quite touch wire 1 at the origin. If both wires are free to move, what hap- pens when currents are sent simultaneously in the + x direction through wire 1 and in the + y direction through wire 2? Note that “clock- wise around origin” refers to an observer look- ing down on an xy plane in which + x is to the right and + y upward. 1. 1 accelerates in the + y direction, 2 in the + x direction. 2. Both wires rotate clockwise around the origin. 3. 1 rotates counterclockwise, 2 clockwise around the origin. correct 4. Neither wire moves. 5. 1 rotates clockwise, 2 counterclockwise around the origin. 6. 1 accelerates in the − y direction, 2 in the- x direction. 7. Both wires rotate counterclockwise around the origin. 8. Both wires accelerate along the direction of current flow. 9. 1 accelerates in the + y direction, 2 in the- x direction. 10. 1 accelerates in the − y direction, 2 in the + x direction. Explanation: After the currents are turned on, each wire creates a magnetic field which applies a mag- netic force on the other wire. Consider the force on wire 1 first. According to the right- hand rule, the magnetic field from wire 2 goes into the xy plane on the right side while out of the plane on the left side. Consequently, the right half of wire 1 will have a upward force while the left half will have a downward force causing wire 1 to rotate counterclock- wise. Similarly, we find that wire 2 will rotate clockwise. 002 (part 1 of 4) 10.0 points Two wires each carry a current I in the xy plane and are subjected to an external uni- form magnetic field vector B , which is directed along the positive y axis as shown in the figure. R I wire #2 I wire #1 B L L x y What is the magnitude of the force on wire #1 due to the external vector B field? 1. bardbl vector B bardbl = I 2 (2 L + 2 R ) B 2. bardbl vector B bardbl = 2 I R B 3. bardbl vector B bardbl = (2 L + R ) I B 4. bardbl vector B bardbl = 2 I L B 5. bardbl vector B bardbl = parenleftbigg L + R 2 parenrightbigg I B 6. bardbl vector B bardbl = 2 I R L B 7. bardbl vector B bardbl = I ( L + R ) B 8. bardbl vector B bardbl = 4 I ( L + R ) B 9. bardbl vector B bardbl = (2 B L + 2 R ) I 10. bardbl vector B bardbl = 2 I ( L + R ) B correct Explanation: yang (ey942) – ohw17 – turner – (56705) 2 vector F = I integraldisplay dvectors × vector B = I vector ℓ × vector B . For wire 1, vector ℓ ⊥ vector B , so the magnitude of vector F is just F = I ℓ B . For wire #1, ℓ = 2 L + 2 R , sp F = 2 I ( R + L ) B ....
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Ask a homework question - tutors are online | 978 | 3,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-05 | latest | en | 0.870255 |
https://www.maths.usyd.edu.au/ub/sums/puzzlehunt/2013/solutions/A3S2_Timing_Diagram | 1,620,695,430,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00121.warc.gz | 865,368,553 | 2,034 | Home Register Teams Puzzles Solve Rules FAQ Credits Archives
### Solutions for Act III Scene 2 - Timing Diagram
The answer is: jubeat What a bizarre diagram! The title indicates that this puzzle should be interpreted as a timing diagram, meaning that the lines represent things turning on and off. But what things? The key step is to realise that each number represents a cell in a 4x4 square grid, with the first digit being the row number and the second digit being the column number. Drawing up the grid and shading in the "on" cells for the first step of time produces something which could be interpreted as the letter G. This is a good sign, and continuing the process produces a message: The message reads GROUPS SIZE FOUR LETTERS XOR. Here XOR refers to the "exclusive or" operation, where the output is 0 if there are an even number of 1s in the input, 1 otherwise. Noting that there are 24 letters in the message, we can take the letters 4 at a time and expect six outputs. The outputs spell JUBEAT, an arcade game played using a 4x4 grid of panels requiring precision timing (in a rhythmic sense)! Design notes: The numbers down the left hand side of the puzzle were originally the numbers from 1 to 16, but testing showed that this was too unhelpful. Also, this would have meant that the 4x4 grid would be completely unclued. A change was made, as well as a change to the hidden message (the original message was "GROUP INTO FOURS THEN USE XOR", which was ambiguous about what to group). In hindsight this was a good decision. | 349 | 1,540 | {"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-2021-21 | latest | en | 0.949731 |
https://github.com/levand/prolin | 1,490,421,562,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188824.36/warc/CC-MAIN-20170322212948-00434-ip-10-233-31-227.ec2.internal.warc.gz | 804,589,911 | 15,492 | # levand/prolin
Linear Programming library for Clojure
Clojure
Latest commit 1483d65 Sep 4, 2013 bump version
# Prolin
Prolin is a linear programming library for Clojure. It provides idiomatic Clojure APIs to formulate and solve linear programming problems, and a small set of utilties for some common transformations of LP objective functions and constraints.
It uses the two-phase Simplex algorithm provided by Apache Commons Math as its internal LP solver.
There are several value propositions to using Prolin over Commons Math directly:
• Idiomatic Clojure API
• Allows the use of arbitrary values (anything with equality semantics) to identify variables
• Modular, protocol-based design to allow for easy extension to alternative solver implementations or alternative representations of constraints and polynomials.
## Usage
The API is centered around a few key protocols.
#### Polynomials
An instance of `prolin.protocols.LinearPolynomial` represents a linear polynomial. LinearPolynomials are a key building block of linear programming.
```(defprotocol LinearPolynomial
"Representation of a linear polynomial (a polynomial of degree
one). A linear polynomial consists of:
- Any number of variables, each with a numerical coefficient
- A constant numerical term
The keys representing variables can be any type that supports good
equality semantics."
(variables [this] "Return a map of variable identifiers to coefficients.")
(constant [this] "Returns the constant term of the polynomial"))```
A `prolin.protocols/linear-polynomial` function is provided to construct a LinearPolynomial from a constant number and a variables map.
An implementation of `LinearPolynomial` for `java.lang.String` is provided, allowing strings such as `"x + y - 4"` or `"3x + 4y - 2z"` to be used anywhere you want a polynomial. Note that the parser is not sophisticated and is provided mostly for experimentation and testing; it will only work for basic equations or inequalities (not, for example, equations with parenthesis, multiple constant terms, etc.)
The `prolin.polynomial` namespace contains utility functions for:
• Subtracting polynomials
• Multiplying polynomials by a scalar
• Constructing a 'zero' polynomial with the given variables
• Instantiating a polynomial by plugging in numbers for each of its variables
#### Constraints
Linear constraints are linear equalities or inequalities that are used to restrict the 'feasible region' of a linear programming problem.
Constraints are represented as the `prolin.protocols.Constraint` protocol, to allow callers to define implementations that are the best fit for a particular problem.
```(defprotocol Constraint
"Representation of a linear constraint as an (in)equality"
(relation [this] "Return one of '>=, '<=, or '=")
(polynomial [this]
"Returns a LinearPolynomial representing the variables,
coefficients and constant term of the (in)equality, when it is put
in one of the forms:
a[0]x[0] + ... + a[n]x[n] + c = 0
a[0]x[0] + ... + a[n]x[n] + c <= 0
a[0]x[0] + ... + a[n]x[n] + c >= 0
Any linear (in)equality can be algebraically manipulated to this
form without any loss of generality, and it is this form that is
used to represent all linear constraints internally. See
'prolin.polynomial/subtract' for a function to help transform
arbitrary (in)equalities to this format."))```
A `prolin.protocols/constraint` constructor is also provided, to construct a constraint directly from a polynomial and its relation to 0.
Additionally, `Constraint` is extended to `java.lang.String` to allow Strings such as `"x = y"`, `"3x + y => 4"` to be used anywhere you want a Constraint. Again, note that the parsing of such strings is naive and intended only for experimentation and testing.
#### Solving
A solution algorithm is provided by an instance of the `prolin.protocols.Solver` protocol.
Only one implementation of `Solver` is currently provided; one based on the Apache Commons Math `SimplexSolver`. You can obtain an instance of this solver by calling `prolin.commons-math/solver`. `solver` optionally takes an options map containing the following keys:
``````:epsilon - Amount of error to accept for algorithm convergence. (double value)
:max-ulps - Amount of error to accept in floating point comparisons. (int value)
:cutoff - Values smaller than the cutOff are treated as zero." (double value)
``````
If you wish to implement `Solver` for an alternative linear programming solver or algorithm, see the protocol definition in `prolin.protocols`.
Once you have an instance of `Solver`, you can invoke the `prolin/optimize` function, which takes a solver, an objective function (as a `LinearPolynomial`), a collection of `Constraints`, and a boolean (true to minimize the objective, false to maximize it.)
The `prolin/maximize` and `prolin/minimize` functions have the same signature, but eliminating the final boolean flag.
### Examples
```(require '[prolin :as p])
(require '[prolin.protocols :as pp])
(require '[prolin.commons-math :as cm])
;; Maximize x
(p/optimize (cm/solver) "x" #{"x <= 5", "x >= -2"} false)
;; => {"x" 5.0}
;; Same as above
(p/maximize (cm/solver) "x" #{"x <= 5", "x >= -2"})
;; => {"x" 5.0}
;; Now minimizing
(p/minimize (cm/solver) "x" #{"x <= 5", "x >= -2"})
;; => {"x" -2.0}
;; Using more than one variable
(p/maximize (cm/solver) "x" #{"2x = y", "y <= 5" })
;; => {"x" 5.0, "y" 2.5}
;; Same as above, but constructing objective & constraints directly,
;; instead of using the String implementations
(p/maximize (cm/solver)
(pp/linear-polynomial 0 {:x 1})
#{(pp/constraint '= (pp/linear-polynomial 0 {:x 2 :y -1}))
(pp/constraint '<= (pp/linear-polynomial -5 {:y 1}))})
;; => {:x 5.0, :y 2.5}
;; Throws an ex-info with a :reason of :no-solution if it can't be solved
(p/maximize (cm/solver) "x" #{"x = 3", "x = 4" })
;; => Exception!
;; Throws an ex-info with a :reason of :unbounded if the solution
;; is unconstrainted
(p/maximize (cm/solver) "x" #{"x >= 1"})
;; => Exception!
```
## Development
To run the `clojure.test` tests, run `lein test` from the project directory.
To run the `clojure.test.generative` tests, run `lein with-profile test generative`. | 1,538 | 6,182 | {"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} | 2.53125 | 3 | CC-MAIN-2017-13 | latest | en | 0.831413 |
https://chromotopy.org/blog/convergence-quickie | 1,553,431,725,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203448.17/warc/CC-MAIN-20190324124545-20190324150545-00177.warc.gz | 452,821,253 | 3,306 | A Quickie on Convergence
The item on today’s agenda is the simple but surprising fact: almost everywhere convergence cannot be topologized! At least, it cannot be topologized in general (and not in many of the nice spaces in which an analyst or probabilist wishes to work).This was rather surprising to me when I first encountered the result (in Durrett’s wonderful probability book, which I am following for this exposition), since the other common notions of convergence in analysis (pointwise, uniform, in measure, strong/weak, etc) can often not only be topologized but in fact metrized or even yielded from a norm. And after all, point-set topology is so ugly that it seems like it should be able to handle almost everywhere convergence. So what goes wrong?
The basic tool we will use to approach this problem is the infamous Borel-Cantelli lemma, which states that if we have a measure $\mu$ and measurable sets $A_n$ then $% $ implies $\mu(\lim_{m \to \infty} \cup_{n=m}^\infty A_n) = 0$ (this last part can be naturally interpreted as the statement that the set of infinitely recurrent points in the sequence $A_n$ has measure 0). This is a fairly standard result, and perhaps the first application one sees is that the Borel-Cantelli lemma yields the following: Given a sequence $f_n$ of functions that converge in measure, one can extract a subsequence $f_{n_i}$ that converges almost everywhere.
The method for establishing this result is fairly typical of such arguments: we rely on diagonalization along with the control that Borel-Cantelli gives us. Let $f_n$ be a sequence that converges in measure to $f$. This means that for any $n$ we have a $f_{m_n}$ with $% 1/n) < 2^{-n} %]]>$ . Applying Borel-Cantelli to the sequence of sets $A_n = \{x : \mid f_{m_n}(x) - f(x)\mid > 1/n\}$ yields $\mu(\limsup_m \cup_{n=m}^\infty A_n) = 0$. But this is simply saying that the set of points on which $f_{m_n}$ doesn’t converge to $f$ has measure 0.
The converse is also true: if $f_n$ is a sequence of measurable functions such that every subsequence has a further subsequence that converges almost everywhere, then $f_n$ converge in measure. This follows from a more general result, that if a sequence of elements $y_n$ in a topological space has the property that every subsequence has a convergent subsequence, then $y_n$ is convergent (of course the limit is not unique, but the point is that if it didn’t converge, we could extract a bad subsequence, contrary to our assumption). This general result applies to our specific case because convergence in measure is not only topologized but in fact on finite measure spaces it can be induced by a pseudometric (on sigma-finite spaces you should get a topology induced by a family of pseudometrics, I think). I omit the proof of this as it would distract from the discussion.
So now what have we done? Well, we’ve asserted that in a topological space, a sequence converges iff every subsequence has a further convergent subsequence. So what of almost everywhere convergence? There are many examples of sequences that converge in measure but not almost surely, e.g. the family $f_{2^k+l} = 1_{[l/2^k, (l+1)/2^k)}$ of indicator functions. Every subsequence of this also converges in measure, and by our above remarks has a further subsequence that converges almost everywhere. But by assumption $f_n$ does not converge a.e., so it follows that a.e. convergence cannot be convergence in any topology.
I’d like to point out briefly that this is not something that can simply be remedied by considering functions under almost everywhere equivalence – such a modification would simply reduce the number of possible limit points of a sequence, rather than allow for any new ones. No, sadly this awkwardness of a.e. convergence is just a manifestation of what I imagine must be pretty core set-theoretic issues. For more things along this line, check out this MO post. And expect more posts to come in the next few months. I’ve got some ideas bouncing around in my head… spectral theory, ergodic theorems, and harmonic analysis. | 972 | 4,085 | {"found_math": true, "script_math_tex": 22, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-13 | latest | en | 0.931819 |
https://ask.truemaths.com/question/an-observer-1-5-m-tall-is-30-m-away-from-a-chimney-the-angle-of-elevation-of-the-top-of-the-chimney-from-his-eye-is-60-find-the-height-of-the-chimney/ | 1,675,253,957,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00275.warc.gz | 126,913,883 | 29,840 | • 0
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# An observer 1.5 m tall is 30 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 60•. Find the height of the chimney
• 0
An important and exam oriented question from trigonometry
Topic- Height and distance
An observer 1.5 m tall is 30 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 60•. Find the height of the chimney.
Book RS Aggarwal, Class 10, chapter 5C, question no 3
Share | 137 | 473 | {"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} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.812474 |
https://www.onlinemathlearning.com/logarithms.html | 1,719,137,883,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00363.warc.gz | 823,586,119 | 10,630 | # Logarithmic Functions
In these lessons, we will look at how to evaluate simple logarithmic functions and solve for x in logarithmic functions.
### What Are Logarithmic Functions?
We can think of logarithmic functions as the inverse of exponential functions.
The following diagram shows how logarithm and exponents are related. Scroll down the page for examples and solutions.
### How To Solve For X In Logarithmic Equations?
Equations of the form x = loga y can be solved (for any of the three variables y, a or x) by first writing them in exponent form. We must be careful to check the answer(s) to see whether the logarithm is defined.
Take note of the following:
• Logarithms of a number to the base of the same number is 1, i.e. loga a = 1
• Logarithms of 1 to any base is 0, i.e. loga 1 = 0
• Loga 0 is undefined
• Logarithms of negative numbers are undefined.
• The base of logarithms cannot be negative or 1.
Example:
Calculate the value of each of the following:
a) 1og2 64
b) log9 3
c) log4 1
d) log6 6
e) log8 0.25
f) log3–9
Solution:
a) Let x = log2 64
2x = 64
x = 6
b) Let x = log9 3
9x = 3
x =
c) Let x = log4 1
4x = 1
x = 0
d) Let x = log6 6
6x = 6
x = 1
e) Let x = log8 0.25
8x = 0.25
f) Let x = log3– 9
3x = – 9
Since it is not possible for 3x to be negative, log 3–9 is undefined.
Example:
Solve logx 4 = 2
Solution:
logx 4 = 2
x2 = 4
x = 2 or –2
Since x is the base, x > 0 and x ≠ 1; so x = –2 is rejected and the only solution is x = 2
Example:
Solve log 3 x = 2
Solution:
log 3 x = 2
32 = x
x = 9
Example:
Solve log x (4x – 3) = 2
Solution:
log x (4x – 3) = 2
x2 = 4x – 3
x2 – 4x + 3 = 0
(x -1)(x – 3) = 0
So, x = 1 or 3
For the logarithm to be defined, the only solution is 3.
How to solve a logarithmic equation using properties of logarithms?
Just as we can use logarithms to access exponents in exponential equations, we can use exponentiation to access the insides of a logarithm. Solving logarithmic equations often involves exponentiating logarithms in order to get rid of the log and access its insides. Sometimes we can use the product rule, the quotient rule, or the power rule of logarithms to help us with solving logarithmic equations.
This video shows how solve a logarithmic equation using properties of logarithms and some other algebra techniques.
Example:
Solve 2log3x - log3(x + 6) = 1
How to solve a Logarithmic Equation with Multiple Logs?
When given a problem on solving a logarithmic equation with multiple logs, students should understand how to condense logarithms. By condensing the logarithms, we can create an equation with only one log, and can use methods of exponentiation for solving a logarithmic equation with multiple logs. This requires knowledge of the product, quotient and power rules of logarithms.
Example:
Solve log7(x + 4) - log7(x - 4) = log7(5)
#### Techniques For Solving Logarithmic Equations
Examples:
a) log3(x + 1) = 2
b) log5(3x - 8) = 2
c) log(x + 2) + log(x -1) = 1
d) log x4 - log 3 = log(3x2)
More examples on solving logarithmic equations
Examples:
log2x + log2(x - 3) = 2
log(5x - 1) = 2 + log(x - 2)
ln x = 1/2 ln(2x + 5/2) + 1/2 ln 2
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Describe functional relationships for given problem situations, and write equations or inequalities to answer questions arising from the situations. | 1,183 | 5,142 | {"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} | 4.4375 | 4 | CC-MAIN-2020-05 | latest | en | 0.751187 |
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# M21-35
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Math Expert
Joined: 02 Sep 2009
Posts: 58381
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16 Sep 2014, 01:15
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Difficulty:
55% (hard)
Question Stats:
66% (02:05) correct 34% (02:01) wrong based on 127 sessions
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If $$x$$ is a positive integer, what is the remainder when $$x$$ is divided by 3?
(1) The remainder when $$x$$ is divided by 4 is 1
(2) The remainder when $$x$$ is divided by 7 is 6
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16 Sep 2014, 01:15
2
7
Official Solution:
(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.
(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.
(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.
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### Show Tags
03 Jul 2015, 22:40
2
1
Bunuel wrote:
Official Solution:
(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.
(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.
(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.
Thanks for the explanation Bunuel. I too used the same method.
However, I was wondering if there is any faster method to determine the two numbers (13 and 41).
In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.
Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?
Math Expert
Joined: 02 Aug 2009
Posts: 7959
### Show Tags
03 Jul 2015, 23:03
11
11
samuraijack256 wrote:
Bunuel wrote:
Official Solution:
(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.
(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.
(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.
Thanks for the explanation Bunuel. I too used the same method.
However, I was wondering if there is any faster method to determine the two numbers (13 and 41).
In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.
Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?
Hi,
The faster method in these Qs is to find the first common multiple and, thereafter, add the LCM of two numbers..
In this case when we have got the first number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41..
the next 41+28=69 and so on...
so 13 gives remainder 1
41 gives remainder 2
69 gives remainder 0... and so on
hope it helped..
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### Show Tags
03 Jul 2015, 23:10
2
1
samuraijack256 wrote:
Bunuel wrote:
Official Solution:
(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.
(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.
(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.
Thanks for the explanation Bunuel. I too used the same method.
However, I was wondering if there is any faster method to determine the two numbers (13 and 41).
In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.
Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?
another way would be to equate two eq from two statements..
$$x=4q+1$$, $$x=7p+6$$...
$$4q+1=7p+6$$....
$$4q=7p+5$$....
now substitute p as 1,3,5... as p has to be odd to make the eq even ..
where we getan integer value of q... those are the numbers we rae looking for..
for ex
p=1,q=3... so number=4q+1=13..
p=3, q is not an int..
p=5, q=10 number =4q+1=41.. and so on
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Joined: 11 Nov 2014
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### Show Tags
03 Jul 2015, 23:22
1
chetan2u wrote:
samuraijack256 wrote:
Bunuel wrote:
Official Solution:
(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.
(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.
(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.
Thanks for the explanation Bunuel. I too used the same method.
However, I was wondering if there is any faster method to determine the two numbers (13 and 41).
In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.
Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?
Hi,
the fatser method in these Q is to find the first common multiple and thereafter add the LCM of two numbers..
in this case when we have got the fiest number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41..
the next 41+28=69 and so on...
so 13 gives remainder 1
41 gives remainder 2
69 gives remainder 0... and so on
hope it helped..
This is exactly what i had been looking for. Thanks Chetan!
Earlier, I too had tried looking at this from an LCM perspective, but after a few steps, I got confused and let it go.
I looked through your other proposed method as well (the one involving equating the two expressions), but I would prefer your first suggestion for the ease of application ( identifying if the variable has to be even/odd will again take a few 10s of seconds ).
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### Show Tags
22 Nov 2015, 07:15
am i correct that as long as no limit is set on how far we can go calculating the potential values of n the answer will always be E? e.g. if we were set some limit: say 2 digit numbers or even etc - then we need to be accurate and find the values?
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14 Oct 2016, 11:59
Thinking in retrospect how the time for this question could have been reduced.
Clearly either statement is not sufficient on its own, which can be quickly tested..
Now for together:
x = LCM of (7,4) * (some variable integer K) + some constant remainder, lets say R
=> x = 28*K + R
We can find R by finding the common number as shown in the official solution.
Now remainder of 28K/3 will depend on the value of K, since 28 is not a multiple of 3.
So no matter what the value of R is, when we divide x by 3, the reminder obtained will depend on K hence we do not have a unique solution.
Bunuel - do you think it is valid line of reasoning. If yes, then we would not need to find R and can save some time.
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28 Aug 2017, 09:34
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 26 Feb 2018
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29 Oct 2018, 17:33
Hi Bunuel- Can you please confirm if my logic is right while doing this exercise?
n = 4k + 1 -> Insuf
n = 7k + 6 -> Insuf
n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).
As 28k is not divide by 3, the Reminder will change.
Math Expert
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Posts: 58381
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29 Oct 2018, 22:09
1
1
efr wrote:
Hi Bunuel- Can you please confirm if my logic is right while doing this exercise?
n = 4k + 1 -> Insuf
n = 7k + 6 -> Insuf
n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).
As 28k is not divide by 3, the Reminder will change.
Correct with small but crucial correction: quotient in all three cases (highlighted) will not be the same so you should not sue one variable for them (use p, q, k instead).
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Joined: 26 Feb 2018
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30 Oct 2018, 09:27
Bunuel wrote:
efr wrote:
Hi Bunuel- Can you please confirm if my logic is right while doing this exercise?
n = 4k + 1 -> Insuf
n = 7k + 6 -> Insuf
n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).
As 28k is not divide by 3, the Reminder will change.
Correct with small but crucial correction: quotient in all three cases (highlighted) will not be the same so you should not sue one variable for them (use p, q, k instead).
Perfect! Thanks for the explanation Bunuel!
Re: M21-35 [#permalink] 30 Oct 2018, 09:27
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# M21-35
Moderators: chetan2u, Bunuel | 3,380 | 10,539 | {"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} | 4.34375 | 4 | CC-MAIN-2019-43 | latest | en | 0.829971 |
https://www.numbersaplenty.com/5021 | 1,718,252,534,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00126.warc.gz | 850,403,345 | 3,013 | Search a number
5021 is a prime number
BaseRepresentation
bin1001110011101
320212222
41032131
5130041
635125
720432
oct11635
96788
105021
113855
122aa5
132393
141b89
15174b
hex139d
5021 has 2 divisors, whose sum is σ = 5022. Its totient is φ = 5020.
The previous prime is 5011. The next prime is 5023. The reversal of 5021 is 1205.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4900 + 121 = 70^2 + 11^2 .
It is a cyclic number.
It is not a de Polignac number, because 5021 - 26 = 4957 is a prime.
Together with 5023, it forms a pair of twin primes.
It is a Chen prime.
It is a plaindrome in base 9 and base 16.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5023) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2510 + 2511.
It is an arithmetic number, because the mean of its divisors is an integer number (2511).
25021 is an apocalyptic number.
It is an amenable number.
5021 is a deficient number, since it is larger than the sum of its proper divisors (1).
5021 is an equidigital number, since it uses as much as digits as its factorization.
5021 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 10, while the sum is 8.
The square root of 5021 is about 70.8590149522. The cubic root of 5021 is about 17.1236656925.
Adding to 5021 its reverse (1205), we get a palindrome (6226).
The spelling of 5021 in words is "five thousand, twenty-one". | 479 | 1,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | latest | en | 0.903924 |
https://nrich.maths.org/public/leg.php?code=57&cl=2&cldcmpid=2430 | 1,508,588,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824775.99/warc/CC-MAIN-20171021114851-20171021134851-00553.warc.gz | 796,933,067 | 9,917 | Search by Topic
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Stage: 3 Challenge Level:
Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large...
Paving Paths
Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
Play a Merry Tune
Stage: 2 Challenge Level:
Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
Sets of Four Numbers
Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
Seven Squares
Stage: 3 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
Times Tables Shifts
Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
Odds, Evens and More Evens
Stage: 3 Challenge Level:
Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions...
Farey Sequences
Stage: 3 Challenge Level:
There are lots of ideas to explore in these sequences of ordered fractions.
Clock Squares
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Square numbers can be represented on the seven-clock (representing these numbers modulo 7). This works like the days of the week.
Differs
Stage: 3 Challenge Level:
Choose any 4 whole numbers and take the difference between consecutive numbers, ending with the difference between the first and the last numbers. What happens when you repeat this process over and. . . .
Converging Means
Stage: 3 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
Millennium Man
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Liitle Millennium Man was born on Saturday 1st January 2000 and he will retire on the first Saturday 1st January that occurs after his 60th birthday. How old will he be when he retires?
Lower Bound
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What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
Intersecting Circles
Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
Pebbles
Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
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What's the greatest number of sides a polygon on a dotty grid could have?
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Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
LOGO Challenge - Sequences and Pentagrams
Stage: 3, 4 and 5 Challenge Level:
Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?
1 Step 2 Step
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Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
What an Odd Fact(or)
Stage: 3 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5? | 2,095 | 8,862 | {"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} | 3.953125 | 4 | CC-MAIN-2017-43 | latest | en | 0.921393 |
http://clay6.com/qa/34508/for-the-curve-sqrt-x-sqrt-y-1-large-frac-at-bigg-large-frac-frac-bigg-is-?show=34509 | 1,558,428,700,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00519.warc.gz | 44,604,557 | 10,777 | Comment
Share
Q)
# For the curve $\sqrt x + \sqrt y=1,\large \frac{dy}{dx}$ at $\bigg(\large\frac{1}{4},\frac{1}{4}\bigg)$ is ____________.
$\begin{array}{1 1}(a)\;1\\(b)\;0\\(c)\;2\\(d)\;3\end{array}$
• Chain Rule :If $z=f(y)$ and $y=g(x)$,then $\large\frac{dz}{dx}=\frac{dz}{dy}-\frac{dy}{dx}$
$\sqrt x+\sqrt y=1$
Differentiating on both sides w.r.t $x$ we get,
$\large\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}=$$0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{\sqrt y}{\sqrt x}$
$\large\frac{dy}{dx}$ at $\big(\large\frac{1}{4},\frac{1}{4}\big)$ is written as
$\large\frac{dy}{dx}=-\frac{\sqrt{\large\frac{1}{4}}}{\sqrt{\large\frac{1}{4}}}$
$\qquad=-1$ | 292 | 668 | {"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} | 4.0625 | 4 | CC-MAIN-2019-22 | longest | en | 0.37377 |
http://mathcentral.uregina.ca/QandQ/topics/5%20foot%20strip | 1,542,549,396,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00426.warc.gz | 215,166,313 | 4,446 | Math Central - mathcentral.uregina.ca
Quandaries & Queries
Q & Q
Topic: 5 foot strip
start over
One item is filed under this topic.
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A 5 foot strip around a building 2009-08-20 From Robert:I am having a problem at work that I am hoping someone can help me figure out. I have a pest control company and we are pretreating the soil for a new hospital being built. The company that we are working has asked to also treat a 5 foot strip around the entire building and for us to get paid we need to figure out what the square footage of this 5 foot strip around the building that we are treating. The structure being built is a hospital and as you can imagine the building has all kinds of nooks and offsets and even some of the wall are cruved more like a circle. I already know the square footage of the building itself because they gave us this information. The square footage of the building is 120,000 SqFt. I would greatly appreciate if someone can help me figure out the square footage of a 5 foot strip around a 120,000 SqFt building who's walls have a lot of insets and odd shapes. I don't if someone can figure out the square footage of the 5 foot strip with the information I gave. I would greatly appreciate any help that you can give me. I need this information for work because we charge for pretreatment of soil by the square footage of the area to be treated and I need this information so that I don't over charge the company we are working for or so I don't cheat myself either. Any and all help we be greatly appreciated because I don't know who else that can help me.Answered by Victoria West.
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Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
about math central :: site map :: links :: notre site français | 406 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-47 | longest | en | 0.966978 |
https://www.gradesaver.com/textbooks/engineering/mechanical-engineering/engineering-mechanics-statics-and-dynamics-14th-edition/chapter-13-kinetics-of-a-particle-force-and-acceleration-section-13-4-equations-of-motion-rectangular-coordinates-problems-page-131/17 | 1,713,620,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00684.warc.gz | 717,958,051 | 13,093 | ## Engineering Mechanics: Statics & Dynamics (14th Edition)
$a=\frac{1}{2}(1-\mu_k)g$
We can determine the required acceleration as follows: $\Sigma F_y=ma_y$ $\implies T-mg=-mg$ $\implies T=m(g-a)~~$ eq(1) Now we apply Newton's second law $\Sigma F_y=ma_y$ $\implies N-mg=0$ $\implies N=mg$ and $\Sigma F_x=ma_x$ $\implies T-\mu_k N=ma$ $\implies T-\mu_k mg=ma~~$ eq(2) We plug in the value of $T$ from eq(1) into eq(2) to obtain: $m(g-a)-\mu_k mg=ma$ This simplifies to: $a=\frac{1}{2}(1-\mu_k)g$ | 191 | 499 | {"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} | 3.421875 | 3 | CC-MAIN-2024-18 | latest | en | 0.561782 |
http://hank.uoregon.edu/wiki/index.php?title=Optical_Tweezers&oldid=2790 | 1,516,333,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00469.warc.gz | 167,534,459 | 8,821 | # Optical Tweezers
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Optical Tweezers
### Background
Scattering Basics - Incident Plane Wave (from Michigan Tech)
Plane wave scattering theory may be used to illustrate gross optical trapping behavior via momentum transfer between the light field and the particle (total momentum is conserved). However, the optical waves employed to trap a particle in a typical optical-tweezers setup are most definitely not planar. For a full description of the scattering process see "Theory of trapping forces in optical tweezers", A. Mazolli, P. A. Maia Neto and H. M. Nussenzveig, Proc. R. Soc. Lond., A 8 December 2003 vol. 459 no. 2040 3021-3041.
When the size parameter $\beta=\frac{2\pi a n}{\lambda_{0}}\gg 1$, where $a$ is the particle radius, $n$ is the index of the medium surrounding the particle and $\lambda_{0}$ is the vacuum wavelength of the trapping light, the results from a geometrical optics treatment holds.
For a geometrical optics treatment see, "Optical tweezers for undergraduates: Theoretical analysis and experiments", M. S. Rocha, Am. J. Phys. 77, 704 (2009).
Example: For a trapping laser of wavelength 635nm and a 2 micron particle in a surrounding medium of water, $\beta\approx14$ so the geometrical optics treatment holds and Mie scattering is the dominant trapping process.
More simply, there are two forces in action on a trapped particle, one of which will be more dominant depending on the size of the particle, the wavelength, etc. One of these forces is the transference of momentum by photons, the other force is the attractive force of the electric field gradient. The first acts somewhat like a strong current of water, the second draws the particle up to the highest point of the gradient, the center of the laser.
### Resources
• Summer 2014 Powerpoint[1]
• [2] Directions on how to use a QPD in an optical tweezer setup.
• [3] Here are some slide prepping instructions from Berkeley.
### Our own setup
• Objective lens alignment & care
• To align the objective lens so that it is mounted parallel to the optical bench surface, use a level on the objective mount.
• Make an oil guard to keep immersion oil off the main chassis of the objective by cutting a small hole in the center of a square of Parafilm and pushing the objective through.
• To remove oil from the objective lens, carefully use a lens tissue around the end of the lens to draw oil off the lens using capillary action without actually touching the delicate output lens.
• Slide Setup
• Microscope Slide Mount
• Stokes' Setup
• Spring 2013 Method (ramped) [4]
• Summer 2014 Method (sinusoidal) [5]
• Stoke's Force Calibration Video (sinusoidal)[6]
### Using NI Vision Assistant
• [7] Image Acquisition/Saving Images. The UI for this program is rather difficult. To exit from video capture mode after capturing images there is a button that says close near the bottom left, underneath the camera options window.
• [8] How to track the microspheres in NI Vision Assistant using pattern matching.
• To get an actual video, you need to run the pictures through another video editing program. There is one on the computer, but there is no video option in NI Vision Assistant itself.
• We actually tried both using pattern matching and brute force point and click methods. Neither worked very well, so we recommend using the QPD to get position measurements. You don't need an actual video for this.
• The pattern matching does work quite well for the purposes of tracking brownian motion. You need to create a pattern matching script and then run it through batch processing.
• During batch processing, pay attention to the red and green tracking squares as the images are being processed. If these squares do not appear, the program is likely not tracking the object position accurately.
• ### Calibration
• Calibrating the laser so that it points directly at the slide is very important. There are two steps.
• First remove the slide so that you have only the objective, then position a pair of pinholes straight above the objective, and adjust the objective and mirrors until the laser can shoot through both pinholes. That's the rough calibration.
• Next, put the slide on and and adjust the focus until you can see the interference patterns the laser makes on the slide. Then adjust the mirrors until the interference patterns are are round. Then the laser should be pointed straight at the slide.
### Our calculations using Brownian Motion
• 2.56 micrometer spheres
• Spring 2014- 4.6 mW Beam [9]
• Summer 2014- 637nm and 980nm lasers [10]
### Calculating Trap Forces Using Stokes' Drag Force
• [11] iPython Calculations
• Beam Power (mW) Escape Velocity (microns/second) Trap Force (pN)
5.5 20.57 0.44
8.5 33.49 0.72
11.7 40.00 0.86
15.3 62.60 1.34
19.0 84.71 1.82
23.0 110.77 2.38
• A simple calculation would say that I need a 10 billion watt laser to achieve a 1 Newton trapping force. We should totally do that. The spheres would be so incredibly trapped.
### Performing Biological Measurements
• Installing Infared Laser
• Preparing bacteria to Analyze
• Current Progress and Issues
### Vortex Wave Retarder
• Summer 2016 Powerpoint[13] | 1,239 | 5,241 | {"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": 5, "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-2018-05 | longest | en | 0.814241 |
https://math.meta.stackexchange.com/questions/12937/should-rings-be-considered-non-commutative-if-the-question-doesnt-imply-they-ar | 1,632,493,958,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00536.warc.gz | 433,251,856 | 41,177 | # Should rings be considered non commutative if the question doesn't imply they are commutative?
A debate in comments was raised by the question
Showing that the only idempotents in $R$ are zero and one
the relevant part of which is
Let $R$ be a ring with $1$ and suppose $R$ has no zero divisors. Show that the only idempotents in $R$ are $0$ and $1$.
One of the answers went straight by assuming the ring $R$ is commutative, which, in my opinion, is too strong an assumption about the question at hand.
The aim of this metaquestion is thus establishing a general policy. It's true that conventions about naming rings are various and contradictory around the world, and I find the answer perfectly acceptable (albeit using a sledgehammer, in the particular case), provided “If the ring is commutative…” is added at the beginning.
To me, for example, rings are not commutative unless stated and I can't get from the question any hint about the OP considering this hypothesis.
• You should probably also add: should rings be considered associative? Should they be rings with $1$ and homs preserving $1$? In any case use determines meaning. And in my experience, in a general level forum like this, ring almost always means commutative ring with 1 and homs preserving 1. Otherwise many answers would be wrong. Mar 4 '14 at 15:47
• @BillDubuque Non associative rings are much more specialized and probably the poster would mention it. I consider your comment not at all constructive. Indeed I avoided mentioning the presence of $1$ (which was stated in the question). Mar 4 '14 at 15:51
• @Bill: I disagree. Care to give an example of an answer that would be wrong, unless you tacitly assume commutativity? IIRC in the ring related questions I've been a party of (admittedly not too many), this has hardly ever been an issue. Either the ring has been implicitly specified (when all and sundry will immediately know whether it is commutative or not), or the OP stated it in the question. Wisely so IMHO. Mar 4 '14 at 16:04
• The tag-wiki for the tag (rings) says, that the tag (rngs) should be used for questions which do not have unity. It also says that rings are not necessarily commutative. (I do not claim that the tag-wikis should be taken as guidance in general - they can of course be changed - but I thought this is worth mentioning.) Mar 4 '14 at 16:04
• @Jyrki Do you also disagree about associativity too? Mar 4 '14 at 16:11
• This probably depends on culture - for example, I would not assume that "ring" means commutative, but I would assume it was associative (I didn't even realize this was up for discussion!). I would probably assume it had $1$ by accident, although I don't think this should be part of the definition. If I was on the ball I would leave a comment to the OP to clarify, but it would be easy to not realize I was adding an assumption (particularly in the case of associativity, where I have never seen a definition omitting it!)
– mdp
Mar 4 '14 at 16:12
• @Bill: No. Amending "my creed" accordingly :-) Mar 4 '14 at 16:12
• @JyrkiLahtonen Do you disagree that by far and away the majority of rings discussed in MSE and sci.math are commutative? Mar 4 '14 at 16:13
• @Bill: Yes. A majority, but not a vast majority. Most of the questions about rings are about subrings of complex numbers or of subquotient of polynomial rings over a field or residue class rings of integers. But there are many questions also about group rings and matrix rings. The context makes it clear in a vast majority of cases. The questions, where there is no strong contextual inference are IMHO in a small minority. Therefore there is no default assumption about commutativity, and in those cases it should be specified, when/if needed. Mar 4 '14 at 16:18
• I'm glad to see that we agree about that. In any case, we cannot force conventions on questioners. They will use whatever they are familiar with. The best we can do is to encourage questioners to be more explicit when there are possible ambiguities due too differences in conventions (and lack of adequate context to unquestionably disambiguate) Mar 4 '14 at 16:31
• And to elaborate on Bill's comment, in some cases, questioners sometimes don't even know about the other conventions, so we can't reasonably expect them to post unambiguous questions. I make it a point to mention that there are two usual conventions when prompting for clarification.
– user14972
Mar 4 '14 at 16:57
• @egreg OK: I think that resolves my comment: thanks Mar 4 '14 at 19:04
• Why not query the OP with a comment if their is a question? Or if you write an answer, clearly indicate what assumptions on the ring one is making. Mar 5 '14 at 20:15
• I see somebody voted to close this as primarily opinion-based. LOL. What do you expect with a question tagged with discussion and policy? Mar 7 '14 at 11:20
• @BillDubuque Obviously rings are associative. Any married person can tell you that. ba dum... tsssss Mar 7 '14 at 17:19
Posting this as an answer to test how widespread my beliefs/definitions are :-)
My creed (largely those of a loyal disciple of Jacobson's BA I-II):
1. A ring is not necessarily commutative.
2. The multiplication of a ring is associative.
3. A ring has a multiplicative neutral element (often denoted by $1$). The lesser structures are rngs.
4. A homomorphism of rings maps the neutral element to the neutral element.
5. But OTOH an integral domain (and a field) is commutative by default.
• I'm currently TA for an algebra course in which 3 and 4 are dropped, and being convinced that this is a good definition (in particular, it means that all ideals are themselves rings). But I'm not fanatical about that. I stand firmly behind 1 and 2.
– mdp
Mar 4 '14 at 16:17
• What conventions prove most convenient for an abstract algebra textbook need not necessarily bear any relation to the conventions that prove most convenient for a general-level math forum. Mar 4 '14 at 16:20
• @MattPressland I find that removing $1$ just shows an artificial similarity between rings and groups, where subring corresponds to subgroup and ideal to normal subgroup. When you start requiring that ring morphisms take $1$ into $1$, this similarity is lost. Mar 4 '14 at 16:28
• @egreg But we don't require this! (For one thing, neither the domain or the codomain need have a $1$!).
– mdp
Mar 4 '14 at 16:29
• @MattPressland Yes, I understand it. Try doing sensible module theory on rings without $1$. ;-) Mar 4 '14 at 16:30
• @egreg No thanks! (It doesn't come up in the course). I understand there are pros and cons. ;)
– mdp
Mar 4 '14 at 16:30
• @BillDubuque While I agree with your sentiment that definition should be determined by usage, I think it would be harmful for the site to diverge from the language used by the wider mathematical community (for example, it would be confusing to new users). While this doesn't help for whether rings have $1$ or not, where there is no general consensus, I'm not aware of anybody who thinks all rings are commutative all of the time, so I think it is best to continue to use "commutative ring" as a special case of "ring".
– mdp
Mar 4 '14 at 16:33
• @Matt I don't propose to attempt to enforce any convention. Rather, I merely point out my observations about such usage in general level math forums such as sci.math and MSE, from a few decades of experience. Personally, in such forums, I adopt conventions that conform to usage, since that proves most convenient. When there is possible ambiguity I usually state my reading, as I did in the question at hand, by saying $R$ is a domain in my answer (domains are commutative by most definitions). Mar 4 '14 at 16:46
• @BillDubuque I am similarly not proposing such a thing. Stating your reading seems like a perfectly good approach (I think I have also done this) although my usual approach is to ask a clarifying question to the OP before answering. Happily, these approaches are not mutually exclusive! My only suggestion that involves actual community action is that, if somebody writes "ring", and it emerges later that they wanted to assume it was commutative, their post should be edited to say "commutative ring" to increase clarity when other people read the post later.
– mdp
Mar 4 '14 at 16:53
• @BillDubuque You have surely noted that the question doesn't say domain, but no zero divisors, which, IMO, is a clear indicator that commutativity is not assumed. Mar 4 '14 at 16:58
• @Matt That seems like a fine approach to me. Usually the context is enough to disambiguate. In this case it was not (I don't think that the mention of idempotents need imply that the ring is noncommutative since idempotents also play a key role in commutative rings e.g. they are intimately connected to the Chinese Remainder Theorem, factorizations into coprimes, etc). In any case, I think we should allows answerers to resolve ambiguity however they find useful to do so (even if they did not correctly infer the author's intent, it often leads to enlightening answers). Mar 4 '14 at 17:00
• @egreg The only thing that is "clear" to me is that you and I happen disagree on some subjective matters about conventions - which has little relevance to the general question posed in this meta thread. In any case, I think that arguing about conventions is one of the most fruitless arguments that one can engage in. Mar 4 '14 at 17:06
If the question is about algebraic geometry or commutative algebra (duh!), you may assume that rings are commutative.
Else you may not.
• And in algebraic geometry, you can even assume they are unital (unless specified otherwise at least). Mar 6 '14 at 8:07
• And also homs preserve $1$. Jul 5 '16 at 21:01
I didn't know that a ring was ever assumed to be commutative. I know that there isn't a clear convention about a ring having an identity. Anyway, I think when we come across questions where the definition is unclear, then either the OP should clarify or an answerer should make clear what definition s(he) is using (and maybe even provide an answer/comment to each case). I guess it all comes down to what you are doing.
As Bill Dubuque says in his comment to the other answer: What conventions prove most convenient for an abstract algebra textbook need not necessarily bear any relation to the conventions that prove most convenient for a general-level math forum. In some cases all rings are assumed to be commutative because that is all you want to consider (See for example: these notes). But clearly in other cases they are not (who would not call the set of $n\times n$ matrices a ring?)
That said I looked up the basic definition of ring various places as to whether or not a ring is in general assumed to be commutative
• Wikipedia says no.
• Hungerford's Algebra says no.
• Rotman's An Introduction to Homological Algebra says no.
• Rotman's Advanced Modern Algebra says no.
• Lang's Algebra says no.
• Dummit and Foote's Abstract Algebra says no.
• Artin's Algebra states that all ring are assumed to be commutative (but states that in general they aren't assumed to be).
• Jacobson's Basic Algebra I says no.
• James Milnes' notes Fields and Galois Theory says no.
• James Milnes' notes Algebraic Number Theory states that all rings are assumed to be commutative.
• Joseph Gallian's Contemporary Abstract Algebra says no.
• I.N. Herstein's Abstract algebra says no.
• J.B. Fraleigh's A first course in abstract algebra says no.
I invite other people to insert references into this list.
• Right: since this list contains many undergraduate mainstays which do not assume commutativity, it is evidence that the "most beginners are only thinking about commutative rings" theory is pretty dubious. Naturally classes focusing on commutative rings alone may do so, but that group does not seem to be representative of the whole population. So for best preparation and least confusion, it is good for answerers to be mindful of this. Mar 5 '14 at 13:53
• Nobody claimed that "most beginners are only thinking about commutative rings". Mar 5 '14 at 16:11
• Dear @BillDubuque : Yes, technically nobody has claimed that here, but somebody has mentioned that they feel now and have felt "for decades" that they are justified in assuming beginners' questions on "forums like this" should be answered as if they had meant "commutative" by default. Presumably that person wants to help the larger half of beginners, so they wouldn't take the inconsistent stance of disagreeing with my first comment. Mar 7 '14 at 17:04
• @rschwieb My remarks mean this: when there is inadequate context to infer if the ring is commutative, I often choose the commutative case, since commutative rings are far more common than noncommutative rings in general level forums like sci.math and MSE. Furthermore, I think it is misguided to attempt to enforce any kind of policy to resolve such ambiguities because not too infrequently beautiful answers result from that extra freedom given to the answerer - allowing them to expound on matters that the OP had no idea to ask about, but are closely related, possibly the essence of the matter. Mar 7 '14 at 17:16
• I like how Milnes has contradictory conventions. Of course, those assumptions are scoped only to those texts, but still... Mar 12 '14 at 4:39
• I think commutative rings are more common when asking about a particular ring. When asking about rings in the abstract without stating commutativity, commutativity is never stated (perhaps there is feedback of OPs saying they meant commutative ring in comments?) Mar 14 '14 at 2:55
I agree with all of Jyrki's conventions in the sense that they are the same as mine.
But, look, they are just conventions. They depend on context and thus should be spelled out. In any course I teach, I surely make sure to explain that my rings contain a multiplicative identity (and if I don't, it soon comes up and someone asks). I don't go into a disquisition about why my convention is the right one and the standard one [although I think it is...]; I just explain.
Whether my rings are commutative or not is not a global convention of mine: in what I am doing, more often than not they are, but plenty often they are not. So I remind people more often which way I mean it to be: e.g. by writing "ring" on the board and saying "commutative ring".
So my answer to the question asked is: it is always better if we don't have to assume. If you as an asker mean your ring to be commutative, I think you should either say so or include the commutative-algebra (or whatever it is) tag. On the other hand, if you are asking about rings which are not obviously noncommutative (e.g. so not if you are asking about ideals in matrix rings), then you should specify whether you would like answers to assume that the ring is commutative, not to assume that, or whether you are open to answers to do both.
In my opinion just writing "Let $R$ be a ring...." and not indicating in any way whether you want it to be commutative or not is not a best practice, because it will confuse some people and may lead to undesired answers. Do we need to press this issue any further than that?
I don't think it makes sense to enforce any convention (not least because there is no enforcement mechanism; also, as the discussion here shows, there is no general agreement on conventions).
Answerers should try to answer questions as best they can; part of this involves interpreting the question as best they can. I think people do this, and one should assume good faith on the part of answerers.
If the OP of the question finds an answer limited in scope (e.g. because it presumes that the rings in question are commutative) they are welcome to clarify their question, ask a new more precise question, leave a comment on the answer, etc.
If another user finds an answer limited in scope and would like to leave a more general answer, they are also free to do so.
More experience/mathematically mature answerers are normally aware of the possibility of different conventions/interpretations, and often mention this. Less experienced users may not be so aware, and so it is not reasonable to expect them to be as attentive to such issues. One way they will gain experience, and awareness, though, is by seeing some of the discussions of such issues in the comments/answers that appear here.
In summary: I don't see that there is any real problem, or anything that needs to be done. | 3,890 | 16,419 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | longest | en | 0.961168 |
https://www.physicsforums.com/threads/game-theory-value-of-a-game.112331/ | 1,511,001,633,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00317.warc.gz | 845,062,648 | 15,405 | # Game theory: value of a game
1. Feb 26, 2006
### mathlete
The problem:
"Player I can choose l or r at the first move in a game G. If he chooses l, a chance move selects L with probability p, or R with probability 1-p. If L is chosen, the game ends with a loss. If R is chosen, a subgame identical in structure to G is played. If player I chooses r, then a chance move selects L with probability q or R with probability 1-q. If L is chosen, the game ends in a win. If R is chosen, a subgame is played that is identical to G except that the outcomes win and loss are interchanged together with the roles of players I and II"
*whew*
Now the question is... if the value of the game is v, show that v=q+(1-q)(1-v)
Now the game tree is so complicated... I really have no idea how to get the value of the game. Is there any easy way to do this that i'm missing?
2. Feb 26, 2006
### Hurkyl
Staff Emeritus
I don't understand the statement of the game.
What happens when player I picks l', and R' gets chosen? Is it now player II's turn? Does "win" always mean a win for player I? et cetera.
If I sat down and tried to teach this game to someone else so we could play, I'd have no idea what the rules are.
Anyways, the analysis should be straightforward.
What is the expected value of the game if player I picks l'?
What is the expected value of the game if player I picks r'?
What is the expected value of the game if player I picks optimally?
Last edited: Feb 26, 2006
3. Feb 27, 2006
### neurocomp2003
are L,R the nodes and l,r are the branches??
4. Oct 11, 2009 | 426 | 1,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-47 | longest | en | 0.940828 |
https://ru.scribd.com/document/357505674/ap-statistics-2017-2018-term-1-assignment-sohee-han | 1,566,041,752,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312128.3/warc/CC-MAIN-20190817102624-20190817124624-00489.warc.gz | 618,490,233 | 62,995 | You are on page 1of 7
# AP Statistics
- Term 1 Assignment -
Ms. Han
shan@daltonschool.kr
HS Room 214
INTEREST PACKET
Students will consider relationships between two quantitative variables, such as the amount of time between eruptions
of Old Faithful and the duration of the previous eruption. As the Old Faithful example illustrates, knowing the
distribution of interval times isnt very helpful if you are trying to predict when the next eruption will occur. Instead,
we look for relationships among variables to help explain patterns and to make predictions. Students will use a
scatterplot to display the relationship between two variables, correlation to measure the strength and direction of a
linear association, and a least-squares regression line to model a linear relationship.
Topic Overview
Scatterplots and Correlation
Least-Squares Regression
Sampling and Surveys
Experiments,
Using Studies Wisely
Essential Questions
Chapter 3: How can we describe the relationship between two quantitative variables?
Chapter 4: How can we design studies: sampling and surveying, performing experiments and how to use
studies wisely?
Reference Materials
The Practice of Statistics (5th edition), by Starnes, Tabor, Yates, and Moore, W. H. Freeman & Co., 2014
Skills List
12. ED. Exploring Data: Describing patterns and departures from patterns
Exploratory analysis of data makes use of graphical and numerical techniques to study patterns and departures from
patterns. Emphasis should be placed on interpreting information from graphical and numerical displays and
summaries.
## Exploring bivariate data
12.ED.14. Analyzing patterns in scatterplots
12.ED.15. Correlation and linearity
12.ED.16. Least-squares regression line
12.ED.17. Residual plots, outliers and influential points
12.ED.18. Transformations to achieve linearity: logarithmic and power transformations
Exploring categorical data
12.ED.19. Frequency tables and bar charts
12.ED.20. Marginal and joint frequencies for two-way tables
12.ED.21. Conditional relative frequencies and association
12.ED.22. Comparing distributions using bar charts
## 12. SE. Sampling and Experimentation: Planning and conducting a study
Data must be collected according to a well-developed plan if valid information on a conjecture is to be obtained. This
plan includes clarifying the question and deciding upon a method of data collection and analysis.
## Overview of methods of data collection
12.SE.1. Census
12.SE.2. Sample survey
12.SE.3. Experiment
12.SE.4. Observational study
## Planning and conducting surveys
12.SE.5. Characteristics of a well-designed and well-conducted survey
12.SE.6. Populations, samples and random selection
12.SE.7. Sources of bias in sampling and surveys
12.SE.8. Sampling methods, including simple random sampling, stratified random sampling and cluster
sampling
Planning and conducting experiments
12.SE.9. Characteristics of a well-designed and well-conducted experiment
12.SE.10. Treatments, control groups, experimental units, random assignments and replication
12.SE.11. Sources of bias and confounding, including placebo effect and blinding
12.SE.12. Completely randomized design
12.SE.13. Randomized block design, including matched pairs design
Generalizability
12.SE.14. Generalizing the results and types of conclusions that can be drawn from observational studies,
experiments and surveys
LESSON OVERVIEW
Learning Objectives
Day Topics Homework
Students will be able to
1 Chapter 1 Quiz
## 2 Course Introduction Picture of what is required for the course and
the AP exam preparation.
3 Chapter 2 Review
4 Chapter 2 Quiz
Chapter 3 Introduction
## Activity: The Case of the Identify explanatory and response variables in
Missing Cookies situations where one variable helps to explain or
influences the other.
5 3.1 Explanatory and Make a scatterplot to display the relationship 1, 5, 7, 11, 13
response variables, between two quantitative variables.
Describe the direction, form, and strength of a
displaying relationships:
relationship displayed in a scatterplot and
scatterplots, describing recognize outliers in a scatterplot.
scatterplots
## 3.1 Measuring linear Interpret the correlation.
association: correlation, Understand the basic properties of correlation,
facts about correlation including how the correlation is influenced by
6 1418, 21
outliers.
Activity: Correlation and Use technology to calculate correlation.
Regression Applet Explain why association does not imply
causation.
3.2 Least-squares Interpret the slope and y intercept of a least-
squares regression line.
regression, interpreting a 2732, 35, 37, 39,
Use the least-squares regression line to predict y
regression line, prediction, for a given x. Explain the dangers of 41, 45
residuals extrapolation.
Calculate and interpret residuals.
3.2 Calculating the
7 equation of the least-
squares regression line,
Explain the concept of least squares.
determining whether a
Determine the equation of a least-squares
linear model is 43, 47, 49, 51
regression line using technology.
appropriate: residual plots Construct and interpret residual plots to assess if
a linear model is appropriate.
Activity: Investigating
Properties of the LSRL
3.2 How well the line fits Interpret the standard deviation of the residuals
the data: the role of s and and r 2 and use these values to assess how well 48, 50, 55, 58
r2 in regression the least-squares regression line models the
relationship between two variables.
3.2 Interpreting computer Determine the equation of a least-squares
8
regression line using computer output.
regression output,
Describe how the slope, y intercept, standard 59, 61, 63, 65, 69,
regression to the mean,
deviation of the residuals, and r 2 are 7178; FRAPPY
correlation and regression
influenced by outliers.
wisdom Find the slope and y intercept of the least-
squares regression line from the means and
standard deviations of x and y and their
correlation.
9 Chapter 3 Review
10 Chapter 3 Quiz
## 4.1 Introduction, The Idea
of a Sample Survey, How Identify the population and sample in a statistical
study.
Identify voluntary response samples and
11 Sample Well: Simple convenience samples. Explain how these 1, 3, 5, 7, 9, 11
Random Sampling sampling methods can lead to bias.
Describe how to obtain a random sample using
Activity: Who Wrote the slips of paper, technology, or a table of random
Federalist Papers? digits.
## 4.1 Other Random
Sampling Methods Distinguish a simple random sample from a 13, 17, 19, 21, 23,
stratified random sample or cluster sample. Give
Sunflowers sampling method.
12
4.1 Inference for
Sampling, Sample Explain how undercoverage, nonresponse,
27, 29, 31, 33, 35
Surveys: What Can Go question wording, and other aspects of a sample
Wrong? survey can lead to bias.
## 4.2 Observational Study Distinguish between an observational study and
an experiment. 3742, 45, 47, 49,
versus Experiment, The
Explain the concept of confounding and how it 51, 53, 55
Language of Experiments limits the ability to make cause-and-effect
conclusions.
Identify the experimental units, explanatory and
13 4.2 How to Experiment response variables, and treatments.
Badly, How to Explain the purpose of comparison, random
assignment, control, and replication in an
Experiment Well, 57, 59, 61, 63, 65
experiment.
Completely Randomized Describe a completely randomized design for an
Designs experiment, including how to randomly assign
treatments using slips of paper, technology, or a
table of random digits.
4.2 Experiments: What Describe the placebo effect and the purpose of 67, 69, 71, 73
14 Can Go Wrong? Inference blinding in an experiment.
for Experiments Interpret the meaning of statistically significant
in the context of an experiment.
Explain the purpose of blocking in an
15 4.2 Blocking experiment. 75, 77, 79, 81, 85
Describe a randomized block design or a
matched pairs design for an experiment.
4.3 Scope of Inference,
83, 8794, 97
16 The Challenges of Describe the scope of inference that is
appropriate in a statistical study. 104; FRAPPY
Establishing Causation
17 Chapter 4 Review
18 Chapter 4 Quiz
20 Term Test
## 21-22 Response Bias Project
23 Project Presentation
Expansion Pack: Please speak to Ms. Han if you are interested in additional exercises.
ASSESSMENT
## Chapter Quizzes 30%
Term Tests (cumulative) 30%
Homework and Homework Quizzes 20%
Response Bias Project 10%
Group Work and In-Class Participation 10%
Chapter Quizzes: The purpose of chapter tests is to keep all students on top of their learning throughout the term,
and also to provide guideline to study for the term test as well as AP exam. Work needs to be shown for each
problem. Incorrect answers but well-written work with simple calculation mistakes will earn partial credit. We
Term Tests: At the end of each term, there will be a cumulative term test. Work needs to be shown for each
problem. Incorrect answers but well-written work with simple calculation mistakes will earn partial credit. We
Homework and Homework Quizzes: Students will be assigned homework every class and they are to have
completed it by the following class unless specific instructions are given. Students should take an average one
hour to complete their homework. Homework will be mostly graded by completion-based. Complete work earned
full point, incomplete work earns half point, and no or little work earns zero point. Late homework will be not
accepted. Students who are absent on the due date of an assignment, they must turn their homework in on the next
following class day to be accepted for credit. Homework must be done independently in which no show of work
will be considered incorrect. In addition, there will be Homework Quizzes on one of the previous night
homework.
Group Work and In-Class Participation: A typical class will be combination of whole class instruction led by
the teacher and group work led by students. Students are expected to pay undivided attention to teacher during
whole class instruction, and active involvement with group mates during group work time. Failure to meet these
expectations will result penalty in participation points.
Important Dates
Chapter 1 Quiz on Wednesday, August 16th
Chapter 2 Quiz on Wednesday, August 23rd
Chapter 3 Quiz on Wednesday, September 6th
Chapter 4 Quiz on Monday, September 26th
Term 1 Test on Friday, September 29th
PROGRESS MONITOR
Use below table to keep track of your progress throughout the school year.
Accomplishment Table
Assessment Title Weight Due Date Points Earned
/ Points Possible
Chapter Quizzes 30%
## Term Test 30%
Homework and Homework
20%
Quizzes
Response Bias Project 10%
Group Work and In-Class
10%
Participation
Overall | 2,473 | 10,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-35 | latest | en | 0.843209 |
https://www.assignmenthelp.net/assignment_help/ideal-gas-equation | 1,632,149,471,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00623.warc.gz | 693,617,992 | 13,609 | # Chemistry Homework Help With Ideal Gas Equation
## Ideal Gas Equation
The simple gas laws relating gas volume to pressure, temperature and amount of gas, respectively,
, T and n constant (Boyle’s law)
, P and n constant (Charle’s law)
, P and T constant (Avogodro’s law)
or or
This is known as ideal gas equation. R is known as gas constant.
Value and Units of R
We know, PV = nRT
R =
In SI-Units : For one mole of gas (n = 1) at NTP
R =
=
= 8.314 JK–1 mol–1.
R = =
= 0.0821 lit-atm K–1 mol–1.
R = 1.99 cal K–1 mol–1.
R = 8.314 × 107 ergs K–1 mol–1.
### Homework Help For Ideal Gas Equation
assignmenthelp.net provides best Online assignment Help service in Ideal Gas equation for all standards. Our Tutor provide their high quality and optimized Tutorial help to fulfill all kind of need of Students.
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• Writing Help | 314 | 1,280 | {"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-2021-39 | latest | en | 0.885063 |
http://www.cfd-online.com/Forums/openfoam/124842-groovy-boundary-condition-pressure.html | 1,444,206,883,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736682947.6/warc/CC-MAIN-20151001215802-00024-ip-10-137-6-227.ec2.internal.warc.gz | 477,999,176 | 16,898 | # groovy boundary condition for pressure
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October 14, 2013, 11:51 groovy boundary condition for pressure #1 Senior Member ahmed Join Date: Feb 2010 Posts: 161 Blog Entries: 1 Rep Power: 7 Hi, i'm using groovy boundary condition to express velocity as a relation of pressure difference of pressure of two regions. the problem is that velocity is vector field while pressure is scalar so, there is a problem to express vector value as a relation of scalar value. i think that if i use gradp instead of p it will be helpful but openfoam can't understand gradp since it is not stored in file. any one can help me figuring out a solution for this problem.
October 14, 2013, 17:29
#2
Assistant Moderator
Bernhard Gschaider
Join Date: Mar 2009
Posts: 3,915
Rep Power: 40
Quote:
Originally Posted by rebel ahmed Hi, i'm using groovy boundary condition to express velocity as a relation of pressure difference of pressure of two regions. the problem is that velocity is vector field while pressure is scalar so, there is a problem to express vector value as a relation of scalar value. i think that if i use gradp instead of p it will be helpful but openfoam can't understand gradp since it is not stored in file. any one can help me figuring out a solution for this problem.
If you look at http://openfoamwiki.net/index.php/Co...her_field_are: then you'll find that snGrad(p) gives you the gradient of p in the normal()-direction
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October 15, 2013, 02:26 #3 Senior Member ahmed Join Date: Feb 2010 Posts: 161 Blog Entries: 1 Rep Power: 7 Dear gschaider, thanks for your reply, i tried this Boundary condition for velocity Code: ``` type groovyBC; valueExpression "(pUpperFlow-pMiddleFlow)/(mu)"; variables "pUpperFlow{patch'upperFlow_to_upperFilter}=snGrad(p);pMiddleFlow{patch'middleFlow_to_upperFilter/middleFlow}=snGrad(p);"; value uniform (0 0 0);``` and this is the error message: Code: ```[2] --> FOAM FATAL ERROR: [3] [3] [3] --> FOAM FATAL ERROR: [3] The expected return type vector is different from the stored result type "scalar" --> FOAM FATAL ERROR: [1] The expected return type vector is different from the stored result type "scalar" [1] [1] [1] From function tmp > ExpressionResult::getResult() [1] [2] The expected return type vector is different from the stored result type "scalar"```
October 15, 2013, 05:30
#4
Assistant Moderator
Bernhard Gschaider
Join Date: Mar 2009
Posts: 3,915
Rep Power: 40
Quote:
Originally Posted by rebel ahmed Dear gschaider, thanks for your reply, i tried this Boundary condition for velocity Code: ``` type groovyBC; valueExpression "(pUpperFlow-pMiddleFlow)/(mu)"; variables "pUpperFlow{patch'upperFlow_to_upperFilter}=snGrad(p);pMiddleFlow{patch'middleFlow_to_upperFilter/middleFlow}=snGrad(p);"; value uniform (0 0 0);``` and this is the error message: Code: ```[2] --> FOAM FATAL ERROR: [3] [3] [3] --> FOAM FATAL ERROR: [3] The expected return type vector is different from the stored result type "scalar" --> FOAM FATAL ERROR: [1] The expected return type vector is different from the stored result type "scalar" [1] [1] [1] From function tmp > ExpressionResult::getResult() [1] [2] The expected return type vector is different from the stored result type "scalar"```
As I said below: snGrad is the gradient IN THE DIRECTION NORMAL TO THE BOUNDARY. So it is a scalar. To get the actual vector multiply it with normal()
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All times are GMT -4. The time now is 04:34. | 1,198 | 4,508 | {"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-2015-40 | latest | en | 0.899159 |
https://stats.stackexchange.com/questions/51046/how-to-check-if-my-regression-model-is-good | 1,656,176,204,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00272.warc.gz | 578,774,371 | 69,032 | # How to check if my regression model is good
One way to find accuracy of the logistic regression model using 'glm' is to find AUC plot. How to check the same for regression model found with continuous response variable (family = 'gaussian')?
What methods are used to check how well does my regression model fit the data?
• You may want to have a look at the r-squared tag and the goodness-of-fit tag.. Feb 28, 2013 at 15:36
• The "Gaussian" family with a linear link is just ordinary least squares (OLS) regression; methods to check such fits are probably discussed in a thousand questions on this site (I do not exaggerate).
– whuber
Feb 28, 2013 at 17:13
• This thread is relevant: stats.stackexchange.com/q/414349/121522
– mkt
Jun 25, 2019 at 5:48
I would suggest a brief search on "linear regression model diagnostics" as a start. But here are some that I would suggest you to check:
Make sure the assumptions are satisfactorily met
• Use scatterplot or component plus residual plot to examine the linear relationship between the independent predictor(s) and the dependent variable.
• Compose a plot with standardized residual versus predicted value and ensure there isn't extreme point with very high residual, and the spread of the residual is largely similar along the predicted value, as well as spreading largely equally above and below the mean of residual, zero.
• You can also change the y-axis to residual$^2$. This plot helps identifying unequal variance.
• Re-examine the study design to ensure the assumption of independence is reasonable.
• Retrieve the variance inflation factor (VIF) or tolerance statistics to examine possible collinearity.
Examine potential influential point(s)
• Check statistics such as Cook's D, DFits, or DF Beta to find out if a certain data point is drastically changing your regression results. You can find more here.
Examine the change in $R^2$ and Adjusted $R^2$ statistics
• Being the ratio of regression sum of squares to total sum of squares, $R^2$ can tell you how many % of variability in your dependent variable are explained by the model.
• Adjusted $R^2$ can be used to check if the extra sum of squares brought about my the additional predictor(s) is really worth the degrees of freedom they'll take.
Check necessary interaction
• If there is a main independent predictor, before you make any interpretation of its independent effect, check if it is interacting with other independent variables. Interaction, if left unadjusted, can bias your estimate.
Apply your model to another data set and check its performance
• You can also apply the regression formula to other separate data and see how well it predicts. Graph like scatter plot and statistics like % difference from the observed value can serve as a good start.
• (+1): Very complete answer! If you're using R, plot.lm can give you most of the diagnostic plots Penguin_Knight mentions.
– Zach
Feb 28, 2013 at 16:31
• @Zach what do you mean with plot.lm in R? Jun 22, 2021 at 10:12
I like to cross-validate my regression models to see how well they generalize to new data. My metric of choice is mean absolute error on the cross-validated data, but root mean squared error is more common and equally useful.
I don't find R2 to be a good metric of how well your model fits the training data, as almost any error metric calculated on the training data will be prone to over fitting. If you must calculate R2 on the training set, I suggest using adjusted R2.
You can use $R^2$ to examine how well your model fits the training data. This will tell you what percentage of the variance in the data are explained by the model.
I suggest using RMSE (root mean square error) of your predictions on your test set when compared to the actual value. This is a standard method of reporting prediction error of a continuous variable.
• @Macro But the question originally asked for a performance metric for an OLS Regression with gaussian errors. He is coming from logistic regression.
– Erik
Feb 28, 2013 at 15:16
• @Erik, thanks, I misread. Anyway, regarding the first part, I don't think $R^2$, in isolation, can be used to "check if my regression model is good", to use the OP's words. Your model could fail miserably to predict effectively on vast majority of the data while still having a high $R^2$. See here for an example - in example (1), there's almost no predictive power but $R^2$ is still high. Feb 28, 2013 at 15:24
• @Macro, I agree with your comments but was aiming for a simple explanation to point the OP in the right direction Feb 28, 2013 at 16:03
I am used to check the functional form of my parameter estimator by plotting a non-parametric (e.g. a kernel regression) or semi-parametric estimation and comparing it to the parametric fitted curve. I think this is in the first step often faster (and perhaps more insightful) than including interaction terms or higher-orders terms.
The R package np provides many nice non-parametric and semi-parametric functions, and its Vignette is well written: http://cran.r-project.org/web/packages/np/vignettes/np.pdf | 1,166 | 5,103 | {"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": 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} | 3.03125 | 3 | CC-MAIN-2022-27 | latest | en | 0.916326 |
https://haravgipdf.com/putnam-and-beyond-pdf/ | 1,726,205,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00581.warc.gz | 259,931,488 | 24,344 | # Putnam And Beyond Pdf
Putnam And Beyond Pdf. Web the book is a compilation of advanced problems (from the putnam exams and various mathematical olympiads, etc), arranged by topic. Web will be nice if you would read chapters 1 and 2 of the book putnam and beyond!
After a brief chapter on the different. Web preview putnam and beyond. Web the book is a compilation of advanced problems (from the putnam exams and various mathematical olympiads, etc), arranged by topic.
### Web A Book By Ra ̆Zvan Gelca And Titu Andreescu That Aims To Prepare Undergraduate Students For The Putnam Competition, A Major Mathematical Event In North America.
Web putnam and beyond takes the reader on a journey through the world of college mathematics, focusing on some of the most important concepts and results in the. The name putnam has become synonymous with excellence in undergraduate mathematics. After a brief chapter on the different.
### Web Preview Putnam And Beyond.
The name putnam has become synonymous with excellence in undergraduate mathematics. A doorstop lled with problems and strategies for the putnam itself: Web putnam and beyond is the fruit of work of the first author as coach of the university of michigan and texas tech university putnam teams and of the international.
### You May Submit An Original Solution Of A Challenging Putnam Problem, Or.
Often putnam problems have extremely interesting generalizations or related problems. The william lowell putnam mathematics competition is a north american math contest for college students, organized by the mathematical association. Using the putnam competition as a symbol, we lay the foundations of.
### Web Putnam And Beyond Takes The Reader On A Journey Through The World Of College Mathematics, Focusing On Some Of The Most Important Concepts And Results In The.
Web putnam and beyond is organized for independent study by undergraduate and graduate students, as well as teachers and researchers in the physical sciences who wish to. Proofs (question 10:) let \$n>1\$ be an arbitrary real number and let \$k\$ be the number of positive prime numbers less than or equal to \$n\$. Web we would like to show you a description here but the site won’t allow us.
### Using The Putnam Competition As A Symbol, We Lay The Foundations Of.
Web will be nice if you would read chapters 1 and 2 of the book putnam and beyond! Web this book takes the reader on a journey through the world of college mathematics, focusing on some of the most important concepts and results in the. Web putnam and beyond. | 535 | 2,564 | {"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-2024-38 | latest | en | 0.904026 |
https://opusmodus.com/forums/tutorials/tutorial-guide/lesson-4-rhythm-repeats-and-random-seeds-r20/ | 1,540,057,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513009.81/warc/CC-MAIN-20181020163619-20181020185119-00026.warc.gz | 785,193,353 | 22,568 | • # Lesson 4. Rhythm, Repeats and Random Seeds
## Annotation
The instruction to repeat is one of the most common and necessary in composing music. Here is the function GEN-REPEAT set to repeat pitches 5 times.
This is the output:
```=> (c4 cs4 fs4 g4 c5 c4 cs4 fs4 g4 c5 c4 cs4 fs4
g4 c5 c4 cs4 fs4 g4 c5 c4 cs4 fs4 g4 c5)```
Sadly, we can’t see the repeat grouping that easily. It’s just one long list! There are no bar lines or divisions in the output. But there is a way of indicating bars coming up in Lesson 5.
Both right and left hands of the piano are playing identical material BUT the right hand has been processed with LENGTH-WEIGHT.
```(length-weight '(3 1) lengths :weight '(3 1):seed 23)
=> (-1/8 -1/8 1/8 1/8 1/8 1/8 1/8 1/8 -1/8 1/8
1/8 1/8 -1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8
1/8 1/8 -1/8 1/8 -1/8)```
You can see the LENGTH-WEIGHT replaces previously 'sounding' lengths with rest-lengths throughout the stream of pitches. The ratio used is (3 1), which indicates roughly 3 sounding lengths to 1 rest. Meanwhile the bass part in the left hand plays steadily each pitch of the repeated figure.
Notice that the LENGTH-WEIGHT expression carries a keyword :seed
`(length-weight lengths :weight '(3 1) :seed 23)`
Without this seed, each time the function was evaluated a different weighting of rest-lengths to note-lengths would be given. Experiment with evaluating the expression without the seed, or simply change the seed itself to see how the output is effected.
## Score
```(setf pitches (gen-repeat 5 '(c4 cs4 fs4 g4 c5)))
(setf transposed-pitches (gen-repeat 5 (pitch-transpose -24 pitches)))
(setf lengths (span pitches '(e)))
(setf lengths-rests (length-weight lengths :weight '(3 1) :seed 23))
(setf piano-righthand
(make-omn
:length lengths-rests
:pitch pitches
:velocity'(mp)))
(setf piano-lefthand
(make-omn
:length lengths
:pitch transposed-pitches
:velocity '(f)))
(setf timesigs (get-time-signature lengths))
(def-score lesson-4
(:key-signature 'chromatic
:time-signature '(5 8)
:tempo 100
:layout (piano-layout 'piano-rh 'piano-lh))
(piano-rh
:omn piano-righthand
:channel 1
:sound 'gm
:program 'acoustic-grand-piano)
(piano-lh
:omn piano-lefthand)
)```
## Notation
Next page Lesson 5. List Processing
Go to the Reference page.
• ### Introduction to Opusmodus
Contents A Contemporary Language for Making Music The Parametric World of Music The Parametric Instrument Learning Opusmodus : A Strategy Important Questions: Necessary Answers
A Contemporary Language for Making Music Composing, like most art-making, is a messy business, It is rarely radio in the head. You don’t turn it on and there it is. A composer goes searching for music. It’s out there somewhere, but it has to be detected, discovered, and then deciphered into music’s own language. To do this requires experiment and imagination. In the 1980s MIDI provided a contemporary language for musical events that let us use computers for recording and editing already conceived ideas. But MIDI is not a natural language, and programming with it is a highly specialist task. Composers want and need a straightforward contemporary language for music that whilst relating to traditional staff notation, and MIDI too, enables the origination of novel ideas and new forms of making. Such a language is parametric: found in and used by Opusmodus.
The Parametric World of Music Musical events belong in a network of parameters: pitch, note duration and rhythm, dynamics, articulation, and at a higher-level tonality, harmony and musical structure itself. They are all connected. In Opusmodus, we are ‘Parametrical’. Increasingly composers create novel musical events by interacting with musical parameters written or ‘found’ through separating them out, processing them, and then putting them back together again. Rhythms are constructed through additive and subtractive processes, pitch aggregates are formulated with magic squares and statistical algorithms, integers, intervals and random numbers are often starting points, ways to ‘make a mark’, to fill the blank page (or screen). Many starting points in music composition are not based on sound at all, but on geometric structure, proportion, chaotic incidence, visual relationships, movement, poetry and prose. Whatever these may be they will need to be pulled somehow onto the musical stave. This remains the format our culture continues to invest in as a notation-led end result, the common currency of most music education, professional performers, ensembles and orchestras. Much new art and media music continues to reach us through such notated scores composed by bringing together those commonplace parametric elements.
The Parametric Instrument With a Parametric Instrument for Composing Music it becomes possible to network musical parameters into inherently variable, adaptive forms that combine into unique and often surprising continuously differentiated fields or systems. This is what Opusmodus does. Musical practice in composition is no longer style-oriented or system-based. It can be everything and anything. Composers can be insatiably curious about the possibilities of phenomena that lie outside music, because so much around us is now understood and able to be captured as data. And so composers need the wherewithal to make conversions of such data to live in the parametric world of music. Opusmodus has the parametric tools to make this happen. Don’t necessarily expect a previous experience with technology to open the door straightaway to what Opusmodus has to offer. This is not about point and click, play and record, copy and paste. It is about thinking and scripting; it is about building expressions made of functions that are able to process or generate one or many musical parameters and provide an output that can be seen and heard, instantly. Opusmodus provides a fast and robust feedback loop for musical ideas.
Learning Opusmodus : A Strategy If you’ve learnt a language there’s a similarity. You might go to a class or know a native speaker, then you can listen, copy and eventually talk. Otherwise you’ll use a CD and a book, or interact with a web-based tutor. At some point you’ll have to work on vocabulary, and maybe learn to write. The language of Opusmodus requires something similar. • Take a look and listen to the example scores.
• Take a Tutorial.
• Browse the Documentation, the vocabulary of Opusmodus.
• Study the score-scripts.
• Modify these scores and start to write your own. The tutorial resources can be accessed from within Opusmodus itself. You’ll find Quick Start, a guide providing the necessary basics. Then there are Lessons: a 30-part collection of score-scripts and text commentaries designed to be opened simultaneously.
Important Questions: Necessary Answers Be sure, you’ll find in all these learning resources something to fire up the imagination. Browse as much as you can, and begin to ask yourself what is it that makes up my musical language? What are the elements and common processes I already use when making a piece of music? Do I know how a piece of my own music is composed? Is it really trial and error, continuous experimentation until it ‘sounds right’ or are there methods, techniques, pathways you’ve already established or invented? Such questioning is a highly recommended exercise. And if you don’t have the answers, learning Opusmodus will prove a unique way into musical literacy! Whatever the answers to these questions, bite the bullet with one of the early tutorial guides. Approach these little score-scripts in a spirit of play. The more time you can devote to playful experimentation before starting on that next commission or project the better. Again, think of learning a foreign language. You may learn enough Italian in a Day with a CD to ‘get by’ but to understand and use the language you have to go further. It’s the same with Opusmodus. Learning takes time, but it will prove such an enriching process, and one that brings together understanding with knowledge: about the music you compose and how you compose it. If you are new to scripting, don’t shy away from the basics. Once you have them you won’t look back and all kinds of possibilities will open up. Next page Reference
opmo
Tutorial Guide 0
• ### Introduction to OMN the language
OMN is designed as a scripting language for musical events. It’s not about sounds themselves, it is about their control and organisation in a musical composition. As a linear script rather than a graphic stave, musical events can be transformed, extended, reorganised by powerful computer algorithms. Some sequencers and score writers provide basic algorithms, but they do not represent the way composers now think about the process of music composition. Composing has become such a multi-faceted process and takes ideas about structure and content from many disciplines: mathematics, astronomy, literature, the visual arts. As such it requires extensive mental resources and experience from the composer. Much of this is still done by hand and eye and brain because although computer systems do exist to help the process along they don’t provide what has become known as the composing continuum. This means that a single workspace and workflow environment has not been generally available that can take in the whole process of composing a piece - from first thoughts to a printed score and reference recording. Wouldn’t it be good to be able to do everything in one place? Most composers acquire a bag full of musical tools to act on musical ideas. These still include those tools Bach used for repetition, inversion, retrograde, transposition, but with computer help musical material can be copied, cut, pasted and generally structured and orchestrated. Since the 1950s composers have been experimenting with tools and processes that take musical transformation into wholly new areas; of random numbers, fractals, statistical distribution, graphical plotting to name just a few. To use such experimental things it is composing with a script that is acknowledged as the most efficient and practical way forward. And to work with a script means working with a language: OMN.
Contents OMN and Musical Notation The Concept The Four Elements Length Pitch Velocity Attribute Repetition Assemble And Disassemble Algorithms The Way Forward
OMN and Musical Notation
The truly original aspect of OMN is that it has been designed to speak directly to traditional musical notation. Everything written in OMN script can be rendered instantly to notation and to a performance simulation. For most composers staff notation remains the common currency they have to work in and with. You couldn’t expect performers to read from a MIDI event display or indeed from OMN script. As the OMN language is laid out and explored we’ll see just how fully the language of music staff notation is mirrored. This is not just in the standard elements of rhythms, pitch and dynamics but in the vast library of musical attributes that cover the way pitches and rhythms are performed by different instruments and voices. So musical notation is always there. Whatever you write there can be an instant ’snippet’ rendered to view alongside your script.
The Concept
Most languages have developed orderings for parts of speech. Romance languages place the verb after the subject, and in the middle of the sentence. Germanic languages tend to conclude sentences with a verb. In music we’re used to the single intersection of pitch position on a stave line with a rhythmic symbol with or without a stem. In developing a right concept for the OMN language much thought was given to choosing the most effective ordering of elements. Culturally our music is one governed by our past experiences, elements of musical tradition gathered through informal and formal musical education, and what is active in the memory. Descartes adage ‘Cogito ergo sum’ (‘I think, therefore I am’) remains an important cornerstone of an individual’s relationship with composing music. It is something known. It is a made thing; it possess architecture. We can say with confidence that we experience music in a hierarchical sequence of time, existence, dynamics and expression. So it is right that the linear ordering of OMN reflects this. In architecture this might be translated as dimension, materials, volume of space, decoration. These are established architectural parametrics able to form the basis for CAD rendering in the new parametric systems architects are now using to allow the conditions surrounding to influence design. OMN is a language wholly sympathetic to parametric composition in music.
The Four Elements
Length
OMN was created to think about the element of TIME first. After all we can be musical without a pitched note being present. If we are going to use the OMN script we need a reference guide to help us whilst we learn the language. What accompanies this introduction is a special dictionary of language terms arranged in the four elements that make up the concept. However, there are some necessary redefinitions required. TIME is a very general element that subdivides in music to rhythm and length. When we describe what makes up a rhythm in notation it is usually a mixture of symbols that have different lengths. So the OMN vocabulary uses the term LENGTH as its general title. (q)
Pitch
The second element of the OMN language is PITCH. Although each piece of music is defined by the length of time, it only starts to EXIST as a proper musical entity when pitch is added. (q c4)
Velocity
The third element of the OMN language is VELOCITY. Staff notation has a set of common symbols that are formed from the first letter of Italian words for degrees of intensity we want to attach to a note or a phrase. In OMN there are 12 such terms ranging from ppppp to fffff. OMN includes many symbols that can only be classed as Dynamics because they are not identified directly with a data value.
(e c4 mp)
Attribute
The fourth element of the OMN language is ATTRIBUTE. The number of general symbols and words used to describe expression in music is vast: tenuto, staccato, legato, trill, fermata etc... Many instruments, particularly those of the string family have their own vocabulary of technical expressive terms: pizzicato, sul ponticello, flautando. Remarkably these can be included in an OMN script and, if your sampler has a string effects library, these expressive instructions can be realised directly.
(e c4 mp trem) Finally, there is SIMULTANEITY possible in the layering of attributes. This is achieved by the + symbol. (q c4 mp trem+fermata)
Repetition
An important fifth element of REPETITION is also present in the OMN language structure. (q c4 q c4)
equals (q c4 =)
Assemble And Disassemble
It is valuable to remember that the composer may need to create material one parameter at a time. OMN allows for discrete parameters to be brought together to make a composite list in OMN. By the same token it may also be necessary to focus on just a single parameter to develop further the argument of a composition. An OMN list can easily be disassembled into its component parts for such work to take place and then made back into an OMN list. (disassemble-omn '(q c4 mp d4 e4 e f4 f g4)) => (:length (1/4 1/4 1/4 1/8 1/8) :pitch (c4 d4 e4 f4 g4) :velocity (mp mp mp f f) :articulation (- - - - -)) (make-omn :length '(q q q e e) :pitch '(c4 d4 e4 f4 g4) :velocity '(mp mp mp f f)) => (q c4 mp d4 e4 e f4 f g4)
Algorithms
OMN script responds directly to the Opusmodus library of algorithmic functions, and with keywords particular elements can be selected to be processed or not. (pitch-transpose 6 '(q c4 mp d4 e4 e f4 f g4)) => (q fs4 mp gs4 bb4 e b4 f cs5)
The Way Forward
This introduction should set you on your way. With what has been covered here, the Tutorial Guide files will demonstrate how closely the OMN language can be integrated with algorithmic composing. In fact, when composing in this way you’ll often only write material in one parameter at a time. Although every function will read an OMN list, it’s often better to keep parameters apart to begin with. You’ll see this clearly in the Tutorial files. There will be some music projects where writing directly in OMN is really necessary. Composing for voice is certainly one medium. There are examples in the How To section to demonstrate word setting with full attention given to syllabic splitting. For more experimental approaches to composing OMN can be integrated with the conversion of integers and intervals into the parameter of pitch. The Tutorials show how this can be achieved with examples that use pitch-class sets to create tone rows. OMN is a way of scripting the whole language of traditional staff notation and modes of experimental and conceptual composition using the tools of parametric modelling. It is a language that responds to the future of music presentation, as notation moves inextricably from the printed page to the backlit digital display. New music technology has focused largely on production and presentation, whereas the conceptualisation and origination of new music requires a very different paradigm. Sequencer and Scorewriters continue to provide valuable ways into composition. Opusmodus provides the 3rd way forward, and one driven by its own notation script: OMN. OMN is perfect for those ‘on the fly’ experiments that all composers make when they are starting out on a project. It is like having a piano close by to try out this or that, but one that always plays what’s written quite flawlessly. What is wonderful about scripting is that those experiments if successful can remain part of the score for the whole progress of the composition. With OMN a composing continuum can be achieved. OMN may look a little hard to decipher at first, but once the logic is understood, be assured, OMN can be read with ease. OMN is the first notation that has been designed from the outset to communicate with MusicXML the de facto standard for communication of notated scores between different software applications. Opusmodus scripts can be converted seamlessly into both Midi and MusicXML. Next page 1st Element - Length
opmo
content_cat_name_15 0
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• Lessons | 4,060 | 18,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-43 | longest | en | 0.809654 |
https://www.icicidirect.com/ilearn/stocks/articles/what-is-margin-against-shares | 1,685,434,755,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00662.warc.gz | 876,156,312 | 31,654 | # What is Margin Against Shares?
24 Dec 2021 0 COMMENT
## Introduction:
Margin Against Shares is when you use the shares you have in your account as collateral to borrow funds from the brokerage. Your broker holds your shares as collateral against credit risk.
For example, say you want to buy 500 shares of a company –XYZ at Rs. 1000 each, However, you don’t have the required funds. However, you have shares of another company ABC worth Rs. 50,000 in your account. You can then borrow the Rs 5,00,000 from your broker by using the shares worth Rs 50,000 in your account as collateral.
The broker also charges interest that is calculated on the basis of the loan given to you. This process of using your shares as collateral is called pledging.
## Calculation of margin:
The margin you receive after pledging your shares is calculated after reducing a certain percentage of the value of shares you have in your account. This is done to protect the broker from risks such as a decline in share price used as collateral. This is known as a haircut.
For instance, the value of the shares of ABC in your account is Rs. 50,000 on the day you decide to borrow margin money. However, while calculating this margin, the broker will reduce a certain percentage from ABC’s current market price. For example, if they reduce 20%, you will receive a margin on Rs. 40,000 worth of ABC’s shares.
Click here to listen to a podcast on the difference between cash and margin
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## Conclusion:
Margin against shares will provide you with greater amount of funds and help you earn higher profits, but it can also be quite risky. You can only gain when the total profit earned is higher than the margin. You will also have to pay interest on the loan provided, till it remains outstanding. If the value of shares you trade does not appreciate, you might suffer further loss even after paying the interest. When you pledge your shares, brokers have the right to liquidate them if the value of your shares fall. So, it is best to engage in margin trading when you are sure that you will make profits.
Disclaimer
ICICI Securities Ltd. ( I-Sec). Registered office of I-Sec is at ICICI Securities Ltd. - ICICI Centre, H. T. Parekh Marg, Churchgate, Mumbai - 400020, India, Tel No : 022 - 2288 2460, 022 - 2288 2470. The contents herein above shall not be considered as an invitation or persuasion to trade or invest. I-Sec and affiliates accept no liabilities for any loss or damage of any kind arising out of any actions taken in reliance thereon. The contents herein above are solely for informational purpose and may not be used or considered as an offer document or solicitation of offer to buy or sell or subscribe for securities or other financial instruments or any other product. Investments in securities market are subject to market risks, read all the related documents carefully before investing. The contents herein mentioned are solely for informational and educational purpose. | 650 | 2,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-23 | longest | en | 0.961728 |
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# Expression For Rolling Load In A Two High Rolling Mill Derived
• ### Deformation Processing Rolling
Flat Rolling Analysis Results without front and back tension p p p p x x d x Stresses on Slab in Entry Zone Stresses on Slab in Exit Zone p p p p x x d x Using slab analysis we can derive roll pressure distributions for the entry and exit zones as h 0 and h b are the same thing 2tan1 ff RR H hh 0 0 2 ...
• ### Formula For Load Calculation Of Rolling Mill
rolling mills equations are derived for the normal roll pressure specific roll load and torque in hot rolling, there have been many reports on rolling load for rolling loads no uniform formula has been established for adopted a 4 high rolling mill
• ### Roll Separating Force an overview ScienceDirect Topics
A two-high rolling mill is assumed to be used with rolls of 250 mm diameter. The elastic modulus of the roll material is taken to be 210 000 MPa and its Poissons ratio is 0.3. The model, developed by Roychoudhury and Lenard 1984, reviewed above, is used in the computations. As implied above, different models predict different magnitudes of ...
• ### rolling mill horsepower calculation
Due to the growing importance of width control in strip and plate mills edge rolling is currently an im-portant process in hot rolling mills Research in edge rolling has been carried out and in the present ar-ticle models for roll force torque and lever arm coefficient are derived using the upper bound method A simple kinematically admissible deformation zone and velocity field...
• ### Maintenance for productivity Rolling models
Roll gap model Mill model Tension practices Model parameters Set-up values Adaption parameters Measured values Thickness, tension, speed Roll force, torque, temperature CVC, HS, position bending Coil-data Operator-trims Scheduling 400 350 300 250 200 150 100 50 0 0 0.5 1 1.5 2 Log deformation Yield stress 3542 Rolling mill applications covered ...
• ### IOM3 Rolls for Metal Rolling Online Seminar Series
Jul 21, 2021 Rolls for Hot Rolling. Four speakers from 3 continents will present developments and field experiences with advanced rolls for hot rolling applications, ranging from plate mills and hot mill roughing stands to early and late finishing stands. Each paper will provide a unique perspective, from varying viewpoints of roll manufacturers, roll users ...
• ### Shape and Gauge Control of Strip in a Cold Rolling Mill
Oct 06, 2020 In general in tandem cold rolling mill two-stand cold reversing mill the stands can be of 4- high type or 6-high type i.e., stands with 6 rolls. The single stand cold reversing mill process can be achieved in particular for stainless steel with stands of 20-high type also known as cluster mill or Sendzimir mill.
• ### Forced transverse vibration of rolls for fourhigh rolling
Chatter in the rolling stack of high velocity tandem mills and temper mills is a widespread problem, and it affects the quality of the finished product and the productivity of the rolling mill.
• ### Formula For Weight Calculation Of Rolling Mill
Load Analysis of Rolls in a Rolling Mill A ... May 2nd, 2018 - Cold Rolled Steel Sheets and Coils with superb press formability to high 6 22. strength steel sheets instrumental in weight reduction Hot Rolling Mill ... EQUATIONS ARE DERIVED FOR THE NORMAL ROLL PRESSURE SPECIFIC ROLL LOAD AND TORQUE IN HOT ROLLING MILLSCONTINUOUS MILL ...
• ### Modeling friction coefficient for roll force calculation
Equations are derived for the normal roll pressure, specific roll load and torque in hot rolling mills, using the condition for plastic deformation in rolling derived by Orowan, together with von ...
• ### expression for roll load roll torque mill horsepower
POWER IN ROLLING IDC-Online. The torque is equal to the product of total rolling load and the effective moment arm. Since there are two work rolls Torque Mt 2P.a Consider two high roll mill as shown in the figure. For one revolution of the top roll the resultant rolling load P moves along the circumference of a circle equal to 2.a Since there are two work rolls involved, the work done is equal to Work done 2 2..a.P 4P..a. Roll Mill
• ### Simulation of Torque during Rod Rolling of HC SS316 at
continuous rolling mill was used in view of the shortcomings of a two-high mill. 2. MATHEMATICAL MODEL 2.1 Rolling Torque The mathematical model used for the simulation is presented below The rolling torque, in N-m is T 2 PR 1 where, coefficient of lever arm P load, N and R undeformed roll radius, m.
• ### 1 Hot rolling and rolling defects 11 Front and back
Gage control in multiple rolling mills is achieved through measurement of strip thickness using x-ray gage and adjusting the strip tension using feedback control system. 1.8 Rolling defects Mill spring is a defect in which the rolled sheet is thicker than the required thickness because, the rolls get deflected by high rolling forces.
• ### Rolling metal forming apratim khandelwaldocx
9 Four High Rolling Mill- It utilizes the principle that a lesser force is required if smaller rolls are used. Hence a major drawback is the risk associated with the deflection of smaller rolls from original position figure at bottom. Fig 8 Four high rolling mill- two main smaller rolls supported by two
• ### The development of the theory of metal rolling and its
mill and control must be based. Rolling theory can be divided into two broad and general cate- gories, that which applies to hot - rolling and that which applies to cold rolling. In hot - rolling the yield stress characteristic of the metal is strain -rate dependent and frictional force between the roll and stock is high.
• ### Formula For Load Calculation Of Rolling Mill
Torque in Hot Rolling Mills Equations are derived for the normal roll pressure specific roll load and torque in hot rolling Free Download Here pdfsdocuments2 com March 19th, 2019 - There have been many reports on rolling load for rolling loads no uniform formula has been established for adopted a 4 high rolling mill for
• ### Composite rolling mill roll and rolling method Nippon
A composite rolling mill roll according to the present invention includes a steel roll shaft and an outer layer provided around the roll shaft, in which the outer layer includes a sintered body including a base metal which is an iron alloy, a fibrous inclusion which consists of a ceramic and has an average diameter of 1 to 30 m and an average aspect ratio of 10 to 500, and a particulate ...
• ### Calculating Power Parameters of Rolling Mill Based on
Making digital twins for rolling processes and mill equipment should begin with the development of mathematical models of the deformation zone. The deformation zone of two-high flat mill rolling have been studied in detail, relevant models are available in many academic papers. However, the same cannot be said about the most complex deformation zones in stands with multi-roll gauge.
• ### A New Temper Rolling Force Model for Dry Thick Strip
Temper rolling or skin pass rolling is one of cold rolling processes composed of a horizontal pass cold rolling mill stand. In this process, the thick strip is subjected to very light reduction 0.54 in thickness in the presence of high friction amplitudes. One of the basic parameters of strip rolling processes is temper rolling force.
• ### Formula For Load Calculation Of Rolling Mill
Formula For Load Calculation Of Rolling Mill ... slip in cold strip rolling using a high rolling mill speed calculation formula the forward slip in strip rolling was defined as ... download citation on researchgate the calculation of roll force and torque in hot rolling mills equations are derived for the | 1,673 | 7,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-21 | latest | en | 0.79212 |
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